HEAT ENGINEERING McGraw-Hill BookGompany Electrical World TheEngineering andMining Journal Engineering Record Engineering Nows Railway Age Gazette American Machinist Signal E,ngin = Maxwell's thermodynamic potential. = coefficient in Nicolson's formula of heat transmission. ^ = external work in making steam. HEAT ENGINEERING CHAPTER I FUNDAMENTAL THERMODYNAMICS Heat is a form of energy and as such it may be measured in any unit of energy. The customary unit is the British thermal unit, B.t.u., although heat may be measured in foot-pounds, ergs, joules, calories, horse-power hours, kilowatt hours or any other unit of energy. The B.t.u. is Jf 80 of the amount of heat neces- sary to raise the temperature of 1 Ib. of water from 32 F. to 212 F. Heat energy is a form of energy due to a vibration or motion of the molecules of a body. .It may be produced by the transfor- mation of other forms of energy into heat, as when mechanical energy of a moving train is changed into heat by brakes. When other forms of energy are produced from heat energy, how- ever, it is found that only a portion of the heat energy may be transformed. This leads at once to the separation of energy into two classes: high-grade energy and low-grade energy. High-grade energy is any kind of energy which may be completely changed into any other form. Mechanical and electrical energy are examples of this while heat energy is low grade since it cannot be changed completely into any other form of energy. For this reason all of the heat energy in a body is not available for trans^ formation. The fractional part of the heat energy which is available for transformation into another form is known as the availability of heat energy. If Q heat units are in a system and if part of these are changed into another form of energy and Q f heat units remain in the system, then Q Q' heat units must have been changed. If this is all that could have been changed. Q-Q' Q represents the fractional part or the availability. It has been shown by Joule and others that when mechanical work is changed into heat or when heat is partially changed into mechan- ical energy there is a definite relation between the work and the 1 2 HEAT ENGINEERING heat in the first case and between the disappearance of heat and the work produced thereby in the second case. This statement is known as the first law of thermodynamics. The numerical relation between heat and work is known as Joule's equivalent, J. It is equal to 778 ft.-lbs. 1 B.t.u. = 778 ft.-lbs. (1) JQ = W (2) For convenience the reciprocal of 'J is used at times, the symbol for this being A. 7 = A W Q = AW (4) Q = Heat in B.t.u. W = Work in ft.-lbs. In French units the value of J is 426 kg. m. = 1 kg. deg. calorie. Now when heat is added to a body it may increase the internal energy of that body and do external work or JdQ = dU + dW (5) This formula is a mathematical statement of the first law of thermodynamics which is only one aspect of the law of the con- servation of energy ; heat added to a body increases the energy of that body and does external work. U is the symbol used to indicate internal energy or intrinsic energy of a body. The capi- tal letters Q, U, W, refer to a body of any weight. If a body of unit weight is considered, small letters are used. Jdq = du + dw (6) Now in the above equation any one of the differential quanti- ties may be zero or have any sign. Thus if dq = 0, no heat is added and such action is known as adiabatic action. If du = 0, there is no change in intrinsic energy and the action is known as isodynamic action, while if dw = 0, no external work is done, and, as will be seen later, there is no change in volume. The signs of these terms may be anything and the equation is true. This equation is one of the important ones in thermodynamics. If at any time the change of intrinsic energy is known on any path or during any change of state and if the work is known during that change, then the sum of these two will be the heat required during this change. If positive, heat must be added ; if negative, heat must be abstracted. FUNDAMENTAL THERMODYNAMICS 3 Now p represents the pressure on unit area (1 sq. ft.) of the surface of the body considered and it is assumed uniform and nor- mal to the surface surrounding the body. If then the total area of the surface of the body isF,pF is the total normal force on the body. If this surface is moved through the normal distance dx, the work done is pFdx = dW but since Fdx = dV pdV = dW (7) or pdv = dw The equation (6) may therefore be written Jdq = du + pdv (8) In this equation du must be measured in the same units as pdv, namely foot-pounds, and dq on account of the constant J must be measured in heat units. The total heat is the integral of dq or The internal energy of a body depends on the condition of the body and not on the method of bringing it to that state and conse- quently the change in intrinsic energy, which is I du, does not depend on the manner in which one state is changed to the other but merely on the states at the beginning and the end of the operation. Hence, if the intrinsic energy at state a is u a and at state b is Ub, the value of the integral du between these two states is Ub u a . This is true whatever the path may be. b pdv represents the area beneath a curve on the pv plane and 1 this area depends on the path considered. Hence the value of this integral can only be told after the curve or path is known. The integral of dq depends therefore on the path since, although I du is independent of the path, I dw or I pdv does depend on the path. A differential whose integral does not depend on the path is called an exact differential because it can be inte- 4 HEAT ENGINEERING grated directly. Hence there must be some functional relation between the independent variables in regard to which it is being integrated, otherwise a path would have to be fixed or some rela- tion would have to be known to give one variable in terms of the other to make the integration possible. Thus Cydx = CdX is not an exact differential since there must be some known rela- tion between y and x before the integration can be performed. Now I dx = x and I (xdy + ydx) = xy are each directly integrable and their definite integral values depend only on the limits and not on the path between the limits and for this reason each quantity behind the integral sign is an exact differential. It is seen that if exact differentials be inte- grated around a closed path their value would be zero since both limits are the same. This is one way of telling whether or not a differential is exact. If the integral on every closed path is zero, the differential is exact. There is one other criterion by which an exact differential may be told or a relation which must exist if a differential is exact, and if known to be a function of two independent variables. Thus if dX = Mdx + Ndy dX is an exact differential if 181 ?-? <> by dx If however dX is known to be exact by some other method, then dM . dN -r must equal (9 ) oy ox The reason for this is seen from the fact that if dX is exact there must be some functional relation X - J(xy) hence 8X j , 8X j dx = " dx + - now but hence FUNDAMENTAL THERMODYNAMICS dX dX -T = M and -r = N 8x dy dxdy dydx _ dM = d 2 X _ dN_ dxdy ~ dy ~ dydx ~ dx This relation between the differential coefficients M and N may be used to determine whether or not the differential is exact, or if known to be exact, the equality of the partial derivative of M and N may be used to determine new relations as will be seen later. The fundamental equation may be written Jq u a + I pdv (10) If q = 0, the line is an adiabatic and r Ub = \p Ja = I pdv (11) This means that work during an adiabatic change is equal to the change of intrinsic energy, or work is done at the expense of intrinsic energy. If ab , Fig. 1, represents an adiabatic path of a substance 'on the pv plane, the area la&2 represents the change of intrinsic energy from a to b. If this is car- ried out to infinity there can be no further area under the curve and the area from a to this point must represent the i v 2 total intrinsic energy at a since FIG. 1. Adiabatic on the pV plane, no heat has been added and work has been done to the point of zero temperature. Although this area is infinite in extent it is not infinite in value since the amount of energy in a body cannot be infinite. This is seen to be true mathematically since the curve approaches the v axis rap- 6 HEAT ENGINEERING idly and the height of the curve is practically zero after a short distance to the right. If u does not change on the curve (the isodynamic) the heat added is equal to the work done. Jq r = I pdv J a (12) If there is no work, I pdv = 0, or dv = 0, and v = constant. In this case Jq = u b u a (13) GRAPHICAL REPRESENTATION OF HEAT ON P.V. PLANE In Fig. 2 let ab be any path. Draw from a and b two adia- batics to infinity and from a draw the isodynamic until it strikes the adiabatic from b at c. 12 v 3 FIG. 2. Graphical representation of heat added. Now la oo = u 26 oo = Ub -I- Jo, Ia62 = I pdv J a r 26 oo -f la62 = Ia6 oo = Ub + | pdv Ia6 oo la o = oo a6 ro = u r I pdz; u a *) a b u a + I pdz; = J a Jq FUNDAMENTAL THERMODYNAMICS 7 This gives the important relation that the area on the pv plane between any path and the two adiabatics from the ex- tremities of the path to infinity is equal to the heat added on the path. This area is infinite in extent and consequently cannot be represented graphically. To make it finite and definite, the iso- dynamic ac was drawn. Now u a = u c . Hence 3c = la c and lab co 3coo = Jq = Ia6c3 or the area beneath any path added to that beneath an adiabatic from the second point of that path to the intersection of the adiabatic and the isodynamic from the first point is equal to the heat added on the path. It will be seen that Iab2 and r = I pdv J a 26c3 = Ub U c = Ub U a This latter statement should be clear since u c = u a . SCALE OF TEMPERATURE When heat is added to a body this body comes into a state in which it will transmit heat to another body with which it had been previously in contact. This ability to transmit heat to another body is determined by a property which is termed tem- perature. The temperature is measured by allowing an instru- ment to come into thermal equilibrium with the body and then noting the effect on the instrument. If one body is at a higher temperature than another it will transmit heat to that body. Suppose ab becomes a line of constant temperature T, such a line is called an isothermal (Fig. 3). Ifa< and b*> are adia- batics, the area ab<* is the heat added on this line ab and is finite in value since it is equal to Iab2 + 26 oo lao and each of these is finite. If q ab is divided by T the result ~ may be called e. If the isothermal a'b' is drawn below ab so that the area aWa r is equal to e and then a second isothermal a"b" is so placed that is equal to e and this is continued, it is found that there 8 HEAT ENGINEERING will be T such isothermals drawn. Each isothermal is so drawn that the small areas are equal and these isothermals determine definite temperatures. These temperatures form a scale and since this results from considerations of absolute units of work and does not depend on any particular substance it is called the Kelvin absolute scale of temperature. The ordinary scale of temperature was first determined by the effects of heat on mercury or other fluids but on account of changes in the rate of expansion of these fluids by which the temperatures were measured, gases were suggested as the media for measure- ment and hydrogen was found to be the most constant for all measurements. There are two scales in common use in this country, the Fahrenheit in which the temperature of melting ice is 32 and that of boiling water under standard conditions is I V 2 FIG. 3. Isothermals for Kelvin's Scale. 212, and the Centigrade scale on which the temperature of melting ice under standard conditions is and that of boiling water under standard conditions is 100. Neither of these start at the true zero of temperature. In Centigrade units the true or absolute zero is at 273 C. and in Fahrenheit units the zero is at 459.6 F. The relative size of degrees in the two systems is given by 1 C = % F It will be shown that Kelvin's Absolute Scale is the same as that determined by the hydrogen thermometer. In looking at the diagram from which the Kelvin scale is fixed it will be seen that the area beneath any of the isothermals and the adiabatics is equal to the heat on the line and that these areas are equal to T times the unit area e. Hence the FUNDAMENTAL THERMODYNAMICS 9 heat added on any isothermal between points of intersection with two adiabatics is proportional to the temperature on that isothermal. This is an important consideration. Q = eT (14) is the expression for the heat added on an isothermal of tem- perature T between two adiabatics and Q' = eT' is true at temperature T' between the same two adiabatics. This leads to the efficiency of the Carnot cycle. CARNOT CYCLE Carnot proposed a cycle made up of isothermal and adiabatic lines in a peculiar form of engine. Imagine a perfect non-con- ducting cylinder with a perfectly conducting head containing a non-conducting piston together with two bodies S and R of infinite FIG. 4. Carnot engine. capacity for heat and a non-conducting plate P (Fig. 4). The fact that S and R are of infinite capacity means that if a finite amount of heat is added or taken away from them their tem- peratures will not change. The body S being at a higher tem- perature, Tij than that of the body R at T z , is known as the source while R is called the refrigerator. If now the cylinder C is placed on the body S and the pressure on the piston rod is made a differential amount less than that exerted by the substance within the cylinder on the piston, the piston will be driven upward. This would mean that work is done by the substance within the cylinder and this would im- mediately cause its temperature to fall. On account of the 10 HEAT ENGINEERING perfect conductor forming the head of the cylinder, heat will flow at once into the substance and keep its temperature con- stant, so that if this operation is allowed to go on, a certain amount of external work would be done, a certain amount of heat would be abstracted from S and the substance in the cylinder would be left in a given condition. If the pressure on the piston rod in this second condition is kept at a differential amount above the pressure exerted on the piston by the sub- stance, the piston will move downward compressing the sub- stance within and thus doing work upon it, which tends to in- crease the temperature causing a flow of heat into S the tem- perature of which cannot change. After the starting point is reached the substance within the cylinder is in its original condition with a restitution of the heat taken from S and the development of the first external work on the substance within. These two actions out and back are isothermal and because (a) they can be imagined to take place in either direction, with (6) external conditions differing by an infinitesimal quantity and because (c) all things external and internal can be restored to the initial conditions by coming back over the path to the original condition by a reversal of directions; such action is known as reversible action. These three conditions must hold for all reversible actions. If instead of returning to the original condition after the addi- tion of Qi heat units from S, the cylinder is slipped to the non- conducting plate and then the substance within the cylinder is allowed to expand by causing the pressure on the rod to be a differential amount less than the pressure on the piston, the substance within can receive no heat and the expansion will be adiabatic. The external work will be done at the expense of or by the intrinsic energy in the substance. Of course the tem- perature within the body must fall and when this reaches the temperature of R, (T 2 ), the operation is stopped and the cylinder is put into communication with R. If before the cylinder is put into communication, the force on the piston rod is increased an infinitesimal amount, the sub- stance within the cylinder will be compressed by external energy applied and this operation will be adiabatic, taking place in the reverse direction over the same path. It is seen that this ful- fills the three conditions mentioned above and hence the adiabatic line is reversible. FUNDAMENTAL THERMODYNAMICS 11 When the cylinder is connected with R and the force on the rod is increased by a differential amount, the substance within is compressed. This tends to increase the temperature which causes a flow through the perfect conductor into the body of in- finite capacity. This causes reversible isothermal compression. Suppose that this is continued until Qz heat units are abstracted to such a point that when the cylinder is removed to P and adiabatic compression to the temperature T\ occurs, it will be in its original condition. This is shown on the pv plane by Fig. 5. In this 1-2 is an isother- mal of temperature Ti on which Qi heat units have been added. Point 2 is arbitrary. 2-3 is an adiabatic on which no heat has been added. The point 3 is fixed by its tem- perature Tz at which the adia- batic expansion must stop. 3-4 is an isothermal of tern- FlG 5 ._ Ca ri^t cycle, perature Tz on which Q 2 heat units have been abstracted and on which the point 4 is fixed so that the adiabatic 4-1 on which no heat is added will end at the original point 1. This is known as the Carnot cycle. REVERSED CYCLE There is no reason why these operations could not occur in the reverse direction. In this case Q 2 heat units would be taken from R and Qi heat units would be given up to S. Consider the path 1, 2, 3, 4, 1 on which the total heat added is Qi + - Q 2 + = d - Q 2 but Q = U b - U a + AW and on the closed path U b = U a since the operation is brought back to the original point. Hence Qi -Q 2 = AW If this takes place in the opposite direction Q 2 - Qi = - AW 12 HEAT ENGINEERING or work AW. must be done from the outside so that Qi heat units may be discharged into the source when a smaller quantity Q 2 has been added from the refrigerator. Of course these must be so since if the substance within the cylinder has been brought back to its original condition and Qi heat units have been added while Q 2 have been abstracted, the difference Qi Q z must have gone into external work. In any case since J CdQ = U 2 -Ui + CdW J (dQ = CdW (15) if C/2 Ui = 0. In any closed cycle (a path ending at the point from which it starts) the algebraic sum of the heats on all paths of the cycle is equal to the work done. This is regardless of the reversibility of the path. The heat supplied on any cycle is Qi and hence its efficiency is For the Carnot cycle this may be simplified, since the heat transfers are on isothermals limited by the same two adiabatics. In this case Qi = eTi [See Eq.(14)] This is the value of the efficiency of the Carnot cycle and it is the expression for availability as will be seen later. Hence be- T" __ r p tween the limits TI and T 2 , heat has an availability * if used on a Carnot cycle. MAXIMUM EFFICIENCY Suppose there is a cycle of efficiency greater than that of the Carnot cycle. Call this cycle, cycle r. If there is a source and a refrigerator of temperatures TI and T 2 , the Carnot cycle and other cycle may be operated between these, and the efficiencies will give the inequality Qlc - Q*c ^ Qlr ~ Q 2 r Qlc < Olr FUNDAMENTAL THERMODYNAMICS 13 Suppose these two engines be connected to the same shaft and the efficient engine be used as the driver to drive the Carnot engine in a reversed direction. In this case Qi r will be taken from S and Q\ c will be given to the source. Neglecting friction the work required by one will just equal the work developed by the other. do - Q 2c = Qir - Q 2r Since the inequality is true Qlc > Qir or more heat is being added to the source than is taken from it. Since the two engines coupled to the source and refrigerator re- ceive no energy or give no energy to the outside (one drives the other) the heat going to the source must come from the re- frigerator, and there exists something with no connection with outside systems which causes heat to flow from a point of low temperature to a point of higher temperature when placed be- tween the two points. This is unthinkable from all experience, hence Qi c cannot be greater than Qi r and the efficiency of the cycle cannot be greater than that of the Carnot cycle. Hence the Carnot cycle is as efficient as any cycle. Suppose now the cycle is reversible, in which case the Carnot cycle might be used as the driver if of greater efficiency than that of the other cycle. Qic - Q2c Qir - Q 2 r Qlc Qir V Qlc - Q 2 c = Qir - Q 2 r Qir > Qlc Or the amount received by the source from the reversed cycle is greater than that taken by the direct Carnot engine. This cannot be so and hence a Carnot cycle cannot have a greater efficiency than that of the reversible cycle. But none can be greater than the Carnot. Since the efficiency of the cycle r cannot be greater and can- not be less than that of the Carnot cycle it must be equal to it and hence all reversible cycles have the same efficiency, which is equal to that of the Carnot cycle. All that can be said of non- reversible cycles is that they cannot have greater efficiencies than that of the Carnot cycle. 14 HEAT ENGINEERING This proof -can be used to show that the efficiency is inde- pendent of the medium used, for if one substance gave a higher efficiency it could be used on a Carnot cycle as a driver while the other substance would be used as the medium of a reversed Carnot cycle. The reasoning would lead to the same absurd result and hence all substances will give the same efficiency theo- retically when operating between the same source and refrigerator. The Carnot efficiency is the maximum efficiency and if the limiting temperatures are Ti and T 2 is the maximum availability. Being the maximum it is that which should be obtained in practice theoretically and hence it is called simply the availability. CONDITIONS FOR AVAILABILITY OF HEAT This leads to the observation that the only way in which heat can be available is to have a difference in temperature necessi- tating two bodies at different temperatures and a substance which can receive and discharge the heat. The three systems are necessary to make heat available. (m _ rji \ -^f, - JQi are available for useful work m and QIY must be rejected. Of course this rejected heat or wasted heat so far as the work ATT is concerned may be used for some valuable purpose, as when it is rejected from a steam engine and used for warming buildings or for heating water for a wash house. This leads to the Second Law of Thermodynamics: SECOND LAW "It is impossible by means of a self-acting machine unaided by any external agency to convey heat from one body to another at higher temperature" or "no change in a system of bodies that can take place of itself can increase the available energy of the system." When the matter is discussed in the manner given above, this law results in the expression m _ rp ^ = Availability or Carnot Efficiency, , dQ dH dS = -r + -r FUNDAMENTAL THERMODYNAMICS 15 is the conclusion from this law when the gain in unavailable energy is considered, as will be seen later. Each of the above expressions results from the second law of thermodynamics. Having seen two paths which are reversible, the isothermal and the adiabatic, it may be well to consider some changes which are irreversible and for that reason cannot truly be represented on the pv plane. LOSS OF AVAILABILITY Suppose heat flows by conduction from one body of high tem- perature to one of low temperature. Such action cannot possibly be assumed to take place in the reversed direction and therefore on account of the first requirement it is non-reversible. If a gas is assumed to discharge freely from a vessel of pressure pi to a place of distinctly lower pressure p%, such action could not be assumed to take place in an opposite manner, and lastly if work be changed into heat as by friction this could not be as- sumed to take place in a reverse direction. In all of these there is a loss of availability. In the first the loss of available heat would be the difference of the available heats at the two temperatures. This is av if To is the lowest possible temperature, and T\ and Tz are the temperatures of the two sides of the conducting surface. In the second case the pressure has been brought nearer to the pressure of the surrounding medium and hence by the time it is brought to this pressure by adiabatic expansion the tem- perature will not be so low as it could have been before. Hence 4 - 9] > 4 - Since To < To' Ti is almost equal to TV Loss in available heat = T Q jf, To' ^r J-l 1\ In the third case if ATT units of work are changed into heat of temperature T\ the available part of this is 16 HEAT ENGINEERING All of it was available as mechanical energy before the change, hence represents the loss of available energy. In the reversible changes or on the reversible cycles the un- available energy before the operation was a Ta Q.jT and this is the amount rejected by the reversible cycle. Hence in this case there is no increase of unavailable energy. ENTROPY Goodenough points out that in all of these cases the non- reversible changes have brought about a loss of available energy and that these expressions are all of the form He then says that the quantity which when multiplied by T , the lowest available temperature, gives the increase of un- available energy due to a non-reversible or other change is called the increase of entropy. This quantity is of the form ^ and refers to one system or to several systems. If one system alone is considered there may be an increase or decrease of entropy due to the change of heat, while if two systems are considered reversible changes lead to no change in unavailable energy and hence no change of entropy, while non-reversible changes lead to an increase of un- available energy and consequently an increase of entropy. Thus if the source alone is considered when Qi heat units are given up to the cylinder of the Carnot engine, there is an un- available amount of heat T >~ taken from this source and 1 1 given to the medium in the cylinder. There is a decrease of entropy of w^ for the source but that of the medium is increased 1 1 by ~- and the sum of these two is zero. This is true if the cycle i i FUNDAMENTAL THERMODYNAMICS 17 is worked in the reverse direction. If there is conduction of Oi heat the loss of entropy will be ~, while the gain of entropy will be 7^-. This latter is gr -I 2 there is a gain in entropy be 7^-. This latter is greater since T 2 is less than TI. Hence -I 2 when the irreversible change takes place. This is the only direction in which this operation can take place, while in re- versible action the operation may take place in either direction. Hence if all bodies or systems taking place in a change be con- sidered together, a process which will take place of itself will be accompanied by an increase of entropy. If there is no change in the entropy the change will take place in either direction. Goodenough shows further that only one part of a system may be considered and then the increase of unavailable energy Q fdQ J T may be accomplished by adding heat Q or if the temperature TI is not constant by adding I dQ. Since this heat may come from internal friction, as well as from the outside, and the symbol Q refers to heat added, H will be used for the heat de- veloped by internal friction. In this case then CdQ , CdH Entropy change = I -^ + I -^ There could be a loss of availability and consequently an in- crease of entropy if there was internal conduction, but this need not be considered as in most problems the system is assumed at a uniform temperature throughout and any change in one part is the same in all parts. If S be assumed for the symbol of entropy Since the amount of unavailable energy is dependent on the energy in a body and its temperature, the entropy, which is the unavailable energy divided by the lowest possible tern- 18 HEAT ENGINEERING perature, is also dependent on these. Hence entropy depends on the state of a body and dS must be an exact differential, giving Ti UTi ENTROPY AROUND A CYCLE Now dH is the heat of internal friction and is positive in all cases. Hence although around a closed cycle I dS = 0, since a return is made to the original point, I -^- must always be positive and therefore in such a case -~~ must be a negative CdQ . \ -^ is negatr quantity. When there is internal friction I -^ is negative around a closed cycle. If, however, dH = there is no friction and 'f = (19) around a closed cycle. This friction when it exists always acts to produce heat no matter in what direction one progresses along a path. It is this which makes a theoretical path of any form on the pv plane an irreversible path. Hence the following statements are made: on any closed cycle. ^=0 C = I on a closed reversible cycle. S* - Si = > I ^ (20) on a closed irreversible cycle (internal friction present). Now dS = ^ + or r r f dH J7US = idO + I FUNDAMENTAL THERMODYNAMICS 19 ENTROPY DIAGRAM TdS is an area between a curve of coordinates T and S and the axis of entropy and it must equal the heat added from the outside if dH = or the line is reversible and if dH does not equal zero it represents the heat added from the outside, plus the heat added by friction. With these coordinates, lines of constant temperature become parallel to the S axis or horizontal while lines of constant entropy are vertical lines. The areas beneath any line represent the heat added from the outside plus the internal friction. If there is no friction (the line is reversible), area beneath the line represents the heat added from the outside. If the lines are reversible the line of constant entropy must be an adiabatic since 'dQ and because T does not equal infinity, dQ must equal zero. This is the condition for an adiabatic. If the lines are adiabatic lines with internal friction the entropy increases due to this friction and hence the lines must progress to the left. CHARACTERISTIC EQUATIONS A substance of unit weight under a definite hydrostatic pressure and at a given temperature will occupy a certain volume and if the pressure and temperature change, the volume will change. In most cases there is for every substance a functional relation between these three properties or 0(p, v, t,) = (21) This is called the characteristic equation of the substance and for each substance there is assumed to be such an equa- tion determined by experiment. This equation being between three coordinates is the equation of a surface. The equations for some of the well-known substances used in heat engines are as follows: For perfect gases pv = BT or (22) pV = MET (23) 20 HEAT ENGINEERING For saturated steam T log p = log k - n log T _ ^ (24) For saturated vapors log p = a -f ba n + c@ n (25) where n = (t t ) For superheated steam v + c = y- - (1 + ap) (26) For other superheated vapors pv = BT - cp n (27) HEAT ON A PATH Now to study any path on which heat is added the path on the surface represented by the characteristic equation could be projected on any one of the three planes, pv, pt or vt. The first one of these is the best since area beneath the path represents external work but further than this there is no reason for its use. Suppose for instance the path is shown in Fig. 6. To find the heat the v path is changed to the broken path of IG. . p ^ ane on e p constant pressure and constant volume lines of differential length as shown. The heat to change the volume by unit amount on the constant pressure line is ( ) and the amount to change the pressure by \ OV I 77 unity on a constant volume line is (-7 ) . In the broken path \0p/ v the change from a to 1 is dv and that from 1 to 6 is dp. These quantities are arbitrary in amount if the path is not fixed and hence they are independent variables. The heat added on a path is *-(),*+* < 28 > It must be remembered that for a definite path only one of these is an independent variable since the path necessitates a relation between the two variables. FUNDAMENTAL THERMODYNAMICS 21 In the same manner, if the path is projected on the plane pt, or vt, the expressions would be or If a pound of substance is referred to, these may be written with small letters. The equations are the fundamental differential equations of heat. The quantities (TT) , etc., which represent the amount of heat added to 1 Ib. of substance to change one of the proper- ties, p, v, or t, by unity when another property, p, v, or t, remains constant, are known as thermal capacities. Since these are ex- pressed in heat units which refer to water their values have the definite names given below f ~~J = c v , specific heat at constant volume. ) = c p , specific heat at constant pressure. ot I p These are the amounts of heat to change the temperature of unit weight by one degree when either the volume or pressure is constant. ~ ) = l v , latent heat of expansion. OV/ i This is the amount of heat added to 1 Ib. of substance at constant temperature to change the volume by unity. It is latent because the temperature does not change. \~~ } = lp, latent heat of pressure change. ( j = n, heat of expansion at constant pressure. I--] = o, heat of pressure change at constant volume. Using these symbols the three equations may be simplified dq = c v dt + l v dv (290 dq = Cpdt + Ipdp (300 dq = ndv + odp (280 22 HEAT ENGINEERING It will be remembered that dq is not an exact differential and hence there is no differential relation between the coefficients of any one of the equations. There are relations which may be made by equating any two of the equations since these are ex- pressions for the same quantity. Hence c v dt + l v dv = c p dt + Ijdp (31) This appears to be an equation with three independent vari- ables. There is always a characteristic equation with p, v, and t, and hence any one of these three is known in terms of the other two. Suppose t is eliminated by Hence r / dt \ , 7~ij / 8t \ r / 8t \ cJ ) + l v \dv + c v ( ) dp = cJ ) + L \ozv p \0>/ 1 - L \dp/ Now this equation is true for any value of dv and for any value of dp if a definite path is not assumed, or in other words, it is true for any path. The only way that this can be true is to have the coefficients of the same variable on each side of the equation equal. Hence or c p c v = M|?) (32) and or 7 II /OO \ c p c v = MTT) (33) \ol/ p Eliminating dv from (31) leads to (34) and /AI\ (33) FUNDAMENTAL THERMODYNAMICS 23 Eliminating dp from (31) leads to and which have been found before. By equating (28') with either (29 ; ) or (30') other relations could be found. (35) (36) (St\ = CJ ) \8pJ v DIFFERENTIAL EQUATIONS AND RELATIONS Now Jdq = du + pdv and for reversible paths (dH = 0) dq = Tds Tds .'. du = Jdq pdv = pdv (37) A. Another quantity used throughout the theory of thermody- namics which has an important use is the heat content. This is not the heat contained in a body but it is that heat plus the product of the pressure and volume. This is the amount of energy which would leave with a substance when it is forced out of a region in which the pressure is kept constant. It is represented by the symbol i and denned by i = A(u + pv) (38) / = Mi . M = weight of the substance. It will be seen that i depends on the state of the substance since u, p } and v are all dependent on the state. It is expressed in heat units. di = Adu + Apdv + Avdp Substituting from (37) di = Tds + Avdp (39) 24 HEAT ENGINEERING Two other quantities known as thermodynamic potentials, F and $, are given by the equations F = Au - Ts = Ai - Ts These are both functions of the state, since u, T, s, and i are fixed by the state. Their differentials reduce to - dF = sdT + Apdv (40) d = Avdp - sdT (41) Now du, di, dF and d are all exact differentials since they depend on the state and hence, as they are expressed as functions of two variables, the partials of the coefficients of the independent variables with regard to the other independent variables are equal. Taking the equations (37), (39), (40) and (41) the follow- ing results are true: ().-'. < '' The four equations above are Maxwell thermodynamic equations. Now dq Y = ds Hence (44) and (45) reduce to * (46) Substituting (46) and (47) equations (29') and (30') become dq = c v dt + AT dv (48) dq - Cj4t - AT dp (49) ot/ p FUNDAMENTAL THERMODYNAMICS 25 Substituting (36), (35), and (33) in (28') this becomes ->(!) > + *(.* (50) Now C r . - 7 || /'QQ\ P Cv L v I I (^cJo; and and /M\ \SfV p Cp C v u iu\ \8pJ v Cp c v Equation (50) becomes (48), (49) and (52) are in the same form since the partial de- rivatives are all derivatives with regard to dT. Equation (51) is an important relation which leads to valuable results. Goodenough derives two other equations which are of value. These follow: du = Jdq pdv using (48) this becomes du = Jc v dt + [^(^) - p]dv (53) di = Tds + Avdp = dq + Avdp using (49) this reduces to di = c p dt - A [T(f}) -v]dp (54) 26 HEAT ENGINEERING du and di are exact differentials. Hence /5c v \ dp 5 z p dp \dv/ t ~ M dt 2 " dt and from (54) (!'),- -, :'.;' <' These equations (55) and (56) are of value in making certain deductions. Equations (46), (47), (48), (49), (50), (51), (52), (53), (54), (55) and (56) are important equations in the development of thermodynamics. It must be remembered that all of these are general equations and refer to any substance. The partial derivatives, such as I } , are found from the characteristic equation of the substance used in any problem. PERFECT GASES ; A perfect gas is a substance which obeys the law of Boyle and the law of Charles. From these two laws the characteristic equation becomes pv = BT (57) or pV = MET (58) p = pressure in pounds per square foot v = volume of 1 Ib. of substance in cubic feet V = total volume of M pounds in cubic feet M = weight in pounds B = a constant T = absolute temperature = Fahr. temp. + 459.6 Now m = weight of 1 cu. ft. of gas. FUNDAMENTAL THERMODYNAMICS 27 If for air, the weight of 1 cu. ft. is found to be 0.08071 Ibs. per cubic foot at atmospheric pressure and at 32 F., B is found to be 2116.3 0.08071 X 491.6 By the law of Avogadro equal volumes of gases at the same pressure and temperature contain equal number of molecules hence, if wt = molecular weight m = Kwt - D _ Po _ const. = Kwt TQ ~~ wt or Bwt = const. = R (60) R = universal gas constant. The weight of oxygen per cubic foot under atmospheric pressure and at 32 F. is 0.089222 Ibs. _ 2116.3 __ B oxygen ~ Q Q89222 X 491.6 Hence R = 48.25 X 32 = 1544 and Now the molecular weight of air is found from the fact that air contains 79 per cent, by volume of N% and 21 per cent, of 02. By Avogadro's Law, considering one molecule to each unit volume, the weight is equal to volume multiplied by the molecular weight. 0.79 X 28.08 = 22.18 0.21 X 32 = 6.72 wt. of 1 vol. = 28.90 Hence R - 15*4 _ #*> ~ 28.9 " The correct value of this is 53.34 showing that the molecular weight of air is 28.93. If there is a mixture of various gases in a given volume there are two points of view to be considered: (a) all of the constituent 28 HEAT ENGINEERING gases occupy the whole volume and each is under a partial pressure such that f the sum of the pressures equals the total pressure, or (6) each gas is under the total pressure but occupies a part of the volume so that the sum of the partial volumes equals the total volume. The first view is known as Dalton's Law and represents what really takes place when a number of gases are mechanically mixed. If pi, p%, p s . . . represent the partial pressures of the various gases which weigh MI, M 2 , M 3 . . . and if all the gases occupy the volume V the following are true: p = PI + p% + PS . . (Dalton's Law) M = Mi + M 2 + M 3 . . . p = M 1 B 1 ^ + V Pp = MB mixture = or R 2MlBl ^mixture ~ This is not as convenient a method as working with volumes since in most cases it is the proportional volumes which are known and not the proportional weights. By Avogadro's Law each unit of volume may be assumed to have one molecule. proportional weight. 2 proportional weights = total weight If this is divided by the sum of the volumes, the weight of unit volume, which by assumption is the molecular weight, is found. Hence ZViwh . J\V ~ Wlmixture WW I 544 & mixture ~ In the above manner by (61), (63) and (64) the value of B for any gas or mixture may be found if the gas is assumed to be a perfect gas. No gas obeys the laws of Boyle and Charles com- pletely and hence the characteristic equation pV = MET is not exactly true for any gas. Under great pressures there is variation, but air, hydrogen, nitrogen, carbon monoxide, meth- FUNDAMENTAL THERMODYNAMICS 29 ane, ethylene, and other hydrocarbons are assumed to follow these laws. Carbon dioxide and steam do not obey the equa- tion, but at times they are handled as if they did. To aid in working out the values of B the following molecular weights are given: Hydrogen H 2 2.016 Carbon monoxide CO 28.0 Oxygen O 2 32.0 Methane CH 4 16.032 Ethylene C 2 H 4 28.032 Nitrogen N 2 28.08 Carbon C 12.0 Ammonia NH 3 17.064 Carbon dioxide CO 2 44.0 Steam H 2 O 18.016 Sulphur S 32.06 DIFFERENTIAL HEAT EQUATIONS FOR PERFECT GASES For a perfect gas pv = MET or pv = BT and the various equations (46), etc., are reduced by the following: Sv\ B It is to be remembered that although t + 459.6 = T U = 5T. ATB dq = c v dt + ' - dv = c v dt + Apdv (48) A TB dq = c p dt - ^dp = c p dt - Avdp (49) dq = c p ^dv+c v ^dp = c p T^ + c v T^ (50) c p -c v = AT~ = AB (51) 30 HE A T ENGINEERING In equation (48) the last term represents the external work, therefore the first term represents the change in intrinsic energy : Jc v dt = du (65) Now du is an exact differential and must be directly integrable. Hence c v dt must also be directly integrable (i.e., with no reference to any path) and therefrom c v for a perfect gas is a constant or a function of t. SPECIFIC HEATS Now experiment proves that c v is a function of t of the form c v = a + bt or c v = a + b(T - 459.6) = a - 459.66 + bT = a' + bT (66) But c p c v = AB c p = AB + a' + bT = a" + bT (67) Now since c p c v = AB c p , c v , and AB must be of the same nature, if c p and c v are specific quantities AB must be a specific quantity. Since c p is the amount of heat to change the temperature one degree and since c v dt du and therefore c v is the amount of internal energy change when the temperature changes one degree on any line, AB must be the amount of external work when the temperature changes one degree at constant pressure. Since wt X B = R, a constant wt X c v = a in + b lu T wt X c p = a + bT are the universal forms of specific heats and these differ by AR. 1544 wt X c - wt X c v = AR = - = 1.9855 The values of the a's and 6's are given below wt X c v = 4.77 + 0.000667* wt X c p = 6.75 + 0.000667* For C0 2 c v = 0.15 + 0.000066* c v = 0.195 + 0.000066* i f for any perfect gas. FUNDAMENTAL THERMODYNAMICS 31 For superheated steam. (Approximate.) c v = 0.324 + 0.000133* c p = 0.435 + 0.000133* Although c p c v = AB is always true the relation C f = k = 1.4 (68) c v is approximately true for ordinary temperatures. Equations (51) and (68) are not both possible for two such equations could only hold for definite values of c p and c v . (68) is approximately true since the value of the quantity b is small and as it gives simpler results to certain problems its use is advisable unless there is a great change in temperature on the path considered. Since c p c v AB and c p = kc v AB = (k- IK or c p : c v : AB = k : I : k 1 ISOTHERMAL AND ISODYNAMIC LINES If in equation (65), T is made constant du will be equal to zero, or u is constant. Hence for perfect gases the isothermal is the same as the isodynamic, Since pv = BT pv = constant is the equation of this line on the pv plane. This is the equation of the rectangular hyperbola. ADIABATIC The adiabatic is such a line that dq 0. From (48). = c v dt + Apdv D/77 c v dt = Apdv = A - - dv c v dt dv 32 HEAT ENGINEERING or Vi\4* but hence ;- 1 or Tv k ~ 1 = const. This is the equation of the adiabatic on the v T plane. To reduce this to the equation for the pv plane, T must be eliminated. Now T -PL ~ B Hence const. or pv k = const. (70) This is the equation of the adiabatic of a perfect gas on the pv plane. Now p Substituting this in (70) gives (BT\ k P \p) '' = const ' k-l p k T = const. (69), (70) and (71) are the projections of the adiabatic line of the surface pv = BT (perfect gas) on the three planes of projection. The line pv n = const, is known as a polytropic and for the per- fect gas this has the three forms on the different planes: pv n = const. (72) n-l ,~ Q N p T = const. v n ~ l T = const. (74) FUNDAMENTAL THERMODYNAMICS 33 THERMODYNAMIC LINES The six important lines used in thermodynamics are the isothermal (constant T) , adiabatic (no heat added from the out- side), isodynamic (constant intrinsic energy), constant pressure, constant volume and polytropic. For a perfect gas these lines on the pv plane have the following equations: Isothermal, pv = constant Adiabatic, pv k = constant Isodynamic, pv = constant Constant pressure, p = constant Constant volume, v = constant Polytropic, pv n = constant FIG. 7. Thermal lines. It is seen that each is of the form pv n = constant in which n has certain values. n = 1 for isothermal and isodynamic n = k for adiabatic n = for constant pressure n = oo for constant volume These are shown in Fig. 7. Since these are all of the same form it will be well to investigate the general case first. 34 HEAT ENGINEERING pv n = const. dq = c v dt + Apdv du = Jc v dt dW = pdv AB C ~ k - = Jc v (T z - TO = ~~ (T z - f f f ,1-n I" I dw = I pdy = const. I v~ n dv = const. ^ t/ t/ti e/0i -^ Bi . const. Since = v n Now Const. = pat)i n 1 n 1 n When _ n = 1 Pzv z = piVi and the expression for work reduces to Work = g- an indeterminate expression. To find the value of the work in this case, the integration must be made by using the substitution for this case. Work = I pdv = const. I = const. log e = piVi\og e (76) since const. P ~ v The value (75) for work holds for all values of n except for n = 1. In this case (76) is the expression for work. Equation (75) may be put in two other forms, thus: n n Pzvz (770 FUNDAMENTAL THERMODYNAMICS 35 Since Ml = Pi y /Pi\ --r or (78) __ k 1 1 n ' for all values of n except n = 1, and q = p&i log e (80) The expressions for the heat become: for n = 1 since there is no change in intrinsic energy on the isothermal. If these do not refer to 1 Ib. but to M Ibs. _Iog. (80') It will be seen from (79') that + ^ -- is proportional to the heat. i i n _ i is proportional to the change of energy, and is proportional to the work IV along the polytropic pv n = const, of a perfect gas. It must be remembered that this is only true for perfect gases and on polytropics. For n = k, the expression (79) becomes _ k-l l-k or the intrinsic energy change *- I - J-^ is equal to minus the fC - 1 work. If the final volume is infinity this in the form of (78) reduces to 36 HEAT ENGINEERING This is the work under an adiabatic to infinity from the point 1 and hence it is the expression for the intrinsic energy at the point 1. The expressions (75) and (76) for work are expressions for areas beneath lines and consequently the expressions are true for all substances expanding on these paths. Since p^Vz = piVi on the polytropic n = 1 there is no change in intrinsic energy on this line for a perfect gas and it is the division line between positive and negative energy changes. On the line n = k there is no heat added and so this is the division line between positive and negative heat. -AW -AC FIG. 8. Lines of division between positive and negative quantities. The constant volume line is the division line between positive and negative work. These are shown in Fig. 8. EQUALITY OF GAS SCALE AND KELVIN SCALE Now pv = BT has been derived from the definition of the perfect gas from which (81) dt v T From (53) which has been derived from absolute considerations du = Jc v dt + \T -IT pldv L ot FUNDAMENTAL THERMODYNAMICS 37 Since du is independent of v for a perfect gas by the experi- ments of Joule and Thompson: (Joules' Law) T ' - p = or (81) and (82) give the same result, hence the absolute Kelvin T of (82) must be the same as the perfect gas T of (81). HEAT CONTENT OF GASES a- P V Since u = k- 1 7 = A fl P V < 84 ) For a change 7 2 - /! = A 7r^-r(p2F 2 - PI^X) (85) ENTROPY OF GASES For reversible lines , dq ds = -~ hence from (48), (49), and (50): s* - 81 = Cc v ~+ CA P T dv = c v log e ^ + AB\og e ^ (86J s 2 - si = c p log e ^ AE log e -^ (87) S 2 - Si = C p loge ~ + C ^ 10 S ~ ( 88 ) These changes in entropy depend only on the states at the beginning and the end of the path. CYCLES AND CROSS PRODUCTS Cycles are the paths showing the changes in the properties of a substance as it undergoes a change. They are often shown on the pv plane as in Fig. 9. A simple cycle is one made up of two 38 HEAT ENGINEERING pairs of similar lines. A cycle of four or more different lines is a complex cycle. The cycles of engines using perfect gases have certain properties provided that they are made up of two pairs of polytropics. If the simple cycle is made up of polytropics, pv n = const., FiF 3 = V 2 V* (89) Pips = Pzpi (90) T,T 3 = T 2 Ti (91) pv n - Const - Const. FIG. 9. Simple cycle of polytropics. This is shown as follows: (89) fCM\\ PIPS = P2P4: C y W If the cross products for pressures and for volumes are equal, those for temperature must be equal. T^ nn T^ '/^ f Q~l ) This latter is true for perfect gases only. For cycles of any substance the cross products of pressures or volumes are equal FUNDAMENTAL THERMODYNAMICS 39 from the geometry of the figure but the cross products of tem- peratures are only true for perfect gases. SATURATED STEAM AND OTHER VAPORS A vapor is a gaseous condition of a substance near its point of liquefaction. When a vapor is in contact with its liquid it is said to be saturated. The characteristic equation for steam (due to Bertrand) is T log p = log k - n log T _ b (92) p = pressure in pounds per sq. in. T = absolute temperature in deg. F. n = 50 32 F. to 90 F. 90 F. to 237 F. 238 F. to 420 F. b = 140.1 b = 141.43 b = 140.8 log k = 6.23167 log k = 6.30217 log k = 6.27756 For other vapors Goodenough gives the Dupre"-Hertz formula with the value of constants from Bertrand. C log p = a - b log T - j, (93) p = pressure in millimeters of mercury T = absolute temperature in degree C. a b c Water 17.44324 3.8682 2795.0 Ether 13.42311 1.9787 1729.97 Alcohol 21.44687 4.2248 2734.8 Sulphur dioxide 16 . 99036 3 . 2198 1604 . 8 Ammonia 13.37156 1.8726 1449.8 Carbon dioxide 6.41443 -0.4186 819.77 These equations are rarely used since tables of the properties of vapors have been constructed giving the pressures and corre- sponding temperatures. From the characteristic equations of saturated vapors it is seen that the pressure and temperature are independent of the volume. When heat is added to 1 Ib. of liquid at 32 F. and of volume v' it is found that the volume in- creases a very slight amount so that dv may be considered as zero, giving dq = c v dt, or better cdt to include the slight amount of work. 40 HEAT ENGINEERING The value of c has been determined experimentally for dif- ferent temperatures, hence if these be plotted as functions of t, the area beneath the curve will give the value of the integral q' = \ cdt (94) This is called the heat of the liquid. It is the amount of heat to raise 1 Ib. of liquid from 32 to some temperature t. For water it is found graphically by plotting c but for other sub- stances the quantity is found by empirical equations of the form q' = a + bt + ct 2 + . . | (95) c = ^JL = 5 _|_ 2ct + Zdt 2 + . . . . (96) After the liquid has been heated to the temperature corre- sponding to the pressure, the addition of heat causes the liquid to boil and some of the liquid is changed into vapor. When all is changed into vapor the volume of 1 Ib. is v" so that the volume has been changed by the amount v" v' = change of volume. If x represents the amount of 1 Ib. which has been changed into steam it is called the quality, (1 x) is the amount which re- mains liquid. The volume of the 1 Ib. of mixture is v = v' + x(v" - v') = (1 - x)v f + xv" (97) V = M[(l - x)v' + xv"} (98) Since v' is small and since (1 x) is small in most cases, (1 x)v' may be neglected, giving V = Mxv" (99) The amount of heat added to change 1 Ib. of liquid from liquid at the boiling point to vapor at this temperature is called the heat of vaporization and is represented by r. r = fdq = Now T is constant and (^7) is independent of v, hence (100) FUNDAMENTAL THERMODYNAMICS 41 This equation may be used to find r if v" v' is known. Since r is usually found by an empirical equation, equation (100) is used to compute (v" v'). For steam r = 970.4 - 0.6550 - 212) - 0.000450 - 212) 2 (101) The total heat added to 1 Ib. of liquid at 32 F. to make it into steam at a given temperature is called the total heat, q" or q" = q' + r (102) For steam q" = 1150.4 + 0.350 - 212) + 0.000333(t - 212) 2 (103) For other substances the equations may be found in tables of properties. Having r, v" v' may be found; since by (92), dp _ nbp dt := T(T - b) v" - ' = - A T _J*P_ ( 104 ) T(T - b) If the volume is changed by v" v' at a pressure p, the external work expressed in heat units is A work = Ap(v" - v') = (105) The internal heat of vaporization is therefore P = r - V (106) This is computed and placed in the tables. If the steam or vapor is such that no liquid is present, x = 1 and the vapor is dry. However, if there is some liquid, x is not unity and the vapor is wet. In either case it is a saturated vapor as the pressure and temperature are related by means of the characteristic equation of saturated vapor. When x is not unity the total heat is q." = q' + xr INTRINSIC ENERGY There is practically no work when the liquid is heated, hence the quantity of heat q' remains in the body and when steam is made the change of intrinsic energy is given by A(u - u 32 ) = q' + r - Ap(v" - v'} = 42 HEAT ENGINEERING Since the intrinsic energy may be measured from 32 F., this atr Vo IXTTM ff on may be written Au = q' + p, or AU = Jlf fa' + p) For wet vapor \' + xp). = q' + xp, or HEAT ON A PATH If heat is added to a vapor on some path as in Fig. 10 q = A(U - U,] (107) (107') = AM [q' 2 .J pd FIG. 10. Path on pV plane. The area beneath the curve is the value of the integral. This may be of any of the forms given earlier. To find x the formula < 108) is used. HEAT CONTENT At times steam tables and charts give the value of the heat content and the method of using this must be clearly understood. i =A(u + pv") = q r + r' - Ap(v" - v') + Apv" = q r + r' - Apv\ Now Apv' is a very small quantity, hence i and q" are almost the same. If, however, i is given i - Apv = Au (109) so that for A(u* HI) may be written and problems solved. FUNDAMENTAL THERMODYNAMICS 43 ENTROPY The entropy change when the liquid is heated is C T dt C T S t - S 32 = c - = i cd(log e T) = s' (110) 1/491.6 -* J|91.6 This entropy of the liquid is found graphically as the area under a curve of c plotted against log T. When the liquid at the boiling point is vaporized The total change of entropy from liquid at 32 F. into dry vapor is Since 32 F. is the datum plane of reference s" = s' + (112) For wet vapor s" = s' + ^ (1120 DIFFERENTIAL EFFECT OF HEAT When a mixture of liquid and vapor has a differential amount of heat added to it, an amount dx of liquid is vaporized, and the liquid (I x) and the vapor x are raised dt degrees. If c' is the specific heat of the liquid and c" that of the vapor, dq = c'(l - x) dt + c" X dt + rdx Now ds = ^ = ^- ~-^P -dt + ^dx This is an exact differential, hence 5 /c'(l -x)+ c"x\ 5 x\ J_ } T ~8T\T T T = T 5T T 2 c" = c' - + (113) as c ' = dT and := ~f~ ~T~~ dq" r /iio/\ c = -Jtfr - m (H3 J 44 HEAT ENGINEERING Since q" = a + &( - 212) - c(t - 212) 2 ^ = 6 - 2c(< - 212) may be substituted and c" may be found. SUPERHEATED VAPOR If the saturated vapor is taken away from its liquid so that additional heat will not vaporize any liquid, the addition of heat will cause the temperature to rise above its saturation value if the pressure is kept constant. This vapor is known as superheated vapor. The equation for superheated steam as reduced by Good- enough from the work of Knoblauch, Linde and Klebe is p = pressure in pounds per square inch v = volume of 1 Ib. in cu. ft. B = 0.5963 log m = 13.67938 n = 5 c = 0.088 a = 0.0006. For other substances the equation of Zeuner is used pv = BT - cp n (115) For high temperatures Mallard and Le Chatelier and Langen give for the specific heat of superheated steam at constant pressure c p = 0.439 + 0.000239* (116) but from (56) according to Goodenough (tep\ AT \5p) T - 2 dt* From (114) 8v B mn .. . 8T = +T^ (l + ap) dc\ 1 Amn(n - FUNDAMENTAL THERMODYNAMICS 45 Using (116) as the form of expression for T Amn(n c m mnn / a = a + 0T + - yUi ~P(l + 2 P From the results of Knoblauch and Mollier, Goodenough re- duces for a and the values a = 0.367 j8 = 0.0001 giving c p = 0.367 + 0.00017 7 + p (I + 0.0003p) ~ (117') log C = 14.42408 7? = pounds per square inch T = degrees absolute F. This value of c p agrees with the results of experiment by Knoblauch and Mollier. Values of this are shown in Fig. 11 as given by Goodenough. The value of the specific heat of other superheated vapors is given below and these are usually taken as constants although this is not true. SPECIFIC HEATS AT CONSTANT PRESSURE AND K'S Superheated ammonia vapor 0.536 k = 1.32 Superheated sulphur dioxide 0.1544 k = 1.26 Superheated carbon dioxide 0.215 k = 1.30 Superheated chloroform 0.144 k = 1.10 Superheated ether 0.462 k = 1.03 If now 1 Ib. of liquid at 32 F. is heated and finally changed into superheated vapor at temperature T sup . the amount of heat required is given by f*T tU p. g,'" = q ' + r + I c p dt (118) J T.at. The last term is found graphically or analytically for different pressures and different amounts of superheat and tabulated or plotted in charts so that q"' may be known. The value of I c p dv may be written as . (mean c p ) (T 8up . - T sat .) HEAT ENGINEERING 0.450 0.425 200 300 400 Degrees of Superheat FIG. 11. Specific heat of superheated steam. 500 FUNDAMENTAL THERMODYNAMICS 47 The value of mean c p is given by rT. UP . I Cpdt mean c p = T jTtai ' T *- sup. -*- sat. The values of mean c p have been compared and plotted in the form of curves by Goodenough. From these curves the table on p. 48 has been constructed: The volume of this superheated steam is computed by (114) and tabulated or plotted. The heat remaining is Au = q f " - Ap(v" f - v') But Au = i - Apv'". (109) /. i = q '" + Apv' i = q"' practically. The entropy of the superheated vapor measured from liquid at 32 F. is (119) T sat . This last term is found analytically or graphically, but in most cases these have been tabulated for steam so that values of s'" may be found for definite conditions. The various quantities for saturated steam mixtures and super- heated steam may be found and these properties are often made into charts. The quantities p, v, t, s, u, i, x are all dependent on the state and, if known, fix the state. Usually any two of them fix the state, so that these two could be used for the coordinates of a diagram. As has been shown before, T and S could be used, and in this the area under a line represents the heat added from the outside if the line is reversible or the heat added from the outside together with friction if the line is non-reversible. Other coordinates such as i and s have been proposed by Mollier. This is known as a Mollier diagram. These will now be explained. T-S AND I-S CHARTS In Fig. 12 the coordinates T-S are used and since heat is added to the water at 32 F. the line of , r dt > = ] C T J 491.6 48 HEAT ENGINEERING -jadns jo Ss 000 co co co odd odd 1C lO C 1C UJ OOOOO CO CO 1C lO MOOOcO ICC TjH T}< 1C O 10 1C dddd O O C5 GO 00 OOOOO 1C 1C iC OOO C 4 } 00 ^t 4 I CM 00 CO t^ CO CO I CO 1C iC odd I odd o I C5COCO I iC 1C iC odd OOiOfN odd iOiOOOiO doodd COOOOiOCO d d d d d S888S? ddddd oco co **< co COIN . CD CD O ooooo ooooo -* Tt< CO CO CO ddddd co co co co co rf< rf Tf T}< ^ OOOOO 00 CO 00 T^ i-H t> t>. CO CO CO 1C iC 1C 1C 1C ddddd lOOCOtNO CO CO iC 1C Tf ddddd ooooo 00CtNO5l^ COCO CO CN CM 1C 3 1C "3 fJ 1C U3 1C 1C >C ddddd ooooo OOCOCTtC C i 5S S3 do I do CM 1 rp ' * 2 ~T *C2 rp \*-"4) r, dt Or = S' 2 + TfT + 1 2 L sup- *, r..,,. Knowing the first point through which the curve must be drawn, the unit volume is found by V v = ~Tur ' M From tables or charts the quality is found for this specific volume. If this cannot be found on account of the limits of these tables v **> The entropy of the first point being known, the quality at the second point is found by (122) and from this the volume, if needed, is V = Mxv" or Mv ' If v can be found from tables or charts, it is easier to get it from the same entropy column or line. The work is then given by Work = Ui - Ut = M(ui - u 2 ) u = i Apv or q' + xp At times the difference between the u's would lead to such an error that it is better to find Vz for a given p 2 and then compute n by the log formula (121) and find work by Work = FUNDAMENTAL THERMODYNAMICS 55 The isodynamic is a line of constant U. U l = U 2 Now Ui = i Apv or q' -f xp .'. ii ApiVi = i z ApzVz or q\ + Xipi = q' 2 + (123) Given the pressure, volume and weight at the first point, the quality is found after determining the specific volume by using the tables and charts if possible; or by using the formula for superheated steam or The value of Ui may then be found and this is equated to u z for the second pressure from which the second quality may be computed and then the volume is found by Vz = Mxzv"z The heat in this case is equal to the work since there is no change of intrinsic energy. The work is found by finding n by formula (121) and then Work , PiZ^PiZ. 1 n The values of s 2 and Si are found and then St - Si = M(SZ - si) The isothermal in the saturated region is the same as the con- stant pressure line and on this Q = M(xz - Xl )r (124) Work = M(xz - xi)t (125) S 2 - Si = M(xz - x,) (126) In the superheated region the curve approaches the rectangular hyperbola and it would be difficult to solve problems on this line if tables were not available. In this case the formula for super- heated steam would have to be used to get the volumes or if tables or charts were used the volumes could be found by finding the degrees of superheat at various pressures by Degree superheat = T T aat , 56 HEAT ENGINEERING and this would fix i, v and s. Then Work Pi7 1 - n Q = U 2 Ui + work S 2 Si = M(SZ Si) On the constant volume line Work = 7 \i ; Uz - Ui = Q = M[q'z + Z 2 p 2 - (q'i + Xi Pi )] Sz Si = M(SZ Si) FLOW OF FLUIDS If a fluid flows through an orifice in which there is a change in , section there is a drop in pressure and as a result there is a change in velocity. In Fig. 17 consider the sections 1 and 2. If the areas are represented by FI and FZ, the pressures by pi and pz } the specific volumes by Vi and t; 2 , the veloc- ities by Wi and w z the following consid- FIG. 17. Orifice for flow era tions must hold when M pounds of of fluid. substance flow per second. Kinetic energy at 1 in ft-lbs. Internal energy at 1 = Mui Work done in pushing substance along Hence total energy at 1 Total energy at 2 = M Now if there is any heat added between these two points, say MJq, the energy at 2 must equal that at 1 plus MJq. Of course if Jq is taken away the sign would change. Hence ^ + + Jq = ^ + uz + or since T" ~ = ^ + ^1 + Q = Jq + J(ti ~ (^2 + (127) (128) If now becomes FUNDAMENTAL THERMODYNAMICS Q\ 57 is so large that Wi is small and if q = 0, the formula 2 7TT = f(*l ~~ *2) or (129) (130) Since q = this action is adiabatic but in the case that there be internal friction, 1 and 2 are not on points on an isoentropic line but a line passing to the right of this line on the T-S plane. In other words, iz is greater than it would have been if there had been no friction. The amount of this increase in iz is usually found by assuming that 2 has the same entropy as 1 and then in- U P FIG. 18. TS diagram for flow of fluids. FIG. 19. pV diagram for flow of fluids. stead of using *i 22 for these points, only a fraction of this is used. In Fig. 18 abcde is equal to ii, since V abcf = q' cdef = xr abcde = q' + abg2e = i% i iz = cd2g xr = i Apv' This assumes no friction, hence d and 2 have the same entropy. By experiment the heat used in friction is found to be y(ii i 2 ), hence this must be subtracted to get the amount of heat left to give the gain in kinetic energy. Hence (131) - y) ii and i 2 are for points on the same entropy line. In Fig. 18 the area cd2g is cut down by the amount hd2k which .isequal to 58 HEAT ENGINEERING y(ii it). This heat stays in the substance and hence at exit i is not the iof.2 but that of 2' which is fixed by making If this value of i v could have been determined the original formula (130) would have been used. It is because iy cannot be found that this method of using a portion of the amount for isoentropic expansion is employed. It must be remembered that the shaded area less the friction loss equals the gain of kinetic energy. 2 2 - y) The quantity z\ iz is the same as the area behind the adia- batic on the pv plane as shown in Fig. 19. Area a!2c = aleb + el2d - c2db = J(ii iz) On account of friction this is reduced as pointed out above. THROTTLING ACTION Suppose now that there is so much friction that there is no gain in kinetic energy although there is a drop in pressure; this is called throttling action. In this case "7% ~"~ f-k ~~ \J *~"~ / v^l ~"~ vfyj 20 2g or t! = iz (132) This means that in throttling action the heat content is con- stant. The point 2' of Fig. 18 then is found to be on a curve of constant heat content. The horizontal lines of the Mollier chart or the constant i curves of the T-S diagrams are throttling curves on these diagrams. Since in perfect gases Ji =^rr PiVi = j-^j MBT l (133) the throttling curves for such are curves of constant temperature. FUNDAMENTAL THERMODYNAMICS 59 VELOCITY OF VARIOUS SUBSTANCES For short tubes and orifices the friction is negligible and the action may be considered isoentropic as well as adiabatic. For liquids, since the temperature change is slight, i = Apv + const. - p z ]V Since Fi = V 2 F = l m [Pi ~ P*]^ = h w* = V20/i (134) This is the usual formula from hydraulics for the velocity of a liquid. k For gases JYi = A; Since This is the formula for velocity and is a maximum when p z = 0. DISCHARGE FROM ORIFICES Now Mv = volume per second = Fw This is zero when p 2 = PI and when p% = 0. Of course the latter is unthinkable because there will always be some weight 60 HEAT ENGINEERING when p z is less than pi. The explanation is that when the pres- sure p 2 falls the pressure which exists in the plane of the orifice where the area is F can never become less than that to give maxi- mum discharge. To find the condition for a maximum value of M, the variable part of the expression is differentiated. For a fixed pi and Vi ,p,x_| /p**i w w is differentiated with regard to the variable p 2 and by equating this to zero, the condition for a maximum is 1 (137) When k = 1.4 for gases this becomes 7)2 = 0.5283pi (138) while for steam, k 1.135, and 7)2 = 0.574pi (139) When p 2 becomes less than the critical value given above for a short mouthpiece, the pressure at this point of area F remains at the critical value. Hence the discharge is constant for all values of p 2 below the critical value. For air this has been proven to be true experimentally by Fliegner who proposes M = 0.53^-^ (140) VT, when 7)2 < 0.53pi and ^ . Lo^ . fet 0.53pi Equation (140) is really a reduction of (136) by substituting (138) for p 2 and reducing. For steam similar experimental results have been found by Napier. Rankine reduced these results to the form M=^ "V. (142) when 7> 2 < 0.574pi or O.Gpi FUNDAMENTAL THERMODYNAMICS 61 when p 2 > 0.6/h The general equation (131) may be used for the velocity as shown above and then M found by after v"% is found for the condition at outflow. CHAPTER II HEAT ENGINES AND EFFICIENCIES A heat engine is any machine in which heat is used to furnish the energy for the production of mechanical work. As examples, the steam engine, the steam turbine, the gas engine and similar machines may be mentioned. To make heat available in one of these engines it must be applied to some substance which undergoes changes. These changes form a cycle and the changes which affect the properties of the substance may be studied on planes of projection by show- ing the successive values that certain properties take. The paths of the change which the substance undergoes are known as a cycle. As pointed out in the chapter on Fundamental Thermo- dynamics Qi heat units are supplied and Q 2 heat units are re- jected, giving Qi Q 2 units of work. All of these machines work between some range of temperature, TI to T 2 , and consequently the availability of the heat is T ' < This is the only part of the heat which could be turned into work and therefore represents the highest possible efficiency. It represents the efficiency of the Carnot cycle for this tempera- ture range and therefore it is sometimes spoken of as the Carnot efficiency of a given cycle. Of course any cycle would have to be a Carnot cycle to have this efficiency, but all that the term means with reference to a particular cycle is the maximum value that might be possible for this cycle working between TI and TV Calling this 171 Ti-T* _ T* ( . 171 = ~TT~ l ~ T, If Qi is the theoretical amount of heat added and Q>2 the theo- retical amount rejected, which cannot be used, ^ = ,3 (3) Vl 62 HEAT ENGINES AND EFFICIENCIES 63 This is the theoretical efficiency of the cycle. The ratio of 773 and 771 shows how close the efficiency of the theoretical cycle approaches the maximum possible efficiency of the cycle. This is called the type efficiency, 772 If this is nearly unity it indicates that the theoretical cycle is almost equal to that of Carnot and hence in theory the cycle is good while a low value of 772 shows that the cycle is a poor one theoretically. For example, the steam engine cycle has a value of 772 above 0.90 while in the case of the gas engine 772 may be less than 0.50. These indicate that an improvement may be ex- pected in the type of cycle used in a gas engine, although for the steam engine there is little hope of bettering the cycle. If for an amount of indicated work AW a the amount of heat Q a is actually required, the actual thermal efficiency 775 is given by AW a ~Q7 =rib If the heat supplied per unit of substance used is q a , be it coal, steam, gas or air, and if M is the amount of substance per indicated horse-power hour, this expression becomes X 33000 X 60 AW a 777 M ~ www ^ w 2546 Q a Mq a ~ Mq a ~ * 5 2546 B.t.u. = 1 h.p.-hr. 42.43 B.t.u. = 1 h.p.-min. The ratio of 775 to 773 is called the practical efficiency, 774, and shows how near the actual efficiency approaches the efficiency demanded by theory and a low value of this means that there have been errors in actually applying the cycle. To make this term larger, jackets, superheated steam, and reheaters have been applied to steam engines. 175 ft ~ * = * (6) If the mechanical efficiency 776 is the ratio of the output to that developed within the machine or shown by the indicator card, this efficiency is found by 64 HEAT ENGINEERING output indicated work AW AW The overall efficiency is then the product of certain of these various efficiencies. output heat supplied (8) The overall efficiency being the product of these, the efficiency may be increased by increasing any of them. In cases such as the gas engine 771 is so great that although 772 and 776 are small the product is greater than that of the steam engine. These various efficiencies will be investigated for different machines. FIG. 20. pV and TS diagrams of cycles. The quantity AW a may be found from the area of the cycle on the pv plane or from the area on the T-S plane in Fig. 20, if the lines are reversible lines. Of course on the pv plane 123456 represents positive work while 6571 represents negative work and hence 723457 represents the network. On the T-S plane, 123456 represents Qi if there is no friction, and 6571 represents Q 2 under these conditions, hence the area of the cycle represents Qi Q 2 or the work done. If, however, there is friction 12345 = Qi + H! and 6571 = (- Q 2 + # 2 ) since the heat removed to the outside from 5 to 7 is more than that shown by the area on account of the heat developed by friction. 