Unit1 lectbyvrk dynofmachinery

DYNAMICS OF MACHINERY
STATIC FORCE ANALYSIS OF MECHANISMS
Define static force analysis.
It is a force analysis in which inertia forces of the machine components are neglected.
(i.e. “ma” is neglected)
Define static equilibrium.
A body is in static equilibrium if it remains in its state of rest or of uniform motion.
Conditions to be satisfied by a body in static equilibrium are:
Vector sum of all the forces acting on the body is zero. ΣF = 0.
Vector sum of the moments of all the forces about any arbitrary point is zero. ΣM = 0.
What is the condition for a body to be in equilibrium under the action of two forces?
The TWO conditions for a body to be in equilibrium under the action of two forces are
(a) The two forces act along the same line;
(b) The two forces are equal in magnitude and opposite in direction.
1

What is the condition for a body to be in equilibrium under the action of two forces
and a torque?
The TWO conditions for a body to be in equilibrium under the action of two forces
and a torque are
(a) The two forces are equal in magnitude, parallel in direction and opposite in sense;
(b) The two forces form a couple which is equal and opposite to the applied torque.
A body is in equilibrium under the action of three forces, out of which two forces are
parallel. What would be the position of third force with respect to the other
forces?
If forces F1 and F2 are parallel, the point of concurrence lies at infinity. Therefore force
F3 is parallel to F1 and F2 .
2

What are the force conventions used in static force analysis of mechanisms?
Force exerted by member “ i” on member “ j “ is denoted by Fij
Consider four members 1,2,3, and 4 connected together as shown. F23 = F32
What are the two principles used in static force analysis?
(a) Principle of super position ; (b) Principle of virtual work.
Define principle of super position.
In linear systems, if number of forces act on a member, the net effect is effect is equal
to the superposition of the effects of individual forces taken one at a time.
Define the principle of virtual work.
If a system is in equilibrium and undergoes an infinitesimal displacement from its
equilibrium position without any lapse of time, then the net work done is equal to
zero.
3

D’Alembert’s Principle – Inertia Force – Inertia Torque
Define D’Alembert’s principle.
D’Alembert’s principle states that when a system (body) is in dynamic equilibrium, the
vector sum of all the external forces and inertia forces acting upon a system of
rigid bodies is zero.
The vector sum of all the external moments and inertia torques acting upon a system
of rigid bodies is also separately zero.
ΣF + FI = 0 ; and ΣT + TI = 0.
What do you mean by inertia force?
Inertia force is a property of matter by virtue of which a body resists any change in
linear velocity (v). FI = – m R aG where m R is mass of the reciprocating body in kg
and a G is linear acceleration of the centre of mass of the body in m/s 2
The negative sign indicates that the inertia force acts in opposite direction to that of
the acceleration. Inertia force acts through the centre of mass of the body.
4

D’Alembert’s Principle – Inertia Force – Inertia Torque
What do you mean by inertia torque?
Inertia torque is a property of matter by virtue of which a body resists any change in
angular velocity ( ω). TI = – IG α where IG is mass moment of inertia of the body
about an axis passing through the centre of mass in kg m 2 , α is angular
acceleration of the body in rad / s 2
The negative sign indicates that the inertia torque acts in opposite direction to that of
the angular acceleration.
What do you mean by equivalent offset inertia force?
Equivalent offset inertia force is a force which replaces both inertia force and inertia
torque by displacing (Offsetting) the line of action of force from the centre of mass
(G) by a perpendicular distance “ h “ such that the torque so produced is equal to
the inertia torque acting on the body.
Offset distance h = TI / FI = (– I G α) / (– m R a G ) = k2 α / a G
5

DYNAMIC FORCE ANALYSIS OF MECHANISMS
What is dynamics of machinery?
Dynamics of machinery deals with forces and their effects while acting upon the
machine parts in motion.
What is a dynamic force?
Each component has a mass and it produces a force while accelerating. This force is
known as dynamic force.
What is dynamic force analysis ?
Dynamic force analysis is the analysis of various forces acting on the components of
machine or mechanism by considering inertia forces due to accelerating parts into
account.
Why dynamic force analysis is necessary?
For high speed mechanisms, inertia forces are high and hence dynamic force analysis
is very important to determine input torque required for balancing.
6

