** ; :Mfc I UNIVERSnYjrCALIFORNIA COLLEGE of MINING DEPARTMENTAL LIBRARY BEQUEST OF SAMUELBENEDICTCHRISTY PROFESSOR OF MINING AND METALLURGY 1885-1914 * * *. , * ' * * * - * * The Compression and Transmission of Illuminating Gas A Thesis Read at the July, 1905, Meeting of the Pacific Coast Gas Association. Some Economics in High Pressure Gas Transmission A Thesis Read at the September, 1906, Meeting of the Pacific Coast Gas Association. By Edward A. Rix Mem. Am. Soc. M. E. Mem. Am. Soc. C. E. Assoc. Mem. Am. Soc. Mining Eng. Mem. Pacific Coast Gas Assn. PUBLISHED BY THE PACIFIC COAST GAS ASSOCIATION SAN FRANCISCO 1907 P KESS OF THE CALKINS PUBLISHING HOUSE 'f THE COMPRESSION AND TRANSMISSION OF ILLUMINATING 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 con- sider the question mainly from the theoretical stand- point, 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 emi- nently practical, because there has been eliminated all of the small variables that are important from a chem- ical standpoint, but which the advancing piston of a compressor cylinder takes little heed of. We are riot 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 com- pressing cylinder, for an added temperature gives an added capacity for saturation, and this probably in- creases in about the same ratio as the volume dimin- ishes during compression. So that for commercial pur- poses we can not 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 303666 THE COMPRESSION AND TRANSMISSION OF GAS gases is intimately associated with temperature; in fact, the power to compress any gas adiabatically 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 convert heat units to foot-pounds. Ex- pressed algebraically, this equation is: L = J W C p (T T ) where J is Joules' equivalent = 772. W the weight in pounds avoirdupois to be com- pressed. (7p is the specific heat of the gas at constant pressure. T is the initial absolute temperature. T is the final absolute temperature. L is the work expressed in foot-pounds. This is the general equation for the compression of any gas adiabatically. In glancing at this equation, the first stumbling block we strike is C P , the specific heat of the gas at con- stant 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 con- stant volume. 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. r- 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 1. 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 pressure remaining constant, or whether the air is con- fined, the volume remaining constant, and the pressure increased. The amount of heat to raise the tempera- ture of a gas 1 under these two conditions is different. THE COMPRESSION AND TRANSMISSION OF GAS 7ABLE J. JULY /905 /sor/Ytft/viL Cu/fy ft, vo ft THE COMPRESSION AND TRANSMISSION OF GAS therefore the specific heat is different. The former is called Specific heat at constant pressure, and the lat- ter Specific heat at constant volume. Referring to Table 1, Figure 3, if we have a cylinder A, containing one pound of gas at atmospheric pres- sure, and a piston P, without weight, but having an area of one square foot, and heat the gas until the tem- perature 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 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 necessary to raise the temperature of the air 1. If the piston was fastened, as in Figure 3, the gas would have required just that less heat to raise it 1 as was required to lift the piston through the distance N N ^ ^* Q^ 0) ^S frs ^) ^ ^ h 1 -|. Jj * : * * 5 N 8 g S i I 5 8 THE COMPRESSION AND TRANSMISSION OF GAS by its corresponding temperature T l will equal R, or to put it algebraically, V.P yP V" P" 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 1 heat unit, and which is always denoted by J, 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, R _ r r Cp Cv 4 from which we have r-r R Cv Cp -- J 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 commer- cial 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 C0 2 , which are inferior components of illuminating gas, as it is now made, show the greatest deviation, but not enough to render their vagaries of moment in the consideration of the power question, consequently all the following data have been calculated on the basis of the simple gas law. PV ~ R Constant. As an example showing how to calculate the specific heat at constant volume, let us take C 2 H 4 . This gas have been calculated on the basis of the simple gas law. values ascribed to Regnault in the references we have at hand. Upon apptying the simple gas equation to the Reg- THE COMPRESSION AND TRANSMISSION OF 3AS 9 i! i h L 1 | 5 5 >o o> Q 0> V| <^ 0> - ^ > > f.kH I 2 S 5 * d M ^> ^ vo Vj 10 THE COMPRESSION AND TRANSMISSION OF GAS 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 show- ing how to determine the specific heat at constant volume and to point out the error. Regnault gives the C p of C H 4 to be .404, and C v to be .173. The weight per cubic foot to be .0780922, or 12.8 cubic feet in one pound at 32 Fah. If, now, one pound, or 12.8 cubic feet, be heated to 1 Fah. and allowed to expand, the simple gas equation P -^= R will give at 32 14.7 X 144 X 12.8 = R = 55. 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' equiv- alent, 772. =.07124 units of heat. 772 Inasmuch as D & = 07124 we have Cv =.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 tempera- ture in compressing gas, and that is why we must be particular about these specific heat factors. = y (gamma), which we shall discuss further C\ on, and which is brought in now simply as additional proof about the figures which we have just obtained for For C 2 HV using Regnault 's values, we have 404 .173 for our values = = THE COMPRESSION AND TRANSMISSION OF GAS 11 * > fv o xS 1 1 1 ; 8 5 I I 5 tM ! sail tt) QQ ? ^ ^ 12 THE COMPRESSION AND TRANSMISSION OF GAS In reading a new book by Travers on the study of gases (page 275), he gives some very interesting calcu- lations to show the limiting values of ^ or -y, C\ His conclusions are that for a monoatomic gas within PV the limits of the simple gas equation -=R, the val- ues of - -- can never exceed 1.667, and the value for a C \- diatomic gas should range about 1.4 and the polyatomic gases still less, until we reach the value of 1, where, of course, there should be no expansion work at all when heat was applied. We can see, therefore, that the value of 2.33 from C\ Regnault's values is an impossibility, the maximum possible value being only 1.667, and C 2 H being the polyatomic gas, its value would be less than 1.4, all of which indicates that our figures =1.214 are approxi- cV mately 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 pressure for gas mixtures, but nothing about the spe- cific 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 g'ases and the methods employed in determining the weights, specific gravities, and specific heats. Column 1 is the chemical symbol for the different components. Column 2 is the percentage by volume of the differ- ent' components. Column 3 gives reliable weights per cubic foot. THE COMPRESSION AND TRANSMISSION OF GAS 13 ! * l II lH 1 1 1 i i N * \/*/?c:AfT XWCIGHJ- ^ ir, 5 I SI 5 t' P * 9 U 1 i V GHT PEff Ct f 3f/lH 60 \0 ? s ^ i ( '0 Ss Of re A 1 I * k THE COMPRESSION AND TRANSMISSION OF GAS 15 X * <; N 1 ' 1 > s J ^ i s M jy ^ ^ $| ^ ^ li N ^ ' > ' xl x NJ H H * x>. IJ ^ >c vft 0\ X ^ N 5 ^ T 5 T \< V "' \1 ' ^ Vc ^ H *h N >> i ? ii ^> i 1 , X) > 5 $ i n ii ^ sJ It i -^^ . 