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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
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Turning Moment Diagram for a Single
Cylinder Double Acting Steam Engine
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Turning Moment Diagram for a Single
Cylinder Double Acting Steam Engine
Turning moment on crankshaft,
FP = Piston effort
Out stroke = curve abc
In stroke = curve cde
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Turning Moment Diagram for a Single
Cylinder Double Acting Steam Engine
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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Turning Moment Diagram for a Single
Cylinder Double Acting Steam Engine
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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Turning Moment Diagram for a Single
Cylinder Double Acting Steam Engine
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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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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Fluctuation of Energy, Fig. 16.1
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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Fluctuation of Energy, Fig. 16.1
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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Fluctuation of Energy, Fig. 16.1
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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Fluctuation of Energy, Fig. 16.1
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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Determination of Max Fluctuation
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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Determination of Max Fluctuation
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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Determination of Max Fluctuation
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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Coefficient of Fluctuation of Energy
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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Coefficient of Fluctuation of Energy
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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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Coefficient of Fluctuation of Speed
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Coefficient of Fluctuation of Speed
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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Energy Stored in a Flywheel
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Energy Stored in a Flywheel
Mean kinetic energy of flywheel
Maximum fluctuation of energy
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Energy Stored in a Flywheel
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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