1. 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
Explanation: For minimum position, h1 = 200 mm
For maximum position,
h2=150 mm
sleeve lift = 2(h1 – h2) = 0.1m.
2. 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
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.
3. 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
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.
4. 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
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.
5. 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
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.
6.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) 5.47
d) 5.0
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
7. When friction acts at the sleeve the maximum angular velocity increases?
a) True
b) False
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
8. The controlling force acting on the governor is also known as ________
a) Centripetal force
b) Centrifugal force
c) Governor effort
d) Piston effort
Explanation: When a governor rotates at a steady speed, an inward force acts on the body which is centripetal in nature. This force is called a controlling force.
9. Controlling force is in the same direction as centrifugal force.
a) True
b) False
Explanation: When a governor rotates at a steady speed, an inward force acts on the body which is centripetal in nature. This force is called controlling force and acts in the opposite direction to that of centrifugal force.
10. If F=m.ω2.r represents the centrifugal force then which of the following expressions represents controlling force.
a) F
b) 2F
c) -2F
d) -F
Explanation: Inward force acts on the body of the governor which is centripetal in nature. This force is called controlling force and acts in the opposite direction to that of centrifugal force.