Governors Questions and Answers

Answer: A

Automobile engines of normal speed are given as having a high coefficient of fluctuation of speed in the key. The coefficient of fluctuation of speed for a flywheel is the ratio of the change in speed to the mean speed. Machines that need smooth running, such as machine tools, weaving machines and belt driven generators, are given smaller values, because a smaller value means a more uniform speed.

Answer: B

The Porter governor is a type of dead weight governor, in which a central dead weight is placed on the sleeve and acts as a load. The Pickering governor is a spring controlled type that is used for small speed variations and the Hartnell and Hartung governors are spring loaded types, so only the Porter governor is a dead weight type.

Answer: B

The new height is 100 mm.

For a Watt governor with massless arms and sleeve, balancing the centrifugal force on a ball against its weight about the pivot gives the governor height

\[ h = \frac{g}{\omega^{2}} \]

independent of the arm lengths. Doubling \(\omega\) divides \(h\) by four: \(400/4 = 100\) mm.

200 mm assumes \(h\) falls in proportion to \(\omega\), not \(\omega^{2}\), and 50 mm divides by eight. 150 mm has no basis. Check with the data: \(\omega^{2} = 9.8/0.4 = 24.5\) rad²/s² at first, and four times that gives \(h = 9.8/98 = 0.1\) m.

Answer: C

The balls are connected by two springs in parallel.

The Wilson-Hartnell governor is a radial, spring-loaded governor. Its two balls are tied together by main springs placed on either side of the spindle; both springs act on the balls at the same time, sharing the load, which is a parallel arrangement. An auxiliary spring acting through a lever is used to adjust the equilibrium speed while the governor is running.

A single spring, springs in series or four springs do not describe this arrangement.

Answer: A

The height is proportional to \(1/N^{2}\).

For a Watt governor, the balance of the centrifugal force and the weight of the ball gives \(h = g/\omega^{2}\). With N in rpm this is

\[ h = \frac{895}{N^{2}}\ \text{m} \]

So doubling the speed reduces the height to a quarter. That is why the Watt governor is useful only at low speeds: at high speed the change in height becomes too small to work the throttle. Height proportional to N, 1/N or \(N^{2}\) contradicts this result.

Answer: A

The height of a Watt governor is \(h = \frac{g}{\omega^{2}}\).

Take moments about the pivot for one ball of mass m rotating at radius r. The centrifugal force \(m\omega^2 r\) acts at height h below the pivot and the weight mg acts at radius r:

\[ \begin{aligned} m\omega^2 r \times h &= mg \times r \\ h &= \frac{g}{\omega^2} \end{aligned} \]

With g in m/s² and \(\omega = 2\pi N/60\), this gives the familiar \(h = 895/N^2\) metres. The height falls as the square of the speed, so \(\frac{g}{\omega^3}\) has the wrong power, and \(\frac{\omega^2}{g}\) and \(g\omega^2\) would make the height grow with speed.

Answer: B

Radius of rotation decreases but height of governor increases.

When the engine slows down, the centrifugal force on the governor balls falls, so the balls move inwards and the radius of rotation decreases. For a Watt-type governor the height is

\[ h = \frac{g}{\omega^2} \]

so a lower speed gives a greater height: the sleeve drops and opens the fuel or steam supply to restore the speed. The options in which the radius increases, or both quantities move together, contradict this relation.

Answer: B

The spring loaded governor is the Hartnell governor.

In the Hartnell governor the centrifugal force of the rotating balls is balanced by a compressed spring acting on the sleeve, instead of by gravity. Being spring controlled, it can run at high speed and in any position.

The Watt, Porter and Proell governors are gravity controlled: the weight of the balls (and of a central load in the Porter and Proell) balances the centrifugal force.

Answer: C

A flywheel absorbs energy when the turning moment is greater than the resisting moment.

When the engine's turning moment exceeds the load's resisting moment, the surplus energy speeds the flywheel up and is stored as kinetic energy. When the turning moment falls below the resisting moment, the flywheel slows and gives that energy back, which keeps speed fluctuation within limits.

When the moments are equal there is no excess to absorb, and a flywheel cannot absorb energy throughout the cycle, as it must also release it.

Answer: C

Sensitiveness is \(\frac{2(N_{1} - N_{2})}{(N_{1} + N_{2})}\).

The sensitiveness of a governor is the range of speed between its extreme positions divided by the mean speed.

\[ \begin{aligned} S &= \frac{N_{1} - N_{2}}{N}, \quad N = \frac{N_{1} + N_{2}}{2} \\ S &= \frac{2(N_{1} - N_{2})}{N_{1} + N_{2}} \end{aligned} \]

The other options invert the ratio or misplace the factor 2. A smaller speed range for the same mean speed means a more sensitive governor.