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IBM application to equilibrium flash calculations

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IBM application to equilibrium flash calculations

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Content IBM APPLICATION TO EQUILIBRIUM FLASH CALCULATIONS A Thesis Presented to the Faculty of the Department of Chemical Engineering University of Southern California In Partial Fulfillment Of the Requirements for the Degree Master of Science by Harold Austin Thompson December 1954 UMI Number: EP41748 All rights reserved IN FO R M A TIO N TO ALL USERS The quality of this reproduction is dependent upon the quality of the copy submitted. In the unlikely event that the author did not send a complete manuscript and there are missing pages, these will be noted. Also, if material had to be removed, a note will indicate the deletion. Dissertation Publishing UMI EP41748 Published by ProQuest LLC (2014). Copyright in the Dissertation held by the Author. Microform Edition © ProQuest LLC. All rights reserved. This work is protected against unauthorized copying under Title 17, United States Code ProQuest LLC. 789 East Eisenhower Parkway P.O. Box 1346 Ann Arbor, Ml 4 8 1 0 6 -1 3 4 6 CLh '55 T V 72- This thesis, written by Harold Austin Thompson under the guidance ofhX&Faculty Committee and approved by all its members, has been presented to and accepted by the School of Engineering in partial fulfillment of the re quirements for the degree of Master of Science in Chemical Engineering 3 I Faculty Committee a L & d t a J L J Chairman TABLE OF CONTENTS CHAPTER. PAGE SUMMARY............................. iii I. EQUILIBRIUM VAPORIZATION Defined..............................• •• 1 Derivation............................. 1 Limitations........ 3 Applications........................... 5 i=n II. METHODS OF SOLVING L = 2/l/(KiVD + l. i=l Trial and Error........................ 6 Aids to Solution.................... 6 This IBM Method......N ............ 7 III. CONVERGENCE Given ....••••••••........ & j Prove ............................ 9 Proof................ 9 j Geometric Interpretation............... 14 IV. STEP BY STEP INSTRUCTIONS To the Engineer.......... 17 To the Operator..................... 23 Ten Example Problems................... 29 V. DISCUSSION Introduction............. 47 The Two Phase Test................ 47 Linear Interpolation........... 49 The Iterative Process.................. 50 The Final Calculation ............. 50 The Planning Chart.............. 56 VI. CONCLUSIONS Comparisons. ...... Electronic Calculators 60 61] TABLE OF CONTENTS (CON’T) CHAPTER PAGE VI. CONCLUSIONS (CON’T) Speeding Convergence.................... 61 In Conclusion........................... 6? BIB LI OGRAPHY ................... 69 LIST OF TABLES TABLE PAGE I. OPERATOR'S DATA SHEET............................. 22 II. RESULTS OF THE FIRST CALCULATION.................. 37 III. RESULTS OF THE SECOND CALCULATION................. 38 IV. RESULTS OF THE ITERATIVE CALCULATIONS............. 39 V. FINAL CALCULATIONS................................. 43 VI. TABLE OF DIFFERENCES............................... 63 VII. TABLE OF DIFFERENCES............................... 66 VIII. TABLE OF DIFFERENCES............................... 67 LIST OF FIGURES FIGURE PAGE 1. GENERAL SHAPE OF $(L) FOR ALL MIXTURES IN THE TWO PHASE REGION..................... 13 2. CONVERGENCE.................................... 15 3. DIVERGENCE........................................ 16 4. AN IBM MARKED SENSE CARD......................... 19 5. AN IBM MARKED SENSE CARD......................... 20 6. CONTROL PANEL..................... 24 1 7. MARK SENSING...................................... 25 I I 8. LINEAR INTERPOLATION................ 51 9. TWO PHASES........................................ 53 i 10. SUBCOOLED LIQUID.................................. 54 11. SUPERHEATED VAPOR................................. 55 1 12. PLANNING CHART................................... 57 CHAPTER I EQUILIBRIUM VAPORIZATION Defined, Given a mixture at conditions such that the lighter components are above their boiling point, but the heavier components are below, the mixture will sometimes se parate into two phases, liquid and vapor. How much of each phase will form and their respective compositions is the subject of this paper. Such a separation is called an equi librium flash; or, sometimes, equilibrium vaporization. The two terms are synonymous. Derivation. In 1&$4, Raoult discovered empirically^^ that the contribution one component made to the vapor pres sure of a liquid mixture was directly proportional to the number of molecules of that component in the mixture. Com bined with Dalton’s discovery that vapor phase mol frac tions equaled partial pressures, we may write the following expressions: Pi = kx« (Raoult’s Law) Pi = Ityi (Dalton’s Law) then, Pi = kXi = Tfyi- (1) 2 where: = partial pressure of component 1 1 in. x^ = mol fraction of nin in the liquid. = mol fraction of 1 1 in in the vapor, k = a proportionality constant. = total pressure of the system. In the case of ideal solutions, the proportionality constant, k, is equal to the vapor pressure of pure lfin at the temperature of the system. Non-ideal solutions require that k be determined experimentally. These values are known as Henry’s Law constants. If the so called ideal solution conditions are assumed, equation (1) may be written as fol lows: 1. p. v* l ri _ vi l i r T (2) where: 1^ = mols of niM in the liquid. L = total mols of liquid. Pi = vapor pressure of pure Min at the system temperature. Vi = mols of “i" in the vapor. V = total mols of vapor. In equation (2), substitute for v^, and for p i/nr- Upon solving for 1 ± 9 the following equation results: ■^i 11 = (KiV/L)+l * Since i=n then, i=n L = 2 f i/(KiV/L)+l. (3) i=l Equation (3) is the form used in this method of solu tion. One advantage this form has is that mols may be used directly, that is, conversion to mol fractions is not nec essary. It was also found more adaptable to automatic cal culation. Limitations. As observed in the derivation, this cal culation is limited to cases where the solution is ideal and the additive volume law applies. While this is usually the case at low pressures, Dalton's Law breaks down under high pressures, thus deviations from RaoultTs Law as appli ed in equation (1) are to be expected. In 1901, Lewis^ conceived the idea of substituting a pseudo-pressure in eq b quation (1) so as to retain the simplicity of the expression but at the same time explain observed phenomina. He called this idealized pressure, "fugacity". Rewriting equation (1) using this idea, we have, Cpi = JtT^i " ^pxi* (br) where: = fugacity of component "i" in the vapor. * fugacity of "i" in the vapor at the temperature and total pressure of the system. jFp = fugacity of "i" at system temperature, but under it’s own vapor pressure at the system temperature. Fugacities may be obtained from PVT data; this has been done by many investigators^. Briefly, the method consists of graphically integrating data represented by a compress ibility chart along lines of constant reduced temperature. Lewis and co-workerscompiled fugacities for methane through normal octane and prepared equilibrium coefficient charts^) from these values. These charts are widely used in the petroleum industry. Their use is limited to pres sures of less than £00-1000 psia^) and at temperatures well 5 removed from the critical region of the mixture. Better values in this region may be obtained from critical data. Applications. Equilibrium vaporization is encountered in many phases of engineering calculations. Nelson^?) has listed some of the applications specifically taken up in his treatise on Petroleum Refinery Engineering. The list is re peated here for the convenience of the reader. 