1 N x N IQ X ^ VJ^^^^V^S?^ 1 ^ || |* IS IN B jso* CAL - 'A/f or tot (cnrff/r c, * S S & 1 11 3 * li i i I li ! ii QX^^^WJQ'O^ Vj 3 ^ i > |, ^ ^ > ,v Hi>^l T ^i T wj ^ ^vj *>4 K Kj ^ i "M | , /GHT renc '00 T 3rS/IA N tn w N * ^ ^* \r\ ^ ^ v ^ 9) N (Jj O^ ^ JS {y\ ^ :* h Q s ;? > % T ^ %I V S ^ 5 5 5 5 ^ ^ ^ ^ 33 ^ ? ^ K 1' &o * 1 ^ \ ^ 1 ^ >. ^ 1 ^ ^ ^ ^ 1 16 THE COMPRESSION AND TRANSMISSION OF GAS 1 S> <5 S 8 3 U I 1 It! e/z/o /sses $ * ? * > < ^ 5 ^ s ^ ^ s ^ * 8.1'H 5 OD THE COMPRESSION AND TRANSMISSION OF GAS 17 t . rt/tTUf?/IL. CAS SUAI CAL DECZ 18 THE COMPRESSION AND TRANSMISSION OF GAS Column 4 gives the specific heat of each component gas as determined by Regnault and others. Column 5 gives the product of the different percent- ages 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 p . Column 7 gives the calculations to find the specific C heat at constant volume and also R and-^ or y for each v 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, anJ California carburetted water gas, and after carefully studying their heat and power properties, as shown in Table 9, we have selected the fuel oil gas made in Oak- land as having the best average properties for the pur- poses 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. Cp =.6884. Cv =.5159. THE COMPRESSION AND TRANSMISSION OF GAS 19 #=133.2. =8467 I : A cubic foot of gas varies in weight according to the altitude or pressure, and also according to the tem- peratures. The law of this variation is expressed as follows : Having given the weight of a gas for any tempera- ture, or any pressure, then the weight at any other temperature or pressure will be as the ratio of absolute temperature or pressure, or W'=W -^. or W^ where W= known weight. 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 Fah., instead of 32 Fah., this cubic foot would weigh 520 .076346 X =.0819 pounds, 460 being the absolute 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- 12 044 sequently gas at this altitude would weigh ' times the weight at sea level. 20 THE COMPRESSION AND TRANSMISSION OF GAS For your convenience it may be well to add here that when the barometric pressure is known, the atmos- pheric pressure is found by multiplying the barometric pressure by .4908, or P = B X -4908. For example. When the barometer is 29.92 the atmospheric pressure 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 , 70 57000 NN* p 100,000,000 m whlch N = altitude in feet. For example. To find the atmospheric pressure at 10,000 feet we have P _ 1A 79 57,000X10,000 (1Q,OQO) 2 100,000,000 P = 14.72 -- 4.7 = 10.02, the atmospheric pres- sure required. The foregoing rules will be all that is necessary to calculate all variations of weights due to pressure, altitude, or temperature, and relative volumes follow exactly the same laws as relative weights. For convenience in many calculations Table 10 is p given herewith, showing the pressure ratios, or - for every pound from 1 to 110, 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 : P V - P V - Constant, or in other words, the product 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 con- tents of receivers, etc. It must always be remembered that in using these rules all temperatures must be alike, or corrections made according to the rules just given. THE COMPRESSION AND TRANSMISSION OF GAS 21 TABLE jo Gauge Gauge fio Giagc /066O27 38 3S3SO2.6 75 //36054- 3 (55305 '3 76 3-7Z/OGO 77 GJS3G073 3-7&9407 78 3 3*7 S 34- 79 6 374/33 ao /4-76/Q9 3 333/SG a/ 6-5/0/37 +S 32 46 83 JO / i /2 34 7+323 f 3S 4- 3333SS 63S03JZ2 37 33 6 336376 4 f 37404 39 2O3Q+32 S3 <# 60&+3/ 30 7/22430 2/S6+S9 7/3O4 4T7 4- 74 J 4SJ- 93 360&+0 57 7 334&3S /> />* T To T , ro -L. Vo JVUf1B# Dirr. JVi/HO/r Oiff. /* /0+66 -+/2 046G BO 63 30O /4 /037& 369 0373 7763 733 /-6 //+7 336 /+7 703O J3+ /3 //fa 3 JO9 /fS3 6436 433 / /39 37 sasz J346 4JO 22 /./73 260 /79 JS33 3f + / +47 ?/ 447 f'S6 SOI 6 /2G36 63 638 433S 64 za /Z3&* S35 366 462/ 33 3 /3S6/ /* 3/6/ 4368 07 3 /337S 04 JJ7S 4/S/ /46 3 + / '3573 /as- 3373 3935 s&a 36> / 3774 /30 3774 3327 /jse J<3 /3962 S<3O 3S62 3675 /J9 4- / 4-/4-Z S73 4J4 3536 / * /43/S /6S +3/S 34/O //7 + 4- A++33 SZ 4433 3233 /OS <+6 / 4645 /J7 464S 3/35 33 <+a /+6O /S 4<90 3O36 34 5 /*3S+ 697 49f4 392 363 6 SS65/ 6/5- S6S/ 603 0* 7 /6266 tt ff66 32* ZZ 8 S6&/0 f&3 6d/3 /O3 /77 3 /732/ -462 732/ /36 /*7 /o / 7733 7733 /773 t\I?'* 28 THE COMPRESSION AND TRANSMISSION OF GAS P 29 4 T p =^ = 2. Then . ='1.1892, or 520 X 1.1892 = 618 abs. or 158 Fah. If the initial temperature has been 100 then 560 X 1.1892 = 666 abs. or 206 Fah. It is readily noted from this that the higher the ini- tial temperature, the higher the final temperature, and it will also be noted that while there is a difference of 40 between the initial temperature, there is a differ- ence of 48 between the final temperatures ; a differ- ence of 8. Inasmuch 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. than 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 consider- able saving, 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 compression. y Consult Table 11, and under the heading - and p opposite the pressure ratio the proper value will be found ; and it must always be remembered that these values of temperature and volumes assume no radia- tion of heat whatever, for when the heat generated by compression has radiated the temperatures and vol- umes are as calculated isothermally. V Please note that ---- is measured from the end of the stroke. The difference given in Table 11 will enable p greater or lesser values of to be conveniently deter- THE COMPRESSION AND TRANSMISSION OF GAS 29 mined by simple rules of proportion. From this table the adiabatic curve can be readily drawn. Refer to Table 1, Figure 2. Let A B be the intake line and C D the line of pres- sure, these lines representing the piston stroke. Divide A B into a decimal scale; beginning at B erect F D at the end of the stroke and divide it into equal values of B D. B D may be the value at sea level or at an altitude or it may be any intake pressure whatever; these rules will always apply. These values of B D may be subdivided into five