23457 = Q 1 - Q 2 + (Hi + # 2 ) = AW + #1 + H 2 HEAT ENGINES AND EFFICIENCIES 65 or the area is greater than the work of the cycle. In any case an irreversible line makes the work less than the area of the cycle on the T-S plane. Looking at the Carnot cycle of Fig. 21, or the cycle of Fig. 20, it is seen that the efficiency is 2345 Q l - Q 2 123456 & If in the Carnot cycle heat is not added on 34 but on some other line 3'4 or 34', the heat added and work developed are decreased by the area of the small triangle. 2345 - 33'4 Hence Eff. = 123456 - 33'4 Subtracting the same thing from numerator and denominator of a fraction less than unity decreases the value of the fraction. This may be seen by remembering that the effect of the removal of the triangular area is greater on the smaller area. Hence the efficiency is decreased. 3 / 8 -- > *^-^. 4 5" 2 n ~ j 5 6 S FIG. 21. Conditions for maximum efficiency. If the heat were removed on a line 5 "2 of Fig. 21 this would cut down the work but would not effect the heat, hence the efficiency would be diminished in this case. From the above it may be said that for a given range of tem- perature heat must be added or taken away at constant tempera- ture if the maximum efficiency is to be obtained and since this efficiency is the values of T 2 and T\ (the limiting temperatures of the range) are to be separated as much as possible. An important point must be borne in mind. Although the 66 HEAT ENGINEERING statement above is absolutely true, it does not follow that a cycle in which heat is added with a varying temperature finally reaching a high value may not be more efficient than one in which heat is added at a constant temperature of lower value. Such cycles may be made necessary by the nature of the medium used. This is shown in Fig. 22. 1234 is less efficient than 5678. What is meant is that given the highest and lowest possible tempera- tures of a cycle, the greatest efficiency would be obtained if all of the heat added were added at the highest temperature and all of the heat removed were abstracted at the lowest temperature. Thus in Fig. 23 the efficiency of the steam engine with saturated . abed ... ,. ... . aWc'd steam is rj while that with superheated steam is rrr~ In eabf eabb g the figure r^ is practically equal to 7 and by adding the FIG. 22. Cycle with varying temperature on heat line. e f 9 FIG. 23. Steam engine cycles with saturated and superheated steam. triangle bb'h to numerator and denominator the expression for the superheated cycle is obtained. This addition having a greater effect on the numerator increases the efficiency. Hence the use of superheated steam, although the heat is not added at constant temperature, does increase the efficiency a slight amount. For other reasons than those mentioned here the use of super- heated steam increases the efficiency of the engine. The effi- ciency would be increased by a greater amount if the heat could have been added on the dotted line d'V. In this case the effi- ciency would have been d'b'c'd ed'b'g HEAT ENGINES AND EFFICIENCIES 67 In some machines as the steam turbine the indicated work cannot be obtained and in such a case the product r/ 5 X r} 6 only can be determined. METHOD OF REPORTING PERFORMANCE OF ENGINES Although the efficiency of an engine tells an exact story, for commercial reasons it is quite common to report the performance of a heat engine in pounds or cubic feet of substance required per indicated or brake horse-power hour. Thus pounds of steam per horse-power hour for an engine or turbine, cubic feet of standard gas per horse-power hour for a gas engine, coal per horse- power hour for a gas or steam engine have all been found in re- ports of tests. This data on the horse-power hour or kilowatt- hour basis is valuable for commercial reasons but because differ- ent gases and coals have different heating values, and steam at different pressures and qualities contains different amounts of heat, these statements are not definite until other data are known. When coal is used in a boiler or producer the coal per unit of output depends on the efficiency of the boiler or producer as well as upon the engine. To separate the losses and find out just where losses occur and just what they are, it is always better to give the efficiency of the heat engine as a percentage using the heat supplied as the base. These other methods are valuable for commercial purposes and the quantities should be reported. A method used in reporting the performance of pumps is by duty. This is the amount of useful work in foot-pounds per (a) 100 Ibs. of coal, (6) 1000 Ibs. of dry steam, or (c) per 1,000,000 B.t.u. The first two methods are not definite although the third method is exact. The duty by the third method, divided by 778,000,000, will give the overall efficiency. Results of a number of tests will now be given: HIGH-SPEED NON- CONDENSING ENGINE Lh.p ............ 130 347.4-213 77l ~347.4+460~ 16 - 7% 1190-1078.5+;*(29.8-15)12.58 ' 1190-181.3 = 14.5% MK" Steam pressure. . . 115.3 Ibs. per 14 ' 5 QA 7 c/ ^2 1 S 7 OQ. / /o sq. in. gauge 1D -' 68 HEAT ENGINEERING Barometer.. . 14.7 Ibs. per 2546 sq. in. *'~ 30. Quality of steam.. 1.00 8.3 sq. in. '~ 30.5[1190- 181.3J " Pressure at end of expan- _ 120 _ sion. . . 15 Ibs. per sq. in. gauge 130 Back pressure. ... 0.3 Ib. per sq. in. gauge Steam per i. h.p. hr. 30.5 Ibs. 42.42 B.t.u. per i.h.p.-min. = Q ~gg = 510 HIGH-SPEED NON- CONDENSING COMPOUND ENGINE I.h.p 130 in = 16.7% B.h.p 115 773 = 14.5% Steam pressure 115.3 Ibs. per sq. 772 = 86.7% in. gauge Barometer 14.7 Ibs. per sq. in. 2546 _ 117C7 775 21.5 X[l 190- 181.3] Back pressure 0.3 Ibs. per sq. in. _ it? m .g| m gauge 14-5 Pressure at end of ex- 115 cc _ iciu 77 e = 757, = 88. 5% pansion... 15 Ibs. per sq. in. gauge Quality of steam .... 1.00 Steam per i. h.p.-hr. . 21.5 Ibs. 42 42 B.t.u. per i.h.p.-min. = -j^r= 362 LOCOMOBILE ENGINE I.h.p.. . 191 391+282-139 Kw. generator. . . . 121.5 1352-1002 773 "1352- 107 = Steam pressure ..... 208 Ibs. per sq. in. (Complete expansion.) gauge Feed Temperature ... 132 F. 28. 1 Barometer . . . 1 4. 7 Ibs. per sq . in. _ 2546 = _ 775 = 9.9[ 1352- 107] Vacuum ........... 1 1.9 Ibs. per sq. in. 20.6 ,_.- 774 = 287l = 7 ' 3 - 4% Degrees superheat. . 282> F. ^ Steam peri.h.p.-hr. . 9.9 Ibs. Eff . of boiler Steam per Ib. coal . . . 8.28 Ibs. Heat of coal ........ 14,099 B.t.u. per Ib. 42 42 B.t.u. per i.h.p.-min. = on** HEAT ENGINES AND EFFICIENCIES 69 PUMPING ENGINE . 861.34 _. 771 ~ 384.2+460"' 839 ' 8 llS9-926+^|(4-1.2)67.8 173 = 1189-76.0 = 26.7% Steam pressure 190.8 Ibs. per sq. 26.7 ??2 QO 17 = ^ /O in. gauge o^./ Barometer 14.8 Ibs. per sq. in. 2546 _ 176 " 10.3711189- 76] - Back pressure. . .1.2 Ibs. per sq. in. 22.1 abs. 7,4 = 2^ Pressure at end of 839.8 . ,, . , 7 /6 = QA1 o=97.4% expansion 4 Ibs. persq. in. abs. 001.0 Quality of steam 0.99 Steam per i.h.p.-hr . . 10.37 B.t.u. per i. h.p.-min. = n 991 = * 92 Duty per million B.t.u. =778,000,000X0.221 X0.974 = 167, 000,000 ft.-lbs. STEAM TURBINE Kw 6257 549-76 771 ~559 + 460~ Steam pressure 203.7 Ibs. per sq. in. 1290-879 abs. ^ 1290- 44 =33% Superheat.. . 165.5 F. 33 ^ = ^Q = 72 ' 5% Barometer.. . 29.92" 2546 176 ~ 11.95 X [1290- 44] xO.746 = 22.9% Backpressure 0.44 Ibs. per sq. in. _22-9 , ^4 DO abs. ***' Steam per kw.-hr ... 11.95 Ib. 42 42 B.t.u. per kw.-min. =^746^0".229 =249 B.t.u. per elec. h.p.-min. =249X0.746 = 186 PRODUCER GAS ENGINE I.h.p 579 2546 775 ~ 0.738 X 0.805 X 14320 = 30% B.h.p 483 483 776 = 579 =83>5% Coal per i.h.p.-hr... 0.805 Heating value of coal 14,320 B.t.u. per Ib. 70 HEAT ENGINEERING Kind of coal ........ Bituminous Efficiency of producer. 73.8% 42.42 B.t.u. per i.h.p.-min. = ' = 141.4 141.4 B.t.u. per b.h.p.-min. = Q~O^K = 169 BLAST-FURNACE GAS ENGINE I-h.p ................... 775 2546 775 = 110.5 B.h.p .................. 565 _5?5_ 776 775 ~ Cu. ft. of gas per i.h.p.-hr. 90.5 Heat value of gas ....... 110.5 B.t.u. per cu. ft. 42.42 B.t.u. per i.h.p.-min. = Q~^KK = 166 1 AA B.t.u. per b.h.p.-min. = 7^5 = 227 U. / o DIESEL OIL ENGINE I-h-P ............. 523 2546 775 ~ 0.37 X 19270 ~ B.h.p ............ 450 450 776 = 523 = 86% Oil per i.h.p.-hr. . 0.37 Ib. Heat value of oil . . 19,270 B.t.u. per Ib. 42.42 B.t.u. per i.h.p.-min. = ^ OKQ = H8 118 B.t.u. per b.h.p.-min. = = 143 O.oo RESULTS OF VARIOUS TESTS Engine tested 110-h.p. Nurnberg gas engine on coke B.t.u, per i.h.p- min. 110 B.t.u. per b.h.p- min. 138 110-h.p. Nurnberg gas engine on anthracite coal. . 210-h.p. Guldner engine on illuminating gas 120 100 150 600-h.p. Ehrhardt engine on coke-oven gas 11,000-kw. Westinghouse turbine 113 136 213 11,000-kw. Curtis turbine 1,500-h.p. locomotive 350 201 10,000-h.p marine engine 246 2,200-h.p. Corliss engine 226 1.000-h.D. air -compressor eneine. . 169 TOPICS Topic 1. What is a heat engine? What is a cycle? What is the general expression for the efficiency of any cycle? What is the expression for the HEAT ENGINES AND EFFICIENCIES 71 efficiency of the Carnot cycle? What does this efficiency represent? Give the meaning of the terms: Carnot efficiency, type efficiency, theoretical efficiency, practical efficiency, actual thermal efficiency, mechanical effi- ciency and overall efficiency. Topic 2. Give the meaning of the symbols: 771, 772, 773, 774, 775, 776, rj. Give the relations between these and the formulae by which each is found. Tell the manner of determining the quantities entering into these formulae. Topic 3. Give the conditions for maximum efficiency in a heat engine and show that although these conditions may not be fulfilled high efficiencies may be obtained. Are these high efficiencies as high as they would be were the conditions for maximum efficiency fulfilled? How are results of tests reported? PROBLEMS Problem 1. An engine using 35 Ibs. of steam per h.p.-hr. at 125 Ibs. gauge pressure, x = 0.98, and with a back pressure of 2.5 Ibs. gauge, has its con- sumption reduced to 30 Ibs. when supplied with steam under the same pressure but superheated 245 F. The back pressure does not change. The barometer is 29.8 in. What is the per cent, saving, if any? Problem 2. A Corliss engine gives the following test results: i.h.p. = 135.6; b.h.p. = 126.0; steam per hour, 3406 Ibs.; pressure at throttle valve by gauge, 135.6 Ibs. per square inch; barometer, 14.68 Ibs. per square inch; pressure at end of expansion by gauge, 20.5 Ibs. per square inch; back pres- sure by gauge, 1.5 Ibs. per square inch. Find the various efficiencies, 77, 776, 775 and 771. Problem 3. A Corliss condensing engine using steam at 174 Ibs. gauge pressure with x = 0.995 and a vacuum of 27 in. with a barometer of 29.8 in. consumes 17.5 Ibs. of steam per kw.-hr. output from generator. Find overall efficiency. By installing a low-pressure turbine and by reducing the vacuum to 28.5 in. and superheating the steam to 200 F. of superheat this consumption is made 13.8 Ibs. of steam per kw.-hr. and the output of the plant has been increased 75 per cent. Is the change of value? Why? Problem 4. A pumping engine gives a delivered duty of 175,000,000 ft.- Ibs. per 1,000,000 B.t.u. when supplied with steam at 165 Ib. gauge pressure with x = 1 and a temperature of hot well of 105 F. How many pounds of steam are used per delivered horse-power hour? If the mechanical efficiency is 95 per cent., what is the steam consumption per i.h.p.-hour? Problem 6. A gas engine has an overall efficiency of 24 per cent. The heat lost in the jacket is 25 per cent, of that supplied by the gas. How much heat is added to the jacket water per hour if the delivered power is 565 kw.? Problem 6. The results of the test of a producer gas engine plant are as follows: Coal burned in 120 hr., 16,000 Ibs.; heating value of coal, 14,450 B.t.u.; gas produced during test, 1,500,000 cu. ft.; heating value of gas, 120 B.t.u. per cubic foot; b.h.p., 120; i.h.p., 150; cooling water, 1,000,000 Ibs.; temperature of inlet water 65 F., outlet water 110 F. Find the efficiency of the producer. Find the overall efficiency, 77. Find the indicated efficiency, 775. Find the heat removed by the jacket water. Find the heat per b.h.p. -hr., per b.h.p.-min., per i.h.p.-hour and per i.h.p.-min. CHAPTER III HEAT TRANSMISSION The phenomenon of the transmission of heat through partitions is very important as it enters into the determination of the areas of radiators for heating systems; of the surface required in con- densers, boilers, evaporators, intercoolers, and for many other engineering structures. Its consideration is important in finding the heat required for warming a building or the amount of re- frigeration to keep a certain cold storage warehouse at a low temperature. The transmission of heat is complicated and al- though much experimentation has been done the exact laws are not completely determined. The following discussion is based on the works of various authors. The results of the authors have been arranged so as to give the student a working knowledge of this important part of applied thermodynamics. To transmit heat from one body to another or from one part of a body to another it is necessary to have a difference of tem- perature. Having this, the heat may be transmitted by one or more of the three methods : radiation, convection and conduc- tion. Radiation is the method of transmitting heat by vibrations of the ether. The hot body starts this vibration which is trans- mitted in all directions through the ether and until another body receives the vibration the energy is not sensible. As soon, how- ever, as a body opaque to these vibrations is placed in the path it becomes heated. In convection heat is applied to some mov- able body and then this energy is conveyed by the actual move- ment of the body with its heat. In this method heat is carried by moving particles of matter. Conduction is the method by which heat travels from the hot portions of a body to another part which is colder. In this method it may be that the more violent vibrations of the particles of the hot portion of the body are gradually transmitted to those in less violent vibration (colder) or according to the later views of the constitution of matter it may be that more electrons are thrown off at higher 72 HEAT TRANSMISSION 73 temperature vibrations and these gradually affect the vibrations of the atoms at the more remote but colder portions of the solid. The laws of transmission by radiation are well known experi- mentally and theoretically. The Stefan-Boltzmann Law for the radiation from a black body is Q = CF(2V - TV) (1) Q = amount of heat per hour in B.t.u. C = a constant = 1.6 X 10~ 9 . F = area radiating heat, in square feet. TI = absolute temperature in degrees F. of hot body. T% = absolute temperature in degrees F. of cold body which must surround the hot body or which must include all rays issuing from the hot body. It is to be remembered that the formula applies to black bodies and then only when the cold surface includes all rays from the hot body. If the surface subtends a solid angle co when it should have subtended a solid angle of 2ir or a hemisphere, the quantity Q is found by multiplying (1) by ZTT The law of radiant energy has been under discussion from the time of Newton in 1690, who proposed a law proportional to the first power of the temperature, until 1879 when Stefan proposed that the law be of the form given in (1) basing his assumption on some experiments of Dulong and Petit, Tyndall and others. After some years (1884) Boltzmann proved that this form was correct theoretically. A black body has to be assumed since a colored body would reflect a certain amount of energy and thus the amount emanating would include this in addition to the amount radiated. For bodies which are not truly black (absorb all radiation falling on them) the law is approximately true. The nature of the surface has also an effect and this should be taken into account. Polished bodies will not radiate as much as rough surfaces. The above law may for convenience be put in the form (2) The above refers to a black body of perfect radiating power. The relative values of different surfaces by Lucke are given in the following table: 74 HEAT ENGINEERING Absorbing power Porous carbon 1 . (X) Glass 0.90 Polished cast iron . 25 Polished wrought iron . 23 Polished steel . 19 Polished brass . 07 Hammered copper . 07 Polished silver 0. 03 The transmission of heat by pure conduction is a simple matter. The law for this is similar to that of the transmission of electric current. The flow of heat is proportional to the differ- ence of temperature and the area of the substance and is inversely proportional to the length of path or thickness. Thus (3) Q = B.t.u. per hour. F = area in square feet. I = thickness in feet. ti temperature at one side in degrees F. t z = temperature at one side in degrees F. C = coefficient of conduction in B.t.u. per hour per square foot per degree F. for 1 ft. thickness. FIG. 24. Transmis- sion surface. Values of C C = C [l + a(t - 32)] t = temperature in degrees F. Substance Co a Substance Co or Air 0.03 to 0,012 0.0011 Cork powdered 0.03 Brass . . . 53.5 0.0011 Glass 0.54 Brick 0.46 Iron 48.5 -0.00012 Carbon 0.8 Limestone 1.35 Carbon dioxide 0.006 Masonry 0.46 Cast iron 40.5 -0.00012 Sandstone 0.87 Concrete 0.46 Steel soft 26.7 Copper 239.0 0.00003 Water 0.292 Cork board . . . 0.17 Wood 0.10 HEAT TRANSMISSION 75 The value of C for gases has been shown by Maxwell 1 to be given by C = Kr,c v K = constant between 0.5 and 2.5. 77 = coefficient of viscosity = the force between two planes separated by distance unity when one plane is moving with unit velocity relative to other. c v = specific heat at constant volume. He also showed that C varies as the % power of the absolute temperature. Nusselt examined insulating materials and found that the conductivity of these substances increased as the absolute temperature. The amount of heat carried by convection depends on the amount of substance involved, its specific heat and its tempera- ture. This is simple to find but the amount of heat which can be abstracted from these particles by a given surface under given conditions is a complex matter and it is this method which will be considered throughout this chapter. This transfer of heat through partitions is the important one in most applications of heat. When heat is transmitted through a surface as in a boiler or as in an indirect heating coil there must be films of substance on each side of the wall which prevent the passage of heat. Suppose, for instance, in a boiler where the temperature of the hot gases on one side is 1500 F. and the temperature of the water on the other is 325 F., the thickness of the steel tube is Y in. and that there are 4 Ibs. of water evaporated per hour per square foot under these conditions. The heat transmitted per square foot per hour = 4 X r 325 = 4 X 889.8 = 3559.2. For the steel wall, by the law of conduction, there results: 26 7 3559 ' 2 - X (<1 ~ ' 2) '' - <* = = 2 ' 78 il. Mag., 1860. 76 HEAT ENGINEERING Now the difference between the water and the gas is 1175 F. and of this there is only a drop of 2.78 F. in the steel wall. A thin scale of Jioo m - on the water side and ^{Q in. of soot on the gas side would account for considerable drop. For the scale, C will be taken as 1. 3559.2 12000 For the soot, C will be taken as 0.1. ' 16 X 12 X 0.1 This accounts for 188 of the 1175, showing that there must be still further resistance at the surface. This of course must be due to films of water and of gas. It is seen from the tables of values of C that water has about thirty times the value of C for air and gas, and hence there will probably be J^i of 987 drop in the water film and 3 %i of 987 drop in the gas film if these are the same thickness. The gas is probably less thick and if the water film is taken as three times the thickness of the gas film the drops in each will be approximately 90 in the water and 897 in the gas film. The thicknesses to give these results are found below assuming C for water 0.292 and 0.009 for the gases. This is the mean of 0.006 and 0.012. X 12 = 0.089 in. X 12 = 0.0272 in. Thus it is seen that a film of small thickness on either side accounts for a great drop which must exist to explain the low conductive powers of the heating surface. If it were not for the films and scale the heat transmitted by the steel tube would be 26 7 Q = -^-(1500 - 325) = 1,505,000 B.t.u. per hour. This would mean an evaporation of 1690 Ibs. of water per square foot of heating surface. Fig. 25 shows the temperature gradient across these surfaces. It appears from the figure and the calculation above that anything which would tend to decrease HEAT TRANSMISSION 77 T the thickness of the films of water or gas would cut down the drop of temperature in these, putting a greater drop in the wall and so causing an increase in the heat transmitted. One method of doing this is to increase the velocity of the substance bringing heat up to the surface or taking heat from the surface. This has been tried and found true. The increase of velocity of the gas or water wipes some of the film away decreasing its thickness. This simple explanation is evident and shows why one would expect an increase in the heat transmitted, if either the velocity of the gas or the velocity of the water were increased across the surface transmitting the heat. In October, 1909, Prof. W. E. Dalby presented to the Institute of Mechan- ical Engineers of Great Britain a paper on Heat Transmission, "the purpose of which was to place before the mem- bers of the Institution a general view of the work which has been done re- lating to the transmission of heat across boiler- heating surfaces." He has given this view and with it a list of 406 references to various papers dealing with this subject. They are listed chronologically, by Authors and by Subjects, and the student is re- ferred to this paper for most of the lit- erature on this important subject. The fact that the velocity of the gases affected the rate of transmission was experimentally known as early as 1848 but there does not seem to be a statement of this until 1874 when Prof. Osborne Reynolds read a paper before the Literary and Philosophical Society of Manchester (Vol. xiv, 1874, p. 9) 1 "On the Extent and Action of the Heating Surface for Steam Boilers." In this very short paper of a few pages, Reynolds points out the im- portance of the subject. He states that the heat carried off from the gas is proportional to the internal diffusion of the fluid at or near the surface, that is, on the rate at which the particles FIG. 25. Temperature gradient. 1 See also The Steam Engine, by John Perry, p. 594. 78 HEAT ENGINEERING which give up their heat diffuse back into the hot gas. This ac- cording to him depends on two things: 1. The natural internal friction. 2. The eddies caused by visible motion which mix up the fluid and continually bring fresh particles into contact with the surface. The first of these is independent of the velocity, and the second term is dependent on the velocity and density of the fluid, the heat transmitted at any point being Q = At + Bmwt (4) where Q = heat per square foot per hour A, B = constants t = difference of temperature m = density of substance, pounds per cubic foot w = velocity along surface in feet per second. He then discusses the quantities A and B and the way in which the heating surface may be of more value by an increase in velocity. He finally closes his paper with the hope of giving a further communication. He does not give the values of A and B. In Reynolds' paper he gave no theoretical discussion nor deri- vation of his formula. In 1898, Professor John Perry gave the following theoretical discussion to prove that transmission varied as the velocity: In a thin film the number of molecules, n, entering the film per square foot of area is equal to the number leaving and the friction force developed by the axial momentum given up by the n mole- cules as they are arrested from velocity w will be given by P(per square foot) oc nw (5) Now since the kinetic energy is proportional to the tem- perature the change in the energy as the surface abstracts heat will be Q oc n (t - 0) (6) t = original temperature of gas 6 = temperature at surface of pipe. Eliminate n from (5) and (6) ) (7) HEAT TRANSMISSION 79 It is known that the force of friction expressed in feet head when a fluid flows through a pipe is* proportional to the square of the velocity w*. To change this to pounds per square foot it must be multiplied by the weight of 1 cu. ft., m, or P oc mw 2 (8) Hence (7) reduces to Q = K'mw(t - 0) (9) where K f = constant of equality. If now, Perry points out, there is a film of gas at the side of the surface of thickness b and conductivity K, there will really be a drop from t' to 9 in the film and only a drop from t to t' in the gas so that Q = K'mw(t - t') = y (f - 0) * jfomtf + * or Q = K'mw t - -- + K'mw K'mwt + ~0 + K'mw K'mw[t - B] 1 + bK'mw (10) K' be called K" the formula will be the same as oK mw ~K~ (9) with a different coefficient. If b decreases as w increases, which is probable, the product bw will be nearly constant and since K is large and m is small for most gases the denominator of the fraction will change little. Hence formula (9) is practically correct. This equation of Perry and that of Reynolds may be changed by remembering that M = mFw where M = total weight per second in pounds F = area of passage in square feet w = velocity in feet per second m = weight per cubic foot. M Hence -=- = mw r 80 HEAT ENGINEERING M and Q = At + B -= Q = K"(t - 0) (100 In Perry's formula the fact that P varied with m and w 2 might have been made to include the fact that P also varies inversely with the mean hydraulic depth, d\. Thus: The hydraulic depth or hydraulic radius is equal to the area of the cross section of a passage carrying a fluid divided by the perimeter of the cross section. Hence Q = K' (t - 0) (11) ai This formula states that the smaller the tube the greater the heat transmission. Dalby shows that in 1888 Ser and in 1897 Mollier gave for- mulae in which the heat transmitted depended on the square root of the velocity of the gases along the tube, while Werner, Halliday, Carcanagues and Brille show that it depends on the velocity. In 1897 T. E. Stanton investigated the "Passage of Heat between Metal Surf aces and Liquids in Contact with Them" (Phil. Trans. Roy-Soc., Vol. cxc A, p. 67) and showed that the transmission of heat varied with the velocity of water, and that although not stated in the paper in words, the formula of Rey- nolds is applicable to the liquid side as well as the gaseous side of a surface. Prof. John T. Nicolson (Junior Institution of Engineers, Jan. 14, 1909, and London Engineering, Feb. 5, 1909, p. 194) and H. P. Jordan (Institution of Mech. Eng. of Great Britain, Dec. 1909, p. 1317) have each proposed formulae for the transmission of heat which are somewhat similar to that of Rey- nolds, but in these as in the works of Stanton the coefficient of transmission depended on the temperature difference. In general, the formula for the transmission of heat through a surface is Q = HEAT TRANSMISSION 81 where Q = B.t.u. transmitted per hour ti = high temperature of fluid on one side in degrees F. t 2 = low temperature of fluid on other side in degrees F. F = surface in square feet K = coefficient of transmission in B.t.u. per hour per square foot per degree F. According to Reynolds K = A + By, or A + Bmw (11') According to Perry K = K", or K"mw (11") or later K = K'" (11"') Finally it has been shown experimentally that K varies with the temperature difference. In the above three values of K it is seen that for a given tube with a given discharge the value M of K would be constant, since mw = -^-. r This part does not depend on the variation of temperature along the pipe. The quantities A, B, K" and K'" may however depend on temperature. In general the temperature along a surface changes from point to point so that t\ 2 is a varying quantity and to apply the values of K determined by experiment it will be necessary to find what is the mean difference in tem- perature if the temperatures of the substance at the ends of the surface are given. The formula to be used for heat transfer is then Q = (K for mean AQ(Mean At)F (12) MEAN TEMPERATURE DIFFERENCE The question then arises: What is the value of mean A? There is evidence to show that K = , .. To determine an ex- pression for mean M for any value of n, two cases will have to be considered, one for any value of n except zero and the other for n = 0. The reason for this is the fact that I x n 'dx = ,_, ^ x n ' +1 for all values of n' except 1. For n f = 1 the value of the 82 HEAT ENGINEERING integral becomes log x. To find the values of mean AZ the two cases are considered. First case: K = constant, or n = dQ = K(t hx - t cx )dF = - 3600 M h c h dt h = 36QQM c c c dt c (13) dQ = heat per hour in B.t.u. for surface of dF sq. ft. K = heat per hour in B.t.u. per square foot per degree t'hx temperature in degrees F. in warm sub- stance at point x t cx = temperature in degrees F. in cool sub- stance at point x M c and M h = weight of substances flowing per second c c and c h = specific heat of substances in B.t.u. per pound per degree F. Temperature of ""^ L^Cool Substance _ ^ __ Transmission- Temperature of~ Warm Substance FIG. 26. Parallel flow. The minus sign before M h is used because dt h is negative as dF increases, measuring from the inlet end of the warm sub- stance, where the temperature is tiA, toward the outlet where the temperature is fa. If the warm and cool substances flow in the same direction the arrangement is called a parallel flow arrangement (Fig. 26) and the temperature of the cold substance increases with dF hence dt c is positive. With the warm substance flowing in the opposite direction from the flow of the cool substance the arrangement is called counter flow (Fig. 27). A minus sign is required before the dt c as the temperature decreases with an increase of dF. If ti c is the temperature of the cool substance at the end cor- responding to the entrance of the warm substance and t^ c is that of the cool substance at the point of exit of the warm substance, the following is true. M h c h (t lh - t 2h ) = + M c c c (t lc - t 2c ) (14) HEAT TRANSMISSION 83 The upper (minus) sign is for parallel flow and the lower (plus) sign is for counter flow in this equation. This of course assumes that all the heat leaving the warm substance goes into the cool substance and hence there is no radiation loss. The following notation may be used: and from (13) it is seen that JlA ~ txh) = + M c C c (t lc - t xc ) / ^ _l J. Vi isxh) "T~ ^lc = txk - t xc = t xh * M eCe txh + M c c c Temperature of Cool Substance Cool ^^. Substance Substance Temperature of Warm Suhstauce FIG. 27. Counter current flow. Now (15) 1 M h c h - M c c c Substituting this in (13) gives 3600 M c c c dF = - F = - d\t x K M h c h 1 M c c c 3600 K M cCc (16) (17) 84 HEAT ENGINEERING H = X(mean At)F = 3QWM h c h (ti 360o;i meanA = Now Substituting for .F its value from (17) M h c h (ti - meanA = , log (18) From (14) ti c - M C C C tih tzh tih ti c (t%h Equation (18) reduces to mean At = tih - At* (180 (19) This is independent of the direction of flow. The form of the expression is the same for parallel or counter current flow. It Counter Current Parallel Current t c Outlet .> t h Outlet t c Outlet '" M k c h t\h (24) This equation gives the value of Q or F if the other is known in terms of K r which is not the transmission per degree per square foot per hour, but the constant which when multiplied by (mean A)~ n gives the coefficient of transmission. The value of K f in (24) as in (12) may vary with any other quantities than temperature and if the value of K' for the conditions of the given 86 HEAT ENGINEERING problem as to velocity, hydraulic radius or density be substi- tuted the expression will be correct. It is well to notice at this point that *^ - gives the value of mean A within J^ of 1 per cent, for any value of n if A^ is less than Ko f ~^~o 2 * The formulae A* 2 or mean At = mean A = need not be employed to find mean AZ if the change in A is slight. If however Ati A 2 is large compared with the mean AZ then the formulae must be used. Having mean A2 for any given problem the heat or surface may be found by or Q = K' X (mean A*) 1 -" X F Q = K X mean AZ X F depending on whether or not K varies with _ e y the temperature. DETERMINATION OF K The method of finding this K will now be considered. FIG. 29. Trans- In Fig. 29 let the various temperatures at mission through tne di v i s i on l mes between the various layers various layers. . J be ti , t , ^2 and t z while ^i represents the temperature of the gas and t 2 represents the temperature of the water. The following results for 1 sq. ft. area: Q = K(ti - h) Cln HEAT TRANSMISSION 87 where c 1 = coefficient of heat conduction for gas film. c 11 = coefficient of heat conduction for soot. c 111 = coefficient of heat conduction for heating surface. c lv = coefficient of heat conduction for scale. c v = coefficient of heat conduction for water film. r, I 11 , P 1 , I, f are the thicknesses in feet of the gas film, soot, heating surface, scale, and water film respectively. Now Q ^ = Zi - ii 1 7 n Q- i l i n ~ *i - v\ It will be remembered that for a perfect gas ra = jf, and this > 1 may be used to find m for any gas. 90 HEAT ENGINEERING For different values of L the values of K are given by the table below. L K 2.0 3.00 m w w w or 1.50 m g w g 1.0 2.00 m w w w or 2.00 m g w g 0.5 1.20 W^WM, or 2.40 m g w g 0.2 0.55 m w w w or 2.73 m^ 0.1 0.29 m w w w or 2M m g w g 0.01 0.03 Wu,^ or 2.99 m w g 0.0 0.0 m w w w or 3.00 m g w g Since air or flue gas at 580 F. weighs about J^ 5 Ib. per cubic foot and a common velocity is 100 ft. per second and hot water which weighs about 60 Ib. per cubic foot may move with a ve- locity of 12 ft. per second, a common value of L will be given by X 100 _ J_ ' " " m w i0 w " 60 X 12 ' 180 and K = 3.00 m g w g (33) Since X is assumed independent of t, Q = 3.00m, W ,^^- 2 F (34) ! ^-At-l 10g 'A^ Before leaving this formula it may be well to point out that Nicolson also suggests bringing into this expression the different diameters of the surface in contact with the gas and the water. Thus dQ = K g m g w g (ti 6}ird g dx = K w m w w w (d t^ird w dx (25) 9- *2 = K g m g w g d g = L , ti e K w m w w w d w e== iT^ + rrL 7 ' 1 *i - e = f(il ~ W (37) ( ll ~ ^ (38) If K g m g w g is known it may be used for the determination of the heat, of L' and (ti t z ) or their mean values are known. HEAT TRANSMISSION 91 Thus Q = K g mgW g i _L_ L f mean (ti t 2 )F This can be done for the water side Hence K = K g m g w g 1 , L , T , 1 r d g since L = ~ L -j- A d w and Kg = 3 according to Nicolson. This result is similar to equation (31) except for the term -r-. The equations (37) and (38) enable one to compute the heat transfer from either the water or the gas side. The value of 2.75 for the coefficient of (33) is recommended by Nicolson. The values of K g = 3 for gas have been determined by Prof. Nicolson from seven sets of experiments in which the tempera- tures of the water and air were varied. This variation however was not sufficient to show the variation of K g with temperature. The values obtained varied from 2.84 to 3.11 but Nicolson did not attempt to show the variation of this with temperature. The value of K w = 6 was determined from Stanton's experiments in which water only was used. In this Stanton points out that K w does vary with the temperature. Nicolson proceeds further to show that K varies with temperature, using superheated steam to transmit heat to air or water and using heated air to transmit heat to water. In addition to these experiments Nicolson discussed certain boiler tests to determine constants for a formula. He uses the formula Q = K(T - e)F (39) and shows that (40) Q = B.t.u. per hour. T = mean temperature of gas along flue in degrees F. 6 = mean temperature of "flue in degrees F. = tem- perature of steam. 4> = \(T+6). 92 HEAT ENGINEERING di = hydraulic mean depth of flue in inches. area _ diam. perimeter ~~ 4 m a = weight of 1 cu. ft. of gas in pounds. w g = velocity of gas in feet per second. F = area in square feet of surface on gas side. M = total weight per second. M This formula is worked out from experiment. It is correct in form if K is of the form K = AM + BM^. Q will be of the form Q = F[AM& 2 + %B(Mi /2 + A^)A^A* 2 1/2 ] or since AZiAZ 2 may be written as At 2 and %(A$i M + A^~) may be written as A^. This reduced to Q = A[M + %BM*]MF This is the form of (39) with the value of K substituted when the value of A is inserted. These are the same and hence the formula is correct in form. If the constants are worked out to fit certain conditions, it will lead to correct results when applied to such conditions- H. P. Jordan in the Transactions of the Institution of Mechan- ical Engineers for Dec., 1909, p. 1317 gives the formula in the form Q = 3600 i 0.0015 + [o. 000506 - 0.00045di + 0.00000165 (^-^)]fr) (T - 0)F (41) This is also seen to be of correct form as is the case of Nicolson's formula. For the range of the experiments the results are usable. Unfortunately these two formulae do not yield the same results for various problems. For instance, using a problem given by Nicolson the differences may be seen. In a Lancashire boiler the flue is 36 in.; 400 Ibs. of coal are burned per hour with 24 Ibs. of air per pound of coal. The gases leave the fire at 2200 F. and at the end of the flue they are 900 F. The steam temperature is 350 F. HEAT TRANSMISSION 93 = 350 F. assumed 400 X 25 _ 3600 X 7 = By (40): i-qcn i -i K = IJgg + ^ V950 X 1.11 X 0.4J = 4.75 + 0.34 = 5.09 By (41): K = 3600 {0.0015 + [0.000506 - 0.00045 X 9 + 0.00000165(950)]0.4} = 2.64 Using the method first proposed by Nicolson and assuming a velocity of 3^ ft. per second for the water, L = - - - 013 K = (2+^013) a4 = 1 - This third method is evidently incorrect due to the temperature effect and there is a great discrepancy between the results from the formula of Nicolson and that of Jordan. If these had been used for a mean temperature of 950 but with 4 in. diameter By (40): K = 4.75 + 0.61 = 5.36. By (41): K = 7.5 rp i n For = 300 and for a number of 4-in. pipes of same area the results would have been as follows: 94 HEAT ENGINEERING 300 1 By (40): X = 200 + 40 V30 X 2 X ' 4 = 1.5 + 0.346 = 1.8 By (41): K = 3600 {0.0015 + [0.000506 - 0.00045 X 1 + 0.00000165 X 300] 04}. = 3600 [0.00172] = 6.2 There is no change in the third method, K being 1.2 as before. This is more nearly the value 1.8 which is for temperatures near the values used in the experiments. The Jordan formula is reduced from a large number of experiments and in none of them does K fall below 4.5. The formula of Nicolson does not agree with that of Jordan and apparently the reason for this may be said to be due to the fact that Nicolson's formula has been derived for large flues with high temperatures and Jordan's has been derived from small flues and for mean temperatures of 300 . Hence it would be well to restrict these to the conditions from which they were derived. The simpler formula for K is evidently in error on account of the effect of temperature. The wide variations make it necessary to compare the results to be obtained from other experiments. H. Kreisinger and W. T. Ray in Bulletin 18 of the Bureau of Mines, U. S. Dept. of Interior, on the Transmission of Heat into Steam Boilers, have shown that the quantity of heat from hot gas to hot water varies directly with the velocity and with the temperature to some power. The efficiency of the heat trans- mission increases with the decrease of diameter of tube. Wilhelm Nusselt (Mit. liber Forschungsarbeiten, Heft 89) in his article on the transfer of heat in tubes has reduced the theoretical form for K with an empirical constant for the value of K in the formula A 4 \4 Q = KF } loge AT 2 0-786 (42) U- \ A / in greater calories per hour per square meter per degree C. HEAT TRANSMISSION 95 coefficient of conduction of gases at wall temperature, greater calories per square meter per hour per degree for 1 meter thickness. w = velocity in meters per second. C p = specific heat at constant pressure for 1 cubic meter at condition of gas in flue. X = coef. of conduction of gases at mean tem- perature of tube. d = diam. of tube in meters. This may be thrown into a different form since C 1 r ^ ^P - CP BT c p = specific heat of 1 kg. of gas. (42) then becomes X /inrtr \ - 786 *^^fcM3B) ^ To change this to K in B.t.u. per square foot per hour per degree the same factor 15.90 is used when the terms have the following meaning : \vaii = coef. of conduction of gas at wall tempera- ture of tube in B.t.u. per hour per square foot per degree F for 1 ft. thickness. w = velocity in feet per second. C p = specific heat at constant pressure for 1 cu. ft. of gas under conditions in tube. X = .coef. of conduction of gas at temp, in tube. d = diameter in feet. The values of X for the different gases are given below, quoted from Nusselt: French English Air . 01894(1 + 228 X 10~ 5 0. . 01287[1 + 127 X 10~ 5 (Z - 32)] CO 2 . 01213(1 + 385 X W~ 5 t). . 00814[1 + 215 X 10~ 5 ( - 32)] Steam 0.0192 (1 +434 X 10~ 5 0. 0.01288U + 241 X KT 5 (f - 32)] Illuminating Gas 0.0506 (1 + 300 X 10~ 5 0. 0.03390[1 + 167 X 10~ 5 (t- 32)] For a 4-in. pipe with 1500 F. gas and a wall at 400 F. with a 96 HEAT ENGINEERING velocity of 50 ft. per second, and assuming gas to be a mean be- tween air and CO 2 , the following results: X = 0.01050[1 + 171 X 10~ 5 (1500 _ 32)] = 0.0371 0.01050U + 171 X 10~ 5 (400 _ 32)] = 0.0171 C P = 0.24 X . 0.0171 /50 X 0.0049\ - 786 ^V 0.0371 / = 15.90 X 0.0171 X 1.265 X 5.55 = 1.91 For a 36-in. flue for the mean temperature of 1550 F. and a wall temperature of 350 F. with m g w g = 0.4 the following results : WgWlgCp = WgCp Xutau = 0.0105[1 + 171 X 10~ 5 X 318] = 0.0162 = 0.0105[1 + 171 X 10~ 6 X 1518] = 0.0377 v - K on Q - 0162 /0.24 X 0.4\ - 786 L5.90 30>214 \ .0377 j ,15.90X0.0162 1.266 For a 4-in. pipe for 1550 F. for gas and 350 F. for wall with w g m g = 0.4, the result is K = i * on - 0162 /- 24 X 0.4\ - 78 ' (M)' 214 V 0.0377 / = 15.90 X 0.0162 X 1.266 X 2.2 = 0.716 These do not check with results of Nicolson nor Jordan for high temperatures, and of course, since based on experiments in which the temperature did not rise to such high values it should not be used in such cases. The formula has been com- puted for values of A of about 20 C. to 80 C. In 1914, F. E. McMullen performed a large number of experi- ments in the Mechanical Engineering Laboratory of the Rensselaer Polytechnic Institute on the transmission of heat from hot air to water through brass and steel pipes. The velocity of the water was varied from 1 to 8 ft. per second, that of the air from 2 to 17 ft. per second. The brass pipes were of %-in. and %-in. outside diameter and the steel pipe was of %-m. diameter. The mean HEAT TRANSMISSION 97 temperature difference was as much as 300 F. The results of this series of experiments showed that K varied inversely as A^ and directly with the density of the air and the velocities of the water and gas. The formula becomes Q = (mean At)F (43) 126m(wq- 1.75)- W* * or Q = K'(meanAO % F K' = 126m(w a - 1.75)'W* (45) m = weight of 1 cu. ft. of air at mean temperature in pounds = - w a = velocity of air in feet per second. w w = velocity of water in feet per second. (meanAO = A From the above discussion of problems it is suggested that Nicolson's second formula be used for boiler problems; that Jordan's, Nusselt's and the Rensselaer formula be used for air coolers and heaters where mean AZ is about 30|0 F. ' STEAM CONDENSERS For transmission of heat through condenser tubes a reference is made to the work of G. A. Orrok, A. S. M. E. Transactions, Vol. xxxii, p. 1138. In this he shows that the constants of heat transmission from the steam to the water depends on the tem- perature difference to the J^ power, on the square root of the velocity of the water, on the square of the steam richness, on the material and its cleanness. In his formula Q = #(mean M)F = ^"(mean ti*F (47) _ JTapVV^ f . . = (meanAO* 98 HEAT ENGINEERING Q = B.t.u. per hour. F = square feet of surface. K = B.t.u. per square foot per hour per degree. K' = 630. a = cleanness factor, 1 to 0.5. p = steam richness factor = f where p s is the pressure corresponding to temperature of steam in condenser and p t = actual pressure in condenser measured in any units. w w = velocity of water in feet per second. M = materials factor = 1.00 for copper, 0.98 for admiralty metal, 0.87 for aluminum bronze, 0.80 for cupro nickel, 0.79 for tin, 0.75 for monel metal, and 0.74 for Shelby steel. The student is also referred to Orrok's article for a bibliography of heat transmission for condensers. AMMONIA CONDENSERS In regard to ammonia condensers this formula should apply, but experimental results reported in the Transactions of the American Society of Refrigerating Engineers for 1907 seem to indicate a much lower result. This is given by a curve in which K for double-pipe condensers is given by K = 130\A^ (50) w w = velocity of water in feet per second and Q = ^(mean At)F (mean At) - BRINE COILS For brine cooling in double pipes the value of K seems to be given by K = 84w b EVAPORATORS AND FEED-WATER HEATERS The formula of Orrok should be applicable to the heating and boiling of water by steam as in evaporators and feed-water HEAT TRANSMISSION 99 heaters. The constants and values have been determined for various steam pressures and for that reason they should be ap- plicable. However this may be, two formulae recommended by Hausbrand will also be mentioned. One of the formulae quoted is that due to Mollier from Hage- mann's experiments reduced to English units : K = 10 + 1 110 + 0.6 (t 8 + ^-^- 2 ) } VW (51) t, = temperature of steam in degrees F. t\ = temperature of liquid at entrance in degrees F. tz temperature of liquid at exit in degrees F. The formula which Hausbrand recommends for the transmis- sion of heat from steam to water which is boiling is that due to Jelinek's experiments. When reduced to English units it becomes d = diameter of tube in feet. I = length of tube in feet. In the above formula the material is copper and steam is on the inside. This formula may be correct for such apparatus in which there is a low velocity of the liquid. This is the only reason for such a formula applying, as all experiments show that the heat varies with the velocity of the liquid. For wrought iron 0.75 of the above values are to be used, 0.5 for cast iron and 0.45 for lead. In evaporators with dilute solutions of 15 per cent, solid matter in solution the transmission is decreased by about 15 per cent, of the above for clean water while, with thick viscous liquids, K is about one-third of that given above. HEAT FROM LIQUIDS TO LIQUIDS The coefficient of transmission from a liquid to a liquid may be found from the formula (29') given by Nicolson in which the constant 6 is used in each term. - (S3) K Qm w w w Qm w w w 3m w w w * ' 100 HEAT ENGINEERING Hausbrand gives the following equation for the coefficient from one liquid to another through brass and copper walls. K m _ 300 JL . + . _J_ (54) This refers to greater calories per square meter per degree C for velocities in meters per second and was calculated by Mollier from Joule's researches. Hausbrand suggests that 66 per cent. of the value of K be used for practical purposes. To change this to English units the following calculations are made where K f refers to French units and K e to English: v 12 Wf = W e X K (55) These two formulae do not give the same results and the later one is recommended for general use. FACTOR OF SAFETY In all of the above expressions for transmission it would be well to reduce the quantity by % to have an excess of surface, or what would be the same thing increase the surface by 33 per cent, in finding the area required in a problem. RADIATORS The coefficient of transmission for direct radiators for heating varies with conditions of the surface and the type of radiators. HEAT TRANSMISSION 101 Carpenter quotes tests varying from 1.23 to 1.97 while Rietschel gives values of K varying from 0.51 to 2.65. A value of K of 1.8 will be used as an average for the computation of the heat transmitted for all types of direct radiators. This gives for such Q = 1.8(f. - t a )F (56) Q B.t.u. per hour. F = square feet of surface. t s = temperature of steam or water in degrees F. t a = temperature of room in degrees F. For indirect heaters the formula may be used in which K = 2 + l.75\/wa. w a = velocity of air across surface in feet per second. ti = temperature of air at inlet. in degrees F. t z = temperature of air at outlet in degrees F.. t s = temperature of steam in degrees F. Experiments have been made and plotted for the determination of the transmission of heat and the resultant air temperatures obtained with coils and cast-iron radiators and these are to be resorted to rather than formulae because the resultant tem- perature is dependent on the velocity and the temperature of the entering air and steam. Since in most problems low pres- sure steam is used the curves constructed from data of the American Radiator Co. and of the Buffalo Forge Co. are given, the first for a cast-iron pin radiator known as a Vento Heater and the second for the sections of pipe coils, each section being four pipes deep. The curves of Figs. 30 and 32 give the result- ant air temperatures when air at zero degrees enters the indirect heater and passes over one, two, three, four, five, six, seven or eight sections at different velocities. The velocities are found by finding the volume of the air at 70 F. and then computing the velocity by dividing this volume by the clear area through which the air passes. Thus the velocity to be used for the curve is not the actual velocity at entrance or exit but is that occurring when the temperature of the air is 70 F. The curves of Figs. 31 and 33 are those giving the average B.t.u. 102 HEAT ENGINEERING per square or lineal foot of the total heater under the conditions given. It will be seen that as the velocity increases the final 200 400 1400 1600 1800 600 800 1000 1200 Velocity in Ft. per Minute FIG. 30. Temperature of air leaving Vento Heater with air at F. and veloc- ity at 70 F. Steam at 5 Ibs. gauge pressure. 2700 200 400 1400 1600 1800 600 800 1000 1200 Velocity in Ft. per Minute FIG. 31. Heat transmitted per sq. ft. of Vento Heater with 5 Ibs. steam and air entering at F. Velocity at 70 F. temperature of the air falls but that the heat transmitted per square foot increases, due to the lower final temperature and the HEAT TRANSMISSION 103 200 400 1400 1600 1800 600 800 1000 1200 Velocity inJFt. per Minute FIG. 32. Temperature of outlet from sectional heaters of four coils each, with air entering at F. Steam at 5 Ibs. gauge pressure. Velocity 1200 1100 1000 900 800 700 600 500 400 300 200 100 5.9 200 400 1800 500 800 1000 1200 1400 1600 Telocity in Ft. per Minute FIG. 33. Heat transmitted per lineal foot of 1 in. pipe for 4 row sections of^oil heaters with 5 Ibs. steam and air entering at F. Velocity 'at 104 HEAT ENGINEERING higher velocity. As the number of sections is increased the final temperature rises by decreasing increments since the tempera- ture difference between the steam and air is less for the succes- sive sections, decreasing the heat transmitted, and the heat per square foot is less for the same reason. Since air often is supplied at a higher temperature than F. it is well to understand how such curves may be used for these conditions. Suppose air at 40 F. and 600 ft. velocity is to be heated in Vento heaters. It will be seen that one section would have been required to give this temperature and if a temperature of 100 F. had been desired three sections would be required to do this with zero air. The first section would bring the air to 40 F. so that only two sections will be required for 40 air. Now with zero air one section at 600 ft. velocity will transmit 1450 B.t.u. per hour per square foot and three sections will transmit 1150 B.t.u. per square foot per hour. The amount transmitted by the last two sections will therefore be 1150 X 3 - 1450 = 2000 or 2 = 1000 B.t.u. per square foot. If two sections had F. air supplied the curves show that 1275 B.t.u. per hour per square foot would have been trans- mitted but in this case, the entering air being at 40 F., there is a smaller A and hence a smaller transmission per square foot per hour. These curves can be used with certainty as they are the results of experiment and they displace any formula for compu- tations. They are better in this case as they show what may be obtained actually in practice. HEAT THROUGH WALLS AND PARTITIONS The heat transmitted through walls and partitions is com- puted by Q = K(t, - t*)F (58) where I V_ V_ 1 (59) ai * c ' T c" ~ ' a 2 Q = B.t.u. per hour. ti = temperature on one side in degrees F. 2 = temperature on other side in degrees F. F = area in square feet. HEAT TRANSMISSION 105 The value of a is found by experiment in the form d = constant depending on condition of air. = 0.82 air at rest in rooms or channels. = 1.03 air with slow motions, as over windows. = 1.23 air with quick motions, as outside of build- ing. e = constant depending on material. T = temperature difference of film of air at wall. Values of e Cast iron ...................................... . 65 Cotton and fabrics .............................. 0. 65 Charcoal ....................................... 0. 71 Glass .......................................... 0.60 Metal, polished ................................. . 05 Masonry ....................................... . 74 Paper ......................................... . 78 Rusted iron .................................... . 69 Water ......................................... 1.07 Wetted glass ................................... 1 .09 Wood ......................................... 0.74 For masonry walls, of thickness I ft. T = 16.2 - 4.00Z (61) For wood T = 1.8 F (62) For glass T = ^fa - t 2 ) (63) The values of c are given below: Air (still) ..................................... 0.03 Brass ......................................... 61 .00 Building paper ................................ . 08 Cork ......................................... 0.17 Cotton and felt ................................ 0.02 Glass ............ ............................. 0.54 Masonry and plaster ........................... 0.46 Sandstone .................................... . 87 Sawdust ...................................... . 03 Slate ........................................ 0.19 Terra cotta .................................... 0.54 Tin .......................................... 35.60 Wood.. 0.12 106 HEAT ENGINEERING Experiments show that the values of c change with the temperature, but for the temperatures used in practice the variation is not great. EFFICIENCY OF HEAT TRANSMISSIONS The efficiency of a heating surface is the ratio of the heat abstracted by the surface from the gases to the amount of heat which the gas could give up to the surface. If TV is the tem- perature of the hot substance entering and T^h is the temperature of the hot substance leaving the tube and T c is the lowest tem- perature of the cool substance in contact with the opposite side of the surface, the expression for efficiency is _ Tih Tzh _ 1 Tzh T c _ - _ A^2 ,,*.<. " T lh -T c ~~ ~ T lh -T c ~ ~ Ah This is true since the heat available in M Ibs. of gas of specific heat c is Qi = Mc(T lh - T e ) and the amount utilized is Qi - Q 2 = Mc(T lh - T 2h ) In a given experiment the temperatures are known but with a given T^ and T c and with a given surface it is necessary to find Tzh in order to ascertain the efficiency. There are two cases to consider: First, if the cool substance changes in temperature along the surface. Second, if the cool substance remains at a fixed temperature as in the case of a boiler. In the first case it has been shown earlier in the chapter that -M h c h dt h = K(t xh - t xc )dF M h c h 1 , ab = ~ -r\ t \ dth & (txh txc) M h Ch 1 dAt dAt TT ~M^~M =const - KKt 1 * M c c c If K is independent of temperature this becomes I + MhCh w c c c i FK - jfij - = loge Ty- HEAT TRANSMISSION 107 Jlc J2c Jlfe Jlc (tzh J2c) _ 1 = " ZTZT 1 ~ 2 l ~ , Hence FlL ,, ,. -- -T = - - = log e T- jT* (66) This shows that the efficiency of a surface increases with the increase of area, with the increase of the constant K or of the difference of Ai, but decreases as Qi increases. If, however, the area F is increased for a fixed length of tube by increasing the diameter, this apparently shows that the efficiency will increase with the diameter. It is known, however, that K varies with the velocity and hence this term would decrease with the square of the diameter so that the product FK would decrease with the increase of diameter and hence the efficiency would decrease. On the other hand, if d is decreased for a given length, the term FK is increased. In the above discussion Qi and (Aii Ai 2 ) have been assumed to be constant but this would not be so for if d were de- creased the velocity would be increased and with it Qi for a given length would be increased, but (Aii Ai 2 ) would increase to a greater degree and the final result would be an increase of efficiency. If K depends on (Ai)~ n , the integration gives 1 r . . . ! K'F Qi n nFK' Aii" - Ai 2 n 1 nFK' Aii Ai 2 /A*, \ U(, v Qi / Ai 2 _ t \ Eff. = 77 = 1 - ^Ep ^ (67) Ai7 " In this as before the efficiency increases with FK' and de- creases with Qi. 108 HEAT ENGINEERING In the second case the temperature is constant on the cold side and the expression then becomes + M h c h dt h = K(t xh - t c )dF + if dt is measured from cold to hot. If K is independent of the temperature this becomes KdF dt h M h c h ~ txh t c T lh - T c F = i ^ e T 2h - T c T ih -T c **L 7F, TFT = Mhch -iZh lc KF 77 = 1 M In this as before the efficiency increases with the product KF and decreases with the quantity M. Thus if the diameter of a pipe is increased the velocity decreases for the same M or M increases for the same velocity. In each of these the exponent is decreased since M would vary with the square of d directly and K inversely. Hence the large diameter means a smaller effi- ciency. If M is increased for a given tube by increasing the ve- locity K will increase in the same ratio and the efficiency may not be changed. If the length is increased F is greater and the efficiency increases. Since this increase of efficiency is logarith- mic, the increase beyond a certain value of F is very slow. If K depends on the temperature the following results (69) This leads to an expression similar to the one used previously. Kreisinger and Ray, in Bulletin 18 of the Bureau of Mines, show by results of tests on an experimental boiler that the effi- ciency decreased with the velocity to a certain point although the quantity of heat increased. In these experiments the value of M increased more rapidly than the coefficient of conduction. Of course if at the same time F were increased to its proper amount the efficiency would increase. In these experiments it was shown that the absorption of heat increased with the in- crease of the initial temperature although the rate of increase HEAT TRANSMISSION 109 of heat gradually diminished. The enlargement of the di- ameter of the tubes reduced the efficiency. Increasing the length of a tube increased the efficiency of the surface. For a boiler Kreisinger and Ray have used the form for the value of K suggested by Perry. K = K'mw Hence M h c h dt = K' M h but m h w h = -~r- r o dt K ' K ' r, = 1 - e K'F X Foe,, (70) This expression for efficiency does not change with change in weights and therefore does not give curves similar to those found by experiment. If the Reynolds form be used M h c h dt h = K'(A + Br this leads to ( A+B ) Fx and This equation yields a curve similar to those found experi- mentally. PROBLEMS A number of problems will now be computed, applying the formulae best suited for the work. Problem 1. A 4-in. boiler tube is used in a return tubular boiler with a grate of six times the flue area. Fifteen pounds of coal are burned per 110 HEAT ENGINEERING square foot of grate per hour with 25 Ibs. of air per pound of coal. The gases entering the tubes are 1350 F. and at exit they have been reduced to 600 F. What is the average value of the heating surface of tubes and how many pounds of water at the boiling point will be evaporated if the pressure is 130.3 Ibs. gauge? The formula to use in this case is that of Nicolson (40). The gases are of practically the same density as air, hence B = 53.35. 7r4 6 X 15 X (25+ 1) nnKOfr ,, Mass of gas per tube = yrrX ^60O == "-0567 Ibs. per second. 1350 -f- 600 Mean temperature of gas = ^ - = 975 F. Temperature of steam and water in boiler = 355.8 F. = H(975 + 355.8) = 665.4 Hydraulic radius = H = 1. Area of tube = ^ = 0.0872. = [3.327 + 0.839] (619.2) = 2570 B.t.u. Now ri45 = 865.2 B.t.u. Hence the pounds of steam per square foot = ogc~o = 2.97 Ibs. This result is an average value per square foot for the whole boiler. The value of the shell directly over the fire increases the average to more than this value. If the temperature entering the flue had been higher this value would be greater. For water tube boilers it is suggested that the area be considered the area between tubes and that the hydraulic radius be taken as the distance between tubes. Problem 2. Thus if the gases enter the tubes at a temperature of 1900 M F. and leave at 600 F. and if the tubes are 3 in. apart and the value of r is the same as before, the following results would hold for a water tube boiler. Mean temperature gases = 1250 F. = 803 Hydraulic radius = 3 in. ~w = 0.65 - 355.8) = [4.015 + 0.613] (894.2) = 4120 B.t.u. Weight of steam = ov = 4.76 Ibs. Problem 3. Air at 350 F. is to be cooled to 75 F. in an intercooler made up of %-in. tubes with the air passing through the tubes and cooled HEAT TRANSMISSION 111 by water entering at 60 F. and leaving at 90 F. Find mean At if (a) K K f is independent of temperature and (6) if K = 777712* In this problem it is evident that a counter-current flow must be em- ployed since the water leaving the intercooler is higher than the air leaving. AZi = 350 - 90 = 260 F. At 2 = 75 - 60 = 15 F. XT OAZ ^ l M io( a" ) =Io( 2 j = "IT Hence the exact methods must be used. Ah - AZ 2 260 - 15 245 (a) Mean At = - - = - ^^ = = 86.2 * 1 20 2.3X1.237 15 81.7 = 95 - 5 Problem 4. Find the heat per square foot for temperature conditions of problem 3 if the water has a velocity of 5 ft. per second over the tube and the air has a velocity of 20 ft. per second. The pressure of the air is 70.3 Ibs. gauge. ra for air at 85 Ibs. abs. and at 35 ^" 5 or 212.5 F. 85 X 144 = 53.35 X 672.5 = ' 341 lbs ' (a) Using (30) and (32) : K = 2 + 0022 X ' 341 X 20 = 20 ' 2 Now Q = K (mean At) = 20.2 (86.2) = 1740 B.t.u. (6) Using (41): f r q Q = 3600 j 0.0015 + 1 0.000506 - 0.00045 X ^ + 0.00000165- (20X0.341) (212.5 - 75) = 21.6(137.5) = 2970 B.t.u. (c) Using (42'): X wa u = 0.01287(1 + 127 X 10~ 5 X 43) = 0.0135 X ga s = 0.01287(1 + 127 X 10' 6 X 180.5) = 0.01583 _ 0.0135 T20 X 0.24 X 0.3411 o.786 10 TiT 4 [ 0701583 J Q = 15.0 X 86.2 = 1293 B.t.u. (d) Using (43) and (44): K - 126 X 0.341 X(18.25)o-(5)M (95.5) W Q = 36.6 X 95.5 = 3490 B.t.u. 112 HE A T ENGINEERING The results do not agree very well and result (d) is very high with (6) next, while result (c) is very low. The reason for the low value of (a) is due to the fact that the data from which equation (32) was deduced holds for smaller differences of temperature. Equation (32) will in general give results on the safe side as the temperature differences for which it holds are small. Problem 5. A sterilizer operates on a counter-current principle with the warm water entering at 212 F. and leaving at 75 F. and the cool water entering at 70 F. and leaving at 202 F. The liquid moves at a velocity of 2 ft. per second. Find the mean At : (a) if K is constant, (6) if K = ~rf^ and (c) arithmetic mean A of At's at the two ends. Ai = 212 - 202 = 10 F. A, = 75 - 70 = 5 F. . (a) Mean A< = ~* = ^ = 7 ' 24 Problem 6. Find K and Q per square foot for problem 5. Use (29) with K" = 6 for water and at 150 F., m w = 61.2: 1 , Qm w w w """ Qm wIVw Q = 367 X 7.24 = 2654 B.t.u. .a, (55) K 60 171 2 1+-7TA/2 Q = 171 X 7.24 = 1240 B.t.u. Problem 7. If the walls are He in. thick in the sterilizer what is the drop in temperature in the copper of the tube to transmit the heat of prob- lem 6? Using equation (3) and the tabular value of C for this: 239.0(1 + 0.00003X150) {fc _ y 16 X 12 '-'- fexISlhsi - ao27 Even in this case the fall of temperature occurs mainly in the films of water. Problem 8. The temperature of the condensing water is 70 F. at inlet and 80 F. at outlet. The temperature of the steam is 105 F. Find mean A and the heat removed per square foot of cooling surface of admiralty HEAT TRANSMISSION 113 metal if the surface is clean and the pressure in the steam space is 1.2 Ibs. absolute, and the velocity of the water is 4 ft. per second. Ai = 105 - 70 = 35 F. A 2 = 105 - 80 = 25 F. |-H(35- 25)1* f 1.25 1* Mean A* = [35^26*] = Ll.560-l.49eJ PS 1.098 * = p t = L200 630 XI X0.914 2 x0.98X\/4 (29.9) H 673 Rtu - Q = 673 X 29.9 = 2010 Problem 9. Steam at 215 F. is used to heat water at 60 F. to 180 F. with the steam inside of copper pipes H in. in diameter and 80 in. long. The velocity for the water is 2 ft. per second. Using Orrok's formula : Mi = 215 - 60 = 155 F. A 2 = 215 - 180 = 35 F. ., fK(155-35)" eo Mean A ' = = l.56 9* f 15 = Ll.878- 630 X 1 X 1 X l.OOA/2 K = Q = 513 X 82 = 42,100 B.t.u. per square foot. Using equation (51) of Hagemann's, = 10 + 441 = 451 Q = 451 X 155 + 35 = 42,800 B.t.u. per square foot. Problem 10. Suppose the pipe in problem 9 was used to boil water at 180, find K and Q. Using equation (52) of Jelinek: K _ _mo == . 1965 HsV* X80 Q = 1965 X (215 - 180) = 68,800 B.t.u. Problem 11. A brine cooler has brine with a velocity of 1 ft. on one side at 20 to 10 F. and ammonia on the other at F. Find K and Q per square foot. In this problem the method will be to use the same constants as those used for steam to water. Using Hagemann's equation (51): 10 + 119 = 129 129 X 15 = 1935 B.t.u. per hour per square foot. 114 HEAT ENGINEERING For 4 ft. velocity this would give for K K = 10 + 238 = 248 Problem 12. An ammonia condenser uses steel pipes with water from 60 to 80 F. and ammonia at 90 F. The water velocity is 4 ft. per second. Using equation (50) : K = 130 X VI = 260 Q = 260 X 23"^ 3 = 260 X 18.3 = 4750 B.t.u. Problem 13. Find the number of sections required and the average heat transmitted per square foot for vento heaters to operate with steam at 220 F. (5 Ibs. gauge) and to heat air from 60 F. to 114 F. when delivered across heater at 1200 ft. per minute. From Fig. 30: Two sections will heat zero air to 60 F. and five sections will heat zero air to 115 F. at 1200 ft. per minute; 5 2=3 sections are required. From Fig. 31: The average transmission for two sections with zero air is 2050 B.t.u. while with five sections 1600 B.t.u. are transmitted. The amount for the last three of the five sections will be 5 X 1600 - 2 X 2050 1Qnnn . Average transmission = ^ = UUU B.t.u. To find the result of this problem by formulae in place of using the experimental curves the following is given: 60 -I- 115 Mean At = 220 - ^ = 132.5 1^ = 9.83 -; Q = 9.83 X 132.5 = 1300 B.t.u. ^Cement Plaster Air Space FIG. 34. Section of wall for cold storage room. Cork foaml Problem 14. Find the value of K for the wall of a cold storage warehouse shown in Fig. 34. T = 16.2 - 4.00 X 3 = 4.2 a for outside = 1.23 + 0.74 + = 2.28 a for air space | = 0<82 + 0<74 + a for inside J = 1.79 42 X 1.23+31 X0.74 1000 42 X 0.82 +31 X0.74 1000 X4.2 X4.2 HEAT TRANSMISSION 115 K 1 i 2 1 J_ 0.67 033 0.17 1 ~T~ n /i i 1 >7(\ T i 7O i n/miniTin/ifti 2.28^0.46^1.79 ' 1.79 ' 0.46 ' 0.17 ' 0.46 r 1.79 1 ~ 0.438 + 4.35 + 0.558 + 0.558 + 1.455 + 1-94 + 0.37 + 0.558 = 0.0978 It will be noted that the amount of air space does not aid in the insulating value of the wall because the air can and will circulate. The value of this lies in the resistance at the two surfaces which amounts to almost as much as 8 in. of brickwork. The various terms of the denominator show the values of the various elements of the wall as heat insulators. Thus 4 in. of cork is as valuable as 11 in. of brickwork and may be used when space is of value. The wall above will transmit 2.35 B.t.u. per square foot per degree difference in 24 hours. For a temperature difference of 60 F. this amounts to 141 B.t.u. per day per square foot. Topics Topic 1. By what methods is heat transmitted? Explain the pecul- iarities of each method. Give the Stefan-Boltzmann Law. For what is this used? How is it applied? Topic 2. What is the law of conduction as stated by a formula? Is the coefficient of conduction a constant? What are the dimensions of this coefficient? What fixes the amount of heat carried by convection? Why is convection of value in engineering problems? Is the heat in these problems carried by convection or conduction? Topic 3. How may it be shown that the films of substance at the surfaces of a partition or plate offer the greatest resistance to flow? What is the effect of velocity on film resistance? Explain why this is true. What was the work of Dalby? What was the work of Reynolds? Topic 4. Prove that Q = K'mw(t - 0) Is this the equivalent of Reynold's expression Q = (A + Bmw)(t - e) M Show that - = rnw F Topic 6. What is meant by hydraulic radius? Determine the mean temperature difference along a surface if K is constant. L Topic 6. What is meant by the constant Kf Prove that if K = J^_ Topic 7. Prove that K = 77 " L i L : 4. c i r C n "^ c m 116 HEAT ENGINEERING To what does this relation reduce for ordinary transmission through thin partitions? Is c a constant? What is the form to which Nicolson reduces this value of K above? 