1. Pipe still outlet temperatures. 2. Tower reboiler temperatures. 3. Composition of materials from separators. 4. Amount of vaporization in pipe still tubes or exchangers. 5. Pressure drop in pipe stills (caused by vaporization). 6. Tower top, bottom and tray temperatures. 7. Heat transfer rates in condensers. &. Sizing of transfer lines, condenser shells, etc. 9. Flash curves of petroleum mixtures. CHAPTER II n METHODS OF SOLVING L fj/UjV/D+l i=l Trial and Error* The most obvious way of solving the summation equation is to assume some value for L, substitute into the right hand member, carry out the arithmetic and compare the L calculated with the L assumed* When the two come out equal, the problem is solved. This could happen on the first trial. This is a rare event, especially if there are very many terms on the right. In the usual case, sever al trials will be necessary to obtain a reasonable check on the assumed value for L. Although this procedure is time consuming, it does have the advantage of being straightfor ward. Aids to Solution. Graphical methods^ of conducting the trial and error solution are available. More recently a method appeared(9) which used the first term of a series ex pansion for estimating the first trial value for L. A meth od of plotting trial failure information simplified the work and reduced the trials necessary. A method for solving flash problems on IBM has been presented but lacked many of the advantages which the method presented in this paper possesses. Briefly, the advantages are as follows: 1. The trial and error part of the solution is carried out with only one deck, whereas the method referred to above requires a new deck for each trial. 2. The approach to the trial and error part is such that far less trials are required. 3. Problems which have no solution, that is, not in the two phase region, are .weeded out with one or two trials. When the solution to the trial and error part of the problem is solved, the same deck is used when calculating the compositions of each phase resulting from the flash. These are the major advantages of this method. They result in a big savings in time, cards, and effort. This IBM Method. The calculating punch is wired to solve the equation as given in the chapter heading. In brief, the method proceeds as follows: Each end of the func tion is tested; this reveals if the mixture has two phases. The results of the first two trials are used for a linear in terpolation for an approximate solution. An iterative pro cedure from this solution converges to the root. Proof of convergence is the subject of the next chapter. CHAPTER III CONVERGENCE Given, An MnM component mixture somewhere in the two phase region, which imposes two conditions, namely, flKl + f2K2 +•••• + fnKn>F* This condition insures that the feed is above its bub ble point. The second condition is, fl/Kl + f2/K2 + •••• + fn/Kn >F* This condition insures that the feed is below its dew point. Also, fx + f2 +....+ fn = F (7) where: f^ = mols of component "i”. = equilibrium constant for component wi" at the system temperature and pressure, F = total mols of feed. The quantities, f^ and are always positive, there fore the right hand member of equation (3) is monotonic in creasing in (0,F). There is only one root in the interval, 0<L<F, where L is the total mols of liquid. Prove* A solution to the summation equation (3) may be obtained by a process of iteration, Ln+^ = where 4>(L) is the right hand side of (3)* Proof« The true value of L satisfies equation (3), that is, L = <|)(L). («) The first approximation gives Lx = ♦(L0). (9) Subtracting (9) from (8), L-Lx = <|>(L)-<|>(L0). (10) By the mean value therom* the right hand side of (10) may be written, (L)-4KL0) = (L-L0)^*(?o). (11) in the interval, Lq*5„*L. *Mean Value Therom: If a function, f(x), is continuous in the interval, a^x^b, and possesses a derivative at every interior point of (a,b), then there exists a point §, such that, (f(b)-f(a))/(b-a) - f»(f); a*%*b. 10 Hence, (10) becomes L - L 1 = ( L - L 0 )4>*<§0 ) . ( 1 2 ) The preceding argument holds for all successive ap proximations, so that L-L2 = (L-LjW'd!). L - L 3 = ( L - L 2 )((»«(5o ) 3 ( 1 3 ) L - L n = ( L - L n _ 1 ) * ' ( f n _ 1 ) . If the product of the above equations is found, and the common factors divided out, the final result is L-Ln = (L-L0)<t>*(50)<t>,(l1)<|>Mf2)-*--lt>M5n_1). (14) Now if the maximum absolute value of ( J ) f(L) is less than one throughout the interval, ( L q , L ) , s o that each of the quantities, ♦•(go) , ♦•(Si) , etc., is not greater than a proper fraction, m, from (14) we obtain | L - L n l s I L — L 0 | mn . ( 1 5 ) Since the right hand member of (15) approaches zero as a becomes large, we can make the error, |L-Ln|, as small as we please by repeating the iteration process a sufficient lumber of times. ^2) 11 Hence, the condition for convergence is < j ) f(L)< 1 in the neighborhood of the desired root. Referring now to our particular < | ) (L), equation (3), we observe the following: 1. AS L^O, 4>(L)-*0. 