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# Machine Dynamics MCQ– Effort and Power of a Porter Governor

For a porter governor, Each arm has a length of 250mm and pivoted on the axis of rotation. Sleeves carry a mass of 25kg and each ball’s mass is 5Kg. Radius of rotation: 150mm at the beginning of lift and 200mm at the maximum speed of governor. Find range in speed neglecting friction in rpm.

a) 25

b) 35

c) 45

d) 15

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Answer: a

Explanation: For minimum position, h1 = 200 mm

N12= (m+M)/m x 895/h1 = 26850 rpm

For maximum position,

h2=150 mm

N22 = (m+M)/m x 895/h2 = 35800 rpm

Range = N2 – N1 = 25 rpm

A Porter governor has equal arms each of length 250 mm and pivoted on the axis of rotation. Each ball has a mass of 5 kg and the central load has a mass on the sleeve of 25 kg. The radius of rotation of the ball is 150 mm when the governor begins to lift and 200 mm when the governor is at maximum speed. Neglecting friction at the sleeve, find the sleeve lift.

a) 0.2m

b) 0.1m

c) 0.5m

d) 200 mm

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Answer: b

Explanation: For minimum position, h1 = 200 mm

For maximum position,

h2=150 mm

sleeve lift = 2(h1 – h2) = 0.1m.

A Porter governor has equal arms each of length 250 mm and pivoted on the axis of rotation. Each ball has a mass of 5 kg and the central load has a mass on the sleeve of 25 kg. The radius of rotation of the ball is 150 mm when the governor begins to lift and 200 mm when the governor is at maximum speed. Neglecting friction at the sleeve, find the governor effort.

a) 44.7 N

b) 22.35 N

c) 89.4 N

d) 50 N

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Answer: a

Explanation: If c = Percentage increase in speed, then

cN1 = N2 – N1 = 25 rpm

c = 25/164 = 0.152

governor effort = c(m+M)g = 44.7 N.

A Porter governor has equal arms each of length 250 mm and pivoted on the axis of rotation. Each ball has a mass of 5 kg and the central load has a mass on the sleeve of 25 kg. The radius of rotation of the ball is 150 mm when the governor begins to lift and 200 mm when the governor is at maximum speed. Neglecting friction at the sleeve, find the power of the governor in N-m.

a) 447

b) 44.7

c) 4.47

d) 5.0

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Answer: c

Explanation: We know that power of a governor is given by

effort of governor x sleeve lift

= 44.7 x 0.1 = 4.47 N-m.

A Porter governor has equal arms each of length 250 mm and pivoted on the axis of rotation. Each ball has a mass of 5 kg and the central load has a mass on the sleeve of 25 kg. The radius of rotation of the ball is 150 mm when the governor begins to lift and 200 mm when the governor is at maximum speed. Considering friction at the sleeve of 10N, find the range of speed in rpm.

a) 31.4

b) 35.4

c) 45.2

d) 15.6

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Answer: a

Explanation: When friction is taken into account

N12 = (mg+Mg-F)/mg x 895/h1

h1 = 0.2m

therefore N1 = 161 rpm

N22 = (mg+Mg+F)/mg x 895/h2

N2 = 192.4 rpm

Therefore range = 31.4 rpm.

A Porter governor has equal arms each of length 250 mm and pivoted on the axis of rotation. Each ball has a mass of 5 kg and the central load has a mass on the sleeve of 25 kg. The radius of rotation of the ball is 150 mm when the governor begins to lift and 200 mm when the governor is at maximum speed. Considering friction at the sleeve of 10N, find the governor effort.

a) 44.7 N

b) 57.4 N

c) 88.4 N

d) 53.8 N

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Answer: b

Explanation: If c is the percentage change in speed

cN1 = N2 – N1 = 31.4

c = 31.4/161 = 0.195

P = c(mg +Mg +F)

= 0.195 (5 × 9.81 + 25 × 9.81 + 10) = 57.4 N.

A Porter governor has equal arms each of length 250 mm and pivoted on the axis of rotation. Each ball has a mass of 5 kg and the central load has a mass on the sleeve of 25 kg. The radius of rotation of the ball is 150 mm when the governor begins to lift and 200 mm when the governor is at maximum speed. Considering friction at the sleeve of 10N, find the power of the governor in N-m

a) 447

b) 44.7

c) 4.47

d) 5.0

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Answer: c

Explanation: We know that the power of a governor is given by the relation

Power = governor effort x sleeve lift

= 57.4 x 0.1 N-m

= 5.74 N-m.

When friction acts at the sleeve the maximum angular velocity increases?

a) True

b) False

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Answer: a

Explanation: If F is the friction force acting on the sleeve, then the value of maximum speed is given

N = (mg + Mg+F )/mg x 895/h2

hence maximum angular velocity increases.