Farm Power

An engine consumes fuel at rate $m_f = 0.002\text{ kg/s}$ with calorific value $CV = 42,000\text{ kJ/kg}$ (total heat input rate $84.0\text{ kW}$). If the engine develops brake power $P_b = 29.4\text{ kW}$, the brake thermal efficiency $\eta_{bth} = \frac{29.4}{84.0} \times 100$ in percent is: