Natural Gas Part 13 ppt

Regarding the cross sectional area of the porous medium A p as a constant, equation 11 can be differentiated and solve simultaneously with equations 2 and 1 to obtain.. Regarding the cro

NPR

pg

W4



0.023417

- E177086.42.32

6 -E289431.54

bQbgG88575.36NP

R



0.023414530

17 -E177086.42.32

5 -E052934.71447.141885750.36

=

=

= A1 1 2 A2 21

Equation (11) can be differentiated and solved simultaneously with the lost head formulas

(equation 2, 3 and 4), and the energy equation (equation 1) to arrive at the general

differential equation for fluid flow in a homogeneous porous media

Regarding the cross sectional area of the porous medium (A p) as a constant, equation (11)

can be differentiated and solve simultaneously with equations (2) and (1) to obtain

_

c v

sin k

Equation (12) is a differential equation that is valid for the laminar flow of any fluid in a

homogeneous porous medium The fluid can be a liquid of constant compressibility or a gas

The negative sign that proceeds the numerator of equation (12) shows that pressure

decreases with increasing length of porous media

The compressibility of a fluid (C f) is defined as:

d 1

NP

R

pg

W4



0.023417

- E

177086

42

.32

6 -

E289431

.5

Tb

zg

pRgd

bQ

bg

G88575

.36

17

E

-177086

42

.32

5 -

E052934

.7

1447

.14

1885750

.36

=

=

= A1 1 2 A2 21

Equation (11) can be differentiated and solved simultaneously with the lost head formulas

(equation 2, 3 and 4), and the energy equation (equation 1) to arrive at the general

differential equation for fluid flow in a homogeneous porous media

Regarding the cross sectional area of the porous medium (A p) as a constant, equation (11)

can be differentiated and solve simultaneously with equations (2) and (1) to obtain

_

c v

sin k

Equation (12) is a differential equation that is valid for the laminar flow of any fluid in a

homogeneous porous medium The fluid can be a liquid of constant compressibility or a gas

The negative sign that proceeds the numerator of equation (12) shows that pressure

decreases with increasing length of porous media

The compressibility of a fluid (C f) is defined as:

d 1

The equation of state for a non ideal gas is:

2pd2

1pd

pp

g d

2 sin 1.621139 5

2

2 1.621139

(Matter et al, 1975) and ( Ohirhian, 2008) have proposed equations for the calculation of the

compressibility of hydrocarbon gases For a sweet natural gas (natural gas that contains CO2

as major contaminant), (Ohirhian, 2008) has expressed the compressibility of the real gas

(Cg ) as:

pf

(22)

For Nigerian (sweet) natural gas K = 1.0328 when p is in psia Then equation (19) can then

be written compactly as:

zRT2KWpC

, zRT

sinM2pB ,M

5pgd

zRT2Wpf621139.1pAA

Kinetic Effect in Pipe and Porous Media

An evaluation of the kinetic effect can be made if values are substituted into the variables that occurs in the denominator of the differential equation (23)

Where C for a pipe is given by, 4

gMdzRT

2KW

C=

Here

sec/lb75.0

W = , d = 4inch = 4/12 ft =0.333333ft ,

psf7128psf4449.5psia45.5

, fluid) theis(air 1.0

z = R=1545 ,g=32.2 ft / sec2, M =28.97.Then,

58628.415044

333333.097.282.32

55015451275.01

an for 1

K=

The equation of state for a non ideal gas is:

2p

d2

1p

d

pp

g d

2 sin 1.621139 5

2

2 1.621139

(Matter et al, 1975) and ( Ohirhian, 2008) have proposed equations for the calculation of the

compressibility of hydrocarbon gases For a sweet natural gas (natural gas that contains CO2

as major contaminant), (Ohirhian, 2008) has expressed the compressibility of the real gas

(Cg ) as:

pf

(22)

For Nigerian (sweet) natural gas K = 1.0328 when p is in psia Then equation (19) can then

be written compactly as:

zRT2KWpC

, zRT

sinM2pB ,M

5pgd

zRT2Wpf621139.1pAA

Kinetic Effect in Pipe and Porous Media

An evaluation of the kinetic effect can be made if values are substituted into the variables that occurs in the denominator of the differential equation (23)

Where C for a pipe is given by, 4

gMdzRT

2KW

C=

Here

sec/lb75.0

W = , d = 4inch = 4/12ft =0.333333ft ,

psf7128psf4449.5psia45.5

, fluid) theis(air 1.0

z= R =1545 ,g =32.2 ft / sec2, M =28.97.Then,

58628.415044

333333.097.282.32

55015451275.01

an for 1

K=

The kinetic effect correction factor is

999183.027128

58628.41504_12p

C_

Example 3

If the pipe in example 1 were to be a cylindrical homogeneous porous medium of 25 %

porosity, what would be the kinetic energy correction factor?

