Engine Governors

The upper arm of a loaded governor are 11 in. long and are pivoted on the axis of rotation. The lower arms are also 11 in. long between pin joints but are pivoted to the sleeve at distances of $1\displaystyle\frac{1}{4}$ in. from the axis of rotation. Each governor ball weighs 12 lb. and is carried on an extension of the lower arm, the ball centre being vertically above and $3\displaystyle\frac{1}{2}$ in. from the pin joint between the upper and lower arms when the radius of gyration is 8 in.

Neglecting friction, find the load required on the sleeve in order to give an equilibrium speed of 270 r.p.m. for this radius of gyration.

The forces acting on one half of the governor are shown in the diagram. Neglecting inertia effects in the upper arms, the reaction at $O$ must be in the direction of $LO$.

Taking moments about the intersection of $OL$ produced and the horizontal through $M$, the reaction at $O$ and the horizontal component at $M$ are eliminated.

$OP=\sqrt{121-64}=\sqrt{57}\;in.$

$LN=\sqrt{121-45.5}=\sqrt{75.5}=8.7\;in.$

The triangles $TLN$ and $LOP$ are similar

$\therefore \;\;\;\;TN=8.7\times\frac{8}{\sqrt{57}}=9.2\;in.$

The centrifugal forces on one of the balls : $=\displaystyle\frac{12}{32.2}\times \displaystyle\frac{8}{12}\left ( \displaystyle\frac{270\times 2\,\pi }{60} \right )^2$

Taking moments about $T$ for the lower arm and ball:

$\frac{W}{2}\times 17.35=\frac{12}{32.2}\times \frac{8}{12}\left ( \frac{270\times 2\,\pi }{60} \right )^2\times 12.2-12\times9,2$

$=2430-110=2320$

$\therefore \;\;\;W=2320\times \frac{2}{17.5}=276\;lb.$

• The load required is $276\;lb.$

A Porter governor is arranged as in the diagram. The rotating weights are each 7 lb. The sleeve weighs 2.5 lb.and actuates a lever which has a weight of 5 lb.,(Its c.g. is as shown)and which must exercise an operating pull $P$ of 6 lb. The frictional effects of the gear reduced to the sleeve must be loaded so that the governor is on the point of moving out from the given position at 180 r.p.m.

Compare the speed at which it would begin to move in from this position. Prove any formula used.

By taking moments about the lever pivot point,it can be seen that the effect of $P$ and the weight of the lever is equivalent to a force at the sleeve of $(F_S)$ .

$F_S\times 5=6\times 14 + 5\times 6$. Therefore, $F_S=22.8\;lb.$

Thus allowing for the weight of the sleeve and friction, the total effective downward force at the sleeve :

This is carried by the tension $T$ in the lower arms and resolving vertically:

$2T\cos30^0=W+29.3$

The centrifugal force on one ball moving round a path of radius 6 in.

$=\frac{7}{32.2}\times \frac{6}{12}\left ( \frac{180\times 2\,\pi }{60} \right )^2=38.6\;lb.$

Taking moments about $O$ for the forces on the upper arm and ball:

Substituting for $T$ from equation (1), $\displaystyle\frac{1}{2}(W+29.3)\times9=401-42$

From which, $W=50.5\;lb.$

When the governor is moving in, friction is reversed whilst the other sleeve forces remain the same, hence

The equivalent Sleeve load =$W+2.5-4+22.8=71.8\:lb.$

Equation (2) now becomes: $\displaystyle\frac{1}{2}(71.8)\times 9=$(Centrifugal force)$\times 10.4-42$

From which the value of the centrifugal force is 35.2 lb.

Thus the centifugal force =$35.2=\displaystyle\frac{7}{32.2}\times \displaystyle\frac{6}{12}\left ( \displaystyle\frac{n\times 2\,\pi }{60} \right )^2$

$\therefore \;\;\;\;n=172\;r.p.m.$

In the bowl $A$ shown in the diagram there are eight steel balls $B$ of $2\displaystyle\frac{1}{4}$ and resting on the balls is a disc $C$ which is 48 lb. in weight.

If the internal surface of the bowl is part of a sphere of 9 in. radius, determine the speed of rotation about the vertical axis at which the balls will move with their centre in a circle of $10\displaystyle\frac{1}{2}$ diameter.

The density of steel is 0.283 lb/cu.in.

The forces acting on each ball are :

Own weight,

Weight of the disc,

The centrifugal force (C.F.),

The angle $theta$ is found from :

The Normal Reaction $R$ is normal to the sphere and resolving at right angles to $R$ gives: $(C.F.)\times \cos\theta=W\times \sin\theta$

Substituting from equations (3) (4) (5) and (6) and re-arranging:

$0.023\,\omega ^2=(6+1.69)\tan41^0\;51'$

$\omega =\sqrt{\left ( \displaystyle\frac{7.69\times0.869}{0.023} \right )}=17.3\;rad/sec.$ or $165\;r.p.m.$

• The speed of rotation is $165\;r.p.m.$

A Watt type governor has an arm of uniform section of length $L$ and mass $m$ and a ball of mass $M$.

Show that when revolving with an angular velocity of $\omega$ it makes an angle $\theta$ to the vertical, where:

$\cos\theta =\frac{M+\displaystyle\frac{m}{2}}{M+\displaystyle\frac{m}{3}}\times\frac{g}{L\,\omega ^2}$

Generalise this for the case where the bar is not of uniform section, its radius of gyration about the point of attachment being $k$ and its distance from its centre of gravity at the point of attachment being $d$.

Generalise also for the case where the bar is of uniform section, but is attached to a point distant $a$ from the axis.

In each case the ball is to be regarded as a point mass.

The three cases described in the question are considered separately as a), b) and c).

a) The mass of an element at a distance $x$ from $O$ $=m\times \displaystyle\frac{dx}{L}$

And the centrifugal force on the element $=\left (m\times \displaystyle\frac{dx }{L} \right )x\,\sin\theta \times \omega ^2$

Taking moments about $O$ for the centrifugal and gravity forces.

$ML\sin\theta\times\omega ^2\times L\cos\theta+\int_{O}^{L}\left ( m\times \frac{dx}{L} \right )x\sin\theta\times\omega ^2\times x\cos\theta=Mg\times L\sin\theta+mg\left ( \frac{L}{2} \right )\sin\theta$

Re-arranging:

$M\sin\theta\cos\theta\times L^2\omega ^2+m\sin\theta\cos\theta\left ( \frac{\phi ^2}{L} \right )\int_{0}^{L}x^2dx=\left ( M+\frac{m}{2} \right )gL\sin\theta$

$\therefore \;\;\;\;L^2\omega ^2\cos\theta\left ( M+\frac{m}{3} \right )=\left ( M+\frac{m}{2} \right )gL$

Or $\cos\theta=\left ( \displaystyle\frac{M+\displaystyle\frac{m}{2}}{M+\displaystyle\frac{m}{3}} \right )\times \displaystyle\frac{g}{L\omega ^2}$

• b) The element of mass of the arm must now be written as $dm$ and the Moments about $O$ gives:

$ML\sin\theta\times\omega ^2\times L\cos\theta+\int_{0}^{L}dm\times x\sin\theta\times \omega ^2\times x\cos\theta=MgL\sin\theta+mgd\sin\theta$

Re-arranging , $ML^2\omega ^2\cos\theta+\omega ^2\cos\theta\int_{0}^{L}x^2\times dm=(ML+md)g$

But, $\displaystyle\int x^2dm=mk^2$

But, $\displaystyle\int x^2dm=mk^2\displaystyle\frac{ML+md}{ML^2+mk^2}\times \displaystyle\frac{g}{\omega ^2}$

• c) The radius of rotation of each element is now $x\sin\theta+a$. Taking Moments about $O$.

$M(L\sin\theta+\alpha)\times\omega ^2\times L\cos\theta+\int_{0}^{L}m\frac{dx}{L}\times (x\sin\theta+\alpha)\times \omega ^2\times x\cos\theta=MgL\sin\theta+mg\frac{L}{2}\sin\theta$

$M(L\sin\theta+a)\omega ^2\cos\theta+\left ( \frac{m}{L^2} \right )\omega ^2\cos\theta\int_{0}^{L}(x\sin\theta+a)xdx =(M+\frac{m}{2})g\sin\theta$

Integrating and re-arranging:-

$\cos\theta=\frac{M+\displaystyle\frac{m}{2}}{M(L\sin\theta+a)+m(\displaystyle\frac{1}{3}L\sin\theta+\displaystyle\frac{a}{2})}\times \frac{g\sin\theta}{\omega ^2}$

The upper and lower ends of the links of a Proell governor are pivoted on the axis of rotation of the governor.

The upper links are each 10 in. long between centres, the lower links are also 10 in. between centre and carry extension arms each 4 in. long and parallel to the axis when the radius of the ball path is 6 in.

Determine the equilibrium speed of the governor for this configuration when each ball weighs 10 lb. and the central load is 80 lb.

Find the alteration required to the weight of the central load if the change of speed in rad./sec. with respect to the change of radius if the ball path in inches is to be 0.65 for the given configuration. What will then be the equilibrium speed?

The centrifugal force on the balls : $= \displaystyle\frac{10}{32.2}\times \displaystyle\frac{6}{12}\omega ^2$

Taking moments about $I$:

$\frac{10}{32.2}\times \frac{6}{12}\omega ^2\times 12=10\times 6+40\times 12$

From which: $\omega =\sqrt{ \displaystyle\frac{540\times 32.2}{10\times6}}= 17.0\; rad./sec.\;\;$ or $\;\;162.5\;r.p.m.$

Let $W lb.$ be the total central weight required, and $r in.$ be the ball-path radius,then by moments about $I$.

Re-arranging:

Differentiating:

$\sqrt{\frac{10}{32.2}\times 12}\times \frac{d\omega }{dr} =\sqrt{10+W}\times[4+\sqrt{100-r^2}]^{-\frac{3}{2}}\left ( -\frac{1}{2}\ \right )\left ( \frac{1}{2} \right )\left ( 100-r^2 \right )^{-\frac{1}{2}}(-2r)$

If $\displaystyle\frac{dw}{dr}=0.65$ when $r=6$ then from equation (8)

$10+W=\frac{10}{32.2\times12}\times 0.65^2\times 12^3\times 2^2\times \frac{64}{2^2}=134$

$\therefore \;\;\;\;\;W=124\;lb.\;\;$ i.e. an increase of 44 lb.

From equation (7)

$\omega =\sqrt{134\times \displaystyle\frac{32.2}{10}}=20.8\;rad./sec.\;\;$ or $\;\;198\;r.p.m.$

• a) $\omega= 17.0\; rad./sec.\;\;$ or $\;\;162.5\;r.p.m.$
• b) $\omega= 20.8\;rad./sec.\;\;$ or $\;\;198\;r.p.m.$

A governor of the Harnell type, having the dimensions shown in the diagram runs at a mean speed of 300 r.p.m. ;each ball weighs 5 lb. and a 3% reduction in speed causes a sleeve movement of $\displaystyle\frac{1}{4}$ in.

If the ball arm is vertical at the mean speed and gravitational effects are ignored, determine the spring stiffness in lb,/in. Neglect the weight of the arms.

By how much must the adjusting nut must be screwed down to render the governor isochronous and what will be the resulting operational speed of the governor?

At the mean speed, Centrifugal Force =$\displaystyle\frac{5}{32.2}\times \displaystyle\frac{4}{12}\left ( \displaystyle\frac{300\times 2\,\pi }{60} \right )^2=51\;lb.$

If $P_1$ is the total spring force, then by taking moments about the pivot point :

Centrifugal Force =$\displaystyle\frac{5}{32.2}\times \displaystyle\frac{4}{12}\left ( \displaystyle\frac{300\times 2\,\pi }{60} \right )=51\;lb.$

$\therefore \;\;\;\;\;P_1=204\;lb.$

At a reduced speed:

A sleeve movement of $\displaystyle\frac{1}{4}$ in. will produce a reduction in ball-path radius of $\displaystyle\frac{1}{2}$ in. and the new centrifugal force will now be :

$\frac{5}{32.2}\times \frac{3.5}{12}\left ( \frac{291\times 2\,\pi }{60} \right )^2=42\;lb.$

Taking moments:

$\left ( \frac{P_2}{2} \right )\times2.5=42\times5$

$\therefore \;\;\;\;\;\;P_2=168\;lb.$

Spring stiffness = $\displaystyle\frac{P_1-P_2}{\displaystyle\frac{1}{4}}=144\;lb./in.$

Let $\omega$ rad./sec. be the isochronous speed . Then at the mid position :

Centrifugal Force =$\left ( \displaystyle\frac{5}{32.2} \right )\left ( \displaystyle\frac{4}{12} \right )\omega ^2=0.5175\:\omega^2$

Taking Moments,

$\therefore \;\;\;\;\;P_1'=0.207\;\omega ^2$

At the lower position, the centifugal force =$\left ( \displaystyle\frac{5}{32.2} \right )\left ( \displaystyle\frac{3.5}{12} \right )=0.452\;\omega ^2$

Taking moments:

$\therefore \;\;\;\;\;P_1'=0.182\omega ^2+36$

Eliminating $P_1'$ by using equations (9) and (10), $\omega =37.2\;rad./sec.=355\;r.p.m.$

$36=0.026\;\omega ^2$

Thus the increase in compression of the spring is given by: $\left ( \displaystyle\frac{P_1'-P_1}{\text{stiffness}} \right )=\displaystyle\frac{83}{144}=0.59 in.$

• The spring stiffness is $144\;lb./in.$