1 1 1 Topic 8. Given: -~ = ~ I- K 6m g w g reduce K = Topic 9. What are the formulae of Nicolson, Jordan and Nusselt? Are these formulas applicable to the same conditions? On what does the heat per square foot per degree per hour depend? Give details and reasons. Topic 10. For what conditions is the Rensselaer formula applicable? For what is Orrok's formula used? On what does the K of Orrok's formula depend? On what for the Rensselaer formula? Topic 11. Is Orrok's formula applicable to ammonia condensers, evaporators and feed-water heaters? Are special formulae given for these forms of apparatus? Give the formulae used for finding K for these. What is the value of K for direct steam radiators? For indirect steam radiators? Topic 12. Explain the method of finding the heat transmitted per square foot of surface per hour in Vento heaters and in pipe coils used as indirect heaters. Topic 13. Explain the method of finding the value of K for a wall or partition. Topic 14. Derive the expression for the efficiency of heat transmission when K is independent of the temperature. Give all steps and discuss the variation of the efficiency of a surface with velocity, diameter of pipe, length, temperature, and heat. Topic 15. Starting with Reynolds' value of K K = K'(A + B reduce the expression for the efficiency of a heating surface. _ = / 17 = M hr.h PROBLEMS Problem 1. A piece of glass is held at 750 F. and a shield covers two- thirds of this. The surface of the glass is 2 sq. ft. The surface of the shield is 20 sq. ft. The shield is held at 125 F. by a water-jacket. How much heat is removed by the water-jacket? Problem 2. A 4-in. boiler tube has gas entering at one end at 1400 F. and leaving at 550 F. with steam at 120 Ibs. gauge pressure. The coal is burned HEAT TRANSMISSION 117 on a grate of eight times the area through the tubes at a rate of 15 Ibs. per hour with 30 Ibs. of air per pound of coal. Find the value of K and the number of B.t.u. per square foot of surface. How many square feet of surface would be required per boiler horse-power? (One boiler horse- power equals the evaporation of 34}4 Ibs. of water at 212 F. per hour into dry steam at 212 F.) Problem 3. A feed-water heater uses steam at 3 Ibs. gauge pressure to heat 6000 Ibs. of water per hour from 60 to 200 F. The water is passed through at a velocity of 2 ft. per second. Find mean At assuming K con- stant. Find K by two formulae. Find the square feet of surface necessary. Of what material will the tube be made? Why? Problem 4. An economizer is used to heat 6000 Ibs. of water per hour from 60 F. to 200 F. by using hot gas at 500 F. in 3-in. iron flues in which the temperature is reduced to 350 F. The velocity of the gas is 25 ft. per second. What is the value of mean At? Find K by Nusselt's formula. Find the amount of surface required. Also use the simple formula suggested by Nicolson in which L is employed. Problem 5. In a condenser the water enters at 50 F. and leaves at 65 F. with a pressure in the condenser of 0.6 Ibs. absolute and a temperature of 80 F. What is the value of mean At for this case? Find K for copper tubes if dirty. Find the square feet per kilowatt of turbo generator if steam consumption is 14 Ibs. with x of exhaust steam equal to 0.95. Problem 6. Find the size of the condenser for an ammonia plant to remove 200,000 B.t.u. per hour with water flowing at 5 ft. per second in the double pipe, entering at 65 F. and leaving at 80 F., when the ammonia is at 100 F. Problem 7. A boiler using a 24-in. flue has gas entering at 1600 F. and leaving at 1000 F. The steam is at 300 F. The velocity of the gas is 100 ft. per second. Find the heat transmitted per hour per square foot of flue. Problem 8. An intercooler receives air at 250 F. and cools it to 80 F. The water enters at 50 F. and leaves at 70 F. Find mean At for rrt parallel and counter-current flow using K = constant and K = ?. Find the heat transmitted per square foot per hour. Problem 9. Find the surface required in an interchanger cooling 7000 Ibs. of water per hour from 220 F. to 80 F. by water entering at 60 F. and leaving at 200 F. Use K = constant. Problem 10. Find the surface required to boil 500 Ibs. of solution at 200 F. by steam at 250 F. if tubes 3 ft. long and 3 in. in diameter are used and the heat of vaporization of the liquid is 750 B.t.u. Problem 11. Assume air at 50 F. and move it with a velocity of 1200 ft. per minute over the Vento heaters or coils to heat it to 105 F. How many sections will it take for the Vento heaters and for the coils? How many square feet of each will be required to heat 200,000 cu. ft. of air per hour? Problem 12. Find the K for a wall composed of 16 in. of brickwork, a 2-in. air space and 12 in. of brickwork with 1 in. of cement plaster. CHAPTER IV AIR COMPRESSORS Compressed air is used for many purposes for which it would be difficult to employ other media. For operating small tools, rock drills, and hoists and for the transmission of power over con- siderable distance it replaces steam with which the loss due to condensation is very excessive. The efficiency of transmission of power by compressed air is not equal to that of electrical transmission nor is it as flexible, yet in certain cases for some reasons it is of value. For cleaning materials with a sand blast FIG. 35. Section of radial blade fan blower. FIG. 36. Rotary blower, section. and for use with the cement gun compressed air is necessary. To force air into a furnace of a boiler or for a blast furnace a compressor of some form is required. Compressed air may be used in air-lift pumps and in direct air pumping. In supplying air to divers, in the making of liquid air and in the operation of street cars compressed air is required. To compress this air several types of compressors are used. For low pressures up to 10 oz. per square inch above or below the atmosphere, centrifugal or fan blowers, Fig. 35, are used. In these, radial or curved blades are driven at a high speed causing a 118 AIR COMPRESSORS 119 flow of air. These are used for ventilation of buildings; for supply of air to boiler furnaces or cupolas; for forges; for suction in an induced draft system or for the conveying of light materials. In forge work Sangster allows 140 cu. ft. of free air per minute per forge at 2 oz. pressure. If an exhaust fan is used 600 cu. ft. of air per minute at % oz. pressure are handled per forge. The air required in cupolas is 40,000 cu. ft. per ton of iron melted. For ventilation 2000 cu. ft. of air are allowed per person per hour. For pressures of from 8 oz. to 7 Ibs. per square inch, rotary blowers such as that shown in Fig. 36 are used. These are run at a sufficient speed to get the necessary discharge in cubic feet. For pressures up to 35 Ibs. turbo compressors of the form shown in Fig. 37 are used although piston compressors are used for this pressure at times. The air at this pressure is used in blast furnaces and converters and the compressors of the piston type for such are known as blowing engines or blow- ing tubs. These are only special forms of air compressor. For higher pressures than 35 Ibs. the piston compressor is in general employed, although the hydraulic compressor or the Humphrey explo- sion compressor could be used. For high pressures it will be shown that efficient work necessitates the compression in one cylinder to a pressure much below that desired and then a further compres- sion of this air to a higher pressure. It may be that a third compression is used before the desired pressure is reached. Each one of these cylinders is known as a stage. The compressor men- tioned above would be known as a two-stage compressor. In general one stage is used to 70 or 80 Ibs. gauge pressure while from that to 500 Ibs. two stages are used, and three stages from 500 to 1500 Ibs. Above this four stages would be used. Fig. 38 shows a two-stage compressor. In this air is sucked into the center of the piston A by the vacuum produced behind the piston when the piston moves to the left, the air flowing through an open- ing left at the periphery B, as shown in Fig. 39. The air on the left of the piston is compressed and after it reaches the pressure 3 7.-Section of turbo blower. 120 HEAT ENGINEERING existing above the mushroom valve at C this opens and allows the air to exhaust into the discharge pipe D and from this into the intercooler E in which the air is cooled by water in the pipes F. The water is caused to circulate back and forth in this inter- FIG. 38. Section of two-stage Ingersoll-Rand compressor. cooler. The air just compressed in cylinder G is forced through the intercooler into the cylinder H where it is compressed to a higher pressure. In this cylinder the inlet valves I are at the top of the cylinder while the discharge valves J are at the lower part of the cylinder head. Both sets of these are mushroom valves. r*^;;SvV;^ ,lyvV viii/VA.'-.v ^'-'- : ; : .- .A'^;-'4v. v .''.-. '^',- -'^ ^ :^' '<7-v'f'-'--'*: v -':^:-.'o: > .:^ ). Enlarged section of L. P. air cylinder of Inger; FIG. 39. Enlarged section of L. P. air cylinder of Ingersoll-Rand compressor. The air is finally sent to the ak storage tank or aftercooler through the pipe K. Fig. 38, which is the Ingersoll-Rand class AA-2 compressor, illustrates one form of two-stage air compressor in which the AIR COMPRESSORS 121 driving steam cylinder is in line with the two air cylinders and the fly wheels are driven by outside connecting rods. The figure illustrates the method of bringing cool air to the compressor from a point outside of the engine room through the conduit L and at M and N are the water jackets to remove some of the heat of compression. Fig. 39 is introduced to show the valves of the low-pressure cylinder. In the periphery of each face of the piston are a series of slots distributed around the piston through which air can pass. Over this is fitted a ring which closes this open- ing and acts as a valve. This is known as the hurricane inlet valve. When the piston moves to the left at the be- ginning of a stroke the right-hand ring is moved from its openings and air can enter behind the right side, the com- pressed air on the left holding closed the valve on the left side. At the end of the stroke the inertia of the ring tends to close the right one as the pis- ton reverses and the left one will open as soon as the pressure of the air in the left-hand clearance space expands to atmospheric pressure. The dis- charge valves are ordinary mushroom FIG. 40. Vertical after cooler and intercooler of Igersoll-Rand Co. valves with a light spring to close them and a tube on the back to act as a guide. Fig. 40 shows the construction of an intercooler, through which the air must pass from one stage to another and give up its heat. By the arrangement of baffle-plates and partitions the air and water are made to take a circuitous path so as to be more efficient in the removal of heat. The moisture which separates 122 HEAT ENGINEERING as the air is cooled is usually caught as shown in the figure so that it will not pass over into the next stage. In many cases an after cooler, Fig. 40, is used after the last stage to remove more of the moisture from the air and to cool it before it passes into the transmission line. In this way the air is of smaller volume and there is less friction. Fig. 41 illustrates the arrangement of the Taylor hydraulic air compressor. In this a system for the flow of water must exist. Suppose the dam A gives a head of H feet and this causes water to flow through the pipes B, C, D, E to the tail race F. The head and friction will cause a certain flow through the system FIG. 41. Taylor hydraulic air compressor. and if the pipe is necked at C the velocity may become so high that a vacuum is formed at this point and air is drawn in through openings placed here. The air enters from G. This air mixes with the water and when the velocity is decreased in H the air separates out and rises to the surface of the water. This air is under pressure due to the head h on the chamber G. The air is taken out through the pipe 7. The Humphrey apparatus is described in Engineering, Vol. LXXXVIII, p. 737, and in " Compressed Air Practice" by Richards. In all of these forms of apparatus the volume of air taken in might be the same while that discharged is determined by its pressure and temperature. To give some idea of the amount of AIR COMPRESSORS 123 air used by tools or machines and the amount handled by the compressor it is customary to reduce the air to some one standard condition. The conditions taken are 14.7 Ibs. absolute pres- sure and 60 F. temperature. This air is known as "free air." At times the temperature is assumed to be that of the atmosphere at the time of use, in which case the term free air refers to air at atmospheric pressure regardless of temperature. With this introduction the thermodynamics of compressed air will be considered. WORK OF COMPRESSION The amount of work required to compress V/ cu. ft. per minute of free air is shown by the diagram of Fig. 42 which assumes no clearance. The line db is the atmospheric line and on account of the friction of the inlet pipe and P valves, the initial pressure p\ is be- low atmospheric pressure. The air is sucked in on the suction line cl and is compressed from 1 to 2 on the line pv n = const, and is then a | _ ^^^_ 6 driven out from 2 to d. The pres- sure at 1 is pi and the volume is V\. The atmospheric pressure is p. and the free air V/ after throttling oc- cupies the volume Vi. The throttling action means constant heat content which for a perfect gas means isothermal action. Hence p a V f = p.V, (1) Some authors say that the air changes its temperature in entering the cylinder, but this cannot be appreciable as air is such a poor conductor of heat that it would take up little or no heat from the walls and the throttling of itself is isothermal. Now the work of compression is = W i This quantity is negative since the work is done on the gas. negative answer means that work is done on the gas. Reducing: 124 Now Hence HEAT ENGINEERING work = n ( ) n \Vi/ (3) If desired p a V f may be substituted for p\V i giving work to compress V/ cu. ft. of free air from pi to p 2 as (4) EFFECT OF CLEARANCE Now if there is clearance the indicator card takes the form shown in Fig. 43. The air remaining in the cylinder, d2', at the end of discharge expands from 2' to 1' on the return stroke preventing air from entering un- til I/ is reached. The amount of air then taken in is V"\ V'i, or V"i - Vi = Vi of the previous discussion. The net work required to drive the compressor is area 1" 2 "2'1' or area l"2"dc - 1'2'dc. This, assuming the same form FIG. 43.-Effect of clearance. of expansion and compression lines, gives Net work with clearance = < This is the same expression as (3) for the compression of V\ cu. ft. of air from pressure p\ to pz, or in other words clearance has no effect on the work required to compress a definite volume of air. The effect of clearance is to decrease the amount of air taken in for a given displacement or to increase the displacement AIR COMPRESSORS 125 for a given amount of air taken in. Thus in Fig. 42 the air taken in is Vi and the displacement D = Vi, but in Fig. 43 the air taken in is Vi but the displacement D = V\ + 1'c 2'd, a quantity greater than V\. In other words the displacement has been increased. This increase of displacement means a larger cylinder in stroke or area of piston and hence there will be more friction and consequently more work will be required to drive the compressor, but the work indicated on the air for the com- pression of a certain amount is the same with or without clearance. The volume 2'd represents the volume of the clearance space and since this is usually expressed as a percentage of the dis- placement it may be represented as ID, where I is the percentage clearance and D is the displacement. Now pi Hence V i = D + ID - ID (6) + I I ( j n i The term 1 + I I j n is called the clearance factor. It may be represented by KI. This term is only used to find the displacement if Vi is given or Vi if D is given. The clearance ratio I is known. EFFECT OF LEAKAGE If there is leakage around the piston, piston rod and through the valves the volume Vi required to deliver a given amount of free air V/ must be increased so as to care for this leakage. If the amount of free air delivered is expressed as a percentage of the free air taken in or as a ratio to the free air taken in, the per- centage is called the leakage factor /. Actual V\= - ^-7 Although the leakage is effective during the compression, giv- ing less work as the compression is carried higher, it may be considered that this loss occurs at the upper pressure only and 126 HEAT ENGINEERING Ft Vt consequently increases the work by causing y- or -4- to be substituted for Fi or F/. Thus WOrk = r- pi -^~ 1 ( ) n I = p a 1 I J n n 1 / L Vpi/ J n 1 j L \pi' (7) The effect of leakage is to increase the work. VOLUMETRIC EFFICIENCY The volumetric efficiency is denned as the ratio of the free air delivered to the displacement of the compressor. If the actual free air is used this is the actual volumetric efficiency while if the indicated free air is used, the indi- cated volumetric efficiency is obtained. Thus if ab is the atmospheric line from actual compressor, indicated volu- }; AA -,r ~^ metric efficiency or apparent volumetric FIG. 44. Card from com- dng volumetric Affi ^ PTW . v = *H = W. D ab Now '- xy = . - since the pressure is lowered by throttling action in which T is constant. The temperature of the cylinder walls may change the temperature of the air slightly but this is so slight that the air is considered to be at inlet temperature. Substituting for xy its value, the following is obtained Apparent vol. eff. = ^ = ^ [l + I - I (^J *J (8) actual Vf / X ind. free air True vol. eff. = ^ - = - ^ True vol. eff. = X leakage factor X clearance factor. (10) HORSE-POWER AND POWER OF MOTOR If Fi cu. ft. are required per minute the expression (7) gives the work per minute in foot-pounds. Consequently dividing the expression by 33,000 gives the horse-power shown by the air AIR COMPRESSORS 127 card. This result must be divided by the efficiency of the air compressor, about 85 per cent, to 95 per cent, depending on size, before the horse-power to apply to the compressor is determined. To find the horse-power applied to the motor, be it a steam engine or an electric motor, the above result must again be divided by the efficiency of the motor. This may be about 90 per cent. Hence, horse-power applied to compressor forFi cu. ft. per minute _J __ PiVi fi (P^\ n -~ "n- 1 eff. X/X 33000 L 1 " \p indicated horse-power of engine drive n ___ _ 1 _ n 1 eff. compressor X eff. of engine TEMPERATURE AT THE END OF COMPRESSION If the temperature is TI at the beginning of compression the temperature at the end of compression is t-i This is seen from the following pzVf BT 2 \ n COMPRESSION CURVE The compression curve desired for air compression depends on the way in which the air is to be used after compression. Air is such a poor conductor of heat, and at 60 or 100 revolutions per minute the action of the compressor is so rapid, that expan- sion in an engine or compression in the compressor practically takes place along an adiabatic pF 1 - 4 = const. If air is compressed along an adiabatic 1-2, Fig. 45, and is expanded along an adiabatic in the engine this adiabatic will be 128 HEAT ENGINEERING 2-1, if there is no leakage nor cooling between the compressor and engine. If, however, the air is stored in a tank for some time before using in the engine the air at a high temperature T 2 is cooled to the original temperature TV This causes the volume to decrease so that the volume occupied by the air in the cylinder of the engine is Vz where 2' lies on the isothermal 12'. The expansion line in the engine is now 2'!' and the area 122'!' represents the loss of work due to the cooling in the tank. Al- though 1-2 is the best line for compression if the air is to be used before it can cool so that it will expand in the engine along 2-1 giving no loss, it is evident that 1-2' would be the better line if the air is to be stored before using, since the temperature along this line is constant. Hence it is often stated that isothermal compression is the ideal and best method of compression. This is true, and true only, if storage of air is to be employed in the system, for other cases adiabatic compression may . const, be the best method. If isothermal com- pression is used the work of compression becomes 12'dc, resulting in a saving in work of the area 122'. The loss when FIG. 45. Saving due to cooling on compression, the expansion in the air engine takes place is then 12'!' instead of 122T. The expression for the work with isothermal compression without clearance is Work = pl Vi log e - ptV't + p% but p^V'z = piVi :. Work = P1 V! loge- = - pl Vi loge - (14) To approach this isothermal line in compression a water jacket is placed around the cylinder to remove heat, or water or oil is sprayed into the cylinder to reduce the heat. These methods are not very effective since n is changed only from n = 1.4 to n = 1.35. This saving is slight. The reason for this, as stated before, is the fact that the air is a poor conductor and also it is in contact with the cylinder walls a very short time. HEAT REMOVED BY JACKET The heat removed by the jacket is made up of two parts, that during the part of the stroke 1-2 and that during the part 2-d. AIR COMPRESSORS 129 const. An expression may be written for the first part but no expression can be written for the part during the time that the piston moves from 2 to d. The temperature difference between the air and the jacket water is greatest dur- ing this time, but the cooling surface in contact with the air is decreasing and this effect would tend to decrease the cooling effect while the greater tempera- ture difference would increase the effect. If the effect is assumed to be the same as that during the portion of the stroke 1-2, the total effect may be found by multiplying the effect Fi 1 FIG. 46. Card from com- pressor with jacket. during 12 by ~r V or Now heat removed on line 1-2 = k- 1 1-n From the above and by considering the leakage factor /, the following is obtained: Heat removed by jacket iy\ J/* ~~| /r\ ~\T I* frf\ i r j /P]\il_a - (15) SAVING DUE TO JACKET The work done by the compressor when the exponent is changed from k of the adiabatic to n is Work= n - 1 J L \p! The work for adiabatic compression is Work = -1 / The saving due to the jacket is the difference of these or 130 HEAT ENGINEERING WATER REQUIRED FOR JACKET If the heat removed by the jacket per minute is found, the water required is determined by assuming the possible range of temperature and then finding the water by (17) ~ _ heat per minute from jacket in B.t.u. Cr = - f - f q o q i G weight of water per minute. q' = heat of liquid at outlet. q'i = heat of liquid at inlet. MULTISTAGING AND INTERMEDIATE PRESSURES Although jacketing is used the slight change in the exponent does not give a great saving in work and moreover for high pres- sures T r 2 becomes so great even with a jacket that the lubricating oil is apt to ignite and cause an explosion. To prevent this and to save work the air is compressed to a pv n m Const. FIG. 47. Two-stage compression. pressure less than that finally desired and discharged from the cylinder into a chamber containing a number of tubes carrying cold water. This chamber is called an intercooler and the tubes are made of such an area and arranged in such a manner that the air is brought to its original temperature at 1 before it is AIR COMPRESSORS 131 sent to a second cylinder of proper volume in which it is com- pressed to its final pressure. This method of compressing in two cylinders of different sizes is known as two-stage compression. The action is shown in Fig. 47. Vi cu. ft. of air are drawn into the low-pressure cylinder and compressed to a pressure p'z. The (7/2\ n 1 ) is then discharged from this cylinder and through the intercooler until its temperature is reduced to TV The air is then drawn into the second cylinder which must be of such a volume as to take the air which occupies the volume V'\ found on an isothermal through 1 at a pressure pV The air is then compressed to a pressure p%. The work in the two cylinders is given by ^ r ip'z\ n ~ 1 ~i Work in low-pressure cylinder = ^-r piVA 1 ( ) n n r / #2 \ re ~i~| Work in high-pressure cylinder = _ ., pYF'il 1 Mr*] w Now p l V l Hence (18) The only variable in this expression is p' 2 . Hence to find the condition for minimum work, the first derivative with respect to p'z must be equated to zero. r 'W 1 /T)' \ (total work) = - ~rPiFi ( "I n 1^ n \vi/ p pi Pl' Pi \P2 P2 = 1 PlP2 . P'2 = VjhP* (19) That is, the work is a minimum if p r 2 is a mean proportional between pi and p 2 . 132 HEAT ENGINEERING Substituting this value of p' 2 in (13) the following results: If there are three stages, proceeding in the same manner, the equation becomes Total work = Pl7 3 1 ' \pi/ in which there are two variables, p' 2 and p" 2 . The conditions for a maximum or minimum are - (total work) p // 2 = and r- (total work) p > 2 = 0. These give p' 2 = Vpip"2 and p'j = = P1 2 P2 (21) (22) < Pi 2 p 2 For m stages there will be m 1 intermediate variable pres- sures in the expression for total work and there will be m 1 partial derivatives to equate to zero giving, in the same manner as above, p' 2 = V^S or ^=(^} (23) P's Pi (24) P2 It is to be noted that the ratio of pressures on each stage is the same and each ratio of pressures on the various stages is equal to ( ) m. If these are substituted in the expression for total work the equation becomes Total work for m-stage compression = mn f- /z> 2 \ 1 /25\ AIR COMPRESSORS 133 This is the general expression for the work of an m stage com- pressor on the compression line pv n = const., for any value of n except 1. INTERMEDIATE TEMPERATURES Since the ratios of pressures on each stage are equal to ( j m it follows that the temperatures at the end of compression, T 2 , are all the same and equal to For a single-stage compression between pi and p 2 T, - Ti (& - HEAT REMOVED BY INTERCOOLER In the intercooler the temperature is reduced at constant pressure from T 2 to TI and the heat is given by Heat from intercooler in foot-pounds = JMc p (T 2 TI) T , - / r>rp C p (l 2 1 i If leakage is considered this becomes AMOUNT OF WATER FOR INTERCOOLER In the above section the heat removed by the intercooler has been determined. The amount of water per minute required for the removal of this heat is found by assuming the temperature allowable at inlet and the temperature at outlet desirable and then computing by the formula, ~ _ heat per minute from intercooler in B.t.u. q'o ~ q'i G = weight of water per minute. q'o = heat of liquid at outlet. q'i = heat of liquid at inlet. 134 HEAT ENGINEERING The area of the surface of intercooler is found by methods of Chapter III. WATER REMOVED IN INTERCOOLER If air at pressure p a has a relative humidity p a and the weight of 1 cu. ft. of moisture to saturate the air at the temperature T\ is m , the total amount of moisture entering per minute is M w = p a m a V f When this is sent to the intercooler and cooled to temperature Tij after it is compressed to pressure p' 2 and volume F' 2 or ^-f F/, the amount of moisture held in the air, if saturated, will be M' w = ra 2 -/- Vf. P2 This is less than M w in most cases, so that the amount of moisture precipitated is M w - M' w If M' w is greater than M w the ratio vW^ will give the relative jJ/J. iff humidity of the air leaving the intercooler. This same method can be used to compute the moisture removed by the after cooler. EFFECT OF LEAKAGE IN MULTISTAGE COMPRESSIONS The leakage in a multistage compressor is a variable quantity, the leakage making the amount of air handled by the various stages different. Thus if 3 per cent, is the leakage in a single stage the leakage in a three-stage compressor might be 1 - 0.97 X 0.97 X 0.97 = 0.088 = 0.09 approximately. That is, the amount of air to be handled by the lowest stage would y ' YI be /TqT, that by the second stage T^TT and that by the third would Fi be TTq^. Of course these differing amounts of air would change the theoretical discussion above but to a slight degree. Because of the similarity of relations for various stages the same pressure ratios will be used although the works on the various stages will not be the same. Hence in solving various problems the ex- AIR COMPRESSORS 135 pression for total work will be used as derived before substituting TT for Vi, and using as / the mean value of the various fs for the m stages.' In computing the displacement of the various stages the correct value of / for each stage will be used. DISPLACEMENT OF CYLINDERS The displacement of each cylinder can be computed after the leakage factor and clearance factor are known for the cylinder. The leakage factor has been discussed in the previous section, and if 3 per cent, per cylinder is assumed the various leakage factors up to four stages may be taken as 88 per cent., 91 per cent., 94 per cent, and 97 per cent. The clearance factor is given by or = 1 +l-l~ (30) \PI/ If the clearance is the same on each stage the clearance factor will be the same for each stage. If, as is often the case, the value of I is small for the low-pressure cylinder and gradually increases, the value of K t will gradually become larger for the high-pressure cylinders. Equations (6) and (30) show that as or becomes greater K t becomes less and so the effect of pressure on this factor is quite noticeable. With very high pressures on any stage the volume of expanded air at the end of the expansion part of the stroke is so great that only a small quantity of air will be drawn in. This is another reason for multistaging. Having KI and / for any stage the displacement is found by n - ~ In other words, the free air to be handled multiplied by the ratio of the pressure of the atmosphere to the initial pressure on the stage considered gives the volume of the free air when at the pressure of this stage and this divided by the volumetric efficiency for this stage, f x K ix , gives the displacement per minute if V/ is the free air per minute. 136 HEAT ENGINEERING If now the number of revolutions per minute is known and if the piston speed per minute allowable is assumed the length of stroke is known and then the area can be found after it is decided whether or not the compressor is double acting. Piston speed = 2LN = 200 to 700 ft. per minute. (32) D = (F h + F C )LN. (33) Fh = area of head end of air piston in square feet. F c = area of crank end of air piston in square feet. L = length of stroke in feet. N = number of revolutions per minute. SIZE OF INLET AND OUTLET PIPES AND VALVES The valve and pipe areas are such that the velocity of air is from 3000 to 6000 ft. per minute. Although these are high the loss in pressure is not excessive. The suction valve is open during a longer time than the discharge valve and for that reason it seems to be necessary to use larger areas on the discharge valves. On the other hand, the effect of the drop is more notice- able at the lower suction pressure and therefore the suction valve must have a large area. The valves are of about the same area. This area of each set may amount to 8 per cent, of the piston area for piston speeds of 300 ft. per minute while for speeds of 700 ft. per minute 12 per cent, might be used. The pipes connecting cylinders or carrying air to or from the compressor should be designed from the allowable velocity if short, while for long pipes the drop in pressure, to be considered later, should determine the size. PERCENTAGE SAVING DUE TO MULTISTAGING OVER A SINGLE STAGE n-l- Work on single stage = _ .. ^ 1 f j Work on multistage = .'. per cent, saving = 100 - 100 ^ ~ P^ (34) Jm I-, /?>2\-^-| AIR COMPRESSORS 137 This saving is equal to the area Ii22'2i, Fig. 48, and of course it is the difference between the work of single-stage compres- sion and two-stage compression. The saving is equal to n n-1 f $- IT f (35) UNAVOIDABLE LOSS OF COMPRESSION ON TWO STAGES The area 12ilil', Fig. 48, represents an area which cannot be regained by a single-stage engine and may therefore be called the unavoidable loss. It is equal to 12ied 1'lied. FIG. 48. Saving due to multistaging. This is divided by / to give the total unavoidable loss with leakage. For three-stage compression the loss is really the work 138 HEAT ENGINEERING on two stages minus the work returned from one stage between pressures p" 2 and pi. Loss . pi pi This is shown in Fig. 49. FIG. 49. Losses and gains for three-stage compression. WORK ON AIR ENGINE Air engines are usually of one stage and as the air expands its temperature falls so that at the end of expansion the moisture in the air freezes and in many cases clogs up the exhaust. To prevent this heat may be applied to the exhaust pipe, multistag- ing may be used as shown in Fig. 51 or the air may be initially heated as shown in Fig. 50 by the dotted lines. This latter method produces such an increase in the work done that it will be carefully investigated later. The pressures between which the air engine operates are p'z and p'i. These are different from p 2 and pi because there is a drop in pressure due to friction in the pipe line carrying air to the engine and in the valves entering the engine, thus changing Pz to p'z. At exhaust the back pressure must be above the atmos- pheric pressure. If there is complete expansion in the engine and if the exhaust valve is so timed that the compression is complete, there is no effect of clearance on the work. Complete expansion or compres- sion means the carrying out of these actions until the final pres- sure is reached as shown in Fig. 52. The effect of clearance on the displacement is the same as that in the compressor and the AIR COMPRESSORS 139 same formula is used. In all theoretical discussions complete expansion is assumed and, for the present, the case of work or quantity of air for an engine with incomplete expansion and compression will not be discussed. The expression for work of the engine without clearance becomes, from Fig. 50, FIG. 50. Single-stage engine card FIG. 51. Two-stage engine with expansion line after preheating. card. ~ -- p - p'\V r \ 1 K Work of single-stage engine = In this case p'zV'z have been factored out because it is these two terms which are known in the case of the engine since the air is here at the original temperature TI; while at the lower pressure the temperature is low and would have to be computed by - jfc-1 k FIG. 52. Card with com- plete expansion and com- , - , T7/ T T i f i pression in engine, before piVi could be found. It is to be observed that (^r)~fc~ is less than unity so that the \P 2/ expression for work is positive, and moreover p'zV* = piV\ or p a V f so that either of these may be substituted, giving k r /P'I\ k- i-i Work on single-stage engine = - 7 r p a Vf\ 1 (r) * (38) K ~~ L L \p 2' -1 For a multistage engine the work becomes (39) 140 HEAT ENGINEERING If leakage is considered this reduces the work, giving The intermediate pressures are found as before LOSS DUE TO COOLING AFTER COMPRESSION The air is compressed to a temperature T 2 , for single or multiple staging, and on storing in the tank or pipe line its temperature is reduced to T\. The work which could have been done in one stage at the temperature T z would be given by while after cooling to TI the work to be obtained is Wo* -j^ Hence the loss is Loss of work due to cooling In this the first bracket is due to the compressor, while the second refers to the engine. UNAVOIDABLE LOSS DUE TO EXPANSION LINE Another unavoidable loss is due to the fact that the expansion line in the engine is of the form pV h = const, while that used on the compressor has been of the form pV n = const. This means that with a single expansion engine the area shown in Fig. 53 by abc is unavoidably lost. This computation is made before cooling the air to the temperature TI and while it is at the point This loss is given by AIR COMPRESSORS 141 n-l Since n f tpi\~r\ ^^Il 1 W n J (42) Const. FIG. 53. Unavoidable loss from difference in values of n in compressor and engine. LOSS OF PRESSURE IN PIPE LINE As the air flows through a pipe the pressure is decreased due to friction. There is only a slight change in velocity and hence this action is that of throttling in which the heat content and consequently the temperature of the perfect gas, air, remains constant. Now it is found that the drop in head when a fluid flows along a pipe line varies with the length of the pipe, with the square of the velocity and inversely with the diameter. This is usually expressed in feet of head of substance flowing. Since the velocity w of the air is only constant over a differential length of pipe due to the increase of volume as the air expands, the differential drop in pressure is all that can be expressed. Ctrl = ~y p: ul d 2g In this : 6 is a constant and equal to , 0.0274 , 0.00145 0.012 1 0.0124 + - - H ^ + , = 0.02 approximately. (43) w a CLW w = velocity in feet per second. d = diameter of pipe in feet. 1 Given by Green Economizer Co. 142 HEAT ENGINEERING g = 32.2. ft. per sec. per sec. dl = differential length in feet. dh = differential drop in head in feet of air. (dh is negative because it decreases as dl increases.) jy Now -g7p dh = dp T) ^nF = weight of 1 cu. ft. of air MET ^ 1 and w = - X j F = area of pipe in square feet. where ,, . , , */ . . . M weight of air per second in pounds. p b M 2 B 2 T* dl Hence - dp = r b M*BT - J Pdp-tj* (T is constant) Pi 2 - p 2 2 _ b M 2 BT bM 2 BT 2 = d F* 2g L ~ ** = d (44) 2d (45) Mean b" = 28 In the equations (44) and (45), pi = original pressure in pounds per square foot, p 2 = final pressure in pounds per square foot, L = length in feet in which 4 drop occurs. The formula may be used to find p 2 , M, or F, depending on what quantities are given. The formula . M 2 L '. Pi 2 - 2>2 2 = b" -- can be changed to refer to the volume of free air at 60 F. instead of weight by dividing by the square of 53.35 X 520 14.7 X 144 glvmg (46) AIR COMPRESSORS 143 Richards suggests &'" = Mooo> if pi and p 2 = pressure in pounds per square inch. Vf = cubic feet of free air per minute. L = length of pipe in feet. D = diameter of pipe in inches. LtH _ ~ 2000 * The expression (44) may be simplified as follows: Pi + P* = p = mean pressure z BT = mean volume of 1 Ib. P ,~ or Head drop in feet of air = ,~ m (47) where w m is the mean velocity since MET V Equation (47) is the formula used in hydraulics where the velocity is uniform. Equations (44), (45) or (46) may be used to find the length of pipe for a given drop and quantity by weight or volume for a given pipe, or D if the allowable drop in a certain length at a given discharge is known, or lastly the drop for a given discharge through a certain pipe of given length. The value of b is obtained by successive approximations in the formula (43) if not known from the given conditions. LOSS DUE TO LEAKAGE FROM TRANSMISSION LINE The leakage from the transmission line may amount to a large percentage of the air delivered if the line is not tight through- out its entire length. The leakage through a number of small 144 HEAT ENGINEERING holes is larger than one would expect. The amount of leakage is found by Fliegner's formulae, depending on the pressure. M=0.53 -^L for p 2 < 0.5 Pl M = 1.060 F al r ~ 2 for p* > 0.5 Pl After M is found this may be reduced to free air by The loss is proportional to the quantity of air. Vfi Loss as per cent, of what should be obtained = 100 -y- LOSS DUE TO THROTTLING The friction action of the pipe line is the equivalent of throttling action and moreover in many cases air is stored in tanks under very high pressure to be used in engines or air motors at a re- duced pressure after passing through a reducing pressure valve. The action of this reducing pressure valve is throttling action and hence the temperature of the air remains constant during this action; p'zV'z is therefore the same as piVi or p a V/. The pressure has been reduced. The available energy should have been IP'--*- but by throttling to p'% this is reduced to or Loss due to throttling = W ' - W" = 1 <> This reduction has been due to the reduction of the upper pressure to a point near the lower pressure. AIR COMPRESSORS 145 GAIN FROM PREHEATING If air at a pressure p' z and temperature 7\ is heated to the temperature T"' 2 so that its volume is changed from V 2 to ~F" 2 , the heat added will be AJQ = JMc p (T" z - T,) p' 2 (7" 2 _ F' 2 ) (49) FIG. 54. Gain from preheating air. The increased work will be the area 2'2"1"1', or (50) The efficiency of this heat used in preheating is therefore k - 1 " 2 - 7' 2 ) -(&)*? (51) 1 2 If now this efficiency of preheating, 1 f^r) * ' is higher \p 2' than the overall efficiency of the system, i.e., the ratio of the work of the air motor to the work required for the compressor, it will pay thermodynamically to preheat. Of course the necessity for a warm exhaust or the possibility of increasing the output of a 10 146 HEAT ENGINEERING given quantity of air may make it advisable to use preheating, although the efficiency of this is not as great as the efficiency of the system. POWER AND DISPLACEMENT OF AIR ENGINE OR MOTOR In the case of complete expansion, the expression for work has been given. Work for single stage = ~j p a fV f [ 1 - (^) HTJ (38) Work for single stage after preheating = w^wk-ffi-?] (52) Work for m-stage expansion = Since in these the volumes represent quantity per minute, the indicated horse-power may be found by dividing by 33,000 and this quantity multiplied by the mechanical efficiency of the motor or engine will give the delivered horse-power. This efficiency may be taken at 75. to 90 per cent. If on the other hand the power required is known as well as the pressure available, the indicated power can be found from the assumed efficiency and then the work per minute. From the equations above, the volume of free air necessary for the engine or the amount of compressed air can be computed. The next step is to find the displacement of the engine. If Vf is known, the amount of air per minute at the upper pressure is given by 7' 2 = ^ (53) If V' 2 is known this can be used to find V'\, the amount of air between the ends of the expansion and compression lines: -tf V'i V'i Displacement = -~- = If there is leakage, D = (56) AIR COMPRESSORS 147 Now all the above determinations are for complete expansion and compression. If, however, there is incomplete expansion or compression other calculations must be made. Let - be the proportional value cut-off and x be the value of compression expressed in terms of the displacement and assume that there is a clearance of ID. The work is found as the area of the card after the pressures at 3 and 6 (Fig. 56) are found. FIG. 55. Diagram for air engine FIG. 56. Diagram from air engine with complete expansion and com- with incomplete expansion and com- pression, pression. = p't Pi -i w , . Work per minute = I J 33000 X D.H.P. 1 (57) (58) mech. eff. 1 -A; -s]Z>- 1 -k pj, l-k (59) In this equation D is the unknown quantity and may be found for any given power required. The quantity in brackets is the mean effective pressure of the card. The amount of free air 148 HEAT ENGINEERING required for this motor is ascertained by determining the differ- ence in the weights of air at 2 and 5 and then reducing this to volume. If there is leakage, D is made D *- / in finding volume. = ~ BT l P D x)D BT *'6!8 (60) PC T z and T 5 are not definitely known on account of the Action of the cylinder walls. On a test they could be determined, since the temperature at the beginning of compression is the same as that of the exhaust and from this the mass in the clearance space would be known. From the air supply the mass entering would be known and consequently the total mass at the point of cut- off or release. Since the pressures and volumes at these points are known, as well as the masses, the temperatures could be found. Although not strictly correct, T$ will be assumed the same as T 3 and T 2 will be assumed equal to TV This gives (61) With either of the cases above if D is known the stroke and diameter can be found as in the case of the compressor. Assume 2LN = 300 to 700 or 1000. Assume N and find L. Now D = (F h + F e )LN. From this F h and F c may be found. The output of the air motor is usually given in a problem and from this the i.h.p. may be found. = i.h.p. (62) . ... mech. eff. From this the size of the motor and the amount of free air may be found after the pressure limits and events of the stroke are assumed. After this is accomplished the drop in pressure in the supply line is found and finally the pressures, free air for the AIR COMPRESSORS 149 compressor, size of compressor and the power to drive the same. The overall efficiency is found by Overall eff. = ^ h.p. output rf eogSne (63) h.p. required to drive compressor This efficiency is found to be about 40 per cent. when worked out, although with leaks a lower efficiency is obtained. To show vari- ous values of efficiency a number of problems will be worked out later. FAN BLOWERS The turbo compressors and fan blowers not only give a com- pression of the air but in addition the velocity of the air at discharge is so great that there is an additional term for the gain of kinetic energy. In the turbo compressors the cooling is practically continuous, although it may be considered as an ra- stage compressor of a value of n of almost unity on account of the cooling effect of the metal and the water jacket. In this case the expression for work is (, For the fan blower the action is so rapid and the path so short that the action is assumed adiabatic and the expression is In this pz and pi differ by a small quantity. GOVERNING The fan and turbo compressor are of value because the quantity of air may be varied with the need, the pressure changing with the quantity and the power changing within known limits so that motors may be provided. With displacement compressors work- ing at a fixed pressure, the simple way of changing the quantity is to change the speed of the compressor. This may be accomplished with steam engine drives by a slight throttling of the steam. The point of cut-off is not changed and the pressure is only slightly changed because for a given delivery pressure for the air a definite steam pressure is required to give sufficient area on the steam card. The slight increase of pressure will overcome the friction 150 HEAT ENGINEERING and speed up the machine. When, however, the speed of the motor cannot be varied, as is the case with certain electric motors, the varying demand for air must be met in other ways. Among the ways suggested to care for a varying quantity of air the following may be mentioned : (a) delayed closing of inlet valve, (6) delayed closing of discharge valves, (c) throttling air on suc- tion and (d) changing clearance. These methods are all under the control of the governor. All except method (c) mean no change in efficiency. Fig. 57 shows methods (a), (6) and (c) Clearance \ Suction Valve Held Open for a Certain Time Ckara Discharge Valve Held Open for a Certain Time Suction Pipe Throttled Change in Clearance FIG. 57. Methods of changing the discharge from an air compressor of fixed speed. as well as three cards for three different clearances. The clear- ance is varied by automatically connecting different chambers to the cylinder heads as the pressure rises. This is controlled by the governor as the pressure rises, necessitating a reduction in the quantity of air; the increase of clearance will cut down the quantity. When compressors are to be operated at different pressures, the work of the steam cylinder is controlled by a throttle governor, a variable cut-off governor or by a Meyer valve gear. AIR COMPRESSORS MOTORS 151 The engines using compressed air may be of the regular form of engine or its equivalent. Fig. 58 shows the section through a Little David riveter of the Ingersoll Rand Co. In this the throttle valve A is controlled by the handle B which is pressed by the thumb of the operator allowing air to enter. The valve C FIG. 58. Little David riveter of Ingersoll-Rand Co. allows air to enter behind the piston D which is driven at a high speed against the shank E which drives the rivet. The valve C shown in black admits air to the groove at / and exhaust takes place through the passage running from the left end of the piston. After passing through passages in the valve the exhaust leaves through the outlet to the left of C. When G is uncovered air FIG. 59. Ingersoll-Rand air drill (Little David). rushes through a small port and actuates the valve so that air is cut off from / and the port H is connected to outlet and the air to the right of the piston is discharged. A small port at this time discharges air into the large exhaust passage which is now cut off from the atmosphere and this forces the piston to the right. 152 HEAT ENGINEERING When the piston passes H the air remaining to the right and an amount which constantly leaks through a passage to the right of / is compressed, cushioning the piston. The increase of pressure here, and that due to a leak into the groove / after this is covered by the piston, causes the valve to reverse and the action to be repeated. Fig. 59 shows the construction of a drill which is the equivalent of four single-acting engines with rotary valves worked from the shaft while Fig. 60 shows a section through a rotary reamer or drill. This is a rotary engine. Air is admitted at A and causes the diaphragm or plate B to turn on the axis C. In Fig. 61 a preheater is shown. In this the passage of the air through the hollow portions of the heating surface raises the tempera- ture to a high point. The fuel may be coal, oil or gas. The arrangement and amount of FIG. 60. Rotary air drill. FIG. 61. Sullivan air preheater. surface in the intercooler is fixed by the principles of Chapter III, and will be discussed later in connection with a definite design. LOSSES IN TRANSMISSION The various losses discussed are shown in Fig. 62. abed = unavoidable loss due to two staging. This may be eliminated if a two-stage engine is used. efgd = loss due to leakage in compressor. gfh = loss due to change of line. AIR COMPRESSORS 153 hfji = loss due to cooling. ijkl = loss due to leakage in line. knop mnql = loss due to throttling. mrst = loss due to high back pressure. FIG. 62. Diagram showing various losses in air transmission. LOGARITHMIC DIAGRAMS Before solving problems it will be well to consider the con- struction of polytropics on logarithmic coordinates. If a curve of the form pV n const, be plotted on the coordinates log p and log V it is found that the equation above takes the form log p + n log V = const. This is of the form x + ny = const., or the curve becomes a straight line and the value of n is the slope of the curve since dx dy = If the coordinates of p and V of a curve are plotted logarithmic- ally with logarithmic coordinates and the points lie in a straight line, as in Fig. 63, the curve must be of the form pV n = const, and the slope of the line is the value of n. The value of n in the figure is to be n = 1.4 In this figure the value of the logarithm extends from to 1 and the numbers from 1 to 10, but the figure is constructed 154 HEAT ENGINEERING beyond that limit since the same variation in logarithm changes the numbers from 10 to 100, 100 to 1000, etc., or 1 to 0.1, 0.1 to 0.01, etc. LogP ido u 0.9 1 \ I? 0.8 \ \ 8.0 80 0.7 \ \, C \ \ 00.0 coo 0.6 \ \ \ \ 60.0 600 \ \ \ 0.4 \ \ N. 140.0 400 V \ \ \ 0.3 \ \ \ 30.0 800 d \ \ \ \ 1 \ 0.2 \ \ \ 1 \ \ \ v \ x \ \ \ 0.1 1.0 \ \ \ lib - 1 C 0.2 0.8 0.4 0.6 0.6 0.7 0.8 0.9 I 1.0 2.00 S.O 4-0 6.0 6.0 7.0 8.0 9.0 1 10 20 30 40 60 60 70 80 90 1C Log V Table from Log Diag. P= 10 F=iOO p . 20 V 61 P . 40 V" 37 P = 60 V 28 P = 100 V" 19 V" 10 1-20 FIG. 63. Logarithmic diagram with repeated coordinates for plotting pV 1 - 4 = const. If in the figure from b at the end of ab, the vertical to c is drawn and then cd is drawn parallel to ab, and after taking d to e and drawing ef with one following fg, a series of lines may be put on a single figure which would require a much larger figure if a single AIR COMPRESSORS 155 line were drawn as in Fig. 64. Fig. 64 is less confusing than Fig. 63, but it requires much more space. Such figures as the above are of value in constructing curves of expansion. If on a logarithmic diagram through the points log pi and log Fi, a line inclined at n = 1.35 be drawn, the values 1000 800 600 400 200 100 80 60 40 7 20 10 \ \ \ \ \ \ \ \ \ \ \ \ N \ > \ \ \ \ \ \ \ .1 0,2 0.4 0.6 0.8 1.0 2.0 4.0 6.0 8.0 1C Log V FIG. 64. Logarithmic diagram with continued coordinates. of p and V for any point may be found and the pV curve drawn with simplicity. To aid in computations of many of the above quantities charts and diagrams have been devised from which results may be obtained rapidly. A set of diagrams arranged by Professor C. R. Richards and Mr. J. A. Dent, based on temperature entropy logarithmic diagrams, is of great value. See Bulletin 63 of the Engineering Experiment Station, University of Illinois. PROBLEMS To apply the above formulae, a problem will be assumed as follows: Ten 5-h.p. air motors are to operate at 200 r.p.m. between 60.3 Ibs. gauge pressure and 0.3 Ib. gauge back pressure. Find the size of the motors if the 156 HEAT ENGINEERING piston speed is 200 ft. per minute and find the amount of free air required to drive these assuming a 3 per cent, leakage. Assume first that the com- pression and expansion are complete with a 5 per cent, clearance and sec- ond assume a cut-off at 0.35 stroke and compression at 0.1 stroke with 5 per cent, clearance. Compression of this air is to take place in a two-stage air compressor from 0.2 Ib. gauge pressure to 135.3 Ibs. gauge pressure. The compressor is 3000 ft. from the motors and the drop in the supply main is to be 4 Ibs. Find the size of the main. There are fifty H2-in. holes in the main; find the leakage. The air temperature is 70 F. and the available water is at 60 F. The leakage in the compressors is 3 per cent. Find the air to be taken into the compressor to cover all leakages. Find the various efficiencies, losses and constants of the system. Would it be advisable to preheat the air to 350 F.? In this problem the work will be done by the use of a 12-in. slide rule of Log. Log. form. FIG. 65. Indicator card for air engine with complete expansion and compression. FIG. 66. Indicator card for an engine with incomplete ex- pansion and compression. Problem 1. Indicated horse-power of one motor. I Lh -P- = Tff7.= qk= 6 - 25 Friction 1.25 h.p. Problem 2. Displacement and free air for two cases, (a) Complete expansion and compression. 6.25 X 33,000 = Y~I 14 - 7 X 144 X ' 97 r = 3.5 X 2118 X 0.977/ 1 1 - 0.4 L4 |(40) 78 cu. ft. per minute 14.7 X 78 75 = 15.26 X (^ = 15.26 V, F' 2 Ki = 1 + 0.05 - 0.05 X 0.97 X 48 48 D = 0.892 (53) (54) = 1.05 - 0.158 = 0.892 (30) 52 cu. ft. per minute. (56) AIR COMPRESSORS 157 (6) Incomplete expansion and compression. 1.4 PA = 75 [005"+'l 5 = 75 X d*6 = 19 - 41bs -P er ^are inch (57) roos + o n 1 - 4 P = 15 n n^ = 15 X 4.66 = 69.9 Ibs. per square inch (58) M.e.p. = 0.35 X 75 - 0.9 X 15 + 19.4 X 1.05 - 75 X 0.4 - 15 X 0.15 + 69.9 X 0.05 ~- 0.4 = 26.25 13.5 + 21 = 33.75 Ibs. per square inch 6.25 X 33,000 = D X 144 X 33.75 D = 42.5 cu. ft. per minute 42 *) x 75 x a4 ~ 15 x 144 I O^X ' ~l ir) x al5 J = 2.89 I 30.0 - 3.3J = 77.2 cu. ft. per minute. If leakage is considered *>' - gf - 79 - 5 cu - ft - The loss in power due to leakage is ((Toy 6.25 j =0.19 h.p. It is seen that although the displacement is larger for the complete expan- sion arrangement the amount of free air is slightly less. The difference between the two cases is not great. Problem 3. Size of motors. (a) and (6). N = 200 .'. L = x ft. = 6 in. (a) D = 52 = (A h + Ac] X 200 X If the areas are equal 52 A = 200 sq ' ft ' 52 X 144 200 =37.4sq.m. Assuming a 1-in. piston rod and tail rod Acyi. = 37.4 + 0.7854 = 38.2 sq. in. d = 6.95 in. = 7 in. Cylinder 7 hi. X 6 in. (6) D = 42.5 cu. ft. = 2 X 200 X % X A A = 2QQ X 144 = 30.6 sq. in. Acyi. = 30.6 + 0.7854 = 31.4 sq. in. d = 6.33 in. Cylinder 6.33 in. X 6 in. 158 HEAT ENGINEERING Problem 4. Leakage from pipe line. Pressure at one end = 135.3 + 14.7 = 150 Assumed pressure for leakage = 150 4 ~ 32 X 32 = 0.132 lbs. per second 60 X 0.132 X 53.35X530 V/ * = " 14.7 X~144"~ = 5 CU ' ft> per mmute - Problem 5. Total free air assuming case of complete expansion motors, (a) Air for motor = 78 cu. ft. per minute. (6) Leakage air = 105 cu. ft. per minute. (c) Total delivered free air = 885 cu. ft. per minute. O Q K (d) Total free air required for compressor = ^ Q > = 942 cu. ft. per minute. (0.94 is leakage factor for two stages.) Problem 6. Diameter of pipe line to carry air. , . 885 X 14.7 X 144 M = 60 X 53.35 X 530 = L1 lbs ' per sec nd * For first approximation b = 0.02 /ISO 2 - 146 2 \ , 0.02 X l.l 2 X 53.35 X 530 ( 2 ) 144 = "~ ^2 2 X 32.2 X yg X d 5 d 6 in inches = d & in feet X 12 s = 1050 d = 4.02 in.; use 4-in. pipe. 6 should then be checked 53.35 X 530 X 148 X 144 16.7 ft. per second. - - 0124 X 0.7854 0.0274 0.00145 0.012 = 0.0124 + 0.0016 + 0.0044 + 0.0020 = 0.0204 The value of 0.02 is close. Using equation (46) 885 2 X 3000 D 6 = 990 D = 3.97 in. or 4 in. The drop in pressure will not be 4 lbs. if a 4-in. pipe is used with 6 = 0.0204 ' X 3000 (44) p 2 2 = 22,500 - 1238 = 21,262 p 2 = 145.8 lbs. . ' . Drop = 4.2 lbs. If a one-third increase in this is assumed a drop of 6 lbs. will be cared for. AIR COMPRESSORS 159 Problem 7. Loss due to throttling from 150 Ibs. to 144 Ibs. abs. 0.4 O4 Loss - H.7 X 144 X *[() "- "] (48) = 6,550,000[0.524 - 0.518] = 39,300 ft.-lbs. per min. = 1.19 h.p. Problem 8. Loss due to throttling from 150 Ibs. to 75 Ibs. abs. 04 Loss = 6,550,000 [0^ IA - 0.518 1 (48) = 6,550,000[0.631 - 0.518] = 6,550,000[0.113] = 740,000 ft.-lbs. per min. = 22.5 h.p. Loss if leakage occurs at higher pressure 780 = 22.5 X gg = 19.8 h.p. Problem 9. Loss due to cooling from T 2 to TV 0.35 53 1 = 530(10.33) 7.7 = 530 X 1.355 = 718 F. abs. = 258 F. 0.35 04 ^X0.94X942XU.7X144[ @ >< -l] [l- = 6,550,000[1.355 - 1][1 - 0.518] = 1,122,000 ft.-lbs. per min. = 34.1 h.p. The use of values previously computed reduces the amount of computation. Problem 10. Work on two-stage compression and intermediate pressure. 0.35 = 15,120,000[1 - 1.355] = - 5,370,000 ft.-lbs. per min. = 162.7 h.p. 94 l Q 97 0.955 = - g ~ ~~ = leakage factor for work on two stages p'z = V 150 X 14.5 = 46.6 Ibs. per sq. in. abs. (19) Problem 11. Work on single stage. 0.35 = 7,470,000[1 - 1.833] = 6,220,000 ft.-lbs. per min. = 188.5 h.p. Problem 12. Saving due to double staging. Gain = 188.5 - 162.7 = 25.8 h.p. 160 HEAT ENGINEERING Problem 13. Saving due to jacket if on single stage (16). OA 0.35 0- 147XH4X ^{[l - gg) "] - t[l - (0) '"] j = 1,930, 000 {3.5[1 - 1.95] - 3.86[1 - 1.833]} = 1,930,000 { - 3.32 + 3.218) = - 197,000 ft.-lbs. per min. = 5.98 h.p. Problem 14. Saving due to jacket if on two-stage compressor. Gain = 1,950,000{7[1 - (1.95)^]- 7.72[1- (1.833)]) = 126,500 ft.-lbs. per min. ~ 3.84 h.p. Problem 15. Unavoidable loss in two staging (36) . 0.35 0.35 1.35 x ."-^ , 885 r /150\-2TTr /14.5\-2T~| Loss = 035 X 14 ' 7 X 144 X 0955 L 1 \14^j J L 1 ~ ( ISO/ J = - 7,560,000[0.355][l - j-|^l = 705,000 ft.-lbs. per min. = 21.25 h.p. Problem 16. Loss due to change in expansion line of engine from n to . starting from point at end of compression. n-l k-l = 14.7 X 144 X 885 X 1.355 1 3.86 j"l - 37^33] = 2,540,000(1.755 - 1.706) = 124,500 ft.-lbs. per min. = 3.78 h.p. Problem 17. Loss due to high back pressure in engine. = X 780 X 14.7 X 144 [0.632 - 0.625] = 40,500 ft.-lbs. per min. = 1.23 h.p. Problem 18. Loss due to leakage in compressor. Loss = available work without leakage available work with leakage = [work of compression unavoidable loss] X [1 leakage factor] = [162.7 - 21.25][1 - 0.955] = 6.38 h.p. AIR COMPRESSORS 161 Problem 19. Loss due to leakage in line. Loss = [work of compression unavoidable loss loss due to compressor leakage loss due to change .. .leakage volume of exponent-loss due to cooling] total volume = [162.7 - 21.25 - 6.38 - 3.78 - 34.1]^ = 11.5 h.p. Problem 20. Power to drive compressor. ind. power 162.7 Power = mech eff. = O^ = 18 ' 8 h *- Friction loss = 180.8 - 162.7 = 18.1 h.p. Problem 21. Power of motor for compressor. Motor loss = 200.1 - 180.8 = 19.3 h.p. TABLE OF POWER H.p. Percentage Power supplied motors 200.1 100.00 Loss motors 19.3 9.6 Power supplied compressor 180 8 Loss compressor 18.1 9.0 Power supplied air. . . 162 7 Unavoidable loss 21.25 10.6 Loss from leakage in compressor. 6 38 3.2 Loss due to change from n to k Loss due to cooling after delivery. 3.78 34.1 1.9 17.1 Available power supplied line 97 2 Loss due to leakage in line 11.5 5.7 Loss due to, throttling to 75 Ibs Available power at engine 19.8 65 9 9.9 Loss due to high back pressure . 1 23 6 Loss due to leakage 3 per cent, of 65 9 1 98 1 Available indicated work 62.70 Amount loss in engine 20 per cent. 12 54 6 3 Amount delivered 50.20 25.1 Saving due to jacket. . 3 84 100.00 1 9 Saving by double staging 25 8 12 9 It is to be noted that there is 9.9 per cent, loss due to throttling and 5.7 per cent, due to leakage. These might be saved, giving the overall effi- ciency about 40 per cent, in place of 25 per cent. The low efficiency has been due to the leakage from line and throttling. In good installations 40 per cent, overall efficiency might be reached, ll 162 HEAT ENGINEERING Problem 22. Heat removed by jacket (15). 14 - 7 H X 0.35 = - 1.73 X j X 13,830 X 144 X 0.355 = 436,000 ft.-lbs. per min. = 561 B.t.u. per min. Problem 23. Heat removed by intercooler (28). 0.35 = + 2,400,000 ft.-lbs. per min. = 3090 B.t.u. per min. Problem 24. Water for jacket. Assume water rises from 75 to 90 F. in jacket. P\A 1 G = QQ __ 75 = 37.4 Ibs. per min. Problem 26. Water for intercooler. Assume rise in temperature from 60 to 75 F. 3090 G = 7 K _ fiQ = 206 Ibs. per min. Problem 26. Mean At, assuming counter current flow from 60 to 75 F. for water and 70 to 258 F. for air, using Rensselaer formula. Aii = 258 - 75 = 183 F. A 2 = 70 - 60 = 10 F. Note that this is quite different from ~^~- = 97 F. Problem 27. Surface required and arrangement. Assume water velocity = 3 ft. per sec. Assume air velocity = 10 ft. per sec. 46:6X144X1 (10 - 1.75) . oW Value of K - 126 X 53 . 35(16 4 + 460) X - -^- -X 3* = 16.7 (44) Ch. I , 3090 X 60 Surface = x 66 = 169 sq.ft. Arrangement of ^-in. tubes 12 ft. long with water inside. Area of 1 tube = X X 12 = 2.35 sq. ft. AIR COMPRESSORS 163 169 No. of tubes = ~oc = ^ tubes. No. of tubes in one nest. Area of inside of tube *| X0 ^JJ/'^X jpj- " - 00258 sc l- ft - Amount of water = 206 Ibs. per min. Velocity of water = 3 ft. per sec., 180 ft. per min. OOA No. tubes in parallel = 64 . 5 x 180 x Q.00258 = 7 tubes ' This means 10 nests, or the water will run from. 7 tubes to 7 tubes so as to give a path of 120 ft. in cooler. 14 7 614 Volume of air per min. = (942 X 0.97) ^g X ^Q = 340 cu. ft. Area for air = in y gn = 0.565 sq. ft. Suppose the 7 tubes are made in a case so that the air may pass along. Area of passage 0.565 sq. ft. = 81 sq. in. To this the area of seven H-m. tubes or 3 sq. in. will be added giving 84 sq. in. for passage of air. Problem 28. Size of compressors at 80 r.p.m. with 400 ft. piston speed. K,- 1+0.05- 0.05 = 0.932 885 Low-pressure displacement = nn /i \x n noo = 1010 cu - ^- P er mm - u.yT: /s. u.yoz 1010 = 400 Amean Anet = 2.52 sq. ft. = 388 sq. in. Assume a 3-in piston rod; A = 7.06 sq. in. Agross = 363 sq. in. dcyi. = 21.5 in. T 40Q 9W ff L = 2~><80- Size of compressor 21.5 in. X 30 in. High pressure displacement. Intermediate pressure = \/150 X 14.5 = 46.6 Ibs. Stroke = 30 in. OAQ Area net = ^ = 0.77 sq. ft. = 111.0 sq. in. Area gross = 111.0 + 7.06 = 118.06 dcyi. = 12.3 in. Size of high-pressure compressor 12.3 in. X 30 in. Problem 29. The efficiency of preheating is 0.4 Eff. = 1 - (11^ 1A = 1 - 0.631 = 0.369 With the leakage and throttling giving an efficiency of 25 per cent, on output or (25 + 6.3) = 31.3 per cent, on indicated work of engine it is 164 HEAT ENGINEERING seen that this would pay. If there had been no throttling and the air were at a 150-lb. absolute pressure the efficiency would be q.4 Eff. = 1 - frJLl IA = 1 - 0.518 = 0.482 This, being better than 40 per cent., would pay. In most cases pre- heating will pay. It is well to note in passing that the efficiency of pre- heating depends on the range of pressure. TOPICS Topic 1. What are the different methods of compressing air? Sketch machines for this purpose. Under what conditions is each used? Topic 2. Explain the construction of a two-stage air compressor and give the reasons for the use of such apparatus. What is the purpose of the inter- cooler? Why is moisture apt to collect in the air space of the intercooler? How is it removed ? What is the peculiar form of inlet valve used on the low- pressure piston? Topic 3. For what reason is air for a compressor taken from the atmos- phere and not from the engine room? Why are water-jackets used? Ex- plain the action of the Taylor hydraulic air compressor. Topic 4. Derive the expression: Topic 6. What is the effect of clearance? Prove this. Derive the ex- pression for the clearance factor. Topic 6. Explain what is meant by volumetric efficiency. Derive the formula for true volumetric efficiency in terms of the leakage and clearance. Show how to find the horse-power to drive a compressor. Topic 6a. Why is jacketing of value? Is this true under all conditions? Is jacketing very effective? Derive the expressions for the heat removed by the jacket, the saving by the jacket and the water required for the jacket. Topic 7. What is multistaging? Sketch a figure and show why this is of value. What is the function of the intercooler? How large should it be? Derive the expression for two-stage compression: Topic 8. Using expression of Topic 7, show that the work is a minimum when p'z = Vpipz' What are the conditions for minimum work for a three-stage compressor? Reduce the expression for work to a simple form. To what does the expression reduce for an ra-stage compressor? Topic 9. Derive the formula for the temperature at the end of compres- sion on the line pV n = const, when pi, pz, and T\ are known. What is the value of the intermediate temperatures on a multistage compressor? Show that these are the same on each stage. Topic 10. Derive the formula for the heat removed by the intercooler and that for the amount of water required? What is the effect of leakage in multistage compression? AIR COMPRESSORS 165 Topic 11. Explain method of finding displacement of the various cylin- ders of a multistage compressor to deliver a given amount of air. Topic 12. Derive formulae for the moisture removed from the intercooler and for the condition of the air discharged from it. How are the inlet and outlet pipes and valves designed? Topic 13. Derive a formula for the saving due to multistaging. Derive the expression for the unavoidable loss due to two-stage compression. Topic 14. Derive a formula for the work done in a single-stage engine. What is the temperature at the end of expansion? To what does the expres- sion for work reduce for a two-stage engine? Topic 16. Derive an expression for the loss due to cooling after compression. Topic 16. Derive a formula for the loss due to a difference in the lines of expansion in the engine and of compression in the compressor. Topic 17. Derive the formula for the loss of pressure in a pipe line carry- ing air. Why is T assumed constant? What is the effect of leakage? How is the amount of leakage determined? Topic 18. Derive the expression for the loss due to throttling and one for the gain from preheating. Topic 19. Derive the expressions for the displacement of an air engine with complete expansion and compression and with incomplete expansion and compression to give a certain power. Topic 20. Derive the expression for the work of a fan blower. Explain how the quantity of air from a constant-speed compressor is regulated. Topic 21. Explain by means of a diagram the various losses which enter into an air-transmission system. Give the efficiency in terms of the ratio of two areas. Topic 22. Explain the use of logarithmic diagrams in plotting curves of the form pV n = const. Explain the method of increasing the range of a diagram. Topic 23. Sketch the forms of some of the air motors in common use and explain their action. PROBLEMS Problem 1. Find the power to compress 1000 cu. ft. of free air per min- ute from 14.5 Ibs. absolute to 60 Ibs. gauge in a single-stage air compressor; n = 1.35, leakage = 3 per cent., clearance = 2 per cent. Problem 2. Find the power to compress 1000 cu. ft. of free air per minute from 0.3 Ibs. gauge pressure to 160 Ibs. gauge pressure in a two-stage air compressor with n = 1.35, leakage 3 per cent, in each cylinder, clearance 3 per cent. Find the intermediate pressure. Problem 3. Find the size of the compressors in Problems 1 and 2 if they operate at 120 r.p.m. and are double-acting. Problem 4. Find the temperature at the end of compression in Problems 1 and 2 if the original temperature were 75 F. Problem 5. Find the heat removed in the intercooler for compressor of Problem 2. Find the amount of water required if the water available is at 60 F. and is arranged by counter current to leave at 80F. Find AT 7 assuming that the value of the coefficient of heat transmission is constant. Find the value of K by assuming the velocity of the air to be 30 ft. per second and the velocity of the water as 10 ft. per second and using Nicolsou's formula. 166 HEAT ENGINEERING Problem 6. Using the variable parts of the expression for work find the relative work for single, two-stage and three-stage compression between 14.5 Ibs. and 290 Ibs. absolute. Find the clearance factor for these assuming 3 per cent, clearance in each cylinder. Find the temperature at the end of compression in each case assuming 75 F. as the original temperature. Problem 7. Find the size air main to deliver 3000 cu. ft. of free air per minute when compressed to 145 Ibs. absolute, if the temperature is 75 F. and the drop in pressure allowable is 2 Ibs. in 160 ft. Problem 8. Find the efficiency of transmission if 1000 cu. ft. of free air per minute is compressed to 190 Ibs. gauge pressure, transmitted 800 ft. in a tight main and used after storage in engines at 150 Ibs. pressure in a single-stage engine. Assume all constants needed. Is the result the probable efficiency or the maximum efficiency? Problem 9. Find the various losses in the system described in Problem 8. Problem 10. Three H-in. holes are in a pipe line carrying 1500 cu. ft. of free air per minute compressed to 125 Ibs. gauge pressure and at 80 F. What is the percentage loss due to these holes? Problem 11. The wet bulb reads 70 F. in 80 F. weather when the ba- rometer stands at 29.85 in. This air is compressed to 50 Ibs. gauge pressure in the intercooler at 80 F. Has any moisture been precipitated? If not what is the relative humidity of the air? If precipitation occurs how much precipitation occurs per minute if 1000 cu. ft. of free air are handled per minute? Problem 12. Find the size of an air engine to develop 15 h.p. (a) with complete expansion and compression, and (6) with cut-off at 30 per cent., compression 20 per cent., clearance 5 per cent. p\ = 725 Ibs. absolute, pb = 14.8 Ibs. absolute. R.p.m. = 125. Problem 13. Air at 1500 Ibs. absolute pressure is stored in a tank and when used in an engine it is throttled to 150 Ibs. absolute pressure. The tempera- ture in the tank is 80 F. What is the temperature at the end of complete expansion in the engine to 1 5 Ibs. absolute pressure ? How much of the energy in the tank is available in the engine after throttling? Problem 14. In Problem 13 the air on leaving the throttle valve is heated by a vapor lamp to 450 F. By what per cent, is the power of the engine increased by this? What is the temperature of discharge from the engine? Problem 15. An air compressor of Problem 1 has the compression line reduced to pv 1 ' 35 = constant by the jacket. How much heat does the jacket remove? How much water does this require if the water range is from 70 F. to90F.? Problem 16. One thousand cubic feet of free air per minute is compressed from 15 Ibs. pressure to 150 Ibs. pressure (absolute) on two stages. The original temperature was 60 F. Find the final temperature of discharge. Find the percentage loss if this air is allowed to cool to 60 F. before it is used in the air motors; n = 1.35. Problem 17. How much loss occurs in Problem 15 due to the fact that the expansion in the engine is on the line piP = const, instead of pF 1 - 35 = const. ? Express this as a percentage. Problem 18. Find the values of p and v at several points on the line, p yi.25 _ const., passing through the point p = 165 Ibs. V = 3 cu. ft. by means of a logarithmic diagram. Use pressures of 150, 100, 50 and 25 Ibs. CHAPTER V THE STEAM ENGINE On account of the numerous improvements made upon it during the two centuries of its use, the steam engine was, until quite recently, the most important heat engine using steam. The steam turbine has taken this place at present due to the decreased cost of production, the greater concentration of power, and the good efficiencies over a wider range of load. The engine was developed without any theory of thermodynam- ics by Worcester, Papin, Savery, Newcomen and Watt, and to account for the action of the engine in later times, Rankine in England and Clausius in Germany developed a mathematical theory of heat. The later development of the engine is connected with the names of Corliss, Porter, Reynolds, Willans, Sultzer and Stumpf. The engine has been improved by increasing the Carnot efficiency through the use of. higher initial pressures and superheat and by lowering the back pressure, while the use of jackets, reheaters, superheated steam, four valves, heavy lagging, multiple staging and flow in one direction have all had their effects on its practical efficiency. In the steam engine (Fig. 67) steam containing heat is ad- mitted to the cylinder A by the valve B, so that it moves the piston. The valve B is operated by the valve rod C from an eccentric or crank D so that when the piston E has reached a certain point steam is cut off from the cylinder and the steam is allowed to expand to a point near or at the end of the stroke (release point) when the valve is so moved as to allow the steam to escape to the outside. The steam admitted to the other end or the inertia of the fly wheel causes the piston to return in its stroke forcing the steam out of the cylinder until a point near the end of the stroke (point of compression) is reached when the exhaust is closed and the movement of the piston compresses the steam until the valve again connects this side of the cylinder t<\ the steam chest F and steam pipe G so that steam enters once more. In theory the indicator card 167 168 HEAT ENGINEERING THE STEAM ENGINE 169 170 HEAT ENGINEERING will appear as in Fig. 68. The expansion and compression are complete. The point of cut-off is at 1, release at 2, compres- sion at 3 and admission at 4. The quality of the steam at 3 is the same as that at 2, hence the condition at 4 is the same as that at 1 if the lines 1-2 and 3-4 are assumed to be adiabatics. 