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M. Onur
29.05.2008
PET467E-Analysis of Well Pressure Tests/2007 Spring Semester/İTÜ
Final Examination (Duration 2 hours)
Solutions
Name of the Student:
Student No:
Instructions: Before starting the exam, write your name clearly on the space provided above, and also
on each page of your answer sheets. Equations are provided at the end of the question’s sheet.
Problem 1 (30 points) Answer each of the following questions given below: (Note that each question
may have more than one correct answer. So you should circle all the correct answers for full credit .)
(1.1) The objectives of a well test include (5 points)
(a) To evaluate formation permeability
(b) To evaluate well damage
(c) To evaluate the hydrocarbon reserves
(d) To evaluate the static reservoir pressure
(e) To evaluate formation porosity
(f) To evaluate oil saturation
(1.2) The parameters influencing the permeability calculations are (2 points):
(a) The porosity
(b) The net pay thickness
(c) The PVT parameters
(d) Well damage
(1.3) The skin factor can be influenced by (3 points):
(a) The average reservoir pressure
(b) Mud invasion
(c) The perforations
(d) Well completion type
(e) The fluid
(1.4) Characteristics of constant wellbore storage (WBS) effect include: (2 points):
(a) Δp is a linear function of time
(b) Unit-slope straight line on the log-log plot
(c) Δp is a linear function of log (time)
(d) Δp is a constant
(1.5) What is the characteristic of Infinite Acting Radial Flow (IARF) in a buildup test: (3 points):
(a) Shut-in pressure is a function of log (time)
(b) Shut-in pressure is a function of log ((tp + Δt)/Δt)
(c) A unit slope straight line is observed on the log-log derivative plot.
(d) Shut-in pressure derivative is constant
(1.6) A Horner analysis is strictly valid (3 points):
(a) During the first drawdown period
(b) On the first buildup after a constant rate drawdown period
(c) At the buildup after any number of rates
(d) For reservoirs having closed (no-flow) boundaries
(1.7) The AOF is the (2 points):
(a) Minimum rate of the tested well
(b) Extend of the drainage area
(c) Maximum theoretical rate when pwf = patm
(d) Well productivity index
(1.8) What are the most suitable diagnostic for analysis of a buildup after multiple rate flow periods (3
points):
(a) A semilog MDH plot of shut-in time
(b) Cartesian plot of shut-in pressure vs. shut-in time
(c) Horner plot
(d) log-log pressure and derivative plot based on multi-rate superposition time function
(1.9) What are the characteristics of linear flow regime (2 points):
(a) Δp is a linear function of time
(b) A -1/2 slope straight line on the log-log derivative plot
(c) A +1/2 slope straight line on the log-log derivative plot
(d) Δp is a linear function of square root of time
(1.10)
On the following log-log plot identify the lines: (5 points):
.
Line 1: (a) WBS (b) IARF (c) Sealing Fault (d) Pseudo-steady state
Line 2: (a) WBS (b) IARF (c) Sealing Fault (d) Pseudo-steady state
Line 3: (a) WBS (b) IARF (c) Sealing Fault (d) Pseudo-steady state
Problem 2 (50 points). The data provided in Table 2.1 are pertinent to a buildup test conducted in an
oil well producing at the center of a closed square reservoir. The length and width of the reservoir is
1000x1000 ft.
(a) Log-Log diagnostic plot of buildup pressure change Δp and its derivative vs. shut-in time is shown
in Fig. 2.1. Identify and explain the flow regimes exhibited by the data and their time intervals on
the log-log diagnostic plot (15 points).
Table 2.1. Reservoir, Well and Fluid Data for the buildup test.
Reservoir thickness, h,
50 ft
0.354 ft
Well radius, rw,
viscosity, μ,
0.8 cp
1.0 RB/STB
Formation volume factor, Bo,
Flow rate prior to buildup, q,
500 STB/D
Producing time, tp,
24 hours
1.5x10-5 1/psi
Total compressibility, ct,
Porosity, φ,
0.20
Flowing pressure at the instant of shut-in, pwf,s ................. 4786.30 psi
Initial pressure, pi................................................................ 5000 psi
Well/Reservoir geometry: Well producing at the center of a closed square
reservoir.
Buildup pressure change, Δp = pws(Δt)-pwf,s,
and its logarithmic derivative, psia
1000
WBS Unit-slope
line
100
IARF –
Radial flow
10
No-flow
boundary effects
1
0.1
0.0001
0.001
0.01
0.1
Shut-in time, Δt
1
10
Figure 2.1 Log-log diagnostic plot for the buildup test.
Solution 2.a:
From 0.0004 to 0.002 h, we observe unit slope straight line, indicative of WBS
From 0.3 to 3 h, we observe zero-slope line on the log-log derivative plot, indicative of IARF
After 3 h to the end of the buildup period, we observe boundary effects.
100
(b) Perform Horner analysis of buildup pressure data and determine permeability, skin and the
average reservoir pressure combining Horner analysis by MBH analysis. You should state how
you chose the time intervals to draw your Horner semilog straight line. Horner plot is given in Fig.
