^, ..'%1 -;'-*-'•>-■ mmmmmkmK's^r' 'i m^'j :r';v,:,;::;^,. > -^-^ The Compression and Transmission of Illuminating Gas A Thesis Read at the July, 1905, Meeting of the Pacific Coast Gas Association By E. A. Rix Mem. Am. Soc. Mech. Eng. Mem. Am. Soc. Civil Eng. Assoc. Mem. Am. Soc Mining Eng. Mem. Pacific Coast Gas Assn. San Francisco, Septembrr i, 1^5. The Journal of Electricity Publishing Company , Room S32, Rialto Building San Francisco, Cat. ^z- THE COMPRESSION AND TRANSMISSION OF ILLUMI- NATING GAS. The subject of illuminating gas compression is almost a new one, and the nature of the gas is so entirely different from that of air that we are obliged to consider the question mainly from the theoretical standpoint, backed up by a few indicator cards, which have been furnished us by gas compressors. But you may be assured that all of the data given herewith is eminently practical, because there has been eliminated all of the small variables that are important fiom a chemical stand- point, but which the advancing piston of a compressor cylinder takes little heed of. We are not concerned about the candle power or the commercial utility of a gas, but simply with its weight and composition, and what may happen to it after it leaves the compressor cylinder is not the province of this paper. All gases are sponge like in that they hold various vapors from water vapor to carbon vapors, which they lose to a more or less extent when the sponge is squeezed as in the act of compressing in a cylinder, and what is squeezed out and how much of it is not essential to our discussion, and lies better in the realm of the technical gas engineer. We have assumed, however, that inasmuch as when we compress a gas the temperature rises in a fixed ratio to the pressures, that there is no direct tendency for a gas to change its physical condition in the compressing cylinder, for an added temperature gives an added capacity for saturation, and this probably increases in about the same ratio as the volume diminishes during compression. So that for commercial purposes we cannot be far wrong in assuming the physical condition of the gases as constant during the range of pressures that will be ordinarily met. All phenomena of compression and expansion of gases is intimately associated with temperature, in fact the .517345 4 THE COMPRESSION AND TRANSMISSION OF GAS. power to compress any gas in foot-pounds is simply the difference in temperature between the gas before and after compression, multiplied by its weight in pounds, by its specific heat, and then by Joules equivalent to con- vert heat units to foot-pounds. Expressed algebraically, this equation is: ^=/f^<:'p(^- ^o) where y is Joules equivalent = 772 W = the weight in pounds avoirdupois to be com- pressed. Cp is the specific heat of the gas at constant pressure. T^ is the initial absolute temperature. 7" is the final absolute temperature. L is the work expressed in footpounds. This is the general equation for the compression of any gas. In glancing at this equation, the first stumbling block we strike is C,, the specific heat of the gas at constant pressure, and this must be first determined. After that, we must discover some means of finding T the final temperature. To anticipate a little, it may be stated here that these temperatures are all functions of the ratio of the specific heats of gas at constant pressure, and at constant volumes. It is then our first duty to understand about these two specific heats and to know how to determine them for any gas, and the rest is simple. The specific heat of any substance is the amount of heat one pound of that substance will absorb to raise its temperature 1° Fah., the specific heat of water being i. When a gas is heated two different results may be obtained, depending upon whether the gas is allowed to expand and increase its volume when heated, the pres- sure remaining constant, or whether the air is corfined and the volume remain constant, and the pressure in- creasing. The amount of heat to raise the temperature of a gas 1° under these two conditions is different, there- fore, the specific heat is different. The former is called — Specific heat at constant pressure, and the latter— Speci- fic heat at constant volume. THE COMPRESSION AND TRANSMISSION OF GAS. TABLt: J. July J305 /SOTH£/iliAL CUKYE /'oVe- fl^" CONS 'T /iOIABATIC Cu/f V£ /i Vo ' Z'/" 'COMS'T. /-joJ /^G-jf I 6 THE COMPRESSION AND TRANSMISSION OF GAS. Referring to Table i, Figure 3, if we have a cylinder A, containing one pound of gas at atmospheric pressure, and a piston P, without weight, but having an area of one square foot, and heat the gas until the temperature has risen 1° Fah,, the gas will have expanded by the small amount d, as in Figure 4, and raised the piston. This expansion is 1/460 of the original volume, at 0° Fah. It is evident that inasmuch as the piston has raised and displaced the atmosphere, that work has been done, which must have absorbed heat in addition to that neces- sary to raise the temperature of the air 1°, If the pis- ton was fastened, as in Figure 3, the gas would have