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Turning Moment Diagrams with particular reference to Engines and Flywheels

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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. 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

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The driving torque $\tau$ 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 $\tau_m$ is constant and when $\tau$ > $\tau_m$ the engine will be accelerating and vice versa.

If there are N complete cycles per minute and the engine speed is n r.p.m. Then the power transmitted is:-

$\text{horse-power}=N\int\frac{d\theta}{33,000} =\frac{2\,\pi \,n\,\tau _m}{33,000}$

(1)

$i.e.\;\;\;\;\;\;\tau _m=\text{The mean height of the turning moment diagram}$

(2)

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

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

(3)

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

(4)

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

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

(5)

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

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

(6)

The coefficient of fluctuation of speed is:-

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

(7)

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

In simple cases $\Delta E$ is given by the area of one "loop" intercepted between $\tau$ and $\tau_m$ 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 body when it rotates behaves as if all of its mass were concentrated in a ring at a distance k from the axis of rotation. The radius k is known as The Radius of Gyration of the body. The product $Mk^2$ is known as the Moment of Inertia of the body and given the symbol I.

For a solid disc of diameter D

$k^2=\frac{D^2}{8}$

(8)

For a ring or rim of diameters D and d

$k^2=\frac{D^2+d^2}{8}$

(9)

Examples.

The workings can be seen by clicking onto the button.

Example 1

The mean speed of an engine mis 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. (U.L.)

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Example 2

The variation in turning moment with crank angle for a six-cylinder engine is shown in the diagram (a). The flywheel is equivalent to a mass of 32 lb. concentrated at a radius of 1 ft. and the mean speed of rotation is 1200 r.p.m. Sketch the speed crank-angle curve over a revolution and estimate the difference between the maximum and minimum speeds.

If the crank-angle base can now be assumed to be a time base and also the speed changes to be linear with time, estimate the angle through which the crank advances and falls back relative to an imaginary crank rotating uniformly at the mean speed.

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Example 3

A machine runs at a mean speed of 300 r.p.m. The torque required by the machine increases uniformly from 500 lb.ft. to 2000 lb.ft. whilst the shaft turns through $40^0$ , remains constant for the next $100^0$ , decreases uniformly to 500 lb.ft for the next $40^0$ and then remains constant for the next $180^0$ . This cycle is repeated during each revolution. The power is supplied by a constant torque motor and the fluctuation in speed is to be limited to $\displaystyle \pm\; 3\%$ of the mean speed.

Find (a) The horse power of the motor. (b) The moment of inertia of a suitable flywheel to be fitted to the machine shaft.

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Example 4

An engine has three single-acting cylinders, the cranks are spaced at $120^0$ to each other. The crank effort diagram for each cylinder consists of a triangle having the following values

Find:- (a) The mean Torque

(b) The moment of inertia of the flywheel in $lb.ft.^2$ to keep the speed within $180\pm 3\;r.p.m.$ (U.L.)

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Example 5

The turning moment diagram for a multi-cylinder engine has been drawn to a scale of 1 to 5000 lb.ft. vertically and 1 in. to $30^0$ horizontally, the intercepted areas between the output torque curve and the mean resistance line, taken in order from one end are :- -0.52, +1.24, -0.92, +1.4, -0.85, +0.72, -1.07 sq.in. when the engine is running at 800 r.p.m.

If the weight of the flywheel is 1200 lb. and the total fluctuation from the maximum to minimum speed does not exceed 2% of the mean speed, what is the minimum value of the radius of gyration. (U.L.)

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Example 6

The torque exerted on the crankshaft of an engine is given by the following equation:-

$\tau (lb.ft.)=10,500+1,620\sin2\theta-1340\cos2\theta$

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where $\theta$ is the crank-angle displacement from the inner dead-centre. Assuming the resisting torque to be constant, determine:-

(a) The horse-power of the engine when the speed is 150 r.p.m.

(b) The moment of Inertia of the flywheel if the speed variation is not to exceed $\pm 0.5\%$ of the mean speed.

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Example 7

The equation of the turning moment curve of a three-crank engine is :-

$\tau=1400+4665\sin3\theta$

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where $\theta$ radians is the crank angle. The moment of inertia of the flywheel is $4.5 tons\.ft.^2$ and the mean speed is 300 r.p.m. Calculate:-

(a)The horse-power of the engine.

(b)The total percentage fluctuation of speed of the flywheel if:- (1) The resisting torque is constant (2) The resisting torque is $(1400+2000\sin\theta)\;lb.ft.$

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