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Process Heat Transfer                                           181


                                           5
                                                        5
            Q = 261.4  kcal/kg (650  kg/hr) = l.VxlO  kcal/h (6.75xl0  Btu/h)
                 To calculate the logarithmic-mean temperature difference,  the terminal  tem-
            peratures of the condenser must be fixed.  Because the condensation is essentially
            isobaric,  the  inlet  and  outlet  temperatures  of  the  ammonia  stream  are  41.4°C
            (106.5 °F). From Table 4.1, the inlet cooling-water temperature is 30°C (86.0  °F)
            if  cooling-tower  water  is  used.  Also,  for thermodynamic  considerations  the  exit
            water  temperature  must  be  less  than  41.4°C,  and  it  is  calculated  from  Equation
            4.7.6. If the lower value of the approach temperature difference  of 5 °C (9.0 °F) is
            selected  from  Table 4.4, a low cooling-water  flow  rate will be  needed. Thus,  exit
            water  temperature  is  36.4°C.  Therefore,  from  Equation  4.7.5,  the  logarithmic-
            mean temperature difference,

                    41.4-36.4-(41.4-30.0)
            (At) LM =———————————————      = 7.765 °C  (14.0  °F)
                             5.0
                          hi ——
                             11.4

                 For  isothermal  condensation,  the  logarithmic-mean  temperature  difference
            correction  factor,  F,  equals one.  Therefore,  from  Equation 4.7.3 for the  existing
            heat exchanger, the available overall heat-transfer  coefficient,

                   Q
            u  = ————

                AF(At) LM
                      1.7xl0 5  kcal  1        1      4.183xl0 3   J    1  h
            U 0  =
                         h          46m 2  7.765 °C     1         kcal  3600s
                            2
               0 = 97.5Btu/h-fV -°F.
                From Equation 4.7.11, the clean overall heat-transfer  coefficient,

                  hj  ho  5000 (8000)
                                            3
                                                               2 0
                                                 2
            U oC = ——— = ———————   = 3.077xl0  W/m -K  (542  Btu/h-ft - ?)
                 hj + ho  5000 + 8000
                 Then, from Equation 4.7.10, the available fouling resistance,

                 U oC -U 0  3077-553.5
                                                                      2
                                                    2
                                                 3
            ROA = —————   =———————     =1.482xlO~ m -°C/W (2.60x10" h-ft -°F/Btu)

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