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8. Gas wells
There is no qualitative difference between the oil and gas flow. However, since the gas is
lighter and less viscous than the oil, effects are negligible for oil may be important for gas
flow, and vice versa.
8.1 Physical properties
8.1.1 Molecular model
Many gas properties can be related to a concept of gas as molecules flying around in a room.
The speed, and thus energy, increases with temperature. The distance between the
molecules is very large, so that there is no interaction between them except for random and
elastic collisions. The pressure is caused by collisions between the molecules and the walls.
The general equation of state: pV  nRT can be justified / derived from this model. This
gives good prediction for most gases at low pressures and temperatures well above the dew
point.
Figure 8.1 Molecular gas model
When gas is compressed, the volume decreases so that the distance between the molecules
becomes smaller. There will then be more interactions between them. Van der Waals
included molecular volume and attraction/repulsion. The deviation from the ideal state
equation can be expressed by the z-factor linked to reduced pressure and temperature
(ratios: pressure/critical-pressure and temperature/critical-temperature). For petroleum gases
the z-factor often estimated graphically by the Standings-Katz chart, Standing /1952 /.
Yarborough & Halls extended equation of state / 1974 / give very similar results.
8.1.2 Density
Density equals the ratio between mass and volume. Gas density depends on
molecular weight: M, pressure and temperature. From the state equation we can
express the relationship
 
m nM
pM


V
V
zRT
(8-1)
It is worth noting that even if the gas density is less than the oil density, the gas
density, at reservoir conditions it can still be large compared what we associate with
gas. (As a rule of thumb, the gas density in the reservoir approximately equals the
reservoir pressure in bar: thus at pressure 100 bar, the density will be of the order of
100 kg/m3)
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Compressibility quantifies how density is changed by pressure. For a given
temperature, we can estimate from the general state equation. If the z-factor changes
little with pressure, the compressibility corresponds to the inverse of pressure
cg 
1 d
 dp

T
z d  p
1
dz 
1
   1  p  
p dp  z  T p 
dp  T p
(8-2)
The result means that the compressibility decreases rapidly when the pressure
increases. We have all experienced that a ball becomes harder when pumped up.
The compressibility due to change in temperature at constant pressure can be
derived similarly.
8.2 Inflow
8.2.1 Forcheimer’s equation
The Forcheimer equation combines adds “turbulence” to viscous friction
dp 
 v   v 2
dr k

(8-3)
8.2.2 Inflow performance
Solved for stationary, radial flow of gas, gives the pressure profile
pe2
 go p oTz   1 1  2
re
p oTz  g
  q g
 p r   o
ln q g 
T kh r
T o 2 2 h 2  r re 
2
with re>>rw,
pe2  pw2  Ak qg  A qg2
(8-4)
where
Ak 
A 
p oT z  g ln( re / rw )
To
 go p oT z
To
kh

2 2
2 h rw
g : gas viscosity at averaged flow conditions
z
: z-faktor at averaged flow conditions
Figure 8.2 illustrates the flow characteristics of a gas well, with -factor estimated
with Tek's correlation (6-7). For the sake of comparison, Figure 8.2 also show flow
characteristics calculated for no turbulent pressure loss:  = 0,
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Figure 8.2 Inflow performance
Backpressure equation
An alternative expression of inflow performance is by "backpressure equation"
2
q g  C ( p R2  p w ) n .
introduced by the U.S. Bureau of Mines in 1929:
The parameters: n and C are determined by multi-rate production test.They lack
connection to reservoir and fluid parameters comparable to (8-4).
8.2.3 Incompressible approximation
By using the identity: (pe2–pw2) = (pe–pw) (pe+pw), we can express the flow
characteristics of the gas, (8-4), as
pe  pw 
 g Bg re
 go Bg 2
ln qg  
q
2kh rw
2h 2 rw g
(8-5)
Here B g is the formation factor at average flow conditions. The relations is analogous
to the inflow performance of liquid, with turbulence included.
By equation (8-5) is assumed:  pR2  pw2    pr  pw  pR  pw   2 pR  pR  pw  This
implies the error margin:
p  pw 1 
p 
  1 R
  1  w 
2 pR
2
pR 
For example: if the reservoir pressure is 200 bar and the well pressure 190, the error
neglecting gas expansion becomes:  = 0.5. (1-190/200) = 0.025, ie 2.5%
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8.3 Flow in production tubing
8.3.1 Pressure in static gas column
By integrating the pressure balance: dp   gdh  0 along the pipe, the relationship
between static tubing head pressure and bottom pressure becomes
pth  pwe

g x L
(8-6)
zR T
L : length along tubing
M : gas mol weight : M  28.97   g
g : spesific gravity
Linear approximation
Series development of exponential gives: e x  1  x  x 2 2!  x3 3!  ... Using the 2 first
terms (8-6) can be re-written
pth
1  g x L 
g x L
 g x L  1  g x L 
 1 
 