DYNAMIC FORCE ANALYSIS IN A RECIPROCATING ENGINE
7

8

STATIC AND DYNAMIC BALANCING
9

Angular velocity of connecting rod, ω PC = ω cos θ / [ n 2 – sin 2 θ] ½
Angular acceleration of the connecting rod, α = – ω 2 sin θ [ n 2 – 1] / [ n 2 – sin 2 θ] 3/2
Crank Pin Effort (Force), FT = FQ sin (θ + φ) = [FP / cos φ] sin (θ + φ)
Crank Effort (Torque) , T = FT r = {[FP / cos φ] sin (θ + φ)} r
= FP r { sin θ +[sin 2 θ / [2 (n 2 – sin 2 θ)1/2]]}
Piston Effort [ Force on piston or cross head pin], FP =FL ± FI – RF ; where
FL is Net load on piston, FI is inertia force and RF is resistance offered.
Net load on piston, FL = P1 A1 – P2 A2
Inertia Force, FI = – m R a G = m R ω 2 r [cos θ + (cos 2θ)/n] where m R is mass of the
reciprocating body in kg and a G is linear acceleration of the centre of mass of the
body in m/s 2
10

GAS FORCES – BEARING LOADS – CRANK SHAFT TORQUE
Gas Forces produce the net load on the piston.
Net load on piston, FL = P1 A1 – P2 A2
Inertia Force, FI = m R aG where m R is mass of the reciprocating body in kg and a G
is linear acceleration of the centre of mass of the body in m/s 2
Linear acceleration, a G = ω 2 r [cos θ + (cos 2θ)/n]
where n = l/r = Length of connecting rod / radius of crank
Bearing Loads
Thrust (Force) on crank shaft bearing, FB = FQ cos (θ + φ) = [FP / cos φ] cos (θ + φ)
Normal Force on cylinder wall, FN = FP tan (φ)
Piston Effort [ Force on piston or cross head pin], FP =FL ± FI – RF ; where
FL is Net load on piston, FI is inertia force and RF is resistance offered.
Crank Shaft Torque
Crank Effort (Torque) , T = FT r = [FP / cos φ] sin (θ + φ) r
= FP r { sin θ +[sin 2 θ / [2 (n 2 – sin 2 θ)1/2]]}
11

SHAKING FORCE – SHAKING COUPLE
Engine Shaking Force
Force acting on the crank shaft bearing FB can be resolved into two components,
FB H and FB V . The horizontal component FB H acts along the line of reciprocation and it
is equal to FI .
This force is unbalanced and is called shaking force.
Engine shaking Couple
As the lines of action FN and FBV are away by a distance x = PO, it forms a couple
known as shaking couple.
Shaking Couple CU = FN x = FBV x
12

Equivalent Masses
What do you mean by equivalent dynamic system?
The motion of connecting rod is not linear. It is difficult to find its inertia. Therefore
the inertia is taken into account by considering the mass of connecting rod into
two parts acting at two points.
The system in which the entire mass of the body is replaced by two masses placed at
fixed distance apart is called equivalent dynamic system.
Conditions to be satisfied by the system to be dynamically equivalent are:
(a) Sum of the two masses must be equal to the total mass of the body.
(b) Centre of gravities of the two masses must coincide with that of the body.
(c) Sum of the mass moments of inertia of the masses about their centre of gravity
must be equal to the mass moment of inertia of the body.
13

Equivalent Masses
Let m1 and m2 be the two masses which form equivalent dynamic system (kg)
Let l1 be the distance of mass m1 from G in (m)
Let l2 be the distance of mass m2 from G in (m) on the other side.
Let L be the total distance between the masses m1and m2
The conditions are
(a) m1+ m2 = m
(b) m1l1 = m2 l2
(c) m1l1
2 + m2 l2
2 = m k2
(d) From the above conditions, we get l1 l2 = k2
15