parts, where special accu- racy is required, and their values will also be found in Table 11. p D H representing a ratio of = 2, the correspond- ing value of y o will be found in Table 11 to be .5948, and laying off the value the point S will be found. P V Similarly at G representing = 3 we find y= .4388, and laying this off we find that the point M. P v And then F representing - = 4 has a value for P yo of .3536, and we lay off this value and find point J. Joining the point J M S A we develop the adiabatic curve, and the shape of this curve will depend upon C the length of the card, the value of ~ or v. The Cv equation of the curve is P V y = P' V n or referring to the diagram. M X (M G) y = JLX (J F) y Problem. To determine the power to compress a gas adiabatically. All that precedes this subject has been necessary to its proper understanding, and while possibly the va- rious symbols are well remembered, it will probably 30 THE COMPRESSION AND TRANSMISSION OF GAS be better to group them together, so that they may be readily referred to. P is always the lesser absolute pressure, and conse- quently the intake pressure in compression. We shall take this as 14.7 at sea level, for the 4-inch water pres- sure of the gas will not fill the cylinder at any greater than atmospheric pressure. P is the final absolute pressure. T is the initial absolute pressure, and unless other- wise specified is taken at 60 Fah. or 520 absolute, that temperature being the probable temperature of the gas mains. T is the final absolute temperature. y is the volume at P. V is the volume at P. p> y T' are intermediate pressures, temperatures, and volumes. L is the work expressed in foot-pounds. H P is horsepower. M E P is mean effective pressure, which is always gauge pressure. W is the weight of a unit volume or one cubic foot of our standard gas at 60 Fah. and at sea level, with an absolute pressure of 14.7 Ibs. per square inch, or 2116.8 pounds per square foot, and equals .03063 pounds avoirdupois. J is Joules' equivalent taken at 772 foot-pounds. (7 P is the specific heat at constant pressure = .6884. (7 V is the specific heat at constant volume = .5159. j/is ~ = 1.334. ^-=4. ^i .25 i-.7S j/-i y y J CP = 772 X -6844 = 531.45 foot-pounds. This value is Joules' equivalent for 1 Ib. of gas at constant pressure. / W CP = 531.45 X .03063 = 16.28 foot-pounds = Joules' equivalent for 1 cubic foot of gas. J W CP T 16.28 X 520 = 8465 foot-pounds. THE COMPRESSION AND TRANSMISSION CF GAS 31 JL x P T 70 = 4 X 144 X 1 X 14.7 = 8465 foot-pounds = the intrinsic energy of 1 cubic foot of gas at 60 Fah. and atmospheric pressure at sea level; or to reduce those values of foot-pounds to horsepower, we have y-\ 33000 All of these foregoing quantities are constants to be used in determining the power to compress gas, and as we have said before, are all based on a quantity of 1 cubic foot of our standard gas at sea level and 60 Fah. We mentioned at the beginning of this paper that the power to compress any gas adiabatically might be expressed by the general formula L = J W C p ( T T) , or to put it in another form, Sri) L = J W C*l You now at once recognize the prefix J W C p T as the one for which we have found a value of 8465 foot- pounds. Therefore, for our standard gas we have L 8465 | l J which is a practical formula. T You also recognize that is all you need solve, and these values are all given in Table 11 for the various p values of . We can now understand our first prob- lem. How many foot-pounds are necessary to compress 1 cubic foot of our standard gas to 14.7 pounds gauge pressure ? P 2Q4- 4r = ^ r^ = 2 - Consulting Table 11 we find P 14. / j^ 1 .1892 and 8465 X -1892 = 1601.57 foot- pounds, and the same method may be applied for all pressures. 32 THE COMPRESSION AND TRANSMISSION OF GAS If we use the value of J W C p T in horsepower, we have HP .2564 | 1 ) a perfectly practical for- mula for 1 cubic foot of our standard gas at 60 Fah. and at sea level. Our previous example would then be rendered: L = .2564 X .1892 = .0485 horsepower for 1 cubic foot compressed to 14.7 Ibs. gauge. At 80 Ibs. gauge pressure. j- 6.442 and y o = 1.593. II P = .2564 X -593 = .1520 horsepower per cubic foot, or 15.20 H P per 100. MEAN EFFECTIVE PRESSURES. It will be found that inasmuch as we learn from an indicator w T hat our gas compressor is doing, and inas- much as M E P pressures are quickly determined by a planimeter from an indicator card, that to become familiar with what the M E P should be and compare it with what the compressor is doing is the best prac- tical way of dealing with the subject. We found that L = J W <7 p T (Ll \ and that , J W (7 p T = ^-P T 70 , therefore L must always equal M E P X V, we have ME PXV = -P ft \ M E P =-2 P I -_ 1 1 and since y- 1 V / 2 = 4, we have for our standard gas Take 80 Ibs. gauge pressure. THE COMPRESSION AND TRANSMISSION OF GAS 33 T 1 = .593 as determined in a former exam- ple by Table 11. P = 14.7. M E P = 4 X -593 X 14.7 = 34.86 Ibs. per sq. in. For our standard gas for one cubic foot at sea level (T \ / r T~ l ) = 58 ' 8 \r " HP = .00436 X 58.8 (~ l\ == .2564 (~ the same result we obtained in a former example. INITIAL TEMPERATURES. The general expression for the work of compression being it is evident that so long as -y- remains constant, the power to compress one cubic foot of the same gas is con- stant, but inasmuch as the temperature of the mains is practically constant and about 60 Fah., if our initial temperature from the holder should happen for any reason to be 100 Fah., as it was entering the compres- sor, it is evident that the compressor must make an extra number of revolutions to deliver a fixed quantity into the mains at 60 Fah. than it would if the mains were the same temperature as the gas in the holder, and the ratio would be as the absolute temperatures or v on ' or ^ P er cen ^ additional. In a plant where 250 horsepower is used in compressing the gas, this would mean a saving of 20 horsepower. By passing the gas through a cooler before it reached the compressor would correct the loss. Inasmuch as little water is required for this, and the water is in no wise impaired for other purposes, that this cooling could always be done. Vice versa, if the temperature of the holder was 34 THE COMPRESSION AND TRANSMISSION OF GAS lower than the mains, as in winter, there would be a corresponding gain and some of the otherwise lost heat of compression would be utilized in expanding the gas to a temperature corresponding to the main. In the long run, the gain might balance the loss, if no cooling were done, but it seems a business proposition to save where possible, especially where it costs little or nothing. TWO STAGE COMPRESSION. If we consider the general equation for the work per- formed in compressing any gas, adiabatically L = JWC*(T T) we note that the only variable is T, the final tempera- ture, if our initial temperature remains the same. In other words, the difference between the initial and final temperature determines always the power ex- pended in a compressor, just as it does the power given out by any heat engine. It is evident, then, that the lower we keep the final temperature the less power it takes. Water-jacketing the cylinders accomplishes but little, probably from 3 to 5 per cent, for the reason that gases being such poor heat conductors that, while they are rapidly drawn in and pushed out of the com- pressing cylinder, there is not time for the heat to radiate through the cylinder walls, and only the portion immediately in contact with the cool cylinder walls suffers any reduction of temperature. The water jacket keeps the cylinder walls cool so that lubrication is effective and is valuable for that reason principally. Practically speaking, the compression is adiabatic, or even greater because the pressure in the cylinder is always greater than the receiver on account of the work expended in forcing the gas through the valve openings, and this extra heat generated overruns the adiabatic temperature corresponding to the receiver pressure. The 'water jacket being ineffective, the device of stage compression was inaugurated, where, after the gas was compressed to a portion of the final pressure THE COMPRESSION AND TRANSMISSION OF GAS 35 in a cylinder, it was discharged into an intercooler, its temperature reduced to the initial and then compressed by a smaller cylinder to the final pressure. The work was found to be a minimum when the final temperature of each stage was the same. If we represent the initial pressure by P and the final by P', and volumes and temperatures similarly, we shall have, using our general formula for work ex- pended, L = y^-P V ApS 1 ) for first stage and (T ' \ TJT 1 I f r second stage. f ) We know that before compression P V must equal P F, consequently if L is. desired to equal L', we must have T' /P'\ y-1 -i _ = / _ I , or reducing pj = p- or P 2 = P P' or P = l/P P' In other words, to make the work in two stages equal, and to have the work a minimum, P, the inter- mediate pressure, must be a mean proportional be- tween the initial and the final pressure, the volumes and the piston areas must follow the same law, since we naturally make the strokes alike. For an example, let us take 80 pounds final gauge pressure : P : : ]/P P' or v/14.7 x 94.7 = 37.31 absolute or 22.61 gauge pressure. This makes 36 THE COMPRESSION AND TRANSMISSION OF GAS P_ == 37.31_ 2 54 and f-M- 2 ' 54 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 HP= .2564 (L Referring to Table 11, we find when |i; =2.54, that |j= 1.2624. Then H P = .2564 X -2624 = .06727 for each stage and for both stages, 2 X .06727 = .13454 H P. It will be remembered that we calculated the single stage H P for 80 Ibs. in a former example as .1520. We have then 13.45 H P per 100 cu. ft. two stage against 15.20 H P single stage, a saving of 13 per cent in power. If the maintaining of a low temperature is any advantage 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 Ibs. the maximum strain is 8000 Ibs., if the compressor is single stage and 4522 Ibs. if the compressor is a tandem two stage, a remarkable difference, tending to show that we can build the two stage 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 amount 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 THE COMPRESSION AND TRANSMISSION OF GAS 37 ^ 38 THE COMPRESSION AND TRANSMISSION OF GAS T S 2 III! JJJ ! 51 $i &fte \\ i m $ va'N >! Is *"> g8 m SK 55> j 1 h JK ^^ " NNN THE COMPRESSION AND TRANSMISSION OF GAS 39 valves get, and the less weight of gas enters the cyl- inder on account of the clearance expansion. There are other losses which need not be mentioned here, but these two are sufficient to make the volumetric efficiency of single stage compressors at 80 Ibs. average about 75 per cent. It will be readily seen that the initial cylinder of a two-stage machine at 80 Ibs. will have its clearance losses divided by 2.54, because that will be the relative ratio of pressures and the temperature losses in pro- portion to because that is the absolute temperature ratio. These combined will make the average two-stage com- pressor good for 90 per cent volumetric efficiency in other words, 15 per centr better than a single stage. One can, therefore, afford 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 extra 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 becomes the necessity for using the two-stage machines, and at altitudes above 3000 feet it is prac- tically imperative. Theoretically, an infinite number of stages world give isothermal compression, but practically the losses involved in driving the gas through too many cylin- ders and valves would offset this gain, and we can con- sider 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 40 THE COMPRESSION AND TRANSMISSION OF GAS same, and it naturally followed that it took more power to compress the same weight at higher tempera- tures, because there would be a larger volume and the piston would have to make more strokes. Altitude acts like an increase of temperature in les- sening the density of a gas, but it introduces another element, 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- ture of compression is increasing and more work per unit of gas weight is being done, but the weight is constantly decreasing as we ascend, and the combina- tion of these results is that while it takes less work to discharge any given cylinder full of gas at an altitude, the increased number of strokes necessary to com- press a weight equivalent to a given sea level volume is considerably greater. Table 17. July, 1905 ** e : 2, 6 ; o : ^ S ^ jj Altitude. P po Y Y 1 po F o-l Icl ii!j S B II! |l ii 'S 'S 3 o Sea level.. 6.44 4 14.7 1XH4 .593 5020 5020 14.7 80 10,000 ft.. 9.47 4 10. 1X144 .753 4337 6375 10. 