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8. Gas

There is no qualitative difference between the oil and gas flow. However, since thegas is lighter and less viscous than the oil, the effects are negligible for oil proveimportant for gas flow, and vice versa.

8.1 Physical properties of the gas

8.1.1 Pressure and volumeThe relationship between pressure and volume can be expressed by the generalequation of state

nzRTpV= (8-1)

ased on !an der "aals equation or other extended state equations, one can in

principle express the # factor as a function of reduced pressure and temperature$pressure % critical pressure and temperature % critical temperature. &or gas mixtures, itis common to estimate the so'called pseudo critical pressure and temperature, basedon averaging the critical pressure and temperature of the gas components, or simplyfrom the specific gravity of the composition.

),( pRpR Tpzz= (8-2)

ppR : pseudo reduced pressure

TpR : pseudo reduced temperature

&igure 8.1 illustrates (8')* by the +tanding'at# correlation.&igure 8.) gives the correlation between pseudo critical pressure and temperatureand specific gravity

y formation factor, we understand the ratio between the actual volume and thevolume of the same quantity of gas at atmospheric standard conditions. &romequation (8.1* follows

oo

oo

og z

z

T

T

p

p

V

VB ===

(8-3)

-ompressibility means relative volume change with pressure change. -ompressibilityat constant temperature can be estimated from the general equation of state. f the #factor can be considered constant, compressibility corresponds to the inverse ofpressure

pp

z

dp

d

z

p

dp

dV

Vc

TT

g

11

== (8-4)

The result implies that gas compressibility decreases rapidly when the pressureincreases. "e have all experienced that bicycle tires become harder when we pumpthem up.

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Figur2 8.1 z-factor diagram, by Yarborough & Hall /19!/ "#t"$d"d "%uatio$ of stat"

8.1.2 Density/ensity equals the ratio between mass and volume. 0as density depends on

molecular weight$ , pressure and temperature. &rom the state equation (8.1* wecan express the relationship

zRT

pM

V

nM

V

m=== (8-)

t is worth noting that even if the gas density is less than the oil density, the gasdensity, at reservoir conditions it can still be large compared what we associate withgas. (2s a rule of thumb, the gas density in the reservoir approximately equals thereservoir pressure in bar$ thus at pressure 133 bar, the density will be of the order of133 4g%m5*

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Figur" 8.2 s"udo critical 'r"ssur" a$d t"m'"ratur"

8.1.5 !iscosity

The viscosity is much less gas than for oil, so that the viscous resistance of flow iscorrespondingly less. The viscosity of natural gas can reasonably accuratelyestimated by the correlations of -arr'obayashi'urrows % 167 %, reproduced below.&igure 8.5 gives the viscosity at atmospheric conditions, as a function of the averagemolecular weight (specific gravity 9 )8.6:* &igure 8. gives viscosity increase athigher temperatures and pressures.

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&igure 8.5 !iscosity for natural gas at atmospheric pressure

&igure 8. !iscosity change as a function of pressure

8.1. Temperature changes with pressureThe gas will release, or absorb, heat when the volume is changed. n the reservoir,the heat released is absorbed by the roc4, usually without much temperature change.so that the reservoir temperature may in practice be considered constant ..

"hen gas flows through a no##le, there is little time to transfer heat and the heat

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capacity of gas is small. f no heat is transferred, the change in temperature isexpressed by the adiabatic ;state equation; (process equation*

kii

kVppV = (8-!)

&rom this, and the general equation of state, the adiabatic changes of temperatureand density is expressed as

k

k

i

ip

pTT

1

= (8-")

k

i

ip

p1

= (8-8)

k : adiabatic exponent, for ideal gases: k = cp/cv

cp : specific heat capacity at constant pressurecv : specific heat capacity at constant volume

i : initial density

Ti : initial temperature

8.2 ($flo)

8.2.1 ($flo) '"rforma$c"The &orcheimer equation (

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&igure 8.7 illustrates the flow characteristics of a gas well, calculated from equation

(8'11* above, with 'factor estimated with Te4?s correlation (

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=+=

R

w

R

wR

p

p"

2

"

p2

pp"

&or example$ if the reservoir pressure is )33 bar and the well pressure 163, the error

neglecting gas expansion becomes$ @ 3.7. (1'163%)33* @ 3.3)7, ie ).7A

8.2.2 +ac'r"ssur" "%uatio$2n alternative way of expressing the inflow of a gas well is by the empirical;bac4pressure equation; introduced by the B.+. ureau of ines in 16)6

n

wRg pp#q )(22 = (8-

13)

The bac4pressure equation (8'15* has two parameters$ n and -. These must bedetermined from multi'rate production test.

