Inter
national
J
our
nal
of
Electrical
and
Computer
Engineering
(IJECE)
V
ol.
8,
No.
4,
August
2018,
pp.
2029
–
2037
ISSN:
2088-8708
2029
I
ns
t
it
u
t
e
o
f
A
d
v
a
nce
d
Eng
ine
e
r
i
ng
a
nd
S
cie
nce
w
w
w
.
i
a
e
s
j
o
u
r
n
a
l
.
c
o
m
Study
and
Dimensioning
of
the
T
anks
Dedicated
to
a
Compr
essed
Air
Ener
gy
Storage
system
(CAES)
I.
Rais
1
and
H.
Mahmoudi
2
1
Mohamadia
Engineering
School,
Mohammed
V
uni
v
ersity
,
Morocco
2
Po
wer
Electronic
and
Control
T
eam
(EPCT),
Department
of
Electrical
Engineering,
Morocco
Article
Inf
o
Article
history:
Recei
v
ed:
October
2,
2017
Re
vised:
April
14,
2018
Accepted:
May
3,
2018
K
eyw
ord:
CAES
CAM
Compressor
Reserv
oir
ener
gy
density
char
ging
time
dischar
ging
time
ABSTRA
CT
The
fundamental
idea
of
storage
is
to
transfer
a
v
ailable
ener
gy
During
periods
of
lo
w
demand
,
using
only
a
fraction
of
the
fuel
that
w
ould
be
consumed
by
the
standard
production
machine
(g
as
turbine,
thermal
engine,
etc.).
The
main
role
of
ener
gy
storage
is
therefore
to
introduce
an
ener
gy
de
gree
of
freedom
to
decouple
Consumers
and
the
producer
by
supplying
or
Deli
v
ering
the
dif
ference
between
these
tw
o
po
wers.
In
this
paper
is
this
paper
presents
a
brief
study
and
dimensioning
of
compressed
air
storage
tanks
to
a
h
ybrid
system
wind-PV
.
adopts
the
CAES
system
as
a
storage
agent.
starting
with
the
technical
criteria
on
which
the
choice
of
reserv
oirs
is
based
and
the
mechanical
constraints
that
must
be
tak
en
into
consideration
for
dimensioning
of
the
reserv
oirs
Copyright
c
2018
Institute
of
Advanced
Engineering
and
Science
.
All
rights
r
eserved.
Corresponding
A
uthor:
Ilham
Rais
Po
wer
Electronic
and
Control
T
eam
(EPCT)
Rabat,
Morocco
Email:
ilhamrais@research.emi.ac.ma
1.
INTR
ODUCTION
In
most
isolated
areas,
the
diesel
generator
is
the
m
ain
source
of
electrical
po
wer
.that
poses
immense
technical
challenges
And
financial.
This
generation
of
electricity
is
relati
v
ely
inef
ficient,
v
ery
costly
and
respon-
sible
for
the
emission
of
lar
ge
quantities
of
greenhouse
g
ases
(GHGs).
The
use
of
wind-solar
(JES)
twinning
in
these
autonomous
netw
orks
could
Thus
reducing
operating
deficits
[5][7].
Ho
we
v
er
,
the
profitability
of
the
JES
is
achie
v
ed
on
the
condition
of
obt
aining
a
high
penetration
rate
of
wind
and/or
solar
ener
gy
,
which
is
possible
only
by
using
storage
systems
[8]
the
solution
proposed
which
meets
all
the
technical
and
financial
requirements
while
ensuring
a
reliability
of
electricity
supply
to
the
isolated
sites.
It
is
the
h
ybrid
wind-photo
v
oltaic
system
with
storage
of
compressed
air
[1].
The
use
of
compressed
air
as
an
ener
gy
storage
agent
i
s
as
well
suited
to
wind
and
solar
production
as
it
is
to
di
esels.
The
principal
idea
for
the
storage
of
electrical
ener
gy
By
wind
turbine
plant
and
Photo
v
oltaic
array
as
a
source
of
ener
gy
coupled
with
tw
o
Pneumatic
machines:
The
first
is
a
compressor
dri
v
en
by
an
electric
motor
and
the
second
i
s
an
compressed
air
motor
which
dri
v
es
in
turn
an
alternator
[1].