2.2, and you should perform your analysis on this plot.
1
10
100
1000
10000
5050
100000
5050
p* = 4995 psia
5000
5000
Shut-in pressure, pws, psia
Slope ≈ 25 psi/cylce
4950
4950
p1hr = 4960 psi
4900
4900
4850
4850
4800
4800
1
10
100
1000
Horner Time Ratio, (tp + Δt )/Δt
10000
100000
Figure 2.2 Horner plot for the buildup test.
Solution 2b:
Horner straight line should be drawn in the time interval from 0.3 to 3 hours in terms of shut-in time
(or 9 to 81 in terms of Horner time ratio) as determined from log-log diagnostic plot shown in Fig. 2.1.
slope =
OR
162.6qsc Bμ
kh
(S2.1)
kh =
162.6qsc Bμ
slope
(S2.2)
kh =
162.6 × 500 × 1.0 × 0.8
= 2602 md − ft
25
(S2.3)
k=
kh 2602
=
≈ 52 md
h
50
(S2.4)
Skin factor S is determined from (Note that p1hr = = 4960 psi)
⎡ p − pwf ,s
⎤
⎛ kh ⎞
S = 1.151 ⎢ 1hr
− log ⎜
+ 3.23⎥
2 ⎟
⎝ φct hμrw ⎠
⎣ slope
⎦
⎡ 4960. − 4786.3
⎤
2602
⎛
⎞
= 1.151 ⎢
− log ⎜
⎟ + 3.23⎥
2
−5
25
⎝ 0.2 × 1.5 ×10 × 50 × 0.354 × 0.8 ⎠
⎣
⎦
= 2.03
(S2.5)
We can perform MBH analysis based on MBH charts in combination with Horner
analysis to estimate the average pressure in the reservoir. To use MBH charts, we first
need to compute the dimensionless producing time based on the drainage area given
as follows:
t pDA =
2.637 × 10−4 kt p
φct μA
=
2.637 × 10−4 × 52 × 24
= 0.137
0.2 × 1.5 × 10−5 × 0.8 × 1000 × 1000
(S2.6)
Because it is known that the well is producing at the center of closed square, then
using the appropriate MBH curve for this geometry and using the dimensionless
producing time, we can determine the value of the dimensionless MBH pressure, as
shown in Fig. 2.3. Note that pDMBH =1.5. Using this value and the value of Horner
false pressure determined from Horner plot, p* = 4995 psi, we can compute the value
of average pressure as follows:
p = p*−
70.6qBo μ pDMBH
70.6 × 500 ×1× 0.8 × 1.5
= 4995 −
≈ 4979 psi
kh
2602
Problem 3 (10 points) A reservoir has the following properties: pi = 2500 psi, B = 1.32 RB/STB, μ =
0.44 cp, k = 25 md, h = 43 ft, ct = 1.8x10-5 psi-1, φ = 0.16. In this reservoir, a well (Well A) is
producing with the following flow rate history (see Figure 3.1), while another well (Well B), 660 ft
away from Well A, is shut-in (see Figure 3.2 for Well B’s flow rate history). What is the pressure at
well B at t = 72 hours?
qA, stb/D
500
450
qB, stb/D
250
0 (shut-in)
24
48
t, hour
Figure 3.1 Flow rate history for well A
80
t, hour
Figure 3.2 Flow rate history for well B
You can assume that both wells have the same well radius, and skin at well A is SA = 3 and well B is SB
= 2. Neglect wellbore storage effects at the wells and assume that each well can be represented by a
line source well so that the pressure change at a point r, when the well produces at a constant rate q can
be given by the well known line source solution:
Δp(r , t ) =
70.6qBμ
kh
⎡ ⎛ r2 ⎞
⎤
⎢ E1 ⎜
⎟ + 2S ⎥
⎣ ⎝ 4η t ⎠
⎦
(3.1)
Hint: If we have a single well producing with an n different constant-rate steps in an infinitely large
reservoir; then the pressure (say pwf) at this well can be computed from
n
pi − pwf (t ) = ∑ ⎡⎣ q j − q j −1 ⎤⎦ Δpcu (rw , t − t j −1 )
(3.2)
j =1
where Δpcu is the unit-rate pressure change response and for a line source well, it is given by
Δpcu (rw , t ) =
70.6 Bμ
kh
⎡ ⎛ rw2 ⎞
⎤
⎢ E1 ⎜
⎟ + 2S ⎥
⎣ ⎝ 4η t ⎠
⎦
(3.3)
When using Eq. 3.2, note that you should take q0 = 0 at t0 = 0. The table of exponential integral
function is given at the end of the exam sheet.