 

  .....  1 
pw
zRT
 zRT  2  zRT  2  3  zRT 
2
3
When the parameter group in parentheses is much smaller than 1, the higher order terms may
be neglected and the pressure approximated as
pˆ th  pw   w g x L
(8-7)
-----8.3.2 Producing gas well
Under normal flow conditions in pipes of constant diameter, we can neglect
acceleration. By expressing the gas density at the general state equation, the flow
equation becomes
 8 fm 2 zRT  1
 g x 
 dx  0
dp  
 pdx   2 5
 zRT 
 d   p
(8-8)
Usually, parameter groups in the parentheses "( )" vary so little along the production
pipe that we can consider them as constants. Integration of (8-18) then gives
2
pth2
  g x L 
8 f  z RT

  e zRT  pw2  2
5 



g
d
x
 






2
 g x L
  zR T 
1

e
 


 
2

 2
m

The relationship above express a simple relationship between pressure and flow rate:
2
2
 pth 
 qg 

  Ag  A f 

(8-9)
 pw 
 pw 
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 Mg x L 
Ag  e 
zRT 

2
2
2


8f
8 f  z p0 T 
o 2 zRT




Af  2
g 
 1  Ag    2 g d 5  T 0  1  Ag 
 g xd 5
M
x




Along a production pipe, the parameter groups Ag and Af can usually considered
constants. This means that we can estimate these from production data, by plotting:
 pth / pw 2
versus:
qg / pw 2
The plot should define a straight line. The intersection with the y-axis provides
parameter: Ag, while the slope provides parameters and Af.
Figure 8.6 compares shut-in pressure along the production pipe. Production is set
relatively high: 20 Sm3 / s, which here gives the flow speed between 12 and 14 m / s.
Since the speed increases towards the top, the pressure drops profile curves slightly.
Figure 8.3 Pressure profiles in gas well
8.3.3 Horizontal pipes
For a horizontal pipe, the static pressure contribution is equal to zero: gx= 0. By
letting gx og towards zero, it can be shown (with l’Hopital’s rule) that the friction
parameter goes asymptotically towards
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Af
g x 0
fM T zL  4 p o 



Rd 5  T o 
2
Equation (8-9) then simplifies to the Weymouth equation for pressure loss in
horizontal pipes:


2
2
5
0
   T  p  pi Rd 
q    0  u
 4  p  f zM T L 
2
g
0.5
(8-10)
8.2 Temperature change
Temperature change affects the physical properties of the fluid and may cause
precipitation of liquid, or solid, or evaporation. For gas, pressure drop can lead to
considerable temperature changes. Energy balances are used to keep account of
temperature, heat transfer and work.
8.2.1 Energy balance
The first law of thermodynamics claims that energy can change form, but not
disappear. We can then put up energy accounting. The steam engine (or the Sterling
Machine) is often used as concept. Energy change can then be formulated and
illustrated as
dE  dq  dw
(8-11)
dq: heat supplied | J |
dw: work performed | J |
Figure 8.4: The energy balance
For gases the internal energy (molecular velocity) is mainly related to temperature,
little impacted by pressure (distance between molecules). The relationship is often
reasonably linear as illustrated below, where: cv denotes internal energy pr. mole,
corresponding to the heat capacity at constant volume
1st Law of Thermodynamics is also relevant when fluid flows as illustrated below.
Energy changes by flow through pipe, pumps and compressors can be estimated
from this
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Figure 8.5 Energy accounting for an open system
Energy forms
Molecular energy can be understood as speed and vibration of the molecules and
depens on temperature. The relationship is illustrated below. Slope: cv represents
internal energy pr. mol and corresponds heat capacity at constant volume. As long as
the distance between molecules is large, interactions between molecules can be
neglected. The molecular energy will then be unaffected by pressure. Within defined
temperature range, change in the molecular energy may be approximated:
dEi  ncv dT
(8-12)
Figure 8.6 Internal energy as function of temperature
Mechanical energy associated with the flow can be expressed by Bernoulli's equation.
In energy units
dEm  Vdp  mvdv  mgdh
(8-13)
Energy balance for open systems
For an open system, the energy balance is expressed by putting the energy forms
above into the first law (8-11)
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ncv dT  Vdp  mvdv  mgdh  dq  dw
(8-14)
Work
Work performed by expansion can be expressed:
dw  pdV
(8-15)
Heat transfer
Supplied heat will depend on the temperature difference between the system and
surroundings: T -Ta and the contact surface: A. Fourier law says proportionality, so
that the change in thermal energy may be expressed
dq  UA T  Ta  dt
(8-16)
- Where: U is the heat transfer coefficient | w/(m2K) |, constant for heat conduction
For flow pipe flow at moderate speed, the temperature change is often dominated by
heat conduction, so that the energy balance (8-14) simplifies to: ncv dT  dq . With eq.
(8-16), this gives a change in temperature with time: T  t   Ta  T  to   Ta  e

