Page 206 - Handbook of Energy Engineering Calculations

Determine the amount of heat used to generate 1 bhp · h (0.75 kWh) from:
heat rate, Btu/bhp · h) = (sfc)(HHV), where sfc = specific fuel consumption,
lb/(bhp · h); HHV = higher heating value of fuel, Btu/lb. Or, heat rate =
(0.36)(19.350) = 6967 Btu/(bhp · h) (2737.3 W/kWh).
Compute the heat balance of the engine by taking the product of the

respective heat rejection percentages and the heat rate as follows:

Then the power output = 6967 − 4422 = 2545 Btu/(bhp · h) (999.9

W/kWh), or 2545/6967 = 0.365, or 36.5 percent. Note that the sum of the
heat losses and power generated, expressed in percent, is 100.0.

2. Compute the jacket cooling-water flow rate
The jacket water cools the jackets and the turbocharger. Hence, the heat that

must be absorbed by the jacket water is 800 + 139 = 939 Btu/(bhp · h) (369
W/kWh), using the heat rejection quantities computed in step 1. When the
engine is developing its full rated output of 1000 bhp (746 kW), the jacket
water must absorb [939 Btu/(bhp · h)(1000 bhp) = 939.000 Btu/h (275,221
W).

Apply a safety factor to allow for scaling of the heat-transfer surfaces and
other unforeseen difficulties. Most designers use a 10 percent safety factor.
Applying this value of the safety factor for this engine, we see the total

jacket-water heat load = 939.000 + (0.10)(939.000) = 1.032,900 Btu/h (302.5
kW).
Find the required jacket-water flow from G = H/500Δt, where G = jacket-
water flow, gal/min; H = heat absorbed by jacket water, Btu/h; Δt =
temperature rise of the water during passage through the jackets,°F. The usual

temperature rise of the jacket water during passage through a diesel engine is
10 to 20°F (5.6 to 11.1°C). Using 10°F for this engine, we find G =
1,032,900/[(500)(10)] = 206.58 gal/min (13.03 L/s), say 207 gal/min (13.06