80 T = 390 Fah. at sea level. T = 450 Fah. at 10,000 feet altitude. Table 17, shows a comparison between compressing gas at a sea level and at 10,000 feet altitude single stage compression. The columns 3 to 6, inclusive, comprise the components of the general formula for compressing gas, and it is interesting to note the variable quantities. It will be seen that while one cubic foot of the altitude gas requires less power, the increased volume necessary to produce a common result makes it require 25 per cent more power. It will also be noted that the final temperature is THE COMPRESSION AND TRANSMISSION OF GAS 41 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 formulas were not sufficiently reliable to be of great service, because there was a variation in the results of a given problem of from 1 to 200 per cent. It would seem, however, that six formulas 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 w r ould probably not be far wrong, and particularly as the results do not differ greatly from those obtained by using Cox's computer, and I am informed by- those who have used the com- puter 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 formulas above referred to. Thus, taking the commercial sizes of pipe from 1" to 6", the average variation between the areas of each size is 35 per cent. If we therefore make a practice 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 July 27, 1899. - 0006 * L P' = absolute initial pressure. P" = absolute final pressure. Q = free air equivalent in cubic feet per minute. L = length of pipe in feet. d = diameter on pipe in inches. Practical results from this formula show that it is a 42 THE COMPRESSION AND TRANSMISSION OF GAS little too liberal, and that P' 2 P" 2 = > Us 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. Inasmuch as the specific gravity of gas is always re- ferred to air as 1, it seems right that our gas formula should refer to air and a co-efficient used for each gas. The velocity of different gases through a pipe varies inversely 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 - or V G Where G is the specific gravity of gas. Prefixing this to our original equation, we have in general, for any gas, p/2 _ p//2 _ .o005i/ x < ~^ Or 44.72 |P' 2 P" 2 X 4^-il I I/ a \ L 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 form- ula the value of the largest probable specific gravity, viz., .49, and we have 1/749 ~ -1 and p/ 2 __ p// _ 00035 9L^. (A) d 5 Or | D/2 P//2 V/ fJo Q = 53.45 J" JL- Q is in cubic feet per minute rather than per hour, because all compressors are so rated. Table 12 gives values of P' 2 - - P" 2 for 100 feet of various sized pipes and quantities will be found conve- nient for figuring gas flows in pipes. The values are calculated from equation (A). THE COMPRESSION AND TRANSMISSION OF GAS 43 JUL.Y /3O5 C/fL . Jo A Ke3 /l ff/x. 44 THE COMPRESSION AND TRANSMISSION OF GAS 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' 2 = 10962. (Table 13) P's __ p"2 = 35.04 for 100 feet. (Table 12) Multiplying by 260 for 26,000 feet P'2 __ p"2 9HO. p//2 = p/2 _ (p/2_p// 2 ) = i 0962 _ 9110 = 1852. P" 2 = 1852. P" = 28 pounds. Example II A pipe line 3" diameter and 11,000 feet long. Re- quired to find the quantity of gas that will be delivered at a terminal pressure of 1 pound, the initial pressure being 40 pounds. P' 2 = 2992. (Table 13) P" 2 = 279 (Table 13) p/2 _ p//2 = 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? p/2 _ p// 2 _ 35 04 f or 100 feet (Table 12). Multiply- ing by 110 we have p/ 2 _ p// 2 _ 3854 for ii ?0 00 feet. P"* = 610. (Table 13) p/2 = p//2 + ( p/2 _ p// 2) _ 3854 + 610 == 4464 P' == j/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- THE COMPRESSION AND TRANSMISSION OF GAS 45 sure is 20 pounds. The final pressure must be 6 pounds. What will be the size of the pipe ? P' 2 = 1204. (Table 13) P" 2 = 428. (Table 13) p/2 _ P " 2 = 776 for 5 3?000 feet of pipe or l 464 per 100 feet. Referring to Table 12, we find 4" to be the proper size. SOME CORROBORATIONS. Table 15 gives at Figure 1 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 CD and A B, we shall have the M E P of a perfect card following the actual compression line. This M E P we find to be 17.4 pounds, using the y which we found for Fresno gas, the adiabatic HEP 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 in- take pressure of 11 pounds drawing lines of 50 pounds' pressure at Figure 2 and 60 pounds at Figure 3, and taking the M E P in the same way that we did in Figure 1, we find that the M E P for Figure 2 is 26 pounds, and for Figure 3, 30 pounds. Using the value of y 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 formulas 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 that the jackets were doing but very little good, and possibly because illu- minating gas may be a much poorer conductor of heat than air. 46 THE COMPRESSION AND TRANSMISSION OF GAS TEJi PK G/fS , VOLUflE ffAfJOS, ME/IN fFCTlV ' JLLUM/fi//I JULY J3O5 S/l LEVEL. /?/VO 6O -f- S//VGLE S7/IGE. S1f AT CoflflCffC/AL 3 OO / 74 JO/+ S30 OS70 3 '6 76 39 733 3 063 Of 7030 3 30 /O 3 6/3 . 2/73 /* '03 703 Z++7 S37 S3 ' 3 6 00 /7 3 76 3 /7 *6 S3 3J $63 29 + SO 7 + 306 32+ 32 3375 /O 96 SS-J JS73 SJ 63 38; 3 36 3774 33 7 <+/ 2 33 07 4 ' 3S JS36 #7 / +J/S rf*4A*4 HIGH PRESSURE GAS TRANSMISSION 51 If a kilowatt hour costs one cent, a horsepower will cost practically three-quarters of a cent, one horse- power hour=60x33,000 or 1,980,000 foot pounds for three quarters of a cent, one cent would therefore pro- 000 duce ::p- =2,640,000 foot pounds. . ( o Eight horsepower equals 264,000 foot pounds, con- sequently every 8 horsepower will cost 1-10 of a cent per minute. This gives us the basis for our curve, for if we lay out our sheet in equal divisions of any size and call each one along the vertical line 8 horsepower, we can also make each division represent 1-10 of a cent, and each horizontal division we can conveniently call 10 pounds. If you will now refer to Table 19, the last table which I read at our last meeting, it will be possible to construct the curve, remembering that the table is constructed for 100 cubic feet per minute, the horse- power therein contained must be multiplied by 10, for the 1000 feet capacity we are now considering. Take, for example, 50 pounds gauge pressure, the brake horse-power required for 100 cubic feet is 14.56, and for 1000 would be 145.6. Where the vertical line indicating 50 pounds meets the horizontal line drawn from 145 horse-power will be a point on the curve. Similarly other points can be made, and joining the points together, we shall have a cost and power curve combined which will be very useful in our calculation. I have constructed two of these curves, A and B. A is the curve of single stage compression and B for two-stage compression. Single stage is rarely used beyond 100 pounds pressure, nor two stage below 90 pounds pressure. You will note quite a difference in favor of two-stage compression. For example, at 100 pounds pressure it costs 2.35 cents per 1000 for two- stage and 2.75 cents for single-stage. In even a small plant using 50,000,000 feet per year, the difference would be $200 per annum, which is well worth saving. The two-stage curve may be readily constructed from the single-stage curve by remembering that the 52 SOME ECONOMICS IN intermediate absolute pressure between the stages is a mean proportional between the initial and final abso- lute pressure, and, inasmuch as it takes the same power for each stage, if we double the power required for the first stage we shall have the desired results, thus the intermediate gauge pressure for 200 pounds pressure