During an official tractor drawbar test on a level concrete track, the tractor develops a constant drawbar pull of $P = 18\text{ kN}$ while traveling at a true forward speed of $v = 6.0\text{ km/h}$. The fuel consumed during a 1-hour test run is measured to be $10.0\text{ kg}$. The lower heating value of the diesel fuel is $42.5\text{ MJ/kg}$. The overall thermal efficiency at the drawbar (drawbar thermal efficiency) expressed as a percentage is ________ (round off to 2 decimal places).
Answer Key:
25.41
Step-by-Step Solution & Explanation
```
Drawbar Thermal Efficiency:
Pull P = 18 kN
Speed v = 6.0 km/h = 1.667 m/s
Drawbar Power P_db = P * v
Fuel: 10.0 kg/h, CV = 42.5 MJ/kg
```
**Step-by-step Calculation:**
1. Forward speed in $\text{m/s}$:
$v = \frac{6.0 \times 1000}{3600} = 1.6667\text{ m/s}$
2. Useful drawbar power developed ($P_{\text{db}}$):
$P_{\text{db}} = P \times v = 18000\text{ N} \times 1.6667\text{ m/s} = 30000\text{ W} = 30.0\text{ kW}$
3. Total chemical heat energy supplied per hour ($Q_{\text{in}}$):
$Q_{\text{in}} = m_f \times CV = 10.0\text{ kg} \times 42.5\text{ MJ/kg} = 425.0\text{ MJ}$
$Q_{\text{rate}} = \frac{425.0 \times 10^6\text{ J}}{3600\text{ s}} = 118055.6\text{ W} = 118.056\text{ kW}$
4. Drawbar thermal efficiency ($\eta_{\text{db}}$):
$\eta_{\text{db}} = \frac{P_{\text{db}}}{Q_{\text{rate}}} \times 100 = \frac{30.0\text{ kW}}{118.056\text{ kW}} \times 100 = 25.4117\% \approx 25.41\%$