Flywheel - nptel

Machine Design II

Prof. K.Gopinath & Prof. M.M.Mayuram

Flywheel

A

flywheel

energy when it

is

and the

during

an

inertial

serves

supply the

as

of

energy-storage

a

energy

period

when

device.

reservoir,

storing

is

than

more

the

absorbs

energy

the

requirement

It

of

during

requirement energy

is

mechanical the

and

more

period releases

than

the

supply.

Flywheels-Function need and Operation The main function of a fly wheel is to smoothen out variations in the speed

of

a

shaft

caused

by

torque

fluctuations.

If

the

source

of

the

driving torque or load torque is fluctuating in nature, then a flywheel is usually

called

torque

time

engines

for.

Many

function

with

compressors,

one punch

to or

machines vary two

presses,

have

over

the

cylinders rock

load

patterns

cycle.

are

a

crushers

etc.

that

cause

Internal typical are

combustion

example.

the

the

Piston

other

systems

its

angular

that have fly wheel. Flywheel

absorbs

mechanical

energy

by

increasing

velocity and delivers the stored energy by decreasing its velocity 1 CYCLE

T2 Tm T1

B

A

D

θ

C D

θ

C

ω max ω min

A

B

Figure 3.3.1

Indian Institute of Technology Madras

Machine Design II


Design Approach There are two stages to the design of a flywheel. First,

the

smoothening

amount must

of

be

energy

found

required

and

the

for

(mass)

the

desired

moment

of

degree

of

inertia

needed

caters

the

required

and

safe

to absorb that energy determined. Then

flywheel

moment

of

geometry

inertia

in

must

a

be

reasonably

defined sized

that

package

is

against

failure at the designed speeds of operation.

Design Parameters Flywheel

inertia

(size)

needed

directly

depends

upon

the

acceptable

changes in the speed.

Speed fluctuation The

change

in

the

shaft

speed

during

a

cycle

is

called

the

dividing

it

speed

fluctuation and is equal to ωmax- ωmin

Fl = ωmax − ωmin We

can

normalize

this

to

a

dimensionless

ratio

by

by

average or nominal shaft speed (ωave) . Cf =

ωmax − ωmin ω

Where ωavg is nominal angular velocity

Co-efficient of speed fluctuation

The above ratio is termed as coefficient of speed fluctuation Cf and it is defined as Cf =


ωmax − ωmin ω

the

Machine Design II


Where shaft

ω

is

speed

nominal

angular

desired.

This

velocity,

and

coefficient

is

ωave a

the

design

average parameter

or

mean to

be

chosen by the designer. The smaller this chosen value, the larger the flywheel have to be and more

the

cost

and

weight

to

be

added

to

the

system.

However

the

smaller this value more smoother the operation of the device It

is

typically

machinery

set

and

as

to

a

value

high

as

between 0.20

0.01

for

to

0.05

applications

for

precision

like

crusher

entire

rotating

hammering machinery.

Design Equation The kinetic energy Ek in a rotating system =

( )

1 I ω2 2

Hence the change in kinetic energy of a system can be given as, EK =

1 ⎛ ⎞ Im ⎜ ω2max − ω2min ⎟ 2 ⎝ ⎠

E K = E 2 − E1 ωavg =

( ωmax + ωmin ) 2

(

1 I 2ωavg 2 s E 2 − E1 = Cf Iω2 Ek Is = 2 Cf ωavg

EK =

Thus system

the in

mass

moment

order

to

obtain

determined using the relation


of

inertia selected

)( Cf ωavg )

Im

needed

coefficient

in of

the speed

fluctuation

is

Machine Design II


EK = Is = The Im

above

equation

corresponding

can

to

the

(

1 I 2ωavg 2 s Ek

)( Cf ωavg )

2 Cf ωavg

be

used

known

to

obtain

appropriate

energy

change

energy

Ek

Ek

for

flywheel a

specific

inertia value

coefficient of speed fluctuation Cf,

Torque Variation and Energy The

required

change

in

kinetic

is

obtained

from

the

known

torque time relation or curve by integrating it for one cycle.

