Turning Moment Diagrams and Flywheel - site.iugaza.edu.pssite.iugaza.edu.ps/mhaiba/files/2013/09/Ch16-Turning-Moment... · Chapter 16 Turning Moment Diagrams and Flywheel 10/29/2018

Chapter 16

Turning Moment

Diagrams and

Flywheel

10/29/2018

Mohammad Suliman Abuhaiba, Ph.D., PE1

Turning moment diagram (TMD)

graphical representation of turning

moment or crank-effort for various

positions of the crank

10/29/2018

Mohammad Suliman Abuhaiba, Ph.D., PE

2

Turning Moment Diagram for a Single

Cylinder Double Acting Steam Engine

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../Videos/Chapter 16/How Two-stroke Engines Work.mp4




Turning moment on crankshaft,

FP = Piston effort

Out stroke = curve abc

In stroke = curve cde

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Work done = turning moment × angle turned

work done = area of turning moment diagram

per rev

In actual practice, engine is assumed to work

against the mean resisting torque

Area of rectangle aAFe is proportional to work

done against the mean resisting torque

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When turning moment is positive (engine

torque is more than mean resisting torque) as

shown between points B & C or D & E,

crankshaft accelerates and work is done by

steam.

When turning moment is negative (engine

torque is less than mean resisting torque) as

shown between points C & D, crankshaft

retards and work is done on steam.

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T = Torque on crankshaft at any instant

Tmean = Mean resisting torque

Accelerating torque on rotating parts of

engine = T – Tmean

If (T –Tmean) is positive, flywheel accelerates

If (T – Tmean) is negative, flywheel retards

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Turning Moment Diagram - 4 Stroke Cycle ICE

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../../Videos/Chapter 16/How does a 4 stroke engine work_ What are the 4 strokes_.mp4

../../Videos/Chapter 16/How does a 4 stroke engine work_ What are the 4 strokes_.mp4

../../Videos/Chapter 16/Four Stroke Engine How it Works.mp4

Turning Moment Diagram for a 4

Stroke Cycle ICE

Pressure inside engine cylinder is less than

atmospheric pressure during suction stroke, therefore

a negative loop is formed.

During compression stroke, work is done on gases,

therefore a higher negative loop is obtained.

During expansion or working stroke, fuel burns & gases

expand, therefore a large positive loop is obtained. In

this stroke, work is done by gases.

During exhaust stroke, work is done on gases,

therefore a negative loop is formed.

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Fluctuation of Energy

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Fluctuation of Energy

Fig. 16.1: TMD for a single cylinder double

acting steam engine

When crank moves from a to p,

work done by engine = area aBp

energy required = area aABp

Engine has done less work (= area aAB) than

required

This amount of energy is taken from flywheel

and hence speed of flywheel decreases

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Fluctuation of Energy, Fig. 16.1

From p to q,

work done by engine = area pBbCq

requirement of energy = area pBCq

Engine has done more work than

requirement

Excess work (area BbC) is stored in flywheel

and hence speed of flywheel increases

while crank moves from p to q

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When crank moves from q to r, more work is

taken from engine than is developed

Loss of work = area CcD

To supply this loss, flywheel gives up some

of its energy and thus speed decreases

while crank moves from q to r.

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As crank moves from r to s, excess energy is

again developed given by area DdE and

speed again increases.

As piston moves from s to e, again there is a

loss of work and speed decreases.

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Fluctuations of energy: variations of energy

above & below mean resisting torque line

Areas BbC, CcD, DdE, etc. represent

fluctuations of energy.

Engine has max speed either at q or at s.

Flywheel absorbs energy while crank moves

from p to q & from r to s.

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Engine has min speed either at p or at r.

Flywheel gives out some of its energy when

crank moves from a to p and q to r.

