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                    312                                                    Lawrence K. Wang et al.




































                                Fig. 4. Vapor pressure–temperature relationship. (From US EPA.)




                    on the shell side of the tubes. Depending on the application, the tube and shell side flu-
                    ids may be reversed. The emission stream is assumed to consist of a two-component
                    mixture: one condensible component (VOC, possible HAP), and one noncondensible
                    component (air). Such systems typically exhibit nonisothermal condensation. However,
                    the calculations here assume isothermality, which usually does not introduce large errors.

                    3.2. Estimating Condensation Temperature
                       Sizing the condenser involves steps to determine the surface area of the condenser.
                    The following design procedure applies to a condenser system with a shell and tube
                    heat exchanger, with condensate forming on the shell side. The waste gas stream is
                    assumed to be the VOC–air mixture noted earlier. Calculations for cases involving
                    mixtures of HAPs and supersaturated streams are quite complex and will not be treated
                    here because they are beyond the scope of this chapter. References 5 and 6 contain
                    information on these streams. For this case, estimation of the condensation tempera-
                    ture assumes the gas (air) stream to be saturated with a VOC component. For a given
                    removal efficiency, the partial pressure (in mm Hg) for the contaminant in the exiting
                    stream, P    , can be calculated:
                             partial
                                                   (10.   RE)    
                                                                  
                                               
                                                      − 01
                                     P partial  = 760   −(  −     HAP ×10  −6                (1)
                                                       ×
                                                                        e
                                                           8
                                                                  
                                                   RE 10 HAP    e)
                                               1
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