A multi-disc clutch has three discs on the driving shaft and two on the driven shaft. The outside diameter of the contact surfaces is 240 mm and inside diameter 120 mm. Assuming uniform wear and coefficient of friction as 0.3, find the maximum axial intensity of pressure between the discs for transmitting 25 kW at 1575 r.p.m.
Question : A multi-disc clutch has three discs on the driving shaft and...
Given : n_{1}=3 ; n_{2}=2 ; d_{1}=240 mm \text { or } r_{1}=120 mm ; d_{2}=120 mm \text { or } r_{2}=60 mm ; \mu=0.3 ; P=25 kW =25 \times 10^{3} W ; N=1575 r.p.m . \text { or } \omega=2 \pi \times 1575 / 60=165 rad / s
Let T = Torque transmitted in N-m, and
W = Axial force on each friction surface.
We know that the power transmitted (P),
25 \times 10^{3}=T \cdot \omega=T \times 165 \quad \text { or } \quad T=25 \times 10^{3} / 165=151.5 N – mNumber of pairs of friction surfaces,
n=n_{1}+n_{2}-1=3+2-1=4and mean radius of friction surfaces for uniform wear,
R=\frac{r_{1}+r_{2}}{2}=\frac{120+60}{2}=90 mm =0.09 mWe know that torque transmitted (T ),
151.5=n \cdot \mu \cdot W \cdot R=4 \times 0.3 \times W \times 0.09=0.108 W
\therefore W=151.5 / 0.108=1403 N
Let p = Maximum axial intensity of pressure.
Since the intensity of pressure ( p) is maximum at the inner radius \left(r_{2}\right), therefore for uniform wear
p \cdot r_{2}=C \text { or } \quad C=p \times 60=60 p N / mmWe know that the axial force on each friction surface (W),
1403=2 \pi \cdot C\left(r_{1}-r_{2}\right)=2 \pi \times 60 p(120-60)=22622 p
\therefore p=1403 / 22622=0.062 N / mm ^{2}