Max fluctuation of energy: difference

between max & min energies

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Determination of Max Fluctuation

of Energy

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of Energy

Fig. 16.4: TMD for a multi-cylinder engine

Line AG: mean torque line

a1, a3, a5 = areas above mean torque line

a2, a4, a6 = areas below mean torque line

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of Energy

Let energy in flywheel at A = E

Energy at B = E + a1

Energy at C = E + a1– a2

Energy at D = E + a1 – a2 + a3

Energy at E = E + a1 – a2 + a3 – a4

Energy at F = E + a1 – a2 + a3 – a4 + a5

Energy at G = E + a1 – a2 + a3 – a4 + a5 – a6 =

Energy at A (cycle repeats)

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of Energy

Suppose that greatest of these energies is at B

and least at E

Max energy in flywheel = E + a1

Min energy in flywheel = E + a1 – a2 + a3 – a4

Max fluctuation of energy, E = Max energy –

Min energy = (E + a1) – (E + a1 – a2 + a3 – a4) =

a2 – a3 + a4

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Coefficient of Fluctuation of Energy

q = Angle turned in one rev

q = 2p in case of steam engine and two

stroke ICE

q = 4p in case of four stroke ICE

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n = Number of working strokes per

minute

n = N, in case of steam engines & two

stroke ICE

n = N /2, in case of four stroke ICE

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Type of engine CE

Single cylinder, double acting steam engine 0.21

Cross-compound steam engine 0.096

Single cylinder, single acting, four stroke gas engine 1.93

Four cylinders, single acting, four stroke gas engine 0.066

Six cylinders, single acting, four stroke gas engine 0.031

Table 16.1: CE for steam and ICEs

Flywheel

A flywheel serves as a reservoir

It stores energy during period when supply of

energy is more than requirement

It releases it during period when

requirement of energy is more than supply

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Flywheel

In case of steam engines, ICEs,

reciprocating compressors and pumps,

energy is developed during one stroke and

engine is to run for the whole cycle on

energy produced during this one stroke.

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Flywheel

In four stroke ICE, energy is developed only

during expansion stroke which is much more

than engine load and no energy is being

developed during suction, compression and

exhaust strokes.

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Flywheel

Excess energy developed during power stroke

is absorbed by flywheel and releases it to

crankshaft during other strokes in which no

energy is developed, thus rotating crankshaft

at a uniform speed.

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Flywheel

When flywheel absorbs energy, its speed

increases and when it releases energy,

speed decreases.

A flywheel does not maintain a constant

speed

It reduces fluctuation of speed

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Flywheel

A flywheel controls speed variations caused

by fluctuation of engine turning moment

during each cycle of operation.

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Flywheel

In machines where operation is

intermittent like crushers, flywheel stores

energy from power source during the

greater portion of the operating cycle and

gives it up during a small period of the

cycle.

Energy from power source to machines is

supplied practically at a constant rate

throughout the operation.

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Governor

Governor in engine regulates mean speed of

engine when there are variations in load, e.g.,

when load on engine increases, it becomes

necessary to increase supply of working fluid.

When load decreases, less working fluid is

required.

Governor automatically controls supply of

working fluid to engine with varying load

condition and keeps mean speed of engine

within certain limits.

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Flywheel

Flywheel does not maintain a constant

speed, it simply reduces fluctuation of

speed.

It does not control speed variations caused

by varying load.

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Coefficient of Fluctuation of Speed

Max fluctuation of speed: difference

between max & min speeds during a

cycle

Coefficient of fluctuation of speed: ratio

of max fluctuation of speed to mean

speed

N1 & N2 = Max & min speeds in rpm

N = (N1 + N2)/2

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Cs is a limiting factor in design of flywheel

It varies depending upon nature of service

to which flywheel is employed.

Coefficient of steadiness = reciprocal of Cs

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Energy Stored in a Flywheel

When a flywheel absorbs energy, its

speed increases and when it gives up

energy, its speed decreases.

m = Mass of flywheel

k = Radius of gyration of flywheel

I = Mass moment of inertia of flywheel

about its axis of rotation = m.k2

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Mean kinetic energy of flywheel

Maximum fluctuation of energy

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Radius of gyration (k) may be taken equal to

mean radius of rim (R),

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Example 16.1

The mass of flywheel of an engine is 6.5 tons and

radius of gyration is 1.8 m. It is found from turning

moment diagram that fluctuation of energy is 56

kN.m. If mean speed of engine is 120 rpm, find

max & min speeds.

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Example 16.2

Flywheel of a steam engine has a radius of gyration

of 1 m and mass 2500 kg. The starting torque of

steam engine is 1500 N.m and may be assumed

constant. Determine:

1. Angular acceleration of flywheel

2. Kinetic energy of flywheel after 10 seconds

from start

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Example 16.3

A horizontal cross compound steam engine

develops 300 kW at 90 rpm. The coefficient of

fluctuation of energy as found from turning

moment diagram is to be 0.1 and fluctuation of

speed is to be kept within ± 0.5% of mean speed.

