Question

The tension on the tight side of the flat belt is 800N and the coefficient of friction between

the belt and the pulley is 0.3. The angular speed N, angle of lap and the diameter D of the

pulley are 300rpm, 150o and 30cm respectively. Without the addition of centrifugal tension,

determine;

(a) Initial tension of the belt.

(b) Power transmitted at this speed in kilowatts.

Answer 1

Given :

$T_1=800N$

$\mu = 0.3$

$\theta = 150 \times \frac {\pi}{180}=2.6 rad$

$N = 300 rpm$

Diameter $D = 300 mm$

1) Initial tension of the belt $T_2$

$\frac{T_1}{T_2} = e ^{\mu \theta}$

${T_2} = \frac{T_1} {e ^{\mu \theta} } = \frac{800} {e ^{0.3\times2.6} } =366.72 N$

2) Power transmitted at this speed in kilowatts.

$V = \frac{\pi DN}{60} = \frac{\pi \times 300\times 300}{60} = 4712.38 \frac{m}{sec}$

$P =(T_1-T_2) \times V = (800 - 366.72) \times 4712.38 =2041.78KW$