Page 188 - Handbook of Energy Engineering Calculations
P. 188

Calculation Procedure:


               1. Determine the sources of waste heat available in the typical I-C engine

               There  are  three  primary  sources  of  waste  heat  available  in  the  usual  I-C
               engine. These are: (1) the exhaust gases from the engine cylinders; (2) the
               jacket  cooling  water;  (3)  the  lubricating  oil.  Of  these  three  sources,  the
               quantity  of  heat  available  is,  in  descending  order:  exhaust  gases;  jacket

               cooling water; lube oil.


               2. Show how to compute the heat recoverable from each source
               For the exhaust gases, use the relation, H  = W(Δt)(c ), where W  = rate of
                                                                                 g
                                                                  A
                                                                                                A
               gas  flow  from  the  engine,  lb/h  (kg/h);Δ  t  =  temperature  drop  of  the  gas
               between the heat exchanger inlet and outlet,°F (°C); c  = specific heat of the
                                                                                 g
               gas, Btu/lb°F (J/kg°C). For example, if an I-C engine exhausts 100,000 lb/h

               (45,400 kg/h) at 700°F (371°C) to a HRSG (heat-recovery steam generator),
               leaving the HRSG at 330°F (166°C), and the specific heat of the gas is 0.24
               Btu/lb°F (1.0 kJ/kg°C), the heat recoverable, neglecting losses in the HRSG

               and connecting piping, is H  = 100,000(700 − 330) (0.24) = 8,880,000 Btu/h
                                                 A
               (2602 MW).
                  With an average heat of vaporization of 1000 Btu/lb (2330 kJ/kg) of steam,

               this exhaust gas flow could generate 8,880,000/1000 = 8880 lb/h (4032 kg/h)
               of steam. If oil with a heating value of 145,000 Btu/gal (40,455 kJ/L) were
               used  to  generate  this  steam,  the  quantity  required  would  be
               8,880,000/145,000 = 61.2 gal/h (232 L/h). At a cost of 90 cents per gallon,

               the  saving  would  be  $0.90(61.2)  =  $55.08/h.  Assuming  5000  hours  of
               operation  per  year,  or  57  percent  load,  the  saving  in  fuel  cost  would  be
               5000($55.08) = $275,400. This is a significant saving in any plant. And even
               if heat losses in the ductwork and heat-recovery boiler cut the savings in half,

               the savings would still exceed $100,000 a year. And as the operating time
               increases, so too do the savings.


               3. Compute the savings potential in jacket-water and lube-oil heat recovery
               A  similar  relation  can  be  used  to  compute  jacket-water  and  lube-oil  heat

               recovery. The flow rate can be expressed in either pounds (kg) per hour or
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