Poto?nik P. (Ed.) Natural Gas
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Steady State Compressible Fluid Flow in Porous Media 491
Where
- -
- -
c
y aa x x x
p
cT
d d d
p
x x x
d d d
2 3
(1 0.5 0.35 )
2
2 3
1
5.0 2.0 0.7
6
c
u
p
- x - x
d d
2
5.0 2.0 0.7
6
Other variables in equation (50) remain as defined in equation (49).
Application of the Runge -Kutta algorithm to the down hill differential equation by use of
Darcian lost head (equation (48)) gives
-
p
p y
d
2 2
2 1
(51)
Where
( )
3
d
x6.0-
2
e
x2.2
e
x2.5-
6
2
1
p
)
3
e
x3.0-
2
e
x5.0
e
x-1(
d
p
aa
d
y
++
+=
( )
2
e
x6.0-
e
x2.2-2.5
6
d
p
u
+
)L
1
S-
1
/
(AAp
d
p
aa =
,
Mk
2
p
d
RW
av
T
e
av
zc54679.2
Mk
p
A
RW
1
T
1
zc2
1
/
AAp
′
=
′
=
2
2Msinθ P
1
S
2
z T R
1 1
,
RTz
L 2Msinθ
x
av
av
e
e
Mk
2
p
d
RW
av
T
e
av
zμc2.546479
Mk
P
A
RW
av
T
av
μzc2
d
p
u
′
=
′
=
e
z
av
= Gas deviation factor (z) calculated
with T
a v
and
e
av
p
T T T
av
0.5( )
1 2
e d
p p aa
av p
2_
1
Equation (49) can be written as:
p p f f f
a
p f f f
S L
f W J x x x
BB Z T x x x XX
2 2 3
1
2 3
1 1
(1 0.5 0.3 )
6
(1 0.5 0.3 )
(53)
Where
f
av av f f
XX z T x x
2
(5.2 2.2 0.6 )
( )
1
S≥
a
p
BB if
2
2
p-
3
f
x6.0-
2
f
x2.2
f
x2.5-
6
2
1
p
-
2
i
P
p
J +=
( )
i
S
a
p
BB if
2
2
p-
f
3
x6.0-
f
2
x2 .. 2
f
x2 . 5
6
2
1
p
-
2
1
p-
2
2
P
p
J <+=
RL RL
a
BB
p
gd M gd M
p p
1.621139 0.270110
5 5
6
R
av
T
f
av
z
LsinM2
f
x,
M
5
p
gd6
2
1
psinM2
1
S
==
Natural Gas492
z
a v
f
= Gas deviation factor at the midsection of the porous medium calculated with
T
a v
and
f
av
p
, where
0.5( T T )
av
1 2
T
and
p p
f
p
av
p p
2
1 2
1 2
During isothermal flow in which there is no significant variation of the gas deviation
factors (z) between the mid section and the exit end of the porous medium, equation
(52) can be written as:
- -
- - -
S L
J x x x
p
f f f
f w
p
a
BB z T x x x z T x x
p
f f f f f
2 3
1
1 0.5 0.35
6
2
2 3 2
1 0.5 0.35 5.0 2.0 0.7
1 1 1 1
(54)
The variables in equation (53) remain as defined in equation (52)
Example 6
Suppose the porous medium of example 3b was vertical what would be the dimensionless
friction factor by use of the same pressure as they were in example 3b?
Solution
Here, P
1
= 7128psf, P
2
= 6585.12psf, T
1
=T a v =550
0
R, W=0.75
b / sec, R= 1544,
L
p
= 1800ft, g= 32.2ft / sec
2
,
d
p
=0.066667ft, z
1
= z
a v
f
= 1, since θ = 90
0,
sin 90
0
=1
122812.0
15445501
1800197.282
R
av
T
f
av
z
LsinM2
f
x
=
××
×××
==
The flow is isothermal; z is constant at 1.0 so equation (52) is used.
