Page 189 - Handbook of Energy Engineering Calculations
P. 189

gallons (L) per minute, depending on the designer’s choice.
                  Since water has a specific heat of unity, the heat-recovery potential of the
               jacket water is H  = w(Δt ), where w = weight of water flow, lb/h (kg/h);Δt                 w
                                    w
                                              w
               = change in temperature of the jacket water when flowing through the heat
               exchanger,°F (°C). Thus, if the jacket-water flow is 25,000 lb/h (11,350 kg/h)
               and  the  temperature  change  during  flow  of  the  jacket  water  through  and
               external heat exchanger is 190 to 70°F (88 to 21°C), the heat given up by the

               jacket water, neglecting losses, is H  = 25,000(190 − 70) = 3,000,000 Btu/h
                                                           w
               (879  MW).  During  25  hours  the  heat  recovery  will  be  24(3,000,000)  =
               72,000,000 Btu (75,960 MJ). This is a significant amount of heat which can

               be used in process or space heating, or to drive an air-conditioning unit.
                  If the jacket-water flow rate is expressed in gallons per minute instead of
               pounds per hour (L/min instead of kg/h), the heat-recovery potential, H  =
                                                                                                        wg
               gpm(Δ t)(8.33) where 8.33 = lb/gal of water. With a water flow rate of 50
               gpm and the same temperature range as above, H  = 50(120)(8.33) = 49,980
                                                                           wg
               Btu/min (52,279 kJ/min).


               4. Find the amount of heat recoverable from the lube oil
               During I-C engine operation, lube-oil temperature can reach high levels—in
               the 300 to 400°F (149 to 201°C) range. And with oil having a typical specific

               heat of 0.5 Btu/lb°F (2.1 kJ/kg°C), the heat-recovery potential for the lube oil
               is H   =  w (Δt)(c ),  where  w   =  oil  flow  in  lb/h  (kg/h);Δt  =  temperature
                              o
                     wo
                                                     o
                                      o
               change of the oil during flow through the heat-recovery heat exchanger = oil
               inlet temperature − oil outlet temperature,°F or°C; c  = specific heat of oil =
                                                                               o
               0.5  Btu/lb°F  (kJ/kg°C).  With  an  oil  flow  of  2000  lb/h  (908  kg/h),  a
               temperature  change  of  140°F  (77.7°C),  H   =  2000(140)  (0.50)  =  140,000
                                                                    o
               Btu/h (41 kW). Thus, as mentioned earlier, the heat recoverable from the lube
               oil is usually the lowest of the three sources.

                  With the heat flow rates computed here, an I-C engine cogeneration facility
               can be easily justified, especially where frequent startups and shutdowns are
               anticipated.  Reciprocating  diesel  engines  are  preferred  over  gas  and  steam

               turbines  where  frequent  startups  and  shutdowns  are  required.  Just  the  fuel
               savings anticipated for recovery of heat in the exhaust gases of this engine
               could pay for it in a relatively short time.
   184   185   186   187   188   189   190   191   192   193   194