Question:
A 150 mm wide polyamide flat belt is transmitting 15 kW power through a belt-pulley system. The driving pulley of 150 mm pitch diameter is rotating at 200 RPM. If \( F_1 \) is the belt tension on high tension side, and \( F_2 \) is the belt tension on low tension side, then the difference in belt tensions \( \Delta F = F_1 - F_2 \) in N is _________.
\text{[round off to one decimal place]}
A 150 mm wide polyamide flat belt is transmitting 15 kW power through a belt-pulley system. The driving pulley of 150 mm pitch diameter is rotating at 200 RPM. If \( F_1 \) is the belt tension on high tension side, and \( F_2 \) is the belt tension on low tension side, then the difference in belt tensions \( \Delta F = F_1 - F_2 \) in N is _________.
\text{[round off to one decimal place]}
\text{[round off to one decimal place]}
Show Hint
To calculate the difference in belt tensions, use the power transmission equation \( P = (F_1 - F_2) v \) and the velocity formula for the belt.
Updated On: Dec 26, 2025
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Verified By Collegedunia
Correct Answer: 9545
Solution and Explanation
The power transmitted by the belt is given by:
\[
P = (F_1 - F_2) v,
\]
where \( P = 15 \, \text{kW} = 15000 \, \text{W} \) and \( v \) is the velocity of the belt.
The velocity \( v \) is related to the angular velocity of the driving pulley:
\[
v = \frac{\pi d n}{60},
\]
where \( d = 0.15 \, \text{m} \) is the pitch diameter and \( n = 200 \, \text{RPM} \).
Thus,
\[
v = \frac{\pi \times 0.15 \times 200}{60} = 1.57 \, \text{m/s}.
\]
Now, calculate the difference in tensions:
\[
\Delta F = \frac{P}{v} = \frac{15000}{1.57} \approx 9555.0 \, \text{N}.
\]
Thus, the difference in belt tensions is approximately \( 9545.0 \, \text{N} \).
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