884081.0
3
f
x35.0-
2
f
x5.0
f
x-1 =+
764934.4
2
f
x7.0
f
x0.2-0.5 =+
585191.0
3
f
x7.0
2
f
x0.2
f
x0.5 - =++
RL
a
BB
p
gd M
p
0.270110
5
0.270110 1544 1800
61128325.
5
32.2 0.066667 28.97
M p
S
z T R
2
2
2 sin
2 28.97 1 7128
1
1
1 550 1544
1 1
3466.601
a
BB S then
p
,
1
- - - -
p
J P x x x p
p
f f f
2
2 2 3 2
1
5.0 2.0 0.7
1 2
6
- -
-
2
7128
2
7128 0.585191
6
2
6585.12 12400014
- -
S L
x x x
f f f
919426.8859
2 3
1
1 0.5 0.35
6
3466.601 1800 0.884081 /6
Then
f
p
E
12400014 919426.8859
2
0.75 611283825 1 550 0.884081 1 550 4.764934
1.977439 6
There is a drastic reduction in the f
p
as compared to f
p
= 6.560860 E-6 when the porous
medium was horizontal. The effect of inclination becomes more severe as the porous
medium gets longer.
Equation (40) can be written as:
- - -
- -
f
b
BB W z T x x x Z Tav x x
p av
f f f f f
k
S L
J x x x
p
f f f
2 3 2
1 0.5 0.3 5.2 2.2 0.6
1 1
.
2 3
1
1 0.5 0.3
6
Where
( )
1
S≥
b
p
BBif ,
2
2
p-
3
f
x6.0-
2
f
x2.2
f
x2.5-
6
2
1
p
-
2
1
P
p
J +=
Steady State Compressible Fluid Flow in Porous Media 493
z
a v
f
= Gas deviation factor at the midsection of the porous medium calculated with
T
a v
and
f
av
p
, where
0.5( T T )
av
1 2
T
and
p p
f
p
av
p p
2
1 2
1 2
During isothermal flow in which there is no significant variation of the gas deviation
factors (z) between the mid section and the exit end of the porous medium, equation
(52) can be written as:
- -
- - -
S L
J x x x
p
f f f
f w
p
a
BB z T x x x z T x x
p
f f f f f
2 3
1
1 0.5 0.35
6
2
2 3 2
1 0.5 0.35 5.0 2.0 0.7
1 1 1 1
(54)
The variables in equation (53) remain as defined in equation (52)
Example 6
Suppose the porous medium of example 3b was vertical what would be the dimensionless
friction factor by use of the same pressure as they were in example 3b?
Solution
Here, P
1
= 7128psf, P
2
= 6585.12psf, T
1
=T a v =550
0
R, W=0.75
b / sec, R= 1544,
L
p
= 1800ft, g= 32.2ft / sec
2
,
d
p
=0.066667ft, z
1
= z
a v
f
= 1, since θ = 90
0,
sin 90
0
=1
122812.0
15445501
1800197.282
R
av
T
f
av
z
LsinM2
f
x
=
××
×××
==
The flow is isothermal; z is constant at 1.0 so equation (52) is used.
884081.0
3
f
x35.0-
2
f
x5.0
f
x-1 =+
764934.4
2
f
x7.0
f
x0.2-0.5 =+
585191.0
3
f
x7.0
2
f
x0.2
f
x0.5 - =++
RL
a
BB
p
gd M
p
0.270110
5
0.270110 1544 1800
61128325.
5
32.2 0.066667 28.97
M p
S
z T R
2
2
2 sin
2 28.97 1 7128
1
1
1 550 1544
1 1
3466.601
a
BB S then
p
,
1
- - - -
p
J P x x x p
p
f f f
2
2 2 3 2
1
5.0 2.0 0.7
1 2
6
- -
-
2
7128
2
7128 0.585191
6
2
6585.12 12400014
- -
S L
x x x
f f f
919426.8859
2 3
1
1 0.5 0.35
6
3466.601 1800 0.884081 /6
Then
f
p
E
12400014 919426.8859
2
0.75 611283825 1 550 0.884081 1 550 4.764934
1.977439 6
There is a drastic reduction in the f
p
as compared to f
p
= 6.560860 E-6 when the porous
medium was horizontal. The effect of inclination becomes more severe as the porous
medium gets longer.
