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]}

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.

Answer 1

Correct Answer: 9545

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 \( 9545.0 \, \text{N} \).