θ @ ωmax Tl − Tavg dθ = E K ∫ θ @ ωmin

)

(

Computing the kinetic energy Ek needed is illustrated in the following example

Torque Time Relation without Flywheel A

typical

torque

time

relation

for

example

of

a

mechanical

punching

press without a fly wheel in shown in the figure. In

the

initially during

absence and

can

fly

intermedialty

punching

fluctuation

of

and be

wheel and

stripping

noted.

surplus

or

positive

enery

absorbtion

operations.

A

large

out

the

To

smoothen

enregy or

is

avalible

negative

energy

magitidue speed

of

speed

fluctuation

fly

wheel is to be added and the fly wheel energy needed is computed as illustrated below


Machine Design II


Torque Area +20 073

34 200

Area +15 388 D

A

A

C

B

rms

Average

7 020 0 Shaft angle time t

ω max

ω min

θ Area -26 105

-34 200

Area -9 202

0

360

Figure 3.3.2 Accumulation of Energy pulses under a Torque- Time curve From

Area= E

Accumulated sum =E

Min & max

A to B

+20 073

+20 073

ω

B to C

-26 105

-6 032

ω

C to D

+15 388

D to A

-9 202

min

@B

max

@C

+9 356 +154 Total Energy= E @ωmin- E@ωmin =(-6 032)-(+20 073)= 26 105 Nmm2

Figure 3.3.3


Machine Design II


Torque Time Relation with Flywheel

Torque Cf =0.05

8730

Average

7020

Time t

0

Shaft angleθ

360 Figure 3.3.4

Geometry of Flywheel The geometry of a flywheel may be as simple as a cylindrical disc of solid

material,

wheels

with

wheels

are

a

to

may

hub

solid

requirements changes

or

discs

and disc

and

be

of

rim

connected

of

size of

central

hollow of

circular

the hub

and to hollow wheels with multiple arms.


spoked

and

construction by

spokes

cross

flywheel peripheral

or

section.

increases rim

like

conventional

arms As

Small the

the

connected

fly

energy

geometry by

webs

Machine Design II


b

b

D D0

d

D

do

Figure 3.3.5

b

D0

d

D

a

Arm Type Flywheel Figure 3.3.6

The

latter

arrangement

is

a

more

efficient

of

material

especially

for

large flywheels, as it concentrates the bulk of its mass in the rim which is

at

the

largest

radius.

Mass

at

largest

radius

since the mass moment of inertia is proportional to mr2


contributes

much

more

Machine Design II


For a solid disc geometry with inside radius ri and out side radius ro, the mass moment of inertia I is Im = mk 2 =

m 2 2 (r + r ) 2 o i

The mass of a hollow circular disc of constant thickness t is

m=

(

)

W γ = π ro2 − ri2 t g g

Combing the two equations we can write

Im =

(

)

πγ 4 4 r −r t 2g o i

Where γ is material’s weight density The

equation

is

better

solved

by

geometric

proportions

i.e

by

upon

its

similar

to

assuming inside to out side radius ratio and radius to thickness ratio.

Stresses in Flywheel Flywheel

being

a

rotating

distributed

mass

and

attempts

disc, to

centrifugal pull

it

apart.

stresses Its

acts

effect

is

those caused by an internally pressurized cylinder

σt =

γ 2 ⎛ 3 + v ⎞⎛ 2 2 1 + 3v 2 ⎞ ω ⎜ r ⎟ ⎟⎜ ri + ro − g ⎝ 8 ⎠⎝ 3+ v ⎠

σr =

⎞ γ 2 ⎛ 3 + v ⎞ ⎛ 2 2 ri2 ro2 2 ω ⎜ + − − r r r ⎜ ⎟ o ⎟ i ⎟ g ⎝ 8 ⎠ ⎜⎝ r2 ⎠

γ = material weight density, ω= angular velocity in rad/sec. ν= Poisson’s ratio, is the radius to a point of interest, ri and ro are inside and outside radii of the solid disc flywheel. Analogous

to

a

thick

cylinder

under

internal

pressure

the

tangential

and radial stress in a solid disc flywheel as a function of its radius r is given by:


Machine Design II


Radius

σt Tang. stress

Radial stress σr

Radius The

point

of

most

maximum.

