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.

**η**_{ith} = \(\mathbf{I.P \over {\dot{m_f}\;\times \;C.V}}\)

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,

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

ηith = **37.5 %**