Solution

Here, d p = d  = 0.333333 0.25 = 0.1666667ft

0212 344046

4 166667 0 97 28 2 32

550 1545 1

2 75 0 1 p C

0212.34410461

2p

pC_

The kinetic effect is also small, though not as small as that of a pipe The higher the pressure,

the more negligible the kinetic energy correction factor For example, at 100 psia, the kinetic

energy correction factor in example 2 is:

998341.02)144100(

0212.3441046_

× Simplification of the Differential Equations for Porous Media

When the kinetic effect is ignored, the differential equations for porous media can be

simplified Equation (14) derived with the Darcy form of the lost head becomes:

Making velocity (v) or weight (W) subject of the simplified differential equations

When v is made subject of equation (24), we obtain:

The kinetic effect correction factor is

999183

02

7128

58628

41504_

12

p

C_

Example 3

If the pipe in example 1 were to be a cylindrical homogeneous porous medium of 25 %

porosity, what would be the kinetic energy correction factor?

Solution

Here, d p = d  = 0.333333 0.25 = 0.1666667ft

0212

344046

4 166667

0

97

28 2

32

550 1545

1

2 75

0

1 p

02

7128

0212

34410461

2p

pC

_

The kinetic effect is also small, though not as small as that of a pipe The higher the pressure,

the more negligible the kinetic energy correction factor For example, at 100 psia, the kinetic

energy correction factor in example 2 is:

998341

02

)144

100(

0212

3441046_

× Simplification of the Differential Equations for Porous Media

When the kinetic effect is ignored, the differential equations for porous media can be

simplified Equation (14) derived with the Darcy form of the lost head becomes:

Making velocity (v) or weight (W) subject of the simplified differential equations

When v is made subject of equation (24), we obtain:

s

v Volumetric flux across a unit area of

porous medium in unit time along

flow path, S cm / sec

2sec / cm980.605gravity, todue

on Accelerati

cc / massgm , fluidofDensity

Mass

cc weight / gm

, fluidof weight Specific

coordinate

Vertical

z

scentipoisefluid,

Horizontal and Uphill Gas Flow in Porous Media

In uphill flow, the + sign in the numerator of equation (23) is used Neglecting the kinetic

effect, which is small, equation (23) becomes

B

,M

5pgd

2zTRWp1.621139fp

AA

=

=

An equation similar to equation (33) can also be derived if the Darcian lost head is used The

horizontal / uphill gas flow equation in porous media becomes

Where

Mk

2pd

zTRWc

546479.2

Mk

22d

zTRWc

MkpA

zTRWc

/pAA

Solution to the Horizontal/Uphill Flow Equation

Differential equations (33) and (34) are of the first order and can be solved by the classical Runge - Kutta algorithm The Runge - Kutta algorithm used in this work came from book of (Aires, 1962) called “Theory and problems of Differential equations” The Runge - Kutta solution to the differential equation

( )x,y at x x given thatf

x at 0 y

n

n sub ervals steps

( , ) 1

s

v Volumetric flux across a unit area of

porous medium in unit time along

flow path, S cm / sec

2sec

/ cm

980.605gravity,

todue

on Accelerati

cc /

massgm

, fluid

ofDensity

Mass

cc weight /

gm ,

fluidof

weight Specific

atmrefers,

alonggradient

sq /

,coordinate

Vertical

z

scentipoise

fluid, the

Horizontal and Uphill Gas Flow in Porous Media

In uphill flow, the + sign in the numerator of equation (23) is used Neglecting the kinetic

effect, which is small, equation (23) becomes

B

,M

5p

gd

2zTRW

p1.621139f

pAA

=

=

An equation similar to equation (33) can also be derived if the Darcian lost head is used The

horizontal / uphill gas flow equation in porous media becomes

Where

Mk

2pd

zTRWc

546479.2

Mk

22d

zTRWc

MkpA

zTRWc

/pAA

Solution to the Horizontal/Uphill Flow Equation

Differential equations (33) and (34) are of the first order and can be solved by the classical Runge - Kutta algorithm The Runge - Kutta algorithm used in this work came from book of (Aires, 1962) called “Theory and problems of Differential equations” The Runge - Kutta solution to the differential equation

