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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?
1. 40%
2. 37.5%
3. 62.5%
4. 75%

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Correct Answer - Option 2 : 37.5%

Concept:

Indicated thermal efficiency:

It is the ratio of indicated power (I.P.) and energy in fuel per second.

ηith =  \(\mathbf{I.P \over {\dot{m_f}\;\times \;C.V}}\)

where \(\dot{m_f}\) = mass flow rate of fuel (kg/s) and C.V. = calorific value of fuel (kJ/kg)

Calculation:

Given: 

I.P. = 15 kW, C.V. = 40000 kJ/kg, \(\dot{m_f}\) = 0.001 kg/s

Indicated thermal efficiency, 

ηith = \({I.P. \over {\dot{m_f}\times \;C.V.}}={15 \over {0.001\;\times\;40000}}\;\times100\)

ηith = 37.5 %

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