The mass of the belt is 0.9 kg per metre length and the maximum tension is not to exceed 2000 N. The coefficient of friction is 0.3. The 1.2 m pulley, which is the driver, runs at 200 rpm. Due to belt slip on one of the pulleys, the velocity of the driven shaft is only 450 rpm. Calculate the torque on each of the two shafts, the power transmitted, and power lost in friction. What is the efficiency of the drive?

Subject: Kinematics of Machinery

Topic: Belts, Chains and Brakes

Difficulty: High

Given : $d_{1}$ = 1.2 m or $r_{1}$ =0.6 m ; $d_{2}$= 0.5 m or $r_{2}$ =0.25 m; x =4 m; m =0.9 kg/m;

T =2000N; $\mu$ =0.3; $N_{1}$ =200 rpm. ; $N_{2}$ =450 rpm.

We know that velocity of the belt,

$v=\frac{\pi d_{1}N_{1}}{60}=\frac{\pi\times1.2\times 200}{60}$= 12.57 m/s.

and centrifugal tension, $T_{C}$ = m$v^{2}$ =0.9(12.57)$^{2}$ =142N.

Tension in the tight side of the belt,

$T_{1}$ =T- $T_{C}$ =2000-142=1858N.

We know that for an open belt drive,

$\sin\alpha=\frac{(r_{1}-r_{2})}{x}$ =(0.6-0.25)/4 =0.0875 or α = 5.020

Angle of lap on the smaller pulley,

$\theta$ = 1800-2$\alpha$ =1800-(2×5.020) =169.960

=169.96 ×($\pi$/180) =2.967 rad.

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

Let

We know that

$2.3log\frac{(T_{1}}{T_{2})}=\mu\theta$ =0.3×2.967 =0.8901

$log\frac{(T_{1}}{T_{2})}$ =(0.8901/2.3) =0.387 or $\frac{(T_{1}}{T_{2})}$ =2.438 …(Taking antilog of 0.387)

$T_{2}$ = $T_{1}$/2.438 =(1858/2.438) =762N.

Torque on the shaft of larger pulley

We know that torque on the shaft of larger pulley,

$T_{L}$ = ($T_{1}$ –$T_{2}$).$r_{1}$ = (1858-762)0.6 =657.6N.

Torque on the shaft of smaller pulley

We know that torque on the shaft of smaller pulley,

$T_{S}$= ($T_{1}$ –$T_{2}$).$r_{2}$ = (1858-762)0.25 =274N.

Power transmitted:

We know that the power transmitted,

P = ($T_{1}$ + $T_{1}$)ν = (1858-762)12.57 =13780 W

Power lost in friction:

We know that input power,

$P_{1}$ = ($T_{L}$ ×2$\pi N_{1}$) /60 = (657.6×2$\pi$×200)/60 =13780 W =13.78 kW

And output power

$P_{2}$ = ($T_{S}$ ×2$\pi n_{2}$) /60 = (274×$2\pi$×450)/60 =12910 W =12.91 kW

Power lost in friction = $P_{1}$ - $P_{2}$ =13.78-13.91 =0.87 kW

Efficiency of the drive:

We know that efficiency of the drive,

$\eta$ = (output power)/(input power) = 12.91/13.78 =0.937 or 93.7%