2. As L-*F, (J(L)-^F. Differentiating the right hand side of equation (3) with respect to L, we obtain n <|>»(L) = F ^ f i K j / d C j L - K i F - L ) ? . ( 1 6 ) 1=1 Letting L = 0, n **(Q)-F^ fjKj/lKjF)2. (17) i=l n = F/F2^ W (18) i=l The summation in equation (IS) is given to be greater than F to insure that the feed is below its dew point; see equation (6). Let this be equal to F+h, where h>0. Then equation (IS) reduces to F(F+h)/F^ = 1 + h/F>l. 12 Letting L = F, equation (16) reduces to n $'(?) = F 2 fiKi/(_F)2* (19) i=l n . = F/F22 fiKi* (20) i=l The summation in equation (20) is given to be greater than F to insure that the feed is above its bubble point; see equation (5)« As before, let this quantity be equal to F+h, where h >0. Then equation (20) reduces to 1 + h/F>l. It is obvious that if these conditions at the end points are satisfied, < | > ( L) must cross the line L = $(L) i-n the interval 0^L<F. But the slope of this line is equal to one, hence the slope of ( | > (L) at the intersection must be less than one. This fulfills the condition for convergence. The argument in the preceding paragraph is illustrated graphically on the next page. From this it is clear that the four conditions found to exist at the end points, plus the fact that the function is monotonic increasing and has only one root in the interval is sufficient to establish ; that the iterative procedure converges to the desired root. When L=F, < ( ) ( L) = F $'(L) >1 at L=F Therefore, ( | ) t(L)<l at the intersection the root There is only one root in the inter val, 0<L<'F When L-**0, 1 (L) 1 at L=0. 0 F L Figure No. 1 GENERAL SHAPE OF (j)(L) FOR ALL MIXTURES IN THE TWO PHASE REGION 14 Geometric Interpretation, It is of some value to il- 1 lustrate graphically just what is meant by the convergence condition, and to show how the iterative process converges I to the solution when the absolute value of the slope of the j function of L is less than one. Figure No. 2 is a plot of : some function, $U); notice that $T(L)^ 1 at the intersec tion with the L= 4>( L) line, the location of the desired root.1 I Four iterations are shown approaching the root from the left; to illustrate the fact that the choice of Lo plays i no part in the convergence, two iterations are shown ap- y | proaching the root from the right. i Figure No. 3 is a plot of some function, <(KL); but i i this function has a slope greater than one at the intersec tion. Three iterations are shown which clearly indicate that the error is increasing with each iteration, that is, i ■ the process is divergent. I While the two Figures Nos. 2 and 3 indicate the be- ; havior of the iterative process when the slope of <&(L) is | less than one and greater than one respectively, the reader | is reminded that the absolute sign allows for two more cas- i j es. These cases are when the slope is negative. Because !0(L) is a monotonic increasing function, these cases will i never occur. 15 F L = 4>(L) 0 L" Figure No. 2 CONVERGENCE ♦ ( L ) Figure No. 3 DIVERGENCE CHAPTER IV STEP BY STEP INSTRUCTIONS To the Engineer. Let us assume several flash prob lems are to be solved. It is desired to use the method presented in this paper. What is the first step? The sec ond? And so on. At the risk of oversimplification, these instructions are going to assume nothing is known about IBM equipment. The following steps are designed to fill the gap between the desire to use IBM to the point where one need only wait for the answers to be delivered to one’s desk. To those who have more than an elementary acquaint ance with IBM methods, the following instructions will no doubt leave a desire for a more mathematical approach. The next chapter is designed to fill this need. Step No. 1. Telephone or visit the tabulating depart ment. Ask the supervisor these two questions: 1. Does this department have a 602-A calcula ting punch? 2. Is the reproducing punch equipped with mark sensing? Assuming the answers to both of these questions are, "Yes." proceed to the next step. IS Step No. 2. Obtain some blank mark sense IBM cards and an electrographic pencil. Take enough cards to allow one for each component in each problem to be worked. For example, ten problems with twelve components in each would require one hundred and twenty cards, plus a few for good measure• Step No. 3* Look at Figure No. 4 and No. 5, pages 19 and 20. These are examples of how the cards are to be mark ed. Notice that there are ten columns designed to accept marks. Some cards have as many as twenty seven. Each col umn has a number to identify it, starting with one, through twenty seven. The cards illustrated have only the last ten, namely, eighteen through twenty seven. Mark the molal values in columns 18-22; the K values in 23-27, as illustrated. The decimal location for the molal values may be anywhere around or within the five pla ces provided. The decimal location for the K value is loca ted between columns 25 and 26. This means the range of K values possible is from 999*99 to 0.01. In addition to the data, some additional marks are re quired. These are made in the eleven row, or the "X" row. / A mark in the "X" row of the twentieth column identifies Top Detail Card Identification Detail Card Identification c0dc0dc0dc0dc0d®g^c0dc( ) d< s£s» c0d 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 8 9 10 II 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 e ^ C l D C l D C l D C l D C l ^ C l D C l D C l D C l D 111111111111111111111111111111111111111111111111111 c2dc2d<=£ b> c2dc2dc2dc2dc2dc2dc2d > 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 c3d^ c3dc3dc3dc3d^ c3dc3dc3d 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 c 4 z > c 4 z > c 4 4 z > c 4 z > c 4 z > c 4 d c 4 d c 4 :i' 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 c5dc5dc5dc5dc5dc5dc5dc5dc5dc:5d 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 c6dc6dc6dc6dc6dc6dc6dc6dc6dc6d 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 c8DC§DC§DC§D^0^iC§DC§DC§DC§DC§D , < = 9^>c9DC9DC9DC9DC9dc9D&9 ^r 9 DC9D 9 9 9 9 9 9 9 9 9 9 9 9 9 0 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 5 75 76 77 78 79 80 IBM 5 2 8 8 ms P R IN T E D IN U S A Mols = 13243 K = 39.07 Figure No. 4 AN IBM MARK SENSE CARD h - * vO Detail Card Identification 2 6 1 9 eeecO3<e9JGO^cO= ? c0^cO=?cO=?cO=>sOe i 000000000000000000000000000000000000000000000000000 1 2 3 4 5 6 7 8 9 10 1 1 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51C 1 Z ) C 1 Z ) C \ Z ) C 1 3 C \ 1 1 1 Z ) 111111111111111111111111111111111111111111111111111 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 c3dc3dc3dc3dc3dc3dc3dc3dc3dc3d h c4z) c4z) c4dc4zic4z) c4z) c4z) c4z) c4z) c4Z' ‘ c5dc5dc5dc5dc5dc5dc5^c5d^^::5d c6=> c6dc6dc6’ z> c6dc63c=6dc63c6=) c6=) 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 38 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 5 5 ^ 5 7 58 59 G O 61 62 63 64 65 68 67 68 69 70 7 1 ^ 73 74 75 76 77 78 79 80 IBM 5 2 8 3 ms P R IN T E O IN U S A c8=>c8=>c8=>c8=>c8=>c8=>c8=>®8sc8=>c8=> :9^c= c9DC9DC9DC9DC9DC9DC9DC9Dr9Z)C9D Mols = 700.1 K = Id.50 Figure No. 5 AN IBM MARK SENSE CARD ; the card as a data card, or "detail” card. A mark in the I i "X" row in the twenty-seventh column identifies the card as1 i I ! the first data card of a stack of data cards. ! i I ; I It is necessary to identify the data cards because ! two other kinds will be used; namely, MASTER cards, and trailer cards. As the names imply, data constant for all cards will be on a MASTER card, and the trailer card will follow the deck to receive the answer calculated from all of the cards of a deck. The top data card needs to be id- ; entified, because the calculation is slightly different for ! j this card. i Jot the problem number in the upper left hand corner l I on the first card of each deck. This will be useful when i i | the IBM operator is working with the cards. I i | Step No. 4* Make a data sheet for the operator’s ! convenience similar to the example shown on page 22. Fill in columns one and three with the total mols of feed, and ; one-one hundredth of this value, respectively. The IBM op-; I erator will refer to these numbers when testing the mixture . for two phases. j Step No. 5* Return the marked decks and the data ! i | sheet to the tab room. From this point on, the IBM will do; 22 TABLE I OPERATORf S DATA SHEET Column No. (1) (2) (3) (4) (5) Problem No.. Feed, Mol s. Trailer c c :76-S0 Feed 100 (2)-(3) Comments 1. 