required just that less heat to raise it 1° as was reqnired to lift the piston through the distance d = 1/460 of its volume. The amount of heat required in the first in- stance is called specific heat at constant pressure, and the latter at constant volume. Specific heat of most of the gases at constant pressure has been determined by Regnault and others experiment- ally, and the symbol is C^. The amount of work done in lifting the piston through the distance d is measured the same as the work done by any piston by multiplying the pressure on the piston by the distance passed through. The area multiplied by the distance is the volume, which may be expressed by V. The distance d \s 1/460 at 0° Fah., or may be ex- pressed by y^ lyCt P be the pressure, and P, the foot-pounds of work dene, then VP —=r = R and this is called the Simple Gas Equa- tion, and about it hangs many important deductions. i? is a constant for any gas, because inasmuch as the gas expands uniformly for each 1° of heat, any volume as Fj multiplied by its corresponding P^ and divided by its corresponding temperature Z) will equal R, or to put it algebraically, VP V, P, V" P" T 7; T" = ^ = Constant THE COMPRESSION AND TRANSMISSION OK GAS. « ^ <\i 5 ^ E^ ? » \ V > S ^ S 5 ^ s 1 ^ N 5> s ■ 5- 11 - ^ N N 1 § 1 ^^ 4 N 1 >*• ii 9> c; S! Ci .^? 1 1 'O ^1 ^4 ^ 05 5 1 ■I- Or? 50 "5 'o 1 11 ? ? ^ 51 1 li ^J <^ 1^ -.1 iT ^ KT tn N, s N ^^ 5: ^ ^ ■o (Q ts >0 «$ ^ ^ >0 5^ <2 J ^ N 5 5 ^ !5 1 IS 5 b^ 1" va OS 1 Is t^ •> s ^1 ^ *) <^ ^ ^ ^ § 1 ^\ >l c^ K. % ^ v^ ? 1 s , 5 g \ 1 e ^l ^ I 'Vl N ^ ^ cr > 5> Vj ^ 1 5^ ^ 2 ^ ^ >( ^ ^ (; ) > Q ^ !Q tn ?^^ (I > i 15 ' I 5 9 ^1 4^ i J V 1 ^ 1 i a N ^ (^ 9 i T) ? ^ '^ ^ c •> IT 5 > § N 1 '^ 1^ ^ s ^ sj o > . 1 N ^s' ; ^ 2 § ^ ^ i 8 THE COMPRESSION AND TRANSMISSION OF GAS. R being always in foot-pounds, if we divide it by Joules equivalent 772, which is, as you know, the amount of foot-pounds equal to i heat unit, and which is always denoted by /, we shall have the amount of heat units that were converted into work to raise the piston, and this amount of heat, we know, must be the difference between the specific heat at the constant pressure and the specific heat at constant volume, or, y ' ' from which we have C =C — - \ " / an equation from which the specific heat at constant volume may be determined for any gas within the limits of its stability, and certainly within the commercial pressures you are likely to encounter. For a perfect gas, these specific heats are practically constant; that is, they are not affected by pressure or temperature, but so far hydrogen and air appear to be nearer than any other gases. CO and CO.^, which are inferior components of illuminating gas, as it is now made, shows the greatest deviation, but not enough to render their vagaries of moment in the consideration of the power question, consequently all the following data has been calculated on the basis of the simple gas law. P V „ = R = Constant. As an example showing how to calculate the specific heat at constant volume, let us take C.^ H^. This gas has been selected because of an evident error in the values ascribed to Regnault in the references we have at hand. Upon applying the simple gas equation to the Reg- nault value there was a large discrepancy, and it will be interesting no doubt to make the calculations here, and thus make them serve the double purpose of showing how to determine the specific heat at constant volume and to point out the error. Regnault gives the C^, of Q H^ to be .404, and C^ to be. 1 73. The weight per cubic foot to be .0780922, or 12.8 cubic feet in one pound at 32^ Fah. THE COMPRESSION AND TRANSMISSION OF GAS. N ^ 4 ?: ^ ^ > Is ^ ^ ^ \ •fi »> 1 s 1 N X N 5} 1 ii ^ "i' ^ "^h If) ^ ^" g ^ ^ 11 1 11 1 1 1 1 i^ S 1 '^ i?v <^ 1^ •1- ^ ''^ <5> O tV) N 1 ^ Is S ^ >i- S^ ^ v§ i ^ V3 ^ 1 5 « i "^ 1 1 T ^ t*5 Q ^ ? 1 1 !o ^ g "i 11 ,1 "^ 1 1 1 S^ ^ k. ^ Q '5 1 \ > 3 \ ^ '^ " ^ ^ ^ s ^ ^ r Q vj N. w ^' o4 1 X ? « ^ 5 ^ \ ^ •I- lo THE COMPRESSION AND TRANSMISSION OF GAS. If, now, one pound, or 22.30 cubic feet, be heated to 1° Fah. and allowed to expand, the simple gas equation P V — ™^ = R will give at 32° 22.39 X 144 X 12.8 _ 492 ^^ Fifty-five foot-pounds of work has been performed by the gas in expanding against the atmosphere; to convert this into heat units we divide by Joules equivalent 772. — = .07124 units of heat. 772 Inasmuch as C^ = C|, — — and — = .07124 we have C^ = .404 — .07124 = .3327, instead of .173 as determined by Regnault. The ratio between the two specific heats forms the basis for all the calculations for the relations between pressure, volume and temperature in compressing gas, and that is why we must be par- ticular about these specific heat factors. C -^- = y, which we shall discuss further on, and which is brought in now simply as additional proof about the figures which we have just obtained for C.^ H^. For C.^ //<, using Regnault's values, we have .404 y=~TZZ= 2-33 • 173 for our values .404 r = = 1. 