The flow characteristics based on &orchheimer equation also involves two

parameters that can be determined from the production test$ :Ako'A. . ut these

are also connected to the reservoir and fluid parameters. The bac4pressure equationis purely empirical no such correlations.

8. Flo) i$ 'roductio$ tubi$g

8..1 Flo) "%uatio$Pressure drop %it& statio$ar *#o% i$ pipes is ua$ti*ied as e*ore

02

1 2 =+++ d$vd

%d$gvdvdp $

(8-14)

%&ere

A

mv

= o'

d

A

mdv

2

1=

2t stationary conditions, the mass flow.m , will be constant all along the pipe, whilethe speed varies with pressure. t is therefore convenient to express (8'1* in massterms

081!

2

2

242

2

=++ d$d

m%d$gd

d

mdp $

(8-1)

8..2 r"ssur" i$ static gas colum$f we have a closed gas well, #ero flowrate, the second and fourth terms of (8'17*

become #ero. f we express the gas density at the general state equation, we get

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8

0=

+ d$zRT

g

p

dp $(8-1!)

y integrating (8'1

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9

2

=

TRz

&MgeA $g

( ) ( )g

$

g

o

g

$

% AT

Tpz

dg

%A

M

TRz

dg

%A

=

= 1

81

82

0

0

2

2

2

2

2long a production pipe, the parameter groups 2g and 2f can usually consideredconstants. This means that we can estimate these from production data, by plotting$

( )2/ w'h pp

ersus$ ( )2/ wg pqThe plot should define a straight line. The intersection with the y'axis providesparameter$ 2g, while the slope provides parameters and 2f.

&igure 8'< compares shut'in pressure along the production pipe, calculated by (8'

1:*, and production pressure by (8')3*. Production is set relatively high$ )3 +m5 % s,which here gives the flow speed between 1) and 1 m % s. +ince the speed increasestowards the top, the pressure drops profile curves slightly.

Figur" 8- r"ssur" 'rofil"s i$ gas )"ll

8..! Horiso$tal" ryr

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or et &oriso$ta#t rr, *a##er det statiske #eddet ort:g$= . 5ed #a g$ ' mot $u## , ka$ i

ise (med #6Hopita#6s re'e#) at *rikso$sparametere$ i (8-20) 'r asmptotisk mot

2

0

4

=

o

o

g%

T

p

Rd

&zT%MA

$

7ik$i$' (8-20) *ore$k#es da ti# de$ k#assiske emoutik$i$'e$ *or trkktapet i&oriso$ta#e

rr

( ).0

22

0

02

4

=

&TMz%

Rdpp

p

Tq i.g

(8-21)

8..* 0yst"ma$alys"

5ed komi$aso$e$ a (8-11) o' (8-20), ka$ i uttrkke samme$&e$'e$ me##om trkk o'

strm$i$'

( ) 222 g%gkgeg'h qAAqAApAp += (8-22)

ette ka$ti*iserer trkktap i 'assprodukso$ssstemet, p 'ru$$ a strm$i$'.

stemparametre$e:Ag Ak A A%, er k$ttet ti# parametere o' es#ut$i$'saria#er 'e$$om

re#aso$e$e ut#edet oe$*or.

8.! Flo) through al"s a$d r"strictio$s

"hen gas flows through the valves and restrictions, will usually acceleration forcesdominate. The flow equation from (8'1*, then becomes

0=+ vdvdp (8-24)

y solving this, we can predict the flow capacity and pressure drop

8.!.1 ($com'r"ssibl" a''ro#imatio$"hen the pressure drop across a valve or restriction is less than 7A (pc% piE 3.67*,corresponding to ach$ a F3.)7, we can neglect compressibility and calculate asfor incompressible fluid

8.!.2 om'r"ssibl", subcritical flo)"hen the pressure drop is greater than 7A, we should ta4e into account that the gaswill expand. +ince the speed is high and distance through the valve short, there is

little time to heat conduction. "e can therefore assume adiabatic flow, so thatequation (8')* becomes

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Figur 8- 3ass flo) through orific", calculat"d by diff"r"$t a''ro#imatio$s

The figure shows the formulas for incompressible and compressible fluid flow, aspreviously claimed, provides almost identical results as long as the pressure drop

across the no##le is relatively small (F7A*.