19995 200 2. 46662 467 3. 17730 177 4 • 36650 367 5 * 11436 114 etc • Column No. . (6) (7) (8) Problem No. Trailer cc:69-73 (6}-(3) Comments 1. 2. 3. 4 • 5.. etc. . - _ _ 23 the work. If a printed report is desired, notify the oper ator; however, the answers may be read directly from the data cards, after a little practice. To the Operator. The following directions are for solving equilibrium flash problems on the 602-A calculating punch. The operation breaks down into five main steps. 1. Wiring the 602-A panel, and the reproducing punch panel. A good time to do this is while the engineer is marking the decks. 2. Punching the cards. 3. Testing each problem. Some may not have a solution. 4* Finding the solution of those passing the tests of part three above. 5. Calculation of the composition of each phase for each problem using the solution found in part four above. Step No. 1. Wire the 602-A control panel following Figure No. 6, page 24* Give due consideration to the con trol brush locations. Wire the punch to read MS brushes 18-27* Wire l£-22 to punch in cc: 1-5* Split wire 20 to also punch in ceS 12 CALCULATING PUNCH TYPE 602 A CONTROL PANEL 24 r r m CONTROL READING RISFT CO SELECTORS DIGIT IM P ] |DO O- -( ‘ - -C L LANCE PICKUP a o a IT PICKUP o W M C B4AK PICKUP O O Q o o o P U N C \C O N lW fti EXIT 0 O O 0.0 O ¥3^0 O O O 0 DROPOUT IMPULSE QUOT / COUNTER ENT n i i r a O N O READING CO SELECTOf S o T r o a < 5 OP 5 * OPGPR fA D 0 0 OR O O M U BIFLIER — COUPLE r ' v i o r EXITS TER EXIT E P -* t M II HI INUSIMREAD out o PESIT I O c t O N O E K l t i ? d * 5 >o o C O PUNCHING c c P U N C H * x - o n DO PUNCH I f X COL P P 1ST GOL PUNCH o l o o o IS T C O L STOR 0 i l l TEST o NCM Iiroi READ IN r t : EAD IN STORAGE CONTROL STORAOR PUNCH ooooooooO — 7 P-PU NC H tM lT T E R -l— • — 0 0 o o o o c fc Figure No. 6 CONTROL PANEL 25 Reproducing Brushes oooo oooo 0 0 0 0 DPBC D 0 0 0 •—e- o o o o o o o o o o o o o o o o O O O- o 0—0 Q- ^uluinrt ■# $^»*o o o o o 0 0 0 0 0*0 o o o • 0 - 0 0 0 0 oooo 0 0 0 0 ^ ■Q.. n r> n r ^ n n Punch MakwretsS* 0 O O b O^Q^O O oj OP^OnS O o o o Q , . Punch B riches' o o ! e- •--e--e-*-e--e— • •- (?-• \ 3 0>yO 0 0 0 / 0 ^ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 _ 0 0 0 0 n n - a . -e- 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 o 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 v° 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 18 MS Brush ft 5 5 0 0 27 0 0 o 0 0 0 0 0 0 0 0 0 " o ■» -e--e--• 0 0 0 0 0 0 0 0 0 0 0 0J&" 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Q/0 0 0 0 0 0 0 0 0 0 \ 0 0 0 °Cor g M n g Magnets or 0 0 0 \«H-e--e--e--e- -e- 0 0 # 0 0 0 o ( 0 o 0 0 0 0 0 0 0 0 0 0 0 O O O , *0 0 0 0 0 \oS. 0 0 0 0 0 0 0 0 0£^&*0 0 0 0 0 0 p 0 - Comp irfig~Mai of MS OTTT - 0 0 9 r -e--&>2 #--e^# r 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Figure No. 7 MARK SENSING I Marked columns 18-27 punched into columns 1-10. i j20 is split wired to cc;12 and 17; 27 into cc:ll. _ Also, i and 17* Control X punches are heeded in these columns. | Split wire 27 to also punch in cc: 11. These instructions apply to the 602-A wiring as presented in Figure No. 6. Step No. 2. Punch the cards. Step No. 3* Make two MASTER cards as follows: Field in MASTER cc: 72-73 cc; 74 cc: 79-50 j MASTER No., 1 99 X 01 ! ! MASTER No. 2 01 X 99 | Insert a blank trailer card behind each detail deck. i | Place MASTER No. 1 on top of the entire stack and run i through the 602-A. The MASTER and detail cards will skip out; the trailers will be punched. j Sort out the punched trailers; sort on cc: 74* The * I ! trailers will all have an X punch in cc: 74; all of the | | other cards will go to reject. I ? Insert; a blank card behind each detail deck as before. ! ' i j Place MASTER No. 2 on top of the stack; start through the ! ✓ | 602-A. While calculating, turn attention to the trailers | yielded from the first calculation. 27 Read the number in cc: 76-30 from each trailer card and enter this value in the appropriate column on the data sheet provided by the engineer. This is column No. 2 in the data sheet. Subtract the number in Column No. 3 from the number just entered in Column No. 2. The result should be a positive number. If a negative number should result, the mixture is a superheated vapor and so naturally no liq- uid-vapor split exists. Enter the comment, ,fSuperheated1 1 in Column No. 5 in the data sheet. Again sort out the trailers. Read the numbers from cc: 69-73 from each trailer card and enter this value in the data sheet in Column No. 6. Again, a positive number should result. If any of the remaining decks yield a neg ative difference, these mixtures are subcooled liquid. En ter the comment, "Subcooled" in Column No. 3 for each prob lem with a negative difference in Column No. 7* Remove all decks, from It he stack which are either sup erheated vapor or subcooled liquid. Step No. 4* Make a MASTER card for each of the re maining decks as follows: Field in MASTER cc: 69-73 cc: 76-30 ______MASTER No. "j"_______Col.7f»i»th Row Col.4.wifith Row 28 Precede each deck with itfs respective MASTER; follow each deck with a blank trailer. Place all of the decks in the 602-A hopper and start calculating. Stand by the 602-A. When the first trailer is punch ed and stacked, quickly remove all the cards in the stacker at that time. Place the MASTER off to one side. Place the trailer just punched on top of the deck and a new blank on I the bottom. Replace this deck in the hopper on top of the i i other decks still in the hopper. Notice that the trailers become MASTERS upon being punched. Continue to pull the decks from the stacker immedi ately after the trailers are punched and stacked. Remove the MASTERS, place the punched trailers on top of the deck, place a new blank on the bottom and return to the hopper. When the first deck comes through the second time, compare the old MASTER on the top with the new one on the bottom. Are they the same? If not, discard the old MASTER place the new one on top, place another blank on