3-4 is the compression line of the steam in the clearance space 4-5. If these are adiabatics with M Q pounds of steam on 3-4 and M + M pounds of steam on 1-2, Fig. 69, with no clearance space and with the weight of steam M on the line 1-2 between the same pressures and qualities at the limiting points will have the same area and represent the cycle with no clear- ance. This is true because in theory the clearance steam ex- FIG. 68. Indicator card from en- gine with complete expansion and com- plete compression. FIG. 69. Indicator card with no clearance. pands and contracts along the same line requiring and giving no work. M is called the clearance steam and M is called the working steam, steam supply or boiler steam. CYCLE OF STEAM ENGINE Fig. 69 then represents the theoretical form of cycle used on the steam engine. It is known as the Rankine or Clausius cycle although the former name is employed at times for the cycle shown in Fig. 71. The difference between the Carnot cycle and the Rankine cycle is that in the former cycle the working substance is supposed to be in the cylinder and heat only is added, while in the latter substance is added and with it heat. On the Carnot cycle, if represented by Fig. 68, the quality changes from that at 4 to that at 1, growing greater, while it decreases from 2 to 3, hence 1-2 and 4-3 are not similar adia- batics. For the Rankine cycle the quality on 4-1 is constant, the variable being the mass and this is true on 2-3. In both THE STEAM ENGINE' 171 cases 4-1 and 2-3 are isothermals if saturated steam is used. The heat added on the Carnot cycle is Mrdx and on the other is xrdm, the total amount being the same in each. In Fig. 69 the exhaust steam could be used to heat an equal weight of water to the temperature of exhaust for boiler feed, since this could be done by the heat of the liquid of the steam even if the water supply were at 32 F. With the heat of vapori- zation in addition for the vapor part of the mixture, this and even more could be done. This additional amount is of no value for the operation of the engine although for warming other water it has a great value. The point which must be grasped by the student is the fact that the heat in the exhaust could, if properly used, heat the feed necessary for the working steam to a temperature of the exhaust pressure. For this reason, the datum plane from which heat is to be measured is always taken as at the temperature of the exhaust. HEAT AND EFFICIENCY OF CYCLE The heats added on the different lines are: M(ii - q'o) on 4-1 (1) on 1-2 Mfa - q'o) on 2-3 (2) on 3-4, since M = (Note. q' is used rather than q'% because is used to refer to the datum in all cases.) .'.AW = Qi - Q 2 It is to be remembered that i\ and z' 2 are the heat contents at two points on the same adiabatic and hence they are found in the same entropy column or line. Qi = M[ii - q f ] Theoretical eff. = r? 3 = ?* ~ *' (5) i\ q o This is sometimes called the Rankine efficiency. In some texts the work of pumping the water into the boiler from 4 to 1 is considered giving net work ii - i 2 - A (pi - p z )v' 172 HEAT ENGINEERING The last term is small. It really does not belong to the engine being one of the charges against the boiler. Now, the Carnot eff. = 771 = ^ If M pounds of steam are required per horse-power hour, the work = 2546 B.t.u. and the heat is M(ii q' ). 2546 (6) 175 FIG. 7Q.T-S diagram of Rankine cycle. FIG. 71. Rankine cycle with incom- plete expansion. The T-S diagram for this is an aid in studying the cycle. It is shown in Fig. 70. In this, points 3 and 4 come together. The heat added per pound from 3 to 1 is 6-4-3-1-5 = ii - q' Heat 12 = Heat 23 = 6-3-2-5 = i 2 - q' Work = 64315-6325 = 3412 = ^ - t 2 4-a-l-2 would show the Carnot cycle. If the expansion is not carried to the back pressure line it is called incomplete expansion and the cycle using such is then sometimes spoken of as the Rankine cycle. The best way to study this cycle is by dividing it into two parts by the line 2-6, Fig. 71. Area 5126 = M(i l - z a ) Area 2346 = M[A(p 2 - p 3 > 2 ] AW = 451234 = M[(ii - i 2 ) + A(p z - p s )v 2 ] (7) THE STEAM ENGINE 173 J! - 1 2 + A (p 2 - p 3 )v 2 " 3 = ~ t! - q'o (8) !Ti- To 171 2546 775 ~ Mfe - q'o) 773 for the engine cycle is sometimes called the Rankine efficiency of the cycle or this term is applied to equation (5). In the above equations, i\ and i% are heat contents on the same adiabatic, hence, if the entropy is known for point 1, i* is found at pressure 2 in same entropy column as i\ if there is a table or on the same entropy line from a chart. If there is neither table nor chart which can be used, the formula for the adiabatic is to be used to find the quality at 2 and from this, zV Fi = Mxiv'i (To find xi). 'i + r = ** + (To find xj F 2 = Mx 2 v" z (To find F 2 ). ii = q'i + Wi (To find z'i). ^2 = q'z + ^2^2 (To find z" 2 ). All necessary terms are known for any of these formulae if taken in the order given. The value of x\ being theoretical and therefore the highest possible, is that of the boiler steam and should be so used. STEAM CONSUMPTION From equations (3) and (7) the amount of steam per horse- power hour theoretically required is given by =s 2546 M = -. -- r -7-r- (10) ti - ^2 + A (p 2 - pi) v"i To illustrate the applications of the formulae, suppose an engine uses 26 Ibs. of steam per horse-power hour at 110 Ibs. gauge pressure with x = 0.98 and has a pressure at the end of expansion of 15 Ibs. gauge and a back pressure of 0.5 Ibs. gauge. It is desired to find the various efficiencies. 174 HEAT ENGINEERING From Peabody's Temperature From Marks and Davis 7 Entropy Tables Charts pi = 110 + 14.7 = 124.7 pi = 124.7 ti = 344.2 si = 1.563 81 = 1.563 ^ = 1173 11 = 1172.5 p 7 2 = 29.7 p 2 = 15 + 14.7 = 29.7 s 2 = 1.563 s 2 = 1.563 i 2 = 1065 1 2 = 1065.8 z 2 = 0.897 v 2 = 12.55 v 2 = 12.5 ?' of (0.5 + 14.7) Ibs. = 181.9 t = 213.7 By computation the following results, using the tables of Marks and Davis: p 1 = 129.7 i 2 = 318.4 + 0.98 X 872.7 = 1173.4 81 = 0.4996 + 0.98 X 1.0820 = 1.56 p 2 = 29.7 S2 = 1.56 = 0.3673 + x X 1.3326 1.1927 X * = E3326 = i 2 = 218.2 + 0.895 X 945.6 = 1065 v 2 = 0.895 X 13.89 = 12.42 q' = q' of (14.7 + 0.5) = 181.6 Any of the three sets of values could be used. Using the first ^ 9 344.2 - 213.7 130.5 771 =344.2 + 459.6 = 803^ = 16 ' 2 per Cent %. f A x 1172.5 - 1065.8 + ^(29.7 - 15.2) 144 X 12.55 77o T 173 = 1172.5 - 181.9 139.9 = QQQ-^ = 14.2 per cent. 14.2 772 = T^~2 = 87.2 per cent. 2546 2546 _ 776 ~ 26(1172.5-181.9) ~ 2580 ~ 9.9 174 = .^7-n = 70 per cent. THE STEAM ENGINE 175 The results of the above computations show that the efficiency of the theoretical cycle is more than 87 per cent, of that of the most perfect cycle but that the actual engine only utilizes 70 per cent, of the amount which should be available theoretically. For these reasons there is not much chance of improving the cycle of the steam engine but to make 174, the practical efficiency, greater the cylinder is covered with some non-conducting material, a steam jacket is used or superheated steam may be employed. These all tend to cut down the transfer of heat to and from the cylinder walls and thus cut down the amount of steam required. TEMPERATURE ENTROPY DIAGRAMS The cycle is shown on the T-s diagram, Fig. 72. The free expansion from 2 to 3 cuts off the corner of the figure as shown by 2-3. This figure and Fig. 70 are used to represent FIG. 72. T-S diagram of Rankine cycle with incomplete expansion. 8'8 7 S FIG. 73. Effect of increasing tem- perature range. the Rankine cycle without and with complete expansion. They are not true representations since the lines of Figs. 70 and 72 represent additions of heat and not of substance as is the case on the actual cycle. Because the heat added in the figures on the various lines represents the heat added on the lines with the substance the figures are used to represent this cycle. More- over, the area of the figure 1234 of Fig. 70 or 123456 of Fig. 72 represents the difference between Qi and Qz and hence repre- sents the work. The efficiency is represented by the area of the cycle divided by the total area beneath 4-1 in Fig. 70 or 72 and hence, if the temperature is increased, both of these areas become greater and the efficiency is increased. This is shown by 6i-l" of Fig. 73. 46 1 l // 234 461234 Jjjtl * ~ 846il'7 > 84617 176 HEAT ENGINEERING If, now, the lower temperature is lowered from 3 to 3', the denominator of the fraction is changed a slight amount, 8'4'48, while the numerator is changed by the area 433 '4'. In this way the efficiency is increased in most cases. If the free expansion is so great, as shown in Fig. 73, that the distance 3-4 is small, it maybe that the added area, 433'4', may not be greater than 8'4'48 and the efficiency is not increased by lowering the back pressure. As the free expansion 2-3 becomes less, as 2'-3" (which means that the expansion 1-2 is more nearly complete), the line 3"-4 is so long that the gain in efficiency due to a decrease in back pressure is great. The increase in pressure to give a perceptible increase in temperature for the boiler steam is so great that the pressure must be increased considerably to raise the line 6-1 an appreciable amount. Hence the gain in efficiency from an increase of pressure is not as marked as a slight decrease in the lower pressure of the cycle when the tempera- ture change is more rapid. Thus, to de- crease the back pressure from 15 Ibs. to 2 Ibs. abs. means a change of 87 F. while an increase from 100 Ibs. to 113 Ibs. means an increase of 10 F., from 150 Ibs. to 163 Ibs.. 8 F., while from 200 Ibs. to 213 Ibs. means an increase of less than 6 F. It must also be remembered that a decrease in the back pressure with much free expansion does not necessarily give an increase of efficiency. In order to increase the efficiency by increasing the upper temperature without an increase in pressure superheated steam is used. Theoretically it will be seen that although this does increase the efficiency there is not much gain because this heat is not added at constant temperature. Fig. 74 illustrates this case Q 1 = 9c618 = q\ - q' + r + fcpdt = n - q' Q 2 = 94328 = i<, - q' - A(p 2 - pjv* Qi - Qz = work = 45c6 1234 Now the triangle l-d-b is the part of the figure which repre- sents the increase due to superheat and being added to 5 cbd 234 FIG. 74. T-S diagram of Rankine cycle with superheated steam. THE STEAM ENGINE 111 as well as to 9 c b d 8 it increases the efficiency, having the greater effect on the small numerator. This reasoning is made evident by observing that the efficiency 5 cbd 234 is about the same 95 cbd 8 as the efficiency of the cycle of Fig. 71 with saturated steam and the same pressure ranges and the expression for the latter is made 'into that for Fig. 74 by adding b-l-d to numerator and denominator. EFFECTS OF CHANGES These facts have been shown by the late Prof. H. W. Spangler in his Applied Thermodynamics. The curves of Figs. 75, 76 and 77 show how the theoretical efficiency is increased as the pressures and superheats are changed. op 30 Jl 2U/ 11 40|i 280^ 40^ X gao^ ^ ' ^ oudens acuum ing 28^" M IIOK 0^ 7 ^ Theoretical Effic s s o **. v. - ' Non-Coud Busing _ _ ^ ^ 1 * 125* 150^ 175# 20 Gage Pressure 0" 74" 15" 22 4" 3 Inches Vacuum ) 50 100 150 20< Degrees of Superheat FIG. 75. FIG. 76. FIG. 77. FIG. 75. Effect of change of steam pressure on efficiency, complete ex- pansion and saturated steam. FIG. 76. Effect of change of back pressure on efficiency (150 Ib. steam 1 superheat), complete expansion. FIG. 77. Effect of change of superheat on efficiency (151 Ib. steam to 28.5 in vac.), complete expansion. These figures are all drawn for complete expansion and, al- though in all cases the effect of vacuum increase is greater than the effect of the same change in the initial pressure, the effect is not so marked as that shown in Fig. 76 when there is incom- plete expansion. These figures are all drawn from theoretical considerations. Although the effect of superheat is theoretically slight, its effect on the actual efficiency may be much more pronounced since it cuts down certain losses which occur in the cylinder and therefore its practical effect is very valuable. 12 178 HEAT ENGINEERING TESTS OF ENGINES To actually determine the steam consumption of an engine a test is made while the load is kept constant on the engine. If possible the exhaust steam is condensed in a surface condenser and weighed at regular intervals and at the same time indicator cards are taken and readings are made on the scales .of the Prony brake, or watt meter if the engine is used to drive a genera- tor, on the calorimeter to determine the quality of the steam, on the pressure gauge and barometer, on the temperature and pressure of the exhaust steam, on the temperature of the con- densate and on the revolution counter. From this data, as will be shown, the steam per horse -power hour, the pressure and quality of the steam supply, and the pressure of the exhaust may be found to be used for the actual thermal efficiency. CALORIMETERS The calorimeter used to determine the quality of the steam may be of three forms: the electric, the separating and the throttling calorimeters. The latter is shown in Fig. 78. The form here shown is the Barrus Universal Throttling Calorimeter, so called because by the addition of the drip pot C it may care for steam of any moisture content. In this a sample of steam is taken preferably from a vertical steam pipe A. In horizontal pipes it has been found that much of the moisture separates out and flows along the bottom of the pipe so that a fair sample cannot be obtained. The sampling tube B is a pipe closed at the end and containing a number of small holes around its circum- ference and along its length to take steam from various parts of the pipe. This steam is carried into the drip pot C, which is in reality a separator, where most of the moisture is taken from the steam. The height of the water in the drip pot is shown by the glass gauge D and the pressure is shown by the gauge E. Steam then passes over through a pipe and around a thermometer well containing the upper thermometer F and from here it passes through a small hole in the throttle plate G to the ther- mometer well containing the lower thermometer H and from here to the atmosphere. The mercury gauge / shows the pressure around the lower thermometer well. The throttle plate is held between flanges but insulated from them by some non-conducting material. THE STEAM ENGINE 179 To find the amount of water collected in the drip pot a string is tied around the glass gauge and when the water reaches that level the time is noted. The water in the drip pot is drawn off at any convenient time into a cup of cold water J so that the water level falls below the string. This is done by opening the valve K and putting the end of the drip pipe beneath the water level in the cup so that none of the hot water from the drip pot FIG. 78. Barrus universal calorimeter. can evaporate on reaching a place of atmospheric pressure. The cup is lowered after closing K and after catching all the drip its change of weight will give the weight of water drawn off. The time at which the water again reaches the level of the string is noted and the interval beween the two times of passing the mark will give the time taken for this water to be caught. This is reduced to pounds per minute or second. Call this m pounds. 180 HEAT ENGINEERING To find the weight of steam passed through the throttle hole in the same interval of time, this steam is exhausted into water and condensed or, what is quite customary, the hole is measured and Napier's formula is used. If the separator were perfect the quality of the steam coming from the pipe would be, since M is the weight of dry steam in M -f- in pounds of mixture. This drip pot, if perfect, would be a form of separat- ing calorimeter. This is not a perfect separator and so the quality of the steam leaving is determined by throttling the steam. The steam is throttled from the pressure shown by E to that shown by /. Throttling action means constant heat content. On expanding steam which is practically dry, it becomes superheated. ie = it q'e + x e r e = q'i + r { + fc p dt = i t (12) The thermometer H gives the number of degrees superheat when the saturation temperature corresponding to the pressure at I is found. Since the thermometer may register incorrectly due to the fact that the mercury column projects beyond the top of the thermometer well, the thermometer F is used to check this and determine the error. The thermometer wells are filled with heavy oil or with mercury and the thermometers are inter- changed after each reading so as to equalize errors in the ther- mometers. After this is done the temperature of saturation at the average pressure shown by E is found and the difference between this and the average reading of F is the error due to stem exposure. An error proportional to the amount of exposed mercury column is assumed and the average reading of H is corrected. From this the degrees of superheat are found, then THE STEAM ENGINE 181 ii and, after this, x e is found from equation (13). The weight of dry steam is then xM xM and the actual x is X = -^ '-. (14) M H- m Part of this X is due to radiation from the instrument and to find this correction the instrument is operated with steam slightly superheated or dry steam and the apparent quality, x dry , is found. is the correction to be applied for this instrumental condensa- tion. A common way to run this dry test, as it is called, is to attach the calorimeter to a boiler with banked fires and a shut stop valve in which case the steam in the steam space is dry. Thermometers are sometimes calibrated to read correctly when immersed to a certain depth and if such are used there is no need for the upper thermometer. Such thermometers are called constant immersion thermometers. Suppose the averages of a number of readings taken at five- minute intervals are as given below and the quality is desired. Pi = 112-lb. gauge Pz =2 in. Hg. Error in gauge = + 2 Ibs. Barometer =29.8 in. Ti = 335.20 F. T 2 = 265.2 F. Wt. of drip = 0.2 Ibs. per 10 min. Area of hole = 0.01 sq. in. Exposed column = 120 F. for TL = 100 F. for T 2 . Observed pi 112 Ibs. Error 2 Corrected pi 110 Ibs. Barometer = 14.64 Absolute pi 114.64 T l sat. 337.92 F. Ti observed 335.2 F. Error T l = 2.72 F. 182 HEAT ENGINEERING 100 Error T 2 = 2.72 X ~ = 2.27 F. Observed T 2 265.2 F. Corrected T 2 = 267.47 F. Observed p 2 = 2 in. Hg. = 0.98 Ibs. Barometer = 14.64 Absolute p 2 15.62 Ibs. Temp. sat. = 215. Q7 F. Corrected T 2 = 267. 47 F. Degrees of superheat 52 . 40 F. i for 15.62 Ibs. and 52.40 F. superheat = 1177 B.t.u (from Marks & Davis) q' for 114.64 Ibs. = 308.7 r for 114.64 Ibs. 880 308.7 + 880 x 1177 m = 0.2 Ibs. in 10 min. - ' M _ nun* 0.01 x 6Q x 1Q _ 9 83 0.986X9.83 9.83 + 02^ The object of the drip pot is to care for any large amount of moisture for, if the amount of moisture is over 4 per cent., the equation q' + XT = ii would give an ii so small that the steam could not be super- heated and, of course, the moisture in the low-pressure steam could not be known. Hence the upper x could not be found. The maximum amount of moisture possible with steam super- heated on the lower side of the orifice depends on the pressure but 4 per cent, is the usual amount for pressures used in practice. Wherever the thermometer / shows 212 F. the drip pot is needed as there is too much moisture present. The electric calorimeter uses an electric current to dry the steam and by calibrating the same the amount of power for each per cent, of moisture at a given pressure can be found and from this the quality for the actual power is determined. THE STEAM ENGINE 183 A condenser is not always applicable for the determination of the weight of steam supplied to an engine and in that case the engine is supplied from a boiler or group of boilers which have been blanked off from the others. In this way the weight of boiler feed gives the steam supplied to the engine if the weight of. water left in the boiler at all times is constant. This condition is approximated by keeping the level of the water in the boiler gauge constant but even in such a case the change of temperature or of the rapidity of firing may make this an uncertain guide. To eliminate the effect of this uncertainty such tests should extend over a considerable time, say 4 to 6 hours and in some cases 24 hours. If a condenser is used a test of an hour or less is sufficient for accurate results after the engine is brought to a uniform condition. ANALYSES OF TESTS To study the action of the cylinder walls two methods of analyzing these test results will be examined, one analytical, the other graphical. Hirn's Analysis. Hirn's analysis of the cylinder performance of steam engines consists in operating a test for a certain length of time and from the observations computing the actual perform- ance of the engine. To better illustrate this analysis a case will be computed. From the test the following average results are found : Time of test 60 min. Size of engine 10 X 15 Clearance . 10 per cent. No. of revolutions during test 14,400 Pounds of steam used 2356 Average weight of steam per card, o y 1440Q = 0.0816 Ibs. Average pressure at throttle 105 Ibs. Barometric pressure 14 . 5 Ibs. Average of quality of steam . 99 Average temperature of condensateT 120 F. Average temperature of water leaving condenser! 105 F. Average temperature of water entering condenser! 85 F. Weight of condensing water used 94,240 Ibs. Weight of water per pound of steam 40 Ibs. 184 HEAT ENGINEERING Average results from indicator cards Point of admission at 1 0.0 per cent. Point of cut-off at 2 25 per cent. Point of release at 3 90 per cent. Point of compression at 4 22 per cent. Abs. pressure at 1 48 Ibs. per sq. in. Abs. pressure at 2 107 Ibs. per sq. in. Abs. pressure at 3 37 Ibs. per sq. in. Abs. pressure at 4 16 Ibs. per sq. in. Work area during admission, Wa = 5-1-2-6 2.83 sq. in. Work area during expansion, Wb = 6-2-3-7 3.91 sq. in. Work area during exhaust, We = (3-10-8-7) - (9-4-10-8) -0.98 sq. in. Work area during compression, Wd = 5-1-4-9.. . 0.52 sq. in. 596 78 FIG. 79. Average indicator card from test for Hirn's analysis. The results above have all been averaged from the various readings and cards. The head end and crank end have been averaged together although at times they are worked up sepa- rately the weight of steam being divided between the two ends in proportion to the total volume at the point of cut-off although there is no true foundation for this method. Another method of division may be used as shown by Clayton. See page 210. The first point to compute in Hirn's analysis is the weight of clearance steam per card, M . At the point 4 steam is assumed to be dry. This has been indicated by experiments as far as they have been tried although these experiments were difficult THE STEAM ENGINE 185 to perform and are not so reliable as would be desired. With this assumption 7 4 = My 4 or M - Mo ~ v\ 7 4 = total volume at 4 v'\ = specific volume of dry steam at 4 (15) v" 4 is found from the steam tables. After this the weight of the working steam per card is found. M = weight 2 X No. of revolutions (16) Knowing M and M the quality of the steam at the various events of the cycle may be found if the pressures and volumes at these points are known from the cards. 7r M v'\ (M 7s (17) (18) *' " (M + MoK's (19) From the pressures the values of the heat of the liquid and the internal heat of vaporization may be found and from these the intrinsic energy at all points may be computed. AUi = M (q'i + Xipi) (20) AC/2 = (M + M )(q f 2 + XZPZ) (21) AUs = (M + Mo)(q' 3 + XSPS) (22) P4 ) (23) From the indicator cards the external works during the various periods may be found. Work of admission = W a = 5126 (24) Work of expansion = W b = 6237 (25) Work of exhaust = W c = - 94108 + 73108 (26) Work of compression = W d = - 5149 (27) The energy added to the engine from the outside during admission is Oi = M(q' + XT) (28) 186 HEAT ENGINEERING The energy removed is Q 2 = - M[q' + G(q' d - q' t ) (29) G weight of condensing water per pound of steam. M = weight of steam per card. q', x, and r are for conditions of steam at the throttle valve. q' = heat of liquid at temperature of condensed steam. q'd heat of liquid at temperature of discharge condensing water. q'i = heat of liquid at temperature of inlet condensing water. Now the heat added during any event is found by Q = A[U 2 - U, + W] (30) If the heat coming from the cylinder walls during the different events is represented by Q aj Qb, Q c or Q d , using the equation above, the various quantities are given by the equations Ql + Qa = AU 2 - AVi + AW a (31) Qa = - Qi + AU* - AUi + AW a (32) Q b = AU 3 - AU 2 + AW b (33) Q c = - Q 2 + AC/4 - AU, + AW C (34) Q d = AUi - AU* + AW d (35) If these result in positive quantities heat is given up by the walls while negative values mean that heat is given to the cylinder walls. If these are added together the net amount must be equal to the amount given or taken by the cylinder walls. If there is no source of heat in the cylinder walls they could not give heat so that the sum could not be positive. If negative this heat would finally melt the iron if abstracted for a considerable time. Since the walls do not change in temperature this negative sum must equal the heat radiated from the cylinder and Qa + Qb + Qc + Q d = Qr (36) A check equation for Q r may be determined by blocking the engine, filling the cylinder with boiler steam and measuring the condensation. Then Q r = M r (q' + xr- q' ) (37) If there is a steam jacket there could be a positive sum as heat could then be given up by the walls from the heat in the jacket. In this case / Q a + Q b + Qc + Qd + Qi = Qr_ (38) THE STEAM ENGINE 187 and Qf is determined by finding the weight of steam condensed in the jacket while the engine is running. M 3 - is the condensa- tion due to the heat given to the steam within and i = My (53 B.t.u. TEMPERATURE ENTROPY ANALYSIS The graphical temperature entropy analysis is to be used next and in this the same test is made but the only observations necessary are: Size of engine 10 in. X 15 in. Average pressure 119. 5 Ibs. per sq. in. abs. Average quality . 99 THE STEAM ENGINE 189 Weight of steam 2356 Ibs. Revolutions 14,400 Clearance 10 per cent. Average compression 22 per cent. Average indicator card Fig. 81 This method has been developed by a number of persons. Prof. Boulvin's method has been combined with that of Reeves in reducing the method given below. The average indicator card is constructed by averaging the pressures at ten or more equidistant points along the card or a Temperature in O Cu. Ft. Volume in Cu. Ft. Deg. F FIG. 80. L Temperature entropy diagram for analysis of engine performance. card may be selected on which the m.e.p. is equal to the average m.e.p. of the test. M and M are then found. M = ^- = 0.0089 as tefore. V 4 M Wt 2 X rev. = 0.0816 Ibs. as before. The chart for the Temperature Entropy analysis is now to be drawn. This consists of four quadrants, Fig. 80, a p-v quadrant having a line for 1 Ib. of saturated steam, a T-p quadrant having the line of saturated steam, a T-s quadrant having the liquid line for 1 Ib. of water and the saturation line for 1 Ib. of steam 190 HEAT ENGINEERING and a V-s quadrant which is used for transferring the values of x. This figure is constructed by aid of the steam tables and takes the form shown in Fig. 80. To transfer the indicator card to this figure, a number of points on the mean diagram, Fig. 81, are taken and the heights from the line of absolute zero pressure and lengths from absolute zero of volume are measured and tabulated. The lengths are then multiplied by the scale of volume of Fig. 81 and divided by the product of the scale of Fig. 80 and the value of M + M in order to get the distance to each point if 1 Ib. of steam were present on the expansion line. The heights are multiplied by the scale of 81 and divided by the scale of 80. Thus, in the 2 35 FIG. 81. Average indicator card from test arranged for T-S analysis. problem considered there is a 4-in. card for a 10 X 15 engine with a 40-lb. spring and . M M + Mo = "b.0816 + 0.0089 = 0.0905 Fig. 80 has scales of 2.5 cu. ft. to the inch and 20 Ibs. per square inch to the inch. The cylinder has a displacement of 0.682 cu. ft. with a 4-in. card. The lengths are therefore multiplied by "IT X 00905 X 2^ = ' 754 in order to get the length to lay off in Fig. 80 so as to represent the card for 1 Ib. of steam on the expansion line. The multi- plier for height is merely 40/20 = 2.00. These distances are tabulated to aid in the work as shown. THE STEAM ENGINE 191 Original Analysis Original Analysis card card card card 00 OQ IS i 3 g . 8 s.aag 1.2 S.aa| "5 1 id i*si 5d I*j | P "S fl 2 i fa S' S "| fa ! 2.85 0.40 5.70 0.30 11 0.40 4.4 0.80 3.31 2 2.82 0.90 5.62 0.68 12 0.40 3.4 0.80 2.56 3 2.78 1.23 5.56 0.93 13 0.40 2.4 0.80 1.81 4 2.64 1.40 5.28 1.05 14 0.40 1.2 0.80 0.93 5 2.01 1.90 4.02 1.43 15 0.51 0.9 1.02 0.68 6 1.56 2.40 3.12 1.81 16 0.80 0.6 1.60 0.45 7 1.28 2.90 2.56 2.18 17 1.18 0.4 2.36 0.31 8 1.10 3.40 2.20 2.56 18 1.50 0.4 3.00 0.31 9 0.93 4.00 1.86 3.00 19 2.00 0.4 4.00 0.31 10 0.70 4.20 1.40 3.16 20 2.50 0.4 5.00 0.31 FIG. 82. Construction of T-S diagram from p-v diagram. This method considers only 1 Ib. of steam to be present on the expansion line. The p-v diagram is now laid out from the table giving the fig- ure shown. This is then transferred from the p-v to the T-s quadrant by drawing a constant pressure line until it strikes the 192 HEAT ENGINEERING saturation line in the T-p quadrant and this fixes the tempera- ture of the point A. A vertical line from this temperature fixes a line in the T-s quadrant on which the point will lie. It would be at a if it had a quality of zero and at 6 if of quality 1, and in the p-v quadrant it would be at c if of zero quality and at d if of unit quality. Now v - v' = xv" (41) and s - s' = Y (42) In other words the volume change and the entropy change from liquid to steam depends on x. Hence the point A' in the T-s quadrant must divide a-b in the same way that A in the p-v quadrant bisects c-d. To construct such a point the point of intersection e of a horizontal from b and a vertical from d is joined to the intersection / of a horizontal from a and a vertical from c. The line ef is then the quality transfer line for if a vertical is drawn from A to this line intersecting it in h and then a horizontal is drawn from h to ab, this will fix the point A'. The same pressure line and construction transfers B to B'. If this is carried on in the same manner for all points a figure such as shown in the T-s quadrant, or enlarged in Fig. 83, is found in the T-s quadrant corresponding to the p-v diagram. The scales to which the p-v quadrant has been drawn were 1 in. = 2.5 cu. ft. 1 in. = 20 Ibs. per sq. in. The T-s quadrant was drawn with 1 in. = 0.25 units of entropy 1 in. = 25 F. The area scales are: 2.5 X 20 X 144 1 sq. in. = __ Q = 9.25 B.t.u. (p-v) /7o 1 sq. in. = 25 X 0.25 = 6.25 B.t.u. (T-s) The area of the indicator card on the p-v plane was 7.94 sq. in. and on the T-s plane the area was 11.76 sq. in. These each reduce to 73.5 B.t.u. for the area. The T-s plane for Fig. 83 has been turned through 90. This is the conventional T-s diagram for the indicator card. THE STEAM ENGINE 193 It is now necessary to add other lines to the T-s diagram in order to interpret it. Letters have been added to the diagram at critical points of Fig. 83. The diagram is for 1 Ib. of total steam and this is the amount present on be only hence the figure is true only for this short line. From the point of admission / to b there is a changing weight of steam and from c to e there is a changing weight. The points on these lines are only drawn according to the actual volumes which are proportional to en- tropies because each is proportional to MX. To study the line ef on which . i~ pounds are present, the line corresponding to zero entropy change must be drawn. If AC be drawn so that EC M 0.0816 9Q E B 0.0905 200 F, FIG. 83. T-S diagram of indicator card. then line AC would be the liquid line for the boiler steam only, while the distance CB or the distance between AC and A 'B repre- sents entropy of the liquid for the clearance steam. The distance eF represents the entropy of the clearance steam at the beginning of compression and GH is laid off at this distance from CFA. This line would be the line of compression if the 13 194 HEAT ENGINEERING clearance steam did not change in entropy because the entropy of the liquid CB for this steam plus the entropy of vaporization BH is constant. GH, although not an adiabatic on this figure, is the line of compression if the clearance steam were com- pressed without giving up heat. The line of compression is seen to pass to the left of the line GH which means that heat is taken up by the cylinder walls. The amount of heat is found by drawing fg parallel to GH to the back pressure line ed and then dropping the perpendiculars to h and i at absolute zero of temperature getting the area hgfei as the heat removed in com- pression. To find the heat lost during admission from / to b it is necessary to find the position which should have been occu- pied by the steam entering from the boiler if no heat had been taken from the clearance steam or the steam entering the en- gine. It is necessary to mark the line corresponding to the pres- sure at the throttle valve and draw this on the diagram. This will be the line ECBH. On this line EC is the entropy of the liquid of the boiler steam, CH is the entropy of the clearance steam and in addition to this there must be HJ for the vaporiza- tion of the steam from the boiler. If x is the quality of this steam the distance HJ is equal to (44) ~ M + M t T Since BI is the value of for 1 Ib. of total steam, BI = BH + #7 The point which should have been at / is now at b and the area nJlk is called the loss due to initial condensation. Since the reference line to care for other losses is shifted to gM the point J is moved to K and the loss due to initial condensation is nKtk. The area mnba is called the loss due to wiredraw- ing. There is still a further loss fMmaf which occurs during the early part of admission. The area prbk is the loss to the cylinder walls after cut-off and area prc'o' represents a gain from the cylinder walls, and occ'o' represents a loss during the last part of expansion. ksc'co brs = net gain from cylinder walls during expansion. iedco = heat removed during exhaust and dcu = loss from incomplete expansion. THE STEAM ENGINE 195 Now the line GH is a liquid line for the cylinder feed and hence ieHJl = the heat supplied with the cylinder feed for one pound of total steam, above the temperature of the exhaust. The work is efabcd Hence ieHJl + (ksc'co srb) dbcdef iedco hgfei = heat removed during admission. This will be found to be equal to the area of fMKtkbaf if gf is continued to M and JK is made equal to ME. The various losses are given as follows (numerical values should check with Hirn's analysis) : Area Sq. in. Per cent, of work on T-s B.t.u. Heat supplied above exhaust = ieHJl or hgMKt . . Work = efabcd . 144.98 11.76 0.71 0.30 44.32 2.80 1.64 8.43 4.74 1232.00 100.00 6.04 2.55 376.20 23.80 13.95 717.50 40.40 905.00 73.50 4.45 1.88 277.00 17.50 10.30 527 . 00 29.60 Loss during first admission = fMma Loss during throttling = amnb Loss due to condensation = nKtk Heat added from walls during expansion = ksc'co brs Loss due to free expansion = dcu Heat necessarily removed during exhaust = ieuo . . Heat removed during compression = hgfi The quality of the steam at cut-off is given on Fig. 83 by the ratio vb 28.8 n r x = - - = TT-X = 0.645 wv 44.6 (45) and this can be found in the same manner for any point of the expansion curve. For the compression curve x may be found by drawing from e a line eO so that its distance from the line AC is M i T/T times the distance from A'B to IN. Thus M + Mo BO = BI M t If the ratio of the distance from AC to the compression line to the distance from AC to eO at the same level be found this ratio is x. AC, A'B and eO make an entropy diagram for the compression steam with a bent axis AC. As before mentioned the only true lines from which x can 196 HEAT ENGINEERING be found are be and ef. The other lines are obtained by following a scheme in which the product MX at any point could be found but not either of the terms of the product. Hence one cannot tell what losses occur at various points on this line. There should be an agreement by the two analyses. MISSING QUANTITY AND INITIAL CONDENSATION Steam of quality 0.99 was admitted into a cylinder and after entering the cylinder it was found that the quality at cut-off has been reduced to 0.633. In other words 37 per cent, of the mix- ture is moisture. This has been caused by the action of the cylinder walls. Fig. 84 shows the actual form taken by the indicator card of an engine. The events in most cases take place slowly giving rounded corners; there is a drop due to throttling on the steam line and release occurs before the end of the stroke. If the weight of dry steam shown by the cards, Fig. 84, is found by the formulae be- low and this is subtracted FIG. 84.-Actual indicator card from frOm * he am Unt f steam engine. actually supplied, the differ- ence is called the "missing quantity." It is expressed as a percentage of the steam actually supplied. It is caused by initial condensation of the steam. ^ = ' x - M a apparent weight from diagram. F = area of piston in square feet. L = length of stroke in feet. D = length of card in inches. I = per cent, clearance from Fig. 84. v" = specific volume of dry steam. i, 2 = distances from clearance lines to points in inches. If M equals the real weight taken per card, M-M a M m ~M ~~M~ = m " ( i"> missm g quantity. THE STEAM ENGINE 197 This is almost equal to 1 #2 from either of the previous analyses. This initial condensation is due to the effect of the cylinder walls. As the steam in the cylinder drops in pressure its tem- perature changes and the moisture on the walls of the cylinder from condensation is evaporated removing heat from the metal and moisture and cooling them. During the exhaust this action continues, the walls being so cooled that when fresh steam enters the cylinder some of the steam is condensed and settles on the walls of the cylinder. This enables the walls to absorb heat from the steam at a faster rate than it would were dry vapor in contact with the metal. This condensation produces the miss- ing quantity and the presence of a film on the walls of the cylinder seems to make it easier to carry heat to the walls during admis- sion while during the exhaust the possibility of evaporating this water means that during this time much heat is given up by the walls and discharged with the exhaust steam when it can be of no value for driving the engine. If superheated steam be furnished, it will give up its heat to the cylinder walls but the absence of the moisture film makes this action slower and if the superheat is sufficient the necessary heat may be abstracted before the steam is reduced to the saturated condition or before much condensation occurs. At release the absence of considerable moisture in the cylinder prevents the removal of much heat during the exhaust. Thus the superheated steam prevents the excessive abstraction of useful heat by the metal during admission and its restoration when it cannot be used. It is for this reason that superheated steam is such a valuable medium to use. This accomplishes a greater increase in efficiency than that which is shown by theo- retical considerations. To determine the amount of steam used by an engine it will be necessary to know this missing quantity. Then M a M = z 1 m.q. To find the value of m.q. theoretical and empirical formulae have been proposed. It will be evident that the amount of condensation per stroke must depend on the surface exposed to the steam at cut-off, the temperature range and the time during which this surface 198 HEAT ENGINEERING is exposed to the steam or inversely on the number of times it is exposed to the action of the steam per minute. There have been several formulae proposed. Some are quite complex taking into account many variables. The more complex, al- though they may be correct in theory, are so changed by slight changes in conditions that the possibility of error is as great as in the less complicated formulae. Some of the formulae pro- posed are as follows : m.q. 217 (r - 0.7) , (Perry) (47) m.q. 15(1 + r - 100 log r ~~ (Cotterill) (48) (Thurston) (49) 1 - m.q. m -1- ~ inO-4-7 (Callendar and Nicolson) (50) 100 m.q. = 1 '- (Rice) (51) Pressure a /8 a 60 0.568 0.412 133 80 0.517 0.384 106 100 0.466 0.359 87 120 0.414 0.333 78 140 0.363 0.306 75 In the above formulae, r = ratio of expansion, the volume at end of stroke divided by the volume at cut-off. d = diameter in inches. pi = absolute steam pressure in Ibs. per sq. in. N = r.p.m. I = percentage clearance. m.q. = c[T b - Teldcos- 1 - + [T b - T e ] * (127c + 0.055)s (Marks) (52) c = 0.02. b = abs. temp, at cut-off. THE STEAM ENGINE 199 T e = abs. temp, during exhaust. d = diam. in feet. e = fraction of volume at cut-off. s = stroke in feet. b = fraction of volume at compression. 7 /TT i \ /r*o\ i r-l no IT" I I ^ X I ~" I XAv?L/lvy \OOy m.q. = missing quantity for small compression or (1 x) at cut-off for large compression. N = r.p.m. s = nominal cylinder surface divided by volume, each in feet, '> -\L + d)- T = temperature range from limiting pressures on curve Fig. 85. (Ti-T*) p = abs. pressure at cut-off in Ibs. per sq. in. e = ratio of total volume at cut-off to volume swept out by piston. Absolute Pressure of Top and Bottom Lines of Indicator Cards in Pounds o 8 g ^ i g / 1 I / / / / / / 50 100 150 200 250 300 350 400 45C Values of T FIG. 85. Heck's value of T for different pressure. The basis for the form of these formulae may be given as follows : The surface exposed to the steam at cut-off in square feet is d = diam. in feet. 200 HEAT ENGINEERING L = stroke in feet. - = point of cut-off. F p = surface of passages. The time during which this is exposed is k M. minutes N where k = a fraction less than 1 N = r.p.m. The temperature range for this surface is some function of T. It is not equal to the actual difference in the temperatures at cut-off and compression as the range in temperature in the metal is much less than this. The quantity of steam necessary to give the heat per stroke required by the cylinder walls will be proportional to the surface, time and temperature range, or M. = VF,f(T) (55) To express this as a fraction, m.q., of the steam shown by the indicator card it must be divided by the steam shown by the card. This steam is given by 7* L 4 M a = - r. --- compression steam. (56) V 2 k" Now the specific volume v" 2 = approximately since the curve of saturated steam on the p-v plane is nearly a rectangular hyperbola. (See Fig. 96.) M a = /b'" T TD Hence M 8 Now , 2 C may be called s. F c is practically equal to 2 - f- *-i T! irdL, so that ' s = Y + ^ L a THE STEAM ENGINE 201 This is a function of -3 giving ' m.q. = fc" 5j/(D or ff p/(D (58) If r is written as -, e, being the relative total volume at cut-off 6 this reduces to By examining a number of tests of all sorts of engines and with different speeds, pressures and cut-offs for the same engine, investigators have been led to the empirical forms shown above. What in the simple theory appears as the first power has in general been changed to a fractional power. In all of the simpler formulae the form agrees in the positions of the terms with the k simple theoretical form shown. In place of -^ being used for the time term 7= or -77= has been used. Heck assumes that the VN last two terms of F c are equal to irdl, thus making F c equal to the inside surface of the piston displacement, From evidence shown by those who have derived these for- mulae it appears that the formula of Heck 0.27 Mr m - q - = " w V^ (57) is a good one to use. It is applicable to non-jacketed engines when supplied with dry steam. For this reason it cannot be used when the supply is superheated steam nor for the lower cylinders of multiple expansion engines when the steam supply is quite wet. If, however, this steam is dried by a separator or by reheater coils before entering the other cylinder, it may be used. For jacketed engines it might be used by deducting the surfaces heated by the jacket from the area F c in finding this or the quantity s. This formula shows that the proportional amount of condensed steam varies inversely as the square root of the cut-off getting less as the cut-off becomes greater. This must not be confused 202 HEAT ENGINEERING with the actual amount of steam condensed which will not vary very much as the cut-off changes. The large part of the conden- sation takes place when the steam first enters the cylinder and as the piston moves along the additional amount is not great. The surfaces which are most important in the condensation of steam are the cylinder head and piston. For this reason greater gain is to be expected from the jacketing of the heads than from the jacketing of the barrel of the cylinder. It might pay to supply the hollow piston with steam, making a jacket of the cored spaces. The value of s decreases as the size of the cylinder in- creases giving less condensation in large cylinders than that found in small ones. This is to be applied to the card below, Fig. 86, which is the assumed card in the design of a 20-in. X 24-in. engine to run at 80 r.p.m. The clearance is 7 per cent.; cut-off is at 25 per cent. Pressure at cut-off is 125 Ibs. abs. and back pressure 17 Ibs. abs. 20 = 1+2.4=3.4 12 = 24" 12 L25 = 360 \7 = 217 t T = 360 - 217 = 143 e = 0.25 + 0.07 = 0.32 0.27 From curve, Fig. 85. m ' q ' -- /3.4 \125 X143 X0.32 = 0.219 FIG. 86. Card for proposed 20 X 40 engine. Amount of steam indicated at cut-off is given by = (0.25 + 0.07) |^X 0.32 X 4.36 3.581 144 0.390 Ibs. 'l25 THE STEAM ENGINE 203 c\ QQ N/ A Qi Steam at compression = -- 00 ' - = 0.0615 Zo.oo Indicated steam per hour = (0.390 - 0.0615) X 80 X 2 X 60 = 3160 Ibs. S160 Probable steam per hour ^ pr^TFrk = 4000 Ibs. i u.^iy Mean height = 1.41 in. m.e.p. = 56.4 56.4 X X 314 X 80 3300 Probable steam consumption, H ' p - = 33000 - X 2 = 172 4000 Steam cons. = - = 23.2 This result is low so that the missing quantity will be com- puted by some of the other formulae. By Perry's formula (48) 1 - m.q. 20 X V80 m.q. = 0.271 By Thurston's formula 30* '^ -- = 0.37 1 - m.q. 20 X A/80 m.q. = 0.237 By Cotterill's formula 100 log = 0.293 = 1 - m.q. 20 X V80 m.q. = 0.226 The result of 23.2 Ibs. per hour per horse-power was low for this type of engine so that Heck's formula would not give as good results as Thurston's, Perry's and Cotterill's. If 0.30 is taken for the value of mq. The steam consumption is equal to 204 HEAT ENGINEERING On account of valve leakage this result would be increased slightly. EXPERIMENTS OF EFFECT OF CYLINDER WALLS This influence of the cylinder walls has been experimentally studied by a number of investigators. Callendar and Nicolson, in their paper presented in the " Proceedings of the Institution of Civil Engineers of Great Britain," Vol. cxxxi, gave results obtained by them on a 10^ X 12 Robb engine in 1895 with a flat slide valve behind a pressure plate. The piston displace- ment was 0.601 cu. ft. and the clearance volume was 0.060 cu. ft. The engine was made single-acting so as to study the action in a better manner. To find the temperature of the metal in the cylinder head eight holes were drilled at regular intervals at 1^2 in. from the center of the head but extending to different depths. The thicknesses of metal remaining at the bottom of seven of these holes were 0.01 in., 0.02 in., 0.04 in., 0.08 in., 0.16 in., 0.32 in. and 0.64 in. Along the length of the cylinder at the end of stroke and at 4, 6 and 12 in. from it a pair of holes were drilled in the side, one to within 0.04 in. of inner surface and the other % m - At 2, 8, 10, 14 and 16 in. from the end single holes were drilled leaving ^ m - a ^ bottom. There were also four holes at the middle of the stroke distributed around the circumference of the cylinder extending to within J/ in. of the inner surface and finally three vertical holes two inches deep were drilled along the side of the barrel at 1 in., 7J/ in. and 15 in. for the use of mercurial thermometers or platinum resistance ther- mometers, while in the other holes thermocouples were used. The thermocouples were made by soldering wrought-iron wires at the bottom of the holes. To make the cold junction, the same kind of wire was attached to cast-iron blocks cast from the same ladle as that from which the head was cast. These were immersed in a paraffine bath at 212 F. The potential was measured by a very sensitive galvanometer by balancing the potential on a potential wire. The formula for the voltage in microvolts produced by this couple was found to be given by E = 1692 - 17.86* + 0.0094Z 2 (58) if the cold junction was at 100 C. and the hot junction at ZC. Later one of the couples in the cylinder was used as the cold THE STEAM ENGINE 205 junction and the drop of temperature between these two was measured. This method was somewhat similar to that used by Prof. E. Hall of Harvard (Trans. A. I. E. E., 1891) but his results were not very extensive. To find the temperature of steam the authors used a platinum resistance thermometer in the cylinder 3 in. from the face of the piston and also one in a small %-in. hole in the center of the piston head. To get the temperature at a definite point in the stroke by any couple or platinum thermometer a pair of revolving brushes were attached to the shaft. One brush made contact with a central copper tube and the other with a sector mounted on a circular disc. The sector was one-thirtieth of a circumference in length. The disc could be rotated and by a scale and vernier the position of the crank for any observation could be read. A number of sectors on the disc reduced the amount of motion neces- 36 39 40 FIG. 87. Indicator card marked with points at various sixtieths of a revolu- tion. From 20 X 40 engine. sary. To facilitate the change of circuit to different couples, mercury cups were used. To show the results of these tests, Fig. 88 has been constructed from the card, Fig. 87 by finding the saturation temperature for pressures of the steam at points corresponding to definite crank angles. This temperature from the steam tables is plotted to sixtieths of a revolution, giving the curve. The marks X give points similar to the results shown by the platinum thermometer while the points marked show the temperatures of the ther- mometer in the steam space in the cylinder head. The curves illustrating the variation of temperature of the metal at J 5 in. from the inside surface in the head and that at holes in the side at 4 in. from end are shown to a larger scale above the card. These curves are ideal and are drawn to indicate the results of Callendar and Nicolson. 206 HEAT ENGINEERING The surprising results of the actual tests by Callendar and Nicolson were that at Jf oo in. from the inside surface the variation of temperature was only 4.3 F. at 100 r.p.m., while at ^5 in. from inside the variation was 6 F. at 46 r.p.m. and 4 F. at 73.4 r.p.m. At J<2 in. from the inside the change in temperature due to cyclic variation was practically zero. Area cbd + ale = 1.89 Sq. In. = Temp.byPt.Ih.ermometei Attached to Piston 03 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 Sixtieth of One Revolution I'-loJj Rev. Fid. 88. Diagram of temperature at various positions of piston for dif- ferent movement of crank. Temperatures taken in head, cylinder wall and steam space. (After Callendar and Nicolson.) Along the length of the barrel the first hole situated in the clearance space and exposed to the same steam action as the heads showed 13.5 F. variation at ^5 in. from inside at 44 r.p.m. when the heads at this thickness showed 4.9 F. variation. This amount of 13.5 F. at J^ 5 in. from the surface probably corre- sponds to about 20 F. variation at the surface and this particular result was the largest obtained by the experimenters. The THE STEAM ENGINE 207 variation at 4 in. was found to be 5 F. and at 6 in., 3^ F. It was found that these temperatures begin to rise before the piston reaches these points showing that there was conduction along the barrel and also the steam could leak around the piston as far as the rings. A curious result near the middle of the stroke is the temperature that is shown on the inner surface which is lower than that some distance farther out. This shows that there is a flow of heat from the metal to the steam at this point. The experi- ments seem to show a gradient of 0.55 F. per inch in the head while that on the side wall of the barrel was a variable quantity in an axial direction being greatest at the center of the stroke where it was 9.3 F. per inch. Experiments were made to determine the conductivity of cast iron giving 5.5 B.t.u. per square feet per degree per hour for 1 in. thickness and then the diffusivity, which is the ratio of this conductivity to the thermal capacity of the same amount of metal, was computed. The results of these experiments showed that the effect of moisture in the steam is to increase the condensation while superheat decreases it. The percentage amount of condensa- tion varies inversely as the ratio of expansion making the actual amount of condensation practically constant. The effect of initial pressure is complex so that an increase does not mean an increase in condensation and the same may be said of the change in back pressure. Early compression means a shorter time for cooling and hence decreases the condensation. Jacket- ing reduces the surface which causes condensation and for that reason reduces the amount of condensation. To find the maximum condensation possible, the limiting amount as they called it, the average height of the temperature degree diagram, Fig. 88, is found and when the area above this line is expressed in degrees and sixtieths of a cycle it will give the thermal units per hour per square foot when multiplied by 45. The surface considered for 20 per cent, cut-off on this en- gine which was a Corliss engine was 8.9 sq. ft. on each end. 45 X Area (cbd + abc) X scale X surface of clearance = 45 X 1.89 X 10 X 50 X 8.9 X 2 = Total B.t.u. of condensation per hour = 758,000 (59) This quantity is then multiplied by 1 + VN VN or (60) 208 HEAT ENGINEERING to make it correct for the speed of N revolutions per minute for cut-off or release respectively; while multiplying by 1.4 gives the result for a double-acting cylinder. Thus, from the Fig. 88, the following results for a 20 in. X 42 in. engine at 80 r.p.m. : 758,000 X - ' y X 1.4 = 884,000 B.t.u. per hr. 3 "fr \/80 If the pressure at entrance from the card were 100 Ibs. absolute for which r = 887.6, the steam condensed per hour will be This engine uses 4000 Ibs. of steam per hour so that the missing quantity is 0.25. The actual condensation may be much less than this and, if the mean temperature of the cylinder wall is known, the area above this line to any event gives the actual condensation when handled in the manner shown. The net area to a point beyond the crossing of the mean line gives the difference between the condensation and the reevaporation. The multiplier is different for cut-off and release. VALVE LEAKAGE Callendar and Nicolson found that in their engine much of the apparent missing quantity was due to leakage of steam beneath the valve into the exhaust and this steam never entered the cylinder. The valve leakage of steam was found to depend on the periphery of the valve, the lap and the pressure difference or W = KI(P > - **> (61) s when pi and p 2 were the mean pressures on the two sides in pounds per square inch, s was the lap in inches and I was the periphery in inches. The value of this leakage term was undoubtedly large in the experiments under review but in many engines it may be a small quantity. In the engine used K was equal to 0.02. The platinum thermometer projecting into the steam of the cylinder 3 in. from the piston face indicated (Fig. 88) that THE STEAM ENGINE 209 ICO Ca 120 C. Piston Travel the steam was practically saturated except during compres- sion when it was superheated, while that in the %-in. hole in the piston head was highly superheated during all of the stroke except at cut-off. This was undoubtedly due to the superheating effect of the metal when the pressure was low causing the steam to be superheated still further during compression. George Duchesne in the Revue de Mechanique 1 gives results of an investigation with a hyperthermom- eter which consisted of a multiple silver- platinum couple 0.002 mm. thick. This couple responded rapidly to the changes of temperature. He finds the cycle of Fig. 89 |ioo for the wall, steam and the saturation tern- | so perature of the steam in his engine on a I eo temperature-piston travel diagram. In 40 this it is noted that the wall temperature 20 is usually higher than the steam and that the steam is saturated except during com- pression when it is superheated, becoming ^ Fl ^' 89> R J sults of r , Duchesne on tempera- saturated at the point where it reaches the ture of wall and steam. wall temperature. Of course this metal cycle is true for one point in the cylinder but for another point the cycle would probably be lower. Duchesne states that the compression should not be carried higher than the temperature of the metal of the head for, if this is done, a loop is found on the card, Fig. 90. That this loop is due to excessive condensation was proven by the fact that, when air was used with this amount of compression, the loop was not shown, compression being carried up to a higher pressure. The loop may not be found in high-speed engines when the piston speed is faster than the speed of condensation. This compression, however, does not affect the steam consumption very greatly as has been shown by Prof. John Barr 2 in 1895 and by E. Heinrich 3 in 1913. In these experiments there was a slight change in the steam per indicated horse-power hour 1 Revue de Mechanique, 1897, pp. 925 and 1236. Power, June 28, 1910; May, 1911. 2 Trans. A. S. M. E., 1895, p. 430. 3 Zeit. des Verein Deutscher Ing., Vol. 58, Jan. 3, 1914. 14 FIG. 90. Duchesne 210 HEAT ENGINEERING Pressure orTemperature " Iteam vs Stroke Temperature of Metal vs Stroke FIG. 91. Adams results. with change in compression but this was very slight. Heinrich found that by adding plates and increasing the clearance surface, although the volume remained the same, the water rate was increased, showing the harmful effect of sur- face. In 1895, at Cornell University, Mr. E. T. Adams used a couple in the cylinder wall near the inside surface (Koo in.) and attached it to a galvanometer of short period. He photo- graphed the beam of light from the mirror of the galvanometer on a sensitive paper moved by the piston. This result is shown in Fig. 91, which shows a rapid drop at the opening of re- lease probably due to the evaporation of water from the surface and, as the heat flowed in from the outer metal of the wall, this temperature rose again. STEAM CONSUMPTION BY CLAYTON'S METHOD J. Paul Clayton 1 has studied the results of actual tests of engines and has shown that the expansion lines of steam engines and, in fact, the expansion lines from any engine using an elastic medium are polytropic curves of the form pv n = K unless there is some peculiarity such as leakage or some other disturbance which is not uniform. He then investigated the values of n from different en- gines, finding the value of n by plot- ting a logarithmic diagram of pres- sure and volume as shown for the cards of Fig. 92 taken from a 20 in. X 42 in. engine at 78 r.p.m. with 4 per cent, clearance. The table below shows the method of con- structing the diagram Fig. 93. If the clearance is properly measured and the correct scale of the spring is used, the ex- pansion lines become straight lines on the logarithmic diagram and the slopes of these lines are the values of n since log pi - log p z log v 2 - log vi 1 Trans. A. S. M. E., 34, p. 17. Bulletin 58, Engineering Experiment Station, Univ. of Illinois. Bulletin 65. Spring FIG. 92. Card from engine for Clayton analysis. n = (62) THE STEAM ENGINE 211 The variation of this line from a straight line, as shown in the figure, would indicate leakage. Point Volume in fifths inches from clearance line Logarithm of volume in fifths inches + 1 Pressures in fifths inches from absolute zero Logarithm of pressure in fifths inches + 1 1 0.70 0.846 8.65 .936 2 2.20 1.340 8.40 .924 3 3.88 1.590 8.30 .919 4 5.50 1.740 8.28 .918 5 6.25 1.795 7.97 .902 6 7.47 1.873 6.50 .813 7 9.35 1.970 5.20 .716 8 12.80 2.106 3.80 .579 9 15.85 2.200 3.15 .498 10 18.92 2.276 2.05 .424 11 17.80 2.250 2.18 .338 12 15.20 2.182 1.88 .274 13 12.55 2.099 1.85 .267 14 9.50 1.978 1.80 .255 15 5.70 1.756 1.80 .255 16 3.52 1.546 1.75 .243 17 2.80 1.447 2.00 .301 18 1.75 1.240 3.15 1.498 19 1.25 1.096 4.35 1.638 20 0.70 0.846 7.20 1.864 2.00 1.75 1.25 1.00 0.75 3 4 1.25 1.50 1.75 Logarithm of Volume 2.00 2.25 2.50 FIG. 93. Logarithmic diagram of steam engine indicator card. In preparing the table, distances to the true zero of pressure and of volume have been found and then the distances from these lines to any point have been measured in fifths of an inch. The logarithms of the numbers have been increased by unity to have 212 HEAT ENGINEERING all of the numbers positive for plotting. This only shifts the lines but does not change the slopes. The logarithms are then plotted giving the figure which also shows the true point of closing or opening of the valve as the point at which the straight line begins marks the beginning or ending of the expansion or compression. This is of value as the events, especially compression are difficult to fix accurately on the indi- cator card. On investigating tests on which the quality of the steam in the cylinder could be determined from data and on making such tests on an engine in the laboratory where the quality of steam 0.85 0.95 1.15 1.25 50 70 90 110 Initial Steam Pressure 130 150 FIG. 94. Clayton's relations between n, x, and p for stationary engines. at cut-off could be varied by using wet steam or superheated steam, Clayton showed that there was a variation of the value of n of the expansion line and the quality of steam at cut-off. The value of n varied from 0.70 to 1.34 while the quality at cut- off varied from about 0.4 to 0.9. He showed that there was a relation between them. In steam engines the value of n of the expansion line could be told from the quality at cut-off or, con- versely, knowing n the quality at cut-off could be found. This relation changes with the steam pressure and with the speed but does not vary with the point of cut-off. Since in stationary en- gines the variation with the speeds in use is slight, the only varia- tion considered is the variation with pressure and Fig. 94 has been prepared from a similar figure due to Clayton giving the values THE STEAM ENGINE 213 of x and n at different pressures. For locomotive work, where the pressures do not vary much, Clayton gives a figure similar to Fig. 95 showing the values of x and n at different speeds. The results have been obtained from non- jacketed engines exhausting near the atmospheric pressure and hence the results should only be applied to such. The method, if used with jackets or with high back pressure can only be considered an approximation. The indicator must be connected directly to the cylinder without the use of piping and a correct reducing motion with no stretch nor lost motion must be employed. 0,85 1.25 100 150 200 250 300 R.P.M. FIG. 95. Clayton's relations between N, n, and x for locomotive engines. On actually applying this method, a result close to the correct steam consumption may be found by assuming dry steam at compression. The error may be 4 per cent. The results for Fig. 92 will be computed. Value of n for expansion curve, Fig. 93, = 0.98 Steam pressure, 108 Ibs. abs. x, from Fig. 94, for n = 0.98 and p = 108, is 0.65 Pressure at cut-off, 93 Ibs. abs. Length to cut-off, 1.22 in.; to comp., 0.73 in.; of card, 3.14 in. Total volume at cut-off = TTT, X 2.96 cu.ft. 3.14 ' 1728 Weight of dry steam at 93 Ibs. = 0.2112 Ibs. per cu. ft. w . , ^ , 4 a 2.96X78X60X0.2112 Weight of steam per hour at cut-off = - 0.65 = 4470 Ibs. 214 HEAT ENGINEERING 0.73 314 X 42 Volume at compression = ^rr X 1f70Q =1.76 Pressure at compression = 24 Ibs. abs. Weight of clearance steam per hour = 1.76 X 78 X 60 X 0.0591 = 486 Steam supplied = 4470 - 486 = 3984 Ibs. per hour H.p. = 160 h.p. Steam per i.h.p.-hour = "TTT = 24.8 Ibs. VALUES OF N FOR EXPANSION LINES Clayton has found by logarithmic diagrams that the value of n for the expansion line of the steam engine varied from 0.835 to 1.234 and that the average with saturated steam in the steam pipe was 0.947 while with superheated steam the value was 1.056 and the average of all tests was 1.004. The value of n = 1 which is so common in the theoretical discussion of indicator cards of steam engines is not far from the truth. The ease of construction of the rectangular hyperbola, pv = constant, and the simple results it leads to in practice are good reasons for its common use. Of course it has no theoretical basis and is only used for the reasons given above. In steam engines actually examined by Clayton the expansion line had values of n near 1 in all cases except poppet valve engines, where n rose to 1.3 with highly superheated steam and in the Stumpf Straight Flow Engine with superheated steam where n was 1.2. The compression lines were mostly near n = I although some values much lower were found. In gas engines n for expansion varied from 1.09 to 1.36 while n for compression varied from 1.09 to 1.43. Gtildner, according to Clayton, gives values of n from 1.30 to 1.38 for compression and 1.35 to 1.50 for expansion. The latter is due to hot water. He states that Burst all gives 1.288 as the average for expansion and 1.352 for compression in gas engines. For compressed air locomotives Clayton found n = 1.35 while on air compressors it was 1.26. For ammonia compressors the cards examined gave values averaging n = 1.20. On gas compressors the value is n = 1.14. This is very low. THE STEAM ENGINE 215 While discussing this value of n it is well to remember that, although the equation of the adiabatic of steam is / , xr s + -i = const, or , , xr c p dt + ~T + ) T = c the line is sometimes put in the approximate form pv n = const. n = -JT, according to Rankine (63) y n = 1.035 + O.lz, according to Zeuner (64) n = 1.059 + 0.000315? + (0.0706 + 0.000376p)z according to E. H. Stone (65) p = pressure at point where quality is x x = quality at highest point For superheated steam, p(v + 0.088) 1 - 805 = const. (Goodenough) EXPANSION LINES It will be well at this point to examine the difference between various lines of expansion which may occur in the steam engine cylinder to see what error might be made in assuming one rather than another in the theoretical discussion of indicator cards. The discussion and curves will also indicate how close the lines approach each other. Using the results of Clayton, suppose that an engine with cut- off at Y stroke and with 10 per cent, clearance has steam of a quality of 0.70 at 115 Ibs. absolute pressure and it is required to draw the following lines through this point: (a) Rectangular hyperbola (6) Adiabatic (c) Isodynamic (d) Constant steam weight, x = const. (e) pv n = const, of Clayton, n = 1.02 (/) pv 1 - 2 = const. (0) pv- 8 = const. From Clayton's diagram, n = 1.02. 216 HEAT ENGINEERING The diagram for 1 Ib. of steam will be computed and tabulated below for pressures at 20-lb. intervals to points beyond the actual length of card. The computations for the first points are given. Vi = Mxv" = 1 X 0.70 X 3.876 = 2.71 cu. ft. pi = 115 Ibs. For p 2 = 95 Ibs. abs., (a) F 2 = 2.71 X ^jr = 3.28 cu. ft. _ 0.4881 + 0.70 X 1.1026 - 0.4699 1.1363 = 0.695 F 2 = 0.695 X 4.644 = 3.23 cu. ft. (6 a ) Stone's value of n, n = 1.059 - 0.000315 X 115 + (0.0706 + 0.000376 X 115)0.70 n = 1.102 2.71 = 3.23 cu. ft. x 2 X 808.8 = 309.0 + 0.70 X 797.0 - 294.6 z 2 = 0.708 F 2 = 0.708 X 4.644 = 3.28 cu. ft. (d) V 2 = 0.70 X 4.644 = 3.25 cu. ft. (e) F 2 = 2.71 1 = 3.28 cu. ft. (/) y 2 = 2.71 (i^p = 3.17 cu. ft. TABLE OF VOLUMES Pressure n = l Adiabatic Isody- namic x const. n = 1.02 ra = 1.2 n = 0.8 Theory Stone 115 2.71 2.71 2.71 2.71 2.71 2.71 2.71 2.71 95 3.28 3.23 3.23 3.28 3.25 3.28 3.17 3.41 75 4.15 4.00 4.01 4.15 4.07 4.13 3.88 4.62 55 5.67 5.30 5.31 5.66 5.45 5.60 5.02 6.82 35 8.90 8.00 8.00 8.82 8.32 8.70 7.34 11.95 The tables show the close agreement of the line pv 1 - 102 with the adiabatic and Fig. 96 shows how closely the rectangular hyper- THE STEAM ENGINE 217 bola, the Clayton line, the adiabatic, the isodynamic and the constant steam weight curves come together. The lines vary slightly. If the card abed is worked out for the various lines it is found that although the mean effective pressure for the lines pv - 8 and pv 1 - 2 vary 12 per cent, from the mean of the two, the variation on the lines pv 1 - 02 and pv 1 ' 1 which are the extremes of the remaining lines vary from the hyperbola by 5 per cent., or, 5 per cent, is the maximum variation in general. The value 1.2 120 100 CO JB a 60 S 40 p 1 i I =1.102 Adiabatic 2.0 4.0 6.0 8.0 Volume in Cu. Ft. 10.0 12.0 FIG. 96. Variation in various expansion curves. Constant steam weight curve lies between adiabatic and pv 1 - 02 = const. Isodynamic can not be separated from curve pv = constant. is found on engines using superheated steam and the variation in the m.e.p. for the card with "this line is 14 per cent, from that with the rectangular hyperbola. USE OF RECTANGULAR HYPERBOLA From the above it will be seen that in general the rectangu- lar hyperbola will give areas within 2 per cent, of the actual cards for ordinary engines although with superheated steam or very wet steam the use of the rectangular hyperbola for the ex- 218 HEAT ENGINEERING pansion line may result in an error of less than 15 per cent, in the area of the indicator card. Since the errors due to round- ing of the corners or the intersection of the various cards of a multiple expansion engine may lead to as great errors, the simplicity in the calculations and constructions when this line is assumed makes this curve of value in the preliminary design of an engine. When the probable condition of the steam is known the exact exponent may be used. CONSTRUCTION OF EXPANSION CURVES To draw the rectangular hyperbola the graphical construction shown in Fig. 97 is used. Given the original point 1, a hori- zontal and a vertical line are drawn through 1. If, from a, b and c, on the horizontal line, slanting lines are drawn to the origin FIG. 97. Construction of rectangular hyperbola. of pressure and volume, they will cut the verticals in a', &' and c'. Verticals from a, b and c intersect horizontals from a', b' and c' in a", b" and c" which are points on the rectangular hyperbola. The student may see that this curve fulfills the equation of the hyperbola since Pi = Pa P'a = P"a From similar triangle V a = V c !^_ _ PI V' ~ V' Substituting the values above or P " " . THE STEAM ENGINE 219 The graphical construction usually employed for the poly- tropic, pv n = constant, is open to an accumulative error and for that reason, although correct geometrically any slight error of one point is increased in the later points and hence the best way to construct the curve is to assume a ratio of two pressures, say = r, and to use this to compute all successive points. p* = rp l In this way successive pressures can be read from a slide rule without moving the middle scales by setting the zero of the C scale at the value of r on the D scale. P2 = /VA PI W Since The successive volumes are given by r"v 2 The p's and v's being known, the curve may be plotted quickly. Thus suppose the line pj,i.4 _ constant is desired to pass through p = 125 Ibs. abs. and 2.5 cu. ft. Assume r = 0.5 1 . /I \ IA / 1 \ !