KLEIN’S CONSTRUCTION DIAGRAM
What is the advantage of Klein’s construction ?
In Klein’s construction, velocity diagram and acceleration diagrams of the mechanism are
constructed in the configuration itself. There is no need to draw them separately.
Draw OM perpendicular to PO.
Extend PC to obtain M.
Draw a circle with CM as radius.
OCM is velocity triangle.
Draw a circle with PC as diameter.
K and L be intersection of the two circles.
Join KL. KL intersects PC at Q.
CQNO is acceleration polygon.= ω 2
16

CORRECTION COUPLE – CORRECTION TORQUE
Define correction couple. Why is it required?
To convert a two mass system in to equivalent dynamic system (without altering the
position of point masses), a couple is applied to the two mass system. The couple
which must be applied to the two mass system in the opposite direction to that of
the angular velocity to make the system dynamically equivalent to the original
system is called correction couple (ΔT) . ΔT = m ( k1
2 – k2 ) α
What is the difference between correction couple and correction torque?
Correction couple may be applied to the system by two equal and opposite forces FY
acting through the small end bearing and big end bearing of the connecting rod.
Correcting torque TC is the torque on the crank shaft due to force FY at big end
bearing.
The relationship between correction couple and correction torque is
TC = ΔT (cos θ) / n where n = l/r = length of connecting rod / crank radius
17

TURNING MOMENT DIAGRAM
What is Turning Moment diagram?
Turning moment or crank effort diagram is a graphical representation of the turning
moment or torque or crank effort for various positions of the crank. It is also called
T- θ diagram where T is the torque and θ is the crank angle.
Draw T- θ diagram for a single cylinder single acting steam engine and single cylinder
double acting steam engine.
18

Draw T- θ diagram for a single cylinder Four Stroke IC engine.
20

Draw TM diagram for 8 cylinder 4 stroke engine and turbine.
21

Why does the turning moment of a single cylinder 4 stroke IC engine transfer from
negative to positive during second half of suction and exhaust strokes?
During the suction stroke, piston moves from TDC to BDC. The pressure inside the
cylinder is less than atmosphere. No work is done. However some work is given to
the system for moving the piston from TDC to BDC. Therefore a negative loop is
obtained in the first part of this stroke. The fresh incoming charge makes waste
gases to exhaust from the cylinder. As some work is done by fresh incoming charge
(i.e. by engine) a positive loop is obtained in the second part of this stroke.
During the exhaust stroke, piston moves from BDC to TDC. The piston forces the waste
gases to go outside. Work is done on the gases. Hence negative loop is obtained in
the first part of the stroke. Some of the waste gases exhaust themselves due to
less weight and hence reduces work to be done on the engine. Hence a loop is
obtained during the second part of this stroke.
22

FLYWHEEL OF ENGINES
Define Fluctuation of speed.
Variation of speed above and below the mean speed of the flywheel is called
fluctuation of speed.
Define fluctuation of energy.
Variation of energy above and below the mean resisting torque line is called
fluctuation of energy.
Define maximum fluctuation of energy.
The difference between maximum energy and minimum energy is called maximum
fluctuation of energy.
Write the equation for maximum fluctuation of energy.
Maximum fluctuation of energy, ΔE = I ω2 CS = 2 E CS = m k2 ω2 CS
23

FLYWHEEL OF ENGINES
Define coefficient of fluctuation of speed.
It is the ratio of maximum fluctuation of speed to the mean speed.
CS = (N1 - N1) / [(N1+ N1)/2]
Define coefficient of fluctuation of energy.
It is the ratio of maximum fluctuation of energy to the work done per cycle.
CE = ΔE / W , ΔE = E max – E min
Define coefficient of steadiness.
Reciprocal of coefficient of fluctuation of speed is called coefficient of steadiness.
24

FLYWHEEL OF ENGINES
Why small flywheel is used in a multi cylinder engine?
In multi cylinder engine, total turning moment is the sum of the turning moments of
each cylinder at any instant. Therefore the fluctuations in the turning moment is
low and flywheel required is also small.
Mention the relation between peripheral velocity of flywheel rim and density of
material.
‘ v = * σ/ρ ] ½ where σ is hoop stress or tensile stress due to centrifugal
force in N/m 2 and ρ is density of rim material in kg/m3
Is work done by IC engine is positive?
In IC engine power is produced only during power stroke. Power is utilised during the
other strokes. Work done per cycle is less than work done during power stroke.
Hence net work done is always positive.
25

