WEEKS 14-15

Dynamics of Machinery

• References

Theory of Machines and Mechanisms, J.J.Uicker,

G.R.Pennock ve J.E. Shigley, 2003

DESIGN OF MACHINERY, Robert L, Norton. 1999

DYNAMICS OF MACHINERY, Prof.Dr. Sadettin Kapucu

Theory of Machines, Khurmi, R. et al.; 2005

Prof.Dr.Hasan ÖZTÜRK 1

Prof.Dr.Hasan ÖZTÜRK 2

A flywheel is an energy storage device. It absorbs mechanical

energy by increasing its angular velocity and delivers energy by

decreasing its angular velocity. Commonly, the flywheel is used to

smooth the flow of energy between a power source and its load. If

the load happens to be a punch press, the actual punching operation

requires energy for only a fraction of its motion cycle. If the power

source happens to be a two-cylinder four-cycle engine, the engine

delivers energy during only about half of its motion cycle. More

recent applications under investigation involve using a flywheel to

absorb braking energy and deliver accelerating energy for an

automobile and to act as energy-smoothing devices for electric

utilities as well as solar and wind-power generating facilities. Electric

railways have long used regenerative braking by feeding braking

energy back into power lines, but newer and stronger materials now

make the flywheel more feasible for such purposes

FLYWHEEL

Prof.Dr.Hasan ÖZTÜRK 3

There are two general situations which we will be faced with in mechanical

systems non smooth operation due to fluctuation of speed.

Output= Average of input

Input

Input= Average of output

Output

Generator

IC

Engine Electrical

Motor

Punching

Machine

FLYWHEEL

Prof.Dr.Hasan ÖZTÜRK 4

FLYWHEEL ENERGY: The near figure shows a

flywheel designed as a flat circular disk, attached

to a motor shaft which might also be the derive

shaft for the crank of our linkage. The motor

supplied a torque magnitude Ti which we would

like to be as constant as possible, i.e., to be equal

to the average torque Tavg . The load, on the other

side of the flywheel, demands a torque To which is

time varying as shown in the below figure. The

kinetic energy in a rotating system is:

where I is the moment of inertia of all rotating mass on the shaft. This includes

the I of the motor rotor and of the linkage crank plus that of the flywheel, We

want to determine how much I we need to add in the form of a flywheel to

reduce the speed variation of the shaft, to an acceptable level. We begin by

writing Newton's law for the free-body diagram in the above Figure.

Prof.Dr.Hasan ÖZTÜRK 5

Prof.Dr.Hasan ÖZTÜRK 6

OR

Prof.Dr.Hasan ÖZTÜRK 7

Prof.Dr.Hasan ÖZTÜRK 8

Prof.Dr.Hasan ÖZTÜRK 9

The left side or this expression represents the

change in energy E between the maximum

and minimum shaft ω's and is equal to the

area under the torque-time diagram between

those extreme values of ω. The right side of

equation (2) is the change in energy stored in

the flywheel. The only way we can extract

energy from the flywheel is to slow it down

as shown in equation (1). Adding energy will

speed it up. Thus it is impossible to obtain

exactly constant shaft velocity in the face of

changing energy demands by the load. The

best we can do is to minimize the speed

variation (ωmax- ωmin) by providing a

flywheel with sufficiently large I.

2

min

2

max2

1 IE

the torque-time diagram

…….1

……..2

Prof.Dr.Hasan ÖZTÜRK 10

Example: An input torque-time function which varies over its cycle. The torque

is varying during the 3600 cycle about its average value.

Find the total energy variation over one cycle Nm

Note that the integration on the left

side of equation (4) is done with

respect to the average line of the

torque function, not with respect to the

axis.

Nm-rad

Calculate the average value of

the torque-time function over

one cycle. which in this case is

70.2 Nm.

2

min

2

max2

1 IE

maxw

minw

maxw

maxw

minw

200.73 ( 60.32) 261.05- - =

Prof.Dr.Hasan ÖZTÜRK 11

Prof.Dr.Hasan ÖZTÜRK 12

System Permissible k

Pumps, Shearing machines 1/5 – 1/30

Machine Tools, Textile Machines 1/40 – 1/50

Generators 1/100 – 1/300

Automobiles 1/200 – 1/300

Aircraft Engines 1/1000 – 1/2000

Coefficient of fluctuation of speed

Prof.Dr.Hasan ÖZTÜRK 13

Example: The turning moment diagram for a multicylinder engine has been drawn to

a scale 1 mm = 600N-m vertically and 1 mm = 3o horizontally. The intercepted areas

between the output torque curve and the mean resistance line, taken in order from

one end, are as follows:

+52, -124, +92, 140, +85, -72 and +107 mm2, when the engine is running at a speed

of 600 rpm. If the total fluctuation speed is not to exceed -+1.5 % of the mean, find

the necessary mass of the flywheel of radius of gyration 0.5 m.

max min

s

ave

C 0.03

Given:

srad

rpmN

/84.6260060

2

600

Since the total

fluctuation of speed is

not exceed +- 1.5 % of

the mean speed,

therefore Since the turning moment scale is 1 mm=600N-m

and crank angle scale is 1 mm=3o=3o/180=/60,

therefore

1 mm2 on turning moment diagram

=600*/60=31.42Nm

Prof.Dr.Hasan ÖZTÜRK 14

03.0sk

Given: srad /84.62

1 mm2 =31.42Nm

AEisAatenergytheLet

10772, AFG EEEGatEnergy

52, AB EEBatEnergy

72124, ABC EEECatEnergy

2092, ACD EEEDatEnergy

120140, ADE EEEEatEnergy

3585, AEF EEEFatEnergy

AGH EEEHatEnergy 107,

Maximum Energy

Minimum Energy

A AE E 52 E 120 172

E 172*31.42 5404 Nm

2

aveE E I k

03.0*82.62*5404 2I264.45 kgmI

22 *64.45 kmkgmI

kgm 58.182

Prof.Dr.Hasan ÖZTÜRK 15

Prof.Dr.Hasan ÖZTÜRK 16

Prof.Dr.Hasan ÖZTÜRK 17