During
windy
period
the
turbine
directly
supplies
the
isolated
site
on
electricity
and
Surplus
ener
gy
will
be
used
by
the
electri
c
motor
to
dri
v
e
the
compressor
to
rechar
ge
the
compressed
air
in
the
tri
vial
tanks.
In
the
absence
of
wind
turbine
the
photo
v
oltaics
ener
gy
will
follo
w
the
same
principle.
In
the
absence
of
the
tw
o
ener
gy
sources
compressed,
the
air
will
be
relax
ed
in
the
compressed
air
motor
which
will
dri
v
e
in
turn
the
generator
to
supply
electricity
at
the
isolated
site
.
This
h
ybrid
system
w
ould
act
in
real
time
in
order
to
maintain
the
balance
between
the
po
wer
generated
and
consumed
by
achie
ving
a
remarkable
reduction
in
fuel
consumption
whate
v
er
the
le
v
el
of
the
po
wer
demanded.
the
follo
wing
paper
will
present
a
brief
study
of
the
selection
criteria
of
the
reserv
oirs
and
the
dimensioning
of
this
latters
as
the
most
interesting
elements
in
the
compression
chain
[1][12].
J
ournal
Homepage:
http://iaescor
e
.com/journals/inde
x.php/IJECE
I
ns
t
it
u
t
e
o
f
A
d
v
a
nce
d
Eng
ine
e
r
i
ng
a
nd
S
cie
nce
w
w
w
.
i
a
e
s
j
o
u
r
n
a
l
.
c
o
m
,
DOI:
10.11591/ijece.v8i4.pp2029-2037
Evaluation Warning : The document was created with Spire.PDF for Python.
2030
ISSN:
2088-8708
2.
TECHNICAL
CHARA
CTERISTIC
FOR
THE
CHOICE
OF
RESER
V
OIRS
The
choice
of
tanks
suitable
for
the
proposed
compressed
air
storage
system
does
not
come
at
random
b
ut
there
are
se
v
eral
technical
criteria
which
characterize
them
we
cite
among
them
[4][2]:
ener
gy
density
of
storage
The
fle
xibility
of
location
and
specific
site
requirements.
Storage
capacity
.
The
self-dischar
ge.
The
yield
of
the
storage
system
2.1.
Ener
gy
density
of
storage
An
open
g
as
c
ycle
system
allo
ws
complete
polytropic
e
xpansion
of
air
compressed
from
the
maximum
pressure
to
atmospheric
pressure.
this
allo
ws
full
e
xploitation
of
the
ener
gy
stored
as
compressed
air
.
F
or
a
1
m3
unit
of
v
olume,
the
storage
ener
gy
density
can
be
e
xpressed
by
the
follo
wing
equation
[3]:
W
=
K
nN
P
r
n
1
(1
(
P
a
P
r
)
n
1
nN
)
(1)
with
K
=
2
:
777810
6
is
the
ener
gy
con
v
ersion
cons
tant
in
kWh,
N
is
the
number
of
e
xpansion
stages
of
the
CAM,
P
a
is
the
atmospheric
pressure
and
P
r
is
the
storage
pressure.
The
figure
abo
v
e
gi
v
es
an
idea
of
the
amount
of
0
50
100
150
200
250
0
0.5
1
1.5
2
2.5
3
3.5
x 10
−3
maximum storage pressure (bar)
Energy density (wh / m
3)
N=1
N=2
N=3
N=4
N=5
Figure
1.
v
ariation
of
the
ener
gy
density
as
a
function
of
the
number
of
stage
and
the
storage
pressure
ener
gy
stored
in
a
gi
v
en
v
olume
of
air
.
It
can
be
noticed
that
by
increasing
the
maximum
allo
w
able
pressure
in
the
tank
this
quantity
can
be
increased.
Thus,
the
higher
the
numbe
r
of
st
ages
of
the
air
e
xpansion
in
the
CAM
increases,
the
more
ener
gy
density
per
m
3
increases
and
consequently
the
mechanical
w
ork
(electrical
idem)
de
v
elops
closer
to
its
maximum
v
alue.
In
practice,
the
compressed
air
e
xpansion
process
must
be
stopped
once
the
pressure
in
the
tank
drops
belo
w
the
minimum
pressure,
P
r
m
,
required
for
the
operation
of
the
system.