the bottom and return to the hopper. If they are alike, set the deck to one side. Continue calculating the remaining decks un til the others also become alike. If they do not become alike within ten or twelve passes however, do not carry 29 the iterations any further. The additional accuracy gain ed beyond this point does not justify the effort. Step No. 5* Shift one of the X mark split wires on the reproducting punch board to punch in cc: 22. Run the decks (still preceded by their respective final MASTER) through the punch. Run the decks so altered through the 602-A. Trailers will not be needed this time. Each detail card will be | punched instead of skipping out. ! List the decks on the 402. Add fields 1-5, 69-73, and 76-80; list 6-10. Send the information to the engineer together with the data sheet. Ten Example Problems. The preceding set of instruc tions will take on a great deal more meaning and clarity if some problems are carried through the calculations, step by step. The following ten problems were made up at random, but with the idea in mind of illustrating all of the poss ible outcomes; that is, mixtures not in the two phase re gion as well as some which are in the two phase region. Of those falling in the two phase region, cases representing wide and narrow boiling mixtures are present; also cases where the ratio of liquid to vapor is both large and small. 30 PROBLEM NO. 1 Component Mols K value A 11.11 111.00 B 22.22 22.20 C 9.99 9.99 D g.gg g.gg E 33.33 3.33 F 7.77 0.77 G 6.66 0.66 H 55.55 0.55 I 44.44 0.44 Total 199.95 PROBLEM NO. 2 Component Mols K value z 333.3 100.00 Y 333.3 50.00 X 333.3 25.00 w 333.3 12.50 V 333.3 6.25 u 333.3 3.13 PROBLEM NO, 2 (CON'T) Component Mols K value T 333.3 1.66 S 333.3 0.&3 R 333.3 0.42 Q 333.3 0.21 P 333.3 0.10 0 333.3 0.05 N 333.3 0.02 M 333.3 0.01 ital 4666.2 PROBLEM NO, 3 Component Mols K value Alpha 1324 3.21 Beta 2435 1.23 Gamma 3546 0.87 Delta 4657 0.76 Epsilon 5768 O.65 Total 17730 PROBLEM NO. 4 32 Component Mols K value Ca . 217.6 114.00 Cb 431.0 12.50 Cc 66.9 9.85 Cd 579.3 8.89 Ce 467.7 5.45 Cf 365.4 4.95 Cg 776.9 2.24 Ch 119.8 1.77 Ci 88.2 0.88 Cj 552.2 0.55 Total 3665.O PROBLEM NO. 5 Component Mols K value Chi 445 1.88 Nu 232 1.56 Xi 981 1.01 Mu 9658 0.87 Pi 120 0.73 Total 11436 33 PROBLEM NO. 6 Component Mols K value DOl 2547 53.30 DO 2 643^ 11.55 DO 3 1652 6.40 DO 4 7076 5.25 ' D05 10226 ’ 2.77 Dq6 6991 0.99 D07 * 11154 0.31 DOa 3977 0.76 D09 4333 0.69 DIO 9429 0.39 Total 63319 PROBLEM NO. 7 Component Mols K value L $$79 73.00 M 2939 22130 N 13221 14.30 o 236$ 9.73. P 5457 3.66 Q 1230^ 5.44 PROBLEM NO. 7 (COM’T) Component Mols K value R 4441 3.79 S 2 SO 2.11 T 4369 1.29 U 177 0.99 V 6501 0.37 W 8891 0.7S X 367 0.44 Total 70286 PROBLEM NO. 3 Component Mols K value I 56.41 3.7S II 43.66 2.15 III 27.54 1.64 IV 54.22 1.11 V 232.22 0.90 VI 153.76 0.79 VII 75.31 0.66 VIII 55.46 0.42 Total 693.62 35 PROBLEM NO. 9 jmponent Mols K value Ab 5456 2.33 Cd 656? 0.33 Ef 767 3 0 -7 7 Gh 3739 0.63 Ij 9390 0.45 K1 21234 0.37 Total 59314 PROBLEM NO. 10 Component Mols K value Abe 442 144.50 Def 335 87.60 Ghi 776 22.30 Jkl 556 10.10 Mno 881 7.70 Pqr 707 4.56 Stu 1225 2.11 Vwx 666 1.23 Yza 2323 0.86 Bed 443 0.68 36 PROBLEM NO. 10 (CON’T) Component Mols K value Efg Hi j 9323 992 0.22 0.40 Total 13674 While the engineer is developing the data and marking the decks, the operator is wiring the boards. The wiring time is approximately 45 minutes for both boards. Punching the cards requires very little time. They are punched at the rate of 100 cards per minute. The entire deck for these ten problems should total ninety-two; one card for each component in each problem. At this point, the operator is ready to start with Step No. 3, page 26. The first four parts of Step No. 3 yields ten punched trailers. The numbers in cc: 76-30 are read from the trailers and entered into Column No. 2 In the Operator’s Data Sheet. Table II on the next page is a portion of this Data Sheet with the results of the first calculation enter ed in the appropriate columns. 37 TABLE II RESULTS OF THE FIRST CALCULATION Column No. 1. 2. 3. 4. 5* Problem No. Feed Mols Trailer cc: 76-SO Feed/100 “ ■V 2.-3. Comments 1. 199.95 234 200 +34 2. 4666.2 3971 467 +3504 3. 17730 214 177 +37 4* 3665..O 1S1 367 -1S6 Superhtd 5. 11436 126 114 + 12 6. 63 £19 621 63S >17 Superhtd 7. 702SS 2S9 703 -414 Superhtd S. 69S.62 797 : 699 +9S 9. 59S14 1113 596 +517 10. 1S674 322 1S7 +135 Notice that the decimal is ignored in Columns No. 2 and No. 3. It adds nothing to the calculation-, so it is convenient to drop it. Due to a negative sign in Column No. 4, Problems 4, 6, and 7 are eliminated. These mixtures are superheated vapor. Table III, page 3&, summarizes the results of the sec ond calculation. 3& TABLE III RESULTS OF THE SECOND CALCULATION Column No. 6. 3. 7. s. Problem No. Trailer cc: 69-73 Feed/100 6.-3. Comments 1. 1317 200 +1117 Two phases 2. 4219 467 +3752 Two phases 3. m 177 -3 Subcooled 4. 2314 Superheated 5. 106 114 -d Subcooled 6. 2594 Superheated 7. 7570 Superheated S. SOI 699 +102 Two phases 9. 446 596 -150 Subcooled 10. 793 1&7 +611 Two phases Due to a negative sign in Column No. 7, Problems 3, 5, and 9 are eliminated. These mixtures are subcooled liquid. In accordance with the instructions in Step No. 4, the following MASTERS were prepared: Field in MASTER MASTER Card No. cc: 69-73 cc: 74 cc: 76-30 1. 1117 X 34 2.___ 3752___________X__________ 3504 39 Field in MASTER MASTER Card No. cc: 69-73 cc: 74 cc: 76-30 3. 102 X 93 10. 611 X 135 Table IV, which follows, records the sequence of num bers which resulted from the iterations for each of the fou^ problems. TABLE IV RESULTS OF THE ITERATIVE CALCULATIONS MASTER Card No. cc: 69-73 cc: 76-30 1-1 1117 34 1-2 19313 677 1-3 19221 774 1-4 19114 331 1-5 13996 999 1-6 13363 1127 1-7 13731 1264 1-3 13534 1411 1-9 13429 1566 1-10 13270 1725 1-11 13103 1337 40 After eleven iterations, Problem No. 1 is still con verging. Further iterations will gain only little more ac curacy, so discontinue the iteration cycle. This slow type of convergence is typical of problems where the ratio of V/L is large. TABLE IV (CON’T) RESULTS OF THE ITERATIVE CALCULATIONS MASTER Card No. cc: 69-73 cc: 76- 2-1 3752 3504 2-2 24295 22367 2-3 24339 22323 2-4 24335 22277 2-5 24335 22277 The fifth iteration checks the fourth. The problem has converged. TABLE. IV (C0NTT) RESULTS OF THE ITERATIVE CALCULATIONS MASTER Card No. cc: 69-73 cc: 76-30 41 TABLE IV (CON*T) RESULTS OF THE ITERATIVE CALCULATIONS MASTER Card No. cc: 69-73 cc; 7 6 -6 0 8-2 35070 34792 8-3 34426 35436 8-4 33926 35936 8-5 33411 36451 8-6 32883 36979 8-7 32335 37527 8-8 31960 37902 8-9 31577 38285 8-10, 31188 38774 This is a narrow boiling range mixture. Slow conver gence is typical of these kind of mixtures. TABLE IV (CON’T) RESULTS OF THE ITERATIVE CALCULATIONS MASTER