pcb3013 hw#4 solution
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PCB3013-Well Test Analysis
HW# 4
Prof. Dr. Mustafa Onur, UTP, September 2013
SOLUTIONS
Given Date: October 29, 2013 Due Data: November 07, 2013
Subject: Computation of skin factor and wellbore storage coefficient.
Problem 1 (15 pts): Calculate the additional pressure drop due to skin for a well producing at
2,000 STB/D. Oil formation volume factor is 1.07 RB/STB, viscosity is 1.9 cp, permeability
is 540 md, net pay thickness is 175 ft, skin factor is 5, and porosity is 12%.
Solution 1:
Problem 2 (15 pts): Calculate the flow efficiency for the well in Problem 1, if the average
reservoir pressure ( )p is 1,800 psi and the flowing bottom-hole pressure ( wfp ) is 1,600 psi.
Flow efficiency is defined as:
wf
swf
ideal
actual
fpp
ppp
J
JE
−
∆−−=≡
where Jactual represents the productivity index of the well with skin, whereas Jideal represents
the productivity of the same well without skin (i.e., S = 0). For a well with neither damage nor
stimulation, Ef = 1, for a damaged well, Ef < 1, and for a stimulated well, Ef > 1.
Solution 2:
It is a damaged well as Ef is computed as: Ef = 0.85 or 85%.
( ) ( ) ( ) ( )( ) ( )
( )
psi
Skh
Bqps
38.30
5175540
9.107.120002.141
2.141
=
=
µ=∆
85.0
16001800
38.3016001800
=
−
−−=
−
∆−−=
wf
swf
fpp
pppE
Problem 3 (10 pts): Calculate the apparent wellbore radius for the well in Problem 1, if the
bit diameter is 8 in. Recall that effective (or apparent) wellbore radius is defined as
( )Srr wwa −≡ exp
where rw is the actual wellbore radius based on the bit diameter, and S is the skin factor.
Solution 3:
Problem 4 (10 pts): For radial flow problems, the mechanical skin factor can be related to
radius of skin zone by using the well-known Hawkin’s formula:
−≡
w
s
s r
r
k
kS ln1
where ks and rs denote the permeability and radius of the skin zone adjacent to the wellbore,
whereas k is the permeability of the formation beyond the skin zone. How can we use this
formula realizing that k and S can be determined from pressure transient analysis (e.g, semi-
log or type-curve matching analysis)?
Solution 4: From semilog or type curve matching of pressure data, we can determine the
permeability k and the skin factor S. If we have additional information about the skin (or
invasion) zone, i.e., rs, around the wellbore, then we can use Hawkin’s equation to determine
the permeability of the skin zone. Perhaps, resistivity logs can be source of estimate of the
skin zone rs.
Problem 5 (20 pts): Calculate the wellbore volume and WBS coefficient for a wellbore filled
with a single phase liquid. The well is 2600 ft deep and has 6 5/8”, 24 lb/ft casing (5.921”
ID). The bottomhole pressure is 1,690 psi. If the well is filled with water (cw = 4 x 10-6 psi-1)
what is the value of wellbore storage coefficient?
Solution 5:
ft
e
errs
wwa
3
5
1025.2
12
4
−
−
−
×=
=
=
( )( )( )
bbl
bblft
ft
ft
hrV
5.88
/615.5
497
497
2600122
921.5
3
3
3
2
2
=
=
=
=
=
π
π
( ) ( )psibbl
cVc wbwb
/1054.3
1045.88
4
6
−
−
×=
×=
=
Problem 6 (20 pts): Calculate the cross-sectional area and wellbore storage coefficient for a
wellbore with a rising liquid level. The well is 2600 ft deep and has 6 5/8”, 24 lb/ft casing
(5.921” ID). The bottom-hole pressure is 750 psi. If the well has a column of water of
density 1.04 g/cm3, in it, what is the value of wellbore storage coefficient?
Solution 6:
Problem 7 (10 pts): Assuming constant wellbore storage coefficient model, one can relate the
flowing (or shut-in for buildup) bottom hole pressure to surface (qsc) and sandface (qsf) rates
by the following equation:
( ) ( )( )
C
Btqtq
dt
dp sfscwf
24
−−≡
Suppose that we measure surface rate, but do not measure the sandface rates, discuss how one
can use to compute sandface rates with the measured values of bottom hole pressures using
the above equation.
Solution 7: We numerically differentiate the pwf vs time data to generate the pressure
derivative data, dpwf/dt by using a numerical differentiation algorithm. Once the derivative
data computed, we can use the following equation to compute the sandface rates with the
computed value of the wellbore storage coefficient C from well completion data as:
( )dt
dp
B
Cqtq
wf
scsf
24+=
2
2
191.0
12.2
921.5
ft
rA
=
=
=
π
π
( )( )
psibbl
cmg
ftlbmcmg
Ac
wb
wb
/1056.7
/
/4.62/04.1
191.065.25
65.25
2
3
33
−×=
×
=
ρ=