Question Bank • QB_FP_052 NAT • 2 Marks

Section 3: Farm Power

Farm Power

Subtopic: Power efficiencies and measurement, engine performance curves

Question Statement:

A tractor diesel engine develops a rated torque of $T_{\text{rated}} = 195.3\text{ N}\cdot\text{m}$ at rated engine speed of $2200\text{ rpm}$. When pulled down under heavy drawbar load, the engine develops its peak maximum torque of $T_{\max} = 238.5\text{ N}\cdot\text{m}$ at $1400\text{ rpm}$. The torque reserve percentage ($TR = \frac{T_{\max} - T_{\text{rated}}}{T_{\text{rated}}} \times 100$) of the engine is ________ (round off to 1 decimal place).

Answer Key:

22.1

Step-by-Step Solution & Explanation

Torque reserve ($TR$) represents the engine lugging ability: $TR = \frac{T_{\max} - T_{\text{rated}}}{T_{\text{rated}}} \times 100$ Given $T_{\max} = 238.5\text{ N}\cdot\text{m}$ and $T_{\text{rated}} = 195.3\text{ N}\cdot\text{m}$: $TR = \frac{238.5 - 195.3}{195.3} \times 100 = \frac{43.2}{195.3} \times 100 \approx 22.1198\% \approx 22.1\%$
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