Page 407 - Handbook of Energy Engineering Calculations
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               1,350,000/[(25)(290)] = 186.4 ft  (17.3 m ), say 200 ft  (18.6 m ).
               4. Compute the required quantity of heating steam
               The heat added to the oil = Q = 1,350,000 Btu/h, from step 1. The enthalpy of
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               vaporization  of  200-lb/in   (abs)  (1379-kPa)  saturated  steam  is,  from  the
               steam tables, 843.0 Btu/lb (1960.8 kJ/kg). Use the relation W = Q/h , where
                                                                                                  fg
               W = flow rate of heating steam, lb/h; h  = enthalpy of vaporization of the
                                                                 fg
               heating steam, Btu/lb. Hence, W = 1,350,000/843.0 = 1600 lb/h (720 kg/h).


               Related  Calculations.  Use  this  general  procedure  to  find  the  heat-transfer
               area, fluid outlet temperature, and required heating-fluid flow rate when true

               parallel flow or counterflow of the fluids occurs in the heat exchanger. When
               such a true flow does not exist, use a suitable correction factor, as shown in
               the next calculation procedure.
                  The procedure described here can be used for heat exchangers in power

               plants,  heating  systems,  marine  propulsion,  air-conditioning  systems,  etc.
               Any heating or cooling fluid—steam, gas, chilled water, etc.—can be used.
                  To  select  a  heat  exchanger  by  using  the  results  of  this  calculation
               procedure, enter the engineering data tables available from manufacturers at

               the computed heat-transfer area. Read the heater dimensions directly from the
               table.  Be  sure  to  use  the  next  larger  heat-transfer  area  when  the  exact
               required area is not available.
                  When  there  is  little  movement  of  the  fluid  on  either  side  of  the  heat-

               transfer area, such as occurs during heat transmission through a building wall,
               the arithmetic mean (average) temperature difference can be used instead of
               the LMTD. Use the LMTD when there is rapid movement of the fluids on
               either side of the heat-transfer area and a rapid change in temperature in one,

               or  both,  fluids.  When  one  of  the  two  fluids  is  partially,  but  not  totally,
               evaporated  or  condensed,  the  true  mean  temperature  difference  is  different
               from  the  arithmetic  mean  and  the  LMTD.  Special  methods,  such  as  those
               presented  in  Perry—Chemical  Engineers’  Handbook,  McGraw-Hill,  2007,

               must  be  used  to  compute  the  actual  temperature  difference  under  these
               conditions.
                  When two liquids or gases with constant specific heats are exchanging heat
               in a heat exchanger, the area between their temperature curves, Fig. 2, is a
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