Card No. cc; 69-73 cc; 76-80 10-1 611 10-2 13960 10-3 12569 135 4714 6105 42 TABLE.IV (CONfT) RESULTS OF THE ITERATIVE CALCULATIONS MASTER Card No, cc: 69-73 cc: 76-30 10-4 11254 7420 10-5 10137 8537 10-6 9252 9422 10-7 8605 10069 10-8 8122 10552 10-9 7756 10918 10-10 7538 11136 Slow convergence is typical of peculiar types of mix tures. Although there is a wide range of components pre sent, the bulk of the mixture is made up of the next to the last component which has a K value of 0.40. Three out of four of these problems did not converge very rapidly. This might leave the reader with the impres sion that this is the usual case. On the contrary, the greatest majority of cases met in actual practice converge very nicely. Remember that these eases were made up delib erately to cause trouble to illustrate the possible diffi culties that might arise, and how to deal with them. 43 One of the wires on the reproducing punch panel is shifted to punch an X in cc: 22. This alters the deck in such a way that the calculator recognizes that the itera tive procedure is over. Each detail card is punched with the quantity of liquid and vapor that particular component contributes to the whole. Table V tabulates these values which were punched in the data cards on the final pass through the 602-A. This is how the report will look to the engineer when he receives the answers from the tab room. TABLE V FINAL CALCULATIONS PROBLEM NO. 1 Mols-Feed R value Mols-Vapor MolS-Liquid 11.11 111.00 11.10 0.01 22.22 22.20 22.12 0.10 9.99 9.99 9.S9 0.10 8 .8 8 8 .8 8 8 .7 8 0.10 33.33 3.33 32.32 1.01 7.77 0.77 6 . 8 k 0.93 6.66 0.66 5.75 0.91 55.55 0.55 46.70 8.85 44 TABLE V (CON'T) FINAL CALCULATIONS PROBLEM No. 1 (C0N»T) lols-Feed K value Mols-Vapor Mols-Liquid 44* 1 + 1 + 0 • 44 35.93 3.51 199.95 179.43 20.52 PROBLEM NQ. 2 ols-Feed K value Mols-Vapor Mols-Liquid 333.3 100.00 330.3 3.0 333.3 50.00 327.3 6.0 333.3 25.00 321.5 11.8 333.3 12.50 310.5 22.8 333.3 6.25 290.6 42.7 333.3 3.13 257. a 75.5 333.3 1.66 214.7 118.6 333.3 0.83 153.3 175.0 333.3 0.42 104.7 228.6 333.3 0.21 62.1 271.2 333.3 0.10 32.8 300.5 333.3 0.05 17.2 316.1 333.3 0.02 7.1 326.2 Mols-Feed 333.3 4666*2 Mols-Feed 56.41 43.63 27.54 54.22 232.22 153.73 75.31 55.46 45 TABLE V (CON'T) FINAL CALCULATIONS PROBLEM NO. 2 (CON'T) K value Mols-Vapor 0.01 3.6 2433.5 PROBLEM NO. 3 K value 3.73 2.15 1.64 1.11 0.90 0.79 0.66 0.42 Mols-Vapor 42.39 27.62 15.63 25.50 97.21 59.55 26.02 13.95 Mols-Liquid 329.7 2227.7 Mols-Liquid 14.02 16.06 11.91 2S.72 135.01 94.23 49.29 41.51 693.62 307.37 390.75 46 Mols-Feed 442 335 776 556 SSI 707 1225 666 2323 443 9323 992 13674 TABLE V (CON’T) FINAL CALCULATIONS PROBLEM NO. 10 K value 144.50 37.60 22.30 10.10 7.70 4.56 2.11 1.23 0.36 0.63 0.40 0.22 Mols-Vapor Mols-Liquid 437 329 727 434 73S 533 717 301 849 139 1972 127 7353 5 6 49 72 143 174 503 365 1474 304 7356 365 11321 In the event it is desired to find the exact root of before the final calculation, refer to Chapter VI for two methods for speeding up convergence. I CHAPTER V i ! DISCUSSION I Introduction, The purpose of this chapter is twofold: i i first, to discuss the procedure outlined in stepwise fash- : ion in the preceding chapter; and second, to present%the i control panel planning chart used in the development of the ! control panel presented on page 24* The control panel is : supplimented by a brief description of each program step. The Two-Phase Test. Before attempting to solve a | flash problem, it is time well spent to make certain that the mixture is actually in the two phase region. The for mal way of making this check is to see if the mixture sat isfies the following criteria: n n F < S fiAi, and F < z f± K ± . (21) | i=l i=l 1 The mixture is in the two phase region if the above ; inequalities are satisfied. I Although the control panel presented in this paper is i J not wired to check this directly, it may be done indirect- j ly. In the proof of Chapter III, it was demonstrated that the slope of <f>(L) > 1 at L=0 and L=F to be in the two phase , region. It follows, therefore, that if the slope is less than one at either end point, the mixture is not in the two phase region. j I In particular, if the slope is less than one when L ap-J proaches zero, the mixture is superheated vapor. Reference j } I to equations (6) and (1$) will confirm this. On the other hand, a slope of less than one as L approaches F is indica- tive of a subcooled liquid. I For the reasons stated above, a trial at each end will Iserve as a test for two phases. It will be recalled that the instructions outline a test at L=0.01F, and L=0.99F. J i ! These values were chosen for three reasons, namely; j 1. The approach to L=0 and L=F is fairly close. I * j 2. The trial values are easily calculated; that is, by a shift of the decimal. | I 3» This is roughly the limit of accuracy wired into this calculation. I This is not meant to imply that trials cannot be made |in the regions, i | i 0<L<0.01F, and 0.99F<L<F. IThis is not the case. In general, however, nothing useful 49 will result from trials in these regions. Definite limits cannot be stated because they are different for each mix ture. The rule for determining these upper and lower limits is a simple one, based on the maximum and minimum size of the factor, (KV/L)+1. RULE: If the factors, K, V, and L combine in the expression (KV/L)+1 such that the result exceeds' the value, 9999*9999 for any one of the compon ents in the mixture, an erroneous answer will be calculated. If KV/L is less than 0.0001, zero vapor will be calculated. Linear Interpolation. If the first two trials reveal that the mixture under consideration is, in fact, in the two phase region, the remaining task is to find the ratio of the split. It is known from the first two trials that the split occurs somewhere between 1$ and 99$® This order of accuracy is not sufficient for most practical purposes, so the following steps are taken to obtain a better answer. The first two trials not only confine the split to a smaller region, two values of $(L) result. If <J)(L ) were linear between these two points, the desired root could be found by solving the two linear equations; Even if <|)(L) is not linear between these two points, : this assumption will yield an approximation somewhere in the vicinity of the true root. Instead of solving the equations above, it is easier ' to make use of the following fact; the ratio of V'/L' is 1 almost equal to V/L in Figure No. 8, page 51. If c=0, the i j two ratios are exactly equal, but here c=0.01F. In view of i the fact that this is only an approximation of the real root, this slight error is not important. i i The Iterative Process. The third trial uses the ratio i •V’/L* which results in a new calculated set of values for V and L. These new values make up the ratio for the next trial. The process continues