2 14. .3327 In reading a new book by Travers on the study of gases (page 275), he gives some very interesting calcu- C lations to show the limiting values of " ox y. His conclusions are that for a monoatomic gas within PV the limits of the simple gas equation -^„ = R, he val- C ues of -^t can never exceed 1.667, ^^^ the value for a diatomic gas should range about 1.4 and the polyatomic gases still less, until we reach the value of i, where, of THK COMPRESSION AND TRANSMISSION OK GAS. s T> ^ 'H ^ 5 ^ ^ H 5 <:> 2> % N «o > ^ 1 'H s 5 > ii >. ^ " ^ s 5^ 1 1 II V J^ S ^ M ^ N <> 1 ^ M \ , 1 i( N ^ ? 1- ii 1 1 ii ^ 1 It N (^1 i?N 1 ^ > CO N N "9 > 1 i ^4 ^ 5 « ^ K IJ P N •1- II ^^ k ^ =4 v4 "0 <5 10 ^ § 1 i5 ^ Vi 1 5 1 1) N ^^ * ? 'ia "S N <:> ^ 1 S N. ■^^ ., ? ►^ ■^ t a> 5 > s Pi > ^ > N ^i S V ^ 0) 1 u 3 ^ S^ 5 ^ ^ 1 «i Si ^ 5 ^ 1 lo ^^ 9) >t- '^ » N ^ (0 s a "^ 5§ N >♦• Q N s «0 5 ^ ^ r ? ^ 9 ^ «i Q 1 ly ij s o ^ ,^ ^^ ^ > "r> <:i Vj ^J > ^ ^ <<) S N ^ ^ > ^ il ^ 1 N 1 CI ^ ^ 5 ^ (^ ^ i 12 THE COMPRESSION AND TRANSMISSION OF GAS. course, there should be no expansion work at all when heat was applied. C We can see, therefore, that the value -^ of 2.33 from Regnault's values is an impossibility, the maximum pos- sible value being only 1.667, and Q H^ being the poly- atomic gas, its value would be less than 1.4, all of which . . C indicates that our figures -^ =^ 1.2 14 are approximately correct. It will now be necessary to apply our understanding of these principles and try and determine the values of the specific heats for illuminating gas. There seems to be plenty of data about the specific heat at constant pres- sure for gas mixtures, but nothing about the specific heat at constant volume. Reference is now made to the Tables 2, 3, 4, 5, 6, 7 and 8, which show the composition and heat properties of seven different gases and the methods employed in determining the weights, specific gravities and specific heats. Column I is the chemical symbol for the different com- ponents. Column 2 is the percentage by volume of the different components. Column 3 gives reliable weights per cubic foot. Column 4 gives the specific heat of each component gas as determined by Regnault and others. Column 5 gives the productof the different percentages of the component gases and their weights per cubic foot, or Column 2 multiplied by Column 3. The total sum divided by 100 gives the weight of the gas per cubic foot. Column 6 gives the product of Column 4 and Column 5 for specific heat, being a weight function. We must, in order to get the specific heat of the compound gas, take into consideration not only the percentages of the component parts, but the weights as well, and also the specific heat of each component. The sum of the products in column divided by 100, and then by the weight of one cubic foot of the compound gas, will give the specific heat at constant pressure C^. THE COMPRESSION AND TRANSMISSION OF GAS. 13 ^h -r. I AN l^ s> > ^ •o ^ N ;§ ^ 1 N ^ ^ ?i 0^ ^i f ^ § ^ 5 "^ :j> <^ 5^ ^ \o Vj ^ «i 14 THE COMPRESSION AND TRANSMISSION OF GAS. \ ^ << ^ \ 5 S< N 5^u > fi > 1 N 5 II X 1 k ft; N ii 1 > "0 1 s to ll S Ss 1 il iS^ 1 ^ p X! 1 N 1 1 1 M ^ ? ■s ■si ^ •si ^ ?) Q Na 1^ \ \ ■1. ^ ' ll ^ ^if i> Ni 5> ii Q 1 >0 '0 1^ "3 5 ^'5 1 ^ ^ 5 *5 5 1 ^ ' > "5 $ H ^ ^ 5 i ^ ^ ■«l ?> N 3: 5 ^ j "i ^ ^ § \ ^ ^^ 5 J V <; 5 ^ ^ ^ ^ § i^^ ^ ? ^ > Si ^ si § y § k Q ^ S ^ li ^ ^ 1 ^ § ^ ^ M ^5 'O s ^ 1 vo ^ ^ •^ s 1 ~ 1 ^ ^ <:> 1 THE COMPRESSION AND TRANSMISSION OF GAS. •5 1 X •^ s > t^ 5 1 Si ii > > >»■ 1 1 N > > > Nl 5^ , 1 1 s H ^ 9> > tk^ 5^ S ^ N ^ 'J^ S ? > > n «> :2 ^. ^ C ii ^^i " 5< 'A It Q IT) ^ I ^ § N 1 N Y JO ^ ? k'^ •^ N H- ^ ^5 5 Q 55 ^ ^ > •> ^ \ «o 1 !5 j^ ^ \ ^ S^ ^ «5 ^ i<^ Q v2 ^ w 1^ 1^ ^ ;. ^« ^ 1 t ^ 'Si 1 ^5 s «5 i ft) ^ i ^ ^ 5^ N 1^ § Q ^ ^ p ^»- * >f "0 <0 ^ ^ <0 ^ ^ Si: s5 ^ ^ N N «[) c g 5 5 1^ ^ N ^ ^ i\j Q > 1 N, ^ N 5; U4 >J '^ vl <:i 1 •^ 1 ^ ^ 3 ^ ^ 'it i6 THE COMPRESSION AND TRANSMISSION OF GAS. 5) 1 >J. N> N^ '^ "0^ 1 •^ ^ v^ ^ X II s IS I 1^ ^ is .^ ^ ^ THE COMPRESSION AND TRANSMISSION OF GAS. 17 i 7^ "n^ ^ •^ ^ ~>«^ 1 1 1 N ^ ^ ^ ^ ^ I Y "i <^ !5 « N 1 "^ 5 ^ § ^ ^ ^ § ^ s ^ 5 5 ^ x> > S! <0 N ^ N IS h '^ i >f Y ^ ^ (J 'i 1- <0 I'S iN !5 ^ 5 ^ N 5 s^ 3t ^n S \ s s S N s X u ^ i 1 1 1 > 1 f^ ?; "0 N ^ 1? s 1 > ^ s^ ? ^ ^ ^ 1 > 5 1 >• N § 5 > ■§ 1 1 1 1 1 1 1 9> "0 « ^ i"-^ Si 5^ J5 > s 1 ^ § B 1 ^ ^ ^ ^ 1 ^ C5 "O Q Q *) ^ N !$ 1 > 1 1 ^ N ^ 5 1- (« N ^ «> *« •s ^ "0 •^ ^ 10 >o \ ^ > ^ 5? 1- ^ \ 5 1 N »5 vO ^ »i N 3 $ •^ ^ ^ N ^ ^ •^ ^ =5 «^ •o s n ^ ■S5 <5i i X 1 1 5 J: ^^ ^ 1 !