Question

During testing of a thresher, 100 N.m is recorded at 600 rpm at the main shaft of threshing cylinder. The diameter of threshing cylinder is 140 mm. The cylinder is operated by a V-belt and unit mass of v-belt is 1.0 \(\text{kg}\cdot\text{m}^{-1}\). What will be the maximum tension in the v-belt at the condition of maximum power transmission?

• A. 1429 N
• B. 714 N
• C. 58 N
• D. 19 N

Hint:

For maximum power transmission, always remember the ratio \(T_{\max} = 3 T_c\).
With \(v \approx 4.4\text{ m/s}\), \(T_c \approx 19.4\text{ N}\).
Multiplying by 3 yields \(\approx 58\text{ N}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For flexible belt drives, maximum power transmission occurs when the centrifugal tension is exactly one-third of the maximum allowable tension.

Step 2: Key Formula or Approach:
1. The condition for maximum power transmission is:
\[ T_c = \frac{T_{\max}}{3} \implies T_{\max} = 3 T_c \] 2. The centrifugal tension \(T_c\) is:
\[ T_c = m v^2 \] where \(m\) is the belt mass per unit length, and \(v\) is the belt speed.
3. The belt linear velocity is:
\[ v = \omega r = \frac{2\pi N}{60} \times \frac{d}{2} \]

Step 3: Detailed Explanation:
Given values:
- Rotational speed (\(N\)) = \(600\text{ rpm}\)
- Cylinder/pulley diameter (\(d\)) = \(140\text{ mm} = 0.14\text{ m}\)
- Pulley radius (\(r\)) = \(0.07\text{ m}\)
- Mass per unit length (\(m\)) = \(1.0\text{ kg/m}\)
First, calculate the belt linear velocity \(v\):
\[ v = \frac{2\pi \times 600}{60} \times 0.07 \] \[ v = 20\pi \times 0.07 = 1.4\pi \approx 4.398\text{ m/s} \] Next, calculate the centrifugal tension \(T_c\):
\[ T_c = m v^2 = 1.0 \times (4.398)^2 \approx 19.34\text{ N} \] Now, calculate the maximum tension \(T_{\max}\):
\[ T_{\max} = 3 T_c = 3 \times 19.34 \approx 58\text{ N} \]

Step 4: Final Answer:
The correct option is 3, which corresponds to 58 N.