Page 209 - Handbook of Energy Engineering Calculations

2
the exhaust gas. Steam at 30 lb/in (abs) (206.8 kPa) has a temperature of
250.33°F (121.3°C). Thus, the exhaust-gas outlet temperature from the boiler
will be 250.33 + 75 = 325.33°F (162.9°C), say 325°F (162.8°C). Then H =
e
(12,500)(0.252) (750 − 325) = 1,375,000 Btu/h (403.0 kW).
2
At 30 lb/in (abs) (206.8 kPa), the enthalpy of vaporization of steam is
945.3 Btu/lb (2198.9 kJ/kg), found in the steam tables. Thus, the exhaust heat
can generate 1,375,000/945.3 = 1415 lb/h (636.8 kg/h) if the boiler is 100
percent efficient. With a boiler efficiency of 85 percent, the steam generated
= (1415 lb/h)(0.85) = 1220 lb/h (549.0 kg/h), or (1200 lb/h)/1000 bhp = 1.22
lb/(bhp · h) (0.74 kg/kWh).

Related Calculations. Use this procedure for any reciprocating internal-
combustion engine burning gasoline, kerosene, natural gas, liquefied-
petroleum gas, or similar fuel. Figure 1 shows typical arrangements for a
number of internal-combustion engine cooling systems.

When ethylene glycol or another antifreeze solution is used in the cooling
system, alter the denominator of the flow equation to reflect the change in
specific gravity and specific heat of the antifreeze solution, as compared with
water. Thus, with a mixture of 50 percent glycol and 50 percent water, the

flow equation in step 2 becomes G = H/(436Δt). With other solutions, the
numerical factor in the denominator will change. This factor = (weight of
liquid lb/gal)(60 min/h), and the factor converts a flow rate of lb/h to gal/min
when divided into the lb/h flow rate. Slant diagrams, Fig. 6, are often useful

for heat-exchanger analysis.