§' ^^ IS! ^^ 1 s N 1 1 1 1 1 1 1 1 II ^1 1^ on 5 1^ i8 The compression and transmission of gas. Column 7 gives the calculations to find the specific C heat at constant volume and also R and ~ or y for each gas, and also various factors of y which we will find useful later. Table 9 concentrates Tables 2 to 8, so that we may study them easier. You will note that our results cover quite a field, taking in California fuel oil gas, Massachusetts coal gas, Indiana natural gas, California natural gas, and Cali- fornia carburetted water gas, and after carefully study- ing their heat and power properties, as shown in Table 9, we have selected the fuel oil gas made in Oakland as having the best average properties for the purposes we have in view, and particularly as fuel oil gas is the one you will probably have most to deal with. We may therefore consider our subject as having for a basis a gas with the following properties at 32° Fah Weight per cubic foot, .0323577, Cubic foot in one pound avoirdupois, 30.98. Specific gravity, .4008. c; = .6884. y ^ I-334- y-i y y = .25. y-^ ='• R= 133.2. L = 8467 (^„ = I ) A cubic foot of gas varies in weight according to the altitude or pressure, and also according to the temper- atures. The law of this variation is expressed as follows: Having given the weight of a gas for any temperature, or any pressure, then the weight at any other temper- ature or pressure will be as the ratio of absolute temper- ature or pressure, or W = W ~ or W~ where IV= known weight. THE COMPRESSION AND TRANSMISSION OF GAS. I9 T" and P" the known temperature or pressure and W the desired weight. For example — Our standard gas weights at sea level, or 14.7 pounds absolute pressure, and 32° Fah., .03235 pounds per cubic foot; at 20 pounds gauge, or 34.17 pounds absolute, a cubic foot would weigh .03235 X — -- = .03235 X 2.36 = .076346 pounds, and at 60° 14.7 Fah., instead of 32° Fah., this cubic foot would weigh 520 .076346 X —- = .0819 pounds. 460 being the absolute 460 temperature of O" and 520 the absolute temperature of 60° Fah. = 460 + 60 = 520. Altitudes are nothing more or less than pressures less than sea level, and are treated just the same as pressures above the normal atmospheric. Thus at 5225 feet the absolute pressure is 12.044, con- sequently, as gas at this altitude would weigh — '_z^ times the weight at sea level. For your convenience it may be well to add here that when the barometric pressure is knovyn, the atmospheric pressure is found by multiplying the barometric pressure by .4908, or P" =^ B X .4908. For example — When the barometric is 29.92 the atmospheric pressurre is 29.92 X 4908. or 14,7, the nor- mal sea level pressure. To find the atmospheric pressure when the altitude in feet is given, we have 57000 A^ — N'^ . , . , P" = 14.72 — in which 100,000,000 A^ = altitude in feet. For example — To find the atmospheric pressure a 10,000 feet we have 57,000 X 10,000 v< (10,000)'^ P" = 14.72 — or 100,000,000 P" =^ 1472 — 4.7 = 10.02, the atmospheric pres- sure required. The foregoing rules will be all that is necessary to calcu- late all variations of weights due to pressure, altitud*' 20 THE COMPRESSION AND TRANSMISSION OF GAS. or temperature, and relative volumes follow exactly the same laws as relative weights. For convenience in many calculations Table lo is P given herewith, showing the pressure ratios, or — for every pound from i to no, and the volumes ratios will be inversely as the pressure ratios and consequently the reciprocal of the figures on the table. This might be called a table showing also the rates of Isothermal compression or expansion or Marriotte's law, the general formula for which is: po yo ^ p y_ Constant, or in other words, the pro- duct of any pressure by its volume is always equal to the product of any other pressure by its volume, and this rule will be found useful in determining the contents of receivers, etc. It must always be remembered that in using these rules all temperatures must be alike, or cor- rections made according to the rules just given. ISOTHERMAL COMPRESSION. There are two methods of compressing any gas. First — Where the temperature remains unchanged during compression. This is called Isothermal compres- sion and is the ideal method never realized in practice. Second — Adiabatic compression, which is the kind we meet in practice where the heat developed by compression expands the air being compressed until it follows a dif- ferent law from Marriotte. While Isothermal compression is not practical, it is necessary to know about it and how to make the calcul- ations concerning it. We have found that the volume ratios are inversely as the absolute pressure ratios in Isothermal compression. Consequently if the pressure ratios are i, 2, 3 and 4, ihe corresponding volumes will be i, }4, yi, }(. To show this graphically — refer to Table i. Figure i. Let A B he the line of o pressure or the perfect vacuum line. C D the intake line and we erect pressures ordi- nates G H = 2 X Z>i9ata point //"equal to ^ , ^ ^ and I J— ^ X B D 2X2. point J = Yz A B and if A' = 4 x B D 2iX. sl point K = % oi A B counting all volumes from F B or the end of the piston stroke. THE COMPRESSION AND TRANSMISSION OF GAS. 21 If we join the points C G I E \n a. curved line, it will be the Isothermal or logarithmic curve and it will be noted that the area E FB K= 4 X 14: = I ILBJ^2> X /3 = 1 G M B H= 2 X >^ = I CDBA^\y.