FLYWHEEL OF PUNCHING MACHINE
What is the function of flywheel in a punching machine?
In a punching machine, the torque on the crank shaft is constant and the load varies
during the cycle. The function of flywheel is to reduce the fluctuation of speed to a
minimum.
26

Differences between Flywheel and Governor
FLYWHEEL GOVERNOR
Flywheel does not
maintain constant
speed.
Governor regulates the
main speed when there
is a variation in load.
Flywheel reduces the
fluctuation of speed
caused by variation in
turning moment.
When load increases,
fuel supply is increased.
When load decreases
fuel supply is
decreased.
Flywheel works
continuously from cycle
to cycle.
Governor functions only
when there is a change
in load.
27

28
F
A
C
B
Given: Dimensions AB = mm ; AC = mm ; BC = mm ;
F = N;
Find the Input Torque T in Nm

29
B
F
P
A C
T
F = N; P = N;

30
F = N; P = N; Q = N;
F
B
Q
T
P
A
C

31
Given: Slider crank mechanism; Forces acting on links: FP = 4000 N;FQ = 500 N; FR = 2000 N;
Lengths of links in mm: AB = 300; BC = 455; BE = 175; AD = 155 and L CAB = 450
To find : Forces F14 ; F34 ; F23 ; F32 ;F14 ; and Driving Torque T2 .

For Torque T4
F 34 = F 14 = T4 / h 4a = 52.2 N
F 24 = F 43 = F23= F32 = F12 = 52.2 N
T2a = F 32 h 2a = 14.3 Nm (CCW)
For Torque T3
F43 = F23= 44.8 N
T2b= F 32 h 2b = 1.88 Nm (CCW)
T2 = T2a +T2b= 16.18 Nm (CCW)
Given: In a four link mechanism shown here torque T3 and T4 have magnitudes of
30 Nm and 20 Nm respectively. The link lengths are AD = 800 mm, AB = 300
mm, BC = 700 mm, CD = 400 mm. For the static equilibrium of the mechanism
determine the input torque T2.

33
Unit-1: Formulae
Parameters involved are
x, r, l, l1, θ,φ, p1, p2, A1, A2, FR, mR, mC,
k or R
, n = l / r; sin φ = [sin θ] / n;
, x = r [ (1-cos θ) + (1/2n) sin2 θ]
, v p = ω r [ sin θ + (1/2n) sin 2θ]
, a p = ω2 r [ cos θ + (1/n) cos 2θ]
, ω pc = (ωcos θ) /n) ;
‘ α pc = – (ω 2 sin θ) /n) ;
FL = [p1 (п/4)D2 ] – [p2(п/4)(D – d )2 ]
FI = mR ω2 r [ cos θ + (1/n) cos 2θ]
Fp = FL – FI ± m g – FR
FQ = Fp / cos φ ; FN = Fp tan φ ;
FB = FQ cos (θ +φ) ;
FT = FQ sin (θ +φ) ;
T = FT r
‘ θ , α, σ, ρ, v, D, A, b, t, m, ω , k or
R, T, P, N, I, E, ΔE, Δ ω , ω, W, Cs, CE .
‘ v = √(σ/ρ) = п D N/60; ω = 2 п N/60
T mean = Wcycle / θ cycle ;
θ cycle = 4 п for 4 stroke IC engine;
θ cycle = 2 п for 2stroke IC engine;
Cs = (ω1 – ω 2)/ ω ; ω = (ω1 + ω 2)/ 2
Cs = (N1 – N 2)/ N ; N = (N1 + N 2)/ 2
Cs = (v1 – v2)/ v ; v = (v1 + v2)/ 2
P = Tmean ω
Wcycle = 60 P /(N/2) for 4 stroke cycle
Wcycle = 60 P / N for 2 stroke cycle
CE = ΔE/ Wcycle
‘ ΔE = m v2 Cs = 2 E Cs = m k2 ω2 Cs
= I ω2 Cs
I α = (T – T mean) = m k2 α

