Problem
A boiler generates 10,000 kg/h steam. Steam enthalpy is 2,780 kJ/kg, feedwater enthalpy 420 kJ/kg, fuel use 750 kg/h and GCV 42,000 kJ/kg.
Formula
η = ms(hs−hfw)/(mf×GCV)×100
Solution
Useful heat = 10,000×(2,780−420)=23,600,000 kJ/h. Fuel heat = 750×42,000=31,500,000 kJ/h.
Answer
74.92%
Engineering check
Verify units, magnitude, assumptions and physical meaning before accepting the result.