Flywheel

Flywheel

FLYWHEEL a rotating member that acts as a storage reservoir for energy

when work is not “consumed” at as fast a rate as the power
is supplied. (Faires 1969)

a rotating energy reservoir which absorbs energy from a

power source during a portion of the operating cycle and
delivers that stored energy as useful work during the other
portion of the cycle. (Tordillo 1998)

Figure 9 - 1. Cast iron Flywheel

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Applications

Flywheels are used in punch presses and shears, internal combustion engines,
compressors, reciprocating pumps and steam engines.
Punch press - is a type of machine press used to cut holes in material.

Purpose
When the work being done is greater than the work input, the flywheel gives up some of
its stored energy to supply the deficiency by means of kinetic energy.
This energy is commonly used in moderating speed fluctuations in an engine through its
inertia.

Flywheel Energy Storage System

Like batteries flywheel energy storage system is one of energy storage devices. They
store energy mechanically in the flywheel rotor by rotating the rotor while as chemical batteries
stores energy electrically. When we want to use the stored energy in the rotor, a generator is used
to convert mechanical energy to electrical energy.

Working principle of Flywheel in ICE

When vehicles are cruising at a constant speed like 60 mph, relying only on internal
combustion engines can be sufficiently efficient way. However, when we need to accelerate
vehicles, internal combustion engines tend to use more fuel than needed. Therefore, if we use
both internal combustion engine power and auxiliary power from flywheels or chemical batteries,
we can save considerable amount of energy to get to a certain speed.

For internal combustion engine applications, the flywheel is a heavy wheel mounted on
the crankshaft. The main function of a flywheel is to maintain a constant angular velocity of the
crankshaft.

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Design calculations

Energy stored in the Flywheel

KE = ½ mv2

m = W/g
v = πDN

Maximum speed, v1

V1 = πDN1

Minimum speed, v2

V2 = πDN2

Total weight of flywheel, WT

WT = Wr + Wah

Weight of flywheel rim, Wr

Wr = πD(b)(t)ρ

where: KE – kinetic energy released by flywheel, kN-m or lb-ft

m – Mass of a body (flywheel), kg or lbm

v – Average velocity of the flywheel, m/s or fpm

W – Weight of the flywheel, lbf or kgf

D – Mean diameter of flywheel, in or mm

N – Revolutions per minute of the flywheel

Wah – weight of arms and hub

b – Width of flywheel rim

t – Thickness of flywheel rim

ρ – Density of flywheel material = 7200 kg/m3

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KE in the function of moment of Inertia

The angular velocity of the flywheel necessarily varies, as the power input and
consumption being a function of its moment of inertia.

KE = ½ Iω2

Moment of Inertia, I

I = mr2 - slugs-ft2 = [(lb-s2/ft)(ft2)]

Radius of gyration, r

r = (I/A)1/2

r = D/4

I = πD4/64

The change of kinetic energy, ΔKE

ΔKE = mr2/2 (ω12 – ω22)

ΔKE = w/2g (v12 – v22)

Sample Problem
A flywheel weighing 910 kg has a radius of gyration of 1.524 m. The shaft journals are
304.8 mm in diameter and have a coefficient of friction of 0.06. After the wheel reaches 120
rpm, the driving force is withdrawn and the wheel reduces its speed to 60 rpm. How much
kinetic energy the wheel losses?

Solution:

m = 910 kg ; r = 1.524 m

d = 304.8mm ; f = 0.06

N1 = 120 rpm ; N2 = 60 rpm

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KE = mr2/2 [(4π)2 – (2π)2]

KE = 125.15 kN-m

Coefficient of Fluctuation of Flywheel

Cf = ω1 – ω2 / ω
Cf = n1 – n2 / n
Cf = v1 – v2 / v

Table 9-1. Coefficients of Fluctuation of Flywheel

Driven Machine Cf
DC gen (direct drive) 0.002
AC gen (direct drive) 0.0035
(Punching, shearing, pressing) machines 0.05 – 0.1
Stamp mills, crushers 0.2
Reciprocating pumps, compressor 0.03 – 0.05
Machine tools, looms, paper mills 0.025
Spinning mills (fine & coarse thread) 0.01 – 0.02
Geared drives 0.02
(DME Faires, p. 534)

