A centrifugal clutch transmits 20kW of power at 750 rpm. The engagement of the clutch commences at 70% of the running speed. The inside diameter of the drum is 200 mm and the distance of the center of the mass of each shoe is 40mm from the contact surface. Determine the

1. mass of each shoe
2. net force exerted by each shoe on the drum surface
3. power transmitted when the shoe is worn 2 mm and is not readjusted.

Assume µ to be 0.25, number of shoes equal to 4 and the stiffness of the spring 150kN/m.

Mumbai University > MECH > Sem 5 > Theory Of Machines 2

Marks: 10 M

Year: Dec 2014

Given:

P= 20kW; N = 750rpm; μ=0.25

N1 = 0.7N

Do = 0.2m; R = 0.1m

Ro = Ri+ 40mm; Ri = 0.06m

Solution:

Find mass:

. w = 2πN/60 = 78.54 rad/s

. w1 = 0.7 x w = 55 rad/s

Centrifugal force = Spring force + Engaging Force

Fc = S + F

At 70% of rpm, F = 0

Fc = S

$4Mrw_1^2 =4S \\ M \times Ri \times w^2 = S \\ M \times 0.06 \times 55^2 = S \\ S=181.5M$

Engagement force at full rpm:

Fc = S + F

$Mrw^2 = S + F \\ M \times 0.1 \times 78.54^2 = 181.5M + F \\ F = 435.4M$

P $= Tw \\ =μ \times 4F \times r \times w \\ = 0.25 \times 4 \times 435.4M \times 78.54 \\ 20000 = 34196 M \\ M = 0.59 kg$

F= 435.4 M = 255 N

Power transmitted after wearing:

After wearing, the spring will elongate an extra 2mm. (0.002m). An extra force will be required for this. Say E. This will reduce the force.

$F2 = F-E = F – kx = 255 – 150000 ( 0.002) = -45N$

F2 is negative, hence, no engagement will take place after wearing and hence, no power will be transmitted.