will be 41 pounds. We note from the single- stage curve that 41 pounds requires 128 horse-power, consequently twice this is 256 horse-power, which, laid out on our curve sheet on the 200-pound vertical line, will give us the point N on the two-stage curve, and so on for other points to complete the curve. It must be understood that the horse-powers are for 1000 cubic feet per minute, and the cost will be per 1000, and if you wish to eliminate the element of time just multiply the horse-power by 33,000 and the re- sult will be the foot pounds to compress 1000 cubic feet of gas, and independent of time. If power costs more or less than 1 cent per kilowatt hour, or the quantity to be compressed is greater or less than 1000 cubic feet per minute, the results may be read from the curve by simply using a correspond- ing proportion, for example : The curve shows that 1000 cubic feet can be com- pressed to 20 pounds gauge pressure at the cost of 1 cent, it follows, therefore, that 2000 cubic feet can be compressed to 20 pounds for 2 cents, or if power costs 2 cents per kilowatt hour instead of 1 cent, then only one-half the quantity can be compressed for 1 cent, or double the quantity if power costs but y 2 a cent a kilowatt hour. This method of proportion, however, does not apply to the matter of pressure, for you will note that while a cost of 1 cent gives 20 pounds pres- sure, a cost of 2 cents gives 58 pounds pressure, and a cost of l /2 a cent gives only 8 pounds pressure. In other words, it costs just as much to compress gas from to 8 pounds as it does from 8 to 20 pounds, and just as much to compress from to 20 pounds as from 20 to 58 pounds. It would be well right here HIGH PRESSURE GAS TRANSMISSION 53 to consider this fact, for it has a great bearing on high- pressure transmission. If it was found, for example, that it was costing 1 cent per 1000 to deliver gas through a certain pipe at 20 pounds pressure, and it became necessary to double the pressure in order to supply an increased de- mand, the gas company might consider it inadvisable because it might double the cost. Consulting the curve, it will be seen that the cost for compressing at 40 pounds pressure is only 1.6 cents per 1000 cubic feet instead of 2 cents, as may be imagined, and this fact might justify the increased pressure, and the higher the pressure the more the seeming disproportion. From the curve take a geometrical progression of gauge pressure, 5-10-20-40-80-160-320, and we note the corresponding costs of compression for 1000 cubic feet to be, in cents, .3-.575-1.00-1.6-2.4-3.-3.9, in other words, while the pressure from 5 to 320 has increased sixty- four times, the cost of compression has increased but thirteen times. It must not, however, be hastily inferred that be- cause of this decreasing power ratio that it is econom- ical to compress at high pressure, because it may not be so and depends upon the amount of gas to be pumped, for while the rate for 1000 may be small and make no material difference where a small quantity is pumped, with a large quantity the total amount of the yearly cost of pumping may exceed so materially the interest and depreciation on a larger pipe using a lower pressure that the latter installation will be deemed preferable. The question of whether a large pipe and small pressure, or a small pipe and high pressure shall be used is simply a matter of equating the relative costs of pumping, together with the interest and deprecia- tion on the plant, and, with the curve given herewith, it may be easily determined, and it is to show how to make this determination quite accurately and simple that I have written this paper. It has seemed to me that the cost of attendance, buildings, laying out 54 SOME ECONOMICS IN pipe, etc., for any one problem may be neglected, for while it is an item of cost, it will be practically the same for whatever pipe you may select. We can then make our comparison, using the market cost of pipe and the cost of power only, to which may be added the other costs after the size of the pipe is determined, in order to give the total cost per 1000 for handling the gas. The power and cost curves, as constructed, can be called " Standard" and white prints made from it, and upon these white prints the pipe curves laid out, as will be shown, and this same white print can be used in all cases. Let us then take two examples, one for small quantity and one for large quantity, and before starting at it let us make a general standard formula, which will simplify many of the calculations. No plant will pump less than 10,000,000 cubic feet per year, which is about 1200 per hour, or at a less distance than 10,000 feet, consequently take for a basis : 10,000,000 feet per year=a 10,000 feet of pipe=Z> 1 cent per foot cost of pipe 10% per annum interest and depreciation on the pipe. Then equating these quantities we will find that the pipe cost C for 1000 cubic feet of gas will be 1/10 of a cent. For any other quantity Q, and length of pipe L, and price of pipe P, we shall have: Pipe cost per 1000 C = ~? X ^ o\2 1U Example 150,000,000 cubic feet per year, or 6000 per hour, 50,000 feet of pipe, and power to cost i/o cent per kilowatt hour, substituting in our formula La P 50,000 X 10,000,000 JP = _JP bQ 10 W 10,000 X 50,000,000 10" 10 That is to say that whatever size pipe we select the pipe cost per 1000 cubic feet of gas will be 1/10 the market cost of pipe per foot. HIGH PRESSURE GAS TRANSMISSION 55 Taking the market prices of to-day, then if we should use: 1% pi? 6 ? cos t per 1000 cubic feet, equals ......... 559 iy 2 pipe, cost per 1000 cubic feet, equals ......... 67 2 pipe, cost per 1000 cubic feet, equals ...... . . .894 21/k pipe, cost per 1000 cubic feet, equals. ....... 1.429 3 pipe, cost per 1000 cubic feet, equals ........ 1.875 Having thus blocked out the matter of pipe, we must find what pressure it is necessary to use to pump the gas through these various sized pipes, assuming al- ways that the terminal pressure shall be one pound gauge. You will remember that we developed a formula in my paper, read last year, which may be used here with accurac. is the difference between the squares of the initial and final absolute pressures. Q is the quantity of gas in cubic feet per minute. L is the length of pipe, in feet. D is the pipe diameter, in inches. Substituing in this equation the elements in our prob- lem, we have : _35X100X 100X50,000 _ 175,000 Fl ~~ ~100,00 6r <-i 09? 