Solution 3. From superposition in space, the pressure drop at well B at t = 72 h is given
by the sum of pressure drop caused by the production of well B itself and the pressure
drop caused at well B by the production of well A (which is 660 ft away from well B)
at t = 72 h; i.e.,
Δp(rwB , t = 72) = Δp A (r = 660, t = 72) + ΔpB (rwB , t = 72)
(S3.1)
But Well B is shut-in for all times, then
ΔpB (rwB , t = 72) = 0
(S3.2)
Hence, using Eq. S3.2 in Eq. S3.1 gives:
Δp(rwB , t = 72) = Δp A (r = 660, t = 72)
(S3.3)
Because Well A producing with three different flow rate steps; then applying Eqs. 3.2
and 3.3 for well A with n=3 (see Fig. 3.1) (Note we took n = 3 because t = 72 h is a
time value of interest in the third rate period)
3
Δp A (r = 660, t = 72) = ∑ ⎡⎣ q A, j − q A, j −1 ⎤⎦ Δpcu , A (r = 660, t − t j −1 )
j =1
= 250* Δpcu , A (r = 660, t = 72) + (450 − 250) * Δpcu , A (r = 660,72 − 24)
+ (500 − 450) * Δpcu , A (r = 660,72 − 48)
= 250*
⎞⎤
70.6 Bμ ⎡ ⎛ 6602 ⎞ ⎤
70.6 Bμ ⎡ ⎛
6602
E
(450
250)
*
E
+
−
⎢
⎜
⎟⎥
⎢ 1⎜
⎟⎥
1
kh ⎣ ⎝ 4η *72 ⎠ ⎦
kh ⎢⎣ ⎜⎝ 4η * ( 72 − 24 ) ⎟⎠ ⎥⎦
+ (500 − 450) *
⎞⎤
70.6 Bμ ⎡ ⎛
6602
⎢ E1 ⎜⎜
⎟⎥
kh ⎣⎢ ⎝ 4η * ( 72 − 48 ) ⎟⎠ ⎦⎥
(S3.4)
Now, let’s compute the argument of each exponential integral function in Eq. S3.4
using the input values given. First, we compute the diffusivity constant, and the
values of E1s with the aid of table of E1 function:
η=
2.637 ×10−4 k
2.637 ×10−4 × 25
=
= 5202
0.16 × 1.8 × 10−5 × 0.44
φ ct μ
(S3.5)
6602
6602
=
= 0.291 ⇒ E1 ( 0.29 ) = 0.931
4η *72 4 × 5202*72
6602
6602
=
= 0.44 ⇒ E1 ( 0.44 ) = 0.640
4η * ( 72 − 24 ) 4 × 5202* ( 72 − 24 )
6602
6602
=
= 0.87 ⇒ E1 ( 0.87 ) = 0.274
4η * ( 72 − 48 ) 4 × 5202* ( 72 − 48 )
Using the values of E1s given above and the other input values in S3.4 gives:
Δp A (r = 660, t = 72) =
70.6 × 1.32 × 0.44
70.6 × 1.32 × 0.44
[0.931] + (450 − 250) *
[0.64]
25 × 43
25 × 43
70.6 × 1.32 × 0.44
+ (500 − 450) *
[0.274]
25 × 43
= 14.28 psi
= 250*
Then, the pressure drop at well B at t = 72 h is
Δp(rwB , t = 72) = 14.28 psi
(S3.6)
Hence pressure at well B at t = 72 h is
p ( rwB , t = 70 ) = pi − Δp (rwB , t = 72)
(S3.5)
= 2500 − 14.28 = 2485.72 psi
Problem 4 (10 points) Consider the following well/reservoir system:
L
L/12
Constant pressure boundary
No-flow bdry
Show by a sketch indicating production/injection wells and their locations how we can apply
superposition (or imaging) method to generate the pressure response at the actual well shown in above
figure. Also sketch the log-log diagnostic plot of delta pressure and derivative vs. time that would be
expected to be observed during a constant-rate drawdown test if the well produces under wellbore
storage and positive skin effects. Explain each flow regime by drawing appropriate straight lines and
their slope to be observed.
Solution 4: See figure below for the application of imaging.
L
L/12
Constant pressure boundary
No-flow bdry
Three image wells (one producer shown as red, and two injectors shown as blue) are needed to
represent the pressure drop at the well.
Typical log-log diganostic plot of delta pressure and its derivative for the well/reservoir geometry
given is shown in the figure below:
Log-Log plot: p-p@dt=0 and derivative [psi] vs dt [hr]
Once the well is opened to flow, we will first observe the effects of wellbore storage (unit slope line)
and the hump caused by the positive skin. Then (if not masked by wellbore storage) we should observe
a full radial flow period until the effect of the first boundary is felt. This is indicated by the first flat
line on the derivative. Then there will be a semi radial flow period until the effects of the second
boundary is felt. This is indicated by the second flat line. Afterwards there will be -1 slope line
indicating the influence of the farthest constant pressure boundary.
AUXILIARY EQUATIONS
162.6qBμ
kh
t p + Δt
=
Δt
2.637 x10 −4 kt p
=
φct μA
m=
RH
t pDA
p DMBH =
2.303(p * − p )
m
⎡ p ws,1sa − p wf ,s
⎛ k
S = 1.151⎢
− log⎜
⎜ φc μr 2
m
⎢
⎝ t w
⎣
⎤
⎞
⎟ + 3.23⎥
⎟
⎥
⎠
⎦
MBH Chart