Exam 14/12 2012
Task 1. Gas well
The following data are provided for the Lerchendahl field:
Reservoir temperature
Gas gravity
Inner diameter production tubing
Length production tubing
Reservoir depth:
Standard temperature
Standard pressure
Gas z factor at well conditions
Gas viscosity at well conditions
Interfacial tension water-gas
Solubility water-in-gas at well conditions
Density of produced water:
Distribution parameters for two-phase flow
61 C
0.58
175 mm
2040 m
1850 m
15 C
1.01bar
0.86
0.0168 cP
60 dyn / cm
1.7 gram/Sm3
1010 kg/m3
Co = 1.2
Production test
Gas rate Water rate Bottom well pressure Tubing head pressure
(Sm3/d)
(Sm3/d)
(bar)
(bar)
0
0
153.7
134.2
9.6 · 105
140
149.0
132.8
12.7 · 105
190
146.8
132.2
5
16.8 · 10
250
143.2
131.1
2
3
Inflow performance: pw  pR  26500  qg  1410  qg (units: Pa, Sm /s)
a) Estimate water flux fraction: w=Qw/(Qg+Qw), (a.k.a.: “no-slip holdup) at down hole
well conditions
b) Estimate gas production (Sm3 / d) so that superficial gas velocity at down hole
conditions exceeds the sinking velocity of water droplets. (This is commonly
considered the lower production to prevent accumulation of water in the well).
c) Estimate water holdup in the well at the lower production limit estimated
above. (If you have not found it, you may use: qg = 5 .105 Sm3 / d)
d) The well produces at lower limit. Gas production is quickly increased to 15 .105
Sm3 / d. Estimate stable water production at the new gas rate and how long it
will take till the water production reaches it’s stable level. (If you make
approximations, discuss how these will affect the estimates, in relation to
reality.)
e) Estimate maximum water production, shortly after gas rate has been
increased to : 15 .105 Sm3 / d. (Consider how potential approaches would
affect the estimates.)
Task 2
The Skalle field (illustrated below) has two productive zones. The table below lists properties
Table Skalle
Zone A
Reservoir reference depth
2000 m
Reference pressure
230 bar
Layer height
30 m
Horizontal permeability
200 md
Vertical permeability
20 md
Saturation pressure
35 bar
Oil density, at reservoir conditions 700 kg/m3
Formation factor
1.5 m3/Sm3
Viscosity, at reservoir condionns
2.5 cP
Reservoir temperature
60 C
zone B
2000 m
235 bar
15 m
100 md
10 md
22 bar
750 kg/m3
1.6 m3/Sm3
1.5 cP
60 C
The field will be produced by a horizontal well, illustrated below. Effective wellbore
diameter is assumed: 200 millimeters. The well will be completed with a liner (pipe)
with inner diameter: 120 millimeters. The annulus between zones A and B will be
closed by a swell packer. Skin factor due to formation damage assumed: S = 2
a) Estimate the specific productivity index for each of the zones A and B.
b) Show that if we complete both zones and neglect pressure drop along the well
bore, the well pressure (pw), depending on specific productivity indices (j), interval
length (L), reservoir pressure (pR) and total production (Qt), may be expressed as
pw 
j A LA pRA  jB LB pRB  Qt
j A LA  jB LB
c) Estimate the total production necessary to avoid cross flow between the zones.
d) For rock-mechanical reasons, we cannot accept pressure drawdown above 20 bar
(pR-pw <20bar). This applies to both zones. Estimate maximum production that fulfills
this requirement.
e) Consider the need for inflow control and give your recommendations
Figure 1: Skalle reservoir and planned well
Unit conversions
1 cp
= 10-3 Pas
1 Darcy = 0.9869  10-12 m2
Physical constants, definitions
Standard temperature :288 K
1 bar =
1 dyn/cm
105Pa
=
10-3 N/m
Standard pressure:
1.01 bar
Formulae
Fluid properties (SI, pressure in bar):
Density, gas saturated oil:
l 
o o   g o Rs
Bo
 
Formation factor saturated oil,: Bo  0.9759  9.52  10   g
  o

4
Above saturation pressure:
Gas density :
104 2.81 Rt 3.10 T 171  o 118  g 1102
p 
Bo  Bob  b 
 p
o
pM  g
g 

zRT Bg
 g o po T z
Bg 

g
p T o zo
Formation factor, gas:
Gas solubility:
0.5


 Rs  0.410 T  103 



Rs  5.90  10 3  g 10 2.14 /  o 10 0.00198 T 0.797 p  1.4 
1.205
Viscosity due to dissolved gas:  os  a od  , od : viscosity of gas free oil;
4.406
3.036
;
a
b
0.515
Rs  17.809
Rs  26.7140.338
b