- 4 t) =(0) =2-64 = P 125.0 62.5 31.7 15.7 7.8 3.9 V 2.5 6.6 17.4 46.0 122.0 323.0 MEAN EFFECTIVE PRESSURE To find the probable mean effective pressure for an engine on which the cut-off and limiting pressures are known but for which the clearance and compression are not known as the pre- 220 HEAT ENGINEERING liminary dimensions have not been found, the best procedure is to assume zero clearance and zero compression and to allow for the effect of these by empirical constants. The card abcde, Fig. 98, shows the theoretical card in this case. If r is the apparent ratio of expansion, ab will equal D/r if ed is assumed equal to D. The area is given by D D area = -pi+~pi D_ D area pi , m.e.p. = -p- = - [ 1 + log e r] - p b (66) This is the formula by which the mean height may be found if the line is assumed to be the rectangular hyperbola. For the line pv n = const. m.e.D. = - n for all values of n except 1. If clearance is added to the diagram as shown in the right-hand Fig. 98, the expansion line will be raised and for that reason the g FIG. 98. Theoretical indicator cards, showing effect of clearance and compression. area will be increased. On the other hand, the presence of com- pression in the cylinder reduces this area. The card is then abfcdgh. From a number of cards drawn with different clear- ances, pressures and cut-offs the net result of these two effects was to decrease the area or m.e.p. by 2 per cent. The variation found was from 3 per cent, to 1 per cent. REAL AND APPARENT RATIO OF EXPANSION In Fig. 98, D/on is the apparent ratio of expansion while D/om is the real ratio of expansion. The first is called r and it is the THE STEAM ENGINE 221 reciprocal of the cut-off. If the clearance is ID, the real ratio of expansion is T r = 1 + Ir The actual card ab'c'de'f, Fig. 99, differs from the theoretical card abcdef at several points. In most cases the steam is throttled or wire-drawn during admission. This causes the line ab in high-speed engines to drop about 10 per cent, in pressure although in Corliss engines the line is nearly horizontal. More- over, with slide valves, the slow closing of the valve causes further throttling and causes the corner to be rounded. The expansion line be is of the form pv n = const., the value of n depending on FIG. 99. Actual and theoretical cards compared. the quality at 6. The release takes place at c r before the end of the stroke giving the line c'd. The back pressure line should be that assumed, as the actual value is taken in theory. This is from J to 2 Ibs. above the pressure in the region into which exhaust takes place. At the end of exhaust the back pressure line will rise due to the slow closing of the exhaust valve so that the point of compression e f is raised above e. This gives the line e'f distinct from ef. In actual cards such a point as g is considered as the point of compression but, as was pointed out in the loga- rithmic diagrams, this point is not on the compression line. The ratio of the actual area to the theoretical area is known as the diagram factor. Area (db'c'de'f) Area (abcdef) = diagram factor. 222 HEAT ENGINEERING The value of this for single cylinder engines is about 0.90. The actual m.e.p. is then given by m.e.p. = 0.90 X 0.9s[y (1 + log, r) - p b ] (68) If the two factors are combined in a single factor known as the combined factor, this may be found from the results of an engine test by the following formula: , actual m.e.p. combined diagram factor = -r . ; - theoretical m.e.p. i.h.p. X 33000 2FLN X * 10000 I 8000 I g 6000 o? | 4000 Tests give the value of this to be about 0.85 for high-speed engines. To find the value of the m.e.p. it is necessary to assume or know the values of p\ t pb and r. For ordinary single cylinders the value of pi is usually 90 Ibs. gauge to 125 Ibs. gauge un- less it is desired to get great power or force from a rela- tively small cylinder when a much higher pressure is used. If high pressure is used cut- off must be made very early so that the expansion of the steam may be utilized com- pletely. This, however, FIG. lOO.-Steam consumption curves. CaUS6S the Percentage effect of initial condensation to be felt although the higher temperature range gives a higher theo- retical efficiency. To reduce this loss a later cut-off is used giving an excessive loss due to free expansion. This same reasoning holds for the determination of the best value of r. If r is large the steam is used to advantage expansively but the percentage effect of initial condensation is great while for a smaller value of r this condensation effect is smaller but there is a loss from free expansion. If the results of a test are plotted with indicated horse-power as abscissae and total weight of steam per hour as ordinates, the most efficient point is found at the point of tangency of a straight line from the origin as this gives the smallest ratio of ordinate 150 I.H.P. THE STEAM ENGINE 223 to abscissa which is the steam per horse-power hour. In Fig. 100 this relation is seen for the curve AB. If the ratios of. ordinates to abscissae are found for different points, the steam consumption curve CD is found on which F is the best point. Heck proposes that the weight of steam per hour be divided by the displacement per hour giving the weight of steam used per cubic foot of displacement. In this case the line AB would take the same form because the displacement per hour which is 120FLN . 144 is a constant for Fig. 100. F = area of piston in square inches. L = stroke in feet. N = revolutions per minute. Mean Effective Pres.suj'e FIG. 101. Curves of steam per cu. ft. of displacement for different mean effective pressures. This quantity is called m/, weight per cubic foot of displacement. wt. per hr. Now m f = = const< 144 from which p m = . _ 2 Pm FLN i.h.p. - 33000 i.h.p. 2FLN const. X i.h.p. (70) In Fig. 101 the line AB drawn with p m and ra/ as coordinates is the same curve as used before with a new scale. The weight 224 HEAT ENGINEERING of steam per cubic foot of piston displacement from the indi- cator card, Fig. 102, when the missing quantity is m.q., is 9 1 V b 1 - m.q. (rv" b )(l-m.q.) - c approximately. (71) FIG. 102. Card for the computation of steam weight. Now the steam per indicated horse-power hour is 13750m/ or N X 60 2p m FLN 33000 M l mf = 13750^ (72) (73) That is, straight lines from the origin of Fig. 101 are lines of constant steam consumption; the consumption being equal to 13,750 times the ratio or the inclination of the lines. p m The theoretical amount of steam per cubic foot of cylinder volume is l m < = l = T = k r (74 > since the volume at cut-off is 1/r and the specific weight of steam m = a + bp (75) is if a is neglected because of small value. This results from THE STEAM ENGINE 225 plotting the specific weights and pressures of the steam tables as shown in Fig. 103. This shows clearly that the value of m is practically bp. Now p m = ^ [1 -f log e r] - p b P [1 + log. rj -&-* (76) In other words Mi decreases as the expansion increases since log e r increases faster than r 175 150 & 3 a 100 P 75 0-1 a ! 50 0.05 0.10 0.15 0.20 0.25 0.30 0.35 Weight of 1 Cu. Ft. of Saturated Steam in Lbs. 0.40 FIG. 103. Curve of relation between pressure and weight of 1 cu. ft. of steam. The amount of initial condensation per cubic foot of cylinder displacement varies slightly with r; that is, as p m increases the condensation increases slightly. CD will represent the initial condensation in Fig. 101. The distance from AB to DC will be the amount shown by equation (74). From this figure Mi can be found as well as from (76). 15 226 HEAT ENGINEERING The best point is usually found at about J4 or % cut-off with single cylinders of the ordinary form. While discussing steam consumption it will be well to call attention to the Willans straight line law for the steam con- sumption of throttle-governed engines. For a given ratio of expansion and back pressure p m = cpi - p b 2(c Pl - Pb )FLN i.h.p. = -33000" 2LFNXW fcpi 2LFN 60fcr33000 i.h.p. , M total = - 1 44--X -344- X[ 2FLN + = k' i.h.p. + k"p b = a i.h.p. + b (77) or, the total weight of steam per hour with a throttle valve engine varies with the horse-power on a straight line. STUMPF ENGINE Although in the ordinary engine the best point of cut-off seems to be about one-fourth stroke, in a single cylinder engine developed by Prof. Stumpf in 1910 the cut-off is carried very early since the initial condensation is reduced by the arrangement of the engine. This is the uniflow (unidirectional flow) or straight flow engine shown in Fig. 104. This was invented independently by Stumpf although a similar idea had been developed by Bowen Eaton in 1857, by E. Roberts in 1874 and by L. J. Todd in 1885. In this engine steam is allowed to enter from the valve A behind a long piston B and forces it to the right. After the piston moves a short distance to the right, steam is cut off by the double beat valve and the steam expands to nine-tenth of the stroke when the piston overrides ports C, cut in the cylinder barrel, allowing steam to exhaust through D to the condenser or atmos- phere. The area thus available is so great that the pressure rapidly falls and when the piston returns and covers this the steam has dropped to the pressure of the condenser or atmosphere. The compression then begins and, by properly selecting the clearance, the final pressure is made equal or nearly equal to the boiler pressure. This clearance depends on the back pressure and; if this is liable to change very much, arrangements are made to vary the compression or clearance. Thus, with a condensing THE STEAM ENGINE 227 cylinder, when it is necessary to start at atmospheric pressure until the air pump connected with the engine has produced the proper vacuum, an auxiliary clearance volume is connected to the cylinder and is controlled by a valve so that it may be con- nected when necessary. As the steam expands in the cylinder, it becomes wet due to the work done, as is known from the discussions of the adiabatic for steam. The supply steam in the hollow head D, however, FIG. 104. Cylinder of Stumpf engine with indicator card. warms the steam in the cylinder in contact with the head and keeps it dry or superheats it as the pressure falls. As the steam is discharged outward at the end of the stroke the steam next to the head expands, driving the wet steam before it, and leaves the cylinder practically full of dry steam. As the piston returns the steam compressed is superheated as it starts from a dry or even superheated condition. This steam gives up heat to the piston head so that the steam entering the cylinder meets a piston surface that is warm while the head and walls are heated 228 HEAT ENGINEERING to the temperature of the incoming steam on one side and to a temperature higher than this by the superheated steam on the other side. The part of the cylinder near the center which is in contact with the low pres- sure or exhaust steam is cut off by the piston from contact with the fresh steam. The curves of Fig. 105 made by Stumpf show how the wall temperature (a) varies along the cylinder of a con- densing engine compared with the temperature (6) of the saturated steam corre- sponding to the pressure when the piston is at various FIG. 105. Curve showing temperature of wall and saturated steam for different positions of piston as found by Stumpf. points. The fact that the metal line ig aboye the satu . ration line indicates that superheated steam must be present and this is only possible on compression. The cool central portion of the cylinder is a good feature structurally for lubrication as at this location the piston is moving at its t 20000 n . 20 Ibs. 94 15000 E 15 Ibs 10000 10 Ibs, : 5000 3 5 Ibs. 10* 20* 30* 40 Mean Effective Pressure 60* FIG. 106. Steam consumption and heat per h.p. hr. curves of Stumpf engine. highest speed. Stumpf claims that the steam in this engine does not enter in a disturbed condition and that it flows always toward the exhaust. This, of course, cannot be true for, THE STEAM ENGINE 229 until release, the action is the same as in any engine, but, at release, the steam which has been dried or superheated by the jacketed head drives out the wet steam, whereas, in the ordi- nary return flow engine, this steam is the first to leave. The distribution of the exhaust around the circumference gives an undisturbed discharge although it tends to remove any moisture from the piston face. The jacket head being in the steam supply is a bad feature as this would introduce wet steam into the cylinder unless superheated steam is used. The effect then would be to reduce the amount of superheat only and in one test this amounted to 54 F. or 27 B.t.u. per pound of steam. The chambers E and F are jackets for the barrel and the steam gradually passes from one to the other and with this a drop in pressure causes a drop in temperature. To show the efficiency of this engine, Fig. 106 is presented. This is from a test of a 300 h.p. engine tested with saturated steam and superheated steam of 580 F. at 135 Ibs. absolute pressure. The results are plotted for pounds per horse-power hour and actual B.t.u. per horse-power hour chargeable to the engine plotted against m.e.p. They show the closeness of results with saturated and superheated steam, the slight variation with great change in load and values, which are ordinarily obtained with multiple expan- sion engines. This engine illustrates the great value of expansion when the effects of initial condensation can be eliminated. SIZE OF ENGINES To find the probable size of an engine to deliver a given horse- power, the formula for horse-power is used b.h.p. 2pFLN mech. eff." 33000 p = m.e.p. in pounds per square inch. F = area of piston in square inches. L = stroke in feet. N = revolutions per minute. In this equation p has been determined from the formula (68) after pi, pb and r have been assumed. N is fixed by the purpose for which the engine is to be used or by the valve gear. For Corliss engines, N varies from 60 to 120. For connected valve gears, N may be as high as 450. The allowable piston 230 HEAT ENGINEERING speed, 2LN, then fixes L. This varies from 250 in small engines to 1000 in larger engines. The only unknown is now F by which the diameter is fixed. To apply this, suppose it is desired to find the size of an engine to drive a 125 kw. generator. 0.746 XOO X (79) This number, 1.6, is an important average number to keep in mind for rapid computations. The following assumptions will be made: pi = 125 Ibs. abs., p b 17 Ibs. abs., r = 4, 2LN = 500, N = 275. m.e.p. = 0.90 X 0.98 [^ (1 + 2.3 X 0.602) - 17J = 50.9 Ibs. per square inch 0.91 ft. = 10.9 in. i.h.p. X 33000 _ 200 X 33000 ~ = _ 2pLN ~ 2 X 50.9 X 0.91 X 275 The size of cylinder necessary would therefore be 18 in. di- ameter X 12-in. stroke, the somewhat unusual ratio of diameter to stroke being caused by a desire to keep the piston speed low. The effect of clearance on an engine is largely dependent on the amount of surface exposed. The volumetric clearance does not seem to have much effect on the efficiency. Tests have been made on an engine with varying amounts of clearance and the results have shown little change in steam consumption. The effect of clearance surface as shown by Heinrich has been mentioned on p. 209. General practice, however, is to cut the clearance to a minimum so as to increase the area of the indicator card for a given displacement and to make the steam loss due to improper compression with different cut-offs as small as possible. BEST POINT OF COMPRESSION The effect of compression is very slight as is shown theoretically and by the tests of Heinrich but to determine the best point THE STEAM ENGINE 231 theoretically several methods may be used. The amount of compression to be used with any given cut-off may be found by the following construction of Stumpf. In Fig. 107 the initial pressure is pi, and the back pressure pb, the curves are rectangular hyperbolae. With the dimensions given on the card, Area = Pl - + Pl D [I + -] log e - Dp b [l - x] - p b D[l + x] log e It is desired to eliminate x in terms of the volume of steam taken from the boiler, that is, x"D. xD FIG. 107. Stumpf s method of finding compression point. Hence Dp b (l - x) = Dp b [l + I] - Dp b [l + x] Dp b [l + 1]- D Pl (l + I - x"] and p b D(l + x) log - Pl D(l+ ~- area pi r 7 1"] m.e.p = -p- fc.-f.p,^ +-J 1 + I . 71 + I] 4- + ~ **] - + - JPi Suppose now that the volume, x"D, of working steam re- mains constant and there is a change in the point of cut-off. 232 HEAT ENGINEERING The relation which gives maximum work under this change with constant volume of working steam is given by equating the derivative to zero. d(m.e.p.) d - l l+l l+l^ i 1 +Pl~Pl\Og e - j -p^l+'-x") I +-- - x" Pl I H X" r Pi I Pb l + X l+l l+x 1 , / (80) or the point of cut-off divides a line from the end of the card to the axis of volume in the same proportion as the clearance di- vides a line to the axis of volume from the point of compression. FIG. 108. Webb's method of finding compression point. Another method of finding the point of compression as given by Webb in the American Machinist for 1890 is to carry out the expansion line to the back pressure line as in Fig. 108 and then lay off CB so that the area ABCD is equal to area EFG. The point C is then taken as the end of compression. The basis of this construction is the fact that, by changing the area ABCD to EFG, the card CBEFGH will have the same area as ABEGHC and will use the same weight of steam since this is proportional to the line CB. The card CBEFGH has complete THE STEAM ENGINE 233 expansion and must have complete compression to overcome the effect of clearance. For instance, if the line H'C' were used for the compression line, the area C'CHH' would have to be saved to make up for the additional steam CC r since C'CHH' CBEFGH C'C CB But the area JC'C is not obtained and hence the gain of area JCHH' is not proportional to the increase in the steam quan- tity C'C. If the line were H"C"K, then the saving in steam would be CC" but the lost work would be H"HCKC". This is greater than the proportional reduction H"HCC" by the area CKC" and hence this does not pay. Figs. 107 and 108 are for the same conditions and the results are practically the same. Of course the constructions above are true if the steam is as- sumed dry or proportional in all cases to the length of the line between the compression and expansion lines. The methods of Clark, see his Steam Engine, p. 399, and of Ball, A.S.M.E., xiv., p. 1067, may be referred to by the student. Clark's method is the equivalent of the two methods given above although the results are given in the form of a table. Having the compression desired, the pressures, the cut-off, release and clearance from the design of the cylinder, the probable card for an engine may be drawn and studied. SPEED The speed of an engine affects its efficiency in changing the amount of condensation. As this varies inversely as the cube root of the number of revolutions per minute it would naturally be expected that high-speed engines would be the more efficient. That this is not so is due to the fact that the term s varies inversely as the linear dimension of the engine and hence this term is so small for large engines that it overbalances the effect of the slower speed. For instance, if the piston speed, 2LN, is fixed, the initial condensation will vary approximately inversely as the cube root of N and also inversely as the square root of L. Hence it would pay to make L large and N small. This is the actual result in practice. The large, slow-speed pumping 234 HEAT ENGINEERING engines represent the best type of engines using saturated steam. The effect of superheat has been explained and results given at the end of Chapter II show that this materially affects the condensation and increases the efficiency a greater amount than that estimated by theory. The effect of the jacket on a cylinder is a debated question. The results shown by the committee of the Institution of Me- THE STEAM ENGINE 235 chanical Engineers of Great Britain indicate a distinct gain. Other tests made on large engines show no gain and, in a few cases, a loss. Of course the total steam used by the engine is always considered in these cases. In most instances the important saving has been found on small engines of less than 300 h.p. For large engines the results will vary. Ten to 15 per cent, may be saved on engines of 200 or 300 h.p. One form of jacket which has bee.n used on the type of semiportable engine known as the locomobile is of value. In this unit the engine is mounted on top of the boiler, Fig. 109, and the exhaust gases from the boiler pass around the cylinder, thus heating it to a high tempera- ture. These units have the condensation reduced to such a de- gree by this jacket combined with the use of superheated steam that they give a brake horse-power hour on 10^ Ibs. of steam 1.25 Ibs. of coal or 210 B.t.u. Effect of regulating by throttling steam and changing cut-off may be seen when the formula (53) is noted. In this the quantity s changes slightly with the cut-off and so there is a slight increase in the steam due to this. With throttling, however, there is an amount of available energy lost apparently. When it is remembered that this energy is used in changing the quality of the steam which becomes drier or superheated, it is seen that when the steam enters the cylinder it is in such a condition that the initial condensation is less and hence the loss due to this effect is smaller. So great is this reheating effect with throttling that, even with this steam of reduced availability the steam consumption is as low as with engines in which the cut-off is varied. A number of years ago the author experimented on a small engine, controlling it by automatic cut-off and by throttle governing, and both curves, similar to CZ>, Fig. 100, were coin- cident throughout their range. The curves will indicate the steam for either of these methods and also how the consumption varies with the load. TOPICS Topic 1. Sketch the Rankine cycle of the steam engine with complete expansion and compression and explain the difference between this and the Carnot cycle. Explain the variation in quality on the various lines in these two cycles. Show that clearance has no effect in this case when initial condensation is not present. Topic 2. Sketch the Rankine or Clausius cycles with complete expansion and incomplete expansion on the p-v and T-S planes and derive the various 236 HEAT ENGINEERING expressions for efficiency. Explain why the cards may be drawn with no compression lines. Topic 3. Sketch the T-s diagram of a Rankine cycle with incomplete expansion and discuss the value of increasing the steam pressure, vacuum or amount of superheat. Does the increase of vacuum always pay? Why? Topic 4. Sketch and explain action of the Barrus calorimeter. Give the method of computing results. Explain what is meant by a dry test, how it is made and for what it is used ? What is a separating calorimeter ? What is the formula for x when this instrument is used. What is an electric calo- rimeter? What are constant immersion thermometers? Why are they used? What is the purpose of the upper thermometer of the Barrus calorimeter? TopicS. What is Hirn's analysis? For what is it used? What observa- tions are made? Explain how the quantities M and M are found. Ex- plain how a?i, x z , and x 3 are found. Explain how AUi, AUz, AU 3 and AU* are found. Explain how Qi and Q 2 are found. Topic 6. Given the average indicator card, show how the quantities AW a , AWb, AW C and AWd are found. Give method of changing from square inches to B.t.u. Having the U's, Q's and AW's explain how Q a , Qb, Q c and Qd are found. What check equation may be used for these quanti- ties? How is the new quantity for this check equation found? What differ- ence does a jacket make in these equations? Topic 7. What is the temperature-entropy analysis? What readings are necessary? Explain how the curves of the quadrants are constructed and sketch the diagram. How is the average indicator card found and how is this transferred to the p-v quadrant of the diagram. Topic 8. Explain how to transfer the diagram from the p-v quadrant to the T-s quadrant. Explain the method of finding the various losses. Which lines on the diagram are true? Why can the diagram be used for other lines? Topic 9. What is initial condensation? How large may it be? To what is this due? What is the missing quantity? In what units is it found? Explain how it may be found from a test. Explain what is meant by the terms of the two formulae below : mq 30 V7 - mq 0.27 Topic 10. Derive the formulae: M. = K'Fc~ V't Why is it that it may be said that v" 2 = - ? What is Heck's expression 0.27 fsr fors? Using Heck's formula m.q. = ~j/= \ - discuss the effect of size, pe THE STEAM ENGINE 237 speed, pressure and cut-off on the actual condensation and on the percent- age condensation, m.q. Topic 11. Give a statement of the work of Callendar and Nicolson, Hall, Duchesne and Adams on the effect of cylinder walls. How do Callendar and Nicolson plot their results so that initial condensation may be found? What do their results show in regard to variation of steam temperature in the cylinder, and in a small hole in the head ? What is the nature of the tempera- ture cycle at various points in the cylinder length and at various depths? Which has the greater effect, clearance surface or clearance volume? When does the major part of the initial condensation take place? To what is it due? Topic 12. Explain Clayton's method of finding the quality at cut-off, constructing the diagram from the indicator card. Show how to find the probable steam consumption from the indicator card. Explain how leaks are detected. Explain how to locate the events of the stroke properly. Topic 13. What values of n are to be expected on various machines on compression and expansion? Explain how the following curves may be constructed on the pv-plane: (a) rectangular hyperbola, (6) adiabatic, (c) constant steam weight curve and (d) pv 1 - 05 = const. What conclusions may be drawn from the figure in the book showing the various expansion curves? Topic 14. Explain how to construct the curves pV = const, and pV n = const, through a given point p\V\. Find the area of the indicator card for each of these to a volume F 2 and pressure p 2 if the back pressure is pi. From this find the formulae for the m.e.p. Topic 16. Explain what is meant by the real and apparent ratio of expan- sion. Derive a formula for the real ratio of expansion in terms of the apparent ratio, r, and the clearance, I. What is meant by diagram factor? How is it found? What value does it have? Topic 16. Sketch the curve of total steam consumption and show how to find from this the curve of steam consumption per horse-power hour. Ex- plain why this curve is the same as the curve between ra/ and p m . What is the Willans straight-line law? Prove that it is true. Topic 17. What is the Stumpf engine? Why is it of value? Sketch it. Sketch curves showing the temperatures of the wall and saturated steam for various positions along the length of the cylinder. What conclusion may be drawn from this? Sketch the cards from this engine. How is compres- sion cared for when high back pressure exists in starting a condensing Stumpf engine with a direct-connected air pump? Topic 18. Explain how to find the indicated horse-power required to produce a given kilowatt output from a direct-connected generator. Show how to find the size of an engine to produce this power. Topic 19. Derive the formula for the best point of compression. Topic 20. Sketch the constructions for the best point of compression due to Stumpf and to Webb. What is the effect of speed? Of superheat? Of jackets? What difference is found between governing by throttling and by automatic cut-off? Topic 21. Sketch and explain particular features of the locomobile. 238 HEAT ENGINEERING PROBLEMS Problem 1. Find the various efficiencies of an engine with pi = 120 Ibs. gauge pressure, p 2 = 30 Ibs. gauge pressure, p l = 2 Ibs. gauge pressure if x : = 0.99 and 35 Ibs. of steam are needed per i.h.p.-hr. Problem 2. One engine uses 25 Ibs. of steam per i.h.p.-hr. with steam at 125 Ibs. gauge pressure and of quality 0.995. The back pressure is 2 Ibs. by gauge. By using steam of 200 F. superheat the steam consumption is reduced to 22 Ibs. per i.h.p.-hr. What was the saving? Problem 3. An engine uses 3500 Ibs. of steam per hour, of which 300 Ibs. is used in the jackets and 200 Ibs. in the receiver. The steam supply is at 150 Ibs. gauge pressure with 175 F. superheat. The temperature of the return from the jackets is 320 F., while the return from the receiver is 338 F. and the hot well temperature from the condensate is 95 F. The engine develops 250 h.p. Find the B.t.u. per h.p.-min. Find the actual efficiency. If the mechanical efficiency of the pump and engine combined is 92 per cent., what is the duty of this engine? Problem 4. In Fig. 68 assume that the pressure on the top line is 120 Ibs. absolute and on the lower line is 15 Ibs. absolute. Suppose that the qual- ity on the top line is 0.98, assuming the cycle to be the Rankine cycle, and that it varies from 0.03 to 0.98 if assumed to be the Carnot cycle. Find the qualities at the lower corners. Find the heats on the four lines. Problem 5. In a Rankine cycle with complete expansion the pressure varies from 125 Ibs. gauge to Ibs. gauge with xi = 1.0. Find the efficiencies, ni, 1)2, 173. Increase the upper pressure to 150 Ibs. gauge and, leaving the other quantities unchanged, find 771, 772 and 173. With 125 Ibs. initial gauge pres- sure assume the quality changes to 160 F. superheat; find 171, 772 and 773. Assume the back pressure is changed to a vacuum of 27 in. but with no change in other conditions; find 771, 772 and 773. Problem 6. In a Rankine cycle with incomplete expansion let the initial gauge pressure be 125 Ibs., the pressure at the end of expansion 20 Ibs. gauge and the back pressure is that of the atmosphere. If x\ = 1.0 find the three efficiencies, 771, 772 and 773. Change the initial pressure only to 150 Ibs. gauge and find the efficiencies. Change the initial quality to 160 F. superheat and find the efficiencies. Change the back pressure only to 27 in. and find the efficiencies. Problem 7. The following results were obtained from a test of an engine : Size of engine 10 in. X 14 in. (neglect rod) Time of test 60 min. Clearance 7 per cent. Number of revolutions 15,000 Steam used 3003 Ibs. Average gauge pressure at throttle 112 Ibs. per sq. in. Barometric pressure 14.7 Ibs. per sq. in. Average quality of steam 0.99 Average temperature of condensate 135 F. Average temperature of water leaving 120 F. Average temperature of water entering 75 F. Weight of condensing water 58,100 Ibs. THE STEAM ENGINE 239 Average results from indicator cards Point of admission 0.0% of stroke,. ... 55 Ibs. abs. per sq. in. Point of cut-off 33 . 0% of stroke, .... 114 . 7 Ibs. abs. per sq. in. Point of release. 93.0% of stroke,. ... 44.7 Ibs. abs. per sq. in. Point of compression 20.0% of stroke,. ... 14.7 Ibs. abs. per sq. in. Work of admission 3.22 sq. in. Work of expansion 3.32 sq. in. Work of exhaust 0.70 sq. in. Work of compression 0.46 sq. in. Make Hirn's analysis from the above data. Problem 8. -In the analysis above the following coordinates give the positions of the points in inches on a 4-in. card from the true zero of pressure and volume with a spring scale of 50 Ibs. per inch. Points Volume Pressure Points Volume Pressure 1 0.28 2.48 10 4.28 0.29 2 0.78 2.44 11 3.28 0.29 3 1.28 2.36 12 2.28 0.29 4 1.60 2.28 13 1.08 0.29 5 2.28 1.60 14 0.78 0.41 6 3.28 1.19 15 0.48 0.63 7 3.78 0.94 16 0.28 1.08 8 4.02 0.88 17 0.28 1.78 9 4.18 0.60 18 0.28 2.30 Plot the card and make the T-S analysis. Find the various losses. Problem 9. Find the probable value of the missing quantity for the engine given in Problem 8 by the various formulae of this chapter. Problem 10. If the ratio of connecting rod to crank is 6 to 1 mark the piston positions for every five-sixtieths of a revolution on the indicator card constructed from the data of Problems 7 and 8. Also find crank positions of events of stroke. Construct the Callendar-Nicolson diagram for this card and find the probable initial condensation. Express this as a percentage of the steam given in Problem 7 and compare results with those of Problems 7, 8 and 9. Problem 11. Using data of Problem 8 construct table and diagram used in Clayton's method. Find probable initial condensation and steam consump- tion. Compare this with other results. Problem 12. Find the real and apparent ratios of expansion of the card of problems 7 and 8. Find the diagram factor. Find the horse-power de- veloped in Problem 7. Find the actual steam consumption. Problem 13. Construct a curve of total steam consumption and from it find the curve of steam per i.h.p.-hr. Problem 14. Find the size of a single-cylinder non-condensing engine to drive a 125-kw. generator with initial gauge pressure of 130 Ibs., cut-off at 0.3 stroke and a back pressure of 2 Ibs. gauge. Problem 15. Construct the card for Problem 14 and find the best point of compression for 10 per cent, clearance by Stumpf's method and Webb's method. Release is at 95 per cent, of stroke. CHAPTER VI \ p. c MULTIPLE EXPANSION ENGINES Multiple expansion, which is the use of steam in one cylinder after another, was introduced to cut down the wastes in steam engines. After its introduction it was seen that for structural reasons such an arrangement is of value, as it reduces the sizes of parts of the machine and gives a more uniform turning moment. The reason for the reduction in waste is the fact that there is a series of small ranges in temperature; that some of these ranges act on small surfaces (i.e., in the smaller high-pressure cylin- ders), and lastly that heat ab- stracted by the exhaust steam from the upper stages may be of value for use in the lower stages. If the indicator cards from a two stage or compound engine are taken as shown in Fig. 110, these cards may be used as shown in the preced- ing chapter to study the ac- tion of the cylinder walls, to find the horse-power and to use for any purpose that cards for simple engines are used. However, to get a fuller picture it is often desired to construct what is known as a combined dia- gram. To form this the cards of Fig. 110 from a 10 and 20 X 24 engine are divided by ordinates at regular intervals, say 10, after laying off the clearance at the end of each diagram. The extreme ends of the diagrams are placed on a new axis of volume and the base lines with their ordinates are drawn in after making the card lengths proportional to the volumes swept out by the respective pistons. In most cases since the strokes are equal these are pro- portional to the squares of the diameters. This increases or de- 240 L. .. ,/ \ >s p^ r \ \ *^^. 16* Spri ng 7S FIG. 110. Indicator cards from high and low pressure cylinders of com- pound engine. MULTIPLE EXPANSION ENGINES 241 FIG. 111. Combined cards for com- pound engine. creases all horizontal dimensions. If now a scale be assumed for pressure and the ordinates from the atmospheric pressure line are measured on the cards and reduced to this new scale, the points may be plotted on the new figure and the cards of Fig. Ill drawn. These are both drawn to the same scales of volume and of pressure. The diagram is known as the com- bined diagram. Since, how- ever, the weights of clearance steam in each cylinder are not the same, although the work- ing steam is equal on each, these figures do not represent diagrams for the same total weight of steam on the ex- pansion line. For this reason the expansion line of one cyl- inder does not pass through that of the other. The same is true of the compression lines. These cards show in this figure the relative amounts of work done by each cylinder and the amount lost due to the drop in pressure between the two cylinders. That these diagrams cut each other means noth- ing since the diagrams are not simultaneous for the same piston positions. The first cylinder is known as the high-pres- sure cylinder and the second as the low-pres- sure cylinder. If there are three stages, known as triple expansion, the cylinders are known as high pressure, interme- p IG H2. Combined cards from Woolf com- diate pressure and low pound engine. pressure. In addition to high- and low- pressure cylinders there are first and second in- termediate pressure cylinders in the quadruple expansion engine. 16 242 HEAT ENGINEERING When the cylinders and cranks are so arranged that when the high-pressure cylinder is ready to discharge steam the low-pres- sure cylinder is ready to receive it, the engine is known as a Woolf compound engine, while if the discharge must be stored in a receiver before it can be used, the arrangement is known as a receiver engine. At times several cylinders may be used for any stage, say low, for the purpose of reducing the size. For the diagram of the Woolf engine see Fig. 1 12. The cut-off in the high- pressure cylinder occurs at 1 and the steam expands to 2. At this instant the steam is connected to the low-pressure cylinder in which the pressure is that shown by 12 together with the con- necting pipe in which the pressure was left the same as at 8. Hence the pressure falls from 2 to 3 in the high and rises from 12 to 13 in the low and from that of 8 to that of 3 or 13 in the 8 1 FIG. 113. Combined cards from receiver engine. receiver or connecting pipe. The pressure at 13 is the same as that at 3 except for a drop due to throttling in the passage through the connecting pipe. As the high pressure forces the steam out it enters the low-pressure cylinder in which there is a larger piston moving at the same speed as that of the high-pressure cylinder, and the pressure drops gradually due to this increase of volume. This finally reaches the point of compression on the high pressure cylinder which corresponds to the point of cut-off on the low-pressure cylinder in many cases. The pressure at 7 and that at 4 are about the same. The line from 4 to 5 is the compression and from 5 to 1 there is admission. The steam in the low-pressure cylinder and connecting pipe or receiver now MULTIPLE EXPANSION ENGINES 243 expands from 7 to 8 at which point the cut-off occurs in the low- pressure cylinder. From 8 to 12 the events on the low are the same as those on a single cylinder. The combined cards from a receiver engine are shown in Fig. 113. In this case at the end of expansion the high-pressure cylinder discharges into the receiver in which the pressure is that at 11 if cut-off in the low pressure occurs after compression on the high-pressure cylinder, while the pressure is that of 6 if this compression in the high occurs later than the cut-off in the low. In either case, the pressure in the receiver is usually lower than that at 2 and there is a drop in pressure. The low-pressure cylinder is not in a position to take steam at this point as in most cases the cranks of the engines are at right angles and for that reason the receiver is used. The amount of drop 2-3 not only depends on the pressure in the receiver but also on the relative volume of the receiver and the cylinder. This may be 1 J times the. volume of the high-pressure cylinder. As the piston now returns the steam is compressed into the clearance space and the receiver to the point 4. Here steam enters the low-pressure cylinder in which the pressure is 15. There is a drop from 4 to 5 in the receiver and high-pressure cylinder and a rise from 15 to 9 in the low. The line 3-4 is a rectangular hyperbola with the origin at a distance of the receiver volume from the clearance line. From 5 to 6 the small piston at the middle of its stroke is moving so much faster than the large piston at the beginning of its stroke that the volume decreases and the pressure rises at first. It then usually falls before 6 is reached, 9 agrees with 5, and 10 with 6. Cut-off on the low occurs after compression on the high but before the high pressure reaches the end of its stroke. If it did not occur sooner than this there would be a second admission of steam from the other end of the high-pressure cylinder. The other points are clearly seen. COMPUTATION OF CARDS FOR CONSTRUCTION To compute these various points the volumes of the two cylinders must be known, with the clearances; the volume of the receiver must be known, and finally the points of all the events of each card and the initial pressures. Let V r = receiver volume. V = volume of cylinder at event represented by sub- script including clearance. p with subscript = pressure at point. 244 HEAT ENGINEERING Now from the curve on Fig. 103 it is seen that m = kp ap- proximately, hence the total weight of saturated steam at any point is M = mV = kpV (1) or the weight of steam is approximately equal to the product pV. Hence if quantities of steam of different volumes and at different pressures are connected the resultant pressure mul- tiplied by the sum of the volumes must be equal to the sum of the individual products of pressure and volume, or With this understanding, the following equations hold: i ,, , ,, (known) yf' (Pn unknown) (4) + (In terms of pi,) (5) p,(V t + V r ) + pnVu (In terms of p u ; since , , F 4 + V r + Fi5~ pi 6 V 15 is known) (7) f ee /ON Pl = y r) ( In terms of PiiJ " P" is known) (10) /-,-,\ To apply these a table is first computed for all volumes in pro- portional numbers if not in actual numbers, and from this table the substitution can be made in the equations above. To find the relative position of events a diagram is made with the cranks as shown in Fig. 114. The actual position could be found by using the connecting rods or if infinite rods are assumed the pro- jections may be used. Thus if the compression is to occur at 0.1 stroke on the high-pressure cylinder, the diagram for an in- finite connecting rod would be shown in Fig. 114, with a crank MULTIPLE EXPANSION ENGINES 245 of 0.5. From this it is seen that the low-pressure piston will have moved 0.2 of its stroke when compression occurs in the high-pressure cylinder. The high-pressure piston is at the middle of the stroke when the low-pressure stroke begins. Suppose that the cylinders have a ratio of 1 to 4 and the clearance is 6 per cent, on the high and 5 per cent, on the low. Suppose that the receiver volume is 0.8 times the volume of the high-pressure cylinder; that the cut-off occurs at 0.33 stroke in high-pressure cylinder and 0.45 stroke in low; that compression is at 0.1 stroke from end on high and 0.2 on low and that pi is 150 Ibs. gauge and pb is 2 Ibs. absolute. A L.P. FIG. 114. Diagram by which to find piston position. (Finite or infinite connecting rod.) Call the volume swept out by the high-pressure cylinder 1 and using Fig. 113: V h = I Vi =4 V r = 0.8 Vch = 0.06 X 1 = 0.06 V cl = 0.05 X 4 = 0.20 Fi = 0.33 + 0.06 = 0.39 F 2 = 1.00 + 0.06 = 1.06 = F 3 F 4 = 0.50 + 0.06 = 0.56 = F 5 F 6 = 0.10 + 0.06 = 0.16 V 7 = 0.06 = F 8 V 9 - 0.20 = Fi 5 . Fio = 0.20 X 4 + 0.20 = 1.00 Fii = 0.45 X 4 + 0.20 = 2.00 y 12 = 4 + 0.20 = 4.20 = Fig Fi4 = 4 X 0.20 + 0.20 = 1.00 = 164.7 X 60.6 X 1 = 60.6 p,i X0.8 P4 = P5 = (34.5 1.06 + 0.8 0.43p) 1.86 34.5 = 47 ' 2 0.56 + 0.8 (47.2 + 0.59piQ 1.36 + 10 X 0.2 1.36 + 0.2 = 43.2 + 0.514 p n 246 HEAT ENGINEERING 1 00 PIB = 2 X -^ = 10 (43.2 + 0.514 Pll ) 1.56 Pe = pio = 0.16 + 0.8+1.00 = 34 ' 4 + ' 41 (34.4 + 0.41 P11 )(1.00 + 0.8) Pn 2.00 + 0.8 = 22 ' 1 + ' 264 22.1 3 0736 Pio = p 6 = 46.7 Pb = PQ = 58.7 pi = 64.9 2> 3 = 47.4 P7 = 46.7 x ^ = 125 |;||f * ; J-; . 2 00 Pu = 30 X j^o = 14.3 This method could be used for triple expansion engines or for any arrangement of cylinders. The main point to remember is to reduce the equations for any unknown to terms of one un- known until at last this term occurs on both sides of the equa- tion and the equation gives its value. EQUIVALENT WORK DONE BY ONE CYLINDER On looking at the cards of Figs. Ill, 112 and 113 it will be seen that the total length of the combined diagram represents the volume of the low-pressure cylinder and if the intermediate lines were removed the card would appear as a single cylinder card with very early cut-off. The cylinder would be of the same size as the low-pressure cylinder. Hence it may be said that if the same initial pressure is used in a cylinder of the size of the low-pressure cylinder of a multiple expansion engine as is used in the high-pressure cylinder, and if this has the same total ratio of expansion, then it will develop the same horse-power as the multiple expansion engine. It is well to note what differences occur. With the single cylinder the whole clearance surface of the large cylinder is subject to the full temperature range which would cause excessive condensation unless it is of the Stumpf form of cylinder. The large cylinder would be subject to full high pressure not only requiring heavy walls for the cylinder but causing the rods, pins MULTIPLE EXPANSION ENGINES 247 and bearings to be excessive. The cut-off on the single card oc- curs at about one-tenth stroke, while on each of the separate cards the division of the expansion caused the pressure to be continued to nearly the middle of the stroke on a smaller area giving a more gradual curve of turning moment. On looking at the computations made above it will be seen that if the receiver volume had been made very large there would have been little change in the intermediate lines and were the receiver infinite in volume these lines would be horizontal. Such an assumption is usually made in determining the preliminary relative sizes of the cylinders of a multiple expansion engine and in addition the effect of clearance and compression are omitted in this work. Although in practice the receiver is not more than 1^2 times the volume of the cylinder which discharges into it, it is formed at times by the large exhaust pipe between cylinders. The size seems to have no effect on the operation. DETERMINATION OF RELATIVE SIZES OF CYLINDERS The combined card without clearance for a triple expansion engine with infinite receivers takes the form shown in Fig. 115. The whole card which represents the entire work of the engine could be developed alone by the low-pressure cylinder. The volume of this cylinder is ab. The volume of the intermediate cylinder is cd and that of the high is ef if the expansions are complete to the receiver pres- c sures. If, however, the vol- a umes were cd' and ef there would be the free expansion gf in the high and hd' in the inter- mediate. In many cases free expansion is used as it reduces the sizes of the cylinders and therefore the sizes of the parts. There is not a great loss of work area due to this. The receiver pressures are fixed by the points of cut-off of the lower cylinders. Thus, with a given cut-off, in the high-pressure cylinder, there is a fixed volume of steam and the line ifh of the expansion of this steam is fixed. If now in any cylinder the volume at cut-off is known this must be on the line of ex- FIG. 115. Intermediate pressures with infinite receivers. 248 HEAT ENGINEERING pansion and consequently the pressure is determined. If the cut-off is made later in the intermediate cylinder the point / would be further to the right and consequently the pressure would have to be lower. This would produce less work on the cylinder. Hence the statement is often made that increasing the cut-off of a cylinder reduces the work of that cylinder. If this is done on the high-pressure cylinder the same result follows, as the other cut-offs will be made relatively earlier by this opera- tion and the intermediate pressure will rise. The whole area of the combined card being the work which could be developed by the low-pressure cylinder gives a method of determining the size of the cylinder. The size of this cylinder fixes the power of the engine. The sizes of the other cylinders determine the relative amount of work done by them but do not affect the power of the engine. The m.e.p. for the whole card is found by assuming pi, pb, and R, the total ratio of expansion. .e.p. = /[| (1 + log, R) -p b ]=P (12) / = diagram factor X clearance factor. = 0.65 for naval triple. = 0.70 for naval compound. = 0.80 for Corliss compound and triple land. = 0.90 for pumping engine. Another form, if p r , the pressure at the end of expansion is assumed rather than R, is .e.p. = f[p r (l + log, g) -p t ]=P (13) From this the size may be found by assuming 2LN and finding Ft. 2PZJW m. m Having Ft the areas of the pistons of the other cylinders are found from the ratios of ed' , ef Thus Fi = F ' (15) MULTIPLE EXPANSION ENGINES 249 The lengths cd' and ef are found by obtaining the pressures 7/2 and p"z and drawing the figure or computing the lines cd and ef and allowing for the distances dd' and ff. Thus cd = ab~ (17) Pr ef = ab ^ (18) Pr There are four ways in which to fix the intermediate pressures : (a) equal works, (6) equal ratios of expansion, (c) equal tempera- ture ranges and (d) assumed ratios. (a) For equal work the card is divided into equal areas. Area card = D,[p r (l + log, ^) - p 6 ] = F t (19) Area 1. p. card = D t [p r (l + log e |^) - p 6 ] = F t (20) Area 1. p. card + i.p. card = D,[p r (l + log. ^) - pt] = F 2 (21) Area of three lowest cards = D, [p r (l + log, ^) - p] = F 3 (22) For n stages: p, = -F, (23) /Z' F 2 - ?F ( (24) F, = ?F, (25) In each of these there is only one unknown term p 2 . Hence these pressures may be found. Thus i[p r (l + log e f) - p b ] = p r (l + log, g - P6 (26) (28) In the same manner: L- B (29) 250 HEAT ENGINEERING For the mth receiver: (30) (b) For equal ratios of expansion the following method is used : 7? - Pl K ~ ^ PT R = ri X r 2 X r 3 X ..... = r m r =\/R (31) p'z = rp r (32) p" 2 = r *p r = rp' 2 (33) p 2 = /"> =>p 2 m ~ 1 (34) (c) For equal temperature ranges the temperature range between pi and pb is found and this divided by the number of stages gives the range per stage. If this is added to the lowest temperature successively the various temperatures are found and from these the corresponding pressures are obtained. nr T - = AT (35) iiL T't = T b + AT (36) T" 2 = T b + 2A7 7 = r 2 + AT (37) T m 2 = T b + mAT (38) (d) For assumed ratios of cylinder volumes there is no necessity of finding the intermediate pressure. The usual ratios of practice are: Compound engines D h : DI = 1 : 2 to 1 : 7 Triple expansion D h : D { ; : A = 1 : 2.5 : 6.5 For compound engines a ratio 1 : 4 is quite common. When quick response is needed to a suddenly changing load the small ratios are used; i.e., the larger high-pressure cylinders are used. As an example, suppose it is desired to construct a triple expansion engine to produce 2000 kw. from a generator at 100 r.p.m. with boiler steam at 175 Ibs. gauge pressure with 80 F. superheat, with the pressure at release 5 Ibs. gauge and a back pressure of 2 Ibs. absolute. MULTIPLE EXPANSION ENGINES 251 m.e.p. = 0.90 [9.7(l + log, ^-^ - 2] = 33.0 ihi - 2000 X 1 6 33 ' X 100 X Fl 33000 Fi = 3200 sq. in. d = 64 in. 2LN = 1000 1000 L = 23000 = 5 ft ' = 60 in ' Intermediate pressures and volumes. Method (a) logo's = (I - l) [l - ~] + Iog e (l89.7 X 9.7*)" = - 0.53 + 3.26 = 2.73 p' 2 = 15.4. log, P\ = (| - l) [g] + Iog e (l89.7 X 9.7 W ) M = - 0.26 + 4.25 = 3.99 p\ = 52.1 Fi = ^ 1^4 = 320 X j^ = 2020; cfc = 51 in. F h = 3200 ^ = 595; d h = 28 in. Engine 28-51 and 64 X 60. Method (6) r = l5 = 2.69 p r 2 = 2.69 X 9.7 = 26.1 p\ = (2.69) 2 X 9.7 = 70.2 di = 39 in. d h = 24 in. Engine 24-39 and 64 X 60. 252 HEAT ENGINEERING Method (c) T 1 = 377.5 + 80 = 457.5 !F 2 = 126.2 T' 2 = 236.6; p' 2 = 23.5 T" 2 = 347.0; p\ = 129.4. di = 42 in. d h = 17^ in. Engine 17^-42 and 64 X 60. Method (d) Assume ratio of 1-2-6. di = 45 in. d h = 26 in. Engine 26-45 and 64 X 60. In the above solutions the expansion was complete in the upper two cylinders. If the diameters are made slightly smaller there will be free expansion in each cylinder. The relative size could now be found with the clearance which varies from 10 per cent, or 15 per cent, in slide valve engines to 5 per cent, in ordinary Corliss engines and 2 per cent, in engines with the valves in the heads. The events could be assumed and the volumes of the receivers after which the cards similar to those in Fig. 113 could be computed. The pressures of 90 to 120 used with simple expansion engines are changed in multiple expansion engines to from 150 to 225 Ibs. gauge. The higher pressures are used with triple expansion engines. 150 to 175 Ibs. would be used for compound engines although higher pressures are used with such. At times lower pressures are used with these two stage engines but generally lower pressures do not give sufficient expansion. The higher pressures should give higher efficiencies but the troubles from initial condensation and free exhaust exist here to a great degree. In most cases, however, neither of these effects is so prominent as with single cylinder engines and the steam consump- tion curves are more nearly flat, giving considerable range without much variation in steam consumption. Such steam consumption curves are valuable in any engine or turbine as they give a good efficiency over a great range. The back pressure should be reduced to as low a value as MULTIPLE EXPANSION ENGINES 253 will give a gain on the entropy diagram. Owing to the free expansion in these engines there may be a condition for which an increase of vacuum means a loss. About 28 in. is sufficient for the vacuum to be carried on a multiple expansion engine. In these engines as in the case of the simple engines, the large, slow-speed engines are those which give the highest efficiency. The larger the cylinder, the more efficient the engine. Method (a) is the one usually employed to fix the relative size of cylinders. JACKETING In these engines the effect of jacketing is not always a source of economy except in small sizes. The gain on a triple expan- sion engine is probably larger than that found on a compound engine or on a simple engine due to the fact that heat added to the exhaust steam by the jacket when the moisture on the cylinder walls is vaporized is of use in the lower stages. A number of tests have been made on large pumping engines and it has been found that the engine would consume about the same amount of total steam with or without jackets. REHEATERS Steam as it leaves the cylinder of an engine is likely to be quite wet and hence the moisture deposited on the walls at entrance into the next cylinder would be increased and this might lead to excessive condensation. To cut down this moisture, the receiver is equipped with a reheating steam coil containing high-pressure steam, the receiver is now called a reheating receiver or reheater. The value of the reheater lies in the fact that dry steam is carried into the cylinder. The value of this apparatus is also questioned as some engines on test show an increase of total heat when reheaters are used. Other engines show a decrease in the total heat. The heaters should contain enough heating surface and have a high enough steam temperature to completely dry the steam and to superheat it sufficiently to bring the steam to a dry condition at cut-off. If this can be done, a gain should result. In both jackets and reheaters the apparatus should be thor- oughly drained and the supply of steam should be taken to the apparatus alone. The steam should be hotter than the steam to be heated. The use of steam on its way to the first cylinder for 254 HEAT ENGINEERING a jacket is bad as this may introduce moisture into the cylinder. The use of exhaust steam for a jacket is condemned. The supply of steam should be through pipes of ample size. This apparatus should always be air vented. Heavy lagging of non-conducting material has been found of value in engines in cutting down radiation. GOVERNING The multiple expansion engines are governed by changing the point of cut-off on the high-pressure cylinder alone or by changing it on each cylinder at the same time and in the same ratio. In Fig. 116, the method of regulation by changing the cut-off on the high-pressure cylinder is shown while the method by FIG. 116. Method of governing by changing cut-off on high pressure only. FIG. 117. Method of governing by changing cut-off proportionately on each cylinder. changing the cut-off on each cylinder is shown in Fig. 117. In this latter method the ratios of the V's are all equal if the pressure in the receiver is to remain constant. In the first method the receiver pressure changes from a to a' and from b to 6'. Before the receiver pressure can drop the work done must be represented by the curves shown in Fig. 118. In this it will be seen that the governor has reduced the area by the shaded amount and if this is enough to care for the change in load, the point of cut-off will have to gradually change. The receiver pressures gradually change to their final values and the MULTIPLE EXPANSION ENGINES 255 expansion curve takes the position shown dotted in Fig. 118; the area between the original and final curves being equal to the shaded area. Since this change does not occur at once with re- ceivers of any size the action is sluggish. Of course this change is accomplished in ten or fifteen revolutions but it will not be as steady as that resulting from the method shown in Fig. 117. With small fluctuations either method is good. The method of Fig. 116 cuts down the work on the lower cylinders leaving the high pressure about the same while the method of Fig. 117 affects all in the same way if not by equal amounts. Throttling would be shown in Fig. 119. In this the pressure in the receiver would change and although the total work is FIG. 118. Sluggish action due to cut- off change on high-pressure cylinder only. Shaded area shows decrease of work on first stroke. FIG. 119. Governing by throttling. less, the reduction is mainly on the low-pressure cylinder, as there is little change on the high. This method throws out the equality of works. BLEEDING ENGINES OR TURBINES The use of low-pressure steam for heating water, for use in warming buildings or for other technical applications is carried out in plants having high-pressure steam, by throttling the steam through a reducing pressure valve to the lower pressure. Now although throttling does not mean a loss of heat, there is the loss of available energy if by this is meant the ability to be turned into mechanical work. This may be seen by remembering that it is a difference in pressure which must exist to give ability for work in a cylinder. The production of high-pressure steam is usually more expensive than that of low pressure due to the higher temperature of the steam ; hence after producing this high- 256 HEAT ENGINEERING pressure steam many engineers believe that it should be brought to its low temperature by passing it through an engine or turbine where its available energy may be utilized. After reaching the desired temperature and pressure this steam may be used. If the engine or turbine is connected to a condenser, provision is made to take off a portion of the steam between stages where the pressure is at or near the atmospheric pressure. Were this amount constant the machines could be made to be operated with one quantity to one pressure and with another quantity to a lower pressure. The card from such an application on a compound engine is FIG. 120 . Combined , -n i <-n mi i i cards from a compound shown in Fig. 120. The volume a has engine with steam removed been taken for use outside of the engine, from receiver. Cylinders mv i v i i T_ of equal sizes -*- ne low-pressure cylinder has been as- sumed to be similar to the high-pressure cylinder in this case. If no steam is taken the low-pressure card becomes rectangular as shown by dotted line. REGENERATIVE ENGINES The preceding suggests the regenerative steam engine cycle in which steam is taken from the receiver to raise the feed water to that temperature before entering the boiler. Steam from one receiver brings the water to that temperature and that from the next receiver brings the water and the steam condensed from its tem- perature to the temperature of the next. To see the applica- tion of this suppose that it is applied to a triple expansion FlG 121. Combined cards from a re- engine the diagram of which generative triple expansion engine, is shown in Fig. 121. The decrease in the volume of each card is due to the removal of a certain amount of steam to mix with the feed water to raise it to the temperature of the receiver. There would be two analyses of this depending on whether or not the condensed MULTIPLE EXPANSION ENGINES 257 steam from the feed heaters is pumped into the feed. Of course this should be done. The expansion is assumed adiabatic in theory and 1 Ib. of steam will be assumed on the lower card. The pressures are pi, p 2 , p s , p^ and p . Heat to raise 1 Ib. of water from t to fa = q' 3 q' Amount of steam at quality x 3 to give this = = M 2 * Amount of steam on second card = 1 + M 2 Amount of steam of quality x 2 to raise 1 + M 2 Ibs. of water Amount of steam on first card = 1 + M 2 + MI Heat to raise 1 + M 2 + MI Ibs. of water from 2 to t\ = (1 + M 2 + AfOte'i -q'*) Heat to vaporize (1 + M 2 + MI) Ibs. = (1 + M l + M 2 ) xtfi Total heat from outside = (1 + M 2 + Mi)[ii q' 2 ] = Qi Work done on cycles = (1 + M 2 + M"i)b\ - i 2 ] + (1 + M 2 ) [ii - i s ] + l{[is - ij + A(p 4 - p )v,} = AW The i's are found on the same adiabatic for 1 Ib. of steam from one end to the other. Efficiency = ~^r~ Vi This is compared with efficiency for the same weight in each cylinder. Efficiency = (39) i\ q o and the saving is shown. Let this be applied to a compound engine working between 150 Ibs. absolute and 2 Ibs. absolute with original dry steam and a receiver pressure of 45 Ibs. absolute and a pressure at re- lease in the low of 8 Ibs. absolute. 11 = 1192.6 q f ! = 330.0 1 2 = 1100.0 q' 2 = 243.7 it = 987.0 gr'j = 150.9 v s = 39.9 q't = 94.2 x 2 = 0.92 r 2 = 927.5 . ,, 243.7 - 94.2 M2 = 0^2X-927^ = - 176 17 258 HEAT ENGINEERING Heat = 1.176[1192.6 - 243.7] = 1113 Work = 1.176[1192.6 - 1100] + ifllOO - 987 + ~~ I i O (8 - 2)39.9} = 108.7 + 157.4 = 266.1 Theoretical eff. = -7^ = 0.236 144 Theoretical eff. of ordinary cycle = [1192.6 - 987 + ==3(6 X 39.9)] 778' 250.0 1098.4 1192.6-94.2 0.228 LP.JHb.Bp, 60 FIG. 122. Individual and combined cards from test of pumping engine. This means a saving of 0.8 in 22.8 or 3^ per cent. It will be seen from the expression for the work on each cycle that the low- pressure cylinder is doing 50 per cent, more work than the high- pressure cylinder. Data from the test of a triple expansion engine with its cards given in Fig. 122; MULTIPLE EXPANSION ENGINES 259 Size of pump 20,000,000 gal. per 24 hr. Size of cylinders. ' Steam 30 in., 60 in. 90 in., X 66 in: Water 33 in. X 66 in. Pressures by gauge Steam supply 180 . 2 Ibs. per sq. in. 1st Receiver 25.4 Ibs. per sq. in. 2d Receiver 9.5 in. vacuum Condenser 26 . 7 in. vacuum Water discharge 86 . 9 Ibs. per sq. in. Barometer 14. 86 Ibs. per sq. in. Revolution in 24 hr 28,989 Displacement per revolution 732 gal. Temperatures. Return from condenser 113 F. Water pumped 40 F. Condensing water 40 F. Water leaving condenser 70 F. Outside air 30 F. Quality of steam . 989 Water pumped to boiler, per hour 9,266 Ibs. Boiler leakage, per hour 196 Ibs. Drip in engine room, per hour 135 Ibs. Steam to drip pump, per hour 58 Ibs. Net boiler steam, per hour 8,877 Ibs. Jacket drip, per hour 661 Ibs. 1st Receiver drip, per hour 422 Ibs. 2d Receiver drip, per hour 361 Ibs. Indicated horse-power steam end High pressure 332 . 02 Intermediate 269 . 10 Low pressure 260. 22 Total 861.34 Water end 839 . 8 Delivered horse-power 822 . 9 Steam per i.h.p.-hr 10. 37 Ibs. Heat per i.h.p.-min 193 . 04 B.t.u. Duty per 1000 Ibs. dry steam 181,000,000 TESTING AND ANALYSIS There are a few points which must be brought out in regard to testing multiple expansion engines. The exhaust from these engines is not all at the temperature of 260 HEAT ENGINEERING the condensed steam when reheaters or jackets are used. In such a case the drips from the jacket or reheater are at tempera- tures much higher. These possess the temperature correspond- ing to the steam pressure. Hence if M pounds of steam are used by the main engine, and m/ and m r are the amounts used in the jackets and reheaters, the amount of heat chargeable to the engine is Qi = (M + my + m r )ii Mq irijq'j m r q f r (40) In this q'j and q' r are the heats of the liquid in the jacket and reheater. The efficiency which has been given as _ _ AW 77 ~ (M + my + rn r }(i - q f ) AW is n W * =: (M + my + mr) i - Mq' - mrf, - m r q f r In Hirn's analysis for multiple expansion engines a new device must be used to find the heat delivered to the second cylinder and discharged from the first cylinder. In analyses of multiple expansion engines it is necessary to find Q r , the heat radiated from each cylinder. Having this for the first cylinder, the heat losses, Q a , Qb, and Qd, are found by the formulae and then Q c is given by Qc = Q r - (Q a + Q b + Qd) (43) The heat exhausted from the first cylinder is - AU 3 + AW C - Q c (44) or Q 2 = Q! - Q r - A(W a + W b + W c + W d ) (45) The heat given to the second cylinder Q f , is Q'l = Q 2 + Q h - Qi (46) Qh is the heat given to the steam in the reheater and Qi is the heat lost in radiation from the reheater. This term Qi is found by a radiation test similar to that made on the engine. By observing the condensation from the receiver coil when no steam is passing from one cylinder to the other, the value of Qi is found while the observations during the operation of the engine gives the term Q h . Q h or Q r = M(i - q' ) (47) MULTIPLE EXPANSION ENGINES 261 Equation (45) furnishes the heat supplied the second stage and a similar procedure is used for the lower stages. If a jacket is used on any cylinder, the heat supplied by the jacket is found by weighing the steam condensed. The heat is given by Q i = M(i-q') (48) If an engine is jacketed there is a new term in Q r : Qr = Qa+Q b + Qc+Q d + Q 3 (49) and for multiple expansion engines Q c is given by Qc = Q r - (Q a + Q b + Q d + Q,-) (50) Then Q 2 = Qi + Qi - Qr - A(W a + W b + W c + W d ) (51) BINARY ENGINES Another method of utilizing the waste heat and avoiding the waste due to free expansion on account of the great volume of low-pressure steam is to use the exhaust to volatilize a liquid, such as sulphur dioxide, which has a much higher saturation pressure than steam. In this way the same limits of temperature may be utilized without passing to such low pressures as used on steam engines. Engines using two vapors are known as Binary Engines. In a development of this engine by Professor E. Josse of Berlin, steam from an engine was exhausted into a surface con- denser in which the condensing fluid was volatile S02, which by its evaporation removed heat from the steam and condensed it. The pressure of the steam was 3 Ibs. per square inch. This gave a tem- perature of 141 F. SO 2 at 132 F. is under pressure of 132 Ibs. per square inch so that the S0 2 would boil and produce a pressure of 132 Ibs. absolute while condensing steam at 3 Ibs. pressure. This could be used in a cylinder connected with the same shaft as the steam cylinders and aid in the driving. This cylinder ex- hausted into a condenser cooled by water at 50 F. in which the SO2 condensed at a pressure of 31 Ibs. absolute and a tempera- ture of 66 F. The condensed SO 2 was pumped into the steam con- denser where it was evaporated again and used. In the steam engine Josse developed about 150 h.p. on 250 B.t.u. per horse-power minute and by using the exhaust as de- scribed above, 50 additional horse-power were developed giving 176 B.t.u. per horse-power minute. 262 HEAT ENGINEERING The reason for the 'gain in the binary engine is found in the fact that the effect of initial condensation on a large cylinder or the effect of free expansion have been eliminated. These are practical reasons and not theoretical. The range of tempera- ture has not been increased but the pressure and volume limits have been so changed that these engines have a high practical effi- ciency, 171. ' Josse has installed the engines commercially and has found that they are valuable provided a high load factor can be had. For an intermittent load or for use at intervals the engine is too expensive. TOPICS Topic 1. What are multiple expansion engines? Why were they used? Why are they of value? Explain how the combined diagram is constructed. What is the difference between the Woolf and the receiver compound engines? Topic 2. Sketch the combined cards for a Woolf compound engine. Write the formulae by which the pressures at the various points may be computed. Explain why p a Va + p,V, _ v a +v v ~ pr ' Topic 3. Sketch the combined cards for a receiver compound engine. Write the formulae by which the pressures at the various points may be com- puted. Explain why y = p V a +V v Topic 4. Explain why it may be said that the size of the low-pressure cylinder fixes the power of a multiple expansion engine. On what does the size of the higher pressure cylinders depend? If the size of the high-pressure cylinder is changed what is the effect of this? Topic 6. Derive the formulae for the total work of an n-stage engine and for the work on the low-pressure cylinder. From these find the pressure at entrance to the low-pressure cylinder. Topic 6. Give the method of finding the intermediate pressures for equal ratios of expansion, equal temperature ranges and given ratios of volume. Knowing the pressures, show how the volumes of the various cylinders are found assuming (a) complete expansion and (6) some free expansion. What is the reason for assuming receivers of infinite capacity? Topic 7. Why is a 28-in. vacuum a limit for vacuum in steam-engine operation? What is the effect of jackets? Reheating? Explain what happens when the cut-off is made later on any cylinder and the reason for this effect. Topic 8. Discuss with diagrams the various methods of governing multi- ple expansion engines and show which is the better form. MULTIPLE EXPANSION ENGINES 263 Topic 9. Explain what is meant by bleeding engines or turbines and why this is of value. Explain the action of the regenerative engine. Give the expressions for the various amounts of heat and work and derive the expres- sion for efficiency. Topic 10. Derive the expression for the heat delivered to one of the lower stages of a multiple expansion engine (Q\ or Qzoi Hirn's analysis). Show what effects jackets or reheaters have on this formula. What heat is charge- able to an engine when the drips from the jacket and receivers may be sent back to the feed ? What is a binary engine ? PROBLEMS Problem 1. Find the pressures at the corners and events of the stroke for a compound engine, 8 in. and 18 in. X 24 in. 80 r.p.m., with a re- ceiver of volume equal to twice the volume of the high-pressure cylinder and with 8 per cent, clearance on the high-pressure cylinder and 4 per cent, on the low. Initial gauge pressure is 125 Ibs. and the back pressure is 2 Ibs. absolute. Cut-off is 25 per cent, and compression is at 20 per cent, of the stroke from the end. Cranks at right angles. Problem 2. Construct the combined cards for Problem 1 to scales of 1 in. = 30 Ibs. and 1 in. = 0.435 cu. ft. Find the m.e.p. of each card. Construct the h.p. card to scales of 1 in. = 60 Ibs. and 1 in. = 0.1745 cu. ft. and the l.p. card to scales of 1 in. = 20 Ibs. and 1 in. = 0.884 cu. ft. Find the scale of B.t.u. per square inch of area of card, and of ft. -Ibs. per square inch of card. Find the h.p. of the engine. Problem 3. Find the equivalent m.e.p. to be expected in Problem 1 if the complete ratio of expansion is 10. What is the size of the low-pressure cylinder to develop 350 i.h.p.? Problem 4. Find the size of the low-pressure cylinder of a compound Corliss engine to drive a 2000-kw. generator. The initial pressure is 160 Ibs. gauge, the back pressure is 3 Ibs. absolute, and the pressure at end of expansion in the low-pressure cylinder is 4 Ibs. gauge. Find the size of the high-pressure cylinder by four methods. Problem 6. Find the receiver pressures for a triple expansion engine operating between 175 Ibs. gauge and 2 Ibs. absolute, with a complete ratio of expansion of 22. Problem 6. Construct a p-v diagram for Problem 5 with infinite receiver and no clearance and compression. Construct the T-S diagram for the same cards and show whether or not it would pay to reduce the back pressure to 1 Ib. absolute. Problem 7. Find the sizes of cylinders to be used with a compound engine developing 1000 h.p. if the steam consumption of the engine is 15 Ibs. per 1 h.p.