Page 87 - Handbook of Energy Engineering Calculations
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4. Determine the electrical output of the cogeneration plant
               Since  the  efficiency  of  the  turbine  is  already  factored  into  the  exhaust
               enthalpy of the cogeneration turbine, use the relation, P = W (h  − h )/3413,
                                                                                          s
                                                                                             1
                                                                                                   2
               where the symbols are as defined earlier. Or, P = 60,000(1448 − 1315)/3413
               = 2338 kW.


               5. Compute the total energy output of the cogeneration plant
                                                                                   2
               Assuming that the latent heat of the steam at 200 lb/in  (abs) (1378 kPa) is
               available  for  industrial  process  heating,  the  total  energy  output  of  the
               cogeneration scheme = electrical output + (steam flow, lb/h)(latent heat of the
               exhaust steam, Btu/lb). Since, from the steam tables, the latent heat of steam
                              2
               at  200  lb/in   (abs)  (1378  kPa)  =  834  Btu/lb  (1939.9  kJ/kg),  total  energy
               output of the cogeneration cycle = (2338 kW)(3413) + (60,000)(834) = 58
               MM Btu/h (61.2 MM kJ/h).
                  Since the total energy output of the conventional cycle was 23 MM Btu/h

               (24.3  MM  kJ/h),  the  ratio  of  the  cogeneration  output  vs.  the  conventional
               output = 58/23 = 2.52. Thus, about 2.5 times as much energy is derived from
               the cogeneration cycle as from the conventional cycle.


               6. Find the comparative efficiencies of the two cycles
               The boiler input = (weight of steam generated, lb/h)(enthalpy of superheated

               steam  at  boiler  outlet,  Btu/lb  −  enthalpy  of  feedwater  entering  the  boiler,
               Btu/lb)/(boiler efficiency, expressed as a decimal). Or, boiler input = (60,000)
               (1448 − 200)/0.82 = 91.3 MM Btu/h (96.3 MM kJ/h). The efficiency of the

               conventional  cycle  is  therefore  (23/91.3)(100)  =  25  percent.  For  the
               cogeneration cycle, the efficiency (58/91.3)(l0) = 63.5 percent.


               Related Calculations. This real-life example shows why cogeneration is such
               a popular alternative in today’s world of power generation. In this study the
               cogeneration scheme is more than twice as efficient as the conventional cycle

               —63.5 percent vs. 25 percent. Higher efficiencies could be obtained if the
                                                                                     2
               boiler outlet steam pressure were higher than 1000 lb/in  (abs) (6890 kPa).
               However, the pressure used here is typical of today’s industrial installations

               using cogeneration to save energy and conserve the environment.
                  This  procedure  is  the  work  of  V.  Ganapathy,  Heat  Transfer  Specialist.
               ABCO Industries, Inc.
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