( )x,y at x x given thatf

x at 0 y

n

n sub ervals steps

( , ) 1

p p1 2 22y a (36) Where

ax72.0

2x48.1x96.46

22

3ax36.0

2ax.0ax1

apaaay

a a

) (

+ +

+

+ +

+

=

a x 72 0 a x 96 1 96 4 6

a p

u

+ +

)L2Sp2(AA

ap

R2T2

22psinM22S

,M

5pgd

2RW2T2pf621139.12AA



=

=

RavT

aavz

LsinM2a

,M

5pgd

2RWavTavzf621139.1ap



=

=

Where:

p1 = Pressure at inlet end of porous medium p2 = Pressure at exit end of porous medium

fp = Friction factor of porous medium

θ = Angle of inclination of porous

medium with horizontal in degrees

z2 = Gas deviation factor at exit end of

p a v a = a

p aa

2 2

In equation (36), the component k4 in the Runge - Kutta algorithm was given some weighting to compensate for the variation of temperature (T) and gas deviation factor (z) between the mid section and the inlet end of the porous medium In isothermal flow where there is little variation of the gas deviation factor between the mid section and the inlet end

of the porous medium, the coefficients of x a change slightly, then,

)

2a5.0a25(6

2p

u

)

3a5.0

2a2a5(6

22

3a25.0

2a5.0a1

apaaa

+ + +

+ +

+

+ +

2b5.0b1

bpaa

b72.0

2b48.1b96.46

22

+ +

+

b72.0b96.196.46

2p

+ +

+

p p1 2 22y a (36) Where

ax

72

0

2x

48

1x

96

46

22

3a

x36

.0

2a

x

0a

x1

ap

aaa

y

a a

) (

+ +

+

+ +

+

=

a x

72

0 a

x 96

1

96

4 6

a p

u

+ +

)L2

Sp2

(AA

ap

R2

T2

22

sinM

22

S

,M

5p

gd

2RW

2T

2p

f621139

.1

2AA



=

=

Rav

T

aav

z

Lsin

M2

a

,M

5p

gd

2RW

avT

avz

f621139

.1

ap

p1 = Pressure at inlet end of porous medium p2 = Pressure at exit end of porous medium

fp = Friction factor of porous medium

θ = Angle of inclination of porous

medium with horizontal in degrees

z2 = Gas deviation factor at exit end of

p a v a = a

p aa

2 2

In equation (36), the component k4 in the Runge - Kutta algorithm was given some weighting to compensate for the variation of temperature (T) and gas deviation factor (z) between the mid section and the inlet end of the porous medium In isothermal flow where there is little variation of the gas deviation factor between the mid section and the inlet end

of the porous medium, the coefficients of x a change slightly, then,

)

2a5.0a25(6

2p

)

3a5.0

2a2a5(6

22

3a25.0

2a5.0a1

apaaa

+ + +

+ +

+

+ +

2b5.0b1

bpaa

b72.0

2b48.1b96.46

22

+ +

+

b72.0b96.196.46

2p

+ +

+

Where aa pb = ) L

2S

/2AA

kM

2p

RW2T2zc546479.2

kM

22

RW2T2zc8kMpA

RW2T2zc2/2AA

RWavT

bavzc2.546479

MkPA

RWavT

bavzc2bp

,R2T2z

22sinM22

b av z

L sin M 2 b

Where

z av b = Average gas deviations factors evaluated with Ta v and p a v b

T a v = Arithmetic average Temperature of the porous medium = 0.5(T1+T2),

pavb p22 0 5 aapb

All other variables remain as defined in equation (36) In isothermal flow where there is not

much variation in the gas deviation factor (z) between the mid section and inlet and of the

porous medium there is no need to make compensation in the k4 parameter in the Runge

Kuta algorithm, then equation (37) becomes:

Where:

+ +

+ +

b x 0 b 1 (

b p aa bT y

Where

TzPU

,R2T2z

22sinM22

,M

5pg

RL621139.1apBB

z Average gas deviations factors evaluated with Tav and pavc and

2

22

21

pav

+

=

All other variables remain as defined in previous equations

In isothermal flow where there is no significant change in the gas deviation factor (z), equation (39) becomes:

a

W f BB z T p P

x c x c

Where aa pb = ) L

2S

/2

AA

kM

2p

RW2

T2

zc

546479

2

kM

22

RW2

T2

zc

8k

Mp

A

RW2

T2

zc

2/

2AA

RWav

T

bav

zc

2.546479

MkP

A

RWav

T

bav

zc

2b

p

,R

2T

2z

22

sinM

22

T

b av

z

L sin

M 2

b

Where

z av b = Average gas deviations factors evaluated with Ta v and p a v b

T a v = Arithmetic average Temperature of the porous medium = 0.5(T1+T2),

pavb p22 0 5 aapb

All other variables remain as defined in equation (36) In isothermal flow where there is not

much variation in the gas deviation factor (z) between the mid section and inlet and of the

porous medium there is no need to make compensation in the k4 parameter in the Runge