-hr. when 30 per cent, of the steam is removed from the receiver and the engines develop equal works. The limits of pressure are 150 Ibs. gauge and 3 Ibs. absolute. Problem 8. Find the thermal efficiency of a triple expansion regenerative engine with gauge pressures as follows: Initial high, 170 Ibs.; first receiver, 65 Ibs. ; second receiver, 5 Ibs. ; end of expansion, 8 Ibs. ; back pressure, - 12 Ibs. 264 HEAT ENGINEERING Problem 9. Design a cylinder to deliver 120 i.h.p. at 100 r.p.m. with pi = 130 Ibs. gauge, p v 3 Ibs. gauge, r = 5. Find the probable x at cut- off. Find the probable steam consumption. Find the pressure of SO 2 evaporated by the heat of the exhaust. Find the back pressure in an engine working with this S0 2 if the condensing water is 70 F. Sketch a card for the SO 2 cylinder. Find the volume of SO 2 vapor per minute at release and from this find the horse-power of the SO 2 cylinder and its size at 100 r.p.m. What is the thermal efficiency of the combined engine? At for heat trans- mission is 15 F. CHAPTER VII STEAM NOZZLES, INJECTORS, STEAM TURBINES NOZZLES In Chapter I the formula for the velocity of discharge of any fluid through a passage from pressure p\ to pressure p% was de- rived. This gave (i) ii and iz are the heat contents at the two pressures regardless of whether or not there is internal friction but dependent only on the fact that there is no external heat transfer. It was pointed out that iz for any pressure was difficult to find if there was in- ternal friction and hence the ordinary method was to find the value of iz on a reversible adiabatic (which means no friction) through pi ii, and to subtract from the heat (ii iz), which should have been transformed into kinetic energy, the amount utilized in friction. This gives the amount remaining for change in kinetic energy. If y(ii iz) is the amount of energy used in internal friction against the sides of the passage and between the particles of the fluid, the amount left for the change of kinetic energy is (ii - 12) (1 - y) (2) This quantity represents the change in kinetic energy and gives w* = 223.7-v/fo - 12) (1 - y) (3) Wi ^ when i0i is so small that ^ may be neglected or I Aw 2 .7-^ fe - i 2 )(l - y) + ~- wz = 223. when wi is appreciable. If Wi is 200 ft. per second it is inap- preciable as ^ 1 is 0.8 while the first term of the bracket may be from 50 to 100 or even 250 B.t.u. The importance of this term is determined by the conditions of any problem. 265 266 HEAT ENGINEERING The value of y is determined by experiment. w 2 is measured by the reaction of the jet and weight of the steam or by a pitot tube and from this measured value of w (17) The area in square feet required at this point is given by 3600 (18) m m Aw' d A = relative density = 0.8 This area F d of the combining tube fixes the number of the injector as most manufacturers use the diameter of this point in millimeters as the nominal size or number of the injector. Kneass points out that the ratio -=r has a value between two and f d three. The shape of the delivery tube is made in various forms. It has been suggested to reverse the form advised by Nagle for hose ^l nozzles. Nagle suggested that Length f. ne acceleration be made uniform. FIG. 134. Shape for delivery tube. If the velocity w' d is to be reduced to Wi in a given length the accel- eration is given by o o :i wd Wb wd time length length (19) Wd + Wb STEAM NOZZLES, INJECTORS, STEAM TURBINES 283 If this occurs in a length 20d t , which is a length often used for the delivery tube, the following results: a = (20) The velocity at any point at distance x from the throat is given by the equation w'd 2 w x 2 w'd 2 But Hence x fi - Wb * 1 - ^*i " 20dtl ~ w' d z \ ~ d x * (21) d x = Now in general w b will be equal to about 4 ft. per second and 2 "I w\ about 140 ft. per second so that j-^ = . This is negli- gible. Hence dx = . dt (22) \ l ~m The solution of this is given by the following table. X 1 2 3 4 8 12 16 19 19.75 d t d x 'dt' 1.013 1.027 1.040 1.058 1.138 1.254 1.496 2.118 3.00 DENSITY The value of the density of a jet is found by obtaining the temperature of the discharge when the injector will just waste and 284 HEAT ENGINEERING the weight of discharge in a given time. If then the theoretical amount in this time and with the pressure be divided into the actual amount of water handled, the square of this will give the density. If for instance an injector lifting 16,600 Ibs. of water at 148 F. will just discharge this quantity at 160 Ibs. gauge pressure when the area at the throat is 0.00054 sq. ft. the density may be found thus: The weight of 1 cu. ft. of water at 148 F. is 61.22 Ibs. Hence the head corresponding to 160 Ibs. with a density of A is 160 X 144 376.6 61.22A A ^6 ft. 'jz (23) Weight discharged = 16,600 = -=- X 0.00054 X 3600 X 61.22A IMPACT COEFFICIENT The value of k could be worked out after knowing A and find- ing the various velocities if z is determined experimentally. This has been done and the values found vary from 0.25 to 0.60. INJECTOR DETAILS The value of z is fixed by the various velocities and on account of the high velocity of steam acquired by discharges at even low pressures, exhaust steam could be used for feeding boilers or boiler steam could be used to force water into a chamber at a pressure much higher than boiler pressure. Thus an injector could be used for the hydrostatic test of a boiler. The injector shown in Fig. 131 is a single jet injector although double jets are sometimes used. The first jet lifts the water and forces it into the second nozzle to be driven into the boiler. A lifting injector is one which will lift its supply and drive it while a non-lifting injector is one which will only force water. In injector testing the term maximum capacity means the greatest quantity of water which could be sent through the injector with a given steam pressure and temperature of feed, while the minimum capacity is the least that will be discharged STEAM NOZZLES, INJECTORS, STEAM TURBINES 285 without waste. The range of capacity is the difference of these expressed as a percentage of the maximum capacity. The over- flowing temperature is the highest temperature of operation with- out overflowing when working at a given pressure. The highest pressure obtainable without wasting is called the overflowing pressure. Injectors can work as pumps when operating with other fluids than steam, such as water. In such apparatus water under a great head acquires a velocity so high that it can lift several times its own quantity through* a smaller height. The principle and equation of momentum would be the same for this apparatus. An injector operated by water is known as a jet pump or siphon. To apply the principles and equations above, suppose it is desired to have an injector pump 6000 Ibs. of water per hour at 68 F. into a boiler at 150.1 Ibs. gauge pressure. The steam has a quality of 0.96. Assume no pressure on the suction and a lift of 6 ft. p t = 0.57 X 164.8 = 93.8 w t = 223.7V1160.8 - 1116 = 1495 ft. per sec. w m = 223.7V(1 160.8 - 991.1)0.9 = 2760 ft. per sec. w c = 223.7\/(1160.8 - 917.2)0.9 = 3310 ft. per sec. v t = 4.34 cu. ft. v m = 22.86 cu. ft. (for i = 991.1 + 17.0 = 1008.1 at 14.7 Ibs.). w w = 8.02VM[2.3 X ( - 4) + 34 - 6 = 20.1 ft. per sec. w - 8 02 /(164.8 - 4)144~T~4 2 Wm 8 ' 2 V 60 X 0.8 ' + 60 " 1/2(3310 + 220.1) = (1 + 2)175.5 3310 -351.0 _ Z 351.0 - 20.1 ~ *' yi) 1160.8 + 8.95 [36.1 + ^ X g] - 9.95^ + ^ X 1160.8 + 323 + 0.072 = 9.95g' m + 6.17 q' m = 148.5 t m = 180 F. m m = 61.0 The value of m m A of 60 X 0.8 will not have to be changed as this is as close as the value of A will warrant. It will be seen 286 HEAT ENGINEERING that the expressions for the kinetic energies are so small that they may be omitted. 6000 Mass of steam per sec. = Q ^AA \/ Q nc = 0.187 Ibs. X o.yo Mass of water in suction per sec. = 1.66 Ibs. Mass of mixture per sec. = 1.847 Ibs. Pl . - 4 . = 000543 sq ft 0187X2286 1 847 = 0.000224 sq. ft. 169.6 X61.0X 0.8 0.000543 . . F d 0.000224 This is between 2 and 3. 0.7854 The size is therefore No. 5. d t = 0.314 in. d m = 0.60 in. 00224 >< 144 =0.202 in. =5 mm. FIG. 135. Shape for injector tubes. The table for the delivery tube is as follows: x = 0.202 0.404 0.606 0.808 1.616 2.424 3.232 3.838 d x = 0.202 0.204 0.237 0.205 0.207 0.228 0.254 0.302 0.443 The length of the delivery tube is 3.8 in. The length of the combining tube is 18 X 0.202 = 3.7 in. The length of the entrance to nozzle = 2^ X 0.314 = 0.78 in. The length of the diverging part of nozzle = 4 X 0.312 = 1.25 in. This is shown in Fig. 135. STEAM NOZZLES, INJECTORS, STEAM TURBINES 287 STEAM TURBINES The steam turbine is operated by the force exerted when a steam jet strikes against moving blades. If a jet of a certain cross-section, discharging m Ibs. of steam per second at a velocity of w a strikes a blade which is moving in the direction of the jet with a velocity w b} the steam is reduced to this velocity in the direction of the jet. The acceleration of this substance is (24) if it is assumed that this action takes place in At seconds. During this time the amount of mass acted upon is mAt. For although all of the fluid leaving the jet would never strike the one vane considered, in all forms of apparatus there are other vanes which come into line so that the amount to be considered on the one vane is the mAt which strikes this vane (if desired At would depend on the number of vanes passing the jet per second). The force exerted due to this decrease of velocity is given by the general formula P = ma (25) W a Wb f ^ 77- = m(w a Wb) FIG. 136. Jet impinging on vanes. This expression is in absolute units of force, poundals in the English system or dynes in the French system. To change it to pounds or grams the expression must be divided by g giving m(w a - w p ) If w b is made zero this becomes mw c Q (26) (27) In this latter case there is no work done as the blade is sta- 288 HEAT ENGINEERING tionary. The force is called the force of impact or the impulse of the jet. The nozzle of the j et must feel a force equal to this as some force must be required to produce this flow. This is called the reaction of the jet. The word impulse refers to the force exerted "by the jet when it strikes a body while reaction refers to the force on the body from which the jet issues in virtue of which the jet exists. Since in the case of the stationary blade there is no work done, there is no abstraction of energy and the velocity of the steam should not decrease. An investigation made on water jets showed this to be true. Although the velocity in the original direction is destroyed, the value of velocity of the water was not changed; its velocity in all directions along the plane being equal to the original velocity. FIG. 137. Jet impinging on vane at FIG. 138. Actual and relative angle with path of motion. velocities of the stream over a mov- ing blade. Entrance and exit tri- angles. In equation (26) the numerator represents the change of ve- locity in the direction of motion of the blade. If the plane of movement of the moving blade is not that of the nozzle so that the actual velocity w a is inclined at the angle a to the direc- tion of the motion of the blade Wb, equation (26) becomes P = -(w a cos a - w b ) (28) If in addition to this the substance is thrown off from the blade with an actual velocity w' a inclined at an angle a, then the change in velocity in the direction of motion as shown in Fig. 138, would be W a COS a W'a COS a and hence the force would be P (w a cos a w' a cos a') (29) STEAM NOZZLES, INJECTORS, STEAM TURBINES 289 If there were no motion of the blade and no friction, w a would equal w' a and the force P would be the force exerted in the direc- tion marked w b . If however there is motion of the blade, w a doe's not equal w' a and there is a relation between these and the angles of the nozzles and blades if the best conditions prevail. If w a is the actual velocity from the jet in space and this is directed toward a blade moving with a velocity w b , the motion of the fluid relative to the blade w r is found as the component of w a by the triangle of velocities if w b is the other component. This means that to one standing on the blade and moving with it, the jet would appear to come in the direction w r at the angle @ to the direction of motion; hence if the blade is to receive this stream without impact or shock it should be tangent to this direction. If now the stream is conducted over the blade to the outlet side and sent off relative to the blade at an angle /?' to the direction of motion and with a relative velocity w' T the actual velocity of discharge in space as the resultant of w' r and w b will be found to be w' a at the angle a. Of course it will be seen that as the blade moves away from the nozzle, the jet will not impinge on it but it must be remembered that another blade or vane will be moved up to take the place of the one shown in the figure. The work done by the force P is equal to the force multiplied by the distance moved through in the direction of P or Work = P X distance (30) Distance = w b X M (31) Work = (w a cos a - w f a cos cf)(w b X AO (32) i? but m&t = M M .', Work for M Ibs. = (w a cos a w' a cos a')w b (33) I/ This is the work for M Ib. in any time. If M is the amount per second this will give the work per second. The work may be separated into two parts: M M , w a cos aWb and w a cos a w b and may be said to be equal to the work done at entrance by reducing the velocity to zero minus the work done at exit in giving the fluid its actual discharge velocity. 19 290 HEAT ENGINEERING In other words the reaction of the discharge jet at exit mul- tiplied by the velocity of the blade subtracted from the impulse of the jet at entrance multiplied by the velocity of the blade at entrance is equal to the work per second. This method of statement in two parts is necessary when a turbine is built with radial flow in which Wb is not the same at entrance and at exit as shown in Fig. 139. In this case work = [w b w a cos a w' b w' a cos a'} (34) y In most cases steam turbines are of the axial flow type in which the steam flows across from inlet to outlet in a direction Velocity of Whirl _J Entrance o ^Velocity of Whirl Exit FIG. 139. Radial flow turbine. Veloc- ities w b and w' b are different. FIG. 140. Triangles of flow at entrance and discharge. parallel to the axis and not in a radial direction. Hence w b = w' b . Equation (33) shows that the work per pound of fluid is work per pound = - (w a cos a w' a cos a)w b (35) y Wb cos a and w' b cos a are the components of the actual velocity of the jets in the direction of motion and are called the velocities of whirl. Hence the work per pound is the difference on j- in the velocities of whirl at inlet and outlet multiplied by . y It will be important to remember that if the velocity of whirl at exit from the blade is in the opposite direction to that at en- trance that this difference is really an arithmetic sum since the STEAM NOZZLES, INJECTORS, STEAM TURBINES 291 sign at exit is minus. This is shown clearly by the value of a. In all cases the angle is measured from Wb when the arrows point toward or away from the vertex for w b and w a , or w b and w r . The graphical parts of Fig. 138 can be redrawn in Fig. 140. In this figure w r may be greater or less than w' r or equal to it. If there is no drop in pressure across the vane and if there is no friction W r W' r If there is no drop in pressure and if there is friction w r is greater than w f r while if there is a drop in pressure there would be an increase in velocity due to the addition of heat energy. Thus vf r = ^fcft - + [l - y] (36) This formula would give w' r if there is a drop of pressure and friction. Turbines in which there is a drop in pressure across the moving blade are called reaction turbines or pressure tur- bines while those in which there are no changes in the pressures as the steam passes over the moving blades are known as impulse turbines or velocity turbines. These names do not mean that impulse takes place in one form and reaction in the other. Reaction and impulse are present in each. These are only names borrowed from hydraulic turbines which have the meanings at- tached above. For impulse turbines with friction (37) (37') Now y depends on the velocity of the steam over the blades. The curves of Fig. 141 have been constructed from data given by Moyer, for stationary and movable blades. S refers to the first and M refers to the second. In Fig. 140 there is no necessary relation between a, /? and $' although a is fixed by the velocities and the three angles. The following trigonometric relations must hold between the triangles of entrance and exit: w a sin a /QQ . tan j8 = - (38) Wb w r = w r \/l ^y = fw r 292 HEAT ENGINEERING w r = w b w a sin a sin ft w a cos a w r cos ft W' r = W r \/l y = fw r ' (39) (40) (41) (42) (43) W' r COS ft' = W' a COS OL W b (40') Thus if w a , a, ft and ft' are given in a problem the above equations must be satisfied if there is to be no shock and the equations give the values of the other quantities. A graphical solution is close enough in most problems and the construction W' a = VVr 2 H- W b 2 + 2WbW' r COS ft' sin a = 0.40 0.30 0.20 0.10 M 500 1000 1500 2000 Velocity Relative to Blade 2500 FKJ. 141. Values of y from data given by Moyer. of the triangles of Fig. 140 will give the results of the above equations. To save space the lower triangle of Fig. 140 is turned through 180 and placed so that its apex agrees with that of the upper triangle giving Fig. 142. In many turbines ft and ft' are made supplements of each other giving Fig. 143. In this figure a represents the velocity of whirl at entrance, b represents in a reversed direction the velocity of whirl at exit and c represents the velocity of the blade. Hence to the scale of the figure, the work per pound of steam is c ( a ~^~ ^ Q Q (44) STEAM NOZZLES, INJECTORS, STEAM TURBINES 293 The energy in the steam, as it strikes the blade, is I ..' (45) since d represents the velocity of the jet leaving the nozzle. The efficiency of application of this jet to the blade is spoken of as the kinetic efficiency and is equal to 2c(a + b) * = ~dT In this expression the scale of the figure need not be known as this is a relation between the lengths. FIG. 142. Diagram for in- let and outlet triangles with vertices at same point. Fric- tion on blade. FIG. 143. Triangles with 0' the sup- plement of 0. Friction on blade. MAXIMUM EFFICIENCY If w b or c of Fig. 143 is increased, the quantity a would re- main the same although b would be decreased, so that it might increase the product if c were increased. In any event there would be a change and there must be a velocity which would give the maximum work. To find this there are two cases to consider, first if a is fixed with /? = 180 /?', and second if j8 and 0' are fixed. There is no need of considering the actual value of w a as this is only a matter of scale. Of course it must be of fixed value in the discussion, whatever that value is. work = [w a cos a w f a cos a] W' a COS a 'r COS j8' + work = [w a cos a (w' r cos @' + cos |S = - cos /3'; w' r = fw r (35) (40') (41) 294 HEAT ENGINEERING Hence where Now Hence work = [w a cos a -f- fw r cos /3 t? / = vi - y W r COS /? = W a COS a W work = [(1 + f) {w a cos a u (40) (47) In this the only variable is Wb, hence the work is a maximum for the value of Wb given by equating the first derivative to zero. d work 1 + / r = = - -[W a COS a 2w b ] j dw b g = M W a COS OL (48) FIG. a. No friction. FIG. 6. Friction. FIG. 144. Triangles of discharge for maximum efficiency without friction and with friction. This means that the velocity of the blades should be one- half the velocity of whirl. If / = or there is no friction, this same fact is true. The best result occurs when the wheel is moving at one-half the velocity of whirl. If there is no friction this will cause the absolute velocity of outflow to be perpendicular to the motion of the wheel, while if there is friction there will be a discharge at an angle to this perpendicular as is shown in Fig. 144, a and b. The work in this case becomes: work = cos 2 a - }i w a z cos 2 a] cos 2 a (49) The kinetic efficiency is: work %= ~^7~ 2? COS 2 a (50) STEAM NOZZLES, INJECTORS, STEAM TURBINES 295 If there is no friction / = 1 and this becomes : vj k = COS 2 a (51) The other case to be considered is that in which and 0' are fixed and a may be varied. W h Work = (w a cos a w' a cos a') (35) y Wh = (w r cos ft + w b w' r cos 0' w b ) tj = w r (cos ft - f cos 00 (52) Now cos / cos 0' is a fixed quantity and w b w r are variable quantities. The efficiency in this case is: work _ 2wbW r (cos / cos 0Q ^"""^f Wa 2 (53) >6 2 + w r 2 + 2w b w r cos (420 2tlW r (c080 /C0800 /C , N (54) R (55) COS j or calling 2(cos /3 - / cos 00 = fc'and + ^ + 2 cos = + ^ + 2 cos j8 ) R = I .'. w r = w b (56) and the triangle at entrance is isosceles. = 2a Hence the work becomes: Wh^ work = (cos j8 / cos 00 296 HEAT ENGINEERING ,. T w a sin a w ( Now w b Hence work = sin 2a 2 cos a a 2 (cos )8 / cos 1 + cos Since 4 cos 2 = 2(1 + cos 0) (Q f Qt\ cos p j cos p ; for = 180 - $ and / = 1 (58) One other problem of maximum efficiency should be con- sidered. Suppose that it is desired to operate a given blade at a velocity w bj what should be the angle a and the velocity w a to give the greatest efficiency? From (52) : work = ^- r (cos - / cos 00 (52) t/ In this 1% 0, / and 0' are known and hence the maximum work would occur when w r is as large as it can be made. Since residual energy w' a also increases, the efficiency may not be in- creased. This problem therefore depends on efficiency. work _ 2wbW r (cos (3 f cos ffQ _ , w r ~' ~~ (59) _ _ , ~'' ~~ Now w a 2 = w b 2 + w r 2 + 2w b w r cos (420 This is a maximum when -^ = 0; or fc(^ 6 2 + w r 2 + 2^ 6 w r cos 0) = fc(2w r 2 + 2^ 6 if r cos dlfr w; r 2 = w b 2 (60) or the triangle at entrance is isosceles. This fixes w a by the formula above w a 2 = 2w b z (1 + cos/3), w b (l w " - STEAM NOZZLES, INJECTORS, STEAM TURBINES 297 This is in reality the same proof as the former one since if for a fixed Wb and blade the triangle is isosceles giving w a , for a fixed w a and blade the isosceles triangle should be the most efficient. The three cases have proven that where the angle a is fixed the turbine blade must travel at one -hah* the velocity of whirl, whether there be friction or not, and that if the angles ft and ft' are fixed the best efficiency with either w a fixed or w b fixed is obtained when the inflow triangle is isosceles. This holds whether there be friction or not. In the first case without fric- tion, the outflow will be perpendicular to the movement of the blade while in all other cases the line may be inclined slightly. If, however, the speed of the blade is desired and the angle a is fixed, then the highest efficiency is obtained if the velocity of whirl is made twice the speed of the blade. This fixes w a . IstMoveable 1 5 FIG. 145. Velocity compounding. In the above figures it has been seen that when the steam is used on one blade the velocity of this blade has to be equal to one-half the velocity of whirl if a is fixed and ft and 180 ft' are fixed. This means that the velocity of the wheel would be very high. To reduce this, the velocity of whirl could be decreased by decreasing the velocity from the nozzle. This is obtained by a small drop in pressure in the nozzle. If steam is to be used through a large difference in pressure this would have to be utilized in a series of nozzles and blades. This is called pressure compounding. Another method is to use a series of movable and fixed blades as shown in Fig. 145, utilizing the high kinetic energy of discharge from one blade in successive blades. This is known as velocity compounding. 298 HEAT ENGINEERING The velocity diagrams are combined in Fig. 146, in which = 180 - ff fa = 180 - 0'i In this the work per pound is: work = - \a + b + ai + bi\ FIG. 146. Combined diagram for velocities in a turbine with velocity compounding. The efficiency is: work 2c(a + b (62) FIG. 147. Velocity compounding to give the best efficiency. The value of w b to make this or the work a maximum is desired. It has been shown that the last stage to give a maximum re- sult should have a velocity of blade equal to one-half the velocity of whirl and hence the figure to give the best result should be in the form shown in Fig. 147, in which c is one-half of the velocity of whirl STEAM NOZZLES, INJECTORS, STEAM TURBINES 299 in the last stage. To find the value of c to properly fit value of a of the diagram, continue the lines until they intersect the lower line. The first intercept is c. The second one is 7, since the slanting line w r has been decreased to w' r = fw r . From similar triangles lower intercept w r c " fw r or intercept = 7 The second line is decreased in the same manner Hence the third intercept is h^j or H,j h-j , The fourth intercept is the same as the third. Hence (63) 256 2" 6" 6'" FIG. 148. Construction for determination of c for multistaying with friction. If there were three velocity stages this expression would be (64) To construct this so as to find c graphically: lay off any dis- tance 12, Fig. 148, which is called c; at right angles lay off the distance 13 equal to unity to any scale and 34 equal to /to this scale. Draw an indefinite line from 4 parallel to 12, pro- 300 HEAT ENGINEERING ject 2 on this line at 2' and draw 3-2' to 5. 1-5 is the distance f* (* j> Project 5 to 5', draw 35' to 6 and 16 is equal to ~- Hence [(1 - 2) + (1 - 5) + (1 - 6 ) + (1 - 6)] = These are laid off from 3 to 6"'and line 37 is laid off equal to a, the velocity of whirl. If 7 and the last 6'" are joined and 2" 8 is drawn parallel to 6"' 7 the distance 3 8 will be equal to the correct distance c for Fig. 146. This same construc- tion can be used for any number of stages. To make use of the same angle of inlet into the various steps, the angle a of the second fixed blade is changed from the value FIG. 149. Diagram in which 01 = a. Friction on blades. to a. This is shown by the diagram in Fig. 149 where the line has been swung up to the original direction. In this way ai is made the same as a and by this action a'i has been made larger than before and the efficiency has been increased. The various kinetic efficiencies for these arrangements of blades have been computed and result in the following: Fig. 144a f] k = 81 per cent. Fig. 1446 r)k = 78 per cent. Fig. 147 rjk = 73 per cent. Fig. 149 rj k = 84 per cent. These have been drawn with a = cos " 1 0.9. The axial thrust is due to the difference between the impulse at entrance to the moving blade and the reaction at outlet in the direction of the axis. The force per pound of steam per second is: STEAM NOZZLES, INJECTORS, STEAM TURBINES 301 p = - [w a sin a w' a sin a'], y (65) Without friction this difference is equal to zero as is seen in Fig. 144. If, however, there is friction this formula does not reduce to zero but to a positive quantity. In the case of reaction turbines there is a static difference of pressure between the two sides of the blades which means an axial pressure. The actual forms of turbines will now be examined. The simplest turbine is the DeLaval turbine. Pressure FIG. 150. FIG. 151. FIG. 150. Section through DeLaval rotor. FIG. 151. Sections of DeLaval turbine with curves of pressure and velocity. Fig. 150 shows a section of the DeLaval turbine. This is an impulse turbine in which velocity is generated in a single set of nozzles A attached to a steam chest. The velocity is utilized on one set of blades. To increase the power of the machine a number of nozzles are used. The angle of the nozzle relative to the plane of the blade is made as small as possible, as shown in Fig. 150, so that the efficiency which is proportional to cos 2 a is as large as possible. The peripheral speed of the wheel is very 302 HEAT ENGINEERING great. w a is equal to 3820 ft. per second if the drop in the nozzle is from 150.1 Ibs. gauge to 6 Ibs. absolute. The speed of the wheel for a value of / of 0.95 and cos a = 0.9 would be w b = J X 0.9 X 3820 = 1719 ft. per sec. This would mean 16,400 r.p.m. for a radius to the blades of 1 ft. The pressure drop for the axial distances of nozzle and blade and the absolute velocity changes are shown in Fig. 151. This figure gives a section through the axis and one parallel to the axis through the blades and nozzle. The Curtis turbine is shown in Fig. 152. In this steam enters the nozzle at A and is discharged against moving vanes. The FIG. 152. Section through horizontal Curtis turbine. discharge from these moving vanes is guided by stationary vanes to another set of movable vanes, and after discharge from these it is taken to another set of nozzles B and discharges into a second set of vanes. In the figure shown there are five sets of nozzles, A, B, C, Z>, and E, and to each of these there are two movable and one fixed set of vanes. The pressure drop takes place in five stages, and there is no drop in pressure over the blades. The exhaust space F is connected to the condenser. This action is better shown in Fig. 153 in which a section through the axis STEAM NOZZLES, INJECTORS, STEAM TURBINES 303 and one parallel to the axis of a two-stage turbine are shown side by side following the method used by Moyer. In the figure the pressure is seen to be constant across the sets of vanes, the drop in pressure and consequently the gain in velocity taking place in the nozzles. The turbine is therefore of the impulse type. In some cases the nozzles which are arranged in sets extending Pressure Absolute Velocity Axial Length Section Perpendicular to Radius through Blades FIG. 153. Sections of Curtis turbine with curves of pressure and velocity. over a portion of the circumference are separated into two sets on diametrically opposite parts of the circumference so as to balance the forces in an axial direction. The axial thrust must be balanced by some form of thrust bearing G. An auxiliary valve is sometimes used to admit live steam into a lower stage under heavy loads. In this turbine holes are made through the 304 HEAT ENGINEERING discs to insure the same pressure throughout the moving elements of one stage. For the Rateau turbine the diagrammatic arrangement of parts is shown in Fig. 154. In this turbine there is only one set of blades to each set of nozzles and there is a drop of pressure in each fixed member. There is no change of pressure across the Pressure Absolute Velocity Axial Length A Section Perpendicular to Radius through Blades FIG. 154. Sections of Rateau turbine with curves of pressure and velocity. movable blades so that this turbine is also of the impulse type. As will be seen later the area through which the steam passes as the pressure falls must increase since the velocity relative to the blades is about the same in each set but the specific volume increases. Hence the length of the blades increases. STEAM NOZZLES, INJECTORS, STEAM TURBINES 305 In all of these the steam is admitted to a portion of the circum- ference and as the steam pressure falls the portion of the cir- cumference covered is greater so as to give greater area for the steam. The position of the successive nozzles will be advanced in the direction of flow due to the advance of the steam as it passes over the moving blades. This is called lead. The Parsons turbine is shown in Fig. 155 and diagrammatically in Fig. 156. In this steam is admitted around the complete circumference to a set of fixed blades. These discharge on the movable set and these in turn to a fixed set from which the action is repeated. In each of these sets of fixed or movable blades there is a pressure drop on all blades and hence this type will be called reaction type of turbine. FIG. 155. Parsons turbine. In this turbine the axial thrust is balanced by connecting balancing drums of proper area to the proper parts of the tur- bine and connecting the back of the last drum to the space lead- ing to the condenser. A thrust bearing is used to ensure align- ment. In the type of turbine shown in Fig. 155 a double flow arrangement through the center A balances most of the thrust. To make the turbine more efficient the Curtis element C has been used for the first reception of the steam. The high-pressure steam is discharged from nozzles attached to the steam chest B. The steam from the movable blades C then passes to the fixed and movable blades at D and finally is discharged at E. At this point the steam divides into two parts, one to blades / and the condenser connection F and the other enters the space A at the center of the drum A through holes and thence to holes to the space from which the steam enters a set of blades and finally issues into the condenser from the space H. This rep- 20 306 HEAT ENGINEERING resents one of the more recent forms of Westinghouse Parsons turbine. The blades are mounted on drums A, I and J of different sizes to allow for the changes in volume of the steam. The main steam valve of the ordinary form of Parsons turbine admits steam in puffs controlled by the governor; while under very heavy load the governor begins to operate on a second Pressure Absolute Telocity Axial Length ^ \J ^ ? 1 I 1 1 I 1 I Section Perpendicular to Kuilius through JJlades FIG. 156. Sections of Parsons turbine with curves of pressure and velocity. valve admitting extra steam to the second stage of the turbine. The end thrust, although reduced by the two turbines placed on the same shaft in which the steam passes to right and left, may exist and for that reason a small thrust bearing is used to ensure alignment. The air leakage around the shaft into the steam space which is at the pressure of the condenser is prevented by STEAM NOZZLES, INJECTORS, STEAM TURBINES 307 some form of steam stuffing box. Fig. 156 shows how the pres- sure drops on both movable and fixed blades although the absolute velocity decreases over the movable blade. The decrease of absolute velocity would have been greater on the movable blade were there no drop of pressure here. The great disadvantage of the DeLaval turbine is the high blade speed and the resulting complications. For structural reasons the maximum output of this type is limited to about 300 kw. The absence of packing and the simplicity of the machine do not make up for this great disadvantage. Between the reaction and the other impulse turbines there are advantages on both sides. The reaction turbines are very inefficient on the high-pressure stages, due to the leakage at the ends of the short blades; while in the impulse turbine packing around the shaft between the discs separating the stages is necessary. The com- bination turbine used by the Westinghouse Company in their use of a Curtis impulse stage for the first utilization of the steam and then the use of the reaction blading after a reduction of pressure has been made to combine the advantages of each type. EFFICIENCY The various efficiencies of the turbine may be computed. The nozzle turns the heat (i\ 2 2 )(1 y) into kinetic energy and ii q' heat units have been required per pound. Hence the overall nozzle efficiency is = (ii - z' 2 )(l - y) In some cases there is a series of nozzles in a turbine and if ?i, 92, #3, etc., are the amounts of heat turned into work the average nozzle efficiency would be I i i gl + g2 + 0.23 = ORS/i 561 ' 8 \ " ' 68 X ' 324 " ' 68 I 1 ~ 83L2/ The Mollier chart repared by Marks and Davis will be used for this problem: Pi = 189.7 superheat = 125 F. si = 1.646 '^ = 1297 p c = 0.98 quality = 0.82 s c = 1.646 i c = 918 7? fT il ~ ic 1 04 12 9? ~ 918 K.H. -^ = 1.U4 - - = 95.7 = ^l 2 Reheat = (1 - 0.68) X 98.7 = 31.6 B.t.u. z* 2 = 1297 - 98.7 = 1198.3 iv = 1198.3 + 31.6 = 1229.9 iv = 1229.9 - 98.7 = 1131.2 ii = 1131.2 + 31.6 = 1162.8 iv = 1162.8 - 98.7 = 1064.1 p 2 = 75 Superheat = 32 F. Pi> =75 s = 1.684 superheat = 95 F. p z > =22 x = 0.97 Pi = 22 s = 1.728 superheat 6 F. p 2 " = 5.3 x = 0.93 ii> = 1065.9 + 31.6 = 1095.7 pv = 5.3 s = 1.781 x = 0.965 iv = 1095.5 - 98.7 = 996.8 p 2 '" = 1.0 s = 1.781 x = 0.895 318 HEAT ENGINEERING The final pressure desired was 0.98 Ibs. and this result is as close as can be expected. The value of ii i 2 is correct and w a may now be found. w a = 223.7V98.7 X 0.94 = 2146 ft. per sec. This holds for each nozzle. The thermal efficiency of the nozzles and blades is therefore: iy.(*'i - )n 0.68 X 4 X 98.7 V i, - q > ^1297-69.2 The overall efficiency is found by assuming* nw = 1 - (0.01 + 0.02) = 0.97 rim = 1 - (0.07 + 0.01) = 0.92 rj e = 0.96 rj t = 0.219 X 0.97 X 0.92 X 0.96 = 0.188 = 18.8 per cent. The probable steam per kw.-hr. = 188^1227 8 = 14 ' 8 ' The actual amount guaranteed by the builders was 15.75 Ibs. per kilowatt hour. Total steam for 5000 kw. = 5000 X 14.8 = 74,000 Ibs. per hour. Steam per second = 20.55 Ibs. The nozzles must be investigated for the presence of the throat. Pti = 0.57 X 189.7 = 108 Ptz = 0.57 X 75 = 42.8 p t3 = 0.57 X 22 = 12.55 pn = 0.57 X 5.3 = 3.02 There is a throat in each nozzle and i at these pressures is found from the chart of Marks and Davis. t(for 108.0 Ibs. s = 1.646) = 1230, 104 superheat = quality * (for 42.8 Ibs. s = 1.684) =1180, 42 superheat = quality i(for 12.55 Ibs. s = 1.728) = 1118, 0.97 = quality (for 3.02 Ibs. s = 1.781) = 1059, 0.94 = quality w t = 223.7\/1297 - 1230 = 1830 ft. per sec. tiV = 223.7\/1229.9 - 1180 = 1580 ft. per sec. w t " = 223.7-\/l 162.6 - 1118 = 1493 ft. per sec. w v ,, = 223.7\/1095.7 - 1059 = 1350 ft. per sec. STEAM NOZZLES, INJECTORS, STEAM TURBINES 319 The values of the specific volumes are given below: v t = 4.99 cu. ft. vtf = 10.56 cu. ft. Vtf = 30.2 cu. ft. vtf" = 110.0 cu. ft. v m = 6.24 cu. ft. for p = 75, i = 1297 - 0.94 X 98.7 = 1204 tw = 18.0 cu. ft. for p = 22, i = 1229.9 - 93 = 1136.9 tw = 64.8 cu. ft. for p = 5.3, i = 1162.2 - 93 = 1069.2 v m > = 307 cu. ft. for p = 0.98, i = 1095.5 - 93 = 1002.5 The specific volumes of the steam leaving the blades is equal to the specific volume for the pressures at these points and for the heat content of exit plus the sum of the friction loss in the nozzles and on the blades. The friction loss of the blades plus the residual kinetic energy is equal to (1-iy) or friction equals 1 77 r.e. W r ' 2 where r.e. is equal to s- w a * ID ^ / ^\ ^ From Fig. 158 ~T * (4) = - 065 W a \ df Friction loss on blades = 1 - 0.722 - 0.065 = 0.213 Total friction effect = 0.06 + 0.94 X 0.213 = 0.260 Heat from friction per stage = 98.7 X 0.260 = 24.7 B.t.u. v = 6.51 for p = 75, i = 1198.3 + 24.7 = 1223.0 v* = 18.30 for p = 22, i = 1131.2 + 24.7 = 1155.9 tV' = 65.5 for p = 5.3, i = 1063.9 + 24.7 = 1088.6 ?V" = 303.4 for p = 0.98, i = 996.8 + 24.7 = 1021.5 From Fig. 156 w r > = 4^ X 2146 = 690 ft. per sec. The areas then at the various points in square inches are given as follows: 20.55 X 4.99 X 144 ' = " 183Q " = 8.06 sq. in. 20.55 X 10.56 X 144 = 19 ' 7 m ' 29.55 X 30.2 X 144 '" = " 1493 - = 59.70 sq. in. 320 HEAT ENGINEERING 20.55 X 110.0 X 144 F *" = ~~ = 24L 20.55 X 6.24 X 144 Fm = "2146" = 8 ' 58 Sq * m ' 20.55 X 18.0 X 144 Fm ' = ~ 2146 " = 24 ' 7 Sq * m ' 20.55 X 64.8 X 144 Fm " = 2146 = 89 ' Sq ' m * 20.55 X 307.0 X 144 Fm '" = ' 2146 = 422 * sq ' m ' The outlet area from the last blade of each stage is given by : 20.55 X 6.51 X 144 F -- - 27.9 sq. in. 20.55 X 18.30 X 144 F > = 79.7 sq. in. 20.55 X 65.5 X 144 " = " 69Q ~ = 281-0 sq. m. 20.55 X 303.4 X 144 = ~ 690 ~ = 1300.0 sq. m. For the outlet area from the blade of each stage use specific volumes at about one-third the difference between the volumes for and m, using from Fig. 156, 3 49 w r' = T7^ X 2146 = 1515 ft. per sec. 4.yo 20.55 X 6.36 X 144 F> ~~ =12.4sq.m. 20.55 X 18.17 X 144 20.55 X 65.3 X 144 " = ~ 1515 ~ = 127 - 6 sq- m - 20.55 X 306.9 X 144 F o> = - -, K1 E - = 600.0 sq. in. lolo For the outlet from the fixed blades the velocities are each equal to 1040 ft. per second and the specific volumes may be taken as the mean of those in the last two cases, giving: STEAM NOZZLES, INJECTORS, STEAM TURBINES 321 F f = 18.4 sq. in. Ff = 52.0 sq. in. F f = 186.0 sq. in. Ff" = 870.0 sq. in. If the nozzles of the first stage be made of J^ sq. in. cross section of mouth there would be nine of these on each side of the turbine. These will give an area of 9 sq. in. where 8.58 have been required. If these are made % in. wide the length of each on the circular face will be: iz = 1.95 in. M sin 20 C These will take up about 30 in. of the circumference if there is allowance made for the partitions between each two nozzles. The throats of these nozzles will each be: H X g 1 ^ = 0.47 sq. in. or 0.75 X 0.63 in. For the second stage the mouth may be made 1 sq. in. in area requiring 13 on each side. If made 1 in. wide the length will be TTITT = 2.92 in. 19.7 The throat will contain 26 X 2477 = 20.8 sq. in. If this is 1 in. wide the depth will be 20 ' 8 26 X 0.34 =2.35 in. These will take up about 42 in. on each side. If the areas for the third stage are made 3 sq. in. each there will be sixteen necessary on each side and these may be worked out as before using 1% in. for the width. On the last stage the area of each might be made 9 sq. in. requiring 25 on each side and the width would be 4J^ in. These would take up about 160 in. on a side or 320 in. in the circumference. If this were to use the complete circumference the diameter of the turbine pitch circle would be 102 in. or 8 ft. 6 in. If this is considered too large the width might be made 6 in., this would require 92 in. diameter. The blades of the first 21 322 HEAT ENGINEERING movable set would be slightly higher than the width of the nozzle mouth. They will be determined by their area at discharge. These will have to be investigated from Fo = length X height X sin /3. Now F m = length X height X sin a. Hence since length is the same for nozzle and blades , . , , , Fo sin a helght = h ~ra^ 124 34 h = X = IX 04 7 X jyTT == 1-19 1Q = 1.75 X ~^- X ~ = 2.20 in. The heights of the fixed blades will have to be made longer than the first movable blades because although the specific volume increases slightly, due to friction, there is a marked decrease in axial velocity as seen from Fig. 158. For the outflow edge of the second movable blade the heights are given by: IX 2^7 X Q-yg = 1.45 in. oci o 04 1.75 X ~~ X ~* = 2.41 in. ^4- X X ^~ = 5.98 in. STEAM NOZZLES, INJECTORS, STEAM TURBINES 323 In this it is seen that the blades will increase from %-in. nozzle to 0.9-in. movable blade, 1.02-in. fixed blade and 1.07-in. second movable blade or, to allow for spreading, say % in., 1J^ in., \y in. and 1% in. These for the second stage would be 1 in., 1% in., 1J in. and 1% in.; for the third stage in., 2% in., 2^ in. and 2% in.,* and for the last stage in., 5J2 in., 6 in. and 6J^ in. The angles of the nozzles would be 20, the first movable blade would have angles of 24 at entrance, the fixed blade would have an angle of 33 and the last movable blade would be 52. The outlet angles are supplements of the above as the blades are all symmetrical. The speed of the wheel, from Fig. 158 is X 2146 = 433 ft. per sec. 4.96 The speed with a diameter of 102 in. would be 975 r.p.m. If 1000 r.p.m. be taken then D = 100 in. This is a possibility. The customary speed is 1500 r.p.m. showing that a smaller diameter is used. To use this diameter of 66 in., the heights of the vanes would have to be increased by 50 per cent. This would make the last blades 9% in. long. Rateau Turbine. In this turbine there are many pressure stages and only one velocity stage to each pressure stage. If this is the case and 20 is the value of a and the desired speed of blades is about 500 ft. per second, the velocity across the vanes will be 750 ft. per second which gives / = 0.94 14-0 Q4 17* = ~ ^ -cos 2 a = 0.97 X (0.94) 2 = 0.857 Vs = rjn X rj k = 0.94 X 0.857 = 0.805 Suppose the turbine is to work through the same range of pressures as the Curtis turbine above: 561. 8\ ' 805 /5(jl.8\ u> \831.27 0.805 561. 8\ 0.805X0.324 / _5bl^X \ 831.27 The heat required to give a velocity of 500 ft. with a = 20 C and / = 0.94 is found thus: 324 HEAT ENGINEERING 2w b 1000 -cos^ = 094 The available heat for this is 1065 ii - i* = 0.94 \j^) = 21.5 B.t.u. The amount required per stage is 21.5 1.035 = 20.4 XT u f 4 1297-918 .__ Number of stages = -- = 18 - 7 Use 20 stages. * _ i, = ^* X 1.035 = 19.7 w a = 223.7\/19.7 X 0.94 = 961 ft. per sec. 961 X 0.94 Wb = - o = 452 ft. per sec. a The heat returned at each stage = 19.7 X (I - 0.805) = 3.84 B.t.u. Heat used = 19.7 - 3.84 = 15.06 B.t.u. Actual thermal efficiency = j^y 69 2 = ' 245 Overall efficiency = 0.245 X 0.97 X 0.90 X 0.96 = 0.205 The steam consumption per kilowatt hour would be 041 o M - 0266x11371 ' 14 ' 75 lbs ' Parsons Turbine. In this turbine the reheat factor will have to be worked out from the blade efficiency of one set of blades. In these blades there is a drop in pressure on each set of blades whether they are movable or stationary. If there is the same amount of heat added on each set and if the velocity at entrance is the same into each set there will be the same ve- locity of exit. If for instance, a from the first set of fixed blades is equal to 20 and /3 of the movable blade is made equal to 75, the value of w& would be equal to w a sin a r sin a W k = W, COS a - - Hence for any desired value of w b , w a could be found. STEAM NOZZLES, INJECTORS, STEAM TURBINES 325 W r = w a sin a r = f sin/3 w a sin a sin/3 (73) (74) If now it is desired to use an angle 180 /8' = a for the angle of discharge and have the absolute direction of the jet at the angle a! = 180 /3 = 105, so that the same form of blades, although right and left handed, may be used for each stage, it will be necessary for ^ w' r actual = w a . Fixed Moveable FIG. 160. Blades of a Parsons turbine. Hence w r f above must be increased to w a or w a - - y) teu" = (223.7) ; //sina\ 2 ^ " V sin ft I ^.(Eg*)*] ~ * 2 = (1 - 2/X223.7) 2 - y) (75) (76) Hence if a, /8, and ty& are assumed, w a and ^ r may be found and from w r , f may be selected, then i\ it for the movable blade. 326 HEAT ENGINEERING For the fixed blade the steam enters with the velocity , w/ sin a w a sin a W a = ! - = = ^ = W r sin |8 sin /3 It leaves this fixed blade with the velocity w a , the same as that of the first blade after being affected by friction and gaining kinetic energy from heat. sn sin (1 - i/)(223.7) 2 This does not give quite the same value as the first expression on account of the difference in /. Now the work done on this stage per pound is work = [w a cos a w' a cos a] (77) y w b r i . sin a cos /3~| /r _ ON = w a cos all -f - o (78) g sin |8 cos a J The heat used is *i - i z + i\ - i \ = Q (79) . w b f, tana"! Hence A w a coso: l+i ^ rj s = - ^ _ + ^ _ ^ n -^ (80) For the angles mentioned above and Wb = 300 ft. per second: 300 300 Wa = ^ 0.34 ~ 0.849 *}4 r = 354 X ~** = 124 / = 1.00 354 " V 1 " [ L X 1-0.04 / for the fixed blade is the same as the value used above, hence i\ - i\ = 2.30 STEAM NOZZLES, INJECTORS, STEAM TURBINES 327 300X354 X 0.94 r , 0.361 Work per pound per stage = ~~32~16~~ L 3^73 J = 3400 ft.-lbs. = 4.38 B.t.u. 4.38 Stage efficiency = T^Q = 0.95 = rj a The reheat factor may now be applied in the usual manner for T? S = 0.95. This is, for the same pressures as before, - 311 '' 0.95 X 0324 Total heat available = (1297 - 918) X 1.013 = 384 B.t.u. Total number of stages if velocities are the same on each = 384 ^ = 83 stages. This answer is not given in correct form as it is customary to enlarge the drum as the pressure falls. If the first drum is of diameter Z>i, the second is often made Z> 2 = \/2 D! and D 3 = \/2 D 2 = 2D 1 The speed will be proportional to the diameters and hence if the above is assumed for the middle portion i*,- -212 w b3 = 300V2 = 424 4.6 (2*1 iz) for first stage is ~- = 2.3 a (ii - iz) for last stage is 4.6 X 2 = 9.2 379 Stages in first portion = .. = 55. o X *& 379 Stages in second portion = . ft = 27. o X 4.t) 379 Stages in third po-rtion = 9 n = 14. o x y.^ Total number = 96 stages. This is rather high. A common rule for the total number of stages is AT u f 4 2400000 /01 , Number of stages = - ^ (81) 328 HEAT ENGINEERING In case above: ,, 2400000 N first = - = 54 _ 2400000 M second - (30() ) 2 _ 2400000 N third ' (424) 2 The other quantities may be computed as shown in previous problems. ALLOWANCE FOR CHANGE OF RUNNING CONDITIONS Where turbines are actually tested and one condition is changed without changing other quantities the following allowances are found to hold: From 0-100 superheat, 10 F. increase in superheat decreases steam consumption 1 per cent. From 100-200 superheat, 12 F. increase in superheat decreases steam consumption 1 per cent. From 200-300 superheat, 14 F. increase in superheat decreases steam consumption 1 per cent. For each 1 per cent, moisture the steam consumption will be increased 2 per cent. For an increase in vacuum from 26 in. to 27 in. the steam con- sumption will decrease 5 per cent. For an increase in vacuum from 27 in. to 28 in. the steam con- sumption will decrease 6 per cent. For an increase in vacuum from 28 in. to 28% in. the steam consumption will decrease 3.87 per cent. For an increase in vacuum from 28% in. to 29 in. the steam consumption will decrease 5.75 per cent. For low-pressure turbines the following corrections may be made: Increase in vacuum from 26 in. to 27 in. decreases steam con- sumption 12 per cent. Increase in vacuum from 27 in. to 28 in. decreases steam con- sumption 13.75 per cent. Increase in vacuum from 28 in. to 28% in. decreases steam consumption 8.5 per cent. Increase in vacuum from 28% in. to 29 in. decreases steam consumption 11.25 per cent. STEAM NOZZLES, INJECTORS, STEAM TURBINES 329 For pressure, the increase of pressure of 10 per cent, between 100 and 140 Ibs. gauge pressure decreases steam consumption 2 per cent. The increase of pressure of 10 per cent, between 140 and 180 Ibs. gauge pressure decreases steam consumption 1.95 per cent. The increase of pressure of 10 per cent, between 180 and 200 Ibs. gauge pressure decreases steam consumption 1.90 per cent. For low-pressure turbines 10 per cent, increase in pressure decreases the steam consumption 4 per cent. In any case if the values of the theoretical thermal efficiency for two sets of conditions are computed the ratio of steam con- sumption may be assumed equal to the ratio of the efficiencies and the equivalent steam consumption thus found. COMBINED ENGINE AND TURBINE Since the turbine is of especial value for low pressures and high pressures lead to certain troubles, steam engines have been used to handle the steam first, after which the steam is exhausted into turbines for use at low pressures. The turbines using exhaust steam are known as low-pressure turbines. In these the exhaust steam, at atmospheric pressure from engines or other apparatus, is utilized by the turbine. This combination of engine and turbine has resulted in very low steam consumptions. If the exhaust is not sufficient at times to drive the turbine, a set of blades is placed in the turbine on which high-pressure steam may be used in a high-pressure part in addition to the low-pressure steam. This is known as a mixed flow turbine. TOPICS Topic 1. Derive the formula What is ?/? What is the meaning of i 2 in the above expression and what would it stand for if the bracket (ly) were omitted? How is y found? Topic 2. Explain how the area F x is found for different points of a nozzle with a uniform pressure drop. Explain why a throat exists. What is the critical pressure? Explain the terms: throat, mouth, over expansion, under expansion. What is the effect of the last two? Topic 3. Explain how the pressure in a tube at a given section may be found by experiment and how it may be calculated. Explain how this may be used to compute y. Sketch the form of apparatus used by Stodola and sketch the form of his curves. 330 HEAT ENGINEERING Topic 4. When are orifices or converging nozzles used? Sketch Rateau's forms. What fixes the length of a nozzle from entrance to throat and from throat to mouth? Give the steps taken in the design of a nozzle. Topic 6. Sketch and explain the action of an injector. Topic 6. Give the formulae for the following velocities : (a) at the throat of a steam nozzle of an injector; (6) at the mouth of a steam nozzle of an in- jector; (c) at a point within the combining tube for the steam; (d) at a point within the combining tube for the water; (e) at the throat of the delivery tube; (/) after impact in the combining tube for the mixture. Topic 7. Write the two equations on which the action of the injector is based. What do these equations determine? Topic 8. Derive the formula for the delivery tube: d t d x = Topic 9. Explain the action of a jet on a blade and prove the formulae p = mw g P = -(WaCOS Ct - Wb). y Topic 10. Explain the action of a jet on a curved blade. Sketch the inflow and outflow triangles and prove that work per pound = - (w a cos a w' a cos a)wb What is meant by velocity of whirl? What are impulse and reaction turbines? Topic 11. Sketch the inflow and outflow triangles and give all resulting trigonometric relations. Topic 12. Deduce the condition for maximum efficiency when a is fixed and ft = 180 - ft'. Topic 13. Deduce the condition for maximum efficiency with a fixed value of ft and ft'. Topic 14. Explain the meaning of pressure compounding and velocity compounding. Explain the construction of a one-stage and two-stage ve- locity diagram with friction and without friction. Explain how to find the work and kinetic efficiency from this. Topic 15. Explain by diagrams the peculiar features of the turbines given in text. Sketch the curves showing the variation of pressure and velocity through the turbine. Topic 16. Give the expressions for the various component efficiencies of a turbine and the expression for the overall efficiency. Explain these and the method of computing them. Explain the similarity of thermal action of an engine and steam turbine. Topic 17. Sketch a Mollier chart and on it show why a reheat factor exists and derive the expressions STEAM NOZZLES, INJECTORS, STEAM TURBINES 331 tan 5 = kT = - s Si tan 7 = - -) Topic 18. Outline the method of design of a turbine to give a definite power. PROBLEMS Problem 1. Find the velocity of discharge at throat and mouth of a nozzle operating with dry steam between 155 Ibs. gauge pressure and 70 Ibs. gauge pressure. Find the area of the nozzle to care for 30 Ibs. of steam per minute. Problem 2. Draw a curve representing the change of area from 2 sq. in. to 0.25 sq. in. in 0.5 in. and then an enlargement to 2 sq. in. in 3 in. Find the pressure along this nozzle if the initial pressure is 135 Ibs. absolute and the steam is superheated 230 F. Problem 3. Find the size and shape of a Rateau nozzle to deliver 2500 Ibs. of steam per hour from a pressure of 25 Ibs. gauge to a pressure of 15 Ibs. gauge. Xi = 0.99. Problem 4. Design an injector to feed 10,000 Ibs. of water per hour into a boiler under a gauge pressure of 220 Ibs. The lift is 5 ft. Make sketches of nozzle, combining tube and delivery tube. Problem 6. An injector operating with steam at 175 Ibs. gauge pressure is to force water into a boiler in which the gauge pressure is 275 Ibs. How much water is lifted per pound of steam ? Problem 6. Prove that it is possible to feed a boiler under 120 Ibs. gauge pressure by means of steam at 25 Ibs. absolute pressure. Problem 7. Find the kinetic efficiency of a single-stage, impulse turbine with cos a" 1 = 0.93 without friction and with friction. Find the efficiency if = 45 and 0' = 135. w a = 2700 ft. per sec. Problem 8. Find the kinetic efficiency of a two-stage impulse turbine with cos oT l =0.93. Use no friction for first assumption. w a = 2700 ft. per sec. Assume correct values of / to suit velocities and find the efficiency with friction. Problem 9. Solve Problem 8 assuming that a. = =~: w It happens that the ratio of the pressure exerted by this moisture compared with that at saturation is practically the same, since the pressure of a perfect gas is proportional to the mass when the volume and temperature are constant. Of course this is not a perfect gas and the statement is not absolutely true. . ' p = ' do) CONDENSERS, COOLING TOWERS AND EVAPORATORS 343 To find the relative humidity a hygrometer is used. The common form used in engineering work is that consisting of two thermometers, one of which has its bulb surrounded by a damp piece of wicking. As this is twirled at a high rate in the air the evaporation of the water on this bulb lowers its temperature, the amount of lowering being dependent on the relative humidity. The fall of temperature is then compared with the temperature of the dry bulb and the barometric reading; and from tables or formulae the relative humidity is found. Carrier proposes the formula p' Bar, -p' t-t' p t Pt 2755 - 1.28*' p' = pressure corresponding to wet bulb temperature p t = pressure corresponding to dry bulb temperature Bar. = barometric pressure t = temperature of dry bulb t f = temperature of wet bulb. If now the air enters the cooling tower at a temperature fa and of relative humidity pi the amount of moisture per cubic foot is where mi = weight of 1 cu. ft. of steam at temperature fa. One cubic foot of entering air is considered in problems of the cooling tower as it leads to simple calculations. If the temperature of the warm water is t i} the temperature fa of the air leaving may be taken as 10 to 20 below U but satu- rated on account of the intimate mixture. The water leaving the tower will probably be-above the temperature fa of the en- tering air although it might be equal to it in some cases. Assume fa to be 10 F. or 20 F. below the temperature t of the outlet water. With intimate mixture fa may correspond to t and fa to fa. The moisture per cubic foot leaving the tower is m 2 but the number of cubic feet have been changed on account of the changes of temperature and pressure. Let the weight of air be M. i, __ (bar. - pipOl447i (bar. - p 2 )1447 2 BTi BT 2 Bar. = barometric pressure in Ibs. per sq. in. Pi = relative humidity at 1 344 HEAT ENGINEERING Pi = saturation pressure at 1 TI = absolute temperature at 1 Vi = volume at 1 = 1 cu. ft. The volume of air at outlet corresponding to 1 cu. ft. at entrance is The weight of moisture evaporated per cubic foot of original air is m e = m 2 V 2 pirai (14) If M a = weight of cooling water cared for by 1 cu. ft. of air at entrance, the weight of water remaining is M a - m e The late Prof. H. W. Spangler solved this problem by equating energies at entrance and exit and hence it is necessary to find the energy brought in by each substance above 32 F. I. Energy in water at bottom of tower above 32 F. = (M a - m e ) q' (15) II. Energy in air at entrance above 32 F. = A (bar. - pi??i)144 X 1 TT III. Energy in moisture entering above 32 F. = piwi [ii - AP&'"] = pimiii - Api X 1 (17) i is heat content of superheated steam at pressure pipi and of superheat of ti t s degrees. IV. Energy in water at top of tower above 32 F. = M a q'i (18) V. Energy in air at exit above 32 F. = A (bar. - p 2 )144F 2 TJ no x 14 _ l ~ U M VI. Energy in moisture leaving above 32 F. = m 2 V 2 [q'z + r. 2 ] - Ap 2 V 2 or better U = ra 2 F 2 [^2 + p 2 ]. (20) VII. Work done by air in changing 1 cu. ft. to V 2 cu. ft. at barometric pressure = A bar. X 144 X (V 2 - 1) (21) CONDENSERS, COOLING TOWERS AND EVAPORATORS 345 Now the sum of the energies entering must equal the sum of those leaving plus the work: II + III + IV = 1+ V + VI + VII (22) In this equation the only unknown is M a and this may be found. From M a the number of cubic feet of air required for a given installation may be known and from this the size of the fan, power required, size of tower and other data may be ascertained. This is the best method for solving such a problem as other methods of attack are open to objections since the temperature at which events take place is not known. PROBLEM Suppose air at 70 gives a wet bulb temperature of 58.5 F. and is used in a cooling tower with hot water at 140 F. The barometer is 14.7 Ibs. The air is so mixed that it leaves at 140 F. while the leaving water is cooled to 80 F. How much air would be required to cool 30,000 Ibs. of water per hour. P58-5 = 0.2428 PTO = 0.3627 0.2428 14.7 - 0.2428 70 - 58.5 0.3627 0.3627 * 2755 - 1.28 X 58.5 = 0.668 - 0.171 = 0.497 = 0.5 mi = 0.001152 = 0.000576 m 2 = 0.00814 pz = 2.885 _ 14.7 - 0.5 X 0.3627 601 _ 14.7 - 2.885 X 531 ~ d9 m 2 7 2 = 0.00814 X 1.39 = 0.0113 m e = 0.0113 - 0.000576 = 0.0107 Energy in water at bottom = (M a - 0.0107)48.1 = 48.1M a - 0.515. Energy in air at entrance = == - ^-^ C7 32 = 6.70 C7 32 . Energy in moisture = 0.000576 [19.1 + 1061.8 + 19 X 0.43] - [ - X L778 0.1814 X 1441 = 0.600. (Moisture at entrance is under a pressure of 0.1814 Ib. and at a temperature of 70. 0.1814 Ib. means a saturation temperature of 51 or the moisture is 19 superheated. From the curves of specific heats in Chapter I, c p is 0.43.) 346 HEAT ENGINEERING Energy in water at top of tower = M a X 108.0. _., . . 1 [14.7 - 2.885J144 X 1.4 TT Energy in air at top = ^ ^ C7 32 = 7.65 t/ 32 . Energy in moisture at top = 0.0113[108.0 + 947.5] = 11.9. Work done by changing 1 cu. ft. of volume to 1.4 cu. ft. of volume at atmospheric pressure = ^s X 14.7 X 144(1.4 - 1) = 1.09. I I O 6.70 - C7 32 + 0.600 + M a X 108.0 = 48.1Ma - 0.515 + 7.65 - U 32 + 11.9 + 1.09 1 2 82 M - 56T9- " ' 214 This is the amount of water cared for by 1 cu. ft. of air at entrance, the reciprocal, 4.7 cu. ft., being the number of cubic feet of air required to care for 1 Ib. of hot water. The amount of water evaporated per cubic foot of air has been found to be 0.0107 Ib. This quantity is 5.2 per cent, of the water supplied (0.214 Ib.) per cubic foot of air. In the problem 30,000 Ibs. of water are passed through the cooling tower. This requires 4.7 X 30,000 or 141,000 cu. ft. of air per hour. The evaporation will amount to 5.2 per cent, of the weight of water or 1600 Ibs. of water per hour. Had this cooling tower been supplied with water at a lower temperature the evaporation would have been much less but the amount of air would be greater. With the temperature of the condensing water from a condenser at 140 F. (this temperature is much higher than that found in practice) with a supply temperature of 80 F., 16 Ibs. of water would be required per pound of steam so that 30,000 Ibs. of water would be used with 1880 Ibs. of steam. If this condensed steam were used in a cooling tower it would more than care for the loss by evaporation and if used for boiler feed, the make-up water to be bought would be less than the feed water saved. In all cases where the condensate can be used either for boiler feed or condensing water the cooling tower reduces the bill for water. If the water cannot be used it will be found in most cases that the sum of the make-up water and feed water will be less than the steam necessary with a non-condensing plant. In all cases the heating of the air and the doing of ex- ternal work makes the amount of evaporation in the tower less than the amount of condensation cared for by the cooling apparatus. It will be well to keep in mind that although the air entering the lower part of the cooling tower be saturated with moisture, due to rain or snow, the greater moisture content for the warm CONDENSERS, COOLING TOWERS AND EVAPORATORS 347 air makes evaporation possible even in this case. The warming of the air also removes heat and cools the water. SIZE OF TOWER AND MATS The size of the tower necessary to be used may be found by assuming the air to pass through at a velocity of 700 ft. per min- ute and the area of the mats or cooling surface may be found by allowing 200 B.t.u. per hour per square foot of surface with 10 cooling to 700 B.t.u. per hour per square foot with 35 cooling. This may sometimes be expressed in terms of the water, 1 sq. ft. for 25 Ibs. of water per hour. This is independent of the amount of temperature change. For the problem above the net area of the horizontal cross section of the tower would be Fn = 60 X700 = 3 ' 36 Sq ' ft ' The area of the mats would be 30000(140 - 80) _ Fm ~ 1000* )0 sq.ft. * This is assumed for the large drop of 60 F. F m = ?^P = 1200 sq. ft. In planning this tower care should be taken to prevent the water from falling free. It must be kept in contact with the mats. There must be care in arranging the passages to prevent air from taking any path which does not bring it into contact with the water. Towers cost from $1.00 to $6.00 per kilowatt capacity or as much in some cases as the condenser equipment. The power to drive the fan is about 2 to 5 per cent, of the engine power. In the case above the practical use of the fan is to give velocity only, as the drop in pressure is practically nothing. For this reason the formula for the power of the fan need only consider the kinetic energy of the air, the friction head being considered as equal to the velocity. This gives 141, 000 X 14.7 X 144 /700\ work per minute = 2 X -^TT^^^F^r^i hsr) 60 X 53.35 X 531 \ 60 / 64.4 = 750 ft. -Ibs. per min. 348 HEAT ENGINEERING SPRAY NOZZLES AND PONDS Another device often used to cool water is the spray foun- tain. Nozzles are supplied with hot water from a main and the velocity of discharge is made to divide it into a fine spray. The water is then caught in a reservoir from which it is taken to the condenser. With these nozzles the water may be cooled 15 when the atmosphere is 25 to 30 below the temperature of the hot water while 20 may be obtained with a 40 difference. These nozzles are placed about 8 ft. apart and are usually fitted to 3-in. pipes in which the velocity should be about 5 ft. per second. In this way each nozzle will care for about 60,000 Ibs. of water per hour. In some plants large cooling ponds are used to cool the water by surface evaporation. With these the hot water is discharged at one end of the pond and the cooling water is taken off at the other. Thomas Box, many years ago, recommended that 210 sq. ft. of cooling surface be used per nominal horse- power if the engine was operated for 24 hr. per day. This was with engines using more steam than those of to-day and this might be reduced to say 120 sq. ft. per 24 h.p.-hr. per day, or about 5 sq. ft. per horse-power hour during the day of 24 hr. W. B. Ruggles, in the Journal of the A.S.M.E. for April, 1912, gave a test of a cooling pond of 288,000 sq. ft. of a depth of 5.38 ft. in which he found a transfer of 3.67 B.t.u. per square foot of water surface per hour per degree difference in temperature between air and water, and that 120 sq. ft. of surface per horse-power was sufficient when 16 Ibs. of steam were used per horse-power hour. ACCUMULATORS AND EVAPORATORS Regenerator accumulators are devices used for the retention of surplus heat until needed when an intermittent supply is given off by a machine and it is desired to use this in another piece of apparatus. They were planned by Rateau to be used with low-pressure turbines which were operated by the exhaust steam from engines, the operation of which was intermittent as is the case with rolling mill engines or from steam hammers. One of his forms is shown in Fig. 167. It consists of a vessel A made of steel in which there is a horizontal partition at the center. CONDENSERS, COOLING TOWERS AND EVAPORATORS 349 A number of elliptical flues B.B. are placed in the tank. These receive steam from the pipe D and manifold C and steam is dis- charged through %-in. holes in the flue walls into the water carried in the chambers. This water may enter the flues. The steam heats the water which is introduced from the float box E as soon as the water level is lowered. The steam entering from D may also enter the top of the accumulator by the check G and pass out to the turbine through F. If however the intermittent supply is reduced or stopped momentarily, the pressure in the accu- mulator falls and the warm water begins to evaporate, thus main- taining the steam supply. Of course if the supply of steam is discontinued for a long time, the turbine receives its steam through a reducing pressure valve from the boiler main. Baffle plates are used above the elliptical steam flues to cause the water and steam to mix and to dry out the steam leaving the accu- A 1 B- O O 000 000 000 1 n 1 FIG. 167. Rateau accumulator. mulator. In some Rateau Accumulators large masses of iron in the form of trays are used to absorb the heat and supply that necessary to evaporate the water when the supply is reduced. Evaporators are used for the purpose of concentrating liquors or solutions and for the production of distilled water or other liquid although when used for this latter purpose they are known as stills. Fig. 168 shows one form of evaporator. In this one a liquid to be evaporated is carried to the line A in a tank B con- taining a set of tubes C held between two tube plates. The space D between the tube plates is separated from the remainder of the shell. This space is supplied with steam or some other hot vapor. The vapor gives up its heat to the fluid within the tubes and is condensed, the condensate leaving at E. If the pressure in the chamber above the level of the liquid is such that the boiling temperature of the liquid is below the temperature of the vapor entering the space D from the pipe F, the liquid will boil and its 350 HEAT ENGINEERING vapor may be used to boil liquid in a second chamber in which the pressure is maintained still lower than that in the first cham- ber. This operation may be repeated as shown in the figure, the vapor from the last chamber B" passing to the condenser H in which the pressure is maintained at a low point by the air pump /. In many cases where these are used for the concentration of liquors as in sugar making, the dilute solution enters at J and as it becomes more concentrated it sinks to the bottom of B and is passed over to the uppfer level of K f of the next evaporator. Since the pressure is lower in C than in B this action will take place and be regulated by opening a valve in the line. This action is carried on throughout the system. FIG. 168. Triple effect evaporator. Where three of these are used as shown in Fig. 168, the arrange- ment is known as a triple effect evaporator, a single one as a single effect, two as a double effect, four as a quadruple effect. Assuming the arrangement as shown, suppose M pounds of liquor enter at J. Of this e\ pounds are evaporated and M ei pounds pass on to the second effect to be evaporated. Of this e z are evaporated and M (61 + 62) are passed on and 6 3 are evaporated and M (61 + 62 + 63) are drawn off as a con- centrated solution. In the first effect suppose 1 Ib. of vapor is admitted at F. This is condensed and the condensate is removed at E. This weighs 1 Ib. 61 pounds of vapor are admitted to F f and 61 pounds of vapor are condensed here. 6 2 pounds are introduced and condensed in the third stage. In working out the condensation and evaporation in the differ- CONDENSERS, COOLING TOWERS AND EVAPORATORS 351 ent effects the method of procedure is to start with the one effect and work to the others. In these various tanks the amount of heat loss must be allowed for and if the size of the tanks be assumed the amount of heat loss per hour may be computed for each from the value K for coverings as given in Chapter III. The pressures in the various tanks are assumed so as to give the proper temperature differences. These may be made 20 for each stage. The condensates from the various stills could be discharged through a feed heater and this heat could be added to the feed entering the first effect so that the condensates would leave at the temperature of the fresh liquid. The same could be done in theory with the condensate from the condenser. The operation of the triple effect depends on a difference in the temperatures of the various effects and hence in any problem the initial steam pressure pi and temperature t\ and the final temperature t c would be assumed and with them the various intermediate tem- peratures, from the temperatures of the condenser and the tem- perature of the condensate. For a triple effect, the temperature of the steam in the first effect is ti in the supply and 2 in the dis- charge while 3 is the temperature of the discharge from the second and t c is the temperature of the discharge from the last stage. The temperature of the cooling water and weak liquor is t{ and that of the warm circulating water is t while the con- densate in the condenser is fo. The following conditions hold for the various stages if 10 per cent, of the heat entering is assumed to be lost in radiation. First Stage Heat Entering: With 1 Ib. steam Q = q\ + n (23) With ei + 2 + 3 Ibs. feed Q = (ei + e% + e^q'i + Ifa'i - 3 = ^ c (T, - T} L v \ J- d. J. c) T e ~ T b T b T e w (2) (3) T d - T c - T b T d ' T ~ x " T 1 c id ~ Te Tc T e - T b T b T d T e T d - T c ~ T c ~ ~ T d Since This is reduced from the cross products of temperature. T e T c = T b T d This states that the theoretical efficiency of the Otto cycle, air standard, is equal to the range of temperature on either adiabatic divided by the higher temperature on the adiabatic. T t / V \ k ~ l Now -, - (4) But V c is the clearance volume and Vi is equal to the clear- ance volume plus the displacement, hence Vc ID I V b (l + l)D I + 1 *-i i k-i Hence r; 3 = 1 - x 7 , , , - . , ^ . H 1 - / - V* As I decreases the last term becomes smaller and the efficiency increases. That is, the decrease of clearance increases the INTERNAL COMBUSTION ENGINES 363 efficiency. Of course this increases the pressure at c, hence in- creasing the compression increases the efficiency. The amount of compression is fixed by the allowable temperature at the end of compression and by the pressure at the end of the explosion. If the pressure of compression is excessive the temperature may be high enough to cause premature explosion while if high with a rich gas the explosion pressure is too high. In practice the pressure at the end of compression is 80 to 160 Ibs. per square inch depending on the kind of gas. As before: T d - T b T d ^3 ^2 = ~ AW ATKINSON CYCLE AND DIESEL CYCLE Certain other cycles have been proposed for gas engines and one will be examined for the purpose of application of theory. In Fig. 173 the cycle proposed by Atkinson is shown. In this engine pistons were so connected by linkage to the shaft that they made strokes of varying length. The suction stroke ab is a short stroke followed by a compression stroke be. The explosion cd is followed by a long stroke so that the expansion reduces the pressure to the initial value. The dis- charge stroke ea, which is long, FIG. 173. Atkinson cycle. brings the pistons to their original points. In this case the heat added is = c v (T d - T e ) The heat removed is Q 2 = c p (T e - T b ) Work = Qi - Q 2 = c v [(T d - T c ) - k(T e , T e ~ T b (7) 364 HEAT ENGINEERING Here there is no simple relation between the temperatures at the corners as this is not a simple cycle of poly tropics. Rudolph Diesel in the last decade of the 19th century pro- posed a new cycle of higher efficiency. He recognized the fact that to obtain high efficiencies, the cycle of the gas engine must approximate the Carnot cycle; that the range of temperature on this cycle must be as large as possible and, on -account of the loss of availability in conduction, the engine must have internal com- bustion. He proposed a cycle of isothermal compression from a to b, Fig. 174, followed by adiabatic compression be to such a high temperature that fuel would ignite on being introduced into the cylinder. The fuel was to be introduced at such a rate that its burning would produce just enough heat to make the expan- sion from c to d isothermal. After the fuel was cut off the ex- pansion da became adiabatic. His original paper in the Zeit- FIG. 174. Original Diesel cycle. FIG. 175. Final Diesel cycle. schrift des Vereins Deutscher Ingineure was translated in the Progressive Age in Dec., 1897 and Jan., 1898. This paper gave the results of his work for a number of years. The great pres- sure developed by the final compression and the slight gain of area by the isothermal ignition led him to abandon the upper part of the figure, while the desire to reduce the volume led to cut- ting out the lower end of the figure. Thus modified the Diesel Cycle took the form shown in Fig. 175, in which adiabatic com- pression in a cylinder of small clearance brings the air from a to b at which the temperature is high enough to ignite oil which is injected into the cylinder. If the burning progresses at the proper rate the pressure will be kept constant until cut off at c. From c to d adiabatic expansion takes place followed by exhaust from d to a and finally to e. The suction stroke from e to a INTERNAL COMBUSTION ENGINES 365 charges the cylinder with air. This is a four-stroke cycle. The efficiency is given by c P (T c -T b )-c v (T d -T a ) c p (T c -T b ) T d -T a k(T c -T d ) The high theoretical efficiency is due to the high compression. (8) ACTUAL ENGINES The form taken by an actual engine is shown in Fig. 176. This represents an engine of the Otto form. In this air is drawn into the cylinder by the outward motion of the piston A through the valve B. Gas is admitted by C which is so arranged as to FIG. 176. Ordinary gas engine. 