Page 225 - Handbook of Energy Engineering Calculations

Related Calculations. Use this method for any type of engine—diesel,
gasoline, or gas—burning any type of fuel. Where desired, a simple heat
balance can be set up between the heat-releasing and heat-absorbing sides of
the system instead of using the equations given here. However, the equations
are faster and more direct.

FUEL STORAGE CAPACITY AND COST FOR I-C ENGINES

A diesel power plant will have six 1000-hp (746-kW) engines and three 600-

hp (448-kW) engines. The annual load factor is 85 percent and is nearly
uniform throughout the year. What capacity day tanks should be used for
these engines? If fuel is delivered every 7 days, what storage capacity is
required? Two fuel supplies are available: a 24° API fuel at $0.0825 per

gallon ($0.022 per liter) and a 28° API fuel at $0.0910 per gallon ($0.024 per
liter). Which is the better buy?

Calculation Procedure:

1. Compute the engine fuel consumption
Assume, or obtain from the engine manufacturer, the specific fuel

consumption of the engine. Typical modern diesel engines have a full-load
heat rate of 6900 to 7500 Btu/(bhp · h) (2711 to 3375 W/kWh), or about 0.35
lb/(bhp · h) of fuel (0.21 kg/kWh). Using this value of fuel consumption for
the nine engines in this plant, we see the hourly fuel consumption at 85

percent load factor will be (6 engines)(1000 hp)(0.35)(0.85) + (3 engines)
(600 hp)(0.35)(0.85) = 2320 lb/h (1044 kg/h).
Convert this consumption rate to gal/h by finding the specific gravity of
the diesel oil. The specific gravity s = 141.5/(131.5 + °API). For the 24° API

oil, s = 141.5/(131.5 + 24) = 0.910. Since water at 60°F (15.6°C) weighs 8.33
lb/gal (3.75 kg/L), the weight of this oil is (0.910)(8.33) = 7.578 lb/gal (3.41
kg/L). For the 28° API oil, s = 141.5/(131.5 + 28) = 0.887, and the weight of
this oil is (0.887)(8.33) = 7.387 lb/gal (3.32 kg/L). Using the lighter oil, since

this will give a larger gal/h consumption, we get the fuel rate = (2320
lb/h)/(7.387 lb/gal) = 315 gal/h (1192 L/h).
The daily fuel consumption is then (24 h/day)(315 gal/h) = 7550 gal/day