Energy required punching a plate

KE = ½ Sus (A) t
F = Sus (A)
where:
KE – energy required to punch a plate
F – Force required to punch a hole
t – Thickness of metal plate
d – Diameter of punched hole

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Force required punching a hole

(Machineries Handbook)
F = d (t) 80, tons

where: F – force required to punch a hole, tons

d – Punched hole dia., inches
t – Punched hole thickness, inches
Solved Problem:
How much power is required to punch a 25 mm hole on a 19mm thick plate every 12
seconds if the Sus of the plate is 400 Mpa. The average diameter of the flywheel is 1m. The
normal operating speed is 200 rpm and Cf is 0.1. The rim width is to be 2t. Assuming that the
arms and hub accounts 10% of the Wr concentrated at the mean diameter, calculate the energy
left on flywheel after punching.
Given: D = 25 mm Ø
t = 19 mm
T = 12 sec
Sus = 400 Mpa
D=1m
N = 200 rpm
Cf = 0.1
Solution:

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Punch
press

MOTOR FLYWHEEL PLATE

As – shear area
As = πDt
As = π(25mm)19mm
F = Sus (As)
F = 400 Mpa [π(25mm)19mm]
F = 596.9 kN
KE = ½ F (t)
KE = 0.5 (596.9kN) 19 mm t
KE = 5670.57 N-m
D

Power of the motor, kW

P = E / T = 5670.57 N-m / 12 sec = 470 N-m / s = 0.470 kW
Energy left after punching, KE2
KE = ½ Wf/g (v22)
Energy required during punching, KE1
KE = ½ Wf/g (v12)
Weight of flywheel, Wf
Wf = Wr + Wah

Wah = 10%Wr
Wf = 1.1 Wr
Wf = 32.2 (ΔKE) / Cf (v2)

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v = v1 + v2 / 2
v1 = π(1m)200 rpm = 10.47 m/s
v2 = π(1m)180.9 rpm = 9.47 m/s
1.1 Wr = [(2)9.81m/s2(5670.57N-m)] / [(10.47m/s)2 – (9.47m/s)2]
Wr = 5.067 kN
Wf = 1.1 (5.0667) kN
KE = ½ Wf/g (v22)
KE = [0.5 (5066.7 N)/9.81m/s2][9.47 m/s]
KE2 = 25.500 kN – m

PROBLEM

During a punching process of 158.69 lb flywheel the speed varies from 6 rps to 4.8 rps with 3
feet mean diameter. Determine the kinetic energy released by the flywheel in Btu?

SOLUTION
Wf = 158.69 lbs
N1 = 6 rps
N2 = 4.8 rps
D = 3 ft
g = 9.81 m/s2 = 32.3 ft/s2

Formula
KE = kinetic energy released by the flywheel

𝑊
KE = 2𝑔 [𝑣1 2 − 𝑣2 2 ]

V1 = πDN = π (3) (6) = 56.549 ft/s

V2 = πDN = π (3) (4.8) = 45.239 ft/s

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158.69
KE = 2(32.2) [56.5492 − 45.2392 ]

KE = 2.464[3197.789 – 2046.567]
KE =2836.611 lb-ft

KE = 3.645 Btu

PROBLEM

A plate 190 mm wide and 25 mm thick with strength of 410 Mpa is to be shear along its width.
During shearing process a 400 mm mean diameter flywheel changes its speed from 200 rpm to
180 rpm. Determine the weight of the flywheel to be use in kg?

SOLUTION

w = 190 mm
t = 25 mm
Su = 410 Mpa
Dm = 400 mm
N1 = 200 rpm
N2 = 180 rpm

Formula
Energy released by the flywheel = energy required to punch/cut a metal
𝑊
KE = 2𝑔 [𝑣1 2 − 𝑣2 2 ] - Energy released by the flywheel

E = 0.5(F)t - energy required to punch/cut a metal

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𝑊
[𝑣1 2 − 𝑣2 2 ] = 0.5(F)t
2𝑔
𝑊
[𝑣1 2 − 𝑣2 2 ] = 0.5(410 * 190) 25
2(9.81)

