Page 415 - Handbook of Energy Engineering Calculations
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FIGURE 6 Barometric condenser condensing-water flow rate.



               2. Compute the total cooling-water flow rate required
               Use this relation: total cooling water required, gal/min = (unit cooling-water
               flow rate, gal/min per 1000 lb/h of steam condensed) (steam flow, lb/h)/1000.

               Or, total gal/min = (52)(250,000/1000) = 13,000 gal/min (820.2 L/s) of 60°F
               (15.6°C)  cooling  water.  For  80°F  (26.7°C)  cooling  water,  total  gal/min  =
               (120)(250,000/1000) = 30,000 gal/min (1892.7 L/s). Thus, a 20°F (11.1°C)

               rise in the cooling-water temperature raises the flow rate required by 30,000 –
               13,000 = 17,000 gal/min (1072.5 L/s).


               3. Compute the quantity of air that must be handled
               With a steam flow of 25,000 lb/h (11,250 kg/h) to a barometric condenser,
               manufacturers’ engineering data show that the quantity of air entering with
                                                       3
                                     3
               the  steam  is  3  ft /min  (0.08  m /min);  with  a  steam  flow  of  250,000  lb/h
                                                                      3
                                                                                       3
               (112,500 kg/h), air enters at the rate of 10 ft /min (0.28 m /min). Hence, the
               quantity  of  air  in  the  steam  that  must  be  handled  by  this  condenser  is  10
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