1Gas Flow

Klinkenberg Effect

Molecules may be considered as spheres of diameter on the order of d = 10-3 microns (or m) One micron equals 10-6 meters

Molecules move at approximately the speed of sound (343 m/s in air at 20 oC) and collide randomly (Light travels through air at a speed of approximately 300 000 000 m/s)

The average length of this free movement is inversely proportional to pressure

Effect of Pressure At high pressures

Molecules are packed closely together. The length of movement between collisions becomes small.

Most of the collisions will take place between gas molecules, and the effect of collisions between molecules and the walls of the container are negligible.

At low pressures Internal friction due to collisions

between molecules decreases and the collisions of molecules with the walls of the capillary begin to have more of an effect on flow.

High Pressure

Low Pressure

2Pipe Flow

For flow in a horizontal pipe Velocity profile is parabolic. Fluid particles at the wall of the pipe do not move,

i.e., the velocity at the wall is zero and velocity is maximum along the centerline of the pipe.

At the wall a shear force acts in the direction opposite to the flow direction.

At the pipe entrance, this shear force acts to slow down the fluid velocity at the walls and a sufficient distance from the end, the velocity at the walls is zero.

Low Pressures

As gases become rarefied (molecules separated by large distances), it has been observed that there is slippage at the pipe walls, i.e., velocity is no longer zero. The Klinkenberg effect is nothing more than a porous medium interpretation of this slippage phenomenon. absolute permeability computed experimentally by

flowing gas through a porous medium and applying Darcys Law gives erroneously high values of absolute permeability and must be corrected.

Klinkenberg Permeability

The relation between gas permeability kg(computed from Darcys Law) and true absolute (liquid) permeability k is given by

b is a constant and pm is the mean (average) pressure at which gas permeability was measured.

+=

mg p

bkk 1

3Gas Permeability

For pm in atmospheres and hydrocarbon gases, b is roughly on the order of 1 or considerably smaller. This indicates that if pm is on the order of 50

atmospheres (735 psia), the difference between the absolute permeability to gas and the absolute permeability to liquid is negligible.

The value of b is proportional to the average distance of free movement (denoted by ) and inversely proportional to the radius of the capillary, denoted by r.

Klinkenberg Effect

As noted above as mean pressure increases, and hence b also decrease

As the radius of the capillaries (pore throats), increase b decreases.

As will be discussed later, larger pore throats are associated with larger permeability, thus larger permeabilities give smaller values of b, i.e, a smaller Klinkenberg effect.

Analysis if we measure kg at

different values of pmand plot kg versus 1/ pmwe should get a straight line.

Extrapolation of this line to 1/ pm = 0 gives k

Data from lowest MW gas yields the greatest slope indicating greatest slippage effect.

+=

mg p

bkk 1

4Effect of Absolute Permeability

Line fit through data from 175 core samples.

As permeability increases, the diameter of the pores increases, and the Klinkenberg effect will be reduced.

Steady-State Gas Flow

For liquid flow, we earlier assumed steady-state flow and neglected the variation in formation volume factor (or density) with respect to pressure.

For liquid, we also assumed that viscosity is independent of pressure.

For the flow of a real gas, this is no longer valid as density is strongly dependent on pressure and viscosity depends on pressure.

Engineering Equation of State

or

where M is the molecular weight, R is a constant which depends on the units used and T is (absolute) temperature. Using oil field units, R = 10.732 (psia - ft3 )/ (lb-mole - R), the units of the molecular weight are lb-mass/lb-mole and temperature is in degree Rankine (equals degrees Fahrenheit + 460), n=mass/M is the number of moles, Z is the z-factor.

ZnRTPV = ZRTpM=

5Steady-State

For one-dimensional steady-state flow, and this implies the mass flow rate

is constant so in particular

and is constant.

This applies for all points in the system.

scscqq =qq

scsc

=

0= tp /q

Darcys Law

For linear flow and any absolute system of units

=dlzdg

dldpkAq

scsc

sc

scsc RT

Mp=

TZppT

pT

ZTp

sc

sc

sc

sc

sc==

ZRTpM=

Darcys Law

Eliminating density terms

Assume ideal gas, porous medium, length L, horizontal flow and integrate

=dlzdg

ZRTpM

dldpkA

TZppTq

scsc

sc

dldldppkA

TpTLq

L

scsc

sc

=0

6Evaluation of Integrals

Perform the integration

Substitute into the following and simplify

( ) ( )( ).021

21 22

0

2

0pLpdL

dLdpdL

dLdpp

LL =

=

dldldppkA

TpTLq

L

scsc

sc

=0

Steady-State Flow

The result for the linear steady state horizontal flow case is any absolute system of units or in Darcy units,

( ) ( )

=

LpLpkA

TpTq

scsc

sc0

2

22

Stabilized Horizontal Flow

Ideal gas

Units: qsc = Mscf/d Remaining units are field units; i.e., psia, ft,

md, oR, cp.

( ) ( )

=

LpLpkA

TpTq

sc

scsc

0103.164 22-6

7Non-Ideal Gases

In Reservoir Engineering, you will learn about a technique of transforming gas pressure to real gas pseudopressure that accounts for variation of Z-factor with pressure.

Here we will use an average Z-factor and average viscosity to adjust the Ideal gas equation.

t

( ) ( )

=

LpLpkA

TpZTq

sc

scsc

022

6-103.164

Example

A rectangular sand body is flowing 10MM SCF/day under a downstream pressure of 1000 psia. Standard conditions are 14.7 psiaand 80F. Average deviation factor is 0.80. The sand body is 1000 feet long, 100 feet wide, and 10 feet thick. Porosity is 22 per cent and average permeability to gas at 17 per cent connate water is 125 md. BHT = 160F; = 0.029 cp. Assume steady-state flow.

