Engineering Questions with Answers - Multiple Choice Questions
Machine Dynamics MCQ– Effort and Power of a Porter Governor
1 - Question
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
View Answer
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
2 - Question
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
View Answer
Answer: b
Explanation: For minimum position, h1 = 200 mm
For maximum position,
h2=150 mm
sleeve lift = 2(h1 – h2) = 0.1m.
3 - Question
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
View Answer
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.
4 - Question
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
View Answer
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.
5 - Question
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
View Answer
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.
6 - Question
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
View Answer
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.
7 - Question
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
View Answer
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.
8 - Question
When friction acts at the sleeve the maximum angular velocity increases?
a) True
b) False
View Answer
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.