05? HIGH PRESSURE GAS TRANSMISSION 51 If a kilowatt hour costs one cent, a horsepower will cost practically three-quarters of a cent, one horse- power hour=60x33,000 or 1,980,000 foot pounds for three quarters of a cent, one cent would therefore pro- 000 duce - =2,640,000 foot pounds. . / o Eight horsepower equals 264,000 foot pounds, con- sequently every 8 horsepower will cost 1-10 of a cent per minute. This gives us the basis for our curve, for if we lay out our sheet in equal divisions of any size and call each one along the vertical line 8 horsepower, we can also make each division represent 1-10 of a cent, and each horizontal division we can conveniently call 10 pounds. If you will now refer to Table 19, the last table which I read at our last meeting, it will be possible to construct the curve, remembering that the table is constructed for 100 cubic feet per minute, the horse- power therein contained must be multiplied by 10, for the 1000 feet capacity we are now considering. Take, for example, 50 pounds gauge pressure, the brake horse-power required for 100 cubic feet is 14.56, and for 1000 would be 145.6. Where the vertical line indicating 50 pounds meets the horizontal line drawn from 145 horse-power will be a point on the curve. Similarly other points can be made, and joining the points together, we shall have a cost and power curve combined which will be very useful in our calculation. I have constructed two of these curves, A and B. A is the curve of single stage compression and B for two-stage compression. Single stage is rarely used beyond 100 pounds pressure, nor two stage below 90 pounds pressure. You will note quite a difference in favor of two-stage compression. For example, at 100 pounds pressure it costs 2.35 cents per 1000 for two- stage and 2.75 cents for single-stage. In even a small plant using 50,000,000 feet per year, the difference would be $200 per annum, which is well worth saving. The two-stage curve may be readily constructed from the single-stage curve by remembering that the 52 SOME ECONOMICS IN intermediate absolute pressure between the stages is a mean proportional between the initial and final abso- lute pressure, and, inasmuch as it takes the same power for each stage, if we double the power required for the first stage we shall have the desired results, thus the intermediate gauge pressure for 200 pounds pressure will be 41 pounds. We note from the single- stage curve that 41 pounds requires 128 horse-power, consequently twice this is 256 horse-power, which, laid out on our curve sheet on the 200-pound vertical line, will give us the point N on the two-stage curve, and so on for other points to complete the curve. It must be understood that the horse-powers are for 1000 cubic feet per minute, and the cost will be per 1000, and if you wish to eliminate the element of time just multiply the horse-power by 33,000 and the re- sult will be the foot pounds to compress 1000 cubic feet of gas, and independent of time. If power costs more or less than 1 cent per kilowatt hour, or the quantity to be compressed is greater or less than 1000 cubic feet per minute, the results may be read from the curve by simply using a correspond- ing proportion, for example : The curve shows that 1000 cubic feet can be com- pressed to 20 pounds gauge pressure at the cost of 1 cent, it follows, therefore, that 2000 cubic feet can be compressed to 20 pounds for 2 cents, or if power costs 2 cents per kilowatt hour instead of 1 cent, then only one-half the quantity can be compressed for 1 cent, or double the quantity if power costs but y 2 a cent a kilowatt hour. This method of proportion, however, does not apply to the matter of pressure, for you will note that while a cost of 1 cent gives 20 pounds pres- sure, a cost of 2 cents gives 58 pounds pressure, and a cost of y<2 a cent gives only 8 pounds pressure. In other words, it costs just as much to compress gas from to 8 pounds as it does from 8 to 20 pounds, and just as much to compress from to 20 pounds as from 20 to 58 pounds. It would be well right here HIGH PRESSURE GAS TRANSMISSION 53 to consider this fact, for it has a great bearing on high- pressure transmission. If it was found, for example, that it was costing 1 cent per 1000 to deliver gas through a certain pipe at 20 pounds pressure, and it became necessary to double the pressure in order to supply an increased de- mand, the gas company might consider it inadvisable because it might double the cost. Consulting the curve, it will be seen that the cost for compressing at 40 pounds pressure is only 1.6 cents per 1000 cubic feet instead of 2 cents, as may be imagined, and this fact might justify the increased pressure, and the higher the pressure the more the seeming disproportion. From the curve take a geometrical progression of gauge pressure, 5-10-20-40-80-160-320, and we note the corresponding costs of compression for 1000 cubic feet to be, in cents, .3-.575-1.00-1.6-2.4-3.-3.9, in other words, while the pressure from 5 to 320 has increased sixty- four times, the cost of compression has increased but thirteen times. It must not, however, be hastily inferred that be- cause of this decreasing power ratio that it is econom- ical to compress at high pressure, because it may not be so and depends upon the amount of gas to be pumped, for while the rate for 1000 may be small and make no material difference where a small quantity is pumped, with a large quantity the total amount of the yearly cost of pumping may exceed so materially the interest and depreciation on a larger pipe using a lower pressure that the latter installation will be deemed preferable. The question of whether a large pipe and small pressure, or a small pipe and high pressure shall be used is simply a matter of equating the relative costs of pumping, together with the interest and deprecia- tion on the plant, and, with the curve given herewith, it may be easily determined, and it is to show how to make this determination quite accurately and simple that I have written this paper. It has seemed to me that the cost of attendance, buildings, laying out 54 SOME ECONOMICS IN pipe, etc., for any one problem may be neglected, for while it is an item of cost, it will be practically the same for whatever pipe you may select. We can then make our comparison, using the market cost of pipe and the cost of power only, to which may be added the other costs after the size of the pipe is determined, in order to give the total cost per 1000 for handling the gas. The power and cost curves, as constructed, can be called " Standard" and white prints made from it, and upon these white prints the pipe curves laid out, as will be shown, and this same white print can be used in all cases. Let us then take two examples, one for small quantity and one for large quantity, and before starting at it let us make a general standard formula, which will simplify many of the calculations. No plant will pump less than 10,000,000 cubic feet per year, which is about 1200 per hour, or at a less distance than 10,000 feet, consequently take for a basis : 10,000,000 feet per year=a 10,000 feet of pipe=& 1 cent per foot cost of pipe 10% per annum interest and depreciation on the pipe. Then equating these quantities we will find that the pipe cost C for 1000 cubic feet of gas will be 1/10 of a cent. For any other quantity Q, and length of pipe L, and price of pipe P, we shall have: Pipe cost per 1000 C = ^ X ^ Example 150,000,000 cubic feet per year, or 6000 per hour, 50,000 feet of pipe, and power to cost % cent per kilowatt hour, substituting in our formula p L a P 50, 000 X 10, 000, 000 _P = = P bQ 10 W 10,000 X 50,000,000 10 10 That is to say that whatever size pipe we select the pipe cost per 1000 cubic feet of gas will be 1/10 the market cost of pipe per foot. HIGH PRESSURE GAS TRANSMISSION 55 Taking the market prices of to-day, then if we should iy^ pipe, cost per 1000 cubic feet, equals ......... 