Equation (40) can be written as:
- - -
- -
f
b
BB W z T x x x Z Tav x x
p av
f f f f f
k
S L
J x x x
p
f f f
2 3 2
1 0.5 0.3 5.2 2.2 0.6
1 1
.
2 3
1
1 0.5 0.3
6
Where
( )
1
S≥
b
p
BBif ,
2
2
p-
3
f
x6.0-
2
f
x2.2
f
x2.5-
6
2
1
p
-
2
1
P
p
J +=
Natural Gas494
( )
1
S
b
p
BBif
2
2
p-
3
f
x6.0-
2
f
x2.2
f
x2.5-
6
2
1
p
-
2
2
P
p
J <+=
c RL c RL
b
BB
p
A M
d M
p
p
/ /
2 2.546479
2
6
6
All other variables remain as defined in equation (52). During isothermal flow in which
there is no significant variation of the gas deviation factor (z) between the mid section
and the exit end of the porous medium, equation (54) can be written as:
( )
=
++
+++
3
f
x35.0
2
f
x5.0
f
x1
6
L
1
S
p
J
)
2
f
x7.0
f
x0.2-0.5()
3
f
x3.0-
2
f
x5.0
f
x-1(
1
T
1
Wz
b
p
BB
--
[
k
(55)
Beside the coefficients of
f
x all other variables in equation (55) remain as defined in
equation (54).
Example 7
Compute the permeability of the core of example 4 assuming that the case was vertical.
Solution
From example 4, W= γ Q
Substituting the given values, W=0.00239716gm/sec
Sin θ = sin 90
0
=1.0, M =28.97, L
p
=2cm
f
va
z
= z
1
=1, T
1
=T
a v
=294.4
o
K, A
p
=2cm
2
p
1
= 1.45atm, p
2
=1.0atm,
= 0.02cp.
004794.0
1.824.2941
297.282
R
av
T
f
av
z
LsinM2
f
x
=
××
××
==
The flow is isothermal so equation (55) is used.
023941.0-
3
f
x70.0-
2
f
x0.2
f
x0.5-
990428.4
3
f
x70.0-
2
f
x7.0
f
x2-0.5
995217.0
3
f
x35.0-
2
f
x05.0
f
x-1
=+
=+
=+
018893.0
97.2823
21.8202.0
M
p
A6
RL
1
c2
p
b
BB =
××
××
==
005040.0
1.824.2941
2
45.1197.282
R
1
T
1
z
2
1
psinM2
1
S
=
××
×××
==
,
1
S
b
p
BB >
therefore,
( )
2
2
p-
3
f
x7.0-
2
f
x0.2
f
x0.5 -
6
2
1
p
-
2
1
p
p
J +=
( )
110883.1
2
1-023941.0-
6
2
45.1
-
2
45.1 ==
Substitution of given values into equation (54) gives
[
]
6
0995217.2005040.0
11088.1
990428.4 99521.04.2941002397.0018893.0
k
××
+
+×××
=
=
001672.0110883.1
173888.176200239716.0018893.0
+
××
=0.071734 darcy = 71.734 millidarcy.
Comparing 71.734 md with 72.562md obtained when the core was considered horizontal, it
is seen that inclination has reduced, the calculated permeability (k) by (72.562-
71.734)/72.564 = 1.141093 percent
The longer the core, the more, the effect of inclination.
Example 8
Use the data of example 4 to calculate the dimensionless friction factor (f
p
). Because of
simplicity assume that the core is horizontal.