Indeed,
belo
w
this
pressure
limit,
the
po
wer
deli
v
ered
becomes
so
lo
w
that
the
operation
of
the
system
becomes
inef
fecti
v
e.
The
v
alue
of
P
r
m
depends
on
the
nature
of
the
application.
Therefore,
in
the
case
where
P
r
m
is
greater
than
atmospheric
pressure
P
a
,
une
xploited
ener
gy
,
W
unex
,
will
remain
in
the
air
reserv
oir
.
this
ener
gy
can
be
e
xpressed
as
follo
ws:
W
unex
=
K
nN
P
r
m
n
1
(1
(
P
a
P
r
m
)
n
1
nN
)
(2)
Consequently
,
the
ef
fecti
v
e
ener
gy
density
W
ef
will
be
reduced
and
equal
to
the
dif
ference
between
the
total
a
v
ailable
ener
gy
density
W
and
the
unused
ener
gy
density
W
unex
,
hence:
W
ef
=
W
W
unex
(3)
IJECE
V
ol.
8,
No.
4,
August
2018:
2029
–
2037
Evaluation Warning : The document was created with Spire.PDF for Python.
IJECE
ISSN:
2088-8708
2031
The
Figure
2
sho
ws
the
ef
fecti
v
e
ener
gy
density
as
a
function
of
the
pressure
minimum
of
operation,
P
r
m
,
for
dif
ferent
v
alues
of
the
maximum
pressure
P
r
.
It
is
that
the
lo
wer
the
minimum
pressure,
the
higher
the
ef
fecti
v
e
ener
gy
density
.
0
50
100
1
5
0
2
0
0
2
5
0
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
Minimum
oper
a
t
ing
pr
e
ssu
r
e
(bar
)
E!ective energy density(KWh/
m
3
)
P=
5
0
bar
P=
1
0
0
bar
P=
1
5
0
bar
P=
2
0
0
bar
P=
2
5
0
bar
P=
3
0
0
bar
Figure
2.
influence
of
minimum
operating
pressure
on
the
density
of
ener
gy
F
or
a
good
e
v
aluation
of
the
unused
ener
gy
,
the
pressure
utilization
f
actor
(PUF)
for
an
open
g
as
c
ycle
can
be
defined
as:
P
U
F
=
1
W
unex
W
(4)
It
is
easy
to
deduce
from
this
equation
that
PUF
=
1
if
P
r
m
=
P
a
and
PUF
=
0
if
P
r
m
=
P
.
The
Figure
3
sho
ws
the
v
ariations
of
PUF
as
a
function
of
the
minimum
storage
pressure
for
a
relaxation
which
is
carried
out
in
5
step
.
0
50
100
150
200
250
300
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Minimum operating pressure (bar)
Pressure utilization factor
P=50bar
P=100bar
P=150bar
P=200bar
P=250bar
P=300bar
Figure
3.
V
ariation
of
PUF
as
a
function
of
the
minimum
operating
pressure
3.
MECHANICAL
CONSTRAINTS
Currently
a
v
ailable
materials
can
store
a
high
density
up
to
pressures
of
the
order
of
500
bar;
for
the
proposed
study
we
wil
l
considered
a
spherical
tanks,
relati
v
ely
slim
whose
pressure
inside
has
a
small
thickness
is
supposed
constant,
because
the
storage
pressure
P
r
depends
on
the
maximum
breaking
stress,
,
and
of
t
he
dimensions
of
the
reserv
oir
,
the
e
xpression
that
connects
these
parameters
is
the
follo
wing
[9][6]:
e
r
D
r
=
P
r
4
(5)
W
ith:
e
r
and
D
r
are,
respecti
v
ely
,
the
thickness
and
the
diameter
of
the
storage
tank.
kno
wing
that
R
int
=
D
r
2
is
the
Study
and
Dimensioning
of
the
T
anks
Dedicated
to
a
Compr
essed
Air
...
(I.
Rais)
Evaluation Warning : The document was created with Spire.PDF for Python.
