Page 229 - Handbook of Energy Engineering Calculations

3
3
62.4 lb/ft (999.0 kg/m ); 3855 for saltwater having a density of 64 lb/ft 3
3
(1024.6 kg/m ).
For this pump handling freshwater, hp = (200)(75)/[(3960)(0.70)] = 5.42

hp (4.0 kW). A 7.5-hp (5.6-kW) motor would probably be selected to handle
the rated capacity plus any overloads.
For this pump handling saltwater, hp = (200)(75)/[(3855)(0.70)] = 5.56 hp
(4.1 kW). A 7.5-hp (5.6-kW) motor would probably be selected to handle the

rated capacity plus any overloads. Thus, the same motor could drive this
pump whether it handles freshwater or saltwater.

2. Compute the lube-oil pump capacity
The lube-oil pump capacity required for a diesel engine is found from G =
H/200Δ t, where G = pump capacity, gal/min; H = heat rejected to the lube

oil, Btu/(bhp · h);Δt = lube-oil temperature rise during passage through the
engine,°F. Usual practice is to limit the temperature rise of the oil to a range
of 20 to 25°F (11.1 to 13.9°C), with a maximum operating temperature of

160°F (71.1°C). The heat rejection to the lube oil can be obtained from the
engine heat balance, the engine manufacturer, or Standard Practices for
Stationary Diesel Engines, published by the Diesel Engine Manufacturers
Association. With a maximum heat rejection rate of 500 Btu/(bhp · h) (196.4
W/kWh) from Standard Practices and an oil-temperature rise of 20°F

(11.1°C), G = [500 Btu/(bhp · h)](1000 hp)/[(200)(20)] = 125 gal/min (7.9
L/s).
By using the lowest temperature rise and the highest heat rejection rate, a

safe pump capacity is obtained. Where the pump cost is a critical factor, use a
higher temperature rise and a lower heat rejection rate. Thus, with a heat
rejection, the above pump would have a capacity of G = (300)(1000)/[(200)
(25)] = 60 gal/min (3.8 L/s).

3. Compute the lube-oil pump power input

The power input to a separate oil pump serving a diesel engine is given by hp
= Gp/1720e, where G = pump discharge rate, gal/min; p = pump discharge
2
pressure, lb/in ; e = pump efficiency. For this pump, hp = (125)(80)/[(1720)
(0.68)] = 8.56 hp (6.4 kW). A 10-hp (7.5-kW) motor would be chosen to
drive this pump.
With a capacity of 60 gal/min (3.8 L/s), the input is hp = (60)(80)/[(1720)