Steady State Compressible Fluid Flow in Porous Media 495
( )
1
S
b
p
BBif
2
2
p-
3
f
x6.0-
2
f
x2.2
f
x2.5-
6
2
1
p
-
2
2
P
p
J <+=
c RL c RL
b
BB
p
A M
d M
p
p
/ /
2 2.546479
2
6
6
All other variables remain as defined in equation (52). During isothermal flow in which
there is no significant variation of the gas deviation factor (z) between the mid section
and the exit end of the porous medium, equation (54) can be written as:
( )
=
++
+++
3
f
x35.0
2
f
x5.0
f
x1
6
L
1
S
p
J
)
2
f
x7.0
f
x0.2-0.5()
3
f
x3.0-
2
f
x5.0
f
x-1(
1
T
1
Wz
b
p
BB
--
[
k
(55)
Beside the coefficients of
f
x all other variables in equation (55) remain as defined in
equation (54).
Example 7
Compute the permeability of the core of example 4 assuming that the case was vertical.
Solution
From example 4, W= γ Q
Substituting the given values, W=0.00239716gm/sec
Sin θ = sin 90
0
=1.0, M =28.97, L
p
=2cm
f
va
z
= z
1
=1, T
1
=T
a v
=294.4
o
K, A
p
=2cm
2
p
1
= 1.45atm, p
2
=1.0atm,
= 0.02cp.
004794.0
1.824.2941
297.282
R
av
T
f
av
z
LsinM2
f
x
=
××
××
==
The flow is isothermal so equation (55) is used.
023941.0-
3
f
x70.0-
2
f
x0.2
f
x0.5-
990428.4
3
f
x70.0-
2
f
x7.0
f
x2-0.5
995217.0
3
f
x35.0-
2
f
x05.0
f
x-1
=+
=+
=+
018893.0
97.2823
21.8202.0
M
p
A6
RL
1
c2
p
b
BB =
××
××
==
005040.0
1.824.2941
2
45.1197.282
R
1
T
1
z
2
1
psinM2
1
S
=
××
×××
==
,
1
S
b
p
BB >
therefore,
( )
2
2
p-
3
f
x7.0-
2
f
x0.2
f
x0.5 -
6
2
1
p
-
2
1
p
p
J +=
( )
110883.1
2
1-023941.0-
6
2
45.1
-
2
45.1 ==
Substitution of given values into equation (54) gives
[
]
6
0995217.2005040.0
11088.1
990428.4 99521.04.2941002397.0018893.0
k
××
+
+×××
=
=
001672.0110883.1
173888.176200239716.0018893.0
+
××
=0.071734 darcy = 71.734 millidarcy.
Comparing 71.734 md with 72.562md obtained when the core was considered horizontal, it
is seen that inclination has reduced, the calculated permeability (k) by (72.562-
71.734)/72.564 = 1.141093 percent
The longer the core, the more, the effect of inclination.
Example 8
Use the data of example 4 to calculate the dimensionless friction factor (f
p
). Because of
simplicity assume that the core is horizontal.
Natural Gas496
Solution
psf36.3069
psf1447.1445.1atm45.1
1
p
=
××==
psf80.2116psf1447.14atm1
2
p =×==
2.0.,R
0
530
2
T,1
2
z ===
( )
ft 1254.2/2in54.2/2cm2
p
L ×===
2
sec/ft2.32g,97.28M
==
cm713650.0
p
A128379.1
p
d
1545R,
2
cm4.0
2
cm2.02
p
A
==
==×=
ft023414.0 =
15455301
97.287.141
R
b
T
b
z
M
b
p
b
××
××
==
3
ft/b074890.0
=
sec/
3
ft5
_
E531467.32sec/
3
cm2
b
Q ×==
sec/b6
_
E289431.5
b
Q
b
W ==
2
ftsec/b7
_
E177086.4
2
ftsec/b5
_
E088543.202.0
=
×=
4172824
97.28
5
023414.02.326
0656168.01545621139.1
M
5
p
gd6
RL621139.1
a
p
BB
=
×××
××
==
( )
8E133065.0
53014172824
2
6
_
E289431.56
2
80.2116-
2
36.3069
2
T
2
z
a
p
BB
2
W6
2
2
p-
2
1
p
p
f
=
××××
=
=
The coordinate (R
NP
, f
p
) = (21.385242 , 0.0133065E8) locates very well in a previous graph of
f
p
versus R
NP
that was generated by (Ohirhian, 2008). The points plotted in the graph were
obtained by flowing water through synthetic tight consolidated cores. The plot is
reproduced here as follows.
Plot of f
p
versus R
Np
for Porous Media
Assignment
Use the data of example 4 to calculate the dimensionless friction factor (f
p
) considering the
core to be vertical
Conclusions
(1) The Darcy law as presented in API code 27 has been derived from the laws of
fluid mechanics
(2)
New general differential equations applicable to horizontal, uphill and downhill
flow of gas through porous media have been developed.
(3)
The Runge-Kutta algorithm has been used to provide accurate solutions to the
differential equations developed in this work.
(4)
The solution to the differential equation shows that inclination has the effect of
reducing laboratory measured values of gas permeability and dimensionless
friction factor- the longer a core the more the reduction of measured permeability
/ dimensionless friction factor.
Steady State Compressible Fluid Flow in Porous Media 497
Solution
psf36.3069
psf1447.1445.1atm45.1
1
p
=
××==
psf80.2116psf1447.14atm1
2
p =×==
2.0.,R
0
530
2
T,1
2
z ===
( )
ft 1254.2/2in54.2/2cm2
p
L ×===
2
sec/ft2.32g,97.28M
==
cm713650.0
p
A128379.1
p
d
1545R,
2
cm4.0
2
cm2.02
p
A
==
==×=
ft023414.0 =
15455301
97.287.141
R
b
T
b
z
M
b
p
b
××
××
==
3
ft/b074890.0
=
sec/
3
ft5
_
E531467.32sec/
3
cm2
b
Q ×==
sec/b6
_
E289431.5
b
Q
b
W ==
2
ftsec/b7
_
E177086.4
2
ftsec/b5
_
E088543.202.0
=
×=
4172824
97.28
5
023414.02.326
0656168.01545621139.1
M
5
p
gd6
RL621139.1
a
p
BB
=
×××
××
==
( )
8E133065.0
53014172824
2
6
_
E289431.56
2
80.2116-
2
36.3069
2
T
2
z
a
p
BB
2
W6
2
2
p-
2
1
p
p
f
=
××××
=
=
The coordinate (R
NP
, f
p
) = (21.385242 , 0.0133065E8) locates very well in a previous graph of
f
p
versus R
NP
that was generated by (Ohirhian, 2008). The points plotted in the graph were
obtained by flowing water through synthetic tight consolidated cores. The plot is
reproduced here as follows.
Plot of f
p
versus R
Np
for Porous Media
Assignment
Use the data of example 4 to calculate the dimensionless friction factor (f
p
) considering the
core to be vertical
Conclusions
(1) The Darcy law as presented in API code 27 has been derived from the laws of
fluid mechanics
(2)
New general differential equations applicable to horizontal, uphill and downhill
flow of gas through porous media have been developed.
(3)
The Runge-Kutta algorithm has been used to provide accurate solutions to the
differential equations developed in this work.
(4)
The solution to the differential equation shows that inclination has the effect of
reducing laboratory measured values of gas permeability and dimensionless
friction factor- the longer a core the more the reduction of measured permeability
/ dimensionless friction factor.
Natural Gas498
Nomenclature
pd
Incremental pressure drop
p
d
Incremental length of porous
medium
Q = Volumetric flow rate
v
= Average velocity flowing fluid
K
/
Proportionality constant that is dependent on both fluid and rock properties
k Permeability of porous medium
Absolute viscosity of flowing fluid
Mass density of flowing fluid
g = Acceleration due to gravity
z
Elevation of the porous medium above a datum. The + sign is used where the point of
interest is above the datum the – sign is used where the chosen point is and below the datum
fluid flowing of viscosityEffective
/
p = Pressure
= Specific weight of flowing fluid
v = Average fluid velocity
g = Acceleration due to gravity in a consistent set of units.
d
p
= Incremental length of porous medium
= Angel of porous medium inclination with the horizontal, degrees
dh
L
=Incremental lost head
c
= Dimensionless constant which is dependent on the pone size .
