Answer to Question #191454 in Mechanical Engineering for Brian

Question

the tensions of a flat belt are;right side 800n and slack side 300n. the angular speed n and diameter d of the pulley are 400 rev/min and 30cm respectively. if the coefficient of friction between the belt and the pulley is 0.7 and the mass of the belt is 2.1kg/g,including centrifugal tension, determine (a) initial tension of the belt (b) angle of lap in degrees (c) power transmitted (d) maximum power)

Answer 1

Given:

$T_1=800N$

$μ=0.7$

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

$N = 400 rpm$

Diameter$\ (D) = 300 mm$

1) Initial tension of the belt $T_2$

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

${T_2} = \dfrac{T_1} {e ^{\mu \theta} } = \dfrac{800} {e ^{0.7\times2.6} } =129.62 N$

2) Power transmitted at this speed in kilowatts.

$V = \dfrac{\pi DN}{60} = \dfrac{\pi \times 300\times 400}{60} = 6283.19\ {m/sec}$

$P =(T_1-T_2) \times V = (800 - 129.62) \times 6283.19 =4212.13\ kW$