Find the length of the belt required and angle of contact between the belt and each pulley.

What power can be transmitted by the belt when the larger pulley rotates at 200 rev/min, if the maximum permissible tension in the belt is 1 KN, and the coefficient of friction between the belt and pulley is 0.25?

Subject: Kinematics of Machinery

Topic: Belts, Chains and Brakes

Difficulty: High

Given :- $d_{1}$ = 450 mm= 0.45 m or $r_{1}$ =0.225 m; $d_{2}$ = 200 mm= 0.2 m; or $r_{2}$ =0.1 m; x =1.95 m, $N_{1}$ =200 rpm; $T_{1}$ = 1kN= 1000N ; $\mu$ =0.25

We know that speed of the belt,

$v=\frac{\pi d_{1}N_{1}}{60}=\frac{\pi\times 0.45\times200}{60}$ = 4.714 m/s.

Length of the belt

We know that length of the crossed belt,

$L=\pi(r_{1}+r_{2})+2x+\frac{(r_{1}+r_{2})^{2}}{x}$

=$\pi$(0.225+0.1)+2$\times$1.95+(0.225+0.1)2/1.95

= 4.975 m.

Angle of contact between the belt and each pulley

Let $\theta$= Angle of contact between the belt and each pulley.

We know that for a crossed belt drive,

$\sin\alpha=\frac{(r_{1}+r_{2})}{x}$ = (0.225+0.1)/1.95 = 0.1667 or $\alpha$ = 9.6$^{\circ}$

$\theta$ = 180$^{\circ}$ +2$\alpha$ = 180$^{\circ}$+2 × 9.6$^{\circ}$ = 199.2$^{\circ}$

=199.2 × $\pi$/180 = 3.477 rad.

Power Transmitted,

Let $T_{2}$ = Tension in the slack side of the belt,

We know that

$log\frac{ T_{1}}{ T_{2}} =\mu\theta$

= 0.25 × 3.477 =0.8692

$log\frac{ T_{1}}{ T_{2}}$ =0.8692/2.3 =0.378 or $\frac{ T_{1}}{ T_{2}}$ = 2.387 …(taking antilog of 0.378)

$T_{2}$ = $T_{1}$/2.387 = 1000/2.387 =419N

We know that power transmitted,

$P =( T_{1} – T_{2})ν $= (1000-419)4.714 =2740W = 2.74kw.