until the problem converges. The Final Calculation. When convergence is reached, the ratio of the split is known. In most cases, the com positions of each phase will also be desired. This is done1 !on the final pass through the calculator. The data cards are altered by adding an X which signals a final pass to Figure No. 3 LINEAR INTERPOLATION the calculator. Instead of forming the summations and skipping the data cards out without punching, each card is punched with the individual quantities of liquid and vapor resulting from that particular component. To clarify the procedure further, three of the prob lems will be solved graphically. The problems chosen were two, three, and four because all three possible outcomes are represented; that is, two phases, subcooled liquid, and superheated vapor, respectively. Certain parts of the dia grams will not be drawn to scale in order to bring out the points more clearly. These graphic solutions are presented on the following three pages. Approximation of the root by the linear interpola tion = 22550. True root = 22277. 0.01F 0.99F Figure No. 9 TWO PHASES 54 I i No root in the 177 interval, CKL/<F i i 0.99F 0.01F Figure No, 10 SUBCOOLED LIQUID No root in the interval 0<L<F I No test necessary 0 .OIF 1 36? Figure No. 11 SUPERHEATED VAPOR 56 The Planning Chart. On page 57, the planning chart used in working out this particular problem is presented. Planning charts are almost indispensable when designing a control panel. They are absolutely necessary when debug ging” a board; that is, searching for errors in wiring. It is presented here more in the interest of completeness than necessity. However, it may be of value to those learning to wire this particular IBM calculator, as many of the features not often used are brought into play. The 602-A has a read cycle, and twelve program steps. This problem used eleven of the twelve available. The op erations performed on these steps were as follows: Read Cycle: Trial values of L and V are read into storage units #3L and #3R from the MASTER cards. The card skips out immediately after the read cycle. Detail cards are treated differently on the read cy cle. The values of fj_ and Kj_ are read into counter #6 and storage unit #2L, respectively. A digit, ”1”, is emitted into counter #4-5* Program Step No. 1: L is transferred to storage unit 1 7 flR. V is transferred to counter #1-2-3. PLANNING CHART STOtAOIUMT — ■ ‘ PUNCH UMTS----------- «wff»>ymoNMu«itwf>TymwcH RASTER iRD S M il CARDS .xs XX RO RO Quotient Product RO xx RO itej : t < i - •XXLXX O Divide RO nc xc* ro Subtract Add Skip whet I "X" in 1: RO RO Occurs w: Program ! when "X" th tep t 7 In 1] . "iii r t li to- Punch when NX READ Skip all programs >rec< ding .f JG In 17 3 and 21 Program Step Mo. 2: V times divided by L is form ed in counter #4-5* Since nl! ! was added into this counter on the read cycle, (K^V/L)+1 is formed. Program Step No. 3: This program step is coupled with step number 2 to provide additional exits needed on step number 2. Program Step No. 4: (K^V/L)+1 is transferred to stor age unit #1R. Counters #1-2-3 and #4-5 are reset to zero. Program Step No. 5; fj_ is added into counter #1-2-3. i Program Step No. 6: fj_ is divided by (K^V/L)+1. The quotient, equal to 1^, is formed in counter #4-5* I Program Step No. 7l li, in counter #4-5, is subtract- ■ ed from fj_ in counter #6. The difference is equal to v^. Program Step No. Si The sum of all preceding values of 1 and v for this deck are transferred from storage units #7R and #7L respectively, and added to Vj_ and lj_ just form ed. This step is omitted on the first card of each deck. Program Step No. 9: Counter #1-2-3 is reset to zero. Program Step No. 10: The v and 1 summations are transferred from counters #6 and #4-5 to storage units #7L | and #7R- Counters #6 and #4-5 are reset to zero. Program Step No. 11: The signal to start calculating the next card is given. Trailer cards skip to this program step immediately after the read cycle. The contents of storage units #7L and #7R are punched. CHAPTER VI CONCLUSIONS > Comparisons. The great advantage in using this meth- I od is that the engineering talent is freed to do other work while the problems are being worked by non-technical per sonnel. This, in effect, increases the productivity of the engineering force. The answers are generated in less time, ■ and a printed report of the calculation is available. i To ascertain the order of magnitude of the time sav- ! ings, a simple test was conducted. The same problem was worked by slide rule, and by IBM. The test was repeated i with a different problem. The results of these tests were as follows: i ■ Time Required ! Seconds per component per trial Slide Rule IBM Problem #1. 93 11 Problem #2. 86 13 ■ Advantage: Approximately eight to one. i It should be emphasized at this point that IBM does I not show the greatest advantage with only one or two prob- j 61 lems. Unlike a human, it does not tire, nor slow down, nor start making mistakes. The one-hundred and first problem is worked with the same speed and accuracy as the first. It is a certainty that the advantage shown on the preceding page would skyrocket if the test had been with two-hundred . problems instead of only two. Electronic Calculators. The 602-A is a mechanical type calculator. Compared to the electromic machines, it is very slow. It does have some advantages, greatest among them being the low rental fee. It has storage capacity su- i I perior to itfs closest electronic competitor, the 604. In j 1 view of the storage capacity situation, the 602-A is prob- ! ably the better machine for this problem. The stored program calculators could be turned to this problem very successfully. It is estimated that the IBM 701 could handle all of the calculations performed in Chap ter IV in around seven seconds. Speeding Convergence. In some cases, it might be de sired to carry the iterations far enough so as to obtain convergence, even though the problem is converging very slowly. This could be a very time consuming process under certain circumstances. It is for this reason that the 62 iterations were arbitrarily broken off after ten or eleven cycles. This is justified in view of the K value data, which is accurate to at best, only 15%. Time spent getting any closer than this accuracy justifies is, in a sense, time wasted. However, there might be occasions when greater accur acy is available, and convergence will be justified. In this event, some method for speeding convergence will be necessary. Two such methods are presented in the following paragraphs. Both assume that slow convergence has been recognized, and the calculations have been suspended until something can be done to hasten convergence. The first method which will be discussed uses Newton's interpolation equations.Three equally spaced argu