Applied Mathematical Sciences, Vol. 8, 2014, no. 124, 6181 - 6190
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/ams.2014.47594
On Specific Energy Capacity of
Flywheel Energy Storage
D. V. Berezhnoi
Kazan Federal University
420008 Kazan, Russia
D. E. Chickrin
Kazan Federal University
420008 Kazan, Russia
A. F. Galimov
Kazan Federal University
420008 Kazan, Russia
Copyright © 2014 D. V. Berezhnoi, D. E. Chickrin and A. F. Galimov. This is an open access
article distributed under the Creative Commons Attribution License, which permits unrestricted
use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract
The paper introduces basic methods of computational investigation for specific
energy capacity of flywheel energy storage. In addition to the traditional
estimation of energy capacity on the kinetic energy specific potential energy
estimation of elastic strain is added. The possibilities of the use of various
structural materials in the manufacture of flywheels is analyzed, some
recommendations on the form of flywheel are given. "Extended" estimation of
energy capacity which is used in this paper gives greater variability in the design
of flywheel energy storage. In some cases it allows the reduction of the rotational
speed of the rotor part of the structure.
Keywords: flywheel, specific energy capacity, kinetic and potential energy.
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D. V. Berezhnoi, D. E. Chickrin and A. F. Galimov
1. Introduction
With the development of modern technologies in industry and transport many
mobile devices appear, that is why the problem of energy storage becomes crucial.
Creation of new materials allows not only improving the traditional
electrochemical batteries that already exist, but look for new methods of
accumulation and storage of energy, including mechanical ones [2,3,7]. These
include the so-called static and dynamic mechanical energy storage.
In static mechanical energy accumulators elastic element works on stretching
(compressive), which is preferable, on twisting (shifting), or on bend, but such
batteries’ accumulated specific energy are relatively small. However, they have a
firm place in many machines and mechanisms, due to its valuable properties:
energy storage stability, high efficiency and long life. One of the simple and
promising technologies is flywheel energy storage. Flywheel accumulates energy
in the form of rotational kinetic energy. In other words, it is a massive rotating
body that is used as kinetic energy (inertial energy storage device) storage [8].
However, even modern technologies do not use all the possibilities of a flywheel.
If rotates rapidly, it can accumulate kinetic energy that is easy to increase, and,
furthermore, to use, which turns a flywheel into electromechanical battery.
In this paper we present some results of the estimation of accumulated flywheel
energy, including kinetic and potential energy of strain for different materials.
Also we give some advices on the choice of materials for a flywheel.
2. Evaluation of the potential and kinetic energy accumulated
during disk rotation around its axis
2.1. Common relations
Let us consider the problem of finding the stresses in a homogeneous disk with
density  [5], which rotates with constant angular velocity  . In this case, the
solution does not depend on the angle of rotation  , but depends only on the
current disk radius r . We know that centrifugal force applies at each point of the
rotating body, proportional to the distance to rotation axis. Then the radial
component of the potential mass forces is  2 r and the equilibrium equation can
be written as follows
r
 rr
  rr      2 r ,
r
where  rr is radial stress,   is tangential stress. Ratios between the
components of the strain tensor and the displacement vector components in polar
coordinates can be written as follows:
 rr 
from which
ur
1 u ur
1  1 ur u u 
,  
 ,  r  

 ,
r
r  r
2  r  r
r 
On specific energy capacity of flywheel energy storage
 rr 
due to
6183
ur
u
,   r ,  r  0,
r
r
ur  ur  r  , u  0.
We represent the ratios between stress and strain in accordance with the
generalized Hooke's law. With respect to the plane stress state, these ratios can be
written as follows
 rr 
u
E
E  ur
    
 r
2  rr
2 
r
1 
1    r
ur
E
E  ur

 ,    1   2     rr   1   2  r   r



.

where  is Poisson ratio, E is Young’s modulus of disk material.
Substituting these relations into the equilibrium equation, we get the equation in
terms of displacements
 2ur 1 ur ur
1 2