\Qx\ As found before. /"" V = P' V = Constant, and the figure represents the ideal indicator card for Isothermal compression for four compressions, counting from o, and the above method will always be proper to lay out an Isothermal curve, no matter what the intake pressure may be. To find the work of compression and delivery Isother- mally d^^P" V hyp. log ^ in foot pounds in which P" = Initial pressure absolute. V =Initial Volume P = Final pressure L = Work required. In all of our calculations V° will be taken as one cubic foot. For Example — Haw many foot-pounds of work is re- quired to compress i cubic foot of gas at sea level to eighty pounds gauge pressure. For sea level P" per square foot = 14.7 x 144 = 2116.8 pounds. Then p L= 21 16.8 hyp. log. ^„ Consulting Table 10 we find p -- for 80 pounds gauge = 6.442 the hyperbolic logarithm of which is 1.863. Substituting, we have L = 2116.8 X 1.863 = 3943 foot-pounds. If a table of hyperbolic logarithms is not at hand, it would be well to remember that hyp, log. = common log. X 2,3026. THE COMPRESSION AND TRANSMISSION OF GAS. F/f£:ssufi£: /fAHOs JuLr/aoj T/ible jo Gauge A Qauge P Guage po po / /oceosr 38 3 Sad'OZ6 75 6/0202S «? //seos'f- 39 3-6530S3 76 6y700S2 3 y.ao4-oay ■fO 3 7s/oao 77 G233073 ^ /^r^/os ^y 3 703y0 7 73 63O€y0€ 5 yj'fO/33 42 3 OS^7y34 73 6 374y33 6 /^oajes 43 3325yay 30 e442yeo 7 y4-F6/G9 44 3-333/33 BJ 6 5/0/3 7 y5'4-'f£y^ J7,, = /> J^ = Constant), such as we found in Isoth- ermal compression, we now have a relation /*„ V J = P V^ = Constant, or in other words, the gamma powers of each volume, multiplied by its corresponding pressure, is Constant. This is the equation of the Adiabatic curve, y is the same that we found to be the ratio be- tween the specific heat at constant pressure and that at constant volume. This relation can perhaps be fastened a little easier in the mind by remembering that the equa- tion of the Isothermal curve represents the law of Mar- riotte and the equation of the adiabatic curve represents the Exponential law of Marriotte. 24 THE COMPRESSION AND TRANSMISSION OF GAS. Inasmuch as the power to compress a gas is measured practically by the indicator diagram, and this in turn is compared to the adiabatic curve which is theoretical curve of compression, and inasmuch as we depend upon the value of y to construct this curve, it will be at once seen why we were to particular to discover the relation — - = jj/. Now if Po Vj = P V-' and from the single gas yo po y p equation -^^3— = "-™- = R, by combining these we have all the adiabatic relations between volume pressure and temperature as follows: T" \v ) yp" ) It will always be necessary to use the above formulae in making calculations for pressures, temperatures and volumes, or for power to compress any gas which varies far enough from the standard we have selected to make it necessary, but there is no doubt that for all practical purposes, at least for the present, Table ii, which is calculated for our standard gas, will give the proper values for rapidly and easily calculating any problems connected with compressing illuminating gas. All reference to expansion is purposely omitted, be- cause gas will probably never be used for expansion work in an engine as air is used Assuming that all may not be familiar with just how to arrive at the results as indicated in Table 1 1 , let us P take a ra.ioof -7^^= 2 corresponding to 14.7 pounds gauge V T pressure and discover what are the values of 7^ and -=i, V /P\~ we have 7^ = I — ^ j y y we have already decided from uor standard gas to be 1.334. THE COMPRESSION AND TRANSMISSION OF GAS. 25 II V /P°\ "'** Therefore, — = = . 749 775-= ( -77 ) or since -77 = ~ or . 5 we nave ±* 2 V -p=.5"" or Log y5 = log .5 X .749 Log .5 -^ 1.6989 X .749 = 1-77447 = log V V v -p^ giving value of -r?5 =. .5949 and 77- will be recipro- V cal of „„ or 1. 68 1. r To find the ratio of temperature for this same rate c • ^ T /P\'^ y-i of compression, we have -~^ =( ^^j y - — = .25. Hence: T P Log y„z=.25log. ^ P T Log. p^= .301 X .25 = .07525 = Log. ^ -f-^ = -1892 T = 520 X 1-1892 = 618° absolute or 158° Fah. If T° = 60° Fah. We have then P V V — = 2 -pr= 1.681 p:^=.5949 ^=1.1892 T= 158° Air under the same conditions gives P V'^ ^ V ^ po = 2 -p-= 1.6349 Yo = -^"7 T y- = 1.2226 T = 175° Fah These examples will serve to show how this Table 1 1 was calculated. A few examples will show its use. Problem — To find the final temperature due to adiaba- tic compreSvSion. 26 THE COMPRESSION AND TRANSMISSION OF GAS. P T Opposite — and under the headline — will be found the ratio of absolute temperatures. Example — What is the final temperature due to 14.7 pounds gauge pressure at sea level and 60° Fah. -= = — — =2. Then -^^ = 1.1892, or. 520 x 1.1892 /^, 14.7 1 = 618° abs. or 158° Fah. If the initial temperature has been 100° then 560 x 1. 