medium porous of ndisributio
c
1
= Constant used for conversion of units. It is equal to 1 in a consistent set of units
d
p =
Diameter of porous medium =
d
d= Diameter of cylindrical pipe
Porosity of medium
p
f
= Dimensionless friction factor of porous medium that is dependent
on the Reynolds number of porous medium.
pN
R Reynolds number of isotropic porous medium.
p
A
Cross-sectional area of porous medium
W
= Weight flow rate of fluid
=
b
Specific weight of fluid at P
b
and T
b
b
Q
= Volumetric rate of fluid, measured at P
b
and T
b
P
b
= Base pressure, absolute unit
T
b
= Base Temperature, absolute unit
z
b
= Gas deviation factor at p
b
and T
b
usually taken as 1
G
g
= Specific gravity of gas (air = 1) at standard condition
M = Molecular weight of gas
R = Universal gas constant
1
A = Pipe cross sectional area at point 1
1
v = Average fluid velocity at point. 1
1
= Specific weight of fluid at point 1
2
A = Pipe cross-sectional area at point 2
2
v =Average fluid velocity at point 2
2
= Specific weight of fluid at point 2
T
Absolute temperature
K = Constant for calculating the compressibility of a real gas
p
1
= Pressure at inlet end of porous medium
p
2
= pressure at exit end of porous medium
θ = Angle of inclination of porous medium with horizontal in degrees.
z
2
= Gas deviation factor at exit end of porous medium.
T
2
= Temperature at exit end of porous medium
T
1
=Temperature at inlet end of porous medium
z
a v
= Average gas deviation factor evaluated with T
a v
and p
a v
T
a v
= Arithmetic average temperature of the porous medium given by 0.5(T
1
+ T
2
) and p
a
v
= Average pressure
References
Amyx, J.W., Bass, D.M. & Whitting, R.L. (1960). Petroleum Reservoir Engineering – Physical
Properties, Mc Graw Hill Book Company, pp73-78, New York.
Aires, F, (1962).Differential Equations, McGraw Hill Book Company, New York.
Belhj, H.A., Agha, K.R., Butt, S.D.& Islam, M.R. (2003). Journal of tech. papers, 27
th
Nigeria
Annual Int. Conf. of Soc. of Pet. Engineers.
Brinkman, H.C. (1947) A calculation of the viscous force exerted by a flowing fluid on a
dense swarm of particles. Appl. Soc. Res. A1, pp. 27-34.
Darcy, H. (1856). Les fontainer publigues de la ville de Dijoin, Dalmont,
Forchheimer, P.Z . (1901). Wasserbewegung durch Boden, ZVDI, 45, PP. 1781
Giles, R.V., Cheng, L. & Evert, J. (2009). Schaum’s Outline Series ofFluid Mechanics and
Hydraulics, McGraw Hill Book Company, New York.
LeRosen, A. L. (1942). Method for the Standardization of Chromatographic Analysis, J.
Amer. Chem. Soc., 64, 1905-1907
Matter, L.G., Brar, S.& Aziz, K. (1975). ‘’Compressibility of Natural Gases, Journal of
Canadian Petroleum Technology”, pp. 77-80.
Steady State Compressible Fluid Flow in Porous Media 499
Nomenclature
pd
Incremental pressure drop
p
d
Incremental length of porous
medium
Q = Volumetric flow rate
v
= Average velocity flowing fluid
K
/
Proportionality constant that is dependent on both fluid and rock properties
k Permeability of porous medium
Absolute viscosity of flowing fluid
Mass density of flowing fluid
g = Acceleration due to gravity
z
Elevation of the porous medium above a datum. The + sign is used where the point of
interest is above the datum the – sign is used where the chosen point is and below the datum
fluid flowing of viscosityEffective
/
p = Pressure
= Specific weight of flowing fluid
v = Average fluid velocity
g = Acceleration due to gravity in a consistent set of units.