UA
 t to 
ncv
or
with distance, if we follow the fluid volume flowing through the pipe.
Heat can also be supplied by the heating element, radiation and the like.
Flow equation
Changes in pressure and speed can be quantified by the flow equation. For pipe flow:
(1-4) given below in energy units, consistent with (8-14)
f 2
Vdp  mgdh  mvdv  m
v dx  0
2d
8.4.2 Adiabatic model
Adiabatic approximation implies that work, heat transfer and friction are neglected:
dw = 0, dpf = dq = 0. Energy change by expansion: pdV remain in the system.
Inserted in (8-14), this gives the energy balance for adiabatic flow:
ncv dT  pdV  Vdp  mvdv  mgdh  0
(8-17)
The last 3 members of (8-17) correspond to frictionless flow (1-4). If the fluid is at rest,
the sum of these is zero and (8-17) simplifies: ncv dT  pdV  0 With the general
equation of state, we can eliminate the temperature: nRdT  pdV  Vdp . Combination
gives: cvVdp   cv  R  pdV  0 . Dividing by pV and integrating from some selected initial
state: pi,Vi , provides the adiabatic process equation ( "adiabatic equation of state")
pV k  piVi
k
Where the "adiabatic exponent" is defined as: k   cv  R  cv  c p cv
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(8-18)
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When energy remains in the system, there will only be one free thermodynamic
variable. This can be selected and all others expressed as a function thereof. For
example, by expressing volume by the general equation of state and insert into (8-18),
cooling during adiabatic expansion is estimated
 p
T  Ti  
 pi 
k 1
k
(8-19)
8.4.3 Stability of stationary gas
We previously estimated how the pressure in the stagnant gas change with altitude,
assuming dynamically stable gas column. Figure 8.7 illustrates a small amount of gas
that rises slightly. Dynamic stability will imply that it then sinks back. This will happen
if the density in the ascending gas is greater than what it becomes surrounded by.
For a down sinking gas quantity, the opposite applies.
Figure 8.7 Perturbation of stagnant gas
When a gas volume rises, it expands so that the pressure is equal to the gas that
surrounds it. If the ascending gas preserves temperature, the density be greater if it
was colder: Ti  Ti 1 . It will thus be dynamically stable if the temperature rises with
height
Expansion will affect the temperature. If the ascending gas not exchanging heat, the
energy balance (8-17) becomes: ncv dT  nRdT  mgdh  0 . This provides temperature
gradient

dT
gM
g


dh cv  R c p
M=m/n : mol weight |kg/Kmol|
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(8-20)
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c p : specific heat capacity cp=cv+R |J/(kg K)|
The air heat capacity is about 1005 J / (kg K). This gives the gradient aka “adiabatic
lapse rate”: -dT / dh = 0.01K / m. When the sun warms the soil so that the gradient
becomes larger, air will circulate upward so that the temperature gradient approaches
1 C/ 100 meters.
Heat radiation from the soil may make the air colder near the surface and gradient
positive. Such "inversion" makes air pollution accumulate. This is common in
northern cities during the winter.
----
8.5 Flow through valves and restrictions
8.5.1 Compressible flow
When acceleration forces dominate, the flow equation becomes: dp  vdv  0 . We
neglect heat transfer and express density by the adiabatic equation (8-18) and the
definition:   m / V . Inserted in the flow equation and integrated from upstream
pressure: pi and velocity: vi~0, to outlet pressure and velocity: pc ,vc , this gives
k 1


