Diesel Generator Fuel Rate and Cooling Load at Part Load
A 1.5 MW standby diesel generator is tested at site. At full load its fuel-to-electrical efficiency is 40%; at 75% load the efficiency falls to 38%. The fuel has a lower heating value of 42.5 MJ/kg and a density of 0.85 kg/L. The fuel tank is sized for 8 hours at full load, and the owner wants to know whether the same tank can carry the plant through a 10-hour outage at 75% load. The radiator cools the jacket water, which removes 25% of the fuel energy at full load, with a design water temperature rise of 8 K. Use a specific heat of 4.18 kJ/kg-K for water.
1. Compute the fuel consumption in L/h at full load.
Read the worked answer
Fuel power input = 1.5 MW / 0.40 = 3.75 MW = 3750 kJ/s.
Energy per hour = 3750 x 3600 = 13,500,000 kJ/h.
Mass rate = 13,500,000 / 42,500 = 317.6 kg/h.
Volume rate = 317.6 / 0.85 = 373.7 L/h (specific consumption 0.249 L/kWh).
2. Compute the fuel consumption at 75% load and decide whether the 8-hour full-load tank can supply a 10-hour outage.
Read the worked answer
Electrical output = 0.75 x 1500 = 1125 kW.
Fuel input = 1125 / 0.38 = 2960.5 kW = 10,657,900 kJ/h.
Mass rate = 10,657,900 / 42,500 = 250.8 kg/h.
Volume rate = 250.8 / 0.85 = 295.1 L/h (0.262 L/kWh).
Tank capacity = 373.7 x 8 = 2990 L.
Fuel needed for 10 h at 75% = 295.1 x 10 = 2951 L.
Since 2951 L < 2990 L, the tank can just cover the 10-hour outage at 75% load, with only about 39 L (1.3%) margin. The judgment is that the margin is too thin for a real outage; a refueling plan or a small tank upgrade is prudent.
3. Compute the jacket-water cooling flow required at full load.
Read the worked answer
Fuel power input at full load = 3750 kW.
Jacket heat = 0.25 x 3750 = 937.5 kW.
Q = mcpdT, so m = 937.5/(4.18 x 8) = 28.0 kg/s.
For water, 28.0 kg/s is about 28.0 L/s, or about 101 m3/h. The radiator and pump must be sized for this flow at the design 8 K rise; reducing flow would raise the water temperature and risk derating the engine.