Question Bank • QB_FP_ADV_057 NAT • 2 Marks

Section 3: Farm Power

Tractors and Power Tillers

Subtopic: Traction theory

Question Statement:

A tractor drive wheel operates at a travel reduction (slip) of $S = 12\%$. The input torque applied to the axle is $T = 3200\text{ N}\cdot\text{m}$ at an axle rotational speed of $N = 30\text{ rpm}$. The rolling radius of the wheel is $r = 0.75\text{ m}$. The net drawbar pull developed by the wheel is $P = 3600\text{ N}$. The tractive efficiency of the drive wheel expressed as a percentage is ________ $\%$ (round off to 2 decimal places).

Answer Key:

74.25

Step-by-Step Solution & Explanation

``` Tractive Efficiency Formula: eta_t = (P * v_a) / (T * omega) ``` **Method 1: Power Ratio Calculation** 1. Input axle power ($P_{\text{axle}}$): $P_{\text{axle}} = T \times \omega = 3200\text{ N}\cdot\text{m} \times \left(\frac{2 \pi \times 30}{60}\right) = 3200 \times \pi = 10053.1\text{ W}$ 2. Theoretical linear speed ($v_t$): $v_t = r \omega = 0.75\text{ m} \times \pi\text{ rad/s} = 2.35619\text{ m/s}$ 3. Actual travel speed ($v_a$) with $12\%$ slip ($S = 0.12$): $v_a = v_t (1 - S) = 2.35619 \times (1 - 0.12) = 2.35619 \times 0.88 = 2.07345\text{ m/s}$ 4. Output drawbar power ($P_{\text{drawbar}}$): $P_{\text{drawbar}} = P \times v_a = 3600\text{ N} \times 2.07345\text{ m/s} = 7464.42\text{ W}$ 5. Tractive efficiency ($\eta_t$): $\eta_t = \frac{P_{\text{drawbar}}}{P_{\text{axle}}} \times 100 = \frac{7464.42}{10053.1} \times 100 = 74.25\%$ **Method 2: Direct Traction Ratio Shortcut** $\eta_t = \frac{P \cdot r}{T} (1 - S) \times 100 = \frac{3600 \times 0.75}{3200} \times 0.88 \times 100 = \frac{2700}{3200} \times 0.88 \times 100 = 0.84375 \times 88 = 74.25\%$
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