The belt is 5 mm thick and 80 mm wide. The coefficient of friction between the belt and pulley is 0.3 and the maximum permissible tension in the belting is 1.4MN/m2.

Determine:

1.diameter of the driving pulley,

2.maximum power that may be transmitted by the belting, and

3.required initial belt tension.

Given:- x= 1.2m ; $N_{2}$ = 350 rpm, $N_{2}$ = 140 rpm.; $d_{2}$ = 400 mm =0.4m;

t =5 mm; 0.005m; b = 80 mm= 0.08m; $\mu$ = 0.3, $\sigma$ = 1.4 MN/m$^{2}$ = 1.4 × 106 N/m$^{2}$

1.Diameter of the driving pulley

Let $d_{1}$ = Diameter of the driving pulley,

We know that $\frac{N_{2}}{N_{1}}$ =$\frac{d_{1}}{d_{2}}$or $d_{1}$ = $\frac{N_{2}d_{2}}{N_{1}}$ =(140 ×0.4)/350 =0.16m.

2.Maximum power transmitted by the belting

First of all, let us find the angle of contact of the belt on the smaller pulley (or driving pulley).

Let $\theta$ = Angle of contact of the belt on the driving pulley.

From fig. we find that

$\sin\alpha=\frac{O_{2}M}{O_{1}O_{2}}=\frac{(r_{2}-r_{1})}{x}=\frac{(d_{2}-d_{1})}{2x}=\frac{0.4 - 0.16}{2\times 1.2}=0.1$

$\alpha$ = 5.740

$\theta$ = 1800 -2 $\alpha$ = 1800 -2$\times$5.740 = 168.520

$T_{1}$ = Tension in the tight side of the belt, and

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

Let

We know that

$2.3log\frac{(T_{1}}{T_{2})}=\mu\theta$

$log\frac{(T_{1}}{T_{2})}$ =0.882/2.3 =0.3835 or $\frac{(T_{1}}{T_{2})}$ = 2.42 ….(Taking antilog of 0.38)

We know that maximum tension to which the belt can be subjected,

$T_{1}$ = $\sigma$ × b × t = 1.4×106×0.08×0.005 = 560N

$T_{2}$ =$T_{1}$/2.42 =560/2.42 =231.4N.

Velocity of the belt,

$v=\frac{\pi d_{1}N_{1}}{60}=\frac{\pi\times0.16\times 350}{60}$ = 2.93m/s

Power transmitted, P = $(T_{1} - T_{2})$ν =(560-231.4)2.93 = 963W = 0.963 kW

3. Required initial belt tension

We know that the initial belt tension,

$T_{0}=\frac{(T_{1}+T_{2})}{2}$ = (560+231.4)/2 =395.7N.

a Flat belt type is use it For a belt drive two pulleys 1.2m apart e driver pulley with a diameter 40 cm is routing with speed 350 rpm while diameter of driven pulley is 100 cm coefficient of friction of the contact surface between belt and pulley is 0.3 maximum allowable tension is 600 N ( neglecting centrifugal tension)