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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).

Difficulty: Medium

Correct Answer: 63%

Explanation:

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


Approach
Boiler 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 Answer
63%

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