A 160-HP Diesel engine (peak fuel consumption = 0.04-gal/HP-hr) hydraulic excavator operates on a cycle time of 20 seconds during a 50-min/hr. During the filling of the bucket cycle, the excavator's engine is at full power for 5-seconds. The remainder of the time, the engine operates at half-power. The fuel consumed per hour is most nearly:
Solution:
Step 1: Calculate the Time Factor (TF):
Time Factor = 50/60 x 100 = 83.3%
Step 2: Calculate the Engine Factor (EF):
Filling the bucket = (5 / 20) x 1 power = 0.25
Rest of Cycle = (15 / 20) x .50 power = 0.375
TOTAL 0.625
Operating Factor = Time Factor x Engine Factor = 0.625 x 0.833 = 0.520
Step 3: Calculate the Fuel Consumed
Fuel consumed/Hr = 0.52 x 160-HP x 0.04-gal/HP-hr = 3.33-gal/hr