Question Bank • QB_FM_013 NAT • 2 Marks

Section 2: Farm Machinery

Machine Design

Subtopic: Design and selection of machine elements – gears, pulleys, chains and sprockets and belts

Question Statement:

A V-belt drive is used to transmit power from a tractor PTO shaft to a forage chopper. The groove angle of the pulley is $2\beta = 38^\circ$ (semi-groove angle $\beta = 19^\circ$). The coefficient of friction between the belt and pulley is $\mu = 0.30$, and the angle of wrap on the smaller pulley is $\theta = 160^\circ$ ($2.793\text{ rad}$). The belt has a linear mass density of $m = 0.20\text{ kg/m}$ and operates at a linear speed of $v = 15\text{ m/s}$. If the maximum permissible belt tension is $T_1 = 900\text{ N}$, the maximum power transmitting capacity of the belt in $\text{kW}$ is ________ (round off to 2 decimal places).

Answer Key:

11.85

Step-by-Step Solution & Explanation

1. Centrifugal tension ($T_c$): $T_c = m v^2 = 0.20 \times 15^2 = 45\text{ N}$ 2. Maximum effective tight-side tension: $T_1 - T_c = 900 - 45 = 855\text{ N}$ 3. Tension ratio for V-belt: $\frac{T_1 - T_c}{T_2 - T_c} = e^{\frac{\mu \theta}{\sin\beta}}$ Given $\beta = 19^\circ$, $\sin(19^\circ) = 0.3256$: $\frac{\mu \theta}{\sin\beta} = \frac{0.30 \times 2.793}{0.3256} = \frac{0.8379}{0.3256} \approx 2.5734$ $e^{2.5734} \approx 13.11$ $T_2 - T_c = \frac{855}{13.11} \approx 65.22\text{ N}$ $T_2 = 65.22 + 45 = 110.22\text{ N}$ 4. Power transmitted ($P$): $P = (T_1 - T_2) v = (900 - 110.22) \times 15 = 789.78 \times 15 = 11,846.7\text{ W} \approx 11.85\text{ kW}$
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