Question

Calculate the power transmitted by belt passing over a pulley of 1m in diameter and running at 500 revolutions per minute; when belt tensions in the two sides of the belt are 50kg and 20kg, respectively.

• A. 17.44 hp
• B. 6.98 hp
• C. 10.47 hp
• D. 6.28 hp

Hint:

To perform this calculation quickly during exams:
\[ \text{Power (hp)} = \frac{(T_1 - T_2) \pi D N}{4500} \]
(This is derived from \(75 \times 60 = 4500\) in the denominator).
\[ \text{Power} = \frac{30 \times 3.1416 \times 1 \times 500}{4500} = \frac{47124}{4500} \approx 10.47\text{ hp} \]

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Power transmission via a flat belt drive is based on the difference in tension between the tight side and the slack side of the belt.
The effective pulling force is this tension difference, which, when multiplied by the linear velocity of the belt, determines the total power transmitted.
Key Formula or Approach:
1. Linear velocity of the belt (\(v\)):
\[ v = \frac{\pi D N}{60} \text{ m/s} \]
Where \(D\) is the pulley diameter (m) and \(N\) is the speed in rpm.
2. Work done per second (Power in kgf\(\cdot\)m/s):
\[ \text{Work done/sec} = (T_1 - T_2) \cdot v \]
Where \(T_1\) and \(T_2\) are the tight and slack side tensions in kgf.
3. Metric Horsepower (hp) conversion:
\[ 1\text{ hp} = 75 \text{ kgf}\cdot\text{m/s} \implies \text{Power (hp)} = \frac{(T_1 - T_2) \cdot v}{75} \]

Step 2: Detailed Explanation:
Let us identify the given values from the problem statement:
- Pulley diameter, \(D = 1\text{ m}\)
- Pulley speed, \(N = 500\text{ rpm}\)
- Tight side tension, \(T_1 = 50\text{ kg}\) (or kgf)
- Slack side tension, \(T_2 = 20\text{ kg}\) (or kgf)
Calculate the linear velocity \(v\):
\[ v = \frac{\pi \times 1 \times 500}{60} \approx \frac{3.1416 \times 500}{60} \approx 26.18\text{ m/s} \]
Calculate the effective pulling force (tension difference):
\[ T_1 - T_2 = 50 - 20 = 30\text{ kgf} \]
Calculate the work done per second:
\[ \text{Work done/sec} = 30\text{ kgf} \times 26.18\text{ m/s} = 785.4\text{ kgf}\cdot\text{m/s} \]
Convert this power into metric horsepower:
\[ \text{Power (hp)} = \frac{785.4}{75} \approx 10.47\text{ hp} \]
Therefore, the power transmitted by the belt is 10.47 hp.
This matches Option (C).

Step 3: Final Answer:
The correct option is (C).