A belt drives a pulley of 200 mm diameter such that the ratio of tensions in the tight side and the slack side is 1.2, the maximum tension in the belt is not to exceed 240 kN. The speed of pulley is 60 rpm. Find the safe power transmitted by the pulley.
25 kW
This problem involves calculating the safe power transmitted by a belt drive system. We are given the pulley diameter, the ratio of tensions in the belt's tight and slack sides, the maximum allowable tension, and the pulley's rotational speed.
To find the power transmitted, we first need to determine the difference in tensions between the tight and slack sides of the belt, and the linear velocity of the belt.
We are given the maximum tension T1 and the ratio T1/T2. We can find the slack side tension T2 using the given ratio.
$\frac{T_1}{T_2} = 1.2$
$T_2 = \frac{T_1}{1.2}$
Substituting the value of T1:
$T_2 = \frac{240,000 \text{ N}}{1.2} = 200,000 \text{ N}$
The difference in tension (T1 - T2) is the effective driving force:
Driving Force = $T_1 - T_2 = 240,000 \text{ N} - 200,000 \text{ N} = 40,000 \text{ N}$
The linear velocity of the belt (v) can be calculated from the pulley's diameter (d) and rotational speed (N).
$v = \frac{\pi d N}{60}$
Substituting the given values:
$v = \frac{\pi \times 0.2 \text{ m} \times 60 \text{ rpm}}{60} = 0.2\pi \text{ m/s}$
Approximately, $v \approx 0.2 \times 3.14159 \approx 0.6283 \text{ m/s}$.
The power transmitted (P) by the belt drive is the product of the effective driving force and the belt velocity.
$P = (T_1 - T_2) \times v$
Substituting the calculated values:
$P = 40,000 \text{ N} \times 0.2\pi \text{ m/s} = 8000\pi \text{ W}$
To convert Watts (W) to kilowatts (kW), we divide by 1000:
$P = \frac{8000\pi}{1000} \text{ kW} = 8\pi \text{ kW}$
Using the approximate value of $\pi$:
$P \approx 8 \times 3.14159 \text{ kW} \approx 25.1327 \text{ kW}$
The calculated safe power transmitted is approximately 25.13 kW. Looking at the options provided, 25 kW is the closest value.
The value of initial tension in belts is equal to
The tight and slack sides of a belt connecting two pulleys are having tensions of 25 N and 15 N respectively, while the belt is running at 10 m/s. The power transmitted as
A flat belt drive with pulley of $r = 20$ cm radius is designed to transmit 6.283 kW power at 600 RPM. In the figure, $\tau$ is the corresponding torque. If the coefficient of static friction between the belt and the pulley is 0.3, then the minimum value of the tightening force $F$ (in kN) required to prevent the belt slip is ________.(Rounded off to 2 decimal places)
