Boiler efficiency: correct formula using steam and fuel data Given m = total mass of water evaporated into steam (kg/h), mf = mass of fuel fired (kg/h), C = calorific value of fuel (kJ/kg), h = total heat of the produced steam (kJ/kg), and hf1 = sensible heat of feedwater (kJ/kg), the boiler efficiency η_boiler is:

Introduction / Context:
Boiler efficiency compares useful energy transferred to water/steam with the chemical energy supplied by the fuel. A precise formula is needed in boiler trials and energy audits to quantify performance and benchmark different units or fuels under comparable conditions.

Given Data / Assumptions:

• m = mass of water evaporated into steam (kg/h).
• mf = mass of fuel consumed (kg/h).
• C = lower calorific value of the fuel (kJ/kg), unless otherwise noted.
• h = total specific enthalpy of steam at delivery state (kJ/kg).
• hf1 = sensible heat of feedwater at inlet conditions (kJ/kg).

Concept / Approach:
Useful heat added to water is (h − hf1) per kg of steam. Total useful heat per hour is m * (h − hf1). Fuel energy input per hour is mf * C. Efficiency is useful output divided by fuel input. This aligns with standard test codes and allows corrections for feedwater temperature and steam condition (saturated or superheated).

Step-by-Step Solution:
Useful heat rate Q_out = m * (h - hf1).Fuel heat rate Q_in = mf * C.η_boiler = Q_out / Q_in.Therefore η_boiler = (m * (h - hf1)) / (mf * C).

Verification / Alternative check:
Dimensional check: numerator and denominator are in kJ/h, yielding a dimensionless ratio between 0 and 1 (or 0–100%). This confirms plausibility.

Why Other Options Are Wrong:
Option (b) is the reciprocal; (c) ignores feedwater preheat; (d) adds hf1 instead of subtracting; (e) is dimensionally inconsistent for general cases.

Common Pitfalls:
Using gross calorific value without accounting for latent heat of condensation in flue gases; forgetting to correct h and hf1 to the same reference (kJ/kg).

Final Answer:
η_boiler = (m * (h - hf1)) / (mf * C)