SSC JE Mechanical Engineering 22 Mar 2021 Shift 2 Papers

Question : 117 of 199

Marks: +1, -0

For a single-cylinder four-stroke oil engine, indicated power is 15 kW, the calorific value of fuel is 40000 kJ/kg and fuel consumed is 0.001 kg/s. What will be the indicated thermal efficiency?

40%
37.5%
62.5%
75%

Validate

Solution:

I.P. = 15 kW, C.V. = 40000 kJ/kg,

m_f˙ = 0.001

kg/s
Indicated thermal efficiency,

\η_{\text "ith "}={\text "I.P. "}/{{m}_{{f}} \×\text "C.V "}={15}/{0.001 \× 40000}

\×100

η_\ith = 37.5 %

Go to Question:

Prev Question

Next Question