2. The turning moment diagram (also known as
crank- effort diagram) is the graphical
representation of the turning moment or crank-
effort for various positions of the crank. It is
plotted on Cartesian co-ordinates, in which the
turning moment is taken as the ordinate and
crank angle as abscissa.
3.
4.
5.
6. The fluctuations above and below the mean
torque line in a turning moment diagram is known
as coefficient of fluctuation of energy.
7. Let energy at A = E
Therefore,
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
8. Ce = Max. Fluctuation of energy
Where,
W.D. = Tmean * ɵ
ɵ = 2𝜋 for steam and 2 stroke IC engine
ɵ = 4𝜋 for 4 stroke IC engine
Tmean =
𝑃∗60
2𝜋𝑁
=
𝑃
𝜔
Work done per cycle
9. A flywheel is a mechanical device specifically
designed to efficiently store rotational
energy when the energy supplied is more
than the requirement and to supply it when
there is a deficiency in the energy
requirement.
Applications : IC engine, punching press,
rivetting machine, etc.
10. • Disk type flywheel :
Radius of gyration (k) = r/√2
• Rim type flywheel :
Radius of gyration (k) = r
11. Consider a rim of flywheel as show besides:
Let,
D = Mean diameter of rim in m,
R = Mean radius of rim in m,
A = Cross-section area of rim in m,
ρ = Density of rim material in m3/kg,
N = Speed of flywheel in r.p.m.,
ω = Angular velocity of flywheel in rad.s-1,
v = Liner velocity at the mean radius in ms-1,
σ = Tensile stress or hoop stress in m,
12. Consider a small element of rim as shown in fig. Let it subtend
an angle δƟ at the centre of the flywheel.
Volume of the small element = A*RδƟ
Therefore, mass of the small element,
= dm
= Density * volume
= ρ .A.R.δƟ
And centrifugal force on the element ,acting radially outwards,
= dF
= dm.ω2R
= ρ.A.ω2.R2.δƟ
13. Vertical component of dF
= dF.sinƟ
= ρ.A.ω2.R2.δƟ. sinƟ
Therefore, vertical upward force (P) tending to
burst he rim across diameter XY.
= ρ.A.ω2.R2
0
𝜋
sinƟ. 𝑑𝜃
= ρ.A.ω2.R2 − cos Ɵ 0
𝜋
= 2ρ.A.ω2.R2 . . . (1)
This vertical force will produce tensile stress or hoop stress and
it is resisted by 2P, such that
2P = 2σA . . . (2)
X Y
14. Equating equations (1) and (2) ,
2ρ.A.ω2.R2 = 2σA
σ = ρ. ω2.R2
= ρ.v2
Therefore, v =
σ
ρ
. . . (3)
We know that mass of rim,
m = Volume * density = 𝜋.D.A.ρ
A =
𝑚
𝜋.𝐷.ρ
. . . (4)
From (3) and (4) we can find out the value of mean radius and
cross – section area of the rim.
15. Cs =
𝑁1
−𝑁2
𝑁
=
𝜔1
− 𝜔2
𝜔
=
𝑣1
−𝑣2
𝑣
Where ,
N = mean speed =
𝑁1
+ 𝑁2
2
N1 = Max. speed during the whole cycle
N2 = Min. speed during the whole cycle
v = mean speed =
𝑣1
+𝑣2
2
v1 = Max. linear speed during the whole cycle
v2 = Min. Linear speed during the whole cycle
𝜔 = mean speed =
𝜔1
+ 𝜔2
2
𝜔1 = Max. angular speed during the whole cycle
𝜔2 = Min. angular speed during the whole cycle
16. Let, m = mass of flywheel in kg,
k = radius of gyration of flywheel in m,
I = moment of inertia of flywheel about
its axis of rotation in kgm-3
N1 & N = max. and min speeds during
the cycle in r.p.m.,
𝜔1 & 𝜔2= max. and min. angular speeds during
the cycle in rad/s,
N = mean speed during the cycle,
𝜔 = mean angular velocity during the cycle,
Cs = coefficient of fluctuation of speed.
17. We know that the mean kinetic energy of the flywheel,
E =
1
2
* I.𝜔2 =
1
2
* m.k2.𝜔2 (in N-m)
As the speed of the flywheel changes from 𝜔1 to 𝜔2, the max.
fluctuation of energy,
ΔE = Max. K.E. – Min. K.E.
=
1
2
* I.(𝜔1)2 -
1
2
* I.(𝜔2)2 =
1
2
* I[(𝜔1)2 - (𝜔2)2 ]
=
1
2
* I (𝜔1 + 𝜔2)(𝜔1 − 𝜔2) = I. 𝜔 𝜔1 − 𝜔2
= I. 𝜔2
(𝜔1
−𝜔2
)
𝜔
= I. 𝜔2.Cs
= m. k2. 𝜔2.Cs
= 2.E. Cs (in N-m)
18. What is a punching press ?
A punch press is a type of machine press used
to cut holes in material.
19. Let E1 be the energy required for punching a hole. This
energy depends on the size of the hole punched, the
thickness of the material and the physical properties of
the material.
Let, d1 = diameter of the hole punched
t1 = thickness of the plate,
τ = ultimate shear stress for the plate material.
20. Max. force required for punching,
Fs = Area sheared * Ultimate shear stress
= π.d1.t1.τu
it is assumed that as the hole is punched, the shear force
decreases uniformly from maximum value to zero.
Therefore, Work done or energy required for punching a
hole,
E1 =
1
2
* Fs * t
Assuming one punching operation per revolution, the
energy supplied to the shaft should be equal to E1. The
energy supplied by the motor to the crank shaft during
actual punching operation,
21. E2 = E1
𝜃2
− 𝜃1
2𝜋
Therefore, balance energy required for punching,
ΔE = E1 - E2
= E1 – E1
𝜃2
− 𝜃1
2𝜋
The energy is to be supplied by the flywheel by the decrease in its Kinetic
energy when its speed fall from max. to min.
ΔE = E1 – E2 = E1 1 −
𝜃2
− 𝜃1
2𝜋
𝜃2 and 𝜃1 can be found using the following relation :
𝜃2
− 𝜃1
2𝜋
=
𝑡
2𝑠
=
𝑡
4𝑟
Where, t = thickness of material to be punched.
s = stroke of the punch.
= 2.r
