SSC JE Mechanical Engineering 27 Sep 2019 Shift 2 Paper

Question : 170 of 200

Marks: +1, -0

A boiler produces 1000 kg/hr steam with rise in enthalpy of steam of 2500 kJ/kg. The calorific value of coal used is 30000 kJ/kg and its consumption is 100 kg/hr. What would be the boiler efficiency?

18.33%
73.33%
8.33%
83.33%

Validate

Solution:

Given:
m = 100 kg/hr, CV = 30000 kJ/kg, Q = 1000 kg/hr, Δh = 2500 kJ/kg

\η ={{Q} \×({h}_{{g}}-{h}_{{f}})}/{{m} \×{GCV}}

\η ={1000 \×2500}/{100 \×30000}

η = 0.8333 \or 83.33%

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