 2 
 2 r.
2
r r r
E
r
(1)
It is known, that general solution of equation (1) can be represented as the sum
of homogeneous equation solution ur0 and heterogeneous equation particular
solution u1r .
We search for the general solution of equation in the form ur0  r n . Substitute it
into homogeneous equilibrium equation we obtain
 n2 1 r n2  0 .
Hence n  1 and
ur0  C1r  C2 r 1 ,
where constants C1 and C2 can be defined from boundary conditions. General
solution of heterogeneous equation
u1r  Ar n .
Substitute it into (1) we have
E
 n2  1 Ar n2   2 r  0 ,
1 2
from which n  3, A  
1 2
 2 .
8E
Thus, solution of the equation (1) have the form
ur  C1r  C2 r 1 
1 2
 2 r 3 ,
8E
and stress ratios
 rr 
E
E
3 
E
E
1  3
C1 
C2 r 2 
 2 r 2 ,   
C1 
C2 r 2 
 2 r 2 .
1 
1 
8E
1 
1 
8E
Strain potential energy

V
1
1
 rr  rr     dV 

 rr2   2  2 rr  dV
2E
2 E V
.
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D. V. Berezhnoi, D. E. Chickrin and A. F. Galimov
2.2. Aperture disk
The most simple static conditions for central aperture disc on inner and outer
surfaces are
 rr  r0   0,  rr  r1   0 .
Stress expressions




 rr  r    0 1  k 2 1  r12 r 2   r 2 r12 ,    r    0 1  k 2 1  r12 r 2    r 2 r12 ,
where
r
3 
1  3
 2 r12 ,  
,k  0.
8
3 
r1
0 
Strain potential energy


1
2E
2 b r 1  k
2 2
0 1
2 b r1
   
2
rr
2
  
 2 rr   rdrdzd 
0 0 kr1
2
 1  k  7  6     2k 17  6   
4
2
3E 1   
2
2
We define disk rotation kinetic energy
K
 b  r14  r04   2
4

2
 b 1  k 4  r12 r12  2
4

.
2 b 1  k 4  r12 0
.
3 
The ratio of strain potential energy to kinetic energy has the form
 K
 0 1  k 4  7  6   2   2k 2 17  6   2   3   
3E 1    1  k 2 
2
Since the mass of the disk is
.
m   br12 1  k 2   ,
the ratio of potential energy to mass
 m
2 02 1  k 4  7  6   2   2k 2 17  6   2  
3E  1   
2
,
and the ration of kinetic energy to mass
K m
2 1  k 2   0
3    
.
Stress  rr is positive and reach its maximum when r  r0 r1 :
 rrmax   0 1  k  .
2
Stress   also positive for all values of r and reach its maximum when r  r0 :
max
 
  0  2  (1   )k 2   2 0
3    1    k 2
3 
.
It is always  max   rrmax . Therefore, the strength condition can be written, for
example, according to the first strength theory:
max
 
y ,
or limiting
On specific energy capacity of flywheel energy storage
0 
 y 3   
6185
.
2  3    1    k 2 
Then expressions for strain potential energy, rotation kinetic energy, its ratios
and specific potential and kinetic energy are as follows:

 b y2 r12 1  k 2   3   
2
1  k  7  6     2k 17  6    ,
4
2
2
6 E 1     3    1    k 2 
2
K
 K
 m
 b 1  k 4  r12 y
3    1    k 2
2
2
,
 y 1  k 4  7  6   2   2k 2 17  6   2   3   
2
3E 1    1  k 2   3    1    k 2 
2
 y2  3   
2
1  k  7  6     2k 17  6    ,
4
2
2
2
6 E  1     3    1    k 2 
2
K m
 y 1  k 2 
  3    1    k 2 
2
.
(2)
Thus, specific energy capacity of rotating axis aperture disk is
y
1  k 

  3    1    k 2 
2
e m  K m m 
 y2  3   

E
2
1  k  7  6     2k 17  6    . ,
4
2
2
6 1     3    1    k 2 
2
2
2
2.3. Solid disk
Equations for finding stresses in solid disk  r0  0  have the form
 rr  r    0 1  r 2 r12  ,    r    0 1   r 2 r12  .
In this case we search for radial displacement
ur  C1r 
1 2
 2 r 3 ,
8E
and expressions for stresses are
 rr 
,
E
3 
E
1  3
C1 
 2 r 2 ,   
C1 
 2 r 2 .
1 
8E
1 
8E
The most simple static conditions on outer surface are as follows
 rr  r1   0 ,
and stresses defined according to formulas
 rr  r    0 1  r 2 r12  ,    r    0 1   r 2 r12  .
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D. V. Berezhnoi, D. E. Chickrin and A. F. Galimov
Both stresses are positive and it increase while reaching disk axis. On the disk
axis (when r  0 ) both stresses turn to maximum and become equal to each other
max
 