Find weight of flywheel required, if radius of

gyration is 2 m.

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Example 16.4

Turning moment diagram for a petrol engine is

drawn to following scales: Turning moment, 1 mm

= 5 N.m; crank angle, 1 mm = 1°. Turning moment

diagram repeats itself at every half revolution of

engine and areas above and below mean turning

moment line taken in order are 295, 685, 40, 340,

960, 270 mm2. Rotating parts are equivalent to a

mass of 36 kg at a radius of gyration of 150 mm.

Determine coefficient of fluctuation of speed when

engine runs at 1800 rpm.

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Example 16.4

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Example 16.5

TMD for a mult-icylinder engine has been drawn to

a scale 1 mm = 600 N.m vertically and 1 mm = 3°

horizontally. The intercepted areas between output

torque curve and mean resistance line, taken in

order from one end, are as follows : + 52, – 124, +

92, – 140, + 85, – 72 and + 107 mm2, when

engine is running at a speed of 600 rpm. If total

fluctuation of speed is not to exceed ± 1.5% of the

mean, find the necessary mass of flywheel of

radius 0.5 m.

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Example 16.5

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Example 16.6

A shaft fitted with a flywheel rotates at 250 rpm and

drives a machine. The torque of machine varies in a cyclic

manner over a period of 3 revolutions as shown.

Determine the power required to drive the machine and

percentage fluctuation in speed, if driving torque applied

to shaft is constant and the mass of flywheel is 500 kg

with radius of gyration of 600 mm.

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Example 16.610/29/2018


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Example 16.7

During forward stroke of the piston of double acting steam

engine, turning moment has max value of 2000 N.m when

crank makes an angle of 80° with IDC. During backward

stroke, max turning moment is 1500 N.m when crank

makes an angle of 80° with ODC. Turning moment diagram

for the engine is shown. If crank makes 100 rpm and

radius of gyration of flywheel is 1.75 m while keeping the

speed within ± 0.75% of mean speed, find

1. the coefficient of fluctuation of energy

2. mass of flywheel

3. crank angle at which speed has its min & max values

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Example 16.7

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Example 16.8

A three cylinder single acting engine has its cranks set

equally at 120° and it runs at 600 rpm. Torque-crank angle

diagram for each cycle is a triangle for the power stroke

with max torque of 90 N.m at 60° from IDC of

corresponding crank. Torque on return stroke is sensibly

zero. Determine:

1. power developed

2. coefficient of fluctuation of speed, if mass of flywheel is

12 kg and has a radius of gyration of 80 mm

3. coefficient of fluctuation of energy

4. maximum angular acceleration of flywheel

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Example 16.8

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Example 16.8

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Example 16.9

A single cylinder, single acting, four stroke gas

engine develops 20 kW at 300 rpm. Work done by

gases during expansion stroke is three times work

done on gases during compression stroke, work

done during suction and exhaust strokes being

negligible. If total fluctuation of speed is not to

exceed ± 2 % of mean speed and TMD during

compression and expansion is assumed to be

triangular in shape, find moment of inertia of

flywheel.

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Example 16.9

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Example 16.10

TMD for a 4-stroke gas engine may be assumed to

be represented by four triangles, areas of which

from line of zero pressure are as follows: Suction

stroke = 0.45 × 10–3 m2; Compression stroke =

1.7 × 10–3 m2; Expansion stroke = 6.8 × 10–3 m2;

Exhaust stroke = 0.65 × 10–3 m2. Each m2 of area

represents 3 MN.m of energy. Assuming resisting

torque to be uniform, find mass of rim of a

flywheel required to keep speed between 202 &

198 rpm. Mean radius of rim is 1.2 m.

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Example 16.1010/29/2018


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Example 16.11Turning moment curve for an engine is represented

by the equation, T = (20000 + 9500 sin 2q – 5700

cos 2q ) N.m, where q is angle moved by crank from

IDC. If resisting torque is constant, find:

1. Power developed by engine

2. Moment of inertia of flywheel in kg-m2, if total

fluctuation of speed is not exceed 1% of mean

speed which is 180 rpm

3. Angular acceleration of flywheel when crank

has turned through 45° from IDC

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Example 16.11

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Example 16.12A certain machine requires a torque of (5000 + 500 sinq )

N.m to drive it, where q is angle of rotation of shaft

measured from certain datum. The machine is directly

coupled to an engine which produces a torque of (5000 +

600 sin 2q) N.m. The flywheel and other rotating parts

attached to engine has a mass of 500 kg at a radius of

gyration of 0.4 m. If mean speed is 150 rpm, find:

1. fluctuation of energy

2. total percentage fluctuation of speed

3. maximum and minimum angular acceleration of the

flywheel and the corresponding shaft position.

