Subtopic: Calculation of power, torque, fuel consumption, heat load and power losses
Question Statement:
During a heat balance test on a tractor diesel engine, the fuel consumption rate was $9.0\text{ kg/h}$ with a lower heating value of $42,000\text{ kJ/kg}$. The brake power developed was $35.0\text{ kW}$. The heat carried away by radiator cooling water was $36.0\text{ kW}$, and the heat carried away by the exhaust gas was $24.0\text{ kW}$. The remaining heat energy lost via radiation, convection, and unaccounted losses in $\text{kW}$ is ________ (answer in integer).
Answer Key:
10
Step-by-Step Solution & Explanation
1. Total heat energy supplied by fuel per second ($Q_{\text{in}}$):
$Q_{\text{in}} = \frac{\dot{m}_f \times CV}{3600} = \frac{9.0\text{ kg/h} \times 42,000\text{ kJ/kg}}{3600\text{ s}} = 105.0\text{ kW}$
2. Sum of accounted energy outputs:
$\Sigma Q_{\text{out}} = BP + Q_{\text{cool}} + Q_{\text{exh}} = 35.0 + 36.0 + 24.0 = 95.0\text{ kW}$
3. Unaccounted and radiation losses:
$Q_{\text{rad}} = Q_{\text{in}} - \Sigma Q_{\text{out}} = 105.0 - 95.0 = 10\text{ kW}$