A loaded governor of the Porter type has equal arms and links each 250 mm long. The mass of each ball is 2 kg and the central mass is 12 kg. When the ball radius is 150 mm, the valve is fully open and when the radius is 185 mm, the valve is closed. Find the maximum speed and the range of speed. If the maximum speed is to be increased 20% by an addition of mass to the central load, find what additional mass is required
Answers
Answer:
Given : BP = BD = 250 mm ; m = 5 kg ; M = 25 kg ; r_{1}r
1
=150 mm ; r_{2}r
2
= 200 mm ; F = 10 N
1. When the friction at the sleeve is neglected
First of all, let us find the minimum and maximum speed of rotation. The minimum and maximum position of the governor is shown in Fig. 18.34(a) and (b) respectively
Let N_{1}N
1
= Minimum speed, and
N_{2}N
2
= Maximum speed.
From Fig. 18.34(a),
h_{1}=P G=\sqrt{(B P)^{2}-(B G)^{2}}=\sqrt{(250)^{2}-(150)^{2}}=200 mm =0.2 mh
1
=PG=
(BP)
2
−(BG)
2
=
(250)
2
−(150)
2
=200mm=0.2m
From Fig. 18.34(b),
h_{2}=P G=\sqrt{(B P)^{2}-(B G)^{2}}=\sqrt{(250)^{2}-(200)^{2}}=150 mm =0.15 mh
2
=PG=
(BP)
2
−(BG)
2
=
(250)
2
−(200)
2
=150mm=0.15m
We know that \left(N_{1}\right)^{2}=\frac{m+M}{m} \times \frac{895}{h_{1}}=\frac{5+25}{5} \times \frac{895}{0.2}=26850(N
1
)
2
=
m
m+M
×
h
1
895
=
5
5+25
×
0.2
895
=26850
\therefore \quad N_{1}=164 \text { r.p.m. }∴N
1
=164 r.p.m.
and \left(N_{2}\right)^{2}=\frac{m+M}{m} \times \frac{895}{h_{2}}=\frac{5+25}{5} \times \frac{895}{0.15}=35800(N
2
)
2
=
m
m+M
×
h
2
895
=
5
5+25
×
0.15
895
=35800
\because \quad N_{2}=189 \text { r.p.m. }∵N
2
=189 r.p.m.
Range of speed
We know that range of speed
=N_{2}-N_{1}=189-164=25 r . p . m=N
2
−N
1
=189−164=25r.p.m
Sleeve lift
We know that sleeve lift,
x=2\left(h_{1}-h_{2}\right)=2(200-150)=100 mm =0.1 mx=2(h
1
−h
2
)=2(200−150)=100mm=0.1m
Governor effort
Let c = Percentage increase in speed.
We know that increase in speed or range of speed,
c \cdot N_{1}=N_{2}-N_{1}=25 \text { r.p.m. }c⋅N
1
=N
2
−N
1
=25 r.p.m.
\therefore∴ c=25 / N_{1}=25 / 164=0.152c=25/N
1
=25/164=0.152
We know that governor effort
P=c(m+M) g=0.152(5+25) 9.81=44.7 NP=c(m+M)g=0.152(5+25)9.81=44.7N
Power of the governor
We know that power of the governor
=P x=44.7 \times 0.1=4.47 N – m=Px=44.7×0.1=4.47N–m
2. When the friction at the sleeve is taken into account
We know that \left(N_{1}\right)^{2}=\frac{m \cdot g+(M \cdot g-F)}{m \cdot g} \times \frac{895}{h_{1}}(N
1
)
2
=
m⋅g
m⋅g+(M⋅g−F)
×
h
1
895
=\frac{5 \times 9.81+(25 \times 9.81-10)}{5 \times 9.81} \times \frac{895}{0.2}=25938=
5×9.81
5×9.81+(25×9.81−10)
×
0.2
895
=25938
\therefore \quad N_{1}=161 \text { r.p.m. }∴N
1
=161 r.p.m.
and \left(N_{2}\right)^{2}=\frac{m \cdot g+(M \cdot g+F)}{m \cdot g} \times \frac{895}{h_{2}}(N
2
)
2
=
m⋅g
m⋅g+(M⋅g+F)
×
h
2
895
\therefore \quad N_{2}=192.4 \text { r.p.m. }∴N
2
=192.4 r.p.m.
Range of speed
We know that range of speed
=N_{2}-N_{1}=192.4-161=31.4 \text { r.p.m. }=N
2
−N
1
=192.4−161=31.4 r.p.m.
Sleeve lift
The sleeve lift (x) will be same as calculated above.
\therefore∴ Sleeve lift, x = 100 mm = 0.1 m
Governor effort
Let c = Percentage increase in speed.
We know that increase in speed or range of speed,
c \cdot N_{1}=N_{2}-N_{1}=31.4 \text { r.p.m. }c⋅N
1
=N
2
−N
1
=31.4 r.p.m.
\therefore \quad c=31.4 / N_{1}=31.4 / 161=0.195∴c=31.4/N
1
=31.4/161=0.195
We know that governor effort,
P=c(m \cdot g+M g+F)=0.195(5 \times 9.81+25 \times 9.81+10) NP=c(m⋅g+Mg+F)=0.195(5×9.81+25×9.81+10)N
= 57.4 N
Power of the governor
We know that power of the governor
=P x=57.4 \times 0.1=5.74 N – m=Px=57.4×0.1=5.74N–m