Solution: Exam 12.13.06
Exercise 1
a) Sand production
Reservoir A:
Upper limit for inflow pressure loss: pA = pR-pw = 30 bar
- Gives the upper limit for production: qA =JApA= 20.30 = 600 Sm3/d
Reservoir B
Upper limit for inflow pressure loss: pB =33 bar
- gives upper limit for production: qB =JBpB= 10.33 = 330 Sm3/d
b) Pipe diameter
As estimated above, the reservoir A will cope a maximum inflow pressure loss: pR – pw =
30 bar, which will be reached at a production of 600 Sm3/d. We may use the following as
a basis:
Volume flow, down hole: Q  Bo  qo / 86400  1.5  600 / 86400  1.042  102 m3 / s
At a wellbore pressure of:
pw  200  30  170 bar
Static pressure loss in the production tubing, independent of the pipe dimension:
pg  gh  700  9.811600  1.1107 Pa  110 bar
Pressure loss due to friction, assume close to vertical well; measured depth = true vertical
depth
1 
1
700 2
p f  f v 2  L  0.02
v  1600
2 d
2
d
Well head pressure estimate:
pth  pw  pg  p f  170  105  110  105  p f
Well head pressure estimated for alternative production tubings:
Pipe diameter
Flowing velocity
Pressure loss due to
(m)
(m/s)
friction
(bar)
0.05
5.3
63
Well head pressure
(bar)
-3
0.08
0.1
2.07
1.23
6.0
1.7
54
58
Diameter 0.08 og 0.1 m synes begge akseptable ut fra gitte opplyninger. Sandproduksjon
kan muligens bli et problem i framtida. Valg av diameter: 0.08 meter vil opprettholde
høgere strømningsfart når produksjonen synker, at sand ikkje så lett akkumuleres i
brønnen. Så vi velger her røyrdiameter 80 millimeter og gjør videre berekninger ut fra
dette.
A diameter of 0.08 and 0.1 seem both to be acceptable from the given information. Sand
production will might be a future problem. By choosing a diameter of 0.08 m one is able
to sustain a larger flowing velocity when the production is declining, i.e. sand will not
easily accumulate in the well
(Under other circumstances, a diameter of 0.1 m might be the ”correct choice”. The
important thing is that you make a rational selection and clarify the premises)
c) Choke
With our selection of pipe, we expect a wellhead pressure of 54 bar, resulting in a
pressure difference between x-mas tree and separator: pc= 54-20 =34 bar. Based on the
formula for a one phase liquid flow:
Ac  Q