  rrmax   0 .
Hence tangential stresses equal zero on any plain, including symmetry axis of
the disk  r  0  , then  rr and   are primary stresses. If we have plain stressed
condition, then  zz  0 . We can use IV theory strength condition to estimate value
of stress, which destroy material. In this case we have
1 2  rrmax   max 
2
2
2
max
max
  
  zz    zz   rrmax     
 0   y.

Then expressions for strain potential energy, rotation kinetic energy, its ratios
and specific potential and kinetic energy are as follows:

2 b y2 r12  7  6   2 
3E 1   
K
 K
2 br12 y
3 
2
,
,
 y  7  6   2   3   
3E 1   
 m
2
2 y2  7  6   2 
3E  1   
K m
2
2 y
3    
.
,
.
(3)
Specific energy capacity of rotating solid disk is
e m  K m m 
2
y 2
 y2 2  7  6   

.
  3    E  3 1   2
2.4. Constant-strength disk
If the disk does not have a constant cross-sectional height, we can choose such a
form that circumferential   and radial stresses  rr in all points of the disk will
be constant and equal. In this case, disk section form can be defined according to
the formula [6]

 2 r 2
2
.
h  h0 e
In order to make the stress of the disk (which form we define through the
formula above) constant, it is necessary to apply load on outer surface, which
cause radial load  .
We can use IV theory strength condition to estimate value of stress, which
destroy material. In this case we have
On specific energy capacity of flywheel energy storage
1 2    
1 2   
2
2
    zz    zz     

2
2
  
6187
2

1 2 2  
2
    y.
Then strain potential energy is



1
2E
2 h0 e
 2r 2
2 y
r1
   
0
2 h0 y2 1    r1
0
0

 2 r 2
2 y
e
E
2
rr
2
  
 2 rr   rdrdzd 
rdr 

2 h0 y2 1     y 1  e

E
0
and kinetic energy is

K

2 2 h0 e
2
 2r 2
2 y
r1
 
  r rdrdzd  h0 
0
0
0
2
r1
2
e

 2 r 2
2 y
r dr  h0
3

2 y2 1  e

2 h0 e
 2r 2
2 y
r1
r1
    rdrdzd 2 h   e

 2 r 2
2 y
0
0
2
  2 r12 2 y
0
0
rdr  2 h0 

 1 e
2 2
1
2
  2 r12 2 y

0
,
1   r

0
where the mass
m
  2 r12 2 y
2
2 y 
,
.
Then expressions for strain potential energy, rotation kinetic energy and its
ratios are as follows:
 K
K m

 y 1    1  e

E 1 e

  2 r12 2 y
  2 r12 2 y
1   r
2 2
1

2 y 

 y2 1   
,
 m
E
 y 1 e
  2 r12 2 y

 1 e
1   r
2 2
1
  2 r12 2 y

2 y 
,
.
If angular velocity approaches infinity    , then first and third expressions
become more simple
 K
 y 1   
K m
E
,
y
.

We can wright specific energy capacity of rotating solid cylinder
e m   m K m 
 y2 1     y  y   y



1     .
1 
E

 
E

(4)
6188
D. V. Berezhnoi, D. E. Chickrin and A. F. Galimov
3. Numerical results
Evaluations for three disks (aperture disk, solid disk and constant-strength disk)
were conducted to estimate specific energy capacity of some construction
materials, which mechanical characteristics can be seen in table 1. In table 2 we
have specific energy capacities of disks of different shapes (for potential energy
 m , kinetic energy K m and general energy e m ). The disks made of materials,
which mechanical characteristics can be seen in table 1.
Table 1. Mechanical characteristics of some construction materials.
NN
Young’s
modulus E
Material
 MPa
1
2
3
4
5
High carbon steels
Titanium alloys
Boron carbide
Composites polymer
CFRP
Polyurethane
elastomers (eiPU)
Poisson’s Yield stress
 y  MPa 
ratio 
Density 
 kg m3 
215000
120000
472000
150000
0.3
0.3
0.3
0.25
1155
1245
5687
1050
7900
4800
2550
1600
3
0.48
51
1250
Table 2. Specific energy capacities of disks of different shapes  MJ kg  made of
materials, which can be seen in table 1.
NN
1
2
3
4
5
Thin ring
Solid disk
Constant-strength
disk
 m
K m
e m
 m
K m
em
 m
K m
e m
0,003
0,009
0,087
0,015
1,92
0,073
0,129
1,115
0,328
0,02
0,0756
0,138
1,202
0,344
1,94
0,002
0,005
0,054
0,011
0,821
0,088
0,157
1,35
0,404
0,023
0,090
0,162
1,41
0,414
0,844
0,0006
0,0019
0,019
0,0034
0,361
0,146
0,259
2,23
0,656
0,041
0,147
0,261
2,25
0,659
0,401
4. Analyzing results
The ratio between specific energy capacity e m and specific strength  âð  of
flywheel material have the form
y
e
k
,
m

On specific energy capacity of flywheel energy storage
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where k is flywheel shape coefficient, which represents its effectiveness. We can
find shape coefficient k using expressions (2-4), which coincides with
expressions [1,4].
Structural formulas for disk shapes are
e m  K m Ï m 
y
 y2
 y 2  k ,  
,
1  k ,   
2  k ,   , Ï K 

E
E 1  k ,  
where functions 1  k ,   and 2  k ,   are different for each type of shape.
Analysis of the ratio  K for each flywheel shapes shows, that the ratio of
potential energy to kinetic energy is greater for aperture disk and smaller for
constant-strength. This is due to the fact that the main body of the aperture disk
locates away from the axis of rotation. Under equal conditions that increases
kinetic energy. For constant-strength disk the situation is vice versa.
The same conclusion can be made analyzing the expression for specific potential
energy  m , i.e. accumulated potential energy reaches its maximum for aperture
disk and minimum for constant-strength disk. It is because the more elastic
material works away from rotation axis, the more specific potential elastic energy
accumulated.
However, we may made the opposite conclusion while estimating specific
potential energy K m . Here the disk strength is the main characteristic, which the
highest for constant-strength disk. Full accumulated specific energy K m   m is
greater for disks with mass concentration near rotation axis for all materials,
except rubber. The reason for this is that ratio  âð E becomes too huge and
percentage of specific potential energy in the full specific energy dominates. But
still general accumulated specific energy too small, because expression  âð  is
small. However, calculations show that for rubber-like materials (for instance, for
polyurethane elastomers (eiPU) [9]), energy accumulation derives from potential
strain energy.
5. Conclusion
We should mention in conclusion that we analyze energy capacity of flywheels
of different shapes in this paper. Similar studies have been conducted previously,
but they gave evaluation accumulated rotation kinetic energy only. We calculate
and accumulated elastic strain potential energy. Calculations were made for some
of the classical types of flywheels for which we can obtain the exact value of
energy capacity. The same evaluation of complex forms of the flywheels
(including composite and combined ones) can be carried out in the known
numerical packages of strength analysis, in the ANSYS, in particular.
We need specific strength of the flywheel material and the form coefficient in
order to calculate flywheels’ specific kinetic energy. We use the ratio of ultimate
material tensile strength to its Young's modulus to estimate elastic strain potential
energy. It gives us wide opportunities of flywheel energy storage devices design,
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D. V. Berezhnoi, D. E. Chickrin and A. F. Galimov
i.e. in some situation accumulated strain potential may exceeds greatly kinetic
energy of rotation. At the same time the shape of the flywheel influence on its
energy capacity (energy capacity of strain potential energy) in opposite manner. In
addition, if the flywheel accumulates more strain potential energy (than kinetic
energy), it will make possible to reduce the rotational speed and acceleration. In
its turn that will positively affect safe operation and service life and allow to
abandon the sealed housing, which creates vacuum in the rotation area of the
flywheel (or at least use lower order vacuum).
Acknowledgements. The work was performed according to the Russian
Government Program of Competitive Growth of Kazan Federal University as part
of OpenLab “Makhovik”. The reported study was also supported by RFBR,
research project No. 12-01-00955, 12-01-97026, 13-01-97059, 13-01-97058.
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Received: July 11, 2014