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Process  Heat Transfer                                          163


            where n is the number of segments.  After  solving Equation 4.6 for Q, we find  that

                                   n
                     ——————————————————————                               (4.7)
            Q = U 0 A 0
                     l/(At) LMi  +  l/(At) LM2  +  • • +  l/(At) LMn
                                           •
                 The  expression to the right  of A 0  is an effective  logarithmic-mean tempera-
            ture difference.  Thus,

            Q=U 0A 0(At) LM,eff


            Correction Factor for Non-countercurrent Flow

            It  was  seen  from  the  discussion  of heat  exchangers that  the  fluid  streams  are not
            strictly  countercurrent.  Baffles  on the  shell  side induce  crossflow,  and  in a two-
            tube-pass  heat  exchanger  both  countercurrent  and  cocurrent  flow  occur.  To  ac-
            count  for deviations  from  countercurrent  flow,  the  logarithmic-mean temperature
            difference  is multiplied by a correction factor, F.  Thus,


            Q = U 0 A 0 F(At) LM                                          (4.9)


                 An  equation for the  correction  factor  can be  derived with the  following  as-
            sumptions:

            1.  adiabatic operation
            2.  well mixed shell-side fluid
            3.  the heat-transfer  surface area is the same for each tube pass
            4.  constant overall heat-transfer coefficient
            5.  constant heat capacity
            6.  no phase change for either fluid

                 The  correction  factor  for  a  one-shell-pass  and  a  two-tube-pass  heat  ex-
            changer (a 1-2 heat exchanger), which is derived by Kern [1], is

                                     1-S
                                  i n____
                                  111
                                    ————————
                   2
                 (R +l) 1/2         1-R S
            F=  ——————   —————————————————                               (4.10)
                                          2
                                               2
                  R-l       2-S[(R+l)-(R +l)" ]
                         In ——————————————



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