4-cycle. open after B by means of a shoulder formed by a sleeve attached to the valve stem. These admission valves are opened by the suction of the piston or they may be opened positively by the valve gearing. They are closed by the spring D, when the piston reaches the end of its stroke or when the valve mechanism re- leases them. At the end of the expansion stroke the exhaust valve E is opened by the lever F operated by a cam. The cam is on a cam shaft, operated by a bevel or spiral gear from the crank shaft. The speed of this cam shaft for a four-cycle engine is one-half the speed of the engine shaft. The governor 366 HEAT ENGINEERING operates to throttle the mixture or to prevent gas from entering as will be explained later. The cylinder K contains a water jacket L for the purpose of keeping the temperature of the cyl- inder wall low enough for lubrication and also to prevent seizing. The two-cycle, Mietz and Weiss oil engine, Fig. 177, is similar in action to the four-cycle engine. In this the explosion and expansion of the charge compresses air in the crank case K and when the piston A overrides the port B the burned gases escape to the exhaust pipe M, while at the next moment the FIG. 177. Two-cycle Mietz and Weiss oil engine. ports C are uncovered, admitting the air compressed in the crank case. This compressed air rushes in and is deflected to the head by the projecting finn D so that the burned gases are blown out or scavenged. This reduces the air in the crank case to atmospheric pressure. After the piston passes C, the air in the crank case is rarified, as no air can enter, and after passing B the air in the cylinder is compressed. When the piston travels back far enough to uncover port E air is drawn into INTERNAL COMBUSTION ENGINES 367 the crank case from the base of the engine by the partial vacuum existing there. In the Mietz and Weiss engine kerosene is sprayed into the cylinder near the end of the stroke by the pump G, and vaporized by the hot cylinder head H and finally ignited by the high temperature from compression in the heated ball F at the end. The mixture then explodes and the action de- scribed above is repeated. In small engines using gasolene, the air entering the crank case is drawn through a carbureter in which the air is drawn through gasolene or is mixed with it. This charges the air with fuel and the mixture is ignited at the end of compression by a Scale 120 Ib. . l" Cylinder Card Scale 10 Ib. =!' Crank Case Care FIG. 178. Cards from cylinder and crank case of a two-cycle engine. spark. In large two-cycle engines the fuel gas and air are compressed in piston compressors and admitted to the cylinder at the proper time. Fig. 178 illustrates the form of indicator cards from the cylinder and crank case of the engine above. GOVERNING Internal combustion engines are governed in two principal ways: (a) by the hit and miss system and (6) by the throttled charge system. In the hit and miss method of governing the governor acts on the gas or fuel valve so that no gas is admitted when the speed exceeds a given limit, but as long as the limit is not exceeded the engine receives its full charge. In throttle governing the charge of fuel and air may be throttled or the 368 HEAT ENGINEERING fuel may be throttled alone. In the first of these less fuel mix- ture is introduced into the cylinder although it is of the proper mixture, while in the second case the mixture is changed. In each of these the explosion pressure will be reduced and the work will be made less. Of course the efficiency in throttle governing is reduced. The disadvantage of the hit and miss system is the fact that the speed may fluctuate, due to this method. In the Diesel engine the governor fixes the point at which the oil supply is cut off. IGNITION The charge in most modern gas engines is ignited by an elec- tric spark made by breaking a circuit between two platinum points or else causing a high-tension spark to jump a gap between two points of platinum or tungsten. The method of compressing the charge into hot tube until the mixture comes in contact with red hot iron is rarely used at present and compression to a high temperature for ignition is used in some small engines. This latter method is used exclusively on the Diesel engine. The rapidity of ignition depends on the pressure and the quality of the mixture. Thus with higher pres- sures the ignition is more rapid. In many cases it is found that the ex- plosion is not instantaneous but is continued after the end line as shown in Fig. 179. Such action is called after burning. This after burning is found to take place in weak mixtures and in throttled supply. The efficiency of the engine is found to increase as more air is added with the gas than that required theoretically. There are several theories given for this, a satisfactory one being that as the heat per cubic foot is decreased by the addition of air, the temperature increase is not so great and the loss from the cylinder is made less, moreover this might give a different specific heat and thus, a lower temperature. The after burning may be ex- plained by the fact that in many cases the high temperatures present prevent chemical combination or may lead to dissociation. The amount of dilution by excess air to insure the presence of sufficient oxygen is a matter of experiment. The excess air may INTERNAL COMBUSTION ENGINES 369 be sufficient to make the excess air and inert nitrogen amount to about 5 times the gas and the oxygen required to burn it, although much larger variations have been observed in explosive mixtures. The time of explosion varies with the amount of excess air present being slow with large quantities. A slight excess gives a quick explosion. It has been found at times that the best efficiency was obtained even with 100 per cent, excess air. HEAT TRANSFER TO WALLS Experiments have been made to determine the loss of heat to the cylinder walls in a similar manner to that used for the steam engine. Coker found that there was a cyclic variation of 7 C. or 12.6 F. at a depth of 0.015 in. in the wall of a cylinder of an engine running at 240 r.p.m. In the Seventh Report of the Committee on Gaseous Explosion of the British Association, the results of Dr. Coker and Mr. Scoble are shown. Fig. 180 is taken from this report. 180 270 300 450 540 Angular Fgame Of Crank in Degrees 630 FIG. 180. Cycle in gas temperature from a gas engine after the Gaseous Explosion Committee Report. This curve shows the temperature of the working medium as determined by a platinum couple. The couple was placed in a tube which could be cut off from the cylinder by a valve until the desired time. The high temperature of the cycle was not measured but computed because at this temperature the couple would melt if exposed to the gas. These curves show the great variation in temperatures which takes place in the gas engine. 24 370 HEAT ENGINEERING The results of Coker give a variation from 620 F. to 3960 F. for one test, while for another it extended from 390 F. to 3250 F. The losses from the cylinder walls are due to radiation from the high temperature gas as well as from convection currents. These losses are affected by the density of the gas. An increased density will increase the heat loss. In addition to this fact that the loss is made greater by increasing the density through the increase of compression, the danger of pre-ignition through an increase of the temperature from this cause gives a limit to the possible compression. FUELS The fuels used in internal combustions are inflammable gases and oils although Diesel proposed in his patents to use powdered solid fuels. The gases in common use are natural gas, illuminat- ing gas, producer gas and blast-furnace gas. Natural gas is a product of nature found in certain localities in various parts of the world, usually where oils are found. It is rich in methane. Its heating value is high. T. R. Wey- mouth in the Journal of the A.S.M.E., May, 1912, gives the following average analysis for natural gas of the United States: Methane, CH 4 87.00 Ethane, C 2 H 6 6.50 Ethylene, C 2 H 4 0.20 Carbon monoxide, CO . 20 Hydrogen, H 2 Trace Nitrogen, N 2 5 . 50 Carbon dioxide, CO 2 0.50 Helium, He 0.10 Oxygen, O 2 Trace 100.00 The average heating value of the gas was 887.3 B.t.u. per cubic foot at 29.82 in. and 60 F. The specific gravity com- pared with air was 0.6135. Illuminating gas is only used for small installations as the cost is prohibitive. Its heating value is about 700 B.t.u. per cubic foot under standard conditions of 760 mm. and C. This gas may be the product of distilling bituminous coal, the gas amounting to 30 per cent, of the coal. The residue is coke and unless it can be sold this method of utilizing coal is not ef- ficient. If, however, the illuminating gas is made by enriching INTERNAL COMBUSTION ENGINES 371 producer gas by adding hydrocarbons from crude oil and fixing the mixture, this waste of coke would not occur. If the gas were made for power purposes alone this enrichment would not be made as producer gas is satisfactory for use in gas engines. An analysis of illuminating gas is given below: Hydrogen, H 2 34.3 Methane, CH 4 28.8 Ethylene, C 2 H 4 9.5 Heavy hydrocarbons, C 2 He 1.7 Carbon dioxide, CO 2 0.2 Carbon monoxide, CO 10 . 4 Oxygen, O 2 0.4 Nitrogen, N 2 14.7 100.0 The common form of gas used in power installations is producer gas. This gas is made from any form of coal or lignite in a producer as shown in Fig. 181 or Fig. 182. These represent two forms: the pressure producer and the suction producer. Fig. 181 illustrates one form of pressure producer of R. D. Wood & Co. In this, air is blown into a bed of incandescent fuel A by means of the steam jet blower B. The air burns the carbon to CO2, but on passing through the thick bed of fuel this is reduced to CO and the gas rises to the outlet C. The heat of the fire serves to drive off the volatile matter from the coal. The carbon burning to CO liberates 4450 B.t.u. per pound of carbon and this heat would be lost in the scrubber if not absorbed in some manner. Some of this heat is used to make steam in the top of the producer or in a boiler heated by the exhaust gases on their way to the scrubber. If this steam is passed into the ingoing air, the dissociation of this by the heat of combustion of the fuel will absorb much of this heat. The steam used in the air blast cools the gas as it is disso- ciated into hydrogen and oxygen on passing over the hot coals. This dissociation utilizes part of the heat of combustion of carbon to CO and so utilizes some of that which would be lost in the scrubber. The steam also prevents the fire from clink- ering. The gas now contains CO, H 2 ,O 2 and N 2 as well as some volatile gases. Fresh coal is discharged from the hopper D into the chamber E and from this it is distributed by a special device F which 372 HEAT ENGINEERING produces in a uniform bed. The openings G in the top are for the introduction of slice bars to break up the fuel bed. The openings H on the side are for observation of the fuel bed and for barring the bed when necessary. The producer is capped by a water-cooled top and the sides are lined with fire brick. The ashes drop into a water-sealed base. From the producer the gas passes through the outlet C into a scrubber which consists FIG. 181. Pressure producer of the R. D. Wood Co. of a metal cylinder filled with coke and cooled by water. The gas enters a water seal at the bottom of the scrubber and leaves at the top. From this point the gas may be taken to a tar extractor if the gas contains tar to any extent. This is practi- cally a centrifugal fan over which the gas passes, being thrown to the outer edge and thus causing the heavier tar to be caught by the casing. From this point it is taken to the gas holder. INTERNAL COMBUSTION ENGINES 373 In Fig. 182 the Otto Suction Gas Producer is shown. In this a suction is produced by the engine drawing in gas. This draws air from the atmosphere at A over the hot water in the top of the producer at B and thence through the pipe R to the bottom of the fuel bed C. The water is heated by the gas passing from the producer. It then passes through the down pipe D to the seal box, depositing tar or dirt at the bottom and passing up through the scrubber E; it finally enters the receiver F and then passes to the engine. FIG. 182. Suction producer of the Otto Gas Engine Co. The blower G is used in starting the fire in the producer. The three-way cock H is used to direct the gases and smoke to the chimney before burnable gas is produced in starting a fire. The scrubber E is filled with coke over which water trickles and through which the gas passes. The spherical charging ball at the hopper mouth prevents gas discharging or air entering. An analysis of producer gas from soft coal gave the following: 374 HEAT ENGINEERING Carbon monoxide, CO 20 . 9 Hydrogen, H 2 15.6 Carbon dioxide, CO 2 9.2 Oxygen, O 2 , 0.0 Ethylene, C 2 H 4 0.4 Methane, CH 4 1.9 Nitrogen, N 2 52.0 100.00 This gas gave 156.1 B.t.u. per cubic foot under standard conditions. Since 1900 blast-furnace gas has been used successfully. Apparently it was first used in England in 1895 on a 12 X 30 15-h.p. gas engine and in the same year on an 8-h.p. engine in Belgium; in 1898 a 150-h.p. engine was used; in 1899 a 600-h.p. engine was used in Belgium and exhibited by Cockerill & Co. at the Paris Exposition of 1900. These were followed by larger engines. In 1903 the Lacka wanna Steel Company installed a number of 2000-h.p. engines using blast-furnace gas. In most cases the gas is cleaned by centrifugal washers before it is applied in the gas engine. This gas was formerly used beneath boilers for steam generation but gas engines utilize the heat more effectively. An analysis of blast-furnace gas is given herewith: CO 25.83 CO 2 9.37 CH 4 0.54 H 2 .. 2.96 N 2 61.30 100.00 Heat value = 105 B.t.u. per cubic foot. Crude oil is usually used by the Diesel engine. This oil varies some in composition and heating value. The analysis below is for one form : C 84.9 H.. 13.7 1.4 100.0 Its heating value is 19,000 to 20,000 B.t.u. per pound. Gasolene is the light first distillate from crude petroleum oil. It is easily vaporized and in gas engines it is usually delivered to the air supply by a form of mixer known as a carbureter. Its INTERNAL COMBUSTION ENGINES 375 heating value is about 20,500 B.t.u. per pound when the steam formed is reduced to water. It contains approximately: C , 83.5 percent. H 15.5 per cent. N 2 , S, and O 2 1 . per cent. Kerosene is one of the later distillates of crude mineral oil. It is used on a few engines. It requires a special form of car- bureter or vaporizer, requiring a high temperature to properly volatilize the oil. The analysis of this gives: Carbon 84^ per cent. Hydrogen 14 per cent. Nitrogen and oxygen Itf per cent. The heating value will be about 19,900 B.t.u. for the higher heating value. COMBUSTION OF FUELS The combustion of these various gases is accomplished by the mixing of air with the fuel. To get a mixture which will explode or burn rapidly the quantity of air must be within certain limits, usually the amount necessary for complete combustion plus 15 per cent, will give good results. To find the weight or volume of air to burn any given constituent, reference is made to chemical formulae and Avogadro's Law. Thus to burn 1 volume of CH 4 , 2 volumes of oxygen are required which means 9.54 volumes of air. From this burn- ing 1 volume of CO 2 results and 2 volumes of H 2 O. The water may condense. The weights of these per pound of CH 4 are: 4 Ibs. oxygen, 17.40 Ibs. air, 2% Ibs. CO 2 and 2% Ibs. of water. These statements are seen from the following: CH 4 + 20 2 = C0 2 + 2H 2 By Avogadro's Law: 1 volume CH 4 + 2 volumes O 2 = 1 volume CO 2 + 2 volumes H 2 O. By chemical equivalents: [12 + 4.032] Ibs. CH 4 + 2[16 X 2] Ibs. 2 = [12 + 32] Ibs. CO 2 + 2[2.016 + 16] Ibs. H 2 0. or 1 Ib. CH 4 + 4 Ibs. 2 = 2% Ibs. C0 2 + 2% Ibs. H 2 0. 376 HEAT ENGINEERING Air contains 77 parts by weight of nitrogen and 23 parts oxygen while by volume the proportion is 79 to 21. Hence 2 volumes of 2 oxygen mean TT-^T or 9.54 volumes of air and 4 Ibs. of oxygen O.Zl. 4 mean ^-^ = 17.40 Ibs. of air. U.^o The heat of combustion of CH 4 is 21,566 B.t.u. per pound if the water produced remains as steam or 24,019 B.t.u. per pound if the steam is condensed to water. The former is spoken of as the lower heating value, the latter as the higher. In gas-engine work in England, Germany and America, the lower value is taken in determining the efficiencies, while in France the higher value is taken. It is fairer to use the higher heating value and charge the engine with the heat which is pres- ent with the steam in the exhaust. To find the amount of heat per cubic foot of gas the numbers above would be divided by the volume of 1 Ib. of the gas under certain definite conditions. . The conditions assumed in many cases are 14.7 Ibs. per square inch pressure and 32 F. although for natural gas the conditions are often a pressure of 28.7 in. of mercury and 60 F. For the former case the volume of 1 Ib. of gas is given by: (9) MBT MET __1 1 X mol. wt. > p P mol. wt. 14.7 X 144 359 mol. wt. The specific weight is given by: 1 mol. wt. m = v = For CH 4 : v = = 22.4 cu. ft. This gives the values of 1072 B.t.u. per cubic foot for the greater heating value and 963 B.t.u. for the lower value for CH 4 . Using the methods above for various gases and elements the following table is computed: GAS COMPOSITION To study the action of the gas engine theoretically the actual gas supplied and the amount of air or the products of combustion INTERNAL COMBUSTION ENGIh r#s 377 T< C (NO'HOiNO'-HOO'-i CO ^H : 7 7 i i : : i i i i : : : : OOOOOOOOINO ,V g^ 3 M 8 8 (N C^ t^ S CO t~- 1C dddo'dddoiNO O O O O : : i i i i : : : : 00 (N IN 1-H : i i i i S Z H 'apiqdpns 00 (N OS ^O^iC i i 1 iKSS888 iCCO(NOO 00 *-l W 1C CO r} O (N 00 O CO iC C005COCO<-IOOO-H O IN T-H d d d d C O K a M< c IN o COCOOOCNOOOSIN "5 OS- N IN oooooooo-*o as' 11 co (N 00 OO^O'^OOOOO CD *O O 00 O O t> 00 O O Tj* OS CO CO 00 uoi^snqtnoD jo s^onpojj 378 HEAT ENGINEERING must be known. From these the various temperatures and forms of lines may be determined. From the chemical analysis of the fuel gas and the exhaust gas, the amount of air may be found, as will be shown below. Suppose then that the producer gas given on p. 374 is to be burned in a gas engine with 25 per cent, dilution and it is re- quired to know the amount of air to be admitted, the volume of the mixture, the heat per cubic foot of mixture and the products of combustion. Per cent. Vol. gas Volume air Heat of combus- tion Products of combustion C0 2 2 N 2 H 2 O SO 2 CO H 2 CO 2 ... O 2 20.9 15.6 9.2 0.0 0.4 1.9 52.0 49.9 37.2 7,060 5,130 20.9 9.2 39.3 . 29.5 15.6 . 0.0 5.7 18.1 0.8 . 3.8 . C 2 H 4 ... CH 4 .... N 2 Excess 669 2,037 0.8 1.9 4.52 14.30 52 00 5.82 21.30. 100.0 Air.... 110.9 27.7 14,896 138.6 14,896 32.8 5.82 160.92 20.2 This table has been prepared from table on p. 377 for 100 Ibs. of gas in the following manner: Air for CO = 20.9 X 2.38 = 49.9 High Heat for CO = 2.09 X 337 = 7060 CO 2 from CO = 20.9 X 1.0 = 20.9 N 2 from CO = 20.9 X 1.88 = 39.3 14896 B.t.u. per cubic foot mixture = 1 S8 6 Air per cu. ft. - = 1.386 Vol. of mixture = 2.38 cu. ft. Products of combustion per cubic foot: CO 2 = 0.328 cu. ft. O 2 = 0.058 cu. ft. N 2 =1.609 cu. ft. The values of B, c p and c v for a mixture are important to con- INTERNAL COMBUSTION ENGINES 379 sider. Since the molecular weights of the various constituents are different from each other and since the mixture contains CO 2 and water vapor the expressions for specific heat must be com- puted. The a's of the expression c v = a + bt are different for these various substances and the same is true for a''s and b's. If the values of a, b or a' for 1 Ib. of a gas be di- vided by the volume of 1 Ib. under standard conditions the value of these quantities for the heat to be added to 1 cu. ft. of a substance to change its temperature 1 deg. will be found. This might be called the specific heat of I cu. ft. The reason for using this quantity is the fact that the gas analysis given is usually by volume and the necessity of reducing this to percentage by weight is eliminated. This is done with the heat of combustion to reduce that to the heat per cubic foot. If Vp is the percentage volume or the relative volume of the various gases in a mixture the following equations hold for the mixture: v p = partial volume of any constituent. Mol. Wt . mix - Zfo X^ut.). . (11) a - = ^-* bmix _ ^m . (13) ^ . z^_x^). (14) 2i> \ J -^ : / H 2 (pXff). ix = ~2v ^ ' Cpmix = a mix -f- b mix T. (16) Cvmix = a' mix + b mix T. (17) B ^^~ Mol.Wt.mix 359 E_ ~ Mol.wt. ~ 4.3' (20) 380 HEAT ENGINEERING PROBLEM Suppose the natural gas given on p. 370 was used in a gas engine and it is required to find the amount of air used per cubic foot of gas and the true products of combustion if the exhaust gas analysis by volume is : CO 2 = 8.8 per cent. N 2 = 85.7 per cent. O 2 = 5.5 per cent. These analyses show no water vapor from the burning of the gas nor from that taken in with the air and gas. Suppose the gas is at 80 and saturated while the air is taken at 70 and is 1 A saturated. The pressure of the water vapor in the gas is 0.5056 Ib. per square inch, while that in the air is 0.1814 Ib. per square inch. The barometer is 14.7 Ibs. per square inch. Now from Dalton's Law the moisture percentage by volume is equal to the percentage of the dry air pressure which gives the partial pressure. Hence Per cent, volume of gas entering equal to moisture = 0.5056 14.7 - 0.5056 Per cent, volume of air entering equal to moisture = 0.1814 3.56 per cent. On account of pressure and temperature differences of gas and air the theoretic amount of air per cubic foot must be multiplied by 14.7 - 0.5056 460 + 70 14.7 - 0.1814 460 + 80 _ ~ to give the actual cubic feet of air per cubic foot of gas. The true constituents of the gas are those given on p. 370 multiplied by 0.964 to which 3.56 per cent, moisture is added. These quantities are then multiplied by the quantities of the table on p. 377 to give the various volumes of the products of combustion and the volumes of the air required. The results are expressed as a percentage. If the total air required is multiplied by 1.25 per cent, the moisture from the air is found. The computation is now made. Per cent, volume C0 2 N 2 H 2 Air Heat CH 4 87.00X0.964 C 2 H 4 83.90 19 83.90 0.38 632.0 2.15 167.80 0.38 800.00 2.72 90,100 318 C 2 H 6 6 23 12 46 82.50 18.69 104.25 11,420 CO 0.19 0.19 0.35 0.45 64 N 2 + He 5 41 5 41 CO 2 48 48 H 2 O . 3 60 3.60 100.00 97.41 722.41 190.47 907.42 101,902 INTERNAL COMBUSTION ENGINES 381 Moisture from air = 907.42 X 0.012 = 10.90 Total moisture = 201.37 The volume of the mixture before burning is 100 + 907.42 + 10.9 = 1018.32 while after burning the volume is 97.41 + 722.41 + 201.37 = 1021.19 There is a slight increase in volume. Products of combustion in the original analysis show that there is some air present. The amount of oxygen is 5.5 per cent. This is associated with 5.5 ~- X 0.79 = 20.6 per cent, of nitrogen. The nitrogen in the analysis from combustion must be the difference between 85.7 per cent, and 20.6 per cent. Nitrogen from combustion = 85.7 20.6 = 65.1 per cent. The nitro- gen in the products of combustion give a total of 722.14 parts of which 5.41 have been brought in by the fuel gas, or 5 41 72^14 X 65.1 = 0.49 is the amount of nitrogen chargeable to gas. The nitrogen from the air required to burn the gas is given by : 65.1 0.49 = 64.61 per cent. = volume of nitrogen from air to burn gases. 85 7 The total nitrogen is ~^ times the nitrogen from combustion, or. 85.7 The free a' ' 2 ' 6 65 l X 722.41 = 952 64.61 Total air is 907.4 + 289.6 = 1197. Moisture with free air = 0.012 X 289.6 = 3.48. The products of combustion are : CO 2 97.41 7.42 N 2 952.00 72.40 5 5 O 2 ^-5X97.41 61.10 4.62 H 2 ;": 204.85 15.56 I o . 48 J 1315.36 100.00 As a check, the total volume of the mixture before burning is : Volume gas 100 . Volume air 1 197 . Q Volume moisture 14.4 Total.. . 1311.4 382 HEAT ENGINEERING The theoretical air per cubic foot is ^~ = 9.07 cu. ft. The theoretical air at atmospheric conditions 9.07 X 0.96 = 8.71 cu. ft. The actual amount is 1197 X 0.96 1QQ = 11.49 cu. ft. The heat per cubic foot of mixture entering is V ' ^ - 77.6 B.t.u. | " -I? The specific heat of the gas mixture entering is given by the following ; Volumes Air... 1197.00 CH 4 C 2 H 4 .. CO N 2 + He. CO 2 .. 83.90 0.19 6.23 0.19 5.41 1292 . 92 0.48 0.48 3.60 14.40 18.00 1311.40 H 2 | For the total volume of 1311.20 units the following is true: Va Vb 1292.92 X 0.018 = 23.3 1292.92 X 0.185 = 239.0 0.48X0.0202= 0.0098 0.48X0.807= 0.3875 18.00 X 0.0188 = 0.339 18.00X0.668= 12.0 23.6488 251.3875 Va' 1292.92 X 0.0125 = 16.2 0.48 X 0.0147 = 0.007 18.00 X 0.0132 = 0.238 16.445 VC P = 23.6488 + 251.39 X 10~ 5 T For one unit volume C p = 0.0180 + 0.191 X 10~ 5 T VC V = 16.445 + 251.29 X 10~ 6 T C v = 0.0125 + 0.191 X 10~ 6 T For T = 1000 F. these become VC V = 18.9554 26.1588 C, ~ 18.9554 = 1.38 INTERNAL COMBUSTION ENGINES The B for this mixture of gas and air is found by (18) 383 Air J Volume 1197.00 X Molecular Relative Per cent, weight weight weight 28.8 = 34,420 94.4 16.032 = 1,344 3.7 28.032 = 5 30.048 = 189 0.5 28.0 5 28.02 = 151 0.4 44.0 21 0.1 18.016 = 324 0.9 83.90 X 0.19 X C 2 H 4 C 2 H 6 CO 6.23 X 0.19 X 5.41 X N- + He CO 2 H 2 . 48 X 18.00 X 36,464 100.00 _ 36464 _ IVlOl. Wt. OT ~~ -I o-i -I 1544 40 ~ = 55.7 Va 77.02 X 0.018 = 1.387 7.42 X 0.0202 = 0.1496 15.56 X 0.0188 = 0.2925 T T " m ~ 27.78 For the exhaust gases : N 2 72.40 O 2 4.62 77.02 C0 2 H 2 0...... 77.02 X 7.42 X 15.56 X VC P = 1.829 C p = 0.0182 .. 7.42 7.42 .... 15.56 15.56 1.8291 Va' 77.02 X 0.0125 = 0.965 7.42 X 0.0147 = 0.109 15.56 X 0.0132 = 0.205 100.00 Vb 0.185 = 14.25 0.807 = 5.99 0.668 = 10.35 30.59 + 30.59 X 10~ 5 T + 0.306 X 10~ 5 T 1.279 VC V = 1.279 + 30.59 X 10~ 5 C v *= 0.0128 + 0.306 X 10~ 5 ForT = 2000F.;FC P = 2.441; VC V = 1.891; k = - j. .oy J. 1.29 By the method used for the mixture, the B for the ignited gas is B = 55.7. Although this value of B is the same as that for the unburned mixture, the composition is so different that the expressions for specific heats are not the same. If mixtures on the compression and expansion are those assumed, the values of the various quantities are as follows : For compression : C pm = 0.0180 + 0.191 X 10~ 5 T C vm = 0.0125 + 0.191 X 10~ 5 T For 1000 F., a mean temperature: C p = 0.0199 C v = 0.0144 384 HEAT ENGINEERING 0.0199 = 1.38 0.0144 B m = 55.7 H = 77.6 B.t.u. per cu. ft. v = 12.95 For the exhaust gases : C pm = 0.0181 + 0.295 X 10~ 5 T C vm = 0.0128 + 0.295 X 10~ 5 T For T = 2000 F.: C pm = 0.0239 C vm = 0.0186 0.0239 CLOI86 B m = 55.7 v = 12.95 k = 1.283 TEMPERATURES AT CORNERS OF CARD The various temperatures at the corners of the indicator card of the gas engine cycle are now computed theoretically. In this computation the variation of specific heat will be disregarded at first after which the effect of this variation will be investigated. In a gas engine the exhaust gas which remains in the cylinder warms the incoming gas and changes its temperature. For this reason the amount of gas actually used by an engine cannot be told by the change in volume. Moreover the heat removed by the cylinder jacket during the different events affects the results. Consider the card of Fig. 183. On the suction stroke 5-1 a charge of gas and air is drawn in, mixing with the burned gases of volume F 5 in the clearance space. The temperature of the gases in the clearance space T 5 will be equal to the temperature resulting from the expulsion of the exhaust gases from point 4. The gas which remains in the cylinder at 4 acts on the gas driven out to force it into the atmosphere and hence it may be considered to expand adiabatically in driv- ing out the exhaust. The gas in contact with the piston from 1 to 5 may be considered to be at a constant temperature equal FIG. 183. Theoretic form of in- dicator card. INTERNAL COMBUSTION ENGINES 385 to that due to adiabatic expansion from 4 to 1 if no loss or gain of heat from the cylinder walls be considered. Thus : Hence T, = T, " (23) The mixture of this burned gas of weight M M and the fresh charge will result in a volume of gas V\ at a pressure Pi and at a temperature TV The weight of this mixture of fresh air and gas and the burned gas will weigh M 2 pounds. BT l Although the values of B are not quite the same in these two formulae they may be considered the same in this work. If this gas enters the cylinder from 5 to 1 the work done by the entering gas on the piston, pi(Vi F 5 ), will just equal the work done by the atmosphere in forcing the air into the cylinder so that this need not be considered in equating the energy at 5 plus the energy in the air entering to the energy at 1. (M, - M ) Jc v T a = (24) PS = Pi. Tr T * 0.4 " \BT l BT, I pi(Vi- Vi) _ Pi (Vi _ Vj\ OAT a 0.4 V7 7 ! Tj Now the clearance V% or V& is given as I times the displace- ment of the piston. F 5 = Z [7i - V 6 ] (26) v* = Vl (27) Hence 25 386 HEAT ENGINEERING *- 1 T 6 + lT a Substituting for T& its value from (23), T\ reduces to (28) Now (29) (30) (31) (32) Now the heat added from 2 to 3 is called VH and is equal to VC V (T 3 - Tz) hence VT-f ff *T! + (33) vc v Hence T 4 = Ti + ^ These equations reduce equation (29) to (34) \T 1+ H '\ a k-1 h +lT a L h c, a \ i H [ a + c.rj (35) In this equation the only unknown is TI but the equation is implicit so that its solution is best made by trial. After TI is found the other temperatures may be found in succession. Thus suppose in an engine with 25 per cent, clearance, the out- side gas is at a temperature of 70 F. and the heat per cubic foot is 80 B.t.u. while the value of C v per cubic foot is 0.013. 1.25 - 4 1.9 INTERNAL COMBUSTION ENGINES 387 -P_ 80 if 1.9 JUf [1.9 80 r 0.013 X TiJ [TI + 80 1 1.9 113 * +(0.25X530) 0.013 X 1.9J 10+ 8 ^0.013 X T, T, = Since the last term in the denominator is small the value of TI is practically equal to (1 + l)T a although on account of another term being additive, the value will be slightly smaller. (1 +l)T a = 1.25 X 530 = 662. Hence try TI = 625. 1.25 X 530 X 3855 X 0.593 3855 X 0.593 + 132 If 627 is tried 1.25 X 530 X 3857 X 0.618 3857 X 0.618 + 132 Use 626. T 2 = 626 X 1.9 = 1190 Qft T 3 = 1190 + = 7340 These are all higher than the values found in practice due to the fact that the combustion on the line 2-3 is not always complete at the point 3, that the jacket removes heat from the cylinder and also because the specific heat varies with the tem- perature increasing perceptibly at the higher temperatures. If the heat of combustion from 2-3 be reduced to 75 per cent. of its value the temperature T 3 will be materially decreased. ADIABATICS The compression and expansion lines of the card have been assumed to be adiabatics of the form pyi.4 = const. (26) This assumes that the gases are perfect gases and that c p and Cv are constant. Now the true forms for the specific heat are c p = o + bT (16) and c v = a' + bT (17) 388 HEAT ENGINEERING and the equation of the adiabatic is = c v dt + Apdv = (a + bT)dt + Apdv f + Mt=-AB d ^ (36) In the above expressions the quantities refer to 1 Ib. al- though as will be seen later it might be simpler to have them refer to 1 cu. ft. The integration of this between limits finally gives a' log, ^ + b(T 2 - rj = - AB loge ^ (37) From this T 2 may be found if 7\ and - are known. , 2 , 2 2 log log T - - log TF (38) log ^ log ^ since & = I? p Pi Ti 7 2 In a form involving T 2 and T 7 ! only, this may be written AB log, p + a' loge p + 6( T 2 - !Ti) n = - ^ (39) Of course since must be known to find T 2 and 7\ this form v\ is not necessary, except in cases in which it is desired to know the value of n over a given range of temperatures. This is a /> closer value of n for the range from I ' 2 to TI than the value c v Equation (37) may be written a! log, T+bT + AB log, V = const. (40) a' log, p + a' loge V+ AB \og e V + bT = const. a' loge p + a loge F+ 6T 7 = const. bpp pa/ya^r = pa/yo B = cons t. (41) These equations considering the variation of c v with tempera- ture will cut down the pressure of compression and its tempera- ture and thus affect the expression for efficiency of the air cycle. INTERNAL COMBUSTION ENGINES 389 EFFICIENCY CHANGE The change in efficiency due to this change in c may be com- puted as follows: c v AB n w , (1 - ,3) log. ~ = ~cT- 1 ~c~ L loge \T~^U J ^ 44 ^ This expression would give the fractional variation of effi- ciency -, due to the fractional change in the value of c v , ( -) ^?3 \ Cy I TEMPERATURE AFTER EXPLOSION To find the temperature after burning, on the assumption that no heat is taken away; the intrinsic energy after burning is made equal to the heat per cubic foot plus the intrinsic energy before burning. If there is an assumed loss of heat to the jackets of 20 per cent, of the heat of combustion this same method is used with 80 per cent, of the heating value of the fuel. U 2 = Ui + H X (100 - per cent, loss to jackets) (45) The intrinsic energy at any point is given by C T bT 2 U = I c v dt = a'T+ - (46) Hence a'T 2 + ^ + #(1 - loss to jackets) = a'iT s + | 7Y (41) The values of a'T* -f ^7Y and of a\T 3 + ^TV are sometimes A plotted as shown in Fig. 184 so that for a given temperature T 2 the heat contained in the gas is read from the figure. After add- ing H or a, portion of H to this, the curve for the burned mixture 390 ENGINEERING will give TS corresponding to the heat contained in each pound, or cubic foot, the intrinsic energy. To find the temperature at 4 equation (37) is used. If now the mixture at point 1 is assumed to be at 626 abso- lute, the temperature at the various points may be found by the methods given above. Assume the gases are those mentioned on p. 380 and that the clearance is 25 per cent. Of course 7\ would be worked out by equation (35) if not known. Using Intrinsic Ettergy per Cu.Ft.inB.t.u. g fe 8 / / / / / s /* ^ s^ 500 F lOOO'F 1500F 2000F 2500F 3000F 3500F 4000F Abs. Abs. Abs.. Abs. Abs. A.bs. Abs. Abs. Degrees Absolute FIG. 184. Plotting of intrinsic energy for variable specific heat. data for the mixture on p. 383 in equation (37), the following results : 12.95 X 0.0125 X 2.3 log + 12.95 X 0.191 X 10~ 5 [T 2 - 626] X 55.7 X 2.3 log 5 log X 10~ 5 T z = 2.797 + 0.0417 + 0.309 = 3.148 Assume log T 2 = 3.148 T 2 = 1400 This is of course too large due to the second term. Assume T 2 = 1200 3.080 + 0.080 = 3.160 slightly too large Try T 2 = 1150 3.061 + 0.077 = 3.148 This is correct. INTERNAL COMBUSTION ENGINES 391 The value of n by equation (38) is 1160 . _ lQ g626" +1 g5 _ 0.265 + 0.699 _ 71 1 r f^ r*f\f\ -I 5O log 5 0.699 C This is the same value found for the ratio of - which was given c w as 1.38. That the value of 1150 may be seen to be correct the following computation is made by equation (4). / v .\ n- I /5\ 0.38 n-fip) -026m =1150 \V-2/ \1/ To find T s the correct method is to equate the intrinsic energy at 3 to that at 2 plus the heat of combustion. Since the composi- tion of gas after burning is not the same as that before, the specific heat is not constant during burning nor are a' and 6 constant during this change. Hence the above method is the only one which may be used to find TV a' m T* + |TV + H = a'tft + ^7V 1 2 Q5 12.95 X 0.0125 X 1150 + ~ X 0.191 X 10~ 5 (1150) 2 + 12.95 Zi 10 OF; X 77.6 X 0.80 =12.95 X 0.0128 XT 3 + ^ X0.306X10" 5 TV TV + 8360r 3 = 50,700,000 T 3 + 4180 = 8260 T 3 = 4080 F. T 4 is found by equation (37) used for the determination of TV 12.95 X 0.0128 X 2.3 X log ^2 + 12 .g 5 x 0.295 X 10~ 5 X [4080 - Tt] = ^ X 55.7 X 2.3 log 5 log T 4 -h 10.04 X 10~ 5 T 4 = 3.610 + 0.408 - 0.301 = 3.717 For log T 7 4 = 3.71 T 7 4 = 5150 This, of course, is too large. Try T 4 = 3000 3.478 + 0.302 = 3.780 392 HEAT ENGINEERING Too large. Too small. Too large, which gives n log Try T 4 = 2500 3.399 + 0.252 = 3.651 Try T, = 2750 3.440 + 0.280 = 3.720 Try T, = 2740 3.438 + 0.275 = 3.713 4080 0.172 + 0.699 2745 log5 log 5 0.699 1.23 In the calculation for the value of n on the two lines, it has been found that on the compression line n = 1.38 while on the expansion line n = 1.23. From actual tests as mentioned on page 214, n = 1.288 on compression and 1.352 on expansion. The value of n is found by plotting the logarithms of the volumes and pressures of various points and drawing a straight line the slope of which will give the value of n. These temperatures, 626, 1150, 4080 and 2745, are not much higher than those found in practice. TEMPERATURE ENTROPY DIAGRAM The temperature entropy diagram for a gas-engine cycle may be easily constructed from the indicator diagram by the method 3.00 .S2.00 1.0 1.0 2.0 3.0 4.0 Volume ID Inches 5.0 6.0 FIG. 185. Card prepared for T-S analysis. shown below. The mean indicator card, Fig. 185, is constructed from a number of cards in the same manner as that used in the Hirn's or Temperature Entropy Analysis, and then the lines of absolute zero of volume and of pressure are laid off after meas- INTERNAL COMBUSTION ENGINES 393 uring the clearance and calibrating the indicator spring. A series of points is then marked and the distances from the axes are measured in inches. These are tabulated as shown: Points Pres- sure in inches Vol- ume in inches Px Pi V, vl log-^ Pi log lr *'osff S* - Si T x Ti 1 0.15 6.00 1.00 1.000 0.000 -0.000 -0.000 0.000 .000 2 0.178 5.25 1.18 0.873 0.072 -0.060 -0.084 -0.012 .030 3 0.224 4.43 1.49 0.737 0.173 -0.132 -0.185 -0.012 .098 4 0.282 3.75 1.88 0.625 0.274 -0.204 -0.286 -0.012 .175 5 0.350 3.18 2.33 0.530 0.367 -0.276 -0.385 -0.018 .235 6 0.447 2.68 2.98 0.447 0.474 -0.350 -0.490 -0.016 .335 7 0.563 2.27 3.76 0.378 0.575 -0.423 -0.592 -0.017 .422 8 0.669 2.00 4.45 0.332 0.648 -0.479 -0.671 -0.023 .478 9 1.000 2.01 6.68 0.333 0.825 -0.478 -0.670 0.155 2.260 10 1.260 2.015 8.40 0.335 0.924 -0.476 -0.668 0.256 2.810 11 1.78 2.025 11.86 0.337 1.074 -0.472 -0.660 0.414 4.000 12 2.51 2.040 16.79 0.339 1.224 -0.470 -0.658 0.566 5.680 13 2.00 2.470 13.35 0.412 1.125 -0.385 -0.539 0.586 5.490 14 1.58 2.940 10.50 0.488 1.022 -0.312 -0.437 0.585 5.130 15 1.24 3.55 8.25 0.590 0.916 -0.229 -0.320 0.596 4.860 16 1.13 3.87 7.52 0.643 0.876 -0.192 -0.268 0.608 4.840 17 1.00 4.23 6.67 0.705 0.824 -0.152 -0.213 0.611 4.770 18 0.89 4.65 5.93 0.775 0.773 -0.111 -0.155 0.618 4.590 19 0.795 5.13 5.30 0.855 0.724 -0.068 -0.095 0.629 4.480 20 0.710 5.64 4.73 0.940 0.674 -0.027 -0.038 0.636 4.450 21 0.60 5.70 4.00 0.950 0.602 -0.022 -0.031 0.571 3.800 22 0.50 5.77 3.33 0.960 0.523 -0.018 -0.025 0.498 3.190 23 0.40 5.83 2.67 0.971 0.426 -0.012 -0.017 0.409 2.590 24 0.30 5.89 2.00 0.980 0.301 -0.009 -0.013 0.288 1.960 Now if one of the points, say 1, be taken as the datum point, the ratios of the temperature at various points to the tem- perature of this datum point and the change of entropy from this datum may be figured. Thus considering the point 2: p.Vi This ratio holds to the point of explosion since the unburned mixture after burning is different in volume, due to the change on burning, and it is necessary to change the temperature found by the above formula in the following proportion: vol. of mixture vol. of burned gases T , = 394 HEAT ENGINEERING In practical application the change in volume due to chemical action is so slight that the volume ratio is assumed unity. If the temperature of the point 1 be assumed on the T-S diagram, the height to other points may be found by multiply- T ing by the ratio 7^- If the height to 1 be made 1 in. the ratio 1 1 T -^r will give the heights to the various points. Of course the 1 1 scale of temperature is not known uritil one temperature is known. As will be shown, much data can be determined even if this scale is not known. dv . dp Now ds = c p -f c v S 2 Si = C p log e + C v loge (50) Vi PI Cp . V 2 . , PZ ,_, N -? logic - + logio jj- (51) Cv t/i PI S 2 Si AT Sz Si S 2 Si Now -x~5 = - 2.3 c v const. The constant of this expression only changes the scale of entropy if the expression is plotted as the entropy. Fig. 186 is obtained by this method and the area of the figure which represents the difference between the heat added and that taken away is equal to the work done or the work shown by the indicator card. If the area scale of the indicator card is known in B.t.u. per square inch the area scale of the T-S diagram will be inversely proportional to the areas of the two figures. Scale* = ^ X Scale pt , (53) Scalep.,,. = cu. ft. per in. X Ibs. per sq.ft. per in. 778 Fpv = area of p-v diagram F t s = area of T-S diagram (54) The scale fixes the heat and efficiency of the cycle, although the component scales of temperature and entropy are not known. As soon, however, as one temperature is determined the scale INTERNAL COMBUSTION ENGINES 395 of temperature will be known and from it and the area scale the entropy scale in B.t.u. per degree can be found. The area abed is the heat accounted for by the card. If the curve be is carried out to e so that abef is equal to the heat per card the area dcef represents the reduction in the heat of combustion due to 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 01 10; 9 22 d\ h -0.10 0.0 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 Temperature -Entropy Diagram to Special Scale FIG. 186. Temperature-entropy diagram of the gas engine card. absorption from the cylinder walls and incomplete combustion. The area dcgh represents the heat absorbed from the walls of the cylinder during expansion. The area ibcg represents the work. The theoretical form of the cycle should have been nbem. abik is equal to heat absorbed by wall during compres- sion. 396 HEAT ENGINEERING LOGARITHMIC DIAGRAM In Fig. 187 the logarithmic diagram for the indicator card is drawn and from it the values of the exponents n are found. S0.50 0.1 0.2 0.3 0.4 0.5 0.6 Logarithm of Volume in Inches 0.8 FIG. 187. Logarithmic diagram of gas engine card. TEST OF GAS ENGINE As a closing problem on gas engines the following data is taken from a report on a test of a producer and two-cylinder gas engine reported by Prof. H. W. Spangler in the Journal of the Franklin Institute, May, 1893. The data is as follows: 2 cylinders each 14^ in. X 25 in. Coal by weight Moisture 4 . 20 per cent. Volatile matter (5.80 per cent. CH 4 , 0.73 per cent. H 2 ) 6 .88 per cent. Fixed carbon 80 . 41 per cent. Ash 8.51 per cent. 100.00 Sulphur . 74 per cent. Time of test 525 min. Revolutions 84,425 160.76 r.p.m. Explosions 81,673 155.44 ex.p.m. Mean gas pressure lii/fc in. = 0.062 Ibs. per sq. in. Barometer 14 . 686 Ibs. per sq. in. Gas temperature 75 . 5 F. Room temperature 81 . 3 F. Jacket temperature outlet 99 . 23 F. Jacket temperature inlet 62 . 42 F. Exhaust pyrometer 752 . 6 F. Brake load 1148.5 Ibs. Zero brake load 160 . Brake arm 3.073 ft. B.h.p 92.73 I.h.p Top cylinder 64.36 Bot. cylinder 66.80 131.16 Coal.. ..1069.6 Ibs. INTERNAL COMBUSTION ENGINES 397 Fuel gas by volume CO 2 . . . 2 .... CO.... H 2 .... CH 4 . . . N 2 .. 4.02 per cent. 0.26 25.38 4.51 1.79 64.04 100.00 Exhaust gas by volume CO,.. CO..' lii.-:. . 15.60 per cent. . 2.24 . 0.28 . 81.93 100.00 Cooling water 664 Ib. in 4 min. Gas for ignition tube 840 cu. ft. Relative humidity of air 80 per cent. 230 Ibs. SOlbs. FIG. 188. Average card from test of Otto engine. The following data will be computed: Heat of coal C = 80.41 X 14,544 = H 2 = 0.73 X 61,500 = CH 4 = 5.80 X 24,000 = 11,690 449 1,390 13,529 Total heat supplied 1069.6 X 13,529 = 14,470,600 B.t.u. Carbon per pound of gas : Relative carbon from CO 2 = 4.02 X 12 = 48.24 Relative carbon from CO = 25.38 X 12 = 304.56 Relative carbon from CH 4 = 1.79 X 12 = 21.48 Total relative carbon . . . 374.3 Relative weight of gas From C0 2 4.02 X 44 = 176.5 O 2 0.26 X 32 = 8.3 CO 25.38 X 28 = 710.0 H 2 ' 4.51 X 2 = 9.0 CH 4 1 79 X 16 = 28.6 N 2 64.04 X 28 = 1792.0 100.00 2724.4 398 HEAT ENGINEERING 2724.4 Hence molecular weight of mixture = forT = 27.24 Carbon in 1 Ib. of gas = - 0.1372 Carbon per pound coal From C = 0.8041 X 1 = 0.8041 From CH 4 = 0.0580 X - Pound of gas per pound of coal = Q ^072 = 6-18 Ibs. 359 Volume of 1 Ib. of gas = 07^4 = *3.1. Volume of gas per pound of coal = 6.18 X 13.1 = 81.00 cu. ft. Heating value of gas per cubic foot : High Low From CO = 0.2538 X 337 = 85.5 85.5 From H 2 = 0.0451 X 345 = 15.5 13.1 From CH 4 = 0.0179 X 1071 = , 19.2 17.1 Heat per cubic foot, B.t.u 120.2 B.t.u. 115.7 B.t.u. Heat in gas per pound coal = 81 X 120.2 =9740 B.t.u. 9740 '/Efficiency of producer = 1Q _ OQ = 0.72 (neglecting temperature) = 72%. Indicated work 131.16 X 2546 X QQ- = 2,920,000. 2920000 Indicated thermal efficiency of producer and engine = 1 A /t7rtAnn = 20.2%. l'' UOUU 20 2 Indicated thermal efficiency of engine = ^^ =- 281 = 28.1%. Delivered work = 92.73 X 2546 X ^jj = 2,070,000. 2070000 Delivered thermal efficiency of producer and engine = 14470600 = ^.30 %. v/Delivered thermal efficiency of engine = ~ = 0.199 = 19.9%. Heat in cooling water = 664 X ^ [99.23 - 62.42] = 3,220,000. 3220000 Percentage of heat in coal removed by jacket water = 14470500 = v/^ercentage of heat in gas removed by jacket water = = 30.9 %. Air required per cubic foot of gas: INTERNAL COMBUSTION ENGINES 399 The products of combustion for complete combustion and no dilution are : Per cent, vol. C0 2 N 2 H 2 O Air OO 2 0.0402 0.0402 O 2 . 0026 0.0098 0.0124 oo 0.2538 0.2538 0.4300 0.6080 H 2 0.0451 0.0451 0.1070 CH 4 0179 0.0179 0.1443 . 0358 0.1710 N 2 6404 6404 0.3119 1.2545 0.0809 0.8736 The analysis of exhaust gases shows 15.60 per cent, by volume of CO 2 and 0.23 per cent. CO. Now 1 volume of CO would produce 1 volume of 23 CO 2 . Hence if ^-^ X 0.3119 = 0.0045 cu. ft. be left as CO the propor- tions will be as follows: CO 2 0.3074 CO 0.0045 N 2 1.2464 H 2 O 0.0809 Air Required . 8629 The exhaust gases contain 2.24 per cent, of O 2 and 15.60 per cent. CO 2 hence there must be 2.24 15.60 of O 2 per cubic foot of gas, or 0.044 X 0.3074 = 0.044 cu. ft. 0.21 = 0.21 cu. ft. of air in excess. This gives as the total amount of air per cubic foot 0.21 -f 0.8629 = 1.0739. The amount of air theoretically required is 0.8629 cu. ft. and hence the per cent, excess is 24 ' 2 per cent " The products of combustion without moisture will then be: CO 2 .................... . ____ . ........ 0.3074 cu. ft. CO ........ ............. ... .... ....... 0.0045 cu. ft. N 2 ..................... . ... ... ........ 1 .2464 cu. ft. H 2 O ........................ , ....... .. 0.0809 cu. ft. Air ................................... Q.2100CU. ft. Total ..................... . ......... 1.7492cu. ft. These came from 1 cu. ft. of gas and 1.0729 cu. ft. of air. The gas passes from the wet scrubber to the engine and is therefore 400 HEAT ENGINEERING saturated. Moisture in gas is sufficient to saturate same at 75.5 F. and will exert a pressure of 0.436 Ib. and hence for a barometer of 14.686 Ibs. and a pressure of 0.062 Ibs. gauge or 14.748 Ibs. absolute, the moisture occupying 1 cu. ft. at 0.436 Ib. would occupy 0.436 14.748 0.0296 cu. ft. at atmospheric pressure plus gas pressure. The air at 81.3 F. is 80 per cent, saturated and would exert a pressure of 0.527 X 0.80 = 0.422 Ibs. per sq. in. and therefore the moisture per cubic foot of air under atmospheric condi- tions would be Now the air and gas are not under the same conditions of pressure and temperature and the 1.0739 cu. ft. of air above computed must be reduced to the same pressure and temperature as the atmosphere. The pressure on the gas is 14.748 - 0.436 = 14.312 Ibs. and the air pressure is 14.686 - 0.422 = 14.264 Ibs. The temperatures are 75.5 F. for the gas and 81.3 F. for the air. The amount of air if reduced to the conditions of atmosphere will be : The mixture entering the engine with 1 cu. ft. of gas and its molecular weight are given below : Gas .................................. 1.000 cu. ft. Air ............. ; ..................... 1 .074 cu. ft. H o f gas .................. . 030 cu. ft. ' \ air .................. 0.031 cu. ft. Total ................... . ........... 2.135 The products of combustion per cubic foot of gas and the computation for the mean molecular weight are given below: CO Volume . 0045 V X mol. wt. ... 0.126 N 2 Air H 2 O. 1.2464 0.2100 f 0.0809) \ 0300 [ ... 35.000 ... 6.02 2 67 CO 2 [ 0.0310 J 0.3074 . . . 13.43 Total 1.9112 57.246 Mean molecular weight = . QI 10 == 30.1 INTERNAL COMBUSTION ENGINES 401 The values of the specific heats for the burned mixture are given below : Gas Value of a CO 0.0045 X 0.018 = 0.000081 N 2 1.2464 X 0.018 = 0.022500 Air 0.2100 X 0.018 = 0.003780 H 2 0.1429 X 0.0188 = 0.002690 CO 2 0.3074 X 0.0202 = 0.006280 0.035331 Gas Value of b CO 0.0045 X 0.185 X 10~ 5 = 0.0083 X 10~ 5 N 2 1.2464 X 0.185 X 10~ 5 = 0.2320 X 10~ 5 Air 0.2100 X 0.185 X 10~ 5 = 0.0388 X lO" 5 . H 2 0.1429 X 0.668 X 10~ 5 = 0.1950 X lO" 8 CO 2 . . . 0.3074 X 0.807 X 10~ 5 = 0.2460 X 10" 5 0.7201 X 10~ 5 0.035331 1.9112 t 10-5 - = 0.3762 X 10~ 5 From this the heat in the exhaust gases at 752.6 F. above 75.5 F. is fc Heat = VfC p dt = V[a(T 2 - Ti) + |(IV - TV)] (55) = 1.9112[0.01831(752.6 - 75.5) + - 3762 2 X -(752.6* - 75.5*)] = 1.9112[12.398 - 1.052] = 25.70 Heat in exhaust gases at 75.5 F. due to evaporation of water: Volume of H 2 O = 0.1429 cu. ft. Weight of H,0 - "-I* X 14.7 XU4 = ^^ ^ X 535.5 Heat of vaporization = 1049 X 0.0066 = 6.92 B.t.u. Total heat in exhaust gases per cu. ft. = 32.62 B.t.u. Heat of combustion in carbon monoxide in exhaust gas 0.0045 X 337 = 1.518 B.t.u. Heat Balance per cu. ft. of gas: Heat supplied 120.20 Heat brought in above 75.5 F. by air = 1.070[81.3 - 75.5] [0.018] 0.11 Moisture in gas and air = e^ 43 x 2.67 1.15 Total Heat Supplied 121.46 26 402 HEAT ENGINEERING Heat in indicated useful work (28.1 per cent.) 33.8 B.t.u. Heat in jacket water (30.9 per cent.) 36 . 2 Heat in exhaust gases { \. D . y^ Heat in CO 1.52 Difference.. . 18.96 COMBUSTION OF FUELS FOR BOILERS 27.8% 29.8 26.8 1.3 14.3 100 .~0% The combustion of other fuels may be considered at this point with advantage. This combustion is practically the same as that of gases considered earlier. This union of the elements with oxygen develops heat. The heat of combustion and the air required are found in the table on p. 377 as was done for gases. The solid and liquid fuels are composed primarily of carbon, hydrogen, oxygen and sulphur., The solid fuels are coal, lignite, peat, wood and certain waste materials such as bagasse. These differ in their chemical com- position. This composition is found by chemical analysis. When the analysis is made in a combustion tube giving the percentages of the various constituents, the analysis is called an ultimate analysis. If, however, the analysis gives only the moisture, volatile matter, fixed carbon and ash, it is known as a proximate analysis. This analysis is easily made and is the one made by engineers to place a coal. The analysis is given in percentages of the various constituents as received and on a dry basis, eliminating the moisture. This latter is necessary to reduce the effect of the variable moisture to zero. The results of analyses are given below: ANALYSIS ON DRY BASIS Ultimate analysis Proximate analysis Heat Value C H N S Ash Fixed C Vol. matter Ash Anthracite Semi Bituminous Bituminous Lignite 88.0 81.0 75.0 65.0 61.0 51.0 2.0 4.8 5.0 4.5 6.0 7.6 2.0 4.0 9.0 20.0 28.10 4.0 0.8 1.5 1.5 1.0 1.9 37.0 0.50 0.75 1.5 1.5 6.7 7.95 8.00 8.00 3 85 73 56 44 7.00 20.00 36.00 48.00 8.00 7.00 8.00 8.00 13,750 14,700 13,800 10,700 9,000 9,000 Peat . . Wood 4 At times the analysis is put on an ash- and moisture-free basis. This is expressed in percentages of the constituents with ash and moisture omitted. INTERNAL COMBUSTION ENGINES 403 The constituents unite with the oxygen according to the formulae c + o 2 = co 2 . H 2 + K0 2 = H 2 0. s + o 2 = so 2 . From the atomic weights of the elements, the weights of the oxygen required per pound of carbon, hydrogen or sulphur may be found, together with the products of combustion. Thus 1 Ib. of carbon requires 3 %2 I DS - of oxygen, 1 Ib. of hydrogen re- quires *% or 8 Ibs. of oxygen and 3 %2 Ib. of oxygen are required per pound sulphur. If these quantities are divided by 0.232 the air required per pound of the various substances is known. If C, H and S represent the parts of these substances in 1 Ib. of fuel, the air required is given by: Lbs. of air per Ib. fuel = 11.5 C -f 34.5 H + 4.35 S. This is given also as: Lbs. of air = 11.5 C + 34.5 (H - ^) + 4.35 S. or approximately Lbs. of air = 12 C + 36 # -- The value 5- is substracted from H if is the weight of the o oxygen in the fuel, since it is assumed that this oxygen is united with the proper amount of hydrogen in the form of H 2 O. The heat of combustion has been given by Dulong. Heat per pound = 14,6500+62,100 (H - ~^j (56) The value given by this formula is approximately that given by the bomb calorimeter. The results of burning determined by a calorimeter are given in the table on p. 402. The products of combustion of solid fuels are principally C0 2 , N 2 and excess air. If pure carbon is burned, the CO 2 formed is just equal in volume to the oxygen consumed, so that the maxi- mum amount of C0 2 is 21 per cent, by volume. As there is some hydrogen present and as the water occupies twice the vol- ume of oxygen consumed, each per cent, of hydrogen reduces the percentage CO 2 because, although the water condenses, the nitro- gen of the air left from its burning cuts down the possible per- centage. As soon as oxygen appears in the burned gases it indicates that excess air has been supplied. This is advisable in 404 HEAT ENGINEERING many cases as there is danger in having some CO when the air is not in excess and the loss due to improper combustion is greater than that due to the presence of excess air. This latter represents a loss due to a larger quantity of hot gas sent from a boiler or gas engine. It is shown that at least a 30 per cent, excess is necessary to prevent the formation of CO. The curve of Fig. 189 illustrates the relation between per cent. CO 2 and total air supply as a per cent, of the theoretical air supply. It is found that when the per cent. CO2 is between 10 and 15 per cent, the best results are obtained. In analyzing gases from furnaces a percentage of CO 2 of 15 per cent, with 3 or 4 per cent, of O 2 represents good results. 35* 30* 25* | 15* 10* 0.8 0.6o 3 0.4 50* 25* 50* 75*100 125 150 175 200 225 250 Deficiency of Air Excess of Air FIG. 189. Curve of relation between per cent. CO2 and excess of air. PROBLEM The analysis of the exhaust gases from a boiler gives the exact excess air. Thus for a coal of the following analysis: Per Cent, by Weight C... H 2 .. 2 .. Ash.. and flue gases of the following analysis: 80.0 5.0 10.0 0.5 4.5 100.0 INTERNAL COMBUSTION ENGINES 405 Per Cent, by Volume C0 2 12.0 CO 0.5 O 2 5.0 N 2 82.5 The amount of air may be worked out. The oxygen and carbon in the ex- haust gas may be found by remembering that from Avogadro's Law, equal volumes at the same temperature and pressure contain equal numbers of molecules. Hence each volume may be considered to contain one molecule. The weights of carbon and oxygen of the gas are found as follows : Carbon from CO 2 = 0.12 X 12 = 1.44 Carbon from CO = 0.005 X 12 = 0.06 Total carbon 1 . 50 1.44 Per cent, carbon burned to CO 2 = = 96 per cent. 1.50 c\ r\(\ Per cent, carbon burned to CO = = 4 per cent. 1.50 Oxygen from CO 2 =0.12 X 32 = 3.84 Oxygen from CO = 0.005 X 16 = 0.08 Oxygen from O 2 =0.05 X 32 = 1.60 Total oxygen required 5 . 52 r en Oxygen per Ib. carbon = - = 3.68 JL *OU 1 60 Pounds excess of oxygen supplied per pound carbon = = 1.067 1 .50 Per cent, excess of oxygen in total oxygen shown by analysis = 160 X 100 552 29 per cent. The gas analysis does not show the oxygen required for the hydrogen and sulphur and these must be added to that for the carbon and from the same the oxygen of the coal is subtracted. The oxygen required is given by: Oxygen for carbon as burned and excess oxygen per pound coal' = . 80 X 3 . 68 = 2 . 944 Oxygen for hydrogen =0.05 X8.0 =0. 400 Oxgen for sulphur to SO 2 = . 005 X 1 . = . 005 3.349 Oxygen in coal . 100 Net oxygen per pound coal 3 249 3.249 Air per pound coal = = 13.95 0.232 29 X 2 944 Excess of air in total air supplied = - = 26.25 per cent. 406 HEAT ENGINEERING Products of Combustion : 44 Carbon dioxide 0.8 X 0.96 X = 2 .810 \2i 28 Carbon monoxide 0.8 X 0.04 X = 0.075 Water vapor 0.05 X 9 = 0.450 SO 2 0.005 X 2 = 0.010 Excess air 1.067 X 0.80 X ~~ = 3.670 Nitrogen (13.95 - 3.67) 0.768 = 7 . 890 Ash.., =0.045 Total weight =14.950 This checks the 1 Ib. of coal and the 13.95 Ibs. of air. The moisture brought in by the air may be treated in the same manner as that used in the problem of p. 400. The heat carried up the stack may be computed in a manner similar to that used in finding the heat in the exhaust gases of a gas engine. This quantity when divided by the heat of combustion of coal gives the percentage loss in the stack. This amounts to about 10 or 15 per cent, when there is just sufficient air, while with 100 per cent, excess it amounts to 20 per cent. A deficiency of air means a loss due to incomplete combustion. After the carbon burns to CO 2 a further passage through a bed of hot coals changes the C0 2 to CO and the water from the burning of the hydrogen may be dissociated. On mixing the CO with hot air and having the H 2 and O mix after cooling, in the presence of a flame the gases burn and give out their heat of combustion. With an insufficient or a cold air supply above the fire the CO may not burn. If volatile hydrocarbons are driven off in the form of smoke these must be mixed with air in the presence of incandescent fuel to prevent smoke formation. SURFACE COMBUSTION To cause gas to burn completely with the theoretical amount of air present or a little above the requisite amount of air the method of surface combustion has been suggested. In this a mixture of gas and air is delivered into a tube of a burner and dis- charged into a receptacle filled with small pieces of refractory material. If the mixture is lighted and its velocity through the tube is more rapid than the speed of burning, the flame will INTERNAL COMBUSTION ENGINES 407 not enter the tube and cause an explosion but the mixture will quietly burn and heat the refractory material to brightness unless the heat is removed by some method. Fig. 190 illustrates one form of burner. In the Bone-Schnabel boiler this method is applied, the combustion taking place in the boiler flues which are cither filled with small pieces of a refractory substance or rods of the same. The heat of combustion then heats these substances so that the heat is transmitted to the tube surface by radiation and by conduction so readily that not only is the capacity in- creased but the efficiency is greatly increased due to the low temperature of the exhaust. The combustion is complete. In tests quoted in the Journal of the A.S.M.E. for Jan., 1914, p. 09, G. Neuman re- ports a test on one of these boilers showing an evapora- tion of 16.4 to 30.75 Ibs. per square foot of surface and an FIG. 190. Burner for surface com- bustion. efficiency of 90 per cent. The porous filling not only causes the gas to move at a higher velocity relative to the tube but these heated objects maintain combustion and thoroughly mix the gas and air preventing stratification. TOPICS Topic 1. Sketch the cycles of Beau de Rochas, Atkinson, Otto, Lenoir, and Diesel. Explain the action on each line and reduce expressions for the thermal efficiency of each cycle. Topic 2. From the efficiency of the Otto cycle on the air standard 773 = 1 - T z -T b T d - To Reduce the expressions T h _ FA *-i 1 + Give reasons for the impossibility of reducing the expressions for efficiency of the other cycles by this method. What is meant by air standard? Topic 3. Explain the action of a four-cycle gas engine and a two-cycle oil engine. Draw cards from each. Explain how these engines are governed. Explain the methods of ignition. 408 HEAT ENGINEERING Topic 4. What affects the rapidity of combustion? Draw an indicator card showing slow burning. Sketch a curve showing the variation of gas temperature in a four-cycle gas engine. Is there much cyclic change in the metal of the cylinder walls? Topic 5. Mention the various fuels used in internal combustion engines. Give some of the peculiarities of each. What are gas producers? What are the two general types? Sketch one of them. Topic 6. Derive the expressions for finding the amount of oxygen and air to burn a cubic foot and a pound of C2H4 together with the expressions for the products of combustion. Show how to find the heat per cubic foot if the heat per pound is 21,400 B.t.u. Find the specific heats per cubic foot if c p = 0.404 andc,, = 0.340. On what laws are these based? Topic 7. Derive the formulae B . 1544 Mol. wt., mol. wt. 2 mol. wt. X vol. 2 vol. ~ Topic 8. Derive the formulae for finding the temperatures at the corner of the Otto cycle assuming the value of c v to be a constant. Topic 9. Derive the equation for the adiabatic of a gas when c v a' + bT. Derive the equation for n on this line in terms of the temperatures. Topic 10. Explain the method of finding temperature at the ends of compression and expansion. Topic 11. Explain by formulae the method of finding the temperature after explosion. Topic 12. Explain the construction of the T-s diagram for the gas engine deriving the expressions for the construction of the figure. Topic 13. Explain the construction of the logarithmic diagram of the gas-engine card. What data can be found from this card ? What data must be known to construct the card? Topic 14. Explain the method of finding the amount of dilution from the analyses of the fuel gas and exhaust gas. Topic 15. Explain the method of finding the efficiency of a producer. Topic 16. Explain the method of finding the heat loss in the exhaust gases and in the jacket water. Topic 17. Explain the method of finding the heating value of a gas from it chemical composition. Explain how to find -the heat equivalent of the work. Topic 18. Explain how to find the amount of air per pound of fuel in a boiler test from the coal analysis and gas analysis. What is Dulong's formula? What is surface combustion? Explain the construction of the Bone-Schnabel boiler. PROBLEMS Problem 1. The clearance of an engine is 30 per cent, of its longest stroke. The initial temperature is 150 F. The heat per cubic foot of mixture INTERNAL COMBUSTION ENGINES 409 is 70 B.t.u. The radiation loss during explosion is 35 per cent. Find the temperatures at the corners of Otto cycle and Atkinson cycle assuming the air standard. Find the Carnot efficiency of each, the theoretical efficiency and the type efficiency. Problem 2. A Diesel engine has a final pressure of 600 Ibs. per square inch. What is the clearance to give this from atmospheric pressure at the beginning of the stroke (n = 1.38). If the original volume including clearance is 1 cu. ft. of air at 160 F., find the weight of crude oil given in this chapter which could be burned by this air with 25 per cent, dilution. If 75 per cent, of the heat is available find the temperature and volume after burning, using the air standard. Problem 3. Find the products of combustion of 1 Ib. of crude oil burned in a Diesel engine with 15 per cent, dilution. Find the temperature resulting therefrom with a 25 per cent, loss to the jacket considering the actual com- position of the gas and .assuring the pressure constant. Problem 4. Find the temperature TI if the clearance is 30 per cent, and the gas and air outside has a temperature of 60 F. with c v per cubic foot 014 and the heat per cubic foot 78 B.t.u. Find the other temperatures assuming 20 per cent, loss to jacket. Problem 5. Assume a blast-furnace gas of form given in this chapter mixed with 10 per cent, excess air. Find the cubic feet of air per cubic foot of gas. Find the value of T 2 considering the actual composition of the mixture if T\ = 600 abs. and the clearance is 25 per cent. Find T 3 with 10 per cent. loss. Find T^. Problem 6. Sketch an indicator card from an engine with 25 per cent, clearance and draw the T-s diagram and logarithmic diagram. Find the heat added if the actual efficiency is 28 per cent. Find the scales of the figure if TI is found to be 125 F. What is the maximum value of jPi. Problem 7. The following results are obtained from a test: Coal C 80.0 per cent. CH 2 6.0 per cent. H 2 0.5 per cent. Ash 10 . per cent. Moisture 3 . 5 per cent. Producer Gas Exhaust Gas C0 2 .. ... 4.0 per cent. Co 2 ... ...15.8 per cent. 2 ... . . . 0.5 per cent. 2 .... . . . 2.5 per cent- CO.. . . .25.5 per cent. CO... ... 0.2 per cent- H 2 ... ... 5.0 per cent. N,.... ...81.5 per cent- CH 4 . . . 1.8 per cent. N 2 63.2 per cent. 100.0 100.0 100.0 Time of test 1,200 min. Revolutions 240,000 Explosions 110,000 F. Gas pressure Barometer Gas temperature. . . Air temperature . . . Relative humidity . Exhaust temperature. 800 C Jacket temperature 70 Jacket temperature. . . 120 2 in. water Coal 4,000 Ibs. 29.9 in. Gas 320,000 cu. ft. 80 F. I.h.p 200 70 F. B.h.p v . ... 160 60 per cent. Cooling water. ..!.... 178,000 Ibs. Compute the various losses and efficiencies and make a heat balance. Problem 8. The heating value of 1 Ib. of coal as fired is 14,680 B.t.u. 410 HEAT ENGINEERING The equivalent evaporation from and at 212 F. per pound of coal is 11.10 Ibs. Find the efficiency of the boiler. Problem 9. In Problem 8 the following was found : The flue gas analysis by volume gave 0.18 per cent. CO, 6.49 per cent. CO 2 , 13.09 per cent. O 2 , and the coal analysis by weight was as follows : Moisture ...................... 1.0 per cent. Ash ........................... 6.9 Hydrogen ..................... 2.8 Oxygen ........................ 4.2 Carbon ........................ 83.0 Sulphur ........................ 0.8 Nitrogen ...................... 