700
 1.042  10 2
 1.06  10 4 m2
5
2pc
2  34  10
nozzle diameter: 1.2 cm
(If we choose the larger pipe diameter 0.1 m, so that the well head pressure estimate will
be 58 bar, the nozzle opening will be: 1.0 .10-4 m2 og 1.1 cm.)
One phase liquid flow assumes that the wellhead pressure is above the saturation
pressure. This is the situation in our case, but should be commented, maybe clarified by
calculations.
d) Total production, limited by sand criteria
When both reservoirs are producing together, one of the reservoirs will most likely reach
the limit for sand production before the other. The total production will therefore be
limited by the reservoir first reaching the sand production limit.
When both reservoirs are producing, the pipe between A and B will be filled with liquid
from reservoir B, with a density: 900 kg/m3. This results in a static difference in well
pressure:
pwB  pwA  pg   B ghAB  900  9.81100  8.83 105 Pa  8.8 bar
Assumption 1: Reservoir A i limiting the total production
Well-pressure at reservoir A
Well-pressure at reservoir B
: pwA = 200-30 = 170 bar
: pwB  pwA  pg  170  8.83  178.8 bar
Inflow pressure loss for reservoir B: pRB - pwB= 219-178.8 = 40.2 bar.
Reservoir B is overriding the limit for sand-production: 33 bar.
Assumption 2: Reservoir B is limiting the total production:
Well-pressure at reservoir B
Well-pressure at reservoir A
: pwB = 219-33 = 186 bar
: pwA  pwB  pg  186  8.8  172.2 bar
Inflow pressure loss for reservoir A: pRA-pwA= 200-172.2 = 27.8 bar.
The inflow pressure loss in reservoir A is lower than the limit for sand production: 30 bar.
Production under assumption 2:
Reservoir A: q A  J A  pRA  pwA   20200  177.2  456 Sm3 / d
Reservoir B:
Total:
qB  330 Sm3 / d
qt = qA + qB =786 Sm3/d
e) Initial production capacity
The production capacity limited by sand production is estimated above. We will examine
if it is possible to produce at this rate with the given pressures and flowing conditions.
Since two reservoirs are producing fluid with a different density, the density in the pipe
will be an average value. We assume volumetric proportion of mixture so that:
 Q   B QB
 A A
QA  QB
QA = qABoA=456 . 1.5
= 684 m3/d
.
QB = qBBob=330 1.2
= 396 m3/d
Qt =QA + QB = 684 + 396 = 1080 m3/d
Average density:  
700  684  900  396
 773 kg / m3
1080
Flowing velocity (when: d= 0.08): v 
1080 / 86400
 2.49 m / s
 0.082 / 4
Pressure loss along the production tubing, if we assume a vertical well above the
reservoir layers:
1
773
pt  773  9.811600  0.02
2.49 2 1600  121.3 105  9.6 105  130.9 105 Pa
2
0.08
Well head pressure:
pth = pwA- pt =177.2 - 130.9=46.3 bar
The separator pressure is 20 bar. The well head pressure is larger than this, so we have to
choke back the well not to override the limit for sand production.
The consideration to sand production is limiting the initial production rate.
f) Cross flow
Shut main valve, zero total production, we get
q A  qB  0  J A  pRA  pwA   J B  pRB  ( pwA  p g ) 
Well pressure at reservoir A is then:
pwA 
J A pRA  J B  pRB  p g 
J A  J B 

20  200  10219  8.8
 203.4 bar
20  10
The flow from both reservoirs, when assuming that the injectivity is equal to the
productivity
q A  J A  pRA  pwA   20200  203.4  68 Sm3 / d
qB  J B  pRB  pwB   10219  (203.4  8.8)   68 Sm3 / d
Reservoir B is producing 68 Sm3/d, which will flow into A.
g) Total production to avoid cross flow
If the well-pressure at A does not override the reservoir pressure: 200 bar, fluid will not
flow into reservoir A. With a well pressure of 200 bar. at A, the well-pressure at reservoir
B is then:
pwB  pwA  pg  200  8.83  208.83 bar
This results in zero production from reservoir A, so the total production to avoid cross
flow is then:
qt  qB  J B  pRB  pwB   10219  208.83  102 Sm3 / d
Exercise 2
a) Productivity index
50000
Effective drainage radius: re 

 126 m
Below the productivity index is estimated by
1. Cinco-Ley’ correlation
2. Semi analytic formula from the compendium
a-1 :Cinco-Ley’s correlation
SC
 70 
  
 41 
2.06
1.865
 70 
 
 56 
log 10
70
 4.14
100  0.125
Produktivity index for ”deviated, vertical wells” : J 
2 k h
Bln re rw   3 4  SC 
re 3
126 3

ln

J
rw 4
0
.
125
4
E o 

 3.04
re 3
126 3
J
   4.14
ln   SC ln
0.125 4
rw 4
ln
Inflow efficiency:
Effective productivity index: J  E  J o  3.04  20  60.8 Sm3 / d
a-2. Semi analytical formula from the compendium
J 
2 k h

 o Bo  ln



4re cos  3
h
3
  cos   ln
 0.685   0.233cos   
h
4
 2rw


ln
re 3

rw 4
J

J o  4re cos  3



h
 ln
 ln
  0.233cos  3 


cos


0
.
685


h
4
 2rw



126 3
ln

0.125 4

 3.23
 4 126 cos 70 3
70


3
  cos 70 ln
 0.685   0.233cos 70 
 ln
70
4
 2 0.125



E
Effective productivity index: J  E  J o  3.23  20  64.5 Sm3 / d
We may observe that the difference between the estimates is: 6%. Without making a
standpoint of view of what estimate is the most credible, further calculations are based on
the estimate: J= 60.8 Sm3/d.
b) Well pressure at a production of 1000 Sm3/d
pw  p R 
1
1
q  80 
1000  63.6 bar
J
60.8
c) Liquid fraction
Formation volume factor for gas Bg 
Superficial velocity, gas:
vsg 
Superficial velocity, liquid: vsl 
p o T z 1.01 330