1892 z. 666° Fah. It is readily noted from this that the higher the initial temperature, the higher the final temperature, and it will also be noted that while there is a difference of 40° be- tween the initial temperature, there is a difference to 48° between the final temperatures; a difference of 8°. Iiiasm I :h as the temperature developed during com- pression is at the expense of power, it is evident that it takes more power to compress the same weight of gas at 100° Fah. then at 60° Fah. to the same pressure, all other conditions being similar. It is an axiom, therefore, that the cost of power for compressing gas will be the least when the initial tem- perature is the lowest, and it will be shown later on that cooling before compression will effect a considerable sav- ing, if the gas to be compressed is drawn from the holder exposed to the sun, provided, of course, that cooling water may be had at a small expenditure of power. Problem — To find the volume immediately after com- pression. V Consult Table 11, and under the headingT^ and oppo- P site the pressure ratio -- the proper value will be found; and it must always be remembered that these Values of temperature and volumes assume no radiation of heat whatever, for when the heat generated by compression has radiated the temperatures and volumes are as cal- culated Isothermally, V Please note that ^^ is measured from the end of the THE COMPRESSION AND TRANSMISSION OF GAS. 27 Table JJ /ioiAo/iT/c T/{bl£ ^o/r cjis July /303 T To ^-' Vo JVa ■ £4,38 ■4336 £64 ^S /£366 y33 £366 462/ £33 3 /3/e/ £/'¥■ 3/6/ 4333 £07 3£ / 337S £0^ 3373 4/3/ 736 3¥- / 3S73 73 S ■3373 3333 ysa 36, /377'^ /aa 3 77^ 33£7 73£ JS /336£ /ao 336Z ■3673 739 ^ /^/'¥-£ 773 >4/ ZL = H or 34 THE COMPRESSION AND TRANSMISSION OF GAS. G^f Y ^ \p~)^ ' °^ reducing 7" />' jj-„ = -p OT P' = P"* P' or P = v//^° 7^' In other words, to make the work in two stages equal, and to have ihe work a minimum, P, the intermediate pressure, must be a mean proportional between the ini- tial and the final pressure, the volumes and the piston areas must follow the same law, since we naturally make make the strokes alike. For an example, let us take 80 lbs. final gauge pressure: P = y/ P^ P' or V14.7 X 94.7 = 37.31 absolute 22.61 gauge pressure. This makes ^" 37-31 P ^ iW = '-54' ^"d P' ^ 94.7 P 37T1 " '-^^ and inasmuch as these pressure ratios are the same, the work expended on each stage will be the same and the piston ratio will be 2.54 also. We found for the standard gas that /r/> = .2564(|-„- i) Referring to Table 11 , we find when P . T . pZ = 2.54, that — = 1.2624 Then H P ^ -2564 X .2624 = .06727 for each stage and for both stages, 2 X .06727 = .13454 HP. It will be remembered that we calculated the single stage HP for 80 lbs. in a former example as .1520. We have then 13.45 H P p&r 100 cu. ft. against 15.10 H P^ a saving of 13 per cent, in power. If the maintaining of a low temperature is any advan- tage in gas compression, we have a temperature of 366° Fah. in the single stage compression against 195° Fah. in the two stage, a remarkable difference. Suppose now that we have a cylinder having an area of 100 sq. inches, when we compress to 80 lbs. single stage the maximum strain is 8000 lbs., if the compressor is single stage THE COMPRESSION AND TRANSMISSION OF GAS. 35 and 4522 lbs. if the compressor is a tandem compound, a remarkable difference, tending to show that we can build the compressor very much lighter for the same work. Another point in favor of the two-stage compressor, it has a greater volumetric efficiency. A piston never delivers from a cylinder an u mount of gas equal to its displacement, because clearance spaces are filled with gas at the discharge pressure, which expands in the return stroke of the piston and occupies more or less space according to the ratio of compression and the amount of clearance. The greater the temperature of compression, the hotter the piston and heads and valves get, and the less weight of gas enters the cylinder on account of its expansion. There are other losses which need not be mentioned here, but these two are sufficient to make the volumetric efficiency of single stage com- pressors at 80 lbs. average about 75 per cent. It will be readily seen that the initial cylinder of a two stage machine at 80 lbs. will have its Clearance losses divided by 2.54,' because that will be the relative ratio of pressures and the temperature losses in proportion to — ~ because that is the temperature ratio. These combined will make the average two stage compressor good for 90 per cent, volumetric efficiency — in other words, 15 per cent, better than a single stage. One can, therefore, affi^rd to pay at least 15 per cent, more for a two stage machine than for a single stage machine, the intake cylinders being the same size, and this ettra 15 per cent, will nearly, or sometimes quite, pay for the difference in price. It is evident from the calculations we have made that the efficiency of a two stage machine over the single stage increases directly as the pressure ratios increase, and inasmuch as altitude increases pressure ratios, it is evident that the higher the altitude the more urgent be- comes the necessity lor using the two stage machines, and at altitudes above 3000 feet it is practically impera- tive. Theoretically, an infinite number of stages would give isothermal compression, but practically the losses in- 36 THE COMPRESSION AND TRANSMISSION OF GAS. volved in driving the gas through too many cylinders and valves would offset this gain, and we can consider that two stages will probably be the limit for all ordinary purposes. ALTITUDE COMPRESSION. We found that it took the same power to compress one cubic foot of