2032
ISSN:
2088-8708
internal
radius
of
the
reserv
oir
,
R
ext
is
the
outer
radius
of
the
reserv
oir
of
storage
and
e
r
=
R
ext
R
int
,
the
v
olumes
of
stored
air
and
the
reserv
oir
are
respecti
v
ely
:
V
=
4
3
R
3
int
(6)
V
r
=
4
3
R
3
ext
(7)
The
v
olume
of
the
material
used
in
the
manuf
acture
of
the
tank
is
then:
V
matr
=
V
V
r
=
4
3
(
R
ext
R
int
)(
R
2
ext
+
R
ext
R
int
+
R
2
int
)
(8)
The
h
ypothesis
that
the
reserv
oir
has
a
small
t
hickness
mak
es
it
possible
to
simplify
the
second
term
of
the
pre
vious
equation
in
3
R
2
int
and
the
e
xpression
of
V
matr
becomes
as
follo
ws:
V
matr
=
4
3
e
r
(3
R
2
int
)
(9)
From
the
abo
v
e
equations,
a
relation
between
the
v
olume
of
air
stored,
V
,
as
wel
l
as
the
v
olume
of
materials
used
in
the
manuf
acture
of
the
tank,
V
matr
,
can
be
deduced,
hence:
V
matr
V
=
4
3
e
r
(3
R
2
int
)
4
3
(
R
3
int
)
=
3
e
r
R
int
=
6
e
r
D
r
=
3
P
r
2
(10)
The
relati
onship
between
the
mass
of
the
ma
terials
of
a
spherical
reserv
oir
,
M
matr
,
and
the
ener
gy
stored,
E,
can
then
be
e
xpressed
by
the
follo
wing
relation
M
matr
E
st
=
V
matr
n
(
n
1)
N
(1
(
P
P
a
)
1
n
nN
)
=
3(
n
1)
2
N
nN
n
(1
(
P
P
a
)
1
n
nN
)
(11)
is
the
o
v
erall
ef
ficienc
y
of
the
con
v
ersion
chain
between
the
PV
-wind
h
ybrid
system
and
the
storage
tank
and
E
st
is
the
stored
ener
gy
which
can
be
defined
by:
E
st
=
V
W
=
E
(12)
The
follo
wing
figure
sho
ws
ho
w
the
stored
ener
gy
can
v
ary
as
a
function
of
the
compression
ratio
(equal
to
the
storage
pres
sure)
and
the
properties
of
the
dif
ferent
materials.
The
ratio
(K=
)
is
the
ra
tio
of
the
tensile
st
rength
to
the
density
of
the
material.
Figure
4.
ratio
(reserv
oir
mass
/
stored
ener
gy)
as
a
function
of
the
compression
ratio
for
dif
ferent
properties
of
the
materials
It
is
easy
to
notice
that
the
stored
ener
gy
is
higher
the
higher
the
compression
ratio.
The
best
performance
(maximum
stored
ener
gy
and
minimum
reserv
oir
mass)
is
obtained
with
the
highest
ratio
(K=
),
for
high
tensile
strength
b
ut
relati
v
ely
lo
w
density
.
IJECE
V
ol.
8,
No.
4,
August
2018:
2029
–
2037
Evaluation Warning : The document was created with Spire.PDF for Python.
IJECE
ISSN:
2088-8708
2033
4.
DIMENSIONING
OF
THE
T
ANKS
The
dimensioning
of
reserv
oirs
is
a
v
ery
important
point
in
this
study
.
The
y
should
not
be
too
small
or
too
lar
ge
to
ensure
that
e
xcess
ener
gy
can
be
stored
and
to
limit
congestion
and
engender
a
disproportionate
in
v
estment.
The
sizing
of
the
reserv
oirs
is
conditioned
by
the
double
study
of
the
flo
w
(v
ariable)
of
the
compressor
(and
therefore
of
the
v
olume
that
can
be
stored
in
a
gi
v
en
time)
and
the
compressed
air
flo
w
required
by
the
CAM
to
dri
v
e
the
associated
alternator
and
supply
the
isolated
site
with
electricity
[2][10].
Then
to
calculate
the
v
olume
of
the
tanks
tw
o
methods
will
be
presented
4.1.
First
method
The
first,
a
con
v
entional
method,
is
a
function
of:
the
maximum
pressures,
P
max
,
and
minimal,
P
min
,
allo
w
able
by
the
CAM,
the
desired
autonomy
a,
maximum
air
flo
w
required
to
po
wer
the
compressed
air
motor
,
.
