Page 170 - Handbook of Energy Engineering Calculations
P. 170

21,520/(730 × 23,879) = 0.00124. By substituting in the equation in step 1,
               we have Q = 58.4 (W )(O). This is an important equation because it relates
                                          g
               the  oxygen  consumption  from  the  exhaust  gases  to  the  burner  fuel

               consumption.


               4. Find the fuel input to the HRSG
               The fuel input is given by W  + h  + Q = (W  + W )(h ), where h  and h                      g2
                                                                       g
                                                                              f
                                                  g
                                                                                                 g1
                                                        g1
                                                                                  g2
               are  the  enthalpies  of  the  exhaust  gas  before  and  after  the  fuel  burner,
               respectively; W  = fuel input, lb/h (kg/h); Q = fuel input in Btu/h (kJ/h).
                                  f
                  The  relation  above  requires  enthalpies  of  the  gases  before  and  after  the
               burner, which entails detailed combustion calculations. However, considering
               that  the  mass  of  fuel  is  a  small  fraction  of  the  total  gas  flow  through  the
               HRSG, the fuel flow can be neglected. Using a specific heat for the gases of

               0.31 Btu/lb°F (1297.9 J/kg K), we have, Q = 150,000(0.31)(1575 − 950) = 29
                     6
               × 10  Btu/h (8.49 kW).
                                                                             6
                  The percent of oxygen by volume, O = (29 × 10 )/(58.4 × 150,000) = 3.32
               percent. That is, only 3.32 percent oxygen by volume is consumed and we
               still have 15.00 − 3.32 = 11.68 percent left in the flue gases. Thus, more fuel
               can be fired and the gases will not run out of oxygen for combustion.
                  Typically, the final oxygen content of the gases can go as low as 2 to 3

               percent using 3 percent final oxygen, the amount of fuel that can be fired =
               (150,000)(58.4)(15 − 3) = 105 MM Btu/h (110.8 MM J/h). It can be shown
               through  an  HRSG  simulation  program  (contact  the  author  for  more
               information) that all of the fuel energy goes into steam. Thus, if the unfired

               HRSG were generating 23,000 lb/h (10,442 kg/h) of steam with an energy
               absorption  of  23  MM  Btu/h  (24.3  MM  J/h),  approximately,  the  amount  of
               steam that can be generated by firing fuel in the HRSG = 23 + 105 = 128
               MM Btu/h (135 MM J/h), or 128,000 lb/h (58,112 kg/h) of steam. This is

               close to a firing temperature of 3000 to 3100°F (1648 to 1704°C).


               Related  Calculations.  Using  the  methods  given  in  Section  1  of  this
               handbook, one may make detailed combustion calculations and obtain a flue-
               gas  analysis  after  combustion.  Then  compute  the  enthalpies  of  the  exhaust
               gas  before  and  after  the  burner.  Using  this  approach,  you  can  check  the

               burner  duty  more  accurately  than  using  the  gas  specific-heat  method
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