1.3 100.0 Find the amount of air per pound of coal. Find the excess air. Find the heat value by Dulong's formula and check with value by calorimeter given in Problem 8. Problem 10. In Problem 9 find the composition of the exhaust gases per pound of coal assuming that the air at a temperature of 75 F. has a wet bulb temperature of 67 F. Problem 11. With the results of Problem 10 find the heat carried away in the flue gases due to a temperature of 427 F. and the heat available from the CO present. Express these as percentages of 14,680 B.t.u. Problem 12. With coal at $4.40 per ton what is the cost of producing 1000 Ibs. of equivalent evaporation, using data of Problem 8. Equivalent evaporation from and at 212 F. is equal to the actual evaporation multiplied by the factor of evaporation. Factor of evaporation is the amount of steam evaporated from and at 212 F. by the same amount of heat as will evaporate 1 Ib. of water at the given feed temperature into steam at the given condition for the boiler. This is given f = r 212 In this problem find the factor of evaporation for a gauge steam pressure of 121.3 with the barometer of 29 in. if the feed temperature is 105 F. and the quality of the steam is 65 F. superheat. Find the cost of producing 1000 Ibs. of actual steam. CHAPTER X REFRIGERATION AIR MACHINES The refrigerating machine operates by placing a substance in such a condition that its temperature is above the tempera- ture of a water supply, and after the removal of heat from the substance by this supply the condition of the substance is so changed that it will abstract heat frpm a body of FIG. 191. Air refrigerating machine. low temperature. The temperature of the substance is changed by changing the pressure on the substance. This is accomplished by machines in which a gas such as air or a vapor such as ammonia or carbon dioxide is used. The air machine is shown in Fig. 191. Air is compressed in the cylinder A from a pressure pi to a pressure p^. At the pressure p 2 it is discharged through a 411 412 HEAT ENGINEERING pipe system B which is cooled by water from the supply C. In this case the cooling of the air reduces its volume so that when this air is taken to a second cylinder D it occupies less space. The air is then allowed to expand in the cylinder D to the original pres- sure, doing work at expense of its intrinsic energy, and hence its temperature falls so much that on entering the coil E in the tank or room F, this air will remove heat from the brine in the tank or from the air, if placed in a room. This air is taken back to the compressor after removing heat from the room F. The cycle is shown in Fig. 192. At times the air is discharged into the room F instead of passing through the coil and air is sucked from the room. The system is then called an open system to distinguish it from that of Fig. 191 which is called a closed system. If the temperature of the air sucked from the cool room F is TI and the pressure is pi, the condition of 1 Ib. of air is shown by the point 1 ; the volume 4-1 is sucked in on the line 4-1 if clearance is neglected. This air is compressed along the line 1-2 and the line is practically an adiabatic. The temperature T 2 given by Compressor IS m m /F 2 1 2 = ill k - 1 k (1) Combined Cord FIG. 192. Cycle of the machine. air This air occupies the volume 3-2 and the card 4123 represents the work done on the compressor A. The air is forced out and is cooled off in the cooler B to a temperature T$ fixed by the temperature of the water at the point of discharge of the air. In a countercurrent cooler this may be less than the temperature of the outlet water. It is about 10 F. above the temperature of the water at the point. When this air passes into the expansion cylinder its volume will be shown by the point 5. The admission line is 3-5. The air expands to the lower pressure and its condi- tion is shown by point 6. This action is also adiabatic, so that REFRIGERATION 413 n == Tl Ts = T b ~ (4) It is thus seen that TI is fixed by the temperature of the refrig- erator, being the room temperature in the case of an open system or 10 to 15 F. less for a closed system; while T 5 is fixed by the temperature of the cooling water, T 2 and T 6 are fixed by the pres- sure ratios after TI and T 5 are found. The work done on the compressor is 4123 while that done by the air in the motor is 3564. The difference or 1256 shown by combining the cards is the net work required. This work is also the difference between the heat on the top and bottom lines since there is no heat on the curved lines. AW = c p {[T 2 - T 6 ] - [Ti - T 6 ]} (5) It is seen that heat removed by the cooler is equal to the heat removed from the refrigerator plus the work done on the substance. But the heat to change the volume from 2 to 5 is the heat re- moved by the cooling water while that added from 6 to 1 is that taken from the refrigerator. Hence Qref. = C p [T, - T,] (6) and Q cooler = c p [T 2 - T 6 ] (7) The efficiency or better the refrigerative performance is the amount of refrigeration per unit of work. This is given by Vr = f rrri _ m i TrF T~\\ (8) CU.li i sj [1 1 1 ejj _ 1 tyr rp _ rp i rfi rti rji fji /TT ' nn ' rji rji fji since =r = TTT 1 6 1 I , L 5 I Q K," f! 414 HEAT ENGINEERING Hence (9) T These expressions are equal to the lower temperature on one of the adiabatics divided by the difference in temperature on that adiabatic. The expression is greater than unity. The refrigeration produced by any machine is usually measured in tons of refrigeration per 24 hrs. The amount of heat liberated when 1 Ib. of ice melts is 144 B.t.u. (best value 143.5 B.t.u.), and hence the heat equivalent to 1 ton of refrigeration is 288,000 B.t.u. in 24 hrs. which is 200 B.t.u. per minute. The number of pounds of air required for this refrigeration is 200 Lbs. of air per min. per ton in 24 hrs. = ^ --- ^T = M a (11) Cp[l i 1 6J The amount of cooling by the air cooler is Q c = M a c p [T 2 - T 6 ] (12) The water required for this is given by M a c p [Ts - T,] 3'o -fl' M w = weight of cooling water per min. per ton in 24 hrs. q' = heat of liquid of cooling water at outlet temperature t q'i = heat of liquid of cooling water at inlet temperature t { . The horse-power required for the production of 1 ton per 24 hrs. is h.p. = M a c p (T, - T b - (T, - T 6 )] (14) Qr 778 rj r 33000 (16) The volume of the compressor is Vi while that for the ex- pansion cylinder is F 6 . These are each given by MgBT, M a BT l ~^ Pl Xcl. factor M a BT 6 . _ ?>e X cl. factor REFRIGERATION 415 The above equations are computed for cylinders without clearance and friction, for non-conducting cylinder walls and for dry air. These three things affect the results of computa- tions, but because their effects are seen by comparison with perfect conditions the equations for perfect conditions are given although not used in practice. PROBLEM Assume that a 3-ton machine was desired to operate between 14.7 Ibs. abs. and 58.8 Ibs. abs. and that the temperatures of the cooling water were 50 F. to 65 F. and that for the cool room was 25 F. It will be assumed further that the air from the expansion cylinder is discharged into the room to be -refrigerated. From the above the following is true: T l = 25 F. = 485 abs. T 5 = 50 + 15 = 65 F. = 525 abs. (assuming a counter-current flow) /KgQ\ <^1 T z = 485 T 1A = 485 X 1.486 = 720 abs. = 260 F. T 6 = 525 X = 354 abs. = - 106 F. 485 * p = 720 - 485 = 2 ' 06 Q r = 200 X 3 = 600 B.t.u. per min. 600 Ma. = Q.24[485 - 354] = 19 ' 08 lbs< of air per min> = 19.08 X 0.24[720 - 525] = 892 B.t.u. per min. 778 H.p. = (892 - 600) = 6.86 600 778 also H.p. = 2^3 X 33^ = 6.85 The water required is 892 Mv, = 33 i _ i g i = 59-4 Ibs. per min. Displacements per minute are : T . 19.08 X 53.35 X 485 _ r Vcomp. = 14 7 x 144 -- = C per mm ' v 19.08 X 53.35 X 354 Vexp. = 14.7 x 144 = CU ' per mm ' The machine considered in the problem above is one in which the discharge has been into an open space. Such a system may be called an open system. If now the discharge takes place in a closed pipe line (the closed system) it would be possible to have the pressure pi higher than atmospheric pressure. Sup- pose that pi is made 58.8 Ibs. abs. and p 2 , 235.2 Ibs. abs. so 416 HEAT ENGINEERING that the ratio is the same as before. If under these condi- Pi tions the above problem is computed the results will be the same except the displacements which will be one-fourth of their former values. This is due to the higher pressures. This "dense air machine," as it is called, is used because it decreases the dis- placement of the cylinders and hence the size or speed of the machine. This is the only reason for its use. The problems must now be investigated for clearance and friction. Since friction increases the work of the compressor and decreases the work obtained from the expansion cylinder, the work done by each of these machines must be considered separately, instead of obtaining the net work 1256 as before. The work of the compressor 1234 is the same with or without clearance. It is equal to If (19) MB[Tl - Til If n = k this becomes W comp = MJc^T, - T 2 ] (20) To make this work as small as possible, jackets may be used on the cylinder changing k to n and making n as low as 1.38. FIG. 193. Effect of friction in air machines. If the friction fraction is /, the work must be multiplied by (1 + 7) or 1'2'34 = work of compression = (! + /) n MB[T l - T 2 ] (21) For the expansion cylinder the work obtained is (1 /') times the work without friction. This work is independent of the REFRIGERATION 417 clearance if the expansion and compression are each complete. The work then becomes : 35'6'4 = work of expansion = (1 - f')p Q V 6 g-z~i[ 1 ~ (~)~*~ or work = (1 - f)MJc p [T 5 - T 6 ] (22) The value of n in the expansion cylinder is k as it is desired to have as low a value of TQ as possible and this is accomplished by having as large a value of n as possible. By lagging the expansion cylinder and omitting any jacket this n is made k as no heat is taken away. The difference of the two expressions for work will give the work required from the outside to drive the machine. Of course with n on compression (1.38) different from that on expansion (1.4) the equality of cross products is not true. 7 7 2 and T 6 must each be found. /r> \ n ~ * r. - Ti(?)T (23) (24) The effect of clearance air on the temperatures is important to consider. The air left in the cylinder at the end of compres- sion is at a temperature T 2 . When this expands down to atmos- pheric pressure the temperature is TI so that the fresh air enter- ing at this temperature mixes with it. In the expander the air in the clearance space at temperature T is compressed to the temperature T 5 by complete compression and so mixes with air from the cooler at that temperature. If this compression were not complete this temperature T$ would not be reached and the effect of this would be to give a different temperature of the mixture at the beginning of expansion. This incomplete compression would require an excess amount of air from the supply to fill the clearance space. This free expansion into the clearance space would warm the air in the cylinder and produce a higher temperature at the beginning and at the end of expan- sion. It would mean a decrease in refrigerating effect and would also decrease the work obtained from the expander per pound of air used. 27 418 HEAT ENGINEERING If the expansion were incomplete the temperature at the end of expansion would not be as low as T& due to the pressure range being less than or and as a result the temperature at Pe' Pe Pi the end of free expansion or the discharge temperature would be higher than T&. This reduces the refrigerating effect. The work obtained from a given amount of air has been decreased by the incomplete expansion and compression. For these two reasons the performance is decreased by a very material amount and the endeavor is made to have complete expansion and compression. To eliminate the effect of clearance the compression in the expansion cylinder is such that the pressure at the end of com- pression is equal to pressure of the compressor and the expansion is carried down to the back pressure. The action is then equiva- lent to a cylinder without clearance. The moisture in the air has an effect on the efficiency of air refrigerating machines. The compression of moist air may cause some of the moisture to condense giving out heat and warming the air while a further condensation and even freezing in the ex- pansion cylinder causes the liberation of additional heat reducing the refrigerating effect. Air will be removed in short time. Suppose that air of relative humidity p, temperature T\ and pressure pi occupies the volume V\. The weight of the air is p t i = sat. steam pressure corresponding to The weight of the moisture M m is m\ = weight of 1 cu. ft. of steam at temperature This may also be approximated by the formula M m = f m l V l (26) (27) t. steam ~ 18 For the complete mixture of air and moisture (M a + M m )B mx T=pV (28) 1544 1544 REFRIGERATION 419 After compression to the pressure pz the volume becomes . (29) and 7 T 2 =7 7 1 (30) Now -y^ = weight of moisture in 1 cu. ft. (31) hence, if ra 2 = weight of 1 cu. ft. of saturated steam Mm Y - Pz = relative humidity after compression (32) 1YlV' 9 ]r 9 (39) The moisture which enters the expansion cylinder is practi- cally all condensed and frozen by the cool walls of the cylinder liberating heat represented by C. C = m 3 V f s [c p (T 2 - r 32 ) + r 32 + 144] (40) This is gradually restored during expansion. The volume of the air remaining is P* The air and ice in the cylinder now expand with the gradual return of the heat C from that produced by the ice formation giving: M a c va dt + (m 3 V' s )cidt + Apdv = + dC (42) M a c va + m 3 F / 3 c, dt dV _dC_ B T^ a V ~ BT Q assume dC = ^ - ^- dt 1 3 1 4 ' T s V s C T 3 T + M a A log, r = - Z? (P* V, =: C 1 T [M a C va + WaV'sCt + rr rn + M a AB] log e ^ i 3 1 4 J 4 (43) NOW C va + -4.^ = C Hence T* = MaAB loge S (44) This is solved for 7% by trial, knowing all the other terms. If z Tt is assumed from an approximate value of T 4 from (45) the equation leads directly to T REFRIGERA TION 421 The value of T then gives the heat taken from the refrigerator G M r (T A - 7M M-fi"! r ''-'- a^p \ * 4 * I/ *%*''/' The moisture is not considered at this point as it has been condensed and frozen in the expansion cylinder and does not enter the refrigerator. The work done in the expansion cylinder is best found by the following : r m pdv tJ V3 C (* T * L* W e = p 3 V 3 +| pdv - piV* (47) From (42): ' C M *"+' m ' V * + jr-=T<\ dt (48) M a c va + mzV'zd + y _ T J [T 3 - T,} Since the moisture has been eliminated on the expansion curve p 3 V 3 - p^V 4 = M a B [T 3 - ITJ (49) and j5+= (50) M a c p + m 3 y 3 c i + Ts _ y [T s - T,] (51) The work required from the outside is (1 + f)W. - (1 - f)W. (52) and the refrigerating effect is Q r = M a c p (T l - T,) (53) To apply the above formulae assume that the air enters the compression cylinder at 25 F., with a relative humidity of 80 per cent, and is compressed to 4 atmospheres. The water in the cooler is sufficient to cool this to 65 F. Assume that Vi = 1 cu. ft. p 25 = 0.065 P p = 0.065 X 0.8 = 0.052 [14.7 - 0.052] X 144 X 1 M = - 53.34X485" ~ = 0.082 Ibs. M w = 0.8 X 0.00023 = 0.00018; call this 0.0002 This is a check. 4= 0.372 0.4 = 485 X (4)1-4 = 720 abs. = 260 F. 0.0002 0.372 = 0086" 422 HEAT ENGINEERING or there is not enough moisture to saturate the air. The air is now cooled to 65 F. as in the problem on page 415. 525 7 3 = 0.372 X 720 = 0.272 0.0002 0.272 _ 0.000753 p3 ~ 0.000979 ~ 0.000979 ~ Hence none of the moisture is condensed in the cooling coil. If p 3 were greater than unity there would be some condensation. After the determina- tion of the amount of condensation a recalculation of V s would be necessary as some of the water has been removed from the mixture leaving the air saturated. The moisture having a specific weight of 0.00073 would be saturated at 56 F. so that this moisture is superheated 9 F. Approximate value of TV k-l QA ' T t = T 3 (- k = 525 X () L4 = 354 abs. = - 106 F. C = 0.0002[(65 - 32) ^ + 1071.7 + 144] = 0.2462 . 0.082 X 0/24 + 0.0002 X 0.5 + ^r^J ^ log* = -^ log, 4 0.02119 , T s 0.082 . ~53^T loge Y* = ~778 loge 4 log = - 161 = lo S L45 T*= j = 361 abs. = - 99 F. W e = 778[0.02119][65 - ( - 99)] = 2695 ft.-lbs. 0.4 Wc = ^1 X 14 ' 7 X 144 X 1 [l - (4) 1 > but the elimination of the expansion cylinder has made the appa- ratus simpler. The exact loss due to this may be found later. 426 HEAT ENGINEERING After the pressure is reduced the temperature of vaporization is so low that heat will be abstracted from the surrounding sub- stances, even though they be at a low temperature, and the liquid vaporizes. This is accomplished in the refrigerating coil D. Here the evaporation of the liquid returns either dry vapor or wet vapor to the compressor A and the cycle is repeated. The expressions for the heat on the various lines are now computed. Heat on 1-2 = Q c = heat on 2-3 at constant pressure = i 2 q's (55) Q c = heat on 2-3" at constant pressure = i% q'z" (56) Heat on 3-4, 3-4', 3"-4" or 3"-4'" = Q r = heat on 4-1 = ii i* or ii is (57) But i 3 = i\' .". Qr = ii - is or ii - i 3 " (58) Now ii is the expression for the heat content at either Id or Iw and iz is the heat content at 2w or 2d. It and the other quan- tities may be found in ammonia tables. i = q' + xr J7W. c p dt Tsat. The expression for the heat added on a constant pressure line from a to 6 is i b - i a (59) as is given under Q r and Q c . The equation of the line 12 is the adiabatic s = constant s = s'i + -r if wet -/ 1 or pv k constant if dry. fc = 1.33 for superheated ammonia. Of course if tables for superheated vapor are known, the formula Si = S 2 may be used for the superheated region also. This is really the better equation even in the superheated region. The work required is the algebraic sum of the heats REFRIGERATION 427 AW = Qr ~ Qc = ~ [Qc ~ Qr] AW = [i* is'] - [ii - i 3 '] = iz - ii With friction - AW = (1 + f)(i* - ii) (60) (61) The amounts above are all for 1 Ib. of ammonia and to find the amount of liquid, per minute, hour or day, the amount of re- frigeration in that time is divided by the refrigeration per pound. Thus for T tons of refrigeration per day the weight of volatile liquid per minute is The horse-power required is (62) .- The weight of cooling water per minute is where q' = heat of liquid for condensing water at outlet q\ = heat of liquid for condensing water at inlet. The displacement per minute of the compressor is given by Mv" Mv clearance factor clearance factor The displacement of the cylinder is found by computing the clearance factor as in the case of air com- pressors. The lines of expansion and compression are of the same form. This is shown by Fig. 197, the actual form of the card. If V - . , . , , , , . . for the ammonia on the lower line is known FIG. 197. Actual card from a compressor. The refrigerative effect is given by In many cases where rooms have to be cooled, the ammonia 428 HEAT ENGINEERING or air is not sent through coils in the room but heat is removed from the rooms by circulating cold brine through pipes. The brine is cooled by the air or ammonia which passes through a coil in a tank through which the brine is pumped. The brine is usually a solution of calcium chloride or sodium chloride. The advantage of the brine system over the direct expansion system is the fact that the break of a pipe would not discharge the ammonia into the room and spoil the contents and also the fact that the compressor may be shut down for some time and the cold brine stored in the brine tank may be used to keep the room cool. These formulae may be used for any volatile liquid. Even water has been used although exceeding low pressures must be used to obtain temperatures below 32 F. A problem will be computed using NHs, SC>2 and CO2 for the media assuming water at 50 to 65 F. and a refrigerating room held at 25 F. In this problem it will be assumed that 15 is the necessary difference of temperature for heat transfer and also that wet and dry compression will be tried for the ammonia. For dry compression Xi = 1 while for wet compression # 2 = 1. It will be assumed that " after cooling" reduces T 3 " to 50 + 15 = 65 F. In all cases the volume drawn into the compressor with 1 per cent, clearance will be assumed to be 1 cu. ft. In these problems T 4 = Ti - 25 - 15 + 460 = 470 abs. T 2d = 65 + 15 + 460 = 540 abs. Ty = 50 + 15 + 460 = 525 AMMONIA: Case I. Wet compression. pi = 38.02 Ibs. p 2 = 153.90 Ibs. x 2 = 1 Xl= 0.1025 + 0.9328 + 0.0483 1.2017 it = 557 ii = - 23.2 + 0.902 X 564.4 = 485.0 i 3 ' = 36.5 M = 6^4 = ' 151 Q e = 0.151 [564.4 - 36.5] = 78.6 B.t.u. Q r = 0.151 [485 - 36.5] = 67.8 B.t.u. AW = 78.6 - 67.8 = 10.8 B.t.u. or = 0.151 [564.4 - 485] = 10.8 B.t.u. REFRIGERATION 429 AW with friction = 10.8 X 1.20 = 12.96 67 8 Refrigerating effect without friction = ^^ = 6.23 67.8 Refrigerating effect with friction = ^QQ = 5.22 A7 ft B.t.u. per ft.-lb. of work = 12 96 x 778 = - 00673 B.t.u. of refrigeration per h.p.-hr. = 0.00673 X 33,000 X 60 = 13,200 13200 X 24 Tons of refrigeration per h.p. = 2000 X 144 = (Assume 3 Ibs. of coal per h.p.-hr.) Tons of refrigeration per Ib. coal = Q ' 0/l = 0.0153 O X ^TC Tons of refrigeration per cu. ft. of compressor volume taken in = 67 ' 8 = 0.000235 144 X 2000 Gallons of cooling water per cu. ft. of compressor volume taken in for 15 7ft A rise in cooling water = ., - V~^? = 0.628 lo X o.oo Gallons per ton of refrigeration = n 000235 = 2670 Case II. Dry compression. pi = 38.02 Ibs. p z = 153.90 Ibs. s 2 = Si = 1.1534. This is s 2 for 110 F. superheat. (This may be checked by 0.25 /153 q\ 0.25 =664'ab, Deg. superheat = 664 - 540 = 124 F.) i 2 = 628.4 B.t.u. M = ^ = 0.136 ii = 541.2 is' = 36.5 Q c = 0.136 [628.4 - 36.5] = 80.5 Q r = 0.136 [541.2 - 36.5] = 68.6 AW = 80.5 - 68.6 = 11.9 ATT with friction = 11.9 X 1.20 = 14.3 AC A Refrigerating effect without friction ^^ = 5.76 Aft A Refrigerating effect with friction ^5 = 4.80 14. o AC A B.t.u. per ft.-lb. of work = 1A , ' 77Q = 0.00617 ll.o X / i o B.t.u. of refrigeration per h.p.-hr. = 0.00617 X 33,000 X 60 = 12,200 12200 X 24 Tons of refrigeration per h.p. = 2 QOO X 144 = L 2 Tons of refrigeration per Ib. ccal (assume 3 Ibs. coal per h.p.-hr.) 0.0142 3 X 24 430 HEAT ENGINEERING Tons of refrigeration per cu. ft. of compressor volume taken in AC A = 0.000238 144 X 2000 Gallons of cooling water per cu. ft. of compressor volume taken in for 80 5 15 rise in cooling water = ]g x o 35 = 0.644 0.644 Gallons per ton of refrigeration = Q QQ0238 = ^ Carbon Dioxide. (Wet compression). T 4 = T l = 470 abs. = 10 F. T* d = 540 abs. = 80 F. T 3 = 525 abs. = 65 F. Pi = 362.8 Ibs. p 2 = 936 Ibs. * 2 = 80 s 2 = 0.1511 si = 0.1511 = - 0.0226 + X! 0.2164 zi = 0.80 ti = - 11.23 + 0.80 X 110.12 = 76.86 it = 21.8 i = 0.247 X 0.80 = 0.198 M = 5.05 Q c = 5.05[80 - 21.8] = 294 Q r = 5.05[76.9 - 21.8] = 278.4 AW = 5.05[80 - 76.9] = 15.6 AW with friction = 15.6 X 1.50 = 23.4 (Friction has been increased due to the packing made necessary by the excessive pressure.) 278 4 Refrigerating effect without friction = j^r = 17.8 Refrigerating effect with friction = -^^ =11.9 070 4 B.t.u. per ft.-lb. work = 234x778 = * 0131 B.t.u. of refrigeration per h.p.-hr. = 0.0131 X 33,000 X 60 = 26,000 Tons of refrigeration per h.p. = 2000 v 144 = 2 ' 17 (Assume 3 Ibs. coal per h.p.-hr.) 2 17 Tons of refrigeration per Ib. coal = ' Oyl = 0.0301 O X ^Tt Tons of refrigeration per cu. ft. of compressor volume taken in = 0.000965 144 X 2000 Gallons of cooling water per cu. ft. of compressor volume taken in for 15 C 294 rise in cooling water = 1{ . = 2.35 1O A o.OO or Gallons per ton of refrigeration = Q QQQ965 = 243 REFRIGERA TION 43 1 Sulphur Dioxide. (Wet compression). T, = T l = 470 abs. = 10 F. T. id = 540 abs. = 80 F. 7 7 3 = 525 abs. = 65 F. pi = 13.5lbs. p 2 = 59.65 Ibs. it = 162.1 S2 = 0.302 si = 0.302 = - 0.1746 + xi 0.3625 xi = 0.88 ix = - 6.89 + 0.88 X 169.78 = 142.6 i s = 10.96 vi = 0.88 X 5.96 = 5.25 M = 5^5 = ' 19 Q c = 0.19[162.1 - 10.96] = 28.8 Q r = 0.19[142.6 - 10.96] = 25.0 AW = 0.19[162.1 - 142.6] = 3.71 AW with friction = 3.71 X 1.20 = 4.46 25 Refrigerating effect without friction = x~yr = 6.75 25 Refrigerating effect with friction = j-^ = 5.61 B.t.u. per ft.-lb. work = A AR nQ = 0.00725 4.4:0 X O B.t.u. per h.p.-hr. = 0.00725 X 33,000 X 60 = 14,350 Tons of refrigeration per h.p. = OQQQ y 144 = 1-195 (Assume 3 Ibs. coal per h.p.-hr.) Tons of refrigeration per Ib. coal = ' 2 , = 0.0167 Tons of refrigeration per cu. ft. of compressor volume taken in = OK = 0.000087 144 X 2000 Gallons of cooling water per cu. ft. of compressor volume taken in for a 28 8 15 rise in cooling water = ^ , .' og = 0.23 10 A o.ou 23 Gallons per ton of refrigeration = Q QQQQgf = ^640 ABSORPTION APPARATUS To eliminate as far as possible the moving parts of a refriger- ating apparatus and to utilize waste heat from other machines the absorption type of refrigerating machine has been developed. In this the high pressure is produced by boiling a solution of NH 3 in water and driving off the NH 3 by a steam coil at such a rate that the pressure is sufficient to produce a temperature of condensa- 432 HEAT ENGINEERING tion above the temperature of the supply water. The low pressure is maintained by absorbing the vapor in water at a rate to produce this pressure. The apparatus, Fig. 199, consists of a generator A in which a steam coil B supplies enough heat to warm the liquor of aqua ammonia to such a temperature that its capacity for ammonia is reduced and the ammonia is driven off. This gas is super- heated as it leaves the generator and to use some of the heat above saturation (as the gas must finally be condensed) it is passed over a series of baffle-plates over which the fresh strong liquor passes on its way to the generator B. The part of the apparatus C is called the analyzer. The function of this is to Cooling Water Outlet T D E Brine Outlet Condenser Strong Liquor FIG. 198. Absorption machine. reduce the superheat in the discharge gases. As the gas or vapor of ammonia contains some water vapor which would interfere with the action of the vapor in the expansion valve, and as it would use some of the refrigerating capacity when it absorbs ammonia after its condensation, a rectifier D is used to reduce the water content. This rectifier cools the vapors still further. The heat is absorbed by cold water or by the strong liquor which is pumped from the absorber through tubes in the rectifier on its way to the generator. This substance is at a lower temperature than the vapors, and although low enough to condense most of the water it is not low enough to condense the ammonia. The condensed water is caught in a separator E and as it absorbs ammonia, the aqua liquor thus formed is sent back to the gen- erator. From the rectifier the dry vapor is sent into the con- denser F where the ammonia vapor is condensed by a cold water supply. The liquid is now passed through the expansion valve REFRIGERATION 433 G and enters the expansion coil of the brine cooler H in which the liquid is evaporated by the abstraction of heat from the surrounding brine. If the direct-expansion system is used this heat is drawn from the room. The vapor is drawn into the absorber / by the weak liquor which has been allowed to flow from the generator. The great capacity of this weak liquor for ammonia produces a reduction in pressure of the vapor. This absorption of ammonia generates heat and a coil / through which cool water is circulated is placed in the absorber to keep the temperature at a low point. This low temperature is necessary to keep liquor at a point of great capacity for ammonia. The absorbing power of water decreases as the temperature is in- creased. The strong liquor is pumped from the absorber into the generator at higher pressure by the pump K. To save some of the heat it is passed through the interchanger L in which it takes heat from the hot weak liquor leaving the bottom of the generator. The weak liquor flows to the absorber / in which the pressure is less than that in the generator. This strong liquor is sometimes first passed through the rectifier pipes as it is cool enough to condense the water vapor, and from this coil it passes to the interchanger or exchanger L and then the strong liquor is discharged into the analyzer in which it is still further warmed by the superheated vapors. The arrangement in the figure, how- ever, is such that water is used to cool the analyzer. The strong liquor finally reaches the generator and the boiling drives off the ammonia. The principle of the absorption machine is the same as that of the commpression machine. The vapor is put under such a pressure that its temperature of vaporization is above that of a water supply and this water may remove heat and condense the ammonia; after this it is placed under such a pressure that heat may be abstracted from a region of low temperature. The pressure is produced by the steam coil in place of the compressor and the low pressure is maintained by absorption in place of the suction of the compressor. PROPERTIES OF AQUA AMMONIA The action of ammonia and water (known as aqua ammonia) will now be studied to form a basis for the analysis of the action of this machine. The amount of ammonia or other vapor absorbed by a pound 28 434 HEAT ENGINEERING of water depends on the pressure and temperature of the water. Thus, according to the table prepared by Lucke, water will only absorb 3.1 per cent, of its weight of NH 3 at 191 F. and atmos- pheric pressure, while at 50 Ibs. gauge pressure this temperature for the same concentration could be raised to 280 F. 191.5 F. is the temperature limit at 50 Ibs. pressure for 28 per cent, of absorption. The relation between the pressure, per cent, of ammonia in a solution and the temperature at which this solution will give off ammonia or rather boil has been determined experimentally by Mollier and although plotted in curves these values are represented by Maclntire by the equation ^ = 0.00466z + 0.656 (67) 1 sol. T aa t. = temperature of saturation of ammonia for a given pressure. Tgoi. = temperature at which the solution boils. x = per cent, of NH 3 in solution = per cent, of con- centration. = Ibs. of NH 3 in 1 Ib. of solution. This formula holds to 50 per cent, concentration. Above that concentration the formula does not give correct results but this is not important as such high concentrations are rarely used. The vapor concentration of the liquid being known with the temperature of boiling (T soL ), the temperature (T sat .) cor- responding to the boiling pressure, p sa t', of liquid ammonia is found. This gives the total pressure exerted by the ammonia and water vapors driven off from the liquor. The partial pressures exerted by the two constituents are not well known at present. These partial pressures are proportional to the number of mole- cules present or to the partial volumes of each constituent. To know how these molecules come off from the solution is a diffi- cult problem. Lucke quotes values of partial pressures from Perman. These are limited to 140 F. and to concentrations from 2.5 to 22.5 per cent, by weight of ammonia. The values indicate the manner in which the ammonia vapor pressure varies with the water vapor. Spangler suggests that the water vapor pressure is equal to that for steam at the temperature given multiplied by the ratio of number of the molecules of water vapor to the total number REFRIGERATION 435 of molecules present. He assumes that the concentration of the vapor is the same as the concentration of the liquid. Thus if x is the per cent, solution which is ammonia or the amount of ammonia in 100 Ibs. of liquor the ratio of the number of ammonia molecules and water molecules is x , 100 - x I7 t( 18" Hence if p is the steam pressure for the temperature considered : 100- x 18 1700 - 17 x P 100 - x x^ '" P 1700 + x 18 17 = partial pressure of moisture. (68) This checks with the results of Perman, as given by Lucke, with a close degree of approximation. If 1 Ib. of ammonia be added to a large quantity of water it is found that 893 B.t.u. of heat will be liberated. This is called the heat of complete absorption. If the ammonia is not mixed with a large quantity of water the absorption will be partial. If strong ammonia liquor is added to water it is found that heat is developed by the solution of this in water; the strong solution acting as ammonia. This latter heat is called the heat of complete dilution and the difference between the heat of complete absorption and that of complete dilution is known as the heat of partial absorption. Complete dilution occurs when the ammonia is diluted with 200 times its weight of water. It has been found by Berthelot that the amount of heat de- veloped when a liquor of aqua ammonia is diluted so that the water content is 200 times the weight of the ammonia, if multi- plied by the weight of water per pound of ammonia in the original solution gives a constant or practically constant prod- uct, 142.5. The heat of complete dilution becomes 142.5 142.5a ^ := 100 - x ~ 100-s (69) x x = per cent, weight of NH 3 in 1 Ib. of mixture. Some later results by Thomsen pertaining to much greater dilutions than those of the useable experiments of Berthelot's 436 HEAT ENGINEERING indicate that 142.5 may be used, the constant varying from 154 for 19.27 parts of water, to 148 for 29.36 parts and 113 for 56.33 parts. If now 893 B.t.u. is the heat of complete absorption and 142 5x - is the heat of complete dilution 1UU x 893 irkn '' - = heat of partial absorption. (70) 1LMJ X The heat of partial absorption is the heat liberated when 1 Ib. of ammonia in solution is added to enough water to bring it to a concentration of x. If T7 Ibs. of ammonia are absorbed by - Ibs. of water the heat developed will be <* - - x) If now ammonia were absorbed to change the concentration from x to x', the amount of ammonia then present in the water contained in the original pound of solution of strength x would be _ 100 A 100 - x' This may be seen from the fact that the amount of water in 1 Ib. of a solution of strength x is --JQQ and the amount of ammonia per pound of water in the second condition is 100 - x' Hence the amount of ammonia in ~~ Ibs. of water is 100 - x __?__ *L v 10 ~ x 100 X 100 - x' ~ 100 100 - x' The heat generated by the addition of this amount of ammonia to bring the solution to strength x' is *' , , 100 - x r 142.5s' ioo x The heat developed by the addition of ammonia to change the strength of the solution from x to x' is therefore *' 100 - x r 142.5* I _ _^r 893 _ 142^1 y ~ ioo x ioo - *'L 893 " 100 - *'J ioor* ioo -xJ (74) REFRIGERATION 437 The weight of ammonia added is 100- a X - an _ jc_ - x'\ 100 ~ lOO 100 If the heat is divided by the weight the result will be the heat per pound or , = 893 - ^-, + (75) This is the amount of heat liberated when 1 Ib. of ammonia vapor is absorbed by a liquor of concentration x to change the concentration to x'. The above expression is true for the absorption of ammonia but when 1 Ib. of ammonia is driven from a solution to change the concentration x f to concentration x not only is the heat q necessary but in addition to this, it is necessary to evaporate the water vapor present and to superheat the vapors. The amounts of these quantities will be found in a problem later. The density of the liquor of aqua ammonia varies with the amount of ammonia absorbed decreasing with the concentration. The specific gravity is given for various concentrations x by 4 Q r T 2 r s -i sp. gr. = 1 - x + (76) The strong solution would be lighter and would stay at the top were it not for diffusion. The specific heat of aqua ammonia will be assumed to be unity. When a strong liquor of strength x' is changed to strength x by giving off 1 Ib. of ammonia the number of pounds of strong solution, y, required is given by x' x ^Too ~ ( y "" l ' loo = l i -*- 100 100 - x V = ^r- r=^n (77) 100 100 The weight of weak solution is (y 1). The above discussion gives the necessary properties of aqua ammonia and to apply them as well as to compute the heat trans- fers in the various parts of the apparatus a problem will be solved. 438 HEAT ENGINEERING PROBLEM Find the amount of refrigeration per pound of ammonia driven off per minute from 25 per cent, solution in the generator if the room temperature is to be held at 25 F. and the cooling water varies from 50 to 65 F. How much steam at 20 Ibs. gauge pressure and x = 0.9 would be required per minute per ton of capacity? Find the heat transfer of the various parts of the apparatus together with the amount of water used and a heat balance. Temperatures and Pressures. Temperature of ammonia in condenser with a 15 difference from the hottest water is 65 + 15 = 80 F. Temperature in expansion coil with 15 difference of tempera- ture is 25 - 15 = 10 F. Temperature of steam in generator (at 34.7 Ibs. abs.) = 259 F. Pressure in condenser Ammonia pressure (for 80 F.) = 153.9 Ibs. Steam pressure (for 80 F.) = 0.51b. Total pressure =154. 4 Ibs. Pressure in expansion coil (10 F.) = 38.02 Ibs. Pressure in absorber [38.02 - 0.5 (assumed)] = 37.52 Ibs. Temperature of saturation of ammonia = 9.5 F. Pressure in rectifier [154.4 + 0.5] = 154.9 Ibs. Temperature limit of rectifier (154.9 Ibs.) = 80.4 F. Pressure in analyzer [154.9 + 0.5] = 155.4 Ibs. Pressure in generator [155.4 + 0.5] = 155.9 Ibs. Temperature limits in various parts : Generator. For 155.9 Ibs. pressure the temperature at which a 25 per cent, solution will boil is ^'V 46 = 0-00466 X 25 + 0.656 1 801- T 80l . = ~5~ = 700 abs. = 240 F. This is the lowest temperature possible to drive off the am- monia with 25 per cent, concentration. As the evaporation is carried on the concentration becomes less. To boil the liquor to a smaller concentration requires a temperature higher than 240 F. The heat required to liberate the ammonia from the REFRIGERATION 439 strong liquor reduces the temperature of the weak liquor and makes the temperature of the vapors leaving the generator that required by the formula for the strong liquor. This will be the assumption used in this work. The limiting concentration at this point when the liquid has time to be brought to within 5 of the temperature of steam, or 259 - 5 = 254 F., is = 0.00466* + 0.656 254 + 460 102 x = T-^T = 21.9 per cent. This gives a very small change in concentration. Suppose that the gauge pressure is raised to 30 Ibs. This gives a temperature of 274 F. - 00466 * + ' 656 x = 17.50 per cent. This will be used. Thus the pressure in the condenser fixes the pressure in the generator and this with the concentration fixes the temperature or pressure for the steam used to boil solutions. The temperature limit of the absorber is given by 95-+- 460 '- - = 0.00466 X 25 + 0.656 J-sol. 469.5 0.773 Suppose this is kept at 145 F. For a 15 difference in temperature in the coils of the absorber the water should be kept at 130 F. Of course the water from the condenser could be used here as its temperature is not above 65 F. Limiting conditions leaving rectifier are given by the tempera- ture of condensation of the ammonia. This is 80.4 F. so the temperature at this point may be taken at 90 F. The possible concentration at 90 F. and pressure 154.9 Ibs. is T SOL = TTT = 607 abs. = 147 F. h 0.656 x = 70 per cent. Having these limiting concentrations, the amount of moisture 440 HEAT ENGINEERING at the various points may be found except at the discharge from the analyzer because at that point the temperature is not known as it is due to the mixture of the strong liquor and the liquor from the rectifier. To find this, the amount of strong liquor to give 1 Ib. of ammonia when changed from 25 per cent, concentra- tion to 17.5 per cent, concentration is computed. The formula 100 -a y = x f -x is true for absorption and gives 11.00 Ibs. for y for the conditions above. 100-17.5 y = 25-17.5 = 1L When ammonia is driven off some steam is driven with the ammonia and for this reason the concentration of the remaining liquor is made greater. To find the value of y the conditions leaving the generator must be known. Total pressure in generator ................................ 155 . 9 Ibs. Temperature in generator ................................. 240 F. Concentration ........................................... 25 per cent. (1700 _ 17 V 2^\ 1700 + 25 / = 18 ' 51bs - Partial pressure of ammonia ............................... 137.4 Ibs. Saturation temperature ammonia .......................... 73 . 5 F. Superheat .................................. ; ............ 166 . 5 F. Heat content ammonia ................................... 658 . 1 Specific volume ammonia ................................. 3 . 07 Saturation temperature steam ............................. 224 F. Superheat ............................................ ... 16 F. Heat content steam ...................................... 1162 Specific volume steam .................................... 22 . o ryj Weight of steam with 1 Ib. NH 3 = |^ = ................. 0. 1395 The value of y may now be found 0.250 - 0.175 [y- (l+ 0.1395)] = 1 1 - 0.1995 y = - =10 ' 65 This is weight of liquor at 25 per cent, concentration entering the generator. The weight of weak liquor leaving is (10.65 - 1.1395) = 9.5105 Ibs. of 17.5 per cent, concentration. This passes to the absorber. REFRIGERATION 441 The amount of ammonia absorbed by bringing this solution to a concentration of 25 per cent, is given by 100 [9.5105 (1.00 - 0.175) -==- - 9.5105] = * o 10.461 - 9.5105 = 0.9505 This is the amount of ammonia absorbed but there is still a further amount absorbed by the water in the rectifier and an- alyzer. That leaving the rectifier includes a further amount, although small, which is taken over by the condensed moisture formed from the moisture leaving the rectifier and passing into the condenser and from it into the expansion coils and absorber. The amount of water vapor leaving the rectifier must now be found. Conditions at discharge of rectifier: Total pressure .................................. 154 . 9 Ibs. Temperature ................................... 90 F. Concentration of liquor leaving ................... 70 per cent. Partial steam pressure f"o.696 X ^^^ ^ ^ 7 1 =0.20 Ibs. Partial ammonia pressure ........................ 154 . 7 Ibs. Temperature saturation of ammonia ............... 80.3 F. Superheat ammonia ............................. 9 . 7 F. Heat content ammonia .......................... 564 . 4 B.t.u. Specific volume ammonia ........................ 1 . 99 cu. ft. Temperature saturation of steam ................. 53 F. Superheat steam ................................ 37 F. Heat content ................................... 1100 B.t.u. Specific volume ................................. 1640 From this the amount of moisture leaving per pound of am- monia is - 0.0012 The amount of ammonia absorbed by this amount of water, if condensed, to produce a 25 per cent, concentration is 05 0.0012 X 75 = 0.0004 Hence for every pound of ammonia absorbed 0.9996 Ib. will be absorbed by the weak liquor and 0.0004 Ib. will be absorbed by the water sent in from condenser. Since the actual weak liquor absorbs 0.9505 Ib. per pound of ammonia, the total weight of ammonia absorbed to make the strong liquor is 442 HEAT ENGINEERING 5^-0 9509 Ib 0.9996 " and the weight of strong liquor is ' I ' . 9.5105 + 0.9505 + 0.9509 X 0.0016 = 10.4625 This liquor at 145 F. is passed through the interchanger which is supplied with 9.5105 Ibs. of weak liquor at 240 F. The heat given up by this liquor in reducing its temperature to 150 F. by the countercurrent interchangers is equal to Q = 9.5105 [240 - 150] X 1 = 856 B.t.u. This assumes that the specific heat of the liquor is 1. If a 20 per cent, radiation loss is assumed from the interchanger the temperature of the strong liquor leaving this apparatus and entering the analyzer is This strong liquor is mixed with a small amount of liquor- of higher concentration but at 90 F. from the rectifier. This will decrease the temperature although the heat developed by the solution of the strong liquor will increase this latter temperature. The net effect is to decrease the temperature about 3^. The exact amount of decrease will be found but to get the quantity of liquor from the rectifier for first approximation it will be well to assume a 3J^ drop rather than no drop at all. The gases from the analyzer will be brought to the temperature of the liquors entering the top of the analyzer by the cooling effect of these as the heat of superheat is much less than the heat re- quired to raise the temperature of the liquor to that at the desired discharge into the generator from the analyzer. The conditions of the vapors leaving the analyzer and entering the rectifier are fixed by the temperature and pressure at this point. Pressure ......................................... 155.4 Ibs. Temperature assumed ............................. 207 F. Concentration f^y^:^ = 0.00466z -f 0.6561 x =33.2 Partial pressure steam [ 13.3 X 17( f 7 ~ Q ^ ^f'^] = 8-72 Ibs. Partial pressure ammonia .......................... 146. 7 Ibs. REFRIGERATION 443 Temperature saturation ammonia 77 . 2 F. Superheat ammonia 129 . 8 F. Heat content ammonia 638 . 9 Specific volume ammonia 2.71 Temperature saturation steam 187 F. Superheat steam 20 F. Heat content steam 1149 . 3 Specific volume steam 45 . 1 2 71 The water vapor per pound of ammonia = -r^r = 0.06. 4O.1 Since the water vapor leaving the rectifier per pound of am- monia is 0.0012, the amount of water removed from the rectifier per pound of ammonia leaving the rectifier is 0.06 - 0.0012 = 0.0588 In addition to this the amount of moisture associated with the ammonia absorbed is also condensed, call this latter M. The concentration of the liquor leaving the rectifier for the analy- zer is 70 per cent. Hence 70 [0.0588 + M] ~ | 0.06 = M 0.0588 X -JQ = Jlf[l - gg M = 0.00957 Hence the moisture condensed per pound of ammonia entering the condenser is 0.0588 + 0.0096 = 0.0684 and for 0.9505 Ib. of ammonia this is 0.065. The ammonia absorbed by this is 0.065 x||* 0.1517 The amount of liquor passing from rectifier is then 0.1517 + 0.065 = 0.2167 Ib. This is at a temperature of 90 F. On mixing with 10.4625 Ibs. of strong liquor at 210.5 F. the temperature of the mixture is given by: , _ 10.4625 X 210.5 + 0.2167 X 90 10.6792 The concentration of this mixture is now found. 444 HEAT ENGINEERING Ammonia from strong liquor = 10.4625 X 0.25 = 2 . 61572 Ammonia from rectifier liquor = = 0.15170 Total ammonia ............................. 2 . 76742 2.76742 Concentration = 1Q 5790 = 25.913 per cent. The temperature of such a concentration at a pressure of 155.4 Ibs. is given by: = 0.00466 X 25.91 + 0.656 1 sol- T 80L = 697 abs. = 237 F. Since the temperature of the mixture is below this point the solution would remain of strength given were it not for the heat of solution when the stronger rectifier liquor is diluted in the strong liquor. To change 10.4652 Ibs. of liquor from strength 25 per cent. to 25.91 per cent, requires an amount of ammonia equal to 10.4625 [0.2591 - 0.25] = 0.0955 This should also equal the ammonia lost by the rectifier liquor 0.2167 [0.70 - 0.25913] = 0.0955. The mixture of a strong solution in a weaker solution de- velops heat. This may be considered as the difference between the heat developed when the weak solution is made strong and that required to weaken the strong solution. Heat = = 0.0955[g-|]l42.5 - 0.0955 X 142;5[2.33 - 0.33] = 27.22 This heat is used to raise the temperature of the mixture and part may be used to drive off ammonia if the temperature be- comes greater than the temperature of solution for the actual concentration at the given pressure. The temperature rise for 27.22 B.t.u. is 27.22 _ 10.6792 = 2 ' 55 This gives 208.05 + 2.55 = 210.55 F. REFRIGERATION 445 This is less than 237 F. and hence there is no evaporation. If this were higher it would be well to assume an intermediate temperature a little above the solution temperature for the pressure and concentration and compute the concentration for this temperature. Then the amount of ammonia and water vapor driven off would be computed together with the heat required to do this. If the assumed temperature were correct, this heat together with the heat to raise the solution from 208.7 to the assumed temperature would equal 27.22 B.t.u. If the sum is greater, a lower temperature is assumed and the quanti- ties computed. After several assumed temperatures the correct one may be interpolated. In any case the final answer gives the temperature at the dis- charge from the analyzer and the second value of condensation in the rectifier may be found. The actual conditions at discharge from the analyzer into the rectifier are as follows : Pressure ....................................... 155 . 4 Ibs. Temperature ................................... 210 . 25 F. Concentration .................................. 32 . 3 Partial steam pressure ........................... 9 . 39 Ibs. Partial ammonia pressure ........................ 146.01 Ibs. Temperature saturation ammonia ................. 76 . 9 F. Superheat ammonia ............................. 133 . 3 F. Heat content ammonia .......................... 640 . 9 Specific volume ammonia ........................ 2 . 73 Temperature saturation steam ................. ... 190 F. Superheat steam ................................ 20 . 25 F. Heat content steam ............................. 1151 Specific volume steam ........................... 42 . 4 Water vapor per Ib. of ammonia .................. 0.0644 Water condensed per Ib. of ammonia passing into condenser 0.0644 - 0.0012 = 0.0632 M = 0.0644 [0.0632 + M] ^ <5U = 0.01117 Total moisture = 0.0632 + 0.01117 = 0.0744 per Ib. Moisture for 0.9509 Ib. = 0.0707 Ammonia absorbed = 0.0707 X T^ = 0.165 Liquor from rectifier = 0.165 + 0.0707 = 0.2357 , . x 10.4625 X 210.5 + 0.2357 X 90 Temperature of mixture = - 1Q - = 207.84 2 7807 Concentration of mixture = 1Q 6982 = 25.99 per cent. 446 HEAT ENGINEERING Temperature solution = 696 abs. = 236 F. Ammonia taken up by weak solution 10.4625 [0.2599 - 0.25] = 0.1036 Ammonia given up by strong solution 0.2357 [0.70 - 0.2599] = 0.1037 Heat of change of solution 0.1037 |^- ^ 1 142.5 = 29.55 29 55 Temperature rise = IQ gob 6 = 2 -76 F. Temperature of discharge = 207.84 + 2.76 = 210.6 F. This is below 236 F., hence there is no further change of con- centration and the condition of the entrance into rectifier of 210.25 may be used as the difference in temperature does not change data. The ammonia entering rectifier is therefore 0.9509 + 0.165 = 1.1159 The vapor entering is given by two methods: Vapor = [0.0744 + 0.0012] 0.9509 = 0.0719 Vapor = 1.1159 X 0.0644 = 0.0740 Mean value = 0.073 The liquor at temperature 210.25 F. drops through the analyzer and receives heat from the hot vapors. As this heat is not sufficient to heat the liquor to 240 F., it will be assumed that a steam coil is used to supply the necessary heat which will be computed. The liquor is assumed to leave at 240 F. The conditions for this are given for exit from generator on p. 440. Since it is known that the discharge from the generator is 1 Ib. of ammonia and 0.1395 Ib. of steam, at each point of the appa- ratus the amount of substance is known. This can be put in tabular form: GENERATOR : Entering. 10.65 Ibs. liquor of 25 per cent, concentration at 240 F. Leaving. 1 Ib. ammonia vapor at 240 F. 0.139 Ib. water vapor at 240 F. 9.501 Ibs. of liquor of 17.5 per cent, concentration at 240 F. ANALYZER : Entering. 1 Ib. ammonia vapor at 240 F. 0.139 Ib. of water vapor at 240 F. 10.67 Ibs. liquor of 25.99 per cent, concentration at 210.6 F. REFRIGERATION 447 Leaving. 10.65 Ibs. liquor of 25 per cent, concentration at 240 F. 1.116 Ibs. of ammonia vapor at 210.6 F. 0.074 Ib. of water vapor at 210.6 F. RECTIFIER: Entering. 1.116 Ibs. of ammonia at 210.25 F. 0.074 Ib. of water vapor at 210.25 F. Leaving. 0.9509 Ib. ammonia vapor at 90 F. 0.0011 Ib. water vapor at 90 F. 0.236 Ib. liquor of 70 per cent, concentration at 90 F. CONDENSER : Entering. 0.9509 Ib. ammonia vapor at 90 F. 0.0011 Ib. of water vapor at 90 F. Leaving. 0.9505 Ib. of ammonia liquid at 80 F. 0.0015 Ib. liquor of strength 25 per cent, at 80 F. EXPANSION COIL: Entering. 0.9505 Ib. ammonia liquid at 80 F. 0.0015 Ib. liquor of strength 25 per cent, at 80 F. Leaving. 0.9505 Ib. ammonia at 10 F. 0.0015 Ib. water vapor (strength 25 per cent, assumed) at 10 F. ABSORBER: Entering. 9.501 Ibs. liquor of 17.5 per cent, concentration at 150 F. 0.9505 Ib. ammonia at 10 F. 0.0015 Ib. water vapor at 10 F. Leaving. 10.463 Ibs. liquor of 25 per cent, concentration at 145 F. The heat interchange for each part is found by the difference between the intrinsic energy and the work (or heat content) at entrance and exit plus an allowance for radiation. GENERATOR : Entering. Heat of liquid of 10.65 Ibs. 10.65[240 - 32] = 2218 Heat of solution of liquor 1^893- -0.25 X 10.65| 893- 142.5 X ^ = - 2255 /oj - 37 Leaving. Heat of 1 Ib. ammonia = 1 X 658.1 = 658.1 Heat of 0.1395 Ib. steam = 0.1395 X 1162 = 162.0 Heat of liquid of 9.501 Ibs. liquor . 9.501 [240 - 32] = 1975.0 Heat of solution of liquor [893 - -0.175 X 9.501 893 - 142.5 X = - 1435.0 1360.1 Net heat to be supplied = 1360.1 - (- 37) = 1397.1 448 HEAT ENGINEERING ANALYZER: Entering. Heat of 1 Ib. ammonia = 1 X 658.1 = 658.1 Heat of 0.1395 Ib. steam = 0.1395 X 1162 = 162.0 Heat of liquid of liquor = 10.67 [210.25 - 32] = 1910.0 [OCQQT 893 - 142.5 X 7401] = - 2340.0 390.1 Leaving. Heat of liquid of 10.65 Ibs. = 10.65 [240 - 32] = 2218 [25~l 893 - 142.5 X 75 = - 2255 Heat of 1.116 Ibs. ammonia = 1.116 X 640.9 = 716.0 Heat of 0.074 Ib. steam = 0.074 X 1151 = 85.2 764.2 Net amount added = 764.2 - 390.1 = 374.1 Total amount added in generator and analyzer allowing 20 per cent, for radiation is 120 [1397.1 + 374.1] = 2125 B.t.u. Pounds of steam required at 30 Ibs. gauge and x = 0.9 is given by Mat - 0.9 X927.9 = 2 ' 55 lbs ' RECTIFIER: Entering. Heat of 1.116 lbs. ammonia = 716.0 Heat of 0.074 Ib. steam = 85.2 801.2 Leaving. Heat of 0.9509 Ib. ammonia = 0.9509 X 564.4 = 536.0 Heat of 0.0011 Ib. steam = 0.0011 X 1100 = 1.2 Heat of liquid of 0.236 Ib. liquor = 0.236 [90 - 32] = 13.7 Heat of solution = - 0.7 X 0.236 s93 - 142.5 X 1 = - 92.5 458.4 Net heat abstracted = 801.2 - 458.4 = 342.8 B.t.u. CONDENSER : Entering. Heat of 0.9509 Ib. ammonia = 536.0 Heat of 0.0011 Ib. steam = 1.2 537.2 Leaving. Heat of liquid of 0.9059 Ib. ammonia = 0.9509 X 53.6 = 51.0 Heat of liquid of 0.0011 Ib. water = 0.0011 X 48.1 = 0.0 Heat equivalent of ApV = 0.0 51.0 Net heat abstracted = 537.2 -51 = 486.2 B.t.u. EXPANSION COIL: Entering. Heat of ammonia = 51.0 Heat of water = 0.0 51.0 REFRIGERATION 449 Leaving. Heat of ammonia vapor = 0.9505 X 541.2 = 514.0 Heat of water vapor = 0.0015 [10 - 32] = - 0.0 [25H 893 - 142.5 X 75 I = - 3.4 510.6 (Work of liquid neglected.) Heat abstracted 510.6 - 51 = 459.6 Tons of refrigeration per pound of vapor per minute: = 2.3 tons ABSORBER: Entering. Heat of liquid of liquor = 9.501 [150 - 32] = 1120.0 Heat of solution of liquor = - 0.175 X 9.501[862.7] = - 1435.0 Heat of ammonia = 514.0 Heat of liquor = 3.4 195.6 Leaving. Heat of liquid of liquor = 10.463 [145 - 32] = 1180 Heat of solution = - 0.25 X 10.463 [845.5] = -2215 - 1035 Net heat abstracted = 195.6 - (- 1035) = 1230.6 INTERCHANGER : Entering. Heat of liquid of strong liquor = 10.463 [145 - 32] = 1180.0 Heat of liquid of weak liquor = 9.501 [240 32] = 1975.0 Heat equivalent of work = ^ I 155.4 - 37.5J X 144X10.463 62.5 X 0.913 3157.6 43 / 25 2 35 3 \ [Density = 1 - ^ ( 2 5 - ^ + ^) - 0.913] Leaving. Heat of liquid of strong liquor = 10.463 [210.5 - 32] = 1865 Heat of liquid of weak liquor = 9.501 [150 - 32] = 1120 2985 Heat radiated = 3157.6 - 2985 = 172.6 B.t.u. 2 6^ The work = 2.65 B.t.u = ' 42 X 2 = 0.125 h.p. for 1 Ib. of ammonia per minute with 50 per cent, efficiency. 29 450 HEAT ENGINEERING HEAT BALANCE: Taken in Given out 1397. I 1 Generator . . Analyzer. . . 374.1 354.0 354.0 Rectifier ............. .............. 342.8 Condenser ......... ..... ..... 486.2 Expansion coil .................... 459 . 6 Absorber ...................... . . i| i ............. 1230.6 Pump .......................... :;.. 2.65 . Exchanger .............. ...... 172.6 Total .......................... 2587.4 ........ 2586.2 The cooling water entering the condenser at 50 F. is raised to 65 F. This could then go to the rectifier where its temperature could be raised to 75 F. and finally to the absorber where the temperature could be increased to 130 F. The weights of water would then be: Wt. for condenser per 1 Ib. ammonia liberated = . ' = 32.2 342 8 Wt. for rectifier per 1 Ib. ammonia liberated = ' = 34.3 6 Wt. for absorber per 1 Ib. ammonia liberated = ^.~- = 22.4 The amount of water actually needed is 34.3 Ibs. The surface required in these different sections is found by methods of Chapter III. The other data computed for the other refrigerating machines will be computed for comparison. Tons of refrigeration per Ib. ammonia = OAAA v 144 = 0-0^159 459 6X8 Tons of ice melting capacity per Ib. coal = o 55 v 2000 X 144 = 0>00 ^ (Assuming 8 Ibs. of steam per Ib.) Tons of ice melting capacity per Ib. steam = -^ = 0.0006 120 The brine pump would require 0.125 X -QQ = 0.25 Ib. of steam per minute for 459.6 B.t.u. of refrigeration or 2.55 Ibs. of steam in generator. Hence the exhaust from the pump can be used easily for part of the steam supply. Gallons of cooling water with 15 rise per ton of refrigeration = 34.3 8.35 X 0.00159 REFRIGERATION 451 ajngsaid 8inss8.id (iq-'d-q jad -sqj OS) urBais qj jad uorvBja -8UJ8J JO SUOJ, 3up[8Ui aoi jo uo^ Jad J8}13AY 8m -1000 JO O O 00 O Tt< T}H 00 00 TH i 1 CO CO CO i I CO CO GO 00 O5 O5 O 00 00 CO CO CO iO O iO iO CO i 1 rH O5 O O> 00 t^ 1> O CO Tt< 00 T-I O CO l> Tt< CO to 3ui 801 JO SUO JL x x x x x x >O O O 00 ^7) b- CO CO CO 00 00 O5 (N (N Oi jq--d-q J8d -q t e) BOO -q[ J8d -jq ^ jad A^pBdeo 3ui 801 JO SUOJ, CO CO CO O co ^O Tt< O CO 10 O i 1 1 CO r ( O d d d d d d d d-q jad -jq ^ J8d X^IO'Bd'BO 3UI ^[801 801 JO SUOJL O i-H i- (N ^H jq -d-q aad -3uj8J jo -n- JO n -3UJ8J jo ' Oi c O O O rH O 00000 d d d d d d tiopouj -J8d CO C^ O 2 machine is the objectionable feature of this. TOPICS Topic 1. Sketch and describe the action of an air refrigerating machine. Give expressions for the heat on each line. Explain what is meant by per- formance or refrigerative effect. Topic 2. Sketch and describe the action of a refrigerating machine using a volatile fluid and its vapor. Sketch the T-S diagram and from it give the expressions for the heat on each line. Explain what is meant by perform- ance. Explain what is meant by a ton of refrigeration. Topic 3. -Sketch and explain action of the absorption refrigerating ma- chine. Give the function of each part of this apparatus. Topic 4. Explain method of finding the temperatures and pressures of the various points on the air cycle and ammonia cycle of a refrigerating machine. What is a dense air machine? Give the expressions for work, heat removed by the cooling water in each case and heat removed from the refrigerator. Topic 5. What is the effect of clearance? Show this completely. What is the effect of friction? Give the expressions for the power necessary for air and ammonia machines per pound of substance used. How is the volume of the compressor found? That of the expander? Topic 6. What is the effect of moisture in an air machine? Derive an expression for the work in the expander when moisture is present and when no moisture is present. Derive expressions for the work of the compressor with and without moisture. From the expressions above write the expres- sions for work with and without friction. Topic 7. -What vapors are used for refrigerating machines ? What are the advantages of one over the other? What is meant by wet and dry compres- sion? Which is the better? Why? What is the clearance factor? Sketch T-S diagrams for each and explain action. Why is the expansion valve used ? Write the expressions for work, heat and refrigerating effect. Topic 8. What is aqua ammonia? What is the effect of adding ammonia to water? What is the effect of adding aqua ammonia to water? What fixes the pressure at which ammonia will be given off from aqua ammonia at a certain temperature? What fixes the partial pressures of the ammonia and water vapors discharging from this solution? What is meant by per cent, ammonia or per cent, concentration? Topic 9. What is the heat of complete dilution? Complete absorption? Partial absorption? What is manner of variation of the density of aqua ammonia? What value is assumed for the specific heat of aqua ammonia? REFRIGERATION 453 Topic 10. Outline the analysis of the action of the absorption machine. What data is assumed and what does this fix? PROBLEMS Problem 1. An air machine is closed and operates between 40 Ibs. gauge and 200 Ibs. gauge with cooling water at 60 F. to 75 F. and a room held at 5 F. Find the temperatures at the various corners. Find the horse-power per ton of refrigeration with 10 per cent, friction effect. Find the cubic feet of displacement per minute in each cylinder and the amount of cooling water per minute per ton of refrigeration. Clearance 5 per cent, on each cylinder. Problem 2. An air machine is to operate under conditions of Problem 1 except that it is open and operates to 45 Ibs. gauge pressure. Find the quan- tities asked for in Problem 1. Problem 3. An ammonia machine operates with cooling water from 60 F. to 75 F. and cools a room with direct expansion to 5 F. Find the pressures used. Find the horse-power per ton of refrigeration with 10 per cent, fric- tion effect and clearance 3 per cent. Find the amount of water per minute and the displacement per minute for 1 ton. (a) Solve this with wet com- pression. (6) Solve this with dry compression. Problem 4. Solve Problem 3 using COa as the medium. Problem 5. Solve Problem 3 using SO 2 as the medium. Problem 6. A 25 per cent, solution is made by the addition of ammonia to a 10 per cent, solution of aqua ammonia. How much ammonia has been added per pound of original solution? What is the density of each solution? What heat is developed by this addition of ammonia? Problem 7. A 25 per cent, solution is to be boiled at 170 Ibs. gauge pressure. At what temperature will it boil? At what temperature must this be heated if it is to be reduced to a 10 per cent, solution. What are the partial pressures above the 10 per cent, solution in this case? Problem 8. Find the amount of heat per pound of vapor in the vapors coming from a 10 per cent, solution of aqua ammonia boiling at 170 Ibs. gauge pressure. INDEX Authors, Authorities, Inventors and Investigators Adams, 210 American Rad. Co., 101 Andrew, 360 Atkinson, 363 Avogadro, 27 Ball, 233 Barnett, 359 Barr, 209 Barrus, 178 Barsanti, 359 Beau de Rochas, 360 Bernouilli, 279 Berthelot, 435 Bertrand, 39 Bisschop, 360 Boltzmann, 73 Boulvin, 189 Box, 348 Boyle, 26 Brille, 80 Brown, 359 Buckingham, 314 Buffalo Forge Co., 101 Callendar, 198, 204, 208 Carcanagues, 80 Carnot, 9, 11, 12, 13, 62, 170 Carrier, 343 Charles, 26 Clark, 233 Clausius, 167, 170 Clayton, 210, 214 Clerk, 360 Coker, 369 Corliss, 167 Cotterill, 198 Curtis, 302 Dalby, 77, 80 Dalton, 28 Davis, 174, 317 De Laval, 301 Dent, 156 Diesel, 364 Duchesne, 209 Dulong, 73, 403 Dupre Hertz, 39 Duhring, 356 Fliegner, 60 Foran, 335 Giffard, 275 Gilles, 360 Goodenough, 16, 17, 25, 39, 44, 45, 47, 215 Hagemann, 99 Hall, 205 Halliday, 80 Harn, 183 Hausbrand, 99, 100, 356 Hautefeuille, 359 Heck, 199, 223 Heinrich, 209 Hilliwell, 360 Huygens, 359 Ingersoll Rand Co., 120, 151 Jelinek, 99 Jordan, 80, 92 Josse, 261 Joule, 2, 37, 100 Kelvin, 8, 36, 37 Klebe, 44 Kneass, 277, 279 Knoblanch, 44, 45 Kreisinger, 94, 108 455 456 INDEX Langen, 44, 360 Ray, 94, 108 Le Chatelier, 44 Reeve, 189 Lenoir, 359 Reynolds, 77, 78, 167 Linde, 44 Rice, 198 Lucke, 73, 464 Richards, 156 Ruggles, 348 Mallard, 44 Marks, 174, 198, 317 Sangster, 119 Matteucci, 359 Savery, 167 Maxwell, 24, 75 Scoble, 369 Mclntire, 434 Ser, 80 Me Mullen, 96 Spangler, 177, 344, 396, 434 Mollier, 45, 47, 80, 99, 100, 434 Stanton, 80, 91 Moyer, 274, 292, 303 Stefan, 73 Stodola, 269, 270, 271 Napier, 60 Stone, 215 Newcomen, 167 Street, 359 Newton, 73, 359 Stumpf, 167, 226, 231 Nicolson, 80, 89, 90, 91, 198, 204, 208 Sultzer, 167 Nusselt, 75, 94 Thomsen, 435 Orrok, 97, 334, 335 Thurston, 198 Otto, 360 Tyndall, 73 Papin, 167, 359 Watt, 167 Parsons, 305, Webb, 232 Perman, 434 Werner, 80 Perry, 78, 198 Weymouth, 370 Petit, 73 Willans, 167, 226 Porter, 167 Worcester, 167 Wright, 359 Rankine, 60, 167, 170, 172, 215 Rateau, 272, 304, 349 Zeuner, 215 INDEX Subjects Absorber, 433 Accumulators, 348 Adiabatic, 2, 5, 16, 31, 54 Adiabatic action, 2 Afterburning, 368 Aftercooler, 120 Air, excess, 404 motor, 151 pump, dry, 334 transmission, efficiency, 152 loss, 152 Allowance for turbines, 328 Ammonia, 428 aqua, 434 Analysis, Hirn's, 183 proximate, 402 T-S, 183 ultimate, 402 Analyzer, 432 Aqua ammonia, concentration, 434 density, 437 heat of absorption, 435 partial pressure, 434 properties, 433 specific heat, 437 Ash and moisture, free basis, 402 Availability, 1, 14, 62 conditions for, 14 loss of, 115 Avogadro's law, 27 Axial thrust, 300 Back pressure, effect of, 177 Binary engine, 261 Blades, number of, 327 Bleeding engines, 255 Blowing engines, 119 tubs, 119 Bone-Schnable boiler, 407 Boyle's law, 26 British thermal unit, 1 Calorimeters, 178 Capacities, thermal, 21 Carbon dioxide, 430 Cards, combined, 243 computation for, 243 construction, 243 Characteristic equations, 19 Charles' law, 26 Charts, 7-S, 47 T-S, 47 Coils for heating, 101 Clearance, effect, 124, 417 effect eliminated, 418 factor, 125 ratio, 125 space, 125 steam, 170, 184 Closed system, 412 Combining tube, 281 Combustion, 375, 402, 403 heat, 376 surface, 406 table, 377 Complex cycle, 38 Compound engine, 241 Compounding for turbines, pres- sures, 297 velocity, 297 Compression, best point, 230 curve, 127 dry, 425 method, 128 temperature, 127 wet, 425 Compressors, 118 air, 118 Condensation, initial, 196 maximum, 207 Condenser, ammonia, 98 barometric, 337 contraflo, 337 457 458 INDEX Condenser design, 338 dry tube, 337 jet, 333 steam, 97 surface, 333 uniflux, 337 Conduction, 15, 72, 74 coefficients, 95 factors, 74 Conductivity, 207 Conservation of energy, 2 Constant steam weight curve, 217 universal gas, 27 Consumption, computation, 224 curves, 222 steam, 173, 210 Convection, 72, 75 Converging nozzle, 273 Cooling, loss from, 140 towers, 341 cost, 347 size, 347 Counter flow, 82 Criterion for exact differentials, 4 Critical pressure, 267 temperature, 49 Cross products, 37 Curve, construction, 218 expansion, 215 pressure in nozzle, 269 steam consumption, 222 Curtis turbine, 302, 311 Dalton's law, 28 De Laval turbine, 301, 309 Delivery tube, 282 shape, 282 Determination of K, 86 Diagrams, combined, 240 factor, 221 I-S, 47 Mollier, 47, 52 temperature-entropy, 172, 175 Differential effect of heat, 43 exact, 3 Diffusivity, 207 Dilution, 368 Discharge, 59 Displacement of air compressor, 135 Double bottoms, 356 Drip pot, object of, 182 Dry compression, 425, 429 basis, 402 test, 187 Dulong's formula, 408 Duty, 67 Efficiency, 12 air standard, 362 Carnot, 12, 14, 62 change, 389 conditions for maximum, 66 cycle, 175 equality, 12 kinetic, 293 maximums, 12, 13, 62, 65, 293 mechanical, 63 overall, 63 practical, 63 Rankine, 171, 173 theoretical, 62 transmission of heat, 103 type, 63 volumetric, 126 Effects, single, double, quadruple, 350 End thrust, 306 Energy, conservation, 2 high grade, 1 internal, 2 intrinsic, 41 Engine, air, displacement, 146 power, 146 work on, 138 and turbine, combined, 329 compared, 309 'binary, 261 bleeding, 255 heat, 62 internal combustion, 359 Mietz and Weiss, 366 multiple expansion, 240 Otto, 365 receiver, 242 regenerative, 256 similar to turbine, 309 size of, 229 INDEX 459 Engine steam, 167 steam cycle, 170 straight flow, 170 Stumpf, 226 tests, 67, 68, 178 uniflow, 226 Woolf, 242 work, 246 Entropy, 16, 37, 43, 47, 51 around cycle, 18 diagram, 19 Equality of temperature scales, 36 Equation, Bernouilli, 279 characteristic, 19 differential, 29 fundamental, 21 Equivalent evaporation, 410 Evaporation, equivalent, 410 factor, 410 Evaporators, 98, 349, 350, 353 Exact differential, 3 Expansion, complete, 138 curves, construction of, 218 free, 172, 175 incomplete, 138, 172, 175, 418 line, 140, 215 loss, 140 ratio, 220 Exploring tube, 269 Explosion temperature, 389 Exponent value, 214 External work, 2, 31, 41 Factor diagram, 221 of evaporation, 410 of safety, 100 Fan blowers, 118 power, 149 Feedwater heaters, 98 First law of thermodynamics, 2 Flow, counter and parallel, 82 of fluids, 56 Free air, 123 Friction factors, 292 Fuels, blast furnace gas, 374 coal, 402 crude oil, 374 gasoline, 374 illuminating gas, 370 Fuels, kerosene, 375 lignite, 402 natural gas, 370 peat, 402 produce gas, 371 wood, 402 Gas composition, 376 engine, temperature, entropy- diagram, 392 test, 69, 70, 396 producer, 371-373 Generator, 432 Governing, air compressors, 149 gas engines, 367 steam engines, 235 multiple expansion engines, 254 Graphical representation of heat, 6 Heat, balance, 402, 450 content, 37, 42, 47, 51 engine, 62 graphical representation, 6 internal, 41 of liquid, 40 of vaporization, 40 on line, 6, 7, 35 on path, 20, 42, 52 transmission, 72, 75, 99, 103, 104 efficiency, 103 Hirn's analysis, 183, 260 High grade energy, 1 Humphrey compressor, 122 Hyperbola, rectangular, 55, 217 Hydraulic radius, 80 compressor, 119, 122 Ignition, 368 Impact coefficient, 284 Indirect heaters, 101 Initial condensation, 196 Injector, 275 capacity, 284 density of discharge, 283 details, 281 double jet, 284 experiments, 277 heat equation, 280 lifting, 284 460 INDEX Injector number, 282 overflowing pressure, 285 temperature, 285 size, 282 steam required, 281 theory, 280 Interchanger, 433 Intercooler, 120, 121, 130 heat removed, 133 water for, 133 from, 134 Internal energy, 2 heat of vaporization, 41 Intrinsic energy, 2, 42, 47, 51 change, 35 Irreversible cycles, 15, 18 Isodynamic, 6, 31, 54 action, 2 Isothermal, 7, 10, 31, 55 Jacket, 121, 128, 253 effect of, 234 heat removed by, 128 saving by, 129 water for, 130 Jet, force of, 287 impulse of, 288 reaction of, 288 work of, 289 Joule's equivalent, 2 Kinetic energy change, 265 Lagging, 254 Latent heat, 21 Law, 14 Avogadro's, 27 Boyle's, 26 Charles', 26 Dalton's, 28 First Law of Thermodynamics, 2 Second Law of Thermody- namics, 14 Stefan-Boltzmann, 73 Leakage, effect of, 134 factor, 125 in pipes, 143 of valves, 208 Lines, thermodynamics, 33, 50 gas-engine card, 392 Locomobile, 235 test of, 68 Logarithmic diagrams, 153, 396 Loss curves, 274 of heat 186, 194 Low-grade energy, 1 Mats, 341, 347 Maxwell's equations, 24 Mean effective pressure, 219 Missing quantity, 196 Mixtures, gases, 28 Molecular weights, 29 Moisture, allowances, 328 effect, 418 Motors, air, 151 Mouth of nozzle, 268 Mouthpiece, 272 Multistaging, 130 reasons for, 135 N-value of, 214 Nozzle, 265, 268, 271 converging, 273 spray, 348 steam, 281 velocity, 278 Open system, 412 Orifices, 271, 272 Overexpansion, 268 Parallel flow, 82 Parson's turbine, 305, 324 Partitions, heat transmitted, 104 Paths, 20 Perfect gases, 19, 26 Performance, method of reporting, 67 Polytropic, 32 Ponds, cooling, 348 Potentials, thermodynamics, 24 Power of motor, 126 Preheater, 152 . Preheating, gain from, 145 Pressure, allowances, 328 change, effect of, 176, 177 critical, 267 INDEX 461 Pressure drop in pipes, 141 intermediate, 130 mean effective, 219 partial, 28, 334 and volume of steam, 225 Producer, gas, 371, 373 Properties of aqua ammonia, 433 Proximate analysis, 402 Pumping engine, test, 69, 258 Quadruple expansion engine, 241 Quality, 40 Radiation, 72 factors, 74 Radiators, 100 Rateau, accumulator, 349 orifice, 272 turbine, 304, 323 Ratio of expansion, 220 Receiver engine, 242 Rectangular hyperbola, 55, 217 Rectifier, 432 Refrigerating machines, absorption, 431 air, 411 dense air, 416 displacement, 414, 427 efficiency, 413 open and closed, 412 power, 427 temperature range, 413 vapor, 424 Refrigeration, tons of, 414 power required, 414 Refrigerative effect, 427 engines, 256 performance, 413 Refrigerator, 9 Reheaters, 253 Reheat factor, 312, 323, 324 Relations, differential, 23 trigonometric, 291 Relative humidity, 342 Reversible action, 10 Rieman steam shock, 272 Rotary blowers, 119 Safety factor, 100 Saturated steam, 20 Scale of temperature, 7 Second law of thermodynamics, 14 Shock, steam, 272 Simple cycle, 37 Solutions, dilute, 99 Source, 9 Specific heat, 21, 30 superheated steam, 30 Speed, 233 Stage compressors, 119 Steam, 20, 39 consumption, 173, 210 curves, 222 turbines, 287 velocity, 278 weight curve, 217 working, 185 Stefan-Boltzmann law, 73 Still, Hodges, 353 Superheat allowances, 328 effect of, 176, 177, 234 Superheated vapor, 44 Sulphur dioxide, 431 Taylor hydraulic compressor, 122 Temperature, after compression, 127 after explosion, 389 boiling, 356 critical, 49 entropy analysis, 188 diagram, 392 gas-engine card, 364, 384 gradient, 76 intermediate, 133 Kelvin's absolute scale, 8 mean difference, 81 scale, 7 equality, 36 Tests, analyses, 183, 188 engines, 178 gas engines, 396 pumping engine, 258 Thermal unit, British, 1 capacities, 21 relations of, 22 Thermodynamic lines, 33 Thermodymanics, fundamental, 1 Thermometers, constant immersion, 181 Throat of nozzle, 268 462 INDEX Throttling action, 58, 123, 425 loss from, 144 Thrust, axial, 300 end, 306 Tons of refrigeration, 414 Total heat, 41 Transmission constants, 105 heat, 72 to boiling water, 99 Trigonometric relations, 291 Triple expansion, 241 Tube exploring, 269 Tubs, blowing, 119 Turbine, compared with engine, 309 combined with engine, 329 computations, 309 Curtis, 302 deLaval, 301 efficiency, 293, 307 low pressure, 329 maximum efficiency, 293 mixed flow, 329 Parsons, 305 pressure, 291 Rateau, 304 steam, 167, 287 test, 69 velocity, 291 Turbo compressors, 118 Ultimate analysis, 402 Underexpansion, 268 Universal gas constant, 27 Vacuum allowance, 328 Valves, air, 136 leakage, 208 Vapor, 39 Variables, independent, 20 Velocity, actual, 289 blade, 289 discharge, 59, 265 factors, 292 injector, 278 relative, 289 whirl, 290, 292 Vento heaters, 101 Viscous liquids, 99 Volume and pressure of steam, 225 Volumes, partial, 28 Volumetric efficiency, 126 Walls, effect of, 204, 369 heat transmission, 104 Water velocity, 278 Weights, molecular, 29 Wet compression, 425, 428 Woolf engine, 242 Work, 31 compression, 123 multiple expansion, 246 Working steam, 170, 185 Y, value of, 266 THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO SO CENTS ON THE FOURTH DAY AND TO $I.OO ON THE SEVENTH DAY "OVERDUE. I 945 'T 19 1941 31 OCT 29 1841 Y 19 N-:V A I'^'W- MAY 28 1S4 194? 4Nov'498fl JAN 17 1942 8 13IW57J REC'D LD JAfr 3 .' 7 UNIVERSITY OF CALIFORNIA LIBRARY