d
p
= Incremental length of porous medium
= Angel of porous medium inclination with the horizontal, degrees
dh
L
=Incremental lost head
c
= Dimensionless constant which is dependent on the pone size .
medium porous of ndisributio
c
1
= Constant used for conversion of units. It is equal to 1 in a consistent set of units
d
p =
Diameter of porous medium =
d
d= Diameter of cylindrical pipe
Porosity of medium
p
f
= Dimensionless friction factor of porous medium that is dependent
on the Reynolds number of porous medium.
pN
R Reynolds number of isotropic porous medium.
p
A
Cross-sectional area of porous medium
W
= Weight flow rate of fluid
=
b
Specific weight of fluid at P
b
and T
b
b
Q
= Volumetric rate of fluid, measured at P
b
and T
b
P
b
= Base pressure, absolute unit
T
b
= Base Temperature, absolute unit
z
b
= Gas deviation factor at p
b
and T
b
usually taken as 1
G
g
= Specific gravity of gas (air = 1) at standard condition
M = Molecular weight of gas
R = Universal gas constant
1
A = Pipe cross sectional area at point 1
1
v = Average fluid velocity at point. 1
1
= Specific weight of fluid at point 1
2
A = Pipe cross-sectional area at point 2
2
v =Average fluid velocity at point 2
2
= Specific weight of fluid at point 2
T
Absolute temperature
K = Constant for calculating the compressibility of a real gas
p
1
= Pressure at inlet end of porous medium
p
2
= pressure at exit end of porous medium
θ = Angle of inclination of porous medium with horizontal in degrees.
z
2
= Gas deviation factor at exit end of porous medium.
T
2
= Temperature at exit end of porous medium
T
1
=Temperature at inlet end of porous medium
z
a v
= Average gas deviation factor evaluated with T
a v
and p
a v
T
a v
= Arithmetic average temperature of the porous medium given by 0.5(T
1
+ T
2
) and p
a
v
= Average pressure
References
Amyx, J.W., Bass, D.M. & Whitting, R.L. (1960). Petroleum Reservoir Engineering – Physical
Properties, Mc Graw Hill Book Company, pp73-78, New York.
Aires, F, (1962).Differential Equations, McGraw Hill Book Company, New York.
Belhj, H.A., Agha, K.R., Butt, S.D.& Islam, M.R. (2003). Journal of tech. papers, 27
th
Nigeria
Annual Int. Conf. of Soc. of Pet. Engineers.
Brinkman, H.C. (1947) A calculation of the viscous force exerted by a flowing fluid on a
dense swarm of particles. Appl. Soc. Res. A1, pp. 27-34.
Darcy, H. (1856). Les fontainer publigues de la ville de Dijoin, Dalmont,
Forchheimer, P.Z . (1901). Wasserbewegung durch Boden, ZVDI, 45, PP. 1781
Giles, R.V., Cheng, L. & Evert, J. (2009). Schaum’s Outline Series ofFluid Mechanics and
Hydraulics, McGraw Hill Book Company, New York.
LeRosen, A. L. (1942). Method for the Standardization of Chromatographic Analysis, J.
Amer. Chem. Soc., 64, 1905-1907
Matter, L.G., Brar, S.& Aziz, K. (1975). ‘’Compressibility of Natural Gases, Journal of
Canadian Petroleum Technology”, pp. 77-80.
Natural Gas500
Muskat, M. (1949). “Physical Principles of Oil Production”, p. 142, McGraw-Hill Book
Company, Inc., New York.
Nutting , P.G. (1930). Physical Analysis of Oil Sands, Bulletin of American Association of
Petroleum Geologists, 14, 1337.
Ohirhian, P.U. (2008). A New Dimensionless Friction Factor for Porous Media, Journal of
Porous Media, Volume 11, Number 5.
Ohirhian, P.U. (2009). Equations for the z-factor and compressibility of Nigerian Natural
gas, Advanced Materials Research, Vols. 62-64, pp. 484-492, Trans Tech
Publications, Switzerland
Vibert. T. (1939). Les gisements de bauxite de I’Índochine, G
ẻnie Civil 115, 84.