The
v
olume
will
be
calculated
using
the
follo
wing
formula:
V
=
P
a
a
P
max
P
min
(13)
4.2.
second
method
The
second
method
for
dimensioning
the
storage
v
olume
is
to
calculate
the
loading
and
dischar
ge
time
of
the
tanks,
gi
v
en
that
the
amount
of
air
injecte
d
into
the
tank
is
v
ariable
and
is
a
function
of
the
po
wer
of
the
h
ybrid
system
absorbed
by
the
compressor
and
of
the
air
consumption
of
the
CAM
[11][15].
4.2.1.
Char
ging
time
T
o
f
acilitate
calculation,
it
is
considered
that
the
po
wer
absorbed
to
compress
the
air
is
constant.
The
instantaneous
air
flo
w
rate
can
then
be
e
xpressed
as
follo
ws:
"
=
p
c
E
c
(14)
W
ith:
P
c
is
the
po
wer
consumed
by
the
compressor
.
E
c
the
ener
gy
per
unit
mass
necessary
to
compress
the
air
at
a
gi
v
en
pressure.
On
the
other
hand,
by
ne
glect
ing
the
losses
by
leakage
of
compressed
air
,
the
equation
of
conserv
ation
of
the
mass
and
the
la
w
of
perfect
g
ases
mak
e
it
possible
to
e
xpress
the
flo
w
of
air
entering
the
v
olume
V
r
of
the
storage
tank
as
follo
ws:
"
=
dm
dt
=
V
r
T
r
dp
dt
(15)
By
inte
grating
the
equation
obtained
from
the
preceding
equations
after
ha
ving
replaced
each
term
(
P
c
and
E
c
)
by
its
v
alue,
the
char
ge
time
of
a
compressed
air
reserv
oir
can
be
calculated
from
the
follo
wing
equation
t
ch
=
(
(
ch
+
1)
1
+
ch
)
N
ch
(16)
W
ith
:
ch
=
n
1
N
n
,
=
P
P
a
and
is
the
time
constant
during
the
char
ging
phase,
defined
by:
=
P
a
V
P
c
C
p
r
(17)
V
is
the
v
olume
of
compressed
air
produced,
C
p
is
the
mass
heat
of
the
compressed
air
.
From
the
equation
(17),
The
equation
of
the
time
constant
Becomes:
=
t
ch
N
(
(
ch
+1)
1+
ch
)
(18)
By
replacing
the
v
alue
of
in
the
equation
(16),
the
e
xpression
of
the
product
v
olume
of
compressed
air
becomes
:
V
=
P
c
r
t
ch
P
a
C
p
N
(
(
ch
+1)
1+
ch
)
(19)
Study
and
Dimensioning
of
the
T
anks
Dedicated
to
a
Compr
essed
Air
...
(I.
Rais)
Evaluation Warning : The document was created with Spire.PDF for Python.
2034
ISSN:
2088-8708
4.2.2.
Dischar
ging
time
The
dischar
ge
time
of
a
compressed
air
reserv
oir
may
be
calculated
from
the
same
the
char
ging
time,
replacing
"
,
P
and
E
respecti
v
ely
with
the
parameters
of
the
compressed
air
motor
[13].
"
M
,
the
mass
flo
w
of
compressed
air
consumed
by
the
MA
C.
P
M
,
the
po
wer
pro
vided
by
the
MA
C.
E
M
,
the
ener
gy
resulting
from
the
e
xpansion
of
the
compressed
air
in
the
MA
C.
The
e
xpression
obtained
from
the
dischar
ge
time
is
then
written
as
follo
ws
[16][14].
t
dch
=
(
1+
dch
M
1
+
dch
+
dch
1
+
dch
M
)
N
dch
(20)
W
ith:
dch
=
n
1
nN
,
M
=
P
M
in
P
M
ou
;N
is
the
number
of
e
xpansion
stages
in
the
CAM
and
dch
is
the
time
constant
during
the
dischar
ge
phase,
defined
by
the
follo
wing
e
xpression
dch
=
P
a
V
r
P
M
C
p
r
(21)
V
r
is
the
v
olume
of
compressed
air
tank,
P
M
is
the
po
wer
produced
by
the
compressed
air
motor
.
