effort and power of a governor and numerical related to this topics

Effort and power of governor

1. Effort and Power of a Governor

The effort of a governor is the mean force exerted at the sleeve for a given percentage
change of speed* (or lift of the sleeve). It may be noted that when the governor is running steadily,

there is no force at the sleeve. But, when the speed changes, there is a resistance at the sleeve which
opposes its motion. It is assumed that this resistance which is equal to the effort, varies uniformly
from a maximum value to zero while the governor moves into its new position of equilibrium.

The power of a governor is the work done at the sleeve tor a given percentage change of
speed. It is the product of the mean value of the effort and the distance through which the sleeve
moves. Mathematically,

Power = Mean effort x lift of sleeve
Effort and Power of a Porter Governor
The effort and power of a Porter governor may be determined as discussed below.

Let N = Equilibrium speed corresponding to the configuration as shown in

Fig. 18.33 (a), and
c = Percentage increase in speed.
Increase in speed = c.N

and increased speed = N+c.N=N(1+c)

2. The equilibrium position of the governor at the increased speed is shown in Fig (b).

(a) Position at equilibrium speed. (a) Position at increased speed.
Fig.
We have discussed in Art. 18.6 that when the speed is N r.p.m., the sleeve load is M.g.
Assuming that the angles ot and ß are equal, so that q = 1, then the height of the governor,

= ee & 895 (in metres)
= N? ©

‘When the increase of speed takes place, a downward force P will have to be exerted on the

sleeve in order to prevent the sleeve from rising. If the speed increases to (1 + c) N r.p.m. and the

height of the governor remains the same, the load on the sleeve increases to M ,.g. Therefore
qe M; e 895

A az (in metres) RR,

3. Equating equations (¿) and (ii), we have

m+M,
A Ga or M,=(m+ M(l+c)-m
and M,—M = (m+M) (1 +0? -m-M=(m+M) [d +0®-1] ... di)

In comparing different types of governors, it is convenient to take the change of speed as one per cent.

A little consideration will show that (M, — M)g is the downward force which must be applied
in order to prevent the sleeve from rising as the speed increases. It is the same force which acts on the
governor sleeve immediately after the increase of speed has taken place and before the sleeve begins
to move. When the sleeve takes the new position as shown in Fig. 18.33 (b), this force gradually
diminishes to zero.
Let P = Mean force exerted on the sleeve during the increase in speed or
the effort of the governor.

(M,-M) g _(m+M)[0+ey -Ug

2 2.
_m+M)ll+c’+2c-1g
2

ai P=

=c(m+M)g ..- (iv)

4. .. (Neglecting e, being very small)
If Fis the frictional force (in newtons) at the sleeve, then
P=c(m3+M.3=F)
We have already discussed that the power of a governor is the product of the governor effort
and the lift of the sleeve.

Let x = Lift of the sleeve.
+ Governor power = Px x -(v)
If the height of the governor at speed N is h and at an increased speed (1 + ec) N is a then
x =2(h-h;)
As there is no resultant force at the sleeve in the two equilibrium positions, therefore
m+M _ 895 m+M 895
h= — 2 = Re on mn
m x N?’ and hy re dro? N?
h 1
a=—— or h = E.
h (l+cy (+ cY
x=2(h-h)=2|h- da =2h|1- E
We know that i La y
1+c?+2c-1 2c
In = A i)
| 1+0 +2c | E u)

.. (Neglecting c?, being very small)

5. Substituting the values of P and x in equation (v), we have

| 4c?

(m+M) gh... (vii)

E =c(m+M)gx2h =
Governor power cm IE 6 + 2c 1+2c

Notes : 1. If @ is not equal to ß, i.e. tan B / tan & = q, then the equations (¿) and (ii) may be written as

M
FEA es

h= ua « -- (viii)
m N?
When speed increases to (1 + c) N and height of the governor remains the same, then
m+ Mas 4) 895 .
h=—2__ - (ix)

gs _
m (1+c) N?

