Answer to Question #313144 in Mechanical Engineering for Jhay

Question

4. A pulley 600 mm. in diameter transmits 40 kW at 500 rpm. The arc of contact between the belt and pulley is 144° having a center distance of 1100 mm. Determine: (a) diameter of the driven pulley for open belt drive, (b) speed of the driven pulley, and (c) belt length for open belt connection.

Answer 1

$P =\frac{2\pi NT}{60}$

$T =\frac{40000\times 60}{2\pi\times500 } = 763.94$ Nm

(a) Diameter of the driven pulley for open belt drive

$sin \beta = \frac{D-d}{2C}$

$D = Sin \beta \times 2C +d =sin 144 \times 2200 + 600 = 1893.13$ mm

(b) Speed of the driven pulley

$\frac{n}{N}=\frac{D}{d}$

$n = N\times \frac{D}{d}= 500\times \frac{1893.13}{600}= 1577.61 \approx 1578$ rpm

(c) belt length for open belt connection

$L = 2C+\frac{\pi(D+d)}{2}+\frac{(D-d)^2}{4C} =2\times 1100+\frac{\pi (1893.13+600)}{2}+\frac{(1893.13-600)^2}{4\times1100}=6496.24$ mm