V1 = πDN = π (0.4) (200) = 4.189 m/s

V2 = πDN = π (0.4) (180) = 3.770 m/s
𝑊
[4.1892 − 3.7702 ] = 0.5(410 * 190) 25
2(9.81)

W (0.170 m) = 24343.75 N-m

W = 143 198.529 N
W = 14 597.2 kg

Design Exercises

PROBLEM
A 910 mm mean diameter flywheel, 4.90 kN weight changes its speed form 220 rpm to 200 rpm
during shearing process. What average force is needed to shear at 25 mm thick plate?
Ans. F = 381 kN

SOLUTION

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PROBLEM
A flywheel on a press rotating at 120 rpm is lowered to 102 rpm during a punching process that
requires 0.80 second for the punching portion of the cycle. What angular deceleration the
flywheel experience, rad/s2? Ans. α = -2.34 rad/s2

SOLUTION

PROBLEM
A flywheel weighing 910 kg has a radius of gyration of 1.524 m. The shaft journals are 304.8
mm in diameter and have a coefficient of friction of 0.06. After the wheel reaches 120 rpm, the
driving force is withdrawn and the wheel reduces its speed to 60 rpm. How much kinetic energy
the wheel losses? Ans. 125.15 kN-m

SOLUTION

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PROBLEM
A shear machine requires 345 N-m of energy to shear as steel sheet and has a normal speed of
200 rpm, slowing down to 120 rpm during the shearing process. The flywheel has mean diameter
of 750 mm and weight of 7200 kg/m3. The width of the rim is 300 mm. if the hub and arms of
the flywheel account for 10% of its total weight, find the thickness of the rim, in mm?

SOLUTION

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Name:__________________________________________Date:______Rating:______

SEAT WORK # 1

Instruction: This is an open notes problem exercises. Data presented can solve using tables in
Design of Machine Elements by Virgil Moring Faires. Avoid erasures.

Problem:
Determine the size and mass of the rim of a flywheel for a punching machine which is to punch a
maximum of one, ¾ in. hole in C1020 plate, normalized, ½ in. thick. The mean speed of the
flywheel is to be 150 rpm during punching. Draw the diagram of the problem.

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Name:______________________________________________Date:__________Rating:______
SEAT WORK # 2

Instruction: This is an open notes problem exercises. The problem below is similar to the
sample problem presented above. You may use the data to solve other requirements. Avoid
erasures.

Problem:
The power required to punch a 25 mm hole on a 19mm thick plate is 0.5 kW in 12
seconds. The Sus of the plate is 400 Mpa. The average diameter of the flywheel is 1m. The
normal operating speed is 200 rpm and Cf is 0.1. The rim width is to be 2t. Assuming that the
arms and hub accounts 10% of the Wr concentrated at the mean diameter, find the dimension of
the cast iron flywheel.

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Problem Exercises:
1. What pressure in tons is required to punch a hole of 2 ¼ in. diameter through a 3/16 in.
plate?
2. A flywheel has a rim weight of 16 kg with a mean radius of 76 mm. Coefficient of
fluctuation of flywheel is 41. Determine the centrifugal force, in lbs developed of a cast
iron flywheel at its speed of 1200 rpm.
3. A shear machine requires 345 N-m of energy to shear as steel sheet and has a normal
speed of 200 rpm, slowing down to 120 rpm during the shearing process. The mean
diameter of 750 mm and weight of 7200 kg/m3. The width of the rim is 300 mm. if the
hub and arms of the flywheel account for 10% of its total weight, find the thickness of the
rim, in mm?

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DESIGN PROBLEM # 9

Name: Rating:
Course/Yr: Date:

FLYWHEEL DESIGN

PROBLEM

How much power is required to punch a 25 mm hole on a 19mm thick plate every 12
seconds if the Sus of the plate is 400 Mpa. The average diameter of the flywheel is 1m. The
normal operating speed is 200 rpm and Cf is 0.1. The rim width is to be 2t. Assuming that the
arms and hub accounts 10% of the Wr concentrated at the mean diameter, calculate the energy
left on flywheel after punching.

Punch
press

MOTOR FLYWHEEL PLATE

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DESIGN PROBLEM # 9 continuation

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