0.8  1.47 102
o
o
p T z
63.6 288
q g Bg
d / 4
2

3 10
5


/ 86400 1.47 10 2
 2.9 m / s
  0.152 / 4
1000  1.5 / 86400
 0.98 m / s
d 2 / 4
2


 vsg
vsl 
vsl
vsl 
1   vsg




yl 
 Co
 1  4Co

 Co
 1
2   vo
vo
vo
vo
vo




Liquid fraction:
2
1   2.9
0.98 
0.98
0.98 
 2.9
  
 1.2
 1  4 1.2

 1.2
 1  0.294
2   0.1
0.1
0.1
0.1
0.1




We are predicting the liquid fraction 30% (29.4%) just after the gas injection.
d) Change in liquid fraction, velocity and gradient
Up the pipe, the pressure will decline. At a constant temperature, the reduction in
pressure gives:
-
Gas expansion and evaporation of gas from the liquid; therefore: increasing gas
volume
A certain reduction in liquid volume, due to the evaporation
The increase in gas volume will then always be larger than the reduction in liquid volume
(molecules occupy larger volume in gas phase than in the liquid phase). Since the mass
flow along the pipe is constant, the increase in gas volume and total volume means:
-
decreasing flowing density
increasing flowing velocity
decreasing liquid fraction (flux fraction and in-situ)
Pressure gradient: 
2

dp
1
1 fTP
 g vsg  l vsl vsg  vsl 
 TP g x  fTP m vm  TP g x 
dx
2
d
2 d
Decreasing the average density, implies:
- the pseudo static contribution: TP g x is decreasing up along the pipe.
- the contribution due to friction is increasing when the velocity is increasing up the pipe.
At moderate flowing velocity (3 – 4 m/s), the decreasing density will dominate, so the
pressure gradient (Pa/m) is decreasing upwards the pipe. When the flowing velocity is
sufficiently large, the friction term will dominate, so a further increase in velocity will
increase the pressure gradient.
The temperature has been assumed to be constant. In the real world the temperature will
always decline upwards along the pipe. The drop in temperature will contribute to a
reduction in the evaporation and reduction in expansion. It will contribute in “the
opposite direction” to the pressure loss. Normally the contribution from pressure loss will
dominate, so that the total effect will be as described above. I certain cases (large
pressure, large drop in temperature) the drop in temperature may dominate, so the total
effect will be opposite of what is described above.
e) Production under certain circumstances
We want to investigate if it still is possible to produce by drawdown between reservoir
and separator
63.5  20
 41.8 bar
2
Average temperature: reservoir temperature (333 K), or a little lower since we have some
cooling: we may assume: 320 K
Average pressure for the production tubing: pt 
Gas density:  g 
 go
Bg

1.23  0.6
 34.4 kg / m3
2
2.15  10
Two phase density: TP   g y g  l yl  34.41  0.214   700  0.214  176.8
Two phase friction factor: fTP  cTP f o
 g yl 1  l 2  l 1  yl l 2
cTP 
 m yl 1  yl 
Liquid flux fraction:
l 
vsl
0.98

 0.182
vsg  vsl 4.40  0.98
g  1  0.182  0.818
m   g g  l l  34.4  0.818  700  0.182  155.5 kg / m3
cTP 
34.4  0.214  0.818 2  700  0.786  0.182 2
 0.885
155.5  0.214  0.786
fTP  cTP f o  0.884  0.02  0.0177
2

1
 p  TP g x h  fTP m vm x
2
d
1
155.5
4.40  0.98 2 2000  27.75  10 5  5.31  10 5  33.1  10 5 Pa
 176.8  9.81  1600  0.0177
2
0.15
Well-head pressure
pth  pw  p  60.2  33.1  27.1 bar
The estimated well-head pressure is larger than the separator pressure.
It is therefore possible to produce at the predicted rate.