The
dischar
ge
time
between
2
pressure
le
v
els
can
be
calculated,
as
follo
ws
:
t
dch
P
1
P
2
=
t
dch
P
1
t
dch
P
2
=
[
(
1+
dch
P
1
1+
dch
P
2
)
1
+
dch
+
(
P
2
P
1
)]
N
dch
(22)
5.
RESUL
T
AND
DISCUSSION
5.1.
The
first
method
Figure
8
gi
v
es
an
idea
of
the
dimensioning
of
the
tank
calculated
from
the
equation
13
It
sho
ws
that
o
v
er
the
desired
range
is
greater
the
greater
the
v
olume
of
air
stored
in
the
reserv
oir
must
be
lar
ge.
Thus,
for
an
autonomy
of
2
days,
the
v
olume
necessary
to
store
compressed
air
at
300
bar
,
will
be
of
the
order
of
34
m3.
On
the
other
hand,
this
v
olume
is
enormous
and
o
v
ersized
and
it
will
be
dif
ficult
to
transport,
to
install
a
tank
ha
ving
this
v
olume
in
an
isolated
sit
e.
In
addition,
the
MA
C
will
rarely
operate
at
maximum
flo
w
when
tank
pressure
is
too
lo
w
.
Thi
s
results
from
the
f
act
that
the
tank
will
be
rechar
ged,
once
the
pressure
drops,
using
the
e
xcess
ener
gy
that
is
a
v
ailable
in
this
site
in
a
f
airly
re
gular
manner
.
Therefore,
this
method
of
sizing
the
reserv
oir
will
not
be
adopted
for
the
rest
of
the
calculation.
50
100
150
200
250
300
0
50
100
150
200
250
300
350
400
450
maximum pression of storage (bar)
reservoir vomume (m
3
)
a=1jours
a= 2jours
a=3jours
a=4jours
Figure
5.
ratio
(v
ariation
of
the
tank
v
olume
as
a
function
of
the
maximum
storage
pressure
and
the
autonomy
IJECE
V
ol.
8,
No.
4,
August
2018:
2029
–
2037
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IJECE
ISSN:
2088-8708
2035
5.2.
The
second
method
5.2.1.
Char
ging
time
Figure
6
and
Figure
7
sho
w
the
char
ging
time
of
a
tank
of
300
L
v
olume,
respecti
v
ely
,
as
a
function
of
the
number
of
stages
of
a
compressor
with
5
kW
of
po
wer
and
the
po
wer
absorbed
by
one
compressor
ha
ving
5
compression
stages.
These
figures
sho
w
that
the
more
the
number
of
compressor
stages
increases
or
the
more
the
electric
po
wer
consumed
to
compress
the
air
increases,
the
f
aster
the
char
ging
time
decreases.
A
single-stage
compressor
can
fill
the
300
L
v
olume
in
3.5
hours,
while
approximately
2
hours
will
be
suf
ficient
to
fill
the
same
v
olume
if
the
compression
of
air
is
done
bay
5
stages.
A
time
sa
ving
of
about
43pc.
Thus,
with
a
po
wer
of
20
kW
,
it
will
tak
e
half
an
hour
to
fill
a
tank
of
300
L
whereas
approximately
2
hours
are
required
to
fill
the
same
v
olume
when
the
e
xcess
po
wer
is
5
kW
.
Precious
time
w
as
sa
v
ed,
about
75pc.
This
results
from
the
f
act
that
the
flo
w
of
compressed
air
injected
into
the
reserv
oir
increases
proportionally
with
the
increase
in
po
wer
.
Ho
we
v
er
,
these
results
still
justify
the
choice
of
5
stages
of
compression
which
has
the
adv
antages
of
increasing
the
reserv
e
of
compressed
air
and
also
prolonging
the
autonomy
of
the
o
v
erall
system.