34
Unit-1: Formulae:
x, r, l, l1, θ,φ, p1, p2, A1, A2, FR, mR, mC,
k or R
FL = [p1 (п/4)D2 ] – [p2(п/4)(D – d )2 ]
FI = mR ω2 r [ cos θ + (1/n) cos 2θ]
Fp = FL – FI ± m g – FR
FQ = Fp / cos φ ; FN = Fp tan φ ;
FB = FQ cos (θ +φ) ;
FT = FQ sin (θ +φ) ;
T = FT r

35
Problem1: Given Crank radius r = 0.3 m
Connecting rod length l = 1 m
Speed of engine N = 200 rpm
Find (a) Crank angle θ at which maximum velocity occurs
(b) Maximum velocity (c ) Derive the equation

36
Problem1: Given Crank radius r = 0.3 m
Connecting rod length l = 1 m
Speed of engine N = 200 rpm
Find (a) Crank angle θ at which maximum velocity occurs
(b) Maximum velocity

37
Problem 2: Given: Horizontal steam engine; Length of connecting rod l = 1.2 m
Crank radius r = 0.3 m; Crank angle θ = 30 o
Acceleration of piston a P = 35 m/s 2 ; Frictional resistance F R = 550 N
Net effective pressure P1 – P1 = 5 x 10 5 N/m 2
Diameter of the piston D = 0.3 m ; Mass of reciprocating parts m R = 160 kg
Find (a) Reaction on the cross head guide F R ; (b) Thrust on crank shaft bearing F B
(c ) Torque on crank shaft T

38
Problem 3: Given: Horizontal steam engine
Length of connecting rod l = 1 m
Crank radius r = 0.2 m
Crank angle θ = 30 o
Speed of the engine N = 400 rpm
Net effective pressure P1 – P2 = 0.4 N/mm 2
Diameter of the piston D = 0.4 m
Mass of reciprocating parts m R = 100 kg
Find (a)Inertia force; (b) Gas Force on piston;
(c ) Piston effort; (d) Thrust on cylinder wall;
(e) Thrust on connecting rod
(f) Crank pin effort and (g) Crank effort
Answers: ω = 41.89 r/s; n = 5; sin φ = 0.1; φ =
5.740;
FI = 33.5 kN; FL= 50.26 kN kN; FP = 16.36 kN;
FQ = 16.44 kN; FN = 1.64 kN;
FT = 9.6 kN; T = 1.921 kNm

39
Problem 4: Given: Horizontal steam engine
Length of connecting rod l = 1.2 m
Crank radius r = 0.3 m
Crank angle θ = 60 o
Speed of the engine N = 250 rpm
Net effective pressure P1 – P2 = 0.35 N/mm 2
Diameter of the piston D = 0.5 m
Mass of reciprocating parts m R = 250 kg
Find (a)Inertia force; (b) Gas Force on piston;
(c ) Piston effort; (d) Thrust on cylinder wall;
(e) Thrust on connecting rod
(f) Crank pin effort and (g) Crank effort
Answers:
‘ ω = r/s; n = 4; sin φ = ; φ = 0;
FI = kN; FL= kN kN; FP = kN;
FQ = kN; FN = kN;
FT = kN; T = kNm

40
Problem 5:
Given: Vertical double acting steam engine
Cylinder bore diameter D = 0.3 m
Stroke length L = 0.45 m
Speed N = 200 rpm
Mass of reciprocating parts mR = 225 kg
Piston rod diameter d = 0.05 m
Connecting rod length l = 1.2 m
Crank angle θ = 180+125 =305o
Steam pressure above piston P1 = 30 x 10 3 N/m 2
Steam pressure below piston P2 = 1.5 x 10 3 N/m 2
Find (a) Crank pin effort;
(b) Effective turning moment on crank shaft

42
Problem 6: Given Suspended connecting rod; h = 0.65 m; l1 = 0.625 m; m = 37.5 kg; t p = 1.87s;
Find the masses at small end and big end m1 and m2