gas at any temperature to the same final pressure, provided the initial pressures were the same, and it naturally followed that it took more power to com- press the same weight at higher temperatures, because there would be a larger volume and the piston would have to make more strokes. Altitude acts like an increase of temperature in lessen- ing the density of a gas, but it introduces another ele- ment, viz., change of initial pressure, so that as we reach higher altitudes the pressure ratio is constantly increasing, which means, of course, that the tempera- tures of compression is increasing and more work per unit of gas weight is being done, but the weight is con- stantly decreasing as we ascend, and the combination of these results is that while it takes less work to discharge any given cylinder full of gas at an altitude, the in- creased number of strokes necessary to compress a weight equivalent to a given sea level volume is considerably greater. Table 17. July, 1905. Altitude. P r P" yo T T ^ $«:' 38 THE COMPRESSION AND TRANSMISSION OF GAS. THE COMPRESSION AND TRANSMISSION OF GAS. 39 that while one cubic foot of the altitude gas requires less power, the increased volume necessary to produce a com- mon result makes it require 25 per cent, more power. It will also be noted that the final temperature is quite high in comparison to sea level compression, which speaks loudly for two stage compression. FLOW OF GAS IN PIPES. After reading the report of the committee on "The Flow of Gas in Pipes," for the Ohio Gas Light Associa- tion, as published in the American Gas Light Journal^ April 24, 1905, the general impression would be that the formulae were not sufficiently reliable to be of great service, because there was a variation in the results of a given problem of from i to 200 per cent. It would seem, however, that six formulae out of the nine do not vary 15 per cent., and the three most frequently used do not vary 2^ per cent. If we should accept the largest of these three, called the Pittsburg formula, we would probably not be far wrong, and particularly as the results do not differ greatly from those obtained by using Cox' computer, and I am informed by those who have used the computer that it is perfectly safe. Again, the variation in the areas of those pipe sizes most likely to be used are much more considerable than the variations of any of the six formulae above referred to. Thus, taking the commercial sizes of pipe from i" to 6", the average variation between the areas of each size is 35 per cent. If we therefore make a practic*' of using the pipe that is the nearest size larger than our calculations, we shall have an ample safety factor. For air we have been using a formula developed by Mr. J. E. Johnson, Jr., and published in the American Machinist ]n\y 27, 1899. (Table 17, No. 2) P' = absolute initial pressure P" = absolute final pressure. Q = free air equivalent in cubic feet per minute. 40 THE COMPRESSION AND TRANSMISSION OF GAS. L = length of pipe in feet. d = diameter on pipe in inches. Practical results from this formula show that it is a little too liberal, and that P" — P'" = i^°-^^|,^^ d^ would be nearer the results. The Pittsburg gas formula reduces to the same value when the proper substitutes are made for the relative specific gravities of gas and air. Inasmnch as the specific gravity of gas is always re- ferred to air as i, it seems right that our gas formula should refer to air and a coefficiency used for each gas. The velocity of different gases through a pipe vary in- versely as the square root of their densities, or what amounts to the same thing, their specific gravities or weights compared to air, then the velocities will vary as j: or y/G Where G is the specific gravity of the gas. Prefixing this to our original equation, we have in general, for any gas, p., _ p,H ^ 0005 y/G X %^ d" Or -^v Q =s/—cr s z Inasmuch as certainly for some considerable time crude oil gas will be most extensively used by members of this Association, let us substitute in the above formula the value of the largest probable specific gravity, viz., .49, and we have\/ .^^ - .7 and /"^-/^"% .00035-^ {A Q = 6^^^- ^" X ^' L Q is in cul)ic feet per minute rather than per hour, be- cause all compressors are so rated. Table 12 give's values of/"^ — P"'- iox 100 feet for various sized pipes and quantities will be found conveni- ent for figuring gas flows in pipes. The values are calcu- lated from equation (/?). THE COMPRESSION AND TRANSMISSION OF GAS. 41 Example I — 1000 cubic feet per minute of gas at 90 pounds gauge pressure is discharging into a 4" pipe 26; 000 feet long. Required the terminal pressure. P'' ^ 10962. (Table 13) P'l _ p'n ^ 2^ o^ for 100 feet. (Table 12) Multiplying by 260 for 26,000 feet P'- — P"'^()\\0. P"i ^ p'i—{^p'i _ p"i-) ^ 10962 — 9110 = 1852. P'" = 1852. P" = 28 pounds. Example II — A pipe line 3" diameter and 1 1 ,000 feet long. Required to find the quantity of gas that will be delivered at a terminal pressure of ^ pound, the initial pressure being 40 pounds. P" = 2992 (Table 13) P'" = 279 (Table 13) P'^ — P"' = 