Turning Moment Diagrams

Turning Moment Diagrams with particular reference to Engines and Flywheels.

View other versions (4)

Contents

Introduction
Fluctuations Of Energy And Speed
Disc And Rim Flywheels
Page Comments

Introduction

In any machine there is at least one point where energy is supplied, and at least one other point from which energy is delivered. In an ideal machine no energy would be lost and these two would be equal. In practice this state of affairs does not exist since it is inevitable that some energy is absorbed in over coming friction at the various joints, couplings and bearings.

The ratio of energy out to energy in is known as the Mechanical Efficiency of the machine.

A flywheel is a mechanical device with a significant moment of inertia used as a storage device for rotational energy.

In addition, over a given interval of time, the kinetic and potential energies of each link will change so that either some of the energy supplied will be absorbed increasing the total energy (Kinetic and potential) of the moving parts, or alternatively the energy supplied will be supplemented by a decrease in the total energy of the moving parts.

It should be stressed that over the time taken for the machine to complete one cycle, the net change of energy for each moving part is nil, since at the end of the cycle each part is in the same position and has the same speed as at the beginning of the cycle. However, during the cycle the input of energy or the load on the machine may vary considerably. In most cases this fluctuation is kept to a minimum by the use of a flywheel.

Fluctuations Of Energy And Speed

The driving torque $\inline \tau$ produced by a reciprocating engine fluctuates during any one cycle. The manner in which it varies depends upon the type of engine, number of cylinders, the characteristics of the flywheel etc. It can usually be assumed that the resisting torque due to the load $\inline \tau_m$ is constant, and when $\inline \tau$ > $\inline \tau_m$ the engine will be accelerating; and vice versa.

If there are $\inline N$ complete cycles per minute and the engine speed is $\inline n\;r.p.m.$ Then the power transmitted is $\inline N\displaystyle\int\displaystyle\frac{d\theta}{33,000} =\displaystyle\frac{2\,\pi \,n\,\tau _m}{33,000}$

i.e. $\inline \tau _m$ = The mean height of the turning moment diagram

For any period during which $\inline \displaystyle \tau >\tau _m$ the area cut off on the turning moment diagram represents "excess energy" $\inline \displaystyle \Delta E$ , which will go to increase the speed of the rotating parts.

$\Delta E=\int \left ( \tau -\tau _m \right )d\theta$

$\therefore\;\;\;\;\;\;\; \Delta E=\frac{1}{2}I(\omega {_{1}}^{2}-\omega{_{2}}^{2})$

where $\inline I$ is the moment of inertia of the flywheel and rotating parts and $\inline \omega_1$ and $\inline \omega_2$ are the maximum and minimum speeds during one cycle.

Thus, $\inline \Delta E=\displaystyle\frac{1}{2}(\omega _1-\omega _2)(\omega_1+\omega_2)$

This can be re-written to include the mean speed $\inline \omega_0$ as:

$\inline \Delta E=(\omega _1-\omega _2)\;\omega_0$ (Approximately)

The coefficient of fluctuation of speed is:

$\displaystyle\frac{(\omega _1-\omega _2)}{\omega_0}= \displaystyle\frac{\Delta E}{I\,\omega{_{0}}^{2}}$

(2)

This is usually expressed as a percentage variation from the mean. i.e. $\inline \displaystyle \pm \left ( \frac{\Delta E}{2I\,\omega{_{0}}^{2}}\times 100 \right )\%$

In simple cases $\inline \Delta E$ is given by the area of one "loop" intercepted between $\inline \tau$ and $\inline \tau_m$ , but for multi-cylinder engines a further analysis is necessary. (See Example 5)

Disc And Rim Flywheels

The purpose of a flywheel is to absorb energy when the supply of energy to a machine exceeds the requirement, and to provide energy when there is a deficit.

A diameter of a circle is any straight line segment that passes through the center of the circle and whose endpoints are on the circle. The diameters are the longest chords of the circle.

A body when it rotates behaves as if all of its mass were concentrated in a ring at a distance $\inline k$ from the axis of rotation. The radius $\inline k$ is known as the Radius of Gyration of the body. The product $\inline Mk^2$ is known as the Moment of Inertia of the body and given the symbol $\inline I$ .

For a solid disc of diameter $\inline D$ , $\inline k^2=\displaystyle\frac{D^2}{8}$

For a ring or rim of diameters $\inline D$ and $\inline d$ , $\inline k^2=\displaystyle\frac{D^2+d^2}{8}$

Example:

[imperial]

Example - Example 1

Problem

The mean speed of an engine $m$ is 250 r.p.m. The maximum fluctuation of energy generated in the engine is 850 ft.lb. and the resisting torque is constant. Determine the moment of inertia of the flywheel required to keep the speed within the range 1% above to 1% below the mean speed. State clearly the units in which the moment of inertia is expressed.

It is desired to reduce the coefficient of speed fluctuation by one fifth by bolting a plain cast-iron ring to the side of the flywheel. The ring is to have an outside diameter of 3ft. 4in. and an inside diameter of 2ft. 8in. and to be made of material which weighs 0.28 lb./cu.in. Find the width of the required ring.

Workings

The coefficient of fluctuation of speed $\displaystyle \pm 1\%$ and from equation (1) $\displaystyle =\pm \left ( \frac{\Delta E}{2\;I\;\omega {_{0}}^{2}} \right )$

$\therefore \;\;\;\;\;I=\frac{850\times 100}{2(250\times \displaystyle\frac{2\,\pi }{60})^2}=62.2\:slugs.ft.^2$

Note. For those of you who are not used to the foot slug second system of engineering quantities, a full description can be found on the Codecogs site under References: Engineering: General. However in brief the Slug is that mass which on this planet weighs one pound (lb.)

To reduce the coefficient of speed fluctuation by one-fifth it is necessary to increase $I$ by one-quarter. As this is to be done by adding a ring of thickness $t\;in.$ :

$\frac{62.2\times 144}{4}=\frac{\pi }{4}\left ( 40^2-32^2 \right )t\times\frac{0.28}{32.2}\left ( \frac{40^2+32^2}{8} \right )$

Note. $I$ is now measured in lb.in.^2 and the density of the flywheel material has been converted into slugs/cu.in. since the equation demands the mass of the flywheel rather than its weight.

$\therefore \;\;\;\;\;t=\frac{62.2\times 144\times 8\times 32.2}{\pi \times(40^2-32^2)\times 0.28\times (40^2+32^2)}=1.74 in.$

Solution

The moment of inertia is $62.2\:slugs.ft.^2$
The width is $1.74 in.$