43
Problem 7: Given: Petrol engine
Speed N = 1800 rpm; Crank radius r = 0.05 m; Connecting rod length l = 0.2 m;
Diameter of piston D = 0.08 m; Mass of reciprocating parts mR = 1 kg;
Pressure on the piston P1 = 0.7 N/mm 2
Displacement from IDC = 0.01 m
Find (a) net load on gudgeon pin FP; (b) Thrust on connecting rod FQ; Reaction
between piston and cylinder FN; Engine speed at which the above values become
zero (i.e. FI =FL )

44
Problem 1: Given : Single cylinder single acting four stroke engine
Power developed P = 20 kW; Speed N = 200 rpm; Flywheel mass m = 1000 kg;
Radius of gyration k = 0.6 m;
Work done in power stroke = 3 times work done on compression stroke;
Find Coefficient of fluctuation of speed C S and coefficient of fluctuation of energy C E
Answers: T mean = P/ω = 954.93 Nm
Work done / cycle = T mean x θ cycle= 60 P / (N/2) = 12 kNm
Work done during expansion stroke = 3 (Work done during compression stroke)
(i.e.) (п/2 ) T max= 3 (п/2 ) T min
Net work done / cycle = (п/2 ) T max – 3 (п/2 ) T min = 12
Hence T max = 11459.16 Nm; C E = ΔE / W; ΔE = (1/2) (T max – T min )2 п / T max
ΔE = I ω 2 C S = m k 2 ω 2 C S = 15125 Nm; C S = 0.096

46
Problem 1: Given : Single cylinder single acting four stroke engine
Power developed P = 20 kW; Speed N = 200 rpm; Flywheel mass m = 1000 kg;
Radius of gyration k = 0.6 m;
Work done in power stroke = 3 times work done on compression stroke;
Find Coefficient of fluctuation of speed C S and coefficient of fluctuation of energy C E
Answers: T mean = P/ω = 954.93 Nm
Work done / cycle = T mean x θ cycle= 60 P / (N/2) = 12 kNm
Work done during expansion stroke = 3 (Work done during compression stroke)
(i.e.) (п/2 ) T max= 3 (п/2 ) T min
Net work done / cycle = (п/2 ) T max – 3 (п/2 ) T min = 12
Hence T max = 11459.16 Nm; C E = ΔE / W; ΔE = (1/2) (T max – T min )2 п / T max
ΔE = I ω 2 C S = m k 2 ω 2 C S = 15125 Nm; C S = 0.096

48
Problem 2: Given : Multi cylinder engine ; Scale: Y axis 1mm = 325 Nm; X axis 1 mm = 3 0
Areas above the mean torque line in mm2 in order are
26, +378, - 256, + 306, - 302, + 244, - 380, +261, - 225
Engine speed N = 800 rpm; Total fluctuation of speed ±1.6% of mean speed.
Radius of flywheel R = 0.7 m; Find Mass of the flywheel m

49
Problem 3: Given: Multi cylinder engine; Scale: Y axis 1mm = 500 Nm; X axis 1 mm = 6 0
Intercepted areas between output torque curve and mean resistance in mm2 in order are
-30, +410, - 280, + 320, - 330, + 250, - 360, +280, - 260
Engine speed N = 800 rpm; Stroke length L = 0.3 m
Total fluctuation of speed ±2% of mean speed.
Safe centrifugal stress σ = 7 N/mm2 ; Density of flywheel material ρ = 7200 kg/m3
Find the diameter of flywheel (Radius of flywheel R)

