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 %