What is the upstream pressure?

qsc = 10000 Mscf/d Tsc = 80 oF = 540oR psc = 14.7 psia Average z = 0.80 Area = 1000 ft2

T = 160 oF = 620 oR

( )( )( )( )

( )( ) ( ) ( )

=

100001000

029.01000125

6207.148.0540103.16410000

226- p

( ) ( )

=

LpLpkA

TpZTq

sc

scsc

022

6-103.164

8Solution

Solve the last equation: p(0) = 3301.6 psia What is the inlet gas flowrate? Recall

Flow rate and density vary with position, but product is constant. Flow rate at any point converted to standard conditions gives the same value,

; so

scsc

scsc

qqqq ==

.scq

Solution

Recall

TZppT

MpRT

ZRTpM

sc

sc

sc

sc

sc

==

sc

scsc

pTTZp=

ppq

pTZTpq sc

sc

sc8

7 1035.110540

6208.07.14 ===

givesqq scsc=

Solution

Solving for inlet flowrate

What is the outlet flow insitu flow rate?

Note the large difference in flowrates between inlet and outlet (different from liquid case).

dayftres

pq

34

88

1009.46.3301

1035.11035.1 ===

dayftres

pq

34

88

105.131000

1035.11035.1 ===

9Problem

What is the pressure gradient at the midpoint of the sand?

From Darcys law

We need to find the pressure and density at the midpoint of the sand pack (at L = 500 ft).

dLdpkAq

scsc

=610328.6

Solution

We have

( ) ( )

( )( )( )( )

( )( )( )

( ) ( )

( ) psia 3.24395950316500

6.3301029.01000125

6207.148.0540103.16410

0103.164

226-4

22-6

==

=

=

Lp

Lp

LpLpkA

TpzTq

sc

scsc

Solution

Solving Darcys Law for pressure gradient

( )( )( ) ( )( )

psi/ft029.2100012510328.6540)3.2439(

029.010)620)(7.14)(8.0(

10328.610328.6

6

4

66

==

==

kAq

pTTZp

kAq

dLdp sc

sc

scscsc

10

Non-Darcy Flow

What is it? For laminar flow in a pipe, flow particles move along

smooth paths in layers or laminars. At high velocities, turbulent flow may occur and in

this case streamlines can cross. Darcy flow in porous media is the analogue of

laminar flow in pipes and non-Darcy flow in porous media is the analogue of turbulent flow in pipes.

The more viscous the fluid, the less likely to obtain non-Darcy or turbulent flow.

In practice, non-Darcy flow in porous media is considered only for gas wells.

Forscheimer Equation Forscheimer equation was developed for turbulent

flow in pipes The analogy basically assumes that flow through

pore throats is similar to flow through a pipe For one-dimensional flow in the L direction

Where is the turbulence factor (of dimension one over length) and vl is the velocity.

Note = 0 corresponds to ?

2ll vvkdl

dp +=

= -m k-n; ( = 4.85e+04, m = 5.5, n = 0.5Geertsma (1974)

11

High Velocity Gas Flow

For flow of a gas at high velocities, Darcys Law no longer applies and must be modified.

Darcy Gas Flow (Absolute system of units)

Non-Darcy Flowsc

sc

qbaq

pp 22

21 +=

( ) ( ) akAT

TLpq

Lppsc

sc

sc== 20

22

Analysis, Absolute or Darcy Units

Cartesian plot of left side versus qsc at different rates gives a straight line

Intercept:

Can get permeability from intercept.

scsc

qbaq

pp 22

21 +=

sc

sc

kATTLpa 2 =

Steady-State Flow, Field Units

Rate in Mscf/D. Assumes standard pressure of 14.7 psi and standard temperature of 60 degrees Fahrenheit. Can rewrite equation as

Plot of left side versus q will give a line of slope B and intercept A.

2 2( )1422 (ln( / ) )

e wfsc

e w sc

kh p pq

ZT r r s Dq= + +

scsc

wfe BqAq

pp +=22

12

Steady-State Flow, Field Units

Rewrite as

scsc

wfe BqAq

pp +=22

khsrrTZA we ))/(ln( += 1422

khTDZB 1422=

2 2( )1422 (ln( / ) )

e wfsc

e w sc

kh p pq

ZT r r s Dq= + +

Steady-State Flow, Field Units Rewrite as

Given , we can compute flow rate in Mscf/D for different wellbore pressure to obtain a deliverability plot. AOF (Absolute open flow potential) is the rate when

scsc

wfe BqAq

pp +=22

0)( 222 =+ wfescsc ppAqBq

BppBAA

q wfesc 2)(4 222 ++=

ep

.0=wfp

Deliverability Plot

0

2000

4000

6000

8000

0 1000 2000 3000 4000

Rate, Mscf/D

Flow

ing

wel

lbor

epr

essu

re

13

Effect of non-Darcy Flow

The most important consequence of non-Darcy flow is that we obtain a smaller flow rate for a given drop in pressure squared.

Example

Actual field data from an isochronal test. q in Mscf/D and pressures in psia. A= 228 and B = 0.024

p2 psi2 q (Darcy) q (non-Darcy) 10000 43.86 43.66 50000 219.30 214.46 100000 438.60 420.03 500000 2192.98 1837.55 1000000 4385.97 3264.31

Field ExampleFlowrate versus p^2

0

500

1000

1500

2000

2500

3000

3500

4000

4500

5000

0 200000 400000 600000 800000 1000000 1200000

p^2 psi ^2

q (Darcy)

q (non-Darcy)