Page 175 - Handbook of Energy Engineering Calculations
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entering temperature,°F − gas temperature leaving evaporator,°F). Or, Q   =
                                                                                                         1
               (140,000)(0.99)(0.27)(980  −  408)  =  21.4  MM  Btu/h  (6.26  MW).  The
               enthalpy absorbed by the steam in the evaporator, Btu/lb (kJ/kg) = (enthalpy

               of  the  saturated  steam  in  the  HRSG  outlet  –  enthalpy  of  the  feedwater
               entering the evaporator at 373°F) + (blowdown percentage)(enthalpy of the
               saturated  liquid  of  the  outlet  steam  −  enthalpy  of  the  water  entering  the
               evaporator, all in Btu/lb). Or, enthalpy absorbed in the evaporator = (1199.3

               − 345) + (0.05)(362.2 − 345) = 855.2 Btu/lb (1992.6 kJ/kg). The quantity of
               steam generated = (Q , energy absorbed by the evaporator, Btu/h)/(enthalpy
                                          1
                                                                                                6
               absorbed  by  the  steam  in  the  evaporator,  Btu/lb)  =  (21.4  ×  10 )/855.2  =
               25,023 lb/h (11,360 kg/h).


               2. Determine the economizer duty and exit-gas temperature
               The economizer duty = (steam generated, lb/h)(enthalpy of water entering the

               economizer, Btu/lb – enthalpy − enthalpy of the feedwater at 230°F, Btu/lb)(l
               + blowdown percentage) = (25,023)(345 − 198.5)(1.05) = 3.849 MM Btu/h
               (1.12 MW).
                  The  gas  temperature  drop  through  the  economizer  =  (economizer
               duty)/(gas  flow  rate,  lb/h)  (1  −  heat  loss  percentage)(specific  heat  of  gas,

                                            6
               Btu/lb°F) = (3.849 × 10 )/(140,000)(0.99)(0.253) = 109.8°F (60.9°C). Hence,
               the  exit-gas  temperature  from  the  economizer  =  (steam  saturation

               temperature,°F − exit-gas temperature from the economizer,°F) = (408 − 109)
               = 299°F (148.3°C).


               3. Calculate the constant K for evaporator performance
               In  simulating  evaporator  performance  the  constant  K   is  used  to  compute
                                                                                  1
               revised performance under differing flow conditions. In equation form, K , =
                                                                                                        1
               ln[(temperature  of  gas-turbine  exhaust  gas  entering  the  HRSG,°F  −  HRSG

               saturated steam temperature,°F)/(gas temperature leaving the evaporator,°F −
               HRSG  saturate  steam  temperature,°F)]/(gas  flow,  lb/h).  Substituting,  K ,  =
                                                                                                        1
               ln[(980 − 388)/(408 − 388)]/140,000 = 387.6, where the temperatures used

               reflect design condition.


               4. Compute the revised evaporator performance
               Under the revised performance conditions, using the given data and the above
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