Kuta algorithm, then equation (37) becomes:

Where:

+ +

+ +

b x 0 b 1 (

b p aa bT y

Where

TzPU

,R2T2z

22sinM22

,M

5pg

RL621139.1apBB

z Average gas deviations factors evaluated with Tav and pavc and

2

22

21

pav

+

=

All other variables remain as defined in previous equations

In isothermal flow where there is no significant change in the gas deviation factor (z), equation (39) becomes:

a

W f BB z T p P

x c x c

a p BB

2 W 6

2 2 p -

2 1

(42)

Example 4

The following data came from the book of (Giles et al., 2009) called “theory and problem of

fluid mechanics and hydraulics”

W = 0.75 1b/sec of air, R = 1544, L = 1800ft, d = 4inch = 0.333333ft,

g = 32.2ft/sec2 , z 2 = zava = 1 (air is fluid), T2 = Tav =900F=5500R

(Isothermal flow), p1 = 49.5psia = 7128psf, P2 = 45.73 psia = 6585.12 psf

Pipe is horizontal

(a) Calculate friction factor of the pipe (f)

(b) If the pipe were to be filled with a homogenous porous material having a porosity

of 20% what would be the friction factor (fp )?



195610.8241

97.285333333.02.326

18001544621139.1

0.75 6

2 6585.12 - 2 7128 2

T 2 z BB 2 W

6

2 2 -

2 1

333333.0p )b

62 10934995

97 28

5 1490751

0 2 32 6

1800 1544 621139

1

M

5 p gd 6

RL 621139 1

` a p BB

apBB2W6

22-

21p

4-E3.667626

550 10934995.62

0.756

26585.12-27128

]

cx72.0c1.96x96.4avTbav

z

c336.02cx.0c x 12T2zk

bpBB2W

,M

2p6

RLc576479.2MpA6

RLc2bp

2 z

a p

BB

2 W

6

2 2

p -

2 1

(42)

Example 4

The following data came from the book of (Giles et al., 2009) called “theory and problem of

fluid mechanics and hydraulics”

W = 0.75 1b/sec of air, R = 1544, L = 1800ft, d = 4inch = 0.333333ft,

g = 32.2ft/sec2 , z 2 = zava = 1 (air is fluid), T2 = Tav =900F=5500R

(Isothermal flow), p1 = 49.5psia = 7128psf, P2 = 45.73 psia = 6585.12 psf

Pipe is horizontal

(a) Calculate friction factor of the pipe (f)

(b) If the pipe were to be filled with a homogenous porous material having a porosity

of 20% what would be the friction factor (fp )?



195610.8241

97

.28

5333333

.0

2

326

18001544

621139

195610.824 2

0.75 6

2 6585.12

2

-7128 2

T 2

z BB

2 W

6

2 2

-2 1

333333.0p )b

62 10934995

97 28

5 1490751

0 2 32 6

1800 1544 621139

1

M

5 p gd 6

RL 621139 1

` a p BB

apBB2W6

22-

21p

4-E3.667626

550 10934995.62

0.756

26585.12-27128

]

cx72.0c1.96x96.4avTbav

z

c336.02cx.0c x 12T2zk

bpBB2W

,M

2p6

RLc576479.2MpA6

RLc2bp

R 2 T 2 z

2 2 P sin M 2 2 S



=

R av T

c av z

L sin M 2 c

p -p 

c b

+ + +

]

[

) 2 c x 0 c 2x 5 ( 2 T 2 z

) c 3 25 0 2 c x 0 c x 1 ( 2 T 2 z k

b p BB 2 W

The following problem came from the book of (Amyx et al., 1960) During a routine

permeability test, the following data were obtained

Flow rate (Q) = 1,000ccof air in 500sec

Pressure down stream of core (p2) = 1 atm absolute Flowing temperature (T) = 70 0F

Viscosity or air at test temperature (μ) = 0.02cp Cross-sectional area of core (A p) = 2cm2 Pressure upstream of core (p1) = 1.45 atm absolute Length of core (L p) = 2cm

Compute the permeability of the core in millidarcy

R0530 = 294.40K



Taking the core to be horizontal

p p

2 T 2 z

b P wBB 6

in 1 c ( , M A 6 RL 2 BB

p

pb  

2

_ E 889311 1 97 28 2 6

2 1 82 02 0 2

R 2

T 2

z

2 2

P sin

M 2

2 S



=

R av

T

c av

z

L sin

M 2

p -p 

c b

+

+ +

+

]