2713 for 11,000 feet of pipe or 24.6 for 100 feet. Referring to Table 12, we find value of 23.70 for 420 cubic feet per minute Example III — A pipe line is 11,000 feet long and 4" diameter. The equivalent of 1000 cubic feet is wanted at the end of the line at 10 pounds pressure. What must be the initial pressure? pr,_ p„, ^ ^^ Q^ foj. jQQ fggj. (Table 12). Multiply- ing by no we have P'' —P"' = 3854 for 11,000 feet. P"' = 610 (Table 13) pr, ^ p". ^ (/>'._/.".) ^ 3854 X 610= 4464. P' = v/4464- Referring to Table 13 we find 52 pounds gauge pres- sure to be the initial pressure. Example IV — The equivalent of 200 cubic feet per minute is to be put through a pipe 53,000 feet long. The initial pres- 42 THE COMPRESSION AND TRANSMISSION OF GAS. sure is 20 pounds. The final pressure must be 6 pounds. What will be the size ol the pipe? P" = 1204 (Table 13) P'" = 428 (Table 13) P'^ — P"'^ = 776 for 53,000 feet ot pipe or 1.464 per 100 feet. Referring to Table 12, we find 4" to be the proper size. SOME CORROBORATIONS. Table 15 gives at Figure i a card from the gas cylin- der of a compressor at Fresno, compressing crude oil gas at a pressure of 27 pounds gauge. If we draw the line of 27 pounds pressure and take the M E P with a planimeter, following the curve A B and the straight lines B C — C D qnd A B, we shall have the J/ £■ /* of a perfect card following the actual compres- sion line. This M E P vft^ find to be 17.4 pounds, using the J which we found for Fresno gas, the adiabatic M E P for 27 pounds = 17.58, making a good check on our values. Figures 2 and 3 are from a compressor pumping natural gas at Anderson, Indiana, each having an intake pressure of 1 1 pounds — drawing lines of 50 pounds pres- sure at Figure 2 and 60 pounds at Figure 3, and taking the M E P \\\ the same way that we did in Figure i, we find that the M E P iox Figure 2 is 26 pounds, and for Figure 3, 30 pounds. Using the value oiy which we developed for natural gas and calculating the adiabatic M E P, we find they are 26.30 and 30.85 pounds, respectively, a very satis- factory check, and from these we may fairly conclude that our theories and formulae are reasonable. It will be noted that the line of the compressor curve is very near the adiabatic. even though the compressors were making but 60 to 70 revolutions per minute. An air card would show at least double the separating space. This would appear to show thatthejackets were doing but very little good, and possibly because illuminating gas may be a much poorer conductor of heat than air. The line of compression comes so near the adiabatic that we may well call the compression adiabatic for THE COMPRESSION AND TRANSMISSION OF GAS. 43 Table /4 /a/djcatoj? CAfiDS r/fOM CAs COMF/fEssoffS. July /SOS /ie J /ie^ £6 l^s tV/TMOvrLosid - ■*-76j(£STX v?/^ ' -r« so -^oi/ia/iTK, /V/iru/fyii. (j4s. A/^a£ffSON /a/o /^ATU/rAL 6/ISAN0£/?S0/V/N0 £M /fix. 44 THE COMPRESSION AND TRANSMISSION OK GAS. safety in our calculations — but while the M E P adiaba- tic for any pressure represents the greatest possible power required to compress a gas, a still greater power must be applied — for example look at the Fresno card, Figure i, Table 14 — the area above the 27 pounds line represents work done in overcoming the inertia of the outlet valves in pushing the gas into the main, and this area will be greater or less depending upon the valve area and the size of the discharge openings and the pis- ton speed. It will also be noted that there is an area representing suction work below the line A D, notwith- standing that the gas has a 4" water pressure at holder. This probably indicates that the pipes from the holder to the compressor are too small. Now, if we run a planimeter over the actual area of the card, we find that the real ME P is 19.4, or about 10 per cent, greater than the adiabatic, and this agrees quite well with ordinary air practice, where a safe rule for single stage work is to take the M E P at 10 per cent, above the adiabatic and the two stage M E P the same as the adiabatic. Slow speed, well constructed compressors will do somewhat better, but it is well to calculate on the average type. Now, for brake power to be delivered to a gas com- pressor, we have to allow a mechanical efficiency of the compressor at not to exceed 85 per cent., so that this 15 per cent, loss combined with the 10 per cent, loss in the cylinder points to the fact that we should add 26^ per cent, to the adiabatic H P for the brake power required. The steam engine cards on the Fresno compressor show an M E P reduced to the size of the air cylinder of 20.75 pounds, or 20 per cent, higher than the adiabatic air M E P, but this compressor had a Meyer cut-ofF, which helped its economy considerable. Referring to Table 9, column 17, gives the formula for computing the power to compress one cubic foot of the gas at sea level and 60° Fah. If the calculation be made it will be noted that it takes practically the same power to compress one cubic foot of any of these gases, conse- quently Table 19 may be used generally. 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