Boiler efficiency calculation from operating data: compute percentage efficiency. Data given: Feed water supplied = 205 kg/h Coal fired = 23 kg/h Net enthalpy rise per kg of water = 145 kJ/kg Calorific value of coal = 2050 kJ/kg Calculate the boiler efficiency (percentage).

Given

• Feedwater mass flow = 205 kg/h
• Coal firing rate = 23 kg/h
• Net enthalpy rise per kg water = 145 kJ/kg
• Coal calorific value (CV) = 2050 kJ/kg

ApproachBoiler efficiency = (Useful heat added to water per hour) / (Fuel energy input per hour).

Step-by-step Useful heat rate = (205 kg/h) × (145 kJ/kg) = 29,725 kJ/h Fuel heat input rate = (23 kg/h) × (2050 kJ/kg) = 47,150 kJ/h Efficiency = 29,725 / 47,150 = 0.6304 ≈ 63%

Common pitfalls

• Forgetting to multiply by the hourly mass flow before taking the ratio.
• Confusing equivalent evaporation with efficiency; efficiency always compares useful output energy to input fuel energy.

Final Answer63%