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How to Figure True Temperature Difference in Shell-and-Tube ...

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, '. I0.1 0.3 0.4 0.5 0.6 0.7 0.8 1.0I.80.90.9 110­eUco ....... c:e'"5Q)10­10­e0.8" ~I­~IIu.:..TITI.. . ita~====~j:Jtl ,0.7 T2One divided··shellTwo-tube pass.0.1 0.2 0.31. From Equation 14, tb = 80 +0.6 (40) = 104.2. LMTD <strong>in</strong> the lower tube pass:Hot fluid Cold fluid <strong>Difference</strong> 200 - 104 96 120 - 80 = 40 LMTD = 64R =3.33 X = 0.2Lltm = LMTD X F = 64 X 0.92= 58.93. LMTD <strong>in</strong> the upper tube pass:Hot fluid Cold fluid <strong>Difference</strong>200 - 120 80120 - 104 16LMTD = 39.8R=5.0X=0.167Lltm = LMTD X F =0.882 = 35.14. From Equation 15,(40) (58.9)tb = 80 +58.9 + 35.139.8 )< = 105 0.5 0.6X=(t~- tJ(r,- t 1)5. S<strong>in</strong>ce the calculated value oftb did not match the value used, usethe calculated value <strong>and</strong> go back <strong>to</strong>the start.The f<strong>in</strong>al value of tb is 105.5.6. With the f<strong>in</strong>al value of tb theMID's are:Lower Lltm = 57 <strong>and</strong> upper Lltm= 32.From Equation 16,~".' 40Lltm= = 44.525.5/57 + 14.5/327. F = 44.5/57.6 = 0.773A computer is almost a must <strong>to</strong>calculate the many different po<strong>in</strong>tsrequired for an F chart.A Fortran program can be writtenfor this type of flow. The biggestproblem <strong>in</strong> putt<strong>in</strong>g it on thecomputer is <strong>to</strong> calculate the correctionfac<strong>to</strong>r for divided-flow, onetubepass. When do<strong>in</strong>g it by h<strong>and</strong>,as <strong>in</strong> the above example, use an Fchart. S<strong>in</strong>ce the F chart' was developedby trial <strong>and</strong> error, we donot have an equation for it. There<strong>in</strong>lies the trouble.0.7 0.8 0.9 1.0Solv<strong>in</strong>g Equation .9 for cp0.80.7isn'tpossible by the usual means. Andwithout cp it is impossible <strong>to</strong> calculateF.In the computer program developedby the author, cp is found bytrial <strong>and</strong>' error. To converge on thevalue of cp, use. Equation 17 as afirst approximation:cp=2[ F/(R+O.5) (17)2 - 1.05 X (2 R + 1)After this value is calculated; usethe first derivative method of converg~nce.Fig. 8 is the F chart for one dividedshell <strong>and</strong> two tube passes.The dashed l<strong>in</strong>e extended across thechart shows when the outlet temperatureof the hot side is ~qual <strong>to</strong>the outlet temperature of the coldside. In ,_ compar<strong>in</strong>g this with thenormal one,-shell pass, the low po<strong>in</strong>tis approximately 0.8 while the dividedshell is approximately 0.775.Three F charts for divided-flow,two-tube pass, from two <strong>to</strong> fourTHE OIL AND GAS JOURNAL •SEPTEMBER . 14, 1964113

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