Subtopic: Power efficiencies and measurement, engine performance curves
Question Statement:
A tractor engine delivers a rated torque of $T_{\text{rated}} = 180\text{ N}\cdot\text{m}$ at rated engine speed of $N_{\text{rated}} = 2200\text{ rpm}$. As the tractor encounters a heavy patch of soil, engine speed lugs down to the peak torque speed of $1400\text{ rpm}$, where the engine produces its maximum torque of $T_{\text{max}} = 225\text{ N}\cdot\text{m}$. The percentage torque backup (torque rise) of the tractor engine is ________ $\%$ (answer in integer).
Answer Key:
25
Step-by-Step Solution & Explanation
```
Tractor Engine Torque Backup (Torque Rise):
T_rated = 180 N*m at 2200 rpm
T_max = 225 N*m at 1400 rpm
```
**Step-by-step Calculation:**
1. Definition of torque backup (torque rise percentage):
$\text{Torque Rise } (\%) = \frac{T_{\text{max}} - T_{\text{rated}}}{T_{\text{rated}}} \times 100$
2. Substituting values:
$\text{Torque Rise } (\%) = \frac{225 - 180}{180} \times 100 = \frac{45}{180} \times 100 = 25\%$
High torque backup ($20\text{--}30\%$) allows farm tractors to pull through tough soil spots without stalling or downshifting.