A multi-plate clutch is to transmit 12 kW at 1500 rev/min. The inner and outer radii for the plates are to be 50 mm and 100 mm respectively. The maximum axial spring force is restricted to lkN. Calculate the necessary number of pairs of surfaces if ll = 0-35 assuming constant ‘vyear. What will be the necessary axial force?

The uniform pressure for the necessary axial force  is  W = 945 N

The uniform wear for the necessary axial force is  W = 970.15 N

Explanation:

Solution

Given that:

r₁ = 0.1 m

r₂ = 0.05m

μ = 0.35

p = 12 N or kW

N = 1500 rpm

W = 1000 N

The angular velocity is denoted as  ω= 2πN/60

Here,

ω = 2π *1500/60 = 157.07 rad/s

Now, the power transferred becomes

P = Tω this is the equation (1)

Thus

12kW = T * 157.07 rad/s

T = 76.4 N.m

Now, when we look at the uniform condition, we have what is called the torque that is frictional which acts at the frictional surface of the clutch dented as :

T = nμW R this is the equation (2)

The frictional surface of the mean radius is denoted by

R =2/3 [(r₁)³ - (r₂)³/(r₁)² - (r₂)²]

=[(0.1)³ - (0.05)³/[(0.1)² - (0.05)²]

R is =0.077 m

Now, we replace this values and put them into the equation (2)

It gives us this, 76. 4 N.m = n * 0.35* 1000 N * 0.077 m

n = 2.809 = 3

The number of pair surfaces is = 3

Secondly, we determine the uniform wear.

So, the mean radius is denoted as follows:

R = r₁ + r₂/ 2

=0.1 + 0.05/2

=0.075 m

Now, we replace the values and put it into the equation (2) formula

76. 4 N.m = n *0.35* 1000 N * 0.075 m

n= 2.91 = 3

Again, the number of pair surfaces = 3

However, for the uniform pressure with regards to the number of clutch plates is 3 we can derive the necessary axial force from the equation (2)

76. 4 N.m = 3 * 0.35 * W *0.077 m

W = 945 N

Also, for the uniform wear with regards to the number of clutch plates is 3 we can derive the necessary axial force from the equation (2)

76. 4 N.m = 3 * 0.35 * W *0.075 m

W = 970. 15 N

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