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Example 16.12

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Example 16.13

The equation of turning moment curve of a three

crank engine is (5000 + 1500 sin 3q ) N.m, where

q is crank angle in radians. The moment of inertia

of flywheel is 1000 kg.m2 and mean speed is 300

rpm. Calculate:

1. power of the engine

2. max fluctuation of speed of flywheel in

percentage when

resisting torque is constant

resisting torque is (5000 + 600 sinq ) N.m.

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Example 16.13

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Dimensions of Flywheel Rim D = Mean diameter of rim

R = Mean radius of rim

A = x-sectional area of rim

r = Density of rim material

N = Speed of flywheel, rpm

w = Speed of flywheel, rad/s

v = Linear velocity at mean

radius = w R

s = hoop stress due to

centrifugal force

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Dimensions of Flywheel Rim

Volume of small element = A × R.dq

Total vertical upward force tending to burst the

rim across diameter X Y

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Dimensions of Flywheel Rim

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Example 16.14 TMD for a multi-cylinder engine has been drawn to a scale

of 1 mm to 500 N.m torque and 1 mm to 6° of crank

displacement. The engine is running at 800 rpm. The

engine has a stroke of 300 mm and fluctuation of speed

is not to exceed ± 2% of mean speed. Determine a

suitable diameter and cross-section of flywheel rim for a

limiting value of the safe centrifugal stress of 7 MPa. The

material density may be assumed as 7200 kg/m3. The

width of rim is to be 5 times thickness.

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Example 16.14

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Example 16.15

A single cylinder double acting steam engine

develops 150 kW at a mean speed of 80 rpm. The

coefficient of fluctuation of energy is 0.1 and

fluctuation of speed is ± 2% of mean speed. If

mean diameter of flywheel rim is 2 m and the

hub and spokes provide 5% of the rotational

inertia of flywheel, find mass and cross-sectional

area of flywheel rim. Assume density of flywheel

material (cast iron) as 7200 kg/m3.

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Example 16.16

A multi-cylinder engine is to run at a speed of 600

rpm. On drawing the turning moment diagram to a

scale of 1 mm = 250 N.m and 1 mm = 3°. The speed

is to be kept within ± 1% of mean speed of engine.

Calculate necessary moment of inertia of flywheel.

Determine suitable dimensions of a rectangular

flywheel rim if breadth is twice its thickness. The

density of cast iron is 7250 kg/m3 and its hoop stress

is 6 MPa. Assume that rim contributes 92% of

flywheel effect.

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Example 16.16

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Example 16.17TMD of a four stroke engine may be assumed to be represented

by four triangles in each stroke. Areas of these triangles are as

follows: Suction stroke = 5 × 10–5 m2; Compression stroke = 21

× 10–5 m2; Expansion stroke = 85 × 10–5 m2; Exhaust stroke =

8 × 10-5 m2. All areas excepting expansion stroke are negative.

Each m2 of area represents 14 MN.m of work. Assuming

resisting torque to be constant, determine moment of inertia of

flywheel to keep the speed between 98 & 102 rpm. Also find

size of a rim-type flywheel based on min material criterion,

given that density of flywheel material is 8150 kg/m3;

allowable tensile stress of flywheel material is 7.5 MPa. The rim

cross-section is rectangular, one side being four times the

length of the other.

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Example 16.1710/29/2018


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Example 16.18

An otto cycle engine develops 50 kW at 150 rpm

with 75 explosions per minute. The change of

speed from commencement to end of power

stroke must not exceed 0.5% of mean on either

side. Find mean diameter of flywheel and a

suitable rim cross section having width four times

depth so that hoop stress does not exceed 4 MPa.

Assume that flywheel stores 16/15 times the

energy stored by rim and work done during power

stroke is 1.40 times work done during the cycle.

Density of rim material is 7200 kg/m3.

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Example 16.1810/29/2018


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Documents

Turning Moment Diagrams and Flywheel - site.iugaza.edu.pssite.iugaza.edu.ps/mhaiba/files/2013/09/Ch16-Turning-Moment... · Chapter 16 Turning Moment Diagrams and Flywheel 10/29/2018