Page 36 - Handbook of Energy Engineering Calculations
P. 36

In this calculation, the temperature correction factor 2.15 = absolute flue-
               gas temperature,°R/absolute atmospheric temperature,°R = (600 + 460)/(32 +
               460).  The  total  weight  of  N   in  the  flue  gas  is  the  sum  of  the  N   in  the
                                                   2
                                                                                                   2
               combustion air and the fuel, or 10.4580 + 0.0022 = 10.4602 lb (4.707 kg).


               5. Compute the CO  content of the flue gas
                                       2
               CO ,  wet  basis  =  55.0/387.82  =  0.142,  or  14.2  percent.  CO ,  dry  basis  =
                                                                                            2
                    2
               55.0/(387.2 − 43.5) = 0.160, or 16.0 percent.


               6. Compute the air required with stated excess flow
               The pounds (kilograms) of air per pound (kilogram) of oil with 20 percent

               excess  air  =  (1.20)  (13.6176)  =  16.3411  lb  (7.353  kg)  of  air  per  pound
               (kilogram) of oil burned.


               7. Compute the weight of the products of combustion
               The  weight  of  the  products  of  combustion  =  product  weight  for  perfect

               combustion,  lb  +  (percent  excess  air)(air  for  perfect  combustion,  lb)  =
               14.6173  +  (0.20)(13.6176)  =  17.3408  lb  (7.803  kilogram)  of  flue  gas  per
               pound (kilogram) of oil burned with 20 percent excess air.


               8. Compute the volume of the combustion products and the percent CO                     2

               The  volume  of  excess  air  in  the  products  of  combustion  is  found  by
               converting  from  the  weight  to  the  volumetric  analysis  and  correcting  for
               temperature  as  in  step  4,  using  the  air  weight  from  step  2  for  perfect

               combustion  and  the  excess-air  percentage,  or  (13.6176)(0.20)(359/28.95)
                                                  3
                                    3
               (2.15)  =  72.7  ft   (2.058  m ).  Add  this  to  the  volume  of  the  products  of
                                                                                                   3
                                                                                     3
               combustion found in step 4, or 387.82 + 72.70 = 460.52 ft  (13.037 m ).
                  By using the procedure in step 5, the percent CO , wet basis = 55.0/460.52
                                                                             2
               =  0.1192,  or  11.92  percent.  The  percent  CO ,  dry  basis  =  55.0/(460.52  −
                                                                        2
               43.5) = 0.1318, or 13.18 percent.


               Related Calculations. Use the method given here when making combustion
               calculations for any type of fuel oil—paraffin-base, asphalt-base, Bunker C,
               no. 2, 3, 4, or 5—from any source, domestic or foreign, in any type of furnace

               —boiler, heater, process, or waste-heat. When the air used for combustion
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