ments of L are chosen, and calculated at each of these three points. Either the forward or backward formula is used, depending where the root is found to be bracketed. The quadratic equation resulting is solved, and the itera tions continued from this value of L. When selecting the arguments of L, make the increments equal to about one-ten th of the feed each, then the root will surely be bracket ed. As an example, consider the problem on page 39» This problem did not converge in eleven iterations; as a matter 63 of fact, the first differences are still increasing. This can be seen by inspecting the values in Table IV, page 39* F in this problem is equal to 199*95, so let the in crements be 20.00. This is roughly 10$ of the feed. In the final calculation, pages 43-44, L = 20.52, so let Lj = 20.00. Then L2 = 40.00, and = 60.00. The results of these trials are summarized in the following table: TABLE VI TABLE OF DIFFERENCES L <t>(L) A A 2 20.00 21.65 40.00 40.28 18.63 60.00 56.74 16.46 -2.17 Because the root is bracketed between the last two trial values of L, use the backward formula: <|)(x) = yn + uAyn + u(u+l) A 2yn/21 + .. (21) where: <i>(x) = L 7n = 56.74 u = (L-60J/20. 64 Ayn = 16.46 A 2yn = .2.17 Substituting these values into equation (21), the fol lowing expression results: L = 56.74 + (0.05L-3)(16.46) + (0.05L-3)(0.05L-2)(-2.17J/2. Solving for L using the quadratic formula, L = 42.1$, and -7*43* Using the first root for the next trial results in a 4>(l ) = 42.19, which checks the trial value. Therefore the desired root is equal to 42.19* With the true root available, the error resulting from stopping after eleven iterations may be calculated. 42.19 - 20.52 199^5— x 100 = Referring back to Table VI, if the root had been brack eted between the first two values of L, the forward formula would have been used. The forward formula is as follows: 4>(x) = y0 + uAy0 + u(u-l) A 2y0/22 + .. (22) where: 4>l.x ) . = L .________ _______ ______ 65 y 0 = 2 1 . 6 5 Aj0 = IS. 63 A2y„ = -2.17 O u = (L-20)/20. The other method of speeding convergence is easier to apply, but it doesn’t speed the convergence quite as rapid ly as the method just discussed. This method of speeding convergence arises from the following considerations; that is, the iterative method is converging to the desired root, L, but very slowly. In ad dition to the foregoing, the relationship between the suc cessive values of are as follows: Eliminating r, and solving for L, the following expres sion results: (L-x^) = r(L-x2) = r2(L-x^) = ... (23) where: L - the desired root. r = some ratio, assumed constant xi= L — X£ — • (24) 66 where: Ai = *2“xl A 2 = x3-x2 O A = (x^-x2) - (x2-x-j_) These quantities will be recognized as the first and second differences* For example, consider problem No. 10, on page 42. The iterations were broken off after the tenth, due to slow con vergence. Apply the method just outlined to speed the con vergence. TABLE VII TABLE OF DIFFERENCES Iteration < I> (l) a 10th 9th 8th 10552 366 -148 10918 218 11136 Apply equation (24): L = 10918 - (366)(2l8)/(-148) L = 10918 + 539 = 11457 Using this- value for L, the iterations were started over again. The eleventh, twelfth, and thirteenth itera- j tions resulted in the following values: TABLE VIII TABLE OF DIFFERENCES Iteration A _____ A L_ 11th' 11511 3$ -12 12th 11549 26 13th 11575 Apply equation (24)• L = 11549 - (38)(26)/(-12) L = 11549 + 32 = 11631. Using this value for L, the iterations were started over again. The fourteenth iteration came out equal to the thirteenth, that is, L = 11632. The error resulting in stopping after eleven iteration cycles, then, is 11632 - 11321 - « 1100-2* In Conclusion. It is felt the method j is well designed from the standpoint of speed, convenience, and card economy. The main advantage is the virtual inde- i j pendence from the other machines in the tab room. The | greatest shortcoming is that the operator must remain in at i tendance to shift the trailers. This could be overcome by i j ; a sorting and collating step following each calculation, i t ! but it is about a break even alternative. I 69 BIBLIOGRAPHY 1. Getman and Daniels, TIOutlines of Physical Chemistry”, pg. 130, 1946 ed. John Wiley and Sons, Inc., New ; York. 2. Lewis, G. N., ”Proc. Am. Acad. Arts, Sci.”, 37, 49, (1901). 3. Selheimer, Souders, Smith, Brown, ”Ind. Eng. Chem.”, 26 514, (1932); Lewis and Luke, ”Trans. Am. Soc. Mech. Engrs.”, 54, 55 (1932) and ”Ind. Eng. Chem.”, 25, 725, (1933); Newton, ”Ind. Eng. Chem.”, 27, 302, ■ (1935); Watson and Smith, ”Nat. Pet. News”, July 1936; Morgan and Childs, ”Ind. Eng. Chem.”, 37, 667, (1945). 4* Lewis, W. K., ”Ind. Eng. Chem.”, 23, 257, (1936). 5. NGSMA Data Book, 1951 ed. 6. Robinson and Gilliland, ”Elements of Fractional Distil lation”, pg. 44, 1939 ed. 7. Nelson, W. L., "Petroleum Refining Engineering”, pg. 409, 1949 ed. 3. Nelson, W. L., ”0il & Gas J.”, June 22, 1937, pg. 63, and Nord, M., "Nomographs for Flash Vaporization”, "Anal. Chem.”, 19, 431, (1947). |9. Reilly, P. M., "Petroleum Refiner”, Vol. 30, No. 7, ! _________ pg._132,.Xl951). -....- - ---- 70 BIBLIOGRAPHY (CON'T) 10. Salmon, R., "Petroleum Refiner", Vol. 33, No. 2, pg. 156, (1954). 11. Donnell, and Turbin, "Petroleum Refiner", Vol. 29, No. 10, pg. 102, (1950). 12. Scarborough, J. B., "Numerical Mathematical Analysis”, pg. 201, 1950 ed. 13. Smith K. A., and Smith, R. B., "Vaporization Equili brium Constants and Activity Coefficient Charts”, "Petroleum Processing", December 1949. 14* Scarborough, J. B., "Numerical Mathematical Analysis”, pg. 61, 1950 ed. L*Htvers*fcy of-South#rn California

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Creator Thompson, Harold Austin (author)

Core Title IBM application to equilibrium flash calculations

Degree Master of Science

Degree Program Education

Publisher University of Southern California (original), University of Southern California. Libraries (digital)

Tag engineering, chemical,OAI-PMH Harvest

Language English

Contributor Digitized by ProQuest (provenance)

Advisor Ballard, John H. (committee chair), McFall, Donald B. (committee member), Smatko, Joseph S. (committee member)

Permanent Link (DOI) https://doi.org/10.25549/usctheses-c20-308266

Unique identifier UC11259466

Identifier EP41748.pdf (filename),usctheses-c20-308266 (legacy record id)

Legacy Identifier EP41748.pdf

Dmrecord 308266

Document Type Thesis

Rights Thompson, Harold Austin

Type texts

Source University of Southern California (contributing entity), University of Southern California Dissertations and Theses (collection)

Access Conditions The author retains rights to his/her dissertation, thesis or other graduate work according to U.S. copyright law. Electronic access is being provided by the USC Libraries in agreement with the au...

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