[

) 2

c x

0

c 2x

5 (

2 T

2 z

) c

3 25

0

2 c

x

0 c

x 1

( 2

T 2

z k

b p

BB 2

The following problem came from the book of (Amyx et al., 1960) During a routine

permeability test, the following data were obtained

Flow rate (Q) = 1,000ccof air in 500sec

Pressure down stream of core (p2) = 1 atm absolute Flowing temperature (T) = 70 0F

Viscosity or air at test temperature (μ) = 0.02cp Cross-sectional area of core (A p) = 2cm2 Pressure upstream of core (p1) = 1.45 atm absolute Length of core (L p) = 2cm

Compute the permeability of the core in millidarcy

R0530 =294.40K



Taking the core to be horizontal

p p

2 T 2 z

b P wBB 6

in 1 c ( , M A 6 RL 2 BB

p

pb  

2

_ E 889311 1 97 28 2 6

2 1 82 02 0 2

Then

millidarcy72.56

darcy07256.0

4.29412E889311.13E397163.26

1 - 45 1

Amyx , et al obtained the permeability of this core as 72.5md with a less rigorous equation

Horizontal and Downhill Gas Flow in Porous Media

In downhill flow, the negative (-) sign in the numerator of equation (23) in used Neglecting

the kinetic effect, equation (23) becomes:

5pgd

zTRpf621139.1p

zTR

sinM2p

2p

zTRWc546479.2MkpA

zTRWcPA

zTR

sin M 2 p

Solution to the differential equation for horizontal and downhill flow

The Runge-Kutta numerical algorithm that was used to provide a solution to the differential

equation for horizontal and uphill flow can also be used to solve the differential equation for

horizontal and downhill flow Application of the Runge - Kutta algorithm to equation (47)

,

2

2 sin 11

5pgd

2RWavT

davz1.621139fc

RavT

davz

2MsinθL

d =d

Other variables remain as defined in previous equations

In equation (49), the parameter k4 in the Runge-Kutta algorithm is given some weighting to compensate for the variation of the temperature (T) and the gas deviation factor between the mid section and the exit end of the porous medium In isothermal flow in which there is no significant variation of the gas deviation factor (z) between the midsection and the exit end

of the porous medium, equation (49) becomes

Then

millidarcy72.56

darcy07256

.0

4

2941

2E

889311

13

E397163

.2

6

1 -

45

Amyx , et al obtained the permeability of this core as 72.5md with a less rigorous equation

Horizontal and Downhill Gas Flow in Porous Media

In downhill flow, the negative (-) sign in the numerator of equation (23) in used Neglecting

the kinetic effect, equation (23) becomes:

5p

gd

zTRp

f621139

.1

p

zTR

sinM

2p

2p

zTRWc

546479

2Mk

pA

zTRWc

PA

zTR

sin M

2 p

Solution to the differential equation for horizontal and downhill flow

The Runge-Kutta numerical algorithm that was used to provide a solution to the differential

equation for horizontal and uphill flow can also be used to solve the differential equation for

horizontal and downhill flow Application of the Runge - Kutta algorithm to equation (47)

,

2

2 sin 11

5pgd

2RWavT

davz1.621139fc

RavT

davz

2MsinθL

d =d

Other variables remain as defined in previous equations

In equation (49), the parameter k4 in the Runge-Kutta algorithm is given some weighting to compensate for the variation of the temperature (T) and the gas deviation factor between the mid section and the exit end of the porous medium In isothermal flow in which there is no significant variation of the gas deviation factor (z) between the midsection and the exit end

of the porous medium, equation (49) becomes

- x - x d d 2

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

2ex.2ex.5-6

21

)

3ex.0-

2ex.0ex-1(

dpaad

+ +

+

=

e x 6 0 - e x 2 - 2 5 6

d p

u

+

)L1S-1

/(AAp

dp

, Mk

2 p d

RW av T

e av z c 54679 2

Mk p A

RW 1 T 1 z c 2 1

/ AAp

1 1



, R T z

L 2Msinθ x

RWavTeavzμc2.546479Mk

PA

RWavTavμzc2dp

a p BB if

2 2 -

3 f x 0 -

2 f x 2 f x 5 - 6

2 1 -

2 iP p

( 5 xf 2- 0 3f)- 22 if BBpa iS6

2 1 -

2 1 -

2 2 P p

a BBp

favz

LsinM2fx,M

5pgd6

21sinM21