2k pi   pc  k 
vc 
1    
k  1  i   pi  


(8-21)
Expressed as mass flow : m   c vc Ac , this provides “Thornhill-Craver’s equation”
m  Ac pi
2k M
k  1 zi RTi
2
k 1


 pc  k   pc  k 
  1    
 pi    pi  


(8-22)
The figure below shows the flow as function of downstream pressure. The red graph
shows incompressible flow, almost coinciding with (8-22) for relative pressure drop
<5%.For outlet pressure equal to inlet pressure: pc = pi, zero flow. When the outlet
pressure is zero, zero flow is also predicted by (8-22); obviously unphysical.
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Figur 8.8 Mass flow through orifice, calculated by different approximations
When the pressure drop across a valve or restriction is less than 5% (pc / pi> 0.95),
corresponding to Mach: Ma <0.25, we can neglect compressibility and calculate as
for incompressible fluid
8.5.2 Critical flow
Figure 8.8 shows maximum rate at approximately at 50% pressure drop. Measured
rate matches the prediction up to this, but further reduction of downstream pressure
does not reduce the rate. From (8-22), the maximum estimate by: dm / dpc  0 This is
fulfilled when
k
p*  2  k 1


pi  k  1 
(8-23)
Maximum throughput reached for the pressure ratio (8-23), governed by the adiabatic
exponent. For air at 20 C : k  1.4 . This gives critical pressure ratio: p * / pc = 0.53.
Setting (8-23) into (8-22) provided maximum mass flow reached by reducing the
downstream pressure
k 1
m*  Ac pi
kM
zi RTi
 2  k 1


 k 1
(8-24)
Further reduction of the downstream pressure will therefore not change neither the
pressure in the nozzle, nor mass flow. The speed in nozzle can be estimated by
setting (8-23) into (8-21)
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vc* 
kRTc*
2k RTi

k 1 M
M
(8-25)
This is recognized as sound velocity, ie the velocity of pressure waves. Sound
velocity is approximately constant, but the outlet pressure and density depends on
the upstream pressure. Figure 8.9 illustrates the relationship upstream pressure and
rate, utreknet from (8-22) and (8-24)
Figure 8.9 Choke characteristics, for constant downstream pressure: 20 bar
8.5.3 Supercritical flow
When the downstream pressure falls below the critical ratio: ps  p * the flow out
expand, thus accelerating further and briefly exceeding the sound velocity. This
causes pressure shocks and severe turbulence as illustrated below
Figure 8.10: Critical outflow
To maintain the flow speed above sound velocity, the outlet must be designed so that
the flow can accelerate evenly. This can be sought outlet geometry as illustrated
below
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Figur 8.11: Supercritical outflow, Laval-choke
8.5 Future production
We will ignore here possible influx of water and assuming that the reservoir produces with
gas expansion. We can formally quantify the expansion with compressibility equation
c
1 dV
VR dp
c
VR
TR
Dp
dV
(8-26)
TR
: compressibility
: gas volume of the reservoir (constant)
: reservoir temperature (constant)
: pressure reduction
: volume change
Compressibility derived (8-1) gives the relationship is expressed as
p z
dV  VR d  
z  p
If the pore volume in the reservoir is constant, the change is: dV, equal to the volume
produced. It is convenient to express the production volume at standard conditions (we could
possibly use mass produced, or moles). We can express the production rate of volume change
over time, divided by the formation factor
2
1 dV
To  p d  z 
To d  p
   VR o
qg 
 VR o  
 
Bg dt
p TR  z  dt  p 
p TR dt  z 
(8-27)
Combining (8-27) with the previously derived system relations, we can predict production
profiles. The combination can be done analytically, but often done numerically. This predicts
how various improvements of the production system will affect production rates and earnings.
t
N p   q g dt  VR
ti
T o  pi p 
  
p oTR  zi z 
The reservoir volume must correspond to gas initially in the place: VR  Bgi N 
Combined, this the classical mass balance for the gas reservoirs
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p oTR zi
N
piT o
14
p pi

z zi
 N p  pi  1 pi 
N p
1 
  
N  zi  N zi 

(8-29)
This predicts that the ratio pressure/z-factor: p/z, should decline proportionally to the
cumulative production. (We have previously shown that if a closed reservoir containing small
compressible oil, the pressure will fall proportionally with cumulative production.)
If the measured p/z plots as a straight line against cumulative production, this indicates that
production is driven by the expansion. We can then predict the gas content in the reservoir by
extrapolating it to zero pressure.
Figure 8.12: Production history for the Frigg field
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