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