Question

A multi disc clutch, steel on bronze is to transmit 20 kW at 1440 rpm. The clutch is to be operated in oil with the coefficient of friction 0.08 and the average pressure is 0.3 MPa. Space limitation permits only 230 mm as outside diameter of the clutch. Assuming uniform pressure, determine: (i) Actual axial force required (W’) (ii) Actual maximum pressure (p’max) (iii) Actual average pressure (p’ave) Ratio of (Ri/Ro) can be chosen between 0.5 and 0.75 suitably and logically

Answer 1

P=20 kW, N=1440 rpm, coefficient of friction=0.08 , Average pressure = 0.3 MPa

Ri/Ro=0.5/0.75 ,

So, axial force required

(i)

$F_a= \frac {1}{2} \times \pi\times pD_i\times (D_o-D_i)$

$F_a= \frac {1}{2} \times 3.14\times 0.3\times150\times (150-100)$

$F_a= 3532.5 N$

(ii) We know that, for power transmitted

$P= \frac{2 \times \pi \times N \times T }{60}$

T= 132696.3 N-mm

Dm= $\frac {150 +100}{2}= 125$

Using uniform wear theory

$T=\frac{1}{2}\times n \times \mu \times F_a \times D_m$

132696.3= 0.5 $\times n \times 0.08\times 3532.5 \times 125$

n= 7.5 it will be even number = 8

number of steel disc= 4

number of bronze disc= 4+1= 5

Average pressure

$T=\frac{1}{2}\times n \times \mu \times F_a \times D_m$

$132696.3 =0.5\times 8 \times 0.08\times F_a \times 125$

$F_a=3317.40 N$

(ii) For average pressure

$F_a= 0.5 \times \pi \times 125 \times p(150-100)$

p= 0.338 MPa

(iii)For actual maximum pressure

$F_a= 0.5 \times \pi \times 100 \times p(150-100)$

p= 0.422 MPa