Drawdown Drawdown testing and and Semi Semilo log g anal analy ysis sis Azeb D Habte Jan 2017 PDB 3013 UTP
Lesson Content •
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Semi Semilo log g plot plot of pwf versus log(t) Determination Determination of permeability and skin factor Early, middle and late time pressure behavior
CO1: To To analyze drawdown drawdown and build up test using analytical analytical solutions.
Lesson Content •
•
•
Semi Semilo log g plot plot of pwf versus log(t) Determination Determination of permeability and skin factor Early, middle and late time pressure behavior
CO1: To To analyze drawdown drawdown and build up test using analytical analytical solutions.
Lesson Outcomes •
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Apply semilog semilog straigh straightt line analysis analysis method method to determine permeability and skin factor from the middle time drawdown test data. Estimate wellbore storage coefficient from early time drawdown test data. Estimate reservoir volume from late time drawdown test data.
CO1: To To analyze drawdown drawdown and build up test using analytical solutions.
Drawdown test •
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Is conducted by producing a well at a known rate or rates while measuring changes in bottomhole pressure (BHP) as a function of time. It is designed to determine permeability and skin factor If the pressure transient is affected by outer reservoir boundary, drawdown test can be used to establish the outer limits of the reservoir and to estimate the hydrocarbon volume (reservoir-limit test).
Drawdown Cont’d
Ei function solution
Ebased i-function solution (line-source solution) is first proposed by Matthews and Russell in 1967. It is on the following assumptions: Infinite acting reservoir, i.e., the reservoir is infinite in size. The well is producing at a constant flow rate. The reservoir is at a uniform pressure,
, when production begins.
The solution has the following form:
, + . − where, p(r,t ) = pressure at radius r from the well after t hours t=time, hrs k=permeability, md
Q=flow rate,∞STB/D −
(1)
Logarithmic approximation •
function has the following logarithmic ln(1.781) (2) where,
For x<0.01, the approximation:
Substituting Eq. 2 into Eq. 1 gives:
(3) •
For the bottomhole flowing pressure, i.e., @r= rewritten as: p wf pi
162.6Qo Bo o kt 3 . 23 log 2 kh c r
, at any time, Eq. 3 can be (4)
The skin effect Skin due to damage
Skin due to stimulation
(ks > k)
(ks < k)
The skin effect cont’d Where
141. 2 ∆ ℎ
∆= additional pressure drop due to skin effect
(5)
S= skin factor
Skin factor (S) is a dimensionless variable used to quantify the
magnitude of skin effect.
Incorporating skin into the Ei-Function solution
≤ r ≤ rs . − , + 2
For rw
For r > rs
,
. − +
For r=rw
70. 6 948ϕ + ℎ 2
Log approximation to the Ei-Function For r=rw
.
3.23+0.8686
Determination of permeability and skin factor
Skin and permeability
Skin and permeability cont’d
log log()
Example 1 A well has been produced at a constant rate of 250 STB/D. During the flow period, bottom-hole flowing pressures were recorded as tabulated below. The following rock and fluid properties are known:
=4412 psi
h= 69 ft
=0.8 cp
=3.9%
0.198 ft
17e-6 psi-1
B=1.136 bbl/STB
Estimate permeability and skin factor.
Solution
3582 psi
0.8∗ 1.1 36 7.65 162.ℎ6 162.6 ∗ 250∗ 70∗69 1.151 +3.23 7.65 1.151 44123652 70 0.039∗0.8∗17×10− ∗0.198 +3.23 6.355
Early, Middle and Late Time Pressure Behavior
Lesson Outcomes At the end of this topic, students should be able to •
Describe two main causes of wellbore storage effect.
•
Calculate wellbore storage coefficient (C) from early time region (ETR).
•
Calculate reservoir volume from late time region (LTR).
Typical drawdown test pressure behavior Early time flow is dominated by
wellbore storage. Middle time (transient) flow is a
period where the reservoir act as an infinite. i.e. a semilog plot of Pwf versus t is a straight line which can be used to determine permeability. Late time flow is a flow period
where the pressure data is dominated by boundary effects. Can be used to calculate the size and shape of the reservoir.
Early time region (ETR) pressure behavior The time period when surface production is primarily due to fluids
flowing out of the tubing or tubing-casing annulus is called wellbore
storage dominated flow period. During this period, the reservoir is not producing fluids, and pressure
versus time data do not contain reservoir information. Wellbore storage effect can be caused by either fluid-filled wellbore
(fluid expansion/compression) or falling/rising liquid level.
Early time region (ETR) pressure behavior cont’d During drawdown
During buildup
Early time region (ETR) pressure behavior cont’d
Falling/rising fluid level
Falling/rising fluid level
1/
Early time region (ETR) pressure behavior cont’d
Falling/rising fluid level
Effect of wellbore storage on pressure
Example 2: Wellbore storage calculation The well is 2600 ft deep and has 6.625’’, 24lb/ft casing (5.921” ID). The bottomhole pressure is 1690 psi. If the well is filled with water ( What is the wellbore storage coefficient?
4 ×10− −).
Solution: For fluid filled wellbore
ℎ 5.921 1 2 2 144 0.191 0.191 ∗ 2600 497.154 − − 497.154 ∗ 4 ×10 1.989 × 10 3.542 × 10−
Wellbore storage from pressure transient test data From material balance, the pressure in the
wellbore is directly proportional to time during wellbore storage dominated period of the test
∆ 24
) i s p (
Unit slope line
∆ for drawdown and ∆ @∆= On log-log plot of pressure drop ( ∆ ) versus Where
time, this gives a characteristic straight line of unit slope.
(hrs)
24 ∆
(bbl/psi)
. (hr) 170000 ℎ
Example 3: Wellbore storage calculation from PTT Determine wellbore storage coefficient (C) from the data and table below, which were obtained in a pressure drawdown test on an oil well. q=500 STB/D
rw=0.3 ft
=0.2
h=56 ft
=0.8 cp
Bo=1.2 RB/STB
Ct=1e-5 psi-1
Pi=3000 psia
t, hrs 0.0109 0.0164 0.0218 0.0273 0.0328 0.0382 0.0437 0.0491 0.0546 0.109 0.164 0.218 0.273 0.328 0.382 0.437 0 491
Δp=pi-pwf, psi 24 36 47 58 70 81 92 103 114 215 307 389 464 531 592 648 698
t, hrs 0.437 0.491 0.546 1.09 1.64 2.18 2.73 3.28 3.82 4.37 4.91 5.46 6.55 8.74 10.9 16.4
Δp=pi-pwf, psi 648 698 744 1048 1172 1232 1266 1288 1304 1316 1326 1335 1349 1370 1386 1413
Example 3: Wellbore storage calculation from PTT
∗. . ∆ ./
10000
1000 i s p , p
100
(0.045 hrs, 100 psi)
Δ
10
1 0.01
0.1
1
, hrs
10
100
Middle time region (MTR) pressure behavior During this period, the reservoir is producing fluids, and pressure
versus time data contains reservoir information. The straight line on semilog graph, whose slope is related to effective
permeability of the flowing phase, usually occurs during this period. The straight line is called the ‘correct semilog straight line’
Late time region (LTR) pressure behavior During this period, the pressure transient encounters
reservoir boundary. The semilog curve deviates from the straight line established
during the middle-time region. The bottomhole flowing pressure for a no-flow boundary can
be estimated using:
0. 2 34 162. 6 ℎ ℎ log +0.351+0.869
Late time region (LTR) pressure behavior cont’d Based on late time straight line on Cartesian plot of
0. 2 34 ∗ ℎ . ℎ ∗ ( )
vs. :
The time to reach pseudo steady-state in a cylindrical
reservoir can be estimated from:
(hr)
Example 4: Reservoir volume calculation from PTT Table 2: PTT data The data provided in Table 1 and 2 are pertinent to a reservoir limit test (RLT) conducted in a fully-penetrating oil well producing at a constant rate. Estimate the pore volume (Vp) of the reservoir. Table 1: Rock and fluid properties
q=200 STB/D
rw=0.328 ft
=0.2
h=30 ft
=0.8 cp
Bo=1.25 RB/STB
Ct=1
× 10− psi-1
Pi=5000 psia
t, hr 1.00E-03 1.58E-03 2.00E-03 2.51E-03 3.16E-03 3.98E-03 5.01E-03 6.31E-03 7.94E-03 1.00E-02 1.26E-02 1.58E-02 2.00E-02 2.51E-02 3.16E-02 3.98E-02 5.01E-02 6.31E-02 7.94E-02 1.00E-01 1.26E-01 1.58E-01 2 00E 01
Pwf, psi 4991.26 4987.1 4984.7 4981.8 4978.8 4975.2 4971.8 4968 4965 4961.4 4958.5 4955.5 4953.1 4951.2 4948.8 4948.8 4945.9 4944.8 4943.7 4942.2 4940.7 4939.9 4939
t, hr 3.98E-01 5.01E-01 6.31E-01 7.94E-01 1.00E+00 1.58E+00 2.51E+00 3.16E+00 3.98E+00 5.01E+00 6.31E+00 7.94E+00 1.00E+01 1.26E+01 1.58E+01 2.00E+01 2.51E+01 3.16E+01 3.98E+01 5.01E+01 6.31E+01 7.94E+01
Pwf, psi 4934.4 4933.9 4932.7 4932.4 4930.7 4927.2 4926.8 4925.2 4924.9 4921.9 4922.2 4919.1 4915.5 4912.7 4909.8 4904.1 4897 4889 4880 4868 4852 4833
Example 4: Reservoir volume calculation from PTT Solution:
5000 4980
Step 1: Plot pwf vs t on a Cartesian graph paper. Step 2: Calculate the slope (m*) of the straight line at late time. Step 3: Calculate Vp using the equation ( ) ℎ . ∗
4960 4940 i s p 4920 , f w P 4900
(40 hrs,4880 psi)
4880
(50 hrs,4868 psi)
4860 4840 4820 1.0E- 3 1.0E+1 2.0E+ 1 3.0 E+ 1 4.0E+ 1 5.0E+1
6.0E+1 7.0E+1 8.0E+ 1 9.0 E+ 1
t, hrs
4868 1.2/ℎ ∗ 4880 40 50 234 ∗ 200 ∗ 1.25 4.88 × 10 ℎ 0.1. 2 ∗1 × 10− 0.869 × 10bbl