What

stress

at

that

fragments

can

Since

forces

speed

the also,

interest

causes point

failure

from

explode

the

inside

in

a

where

the

of

flywheel

for

where

is

are

a

the

typically

originated

extremely

stresses

checking

radius

fracture

resulting

causing

instead

is

and

stress the

is

a

tangential

upon

fracture

dangerous

consequences,

function

the

stresses,

the

of

maximum

rotational speed

at

which the stresses reach the critical value can be determined and safe operating factor.

speed

Generally

can some

be

calculated

means

to

or

specified

preclude

its

speed is desirable, for example like a governor. Consequently

ω F.O.S (N) = Nos = ωyield


based operation

on

a

beyond

safety this

Machine Design II


WORKED OUT EXAMPLE 1 A 2.2 kw, 960 rpm motor powers the cam driven ram of a press through a gearing of 6:1 ratio. The rated capacity of the press is 20 kN and has a stroke of 200 mm. Assuming that the cam driven ram is capable of delivering the rated load at a constant velocity during the last 15% of a constant velocity stroke. Design a suitable flywheel that can maintain a coefficient of Speed fluctuation of 0.02. Assume that the maximum diameter of the flywheel is not to exceed 0.6m.

Work done by the press=

U = 20 *103 * 0.2 * 0.15 = 600Nm

Energy absorbed= work done= 600 Nm Mean torque on the shaft:

2.2 *103 = 21.88Nm 960 2*π* 60 Energy supplied= work don per cycle = 2π * 21.88 * 6 = 825 Nm Thus the mechanical efficiency of the system is = 600 η= = 0.727 = 72% 825

There fore the fluctuation in energy is =

E k = Energy absorbed - Energy supplied


Machine Design II


600 − 825 * 0.075 ( 21.88 * 6 * π * 0.15 ) 538.125Nm Ek I=

(

Cf ωavg

)

2

538.125

=

960 ⎞ ⎛ 0.02 ⎜ 2π * ⎟ 60 ⎠ ⎝

2

= 2.6622 kg m2

(

)

π r 2 2 . r − ri .t 2 g o r Assuming i = 0.8 ro π 78500 2.6622 = * 0.304 − 0.244 t 2 9.86 = 59.805t I=

(

∴ t=

)

2 .6622 = 0.0445 59.805

or 45 mm

σt =

r 2 ⎛ 3 + γ ⎞ ⎛ 2 2 1 + 3γ 2 ⎞ r ⎟ ω ⎜ ⎟ ⎜ r + ro − g 3+ γ ⎝ 8 ⎠⎝ i ⎠

78500 2 ⎛ 3 + 0.3 ⎞ ⎛ 2 2 1.9 * 0.242 ⎞ .ω ⎜ ⎟ ⎜ 0.24 + 0.3 − ⎟ 9.81 3.3 ⎝ 8 ⎠⎝ ⎠ 960 ⎞2 ⎛ σ t = 0.543* ⎜ 2π * ⎟ 60 ⎠ ⎝ = 55667N / m 2 σt =

= 0.556MPa or if σ t = 150 MPa 150 *106 = 7961.4ω2 ( 0.4125 )( 0.0376 )( 0.090 )( 0.0331) = 0.548ω2 ω = 16544 rad / sec2 N OS =

ωyield

ω = 164.65


=

16544 32π