Problem 4: Given: 3 cylinder single acting engine ; Angle between cranks θ = 120 0
Engine speed N = 600 rpm; Turning moment diagram for each cycle is a triangle
In power stroke, T max = 90 Nm at 60 0 ; Torque on return stroke = 0
Mass of flywheel m = 12 kg; Radius of gyration of flywheel R = 0.08 m
Find (a) Power developed P; (b) Coefficient of fluctuation of speed C S ;
(c ) Coefficient of fluctuation of speed C E ; (d) Maximum angular acceleration of flywheel αmax
ω = 2 п N/60 = 62.832 r/s ;
Wcycle =3 (1/2)п 90 = 424.115 Nm
T mean = Wcycle / θ cycle = 424.115 / 2п = 67.5 Nm
P = Tmean ω = 67.5 x 62.832 = 4241 W
Energy at B = E – (1/2) (п/6)(67.5-45) = E – 5.89
Enery at
A B C D E F G H
E E -5.89 E +5.89 E -5.89 E +5.89 E -5.89 E +5.89 E
ΔE = E +5.89 – (E -5.89) = 11.78 Nm
ΔE = m v2 Cs = 2 E Cs = I ω2 Cs = m k2 ω2 Cs
11.78 = 12 x 0.082 x 62.832 2 Cs ; Solving Cs = 0.04
CE = ΔE/ Wcycle = 11.78 / 424.115 = 0.028
Maximum angular acceleration of flywheel,
I α max = m k2 α max = (T max– T mean)
12 0.08 2 α max = 90 – 67.5 ;
Solving α max = 292 rad/s

51
Problem 5: Given: Two stroke engine
Torque developed T = 1000 + 300 sin 2θ – 500 cos 2θ
Engine speed N = 250 rpm
Mass of flywheel m = 400kg
Radius of gyration of flywheel R = 0.4 m
Find (a) Power developed P; (b) Total percentage of fluctuation of speed ;
(c ) angular acceleration of flywheel α at θ = 60 0 ;
(d) Maximum angular acceleration of flywheel αmax
Answers: Wcycle = 1000 п Nm;
T mean = Wcycle / θ cycle = 1000 Nm
P = Tmean ω = 26180 W
T = T mean =at θB = 29.5 0 and θC = = 119.5 0
ΔE = 583.08 Nm = I ω2 Cs
Cs = 0.0133;
I α = (T – T mean) = m k2 α
At θ = 60 0 , T = ; α = 7.965 rad/s2
Tmax occurs at 2θ = 149.04 0 , 329.04 0
I α = (Tmax – T mean) = m k2 α ; αmax = 9.11 rad/s2

53
Problem 6: Given: Electric motor drives a punching machine
Maximum speed of flywheel N1 = 250 rpm
Minimum speed of flywheel N2 = 230 rpm
Radius of gyration of flywheel k = 0.5 m
Number of holes per hour = 800
Time for punching t = 1.5 sec/hole
Energy required for punching E1 = 12000 Nm
Find (a) Motor power P; (b) Fluctuation of energy ΔE; (c ) Mass of the flywheel m
Motor power P = energy required / sec =
= Energy required for punchinga hole x No. of holes punched/hr
= 12000 x 800 /(3600 x 1000)= 2.67 kW
Energy supplied during punching operation E2 = P t = 2.667 x 1.5 = 4 000 Nm
Fluctuation of energy of flywheel ΔE = 12000 – 4000 = 8000 Nm
We know ΔE = (1/2) I (ω1
2 – ω2
2) = (1/2) m k 2 (ω1
2 – ω2
2)
Solving, m = 607.93 kg

54
Problem 7: Given: Electric motor drives a punching machine
Maximum speed of flywheel N1 = 225 rpm
Minimum speed of flywheel N2 = 200 rpm
Radius of gyration of flywheel k = 0.5 m
Number of holes per hour = 720
Time for punching t = 2 sec/hole
Torque required for punching E1 = 15 kNm
Find (a) Motor power P; (b) Fluctuation of energy ΔE; (c ) Mass of the flywheel m
Motor power P = energy required / sec =
= Energy required for punchinga hole x No. of holes punched/hr
= 15000 x 720/(3600 x 1000)= 3 kW
Energy supplied during punching operation E2 = P t = 3000 x 2 = 6000 Nm
Fluctuation of energy of flywheel ΔE = 15000 – 6000 = 9000 Nm
We know ΔE = (1/2) I (ω1
2 – ω2
2) = (1/2) m k 2 (ω1
2 – ω2
2)
Solving, m = kg

Unit1 lectbyvrk dynofmachinery

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