Engineering Board Exam Reviewer · Thermodynamics · hard
An engine with a brake thermal efficiency of 30% develops 100 kW. If the heating value of the fuel is 44,000 kJ/kg, what is the fuel consumption?
- 8.18 kg/h
- 27.27 kg/h
- 2.455 kg/h
- 7.58 kg/h
Answer
B. 27.27 kg/h
Solution
mf = W/( × HV) = 100/(0.30 × 44,000) = 0.007576 kg/s × 3600 = 27.27 kg/h.
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