Generator Fuel Consumption With Two Rates
A rented 25-gallon diesel generator powers a jobsite trailer and two 480-volt welders. The rental company's fuel log gives consumption at three points: with no load the unit burns 0.4 gallons per hour; at 8 kW it burns 2.2 gallons per hour; at 12 kW, its rated maximum, it burns 3.4 gallons per hour. Between these points consumption changes at a constant rate - the log is a straight line from 0 to 8 kW and a different straight line from 8 to 12 kW, with a bend at 8 kW. On a typical shift the trailer loads the generator to 6 kW for 4 hours and the welders load it to 10 kW for 3 hours, and the shift ends there. The tank is filled to 25 gallons at the start of the shift. The foreman wants to know whether the unit will run dry before the shift ends, and how much running time would remain at the end of the shift if the welders kept working at 10 kW.
1. Find the rate of change of fuel consumption on each segment of the log and state in plain terms what each number means.
Read the worked answer
Lower segment, 0 to 8 kW: (2.2 − 0.4) ÷ (8 − 0) = 1.8 ÷ 8 = 0.225 gal/hr per kW. Function: f(k) = 0.4 + 0.225k for 0 ≤ k ≤ 8.
Upper segment, 8 to 12 kW: (3.4 − 2.2) ÷ (12 − 8) = 1.2 ÷ 4 = 0.30 gal/hr per kW. Function: f(k) = 2.2 + 0.30(k − 8) for 8 < k ≤ 12.
The second slope is larger (0.30 vs 0.225), meaning each extra kilowatt above 8 kW costs about 33 percent more fuel per hour than a kilowatt below 8 kW. That bend at 8 kW is the practical ceiling for economical loading.
2. How much fuel does the shift actually burn, and how many hours of running time remain at 10 kW after the shift?
Read the worked answer
At 6 kW, use the lower segment: f(6) = 0.4 + 0.225(6) = 0.4 + 1.35 = 1.75 gal/hr. Over 4 hours: 4 × 1.75 = 7.0 gal.
At 10 kW, use the upper segment (10 is above the bend): f(10) = 2.2 + 0.30(10 − 8) = 2.2 + 0.60 = 2.8 gal/hr. Over 3 hours: 3 × 2.8 = 8.4 gal.
Shift total: 7.0 + 8.4 = 15.4 gal from a 25-gallon tank, leaving 25 − 15.4 = 9.6 gal. At 10 kW that is 9.6 ÷ 2.8 = 3.43 hours, about 3 hours 26 minutes. The unit does not run dry during the shift, but a full second shift at the same profile would not fit in one tank.
3. At what load does consumption reach 3.0 gal/hr, and why is it wrong to solve this with the lower-segment equation?
Read the worked answer
Solve on the upper segment, because 3.0 gal/hr lies between 2.2 (at 8 kW) and 3.4 (at 12 kW): 2.2 + 0.30(k − 8) = 3.0 → 0.30(k − 8) = 0.8 → k − 8 = 2.67 → k = 10.67 kW.
Check the answer against the domain: 10.67 is between 8 and 12, so it is valid.
Solving with the lower equation gives 0.4 + 0.225k = 3.0 → k = 11.56 kW, but 11.56 is outside that segment's 0 to 8 kW domain, so the result is meaningless - the generator does not follow that rate above 8 kW. Always test a solution against the interval it came from.