0
50
100
150
200
250
300
0
0.5
1
1.5
2
2.5
3
3.5
4
compression ratio
The filling time of a 300L (h)
N=1
N=2
N=3
N=4
N=5
Figure
6.
v
ariation
of
char
ging
time
as
a
function
of
compress
ion
ratio
and
number
of
stages
for
a
compressor
of
5KW
of
po
wer
0
50
100
150
200
250
300
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
compression ratio
The filling time of a 300L (h)
P=5KW
P=10KW
P=15KW
P=20KW
Figure
7.
v
ari
ation
of
the
char
ging
time
as
function
of
compression
ratio
and
po
wer
consumed
by
the
compressor
5
floor
It
is
easy
to
see
from
this
figure
that
the
amount
of
compressed
air
produced
increases
with
the
increase
in
the
number
of
compression
stages.A
remarkable
v
olume
g
ain
can
be
achie
v
ed
if
the
compressor
has
5
floor
of
compression
is
used
instead
of
a
single
stage
compressor
.
Study
and
Dimensioning
of
the
T
anks
Dedicated
to
a
Compr
essed
Air
...
(I.
Rais)
Evaluation Warning : The document was created with Spire.PDF for Python.
2036
ISSN:
2088-8708
0
5
10
15
20
25
30
0
100
200
300
400
500
600
700
800
900
POWER ABSORBED BY THE COMPRESSOR (KW)
The volume of compressed air produced (L)
N=5
N=4
N=3
N=2
N=1
Figure
8.
v
ariation
of
v
olume
of
air
produced
as
a
function
of
the
number
of
floors
and
of
the
po
wer
consumed
by
the
compressor
5.2.2.
Dischar
ging
time
The
figures
9
and
10
sho
w
the
v
ariations
of
the
di
schar
ge
time,
respecti
v
ely
,
as
a
function
of
the
max-
imum
storage
pressure
and
the
number
of
e
xpansion
stages
in
the
compressed
air
motor
.
The
analysis
of
these
figures
sho
ws
that
increasing
the
number
of
stages
of
a
CAM
serv
es
to
reduce
the
dischar
ge
time
of
the
reserv
oir
and
consequently
to
accelerate
the
restoration
of
t
h
e
stored
ener
gy
in
the
form
of
compressed
air
,
Thus,
the
higher
the
storage
pressure,
the
longer
the
dischar
ge
time.
0
0.5
1
1.5
2
2.5
3
50
100
150
200
250
300
Discharge time (h)
Pressure in a tank of 300bar
P=250bar
P=200bar
P=150 bar
P=100bar
P=50 bar
Figure
9.
v
ariation
of
pressure
of
the
tank
as
a
function
of
maximum
storage
pressure
and
dischar
ge
time
0
0.5
1
1.5
2
2.5
3
50
100
150
200
250
300
Discharge time (h)
Pressure in a reservoir of 300bar
N=1
N=2
N=3
N=4
N=5
Figure
10.
v
ariation
of
pressure
of
the
tank
as
a
function
of
the
dischar
ge
time
and
the
number
of
stages
6.
CONCLUSION
The
fluctuating
and
intermittent
nature
of
rene
w
able
ener
gies
requires
the
strengthening
of
the
control
of
ener
gy
flo
ws
between
supply
and
demand
for
electricity
.
ener
gy
storage
then
constitute
a
rele
v
ant
response
to
IJECE
V
ol.
8,
No.
4,
August
2018:
2029
–
2037
Evaluation Warning : The document was created with Spire.PDF for Python.
IJECE
ISSN:
2088-8708
2037
this
problematic,currently
v
arious
solutions
for
storing
green
electricity
e
xist
(batteries,
storage
by
compressed
air
or
STEP
-
Stations
of
T
ransfer
of
Ener
gy
by
Pumping).
Ho
we
v
er
,
the
y
do
not
permit
massi
v
e
storage
of
the
intermittent
ener
gy
produced
o
v
er
a
long
period
of
time.
this
study
presented
one
of
the
main
e
xisting
means
of
storing,
the
CAES
(Compressed
Air
Ener
gy
Storage)
or
ener
gy
storage
by
compressing
air
which
consists
of
storing
ener
gy
in
the
form
of
compressed
air
,
in
an
under
ground
ca
vity
(for
a
po
wer
of
more
than
100
MW),
or
in
tri
vial
tanks
for
small-scale
storage,
this
is
the
case
presented
in
this
paper
and
then
restore
via
a
turbine
producing
electricity
ag
ain.
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T
ools
Nr
.
9833
8998
03,
2002,
www
.adascopcoairmotors.
com
Study
and
Dimensioning
of
the
T
anks
Dedicated
to
a
Compr
essed
Air
...
(I.
Rais)
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