Multi-disc clutch, consisting of alternate steel and bronze plates, is used to transmit 7.5 kW power at 720 rpm. The inner and outer diameters of contacting surface are 100 mm and 150 mm respectively. The coefficient of friction between the contacting surface is 0.12 . If the intensity of pressure is limited to $0.3 \mathrm{N} / \mathrm{mm}^{2}$ ,

Determine:

The number of steel and bronze plates and

The axial force required to engage the clutch.

Given Data-

Multi disc clutch

Power $(P) =7.5 \mathrm{kW}$

Speed $(N) =720 \mathrm{rpm}$

Coefficient of friction $(\mu)=0.12$

Pressure $\mathrm{P}_{\max }=0.3 \mathrm{N} / \mathrm{mm}^{2}$

Outer Diameter $\left(D_{o}\right)=150=\left(r_{o}=75\right)$

Inner Diameter $\left(D_{i}\right)=100=\left(r_{i}=50\right)$

Solution-

1) Number of friction plates in clutch (i)

$i=m_{1}+m_{2}-1$

2) Power Transmitted (P)

$ P=\frac{2 \pi N T}{60000} $

$\therefore 7.5 \times 10^{3}=\frac{2 \pi \times 720 \times T}{60000}$ $\therefore T=99471.83 \mathrm{Nmm}=99472 \mathrm{Nmm}$

3) Axial Thrust

$W=2 \pi P_{\max } r_{i}\left(r_{o}-r_{i}\right)$

$W=2 \pi \times 0.3 \times 50(75-50)$

$\therefore W=2356.2 N$

4) Torque 'Transmitted

$T=\frac{1}{2} i \cdot \mu \cdot W \cdot\left(r_{o}+r_{i}\right)$

$99472=\frac{1}{2} \times i \times 0.12 \times 2356.2(75+50)$

$\therefore i=5.62 \cong 6$

Therefor Number of Friction Driving Steel Plates $\left(m_{1}\right) \ \& \ $ Bronze Plates $\left(m_{2}\right)$ are

$i=m_{1}+m_{2}-1$

$\therefore i=m_{1}+m_{2}-1$

$\therefore 6=m_{1}+m_{2}-1$

$\therefore m_{1}=4 \quad \& m_{2}=3$

5) Axial Force required to Engage the clutches.

$\therefore T=\frac{1}{2} i . \mu . W .\left(r_{o}+r_{i}\right)$

$\therefore 99472=\frac{1}{2} \times 6 \times 0.12 \times W(75+50)$

$\therefore W=2210.49 N$