Under saturated oil:
 o   os  10 3 0.35  os1.6  0.55  os
Single phase flow in reservoir and well
1
qo
J
Inflow characteristics:
pw  pR 
Flow in pipe:
dp   v d v   g x dx 
Friction factor correlation:
1  2
f v dx  0
2 d
f  0.16  Re 0.172 with: Re 
vd

Horizontal wells, PI for isotropic permeability :
6  kh
Long wells, pseudo steady : J 
 x

h 
h
 o Bo   e  3  ln
 S  
Lw  2rw

 Lw
2kh
J
Short wells, steady:


LD / 
h 
h
 ln
 o Bo  ln

 S  
Lw / 4
Lw  2rw


6  kh
Generally, pseudo steady: J 
 D

h 
h
 o Bo 
f a  3  ln
 S  
Lw  2rw

 2 Lw
0.56
 p  p 
s
1.2
Geometry factor:
fa 
Skin pressure loss:
Lw
L
2
 
 1  Lw L  
L
L

1

0
.
53
 1.15  0.164 




 0.45  Lw L 
D
D
 


q B
ps  o o o S
2 k Lw
Inclined wells
Formulae as for horizontal, with: Lw= length of completed interval
with skin due to inclination: S  0.57 (assuming well through formation height)
Anisotropy
Scaling rules for anisotropic permeability: kˆ  k x k y

Equivalent length of inclined well:
Equivalent angle:
tanˆ 
yˆ  y k x k y  y

L̂w  Lw 1   2  1 cos 2 
Lw sin 
1
 tan
 Lw cos  
evt: cos ˆ 
 Lw cos 
L̂w
Inflow control
Inflow density: qL  jL  pR  pw x 
Specific productivity index: jL  J
Bo
Lw
Pressure drop reservoir – sand face: pR  pw 
1
qL
jL
Pressure drop along production liner, constant qL: pw 
8

2
3
f 2 5 q L Lw
3  d
2
1  L 
Orifice based inflow control: pc    c  qL2
2  nc Ac 
Segregated flow
Stationary crest:
Stationary cone:
qoL xe  x 
o g  w   o 
qo
r
ho2 r   he2 
ln e
o g  w   o  r
ho2 x   he2 
1
1 
v      w  o  g x   w  o  g y tg w
 o w 
- w : stable interface inclination tg w  yw x
Flow in inclined layer:
Transient rate after break through: qoL  
2ko  w   o ghe2 K
o
t
Two phase flow in pipes
Rising/sinking velocity for small bubbles/drops:
K = 1.53 for bubbles. K = 2.75-3.1 for droplets
 g  d    

vo  K 
2


0.25
 g 

vD  0.347 g D 1 
l 

Rise velocity for Dumitrescu bubbles:
Velocities:
vg 
vsg
yg

vsg
vl 
1  yl 
vsl
yl
Drift flux relation:
vg  Co vl  vo
Usually: Co in the range of 1 - 1.4
Stationary liquid fraction:
1
yl  
2
2
 vsg

v
v

 Co sl  1  4Co sl
vo
vo
 vo


v
1  vsg
 
 Co sl  1
2  vo
vo

Flux fraction:
l  vsl / vm
Volume averaged density: TP   g y g  l yl
Flow averaged density:  m   g g   l l
2

1
dp  TP g x dx   g vsg dv g   l vsl dvl  cTP f o m vm dx  0
2
d

v d
fo : 1-phase friction factor, for 2-phase Reynolds number: Re m  m m
Pipe flow relation:
Slip multiplier:
m
 g yl 1  l 2  l 1  yl l 2
cTP 
 m yl 1  yl 
Critical velocity
:
v* 
p

TP cTP
p
l yl 1  yl 
vm  vo  yl  vo yl 2
Co 1  yl   yl
2

vm  vo Co  2Co vo yl  Co  1vo yl

Co 1  yl   yl 2
Superficial velocity, at given total velocity: vsl 
Kinematic wave velocity:
Shock velocity: v f 
vk 
vsl
yl
vm
vsl ,2  vsl ,1
yl ,2  yl ,1
General gas constant
Acceleration of gravity
:8314
Mole weight air
2
:9.81 … m/s
:
28.97 kg/kmole