From cquations (viii) and (ix), we have

ee

M
+H(+9)=
m Zt q)

6. From equations (vifi) and (ix), we have
LE d+q)

M =
eh ae ad+c?

=[-Larpjaro m

or

>
M,_mQreyYy Ma _™
2 l+g 2 l+q
Mi M _mG+c) M a m M
= = 1 - -
or > 2 19 a 2

Se id -11+ Lrar+oy-1]

EA Alaro

Corman e
= 1+q =

2m
fe +M Je . . . (Neglecting c?)

The equation (vi) for the lift of the sleeve becomes,

7. The equation (vi) for the lift of the sleeve becomes,

2e
=(1 h
x=(1+9) (53)

Govemorpower Pons E gm lene gyal E
l+q 1+2c

2c” 4c? | M |
2m+M (+ .h= —— |m+— (+ A
et U+Qle.h= EN E
2. The above method of determining the effort and power of a Porter governor may be followed for any
other type of the governor.

8. Example 1 A Porter governor has equal arms each 250 mm long and pivoted on the
axis of rotation. Each bail has a mass of 5 kg and the mass of the central load on the sleeve is 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. Find the range of speed, sleeve lift, governor effort and power of the
governor in the following cases :

1. When the friction at the sleeve is neglected, and

2. When the friction at the sleeve is equivalent to 10 N.

Solution. Given: BP = BD = 250 mm ; m = Skg ; M = 25 kg; r, =150 mm; 7,= 200 mm;
F=10N
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 (4) respectively.

Let N > Minimum speed, and

N, = Maximum speed.
From Fig. 18.34 (a),

h = PG = (BP)? - (BG)? = (250 - (150)? = 200 mm = 0.2 m

9. From Pig. (0),

hy = PG = Y(BP)” — (BG)? = 50)? — (200)? = 150 mm = 0.15 m

m+M , 895 _5+25, 895 _ 4. 259
m hy 5 0.2

“ON, = 164 rpm.

We know that (NY? =

anal EN SRS BOS e

m m s 05

(a) Minimum po: (b) Maximum position.

10. Range of speed
We know that range of speed
= N,—N,= 189 - 164 = 25 rp.m. Ans.

Sleeve lift
We know that sleeve lift,
x = 2 (h, —h,) = 2 (200 — 150) = 100 mm = 0.1m Ans.
Governor effort
Let c = Percentage increase in speed.

We know that increase in speed or range of speed,
eN, =N,—N, =25r.p.m.
> € = 25/N, = 25/164 = 0.152
We know that governor effort
P = c(m+M)g= 0.152 (5 + 25) 9.81 = 44.7 N Ans.

Power of the governor
We know that power of the governor
= Px = 44.7 x 0.1 = 4.47 N-m Ans.

11. 2. When the friction at the sleeve is taken into account

Cinwas PR PE PI, AA

m.g Ay
59.81 + (25 x9.81—10) _ 895
= = 9
5 x 9.81 Kg oe
= N,~ 161 rpm.
<a (Ny)? = PE + Og + F),, 895
m.g 3
_5X981+ 25 x981+10) 895 _ 37 016

Sx9.81 0.15

ss N, = 1924 rpm.
Range of speed

We know that range of speed

=N,—N, = 192.4 — 161 = 31.4 r.p.m. Ans.

‚Sleeve lift

The sleeve lift (x) will be same as calculated above.

+. Sleeve lift, x = 100 mm = 0.1 m Ans.

12. Governor effort

Let € = Percentage increase in speed.

We know that increase in speed or range of speed,

EN, = N,-N,=314rpm.
= € = 31.4/N, = 31.4/161 = 0.195
We know that governor effort,
P = c(m.g+ Mg + F)=0.195 (5 x 9.81 + 25 x 9.81 + 10) N

57.4 N Ans.

Power of the governor
We know that power of the governor
= Px = 57.4 x 0.1 = 5.74 N-m Ans.