8.4.4 as !alve characteristics

The results above show that we must distinguish between subcritical and critical flow.&or gas wells, it is common to express the output in standard volumemes. This isdirectly related to the mass flow

o

g

g

mq

=

"ncompressi#le flowf the pressure drop is small, the volume flow through the no##le change so little thatwe can often ignore this and approximate pressure drop as for incompressible flow$.

2

i v-(+pp = .

y expressing the density of state equation, we get the mass flow through the no##le

==

i

icc

p

p"

zRT

M2pAvAm

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2s the limit of this approximation, we use 7A pressure drop -(+pp i. >

$u#critical flow

The equation for compressible flow (8'):* provide a maximum at

12

1

k

k

i

p

p k

= +

@&is correspo$ds to approximately at 73A pressure drop. &or larger pressure reductionover the no##le, the formula predicts reduced throughput. This is illogical and not inaccordance with reality. 2fter the maximum according to (8'):*, further pressurereduction does not affect the mass flow.

-ritical flowThe speed through the valve becomes critical when the pressure difference acrossgreater than critical

12

1

k

k

i

p

p k

+

-orresponding speed through the no##le, we will call$ critical speed. y putting (8')8*

into (8')

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Figur" 8.8 Gas al" charact"ristics

8.* Futur" 'roductio$

e %i## i'$ore &ere possi#e i$*#u= o* %ater a$d assumi$' t&at t&e reseroir produces %it&

'as e=pa$sio$. e ca$ *orma## ua$ti* t&e e=pa$sio$ %it& compressii#it euatio$

RTR dp

dV

Vc

1= (8-

31)

c : compressii#itVR : 'as o#ume o* t&e reseroir (co$sta$t)

TR : reseroir temperature (co$sta$t)

3p : pressure reductio$

dV : o#ume c&a$'e

Bompressii#it o* t&e 'as %as deried preious# (8-). C$serted i$ (8-38) 'ies t&e

re#atio$s&ip et%ee$ pressure drop a$d t&e c&a$'e i$ 'as o#ume

=

p

zd

z

pVdV R (8-

32)

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C* t&e pore o#ume i$ t&e reseroir is co$sta$t, t&e c&a$'e is: d5, eua# to t&e o#ume

produced. Ct is co$e$ie$t to e=press t&e productio$ o#ume at sta$dard co$ditio$s (%e cou#d

possi# use mass produced, or mo#es). e ca$ e=press t&e productio$ rate o* o#ume c&a$'e

oer time, diided t&e *ormatio$ *actor

=

==

z

p

d'

d

Tp

TVp

z

d'

d

z

p

Tp

TVd'

dV

Bq

Ro

o

R

Ro

o

R

g

g

2

1(8-

33)

Bomi$i$' (8-33) %it& t&e preious# deried sstem re#atio$ (8-22), %e ca$ predict

productio$ pro*i#es. @&e comi$atio$ ca$ e do$e a$a#tica##, ut o*te$ do$e $umerica##.

@&is makes it possi#e to predict &o% arious improeme$ts o* t&e productio$ sstem %i##

a**ect productio$ rates a$d ear$i$'s.

Here %e %i## #imit our oectie to predict t&e re#atio$s&ip et%ee$ cumu#atie productio$

a$d reseroir pressure. i$te'rati$' (8-40), t&is ca$ e e=pressed as

= zp

z

p

Tp

TVd'q4

i

i

Ro

o

R

'

'

gp

i

(8-

34)

D*te$ %e %i## a#so e=press t&e reseroir o#ume, &o% muc& 'as is co$tai$ed i$ t&e p#ace

4Tp

zTp4BV

oi

iRo

giR == (8-

3)

putti$' (8-3) i$to (8-34), %e otai$ t&e c#assica# mass a#a$ce *or t&e 'as reseroirs

p

i

i

i

ip

i

i 4z

p

4z

p

4

4

z

p

z

p

=

=

11 (8-

3!)

@&e resu#t mea$s t&at t&e re#atio$s&ip et%ee$ pressure a$d +-*actor:p/z,, *a## i$ proportio$

to t&e cumu#atie productio$. (e &ae preious# s&o%$ t&at i* a c#osed reseroir co$tai$i$'

sma## compressi#e oi#, t&e pressure %i## *a## proportio$a## %it& cumu#atie productio$.)

C* t&e measuredp/zp#ots as a strai'&t #i$e a'ai$st cumu#atie productio$, t&is i$dicates t&atproductio$ is drie$ t&e e=pa$sio$. e ca$ t&e$ predict t&e 'as co$te$t i$ t&e reseroir

e=trapo#ati$' it to +ero pressure.

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