Answer to Question #238588 in Mechanical Engineering for Maps

Question

A flat belt is required to transmit 22 kW from a 250 mm diameter pulley running at 1450 rpm to a 355 mm diameter pulley. The proposed belt is 3.5 mm thick. The other system specifications are as follows: Coefficient of friction 0.7, Density of the belt is 1100 kg/m3, Maximum permissible stress is 7 MPa. The distance between the shafts is 1.8 m.

Calculate the width required. 

Answer 1

solution:

$V=\frac{\pi dn}{60\times1000}$

$V=\frac{\pi\times250\times1450}{60\times1000}$

$=18.98m/s=19m/s$

V$\implies$ belt velocity $(m/s)$

$\alpha_s=180-2sin^{-1}(\frac{D-d}{2C})$

$\alpha_s=180-2sin^{-1}(\frac{0.355-0.25}{2\times1.8})$

$=176.66\degree$

in radians; $\alpha_s=(\frac{176.66}{180})\pi=3.08 radians$

according to the power transmitted by the belt

$P=\frac{(P_1-P_2)V}{1000}$

$22=\frac{(P_1-P_2)19}{1000}$

$P_1-P_2=1157.9(N)$

by solving the upper equation

$P_1=1388.2N , P_2=230.3N$

$b=56.66mm=57mm$