A2 - Solutions

Solutions 1

NExT April 2000

Introduction to Well Testing

Solutions 2

NExT April 2000

Exercise 3


Calculate Compressibility for Undersaturated Oil Reservoir

Solution

ct = cf + So co + Sw cw + Sg cg

cf = 3.6 x 10-6 psi-1

Sw = 0.17

Sg = 0

So = 1 - Sw - Sg = 1 - 0.17 - 0.0 = 0.83

From fluid properties correlations,

pb = 2803 psi

co = 1.158 x 10-5 psi-1

cw = 2.277 x 10-6 psi-1


= 3.6 x 10-6 + (0.83) (1.158 x 10-5)

+ (0.17) (2.277 x 10-6) + (0) (?)

= 1.36 x 10-5 psi-1

Solutions 3

NExT April 2000

Exercise 4


Calculate Compressibility for Saturated Oil Reservoir

Solution


cf = 3.6 x 10-6 psi-1

Sw = 0.17

Sg = 0.05

So = 1 - 0.17 - 0.05 = 0.78


pb = 2803 psi

co = 1.429 x 10-4 psi-1

cg = 5.251 x 10-4 psi-1

cw = 4.995 x 10-6 psi-1


= 3.6 x 10-6 + (0.78) (1.429 x 10-4) + (0.17) (4.995 x 10-6)

+ (0.05) (5.251 x 10-4 )

= 1.42 x 10-4 psi-1

Solutions 4

NExT April 2000

Exercise 5Introduction to Well Testing

Calculate Compressibility for Low-Pressure, High-Permeability Gas Reservoir

Solution

ct = cf + So co + Sg cg + Sw cw

cf = 3.6 x 10-6 psi-1

Sw = 0.2

Sg = 0

Sg = 1 - Sw - So = 1 - 0.2 - 0 = 0.8


cg = 8.144 x 10-3 psi-1

cw = 4 x 10-6 psi-1


= 3.6 x 10-6 + (0) ( ? ) + (0.8) (8.144 x 10-3)

+ (0.2) (4 x 10-6)

= 6.52 x 10-3 psi-1

Solutions 5

NExT April 2000

Exercise 6


Calculate Compressibility for High-Pressure, Low-Permeability Gas Reservoir

Solution


cf = 2.0 x 10-5 psi-1

So = 0

Sw = 0.35

Sg = 1 - So - Sw = 1 - 0 - 0.35 = 0.65


cg = 1.447 x 10-4 psi-1

cw = 3.512 x 10-6 psi-1


= 2.0 x 10-5 + (0) ( ? ) + (0.65) (1.447 x 10-4)

+(0.35) (3.512 x 10-6)

= 1.15 x 10-4 psi-1

Solutions 6

NExT April 2000

Radial Flow and Radius of Investigation

Solutions 7

NExT April 2000

Exercise 1


Factors That Affect Radius of Investigation

Solution

Without looking at the notes, choose the correct response to complete each statement. Check your answers by referring to the radius of investigation equation.

A) increases

B) decreases

C) does not affect

1. Increasing viscosity __________________ the radius of investigation.

2. Increasing permeability __________________ the radius of investigation.

3. Increasing formation volume factor __________________ the radius ofinvestigation.

4. Increasing test time __________________ the radius of investigation.

5. Increasing production rate __________________ the radius ofinvestigation.

6. Increasing net pay thickness __________________ the radius ofinvestigation.

7. Increasing porosity __________________ the radius of investigation.

8. Increasing total compressibility __________________ the radius ofinvestigation.

decreases

increases

does not affect

increases

does not affect

does not affect

decreases

decreases

Solutions 8

NExT April 2000

Exercise 2


Calculate Radius of Investigation for an Undersaturated Oil Reservoir

Solution

φ = 0.17

µ = 1.06 cp

ct = 1.36 x 10-5 psi-1 (from Ex. 3, previous section)

ri = 745 ft

k = 250 md

krc948

t2itµφ=

( ) ( ) ( ) ( ) ( )

hrs1.5

25074510x36.106.117.0948

krc948

t

25

2it

=

=

µφ=

−

Solutions 9

NExT April 2000

Exercise 3


Calculate Radius of Investigation for aSaturated Oil Reservoir

Solution

φ = 0.17

µ = 1.185 cp

ct = 1.42 x 10-4, from Ex. 4, previous section

ri = 745 ft

kro = 0.8

k = 250 md

ko = k kro = (250) (0.8) = 200 md

krc948

t2itµφ=

( ) ( ) ( ) ( ) ( )

hrs75200

74510x42.1185.117.0948

krc948

t

24

2it

=

=

µφ=

−

Solutions 10

NExT April 2000

Exercise 4


Calculate Radius of Investigation for a Low-Pressure, High-Permeability Gas Reservoir

Solution

φ = 0.12

µ = 0.01151 cp

ct = 6.52 x 10-3 psi-1, from Ex. 5, previous section

ri = 745 ft

k = 100 md

krc948

t2itµφ=

( ) ( ) ( ) ( ) ( )

hrs

x

k

rct it

47100

7451052.601151.012.0948

948

23

2

=

=

µφ=

−

Solutions 11

NExT April 2000

Exercise 5


Calculate Radius of Investigation for a High-Pressure, Low-Permeability Gas Reservoir

Solution

φ = 0.04

µ = 0.02514 cp

ct = 1.151 x 10-4 psi-1, from Ex. 6, previous section

ri = 745 ft

k = 0.08 md

krc948

t2itµφ=

( ) ( ) ( ) ( ) ( )

hrs76108.0

74510x151.102514.004.0948

krc948

t

24

2it

=

=

µφ=

−

Solutions 12

NExT April 2000

CharacterizingDamage and Stimulation

Solutions 13

NExT April 2000

Exercise 1

Skin Factor and Wellbore Storage

Damage and Skin Factor Calculations

Solution

1.

2.

3.

4.

( ) ( ) ( ) ( )( ) ( ) ( )

psi8.66

111755400

1907.120002.141

khBq2.141

ps

=

=

µ=∆

%6.66

160018008.6616001800

ppppp

Ewf

swff

=

−−−=

−∆−−=

ft10x6.5

e124

err

6

11

swwa

−

−

−

=

=

=

5

2/10012/4

ln

2/Lr

ln

rr

lns

f

w

wa

w

−=

=

=

=

Solutions 14

NExT April 2000

Semilog Analysis for Oil Wells

Solutions 15

NExT April 2000

Exercise 1Semilog Analysis for Oil Wells

Solution

1. First, we plot flowing bottomhole pressure, pwf, as a function of time on semilog paper. The slope, m, of the resulting straight line is the difference between values of pwf one log cycle apart, or

where by convention we use the absolute value of the slope.

2. Effective permeability to oil is estimated using the absolute value of the slope, m, of the line,

( ) ( ) ( ) ( ) cyclepsi1001log100logpsia340,3psia540,3

tlogtlogpp

m12

1wf2wf =−

−=−−=

( )( )( )( )( )( ) md0.8

461008.0136.12506.162

mhBq6.162

k ==µ=

Solutions 16

NExT April 2000


Solution

3300

3350

3400

3450

3500

3550

3600

1 10 100

Time, hrs

Pre

ssu

re, p

si

p1hr = 3540

p100hr = 3340

m = 100 psi/cycle

3. Noting that p1hr = 3,540 psi, we calculate the skin factor.

As we shall discuss later, a positive value of skin factor indicates a flow restriction (i.e., damage) around the wellbore.

( )( )( )( )05.5

23.3365.010178.012.0

8log

100540,3412,4

151.1

23.3rc

klog

mpp

151.1s

26

2wt

hr1i

=

+

×−−=

+

µφ−−=

−

Solutions 17

NExT April 2000


Solution

1. First, construct a semilog plot of shut-in pressure, pws, as a function of the Horner time ratio, (tp + ∆t)/∆t. The calculated Horner time ratios are given in the table below, while the semilog plot is shown on the next slide. The absolute value of the slope of the semilog straight line.

t HTR pws t HTR pws

2 500.5 2615.1 18 56.5 2662.5

3 334 2623.9 24 42.625 2668.6