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 ( wf p ) is 1,600 psi. Flow efficiency is defined as: wf s wf ideal actual f p p p p p J J E - Δ - - = where J actual represents the productivity index of the well with skin, whereas J ideal represents the productivity of the same well without skin (i.e., S = 0). For a well with neither damage nor stimulation, E f = 1, for a damaged well, E f < 1, and for a stimulated well, E f > 1. Solution 2: It is a damaged well as E f is computed as: E f = 0.85 or 85%. ( )( )( )( ) ( )( ) () psi S kh B q p s 38 . 30 5 175 540 9 . 1 07 . 1 2000 2 . 141 2 . 141 = = μ = Δ 85 . 0 1600 1800 38 . 30 1600 1800 = - - - = - Δ - - = wf s wf f p p p p p E

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Page 1: PCB3013 HW#4 Solution

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

Page 2: PCB3013 HW#4 Solution

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

×=

×=

=

Page 3: PCB3013 HW#4 Solution

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=