559 1% pipe, cost per 1000 cubic feet, equals ......... 67 2 pipe, cost per 1000 cubic feet, equals ......... 894 2y 2 pipe, cost per 1000 cubic feet, equals ........ 1.429 3 pipe, cost per 1000 cubic feet, equals ........ 1.875 Having thus blocked out the matter of pipe, we must find what pressure it is necessary to use to pump the gas through these various sized pipes, assuming al- ways that the terminal pressure shall be one pound gauge. You will remember that we developed a formula in my paper, read last year, which may be used here with accurac. p ,__ p , =1d is the difference between the squares of the initial and final absolute pressures. Q is the quantity of gas in cubic feet per minute. L is the length of pipe, in feet. D is the pipe diameter, in inches. Substituing in this equation the elements in our prob- lem, we have : P , P , _ 35X100X100X50,000 n2 _175,000 Fl 100,000 Xd Fl d 5 Then P x 2 XP, 2 equals for 1^4 pipe ....................... 57,370 1V 2 pipe ....................... 23,000 2 pipe ....................... 5,500 2i/ 2 pipe ....................... 1,800 3 pipe ....................... 700 P 2 being our final pressure lib. or 15.7 Ibs. absolute makes P 2 2 =246, and remembering that P! 2 P 2 2 =(P l 2 P l 2 + Pi' 2 we have ikj. pine P./ 2 absolute=57,616 then P l gauge=225 Ibs. 11/2 " " " =23,246 " " " =140 " 2 " " " = 5,746 " " " = 61 " 2V 2 " " " - 2,046 " " " = 31 " 3 ^ = 946 " " " = 16 " Now we are ready to put all this on our curve sheet in order that we may have a graphic representation of 56 SOME ECONOMICS IN the situation. The power cost curve is on a basis of 1 cent per kilowatt hour; if, therefore, we plot any other costs on this standard sheet, they must be in- creased or diminished by the ratio the actual power cost bears to the standard power cost of 1 cent per kilo- watt. Our problem calls for a power cost of y 2 cent per kilowatt hour, consequently we must plot in our pipe costs at double their real amount, for the standard power cost curve is double the cost stated in our prob- lem. Take, then, the 2-inch pipe. We have found the pipe cost to be .894, which, multiplied by 2, equals 1.798, and the initial pressure required is 61 pounds. If then we lay off the point P on the 61 pounds pres- sure ordinate and opposite the cost Line 1.798 or 1.8, we shall have the two-inch pipe established, and similarly we can establish the other points, and joining them by lines we have a pipe cost curve S. C., which shows the cost of pipe per 1000 cubic feet of gas for all sizes from 1*4 to 3 inch. It is evident that the total cost of pumping the gas is the pipe cost plus the power cost, consequently if we add the pipe cost curve to the power cost curve, we shall have a curve of total cost. Take the 2-inch pipe once more, and with dividers measure off the distance W. P., the cost of the pipe line, and add it to the line W. Y., which is the power cost, and we have the point T., as the result of the addition. Do the same for the other sizes of pipe, join these points by lines, and we have the curve D. F. as the final result, showing the combined power and pipe cost per 1000 cubic feet of gas for all the sizes of pipe under consid- eration. It is evident at a glance that the 2-inch pipe shows the least cost, consequently the solution of our prob- lem is 2-inch pipe at 60-pound initial pressure. Total cost per 1000 3.6 cents on a basis of 1 cent per kilowatt hour, but as our power costs !/o a cent per kilowatt hour, the total cost is 1.8 cents per 1000 cubic feet of gas. The graphical method is very satisfactory, for one can see at a glance the relations between the various HIGH PRESSURE GAS TRANSMISSION 57 elements, for example : You will note that it cost prac- tically the same to pump this gas through a l^-inch pipe at 225 pounds pressure or a li/o-inch pipe at 140 pounds pressure or the 3 -inch pipe at 16 pounds pres- sure, which is interesting. PROBLEM TWO. Take 5000 cubic feet per minute, 300,000 per hour or 2,500,000,000 per year, through the same length pipe as in our former problem, and at the same power cost per kilowatt hour. The pipe being the same length and the quantity fifty times greater. We will have P P for pipe cost per 1000 where we had formerly, and taking pipe casing prices up to 12 inches, which was the largest I could get, and assuming them above that size simply for illustration, we find the following prices per 1000 feet of gas for these pipes : 10 inch equals .25 cent per 1000 cubic feet of gas. 12 inch equals .30 cent per 1000 cubic feet of gas. 14 inch equals .40 cent per 1000 cubic feet of gas. 16 inch equals .60 cent per 1000 cubic feet of gas. 18 inch equals .80 cent per 1000 cubic feet of gas. 20 inch equals 1.00 cent per 1000 cubic feet of gas. and the respective pressures necessary to force the gas through these pipes to be: 10 inch equals 53 pounds. 16 inch equals 12 pounds. 12 inch equals 26 pounds. 18 inch equals 8 pounds. 14 inch equals 20 pounds. 20 inch equals 4 pounds. Multiply the pipe cost by 2 and transferring those quantities to our curve sheet, precisely as we did in the other problem, we have a pipe curve, E, showing the pipe cost per 1000 cubic feet, and adding this to the power curve, we have the final result in the curve F, which shows this rather surprising fact, that the 12- inch pipe is the best, and that it will carry this gas at 26 pounds pressure, and at a cost of 8/10 of a cent per 1000 for power and pipe. The curve shows also that the gas can be put through the 10-inch pipe at 53 pounds pressure at the same 58 SOME ECONOMICS IN cost as through the 20-inch pipe at 4 pounds pressure. I believe this graphical method will be of service to you, particularly as these curve sheets can be filed away and always used for quick reference, for the eye can take in the whole relative situation at a glance. I believe this method of handling the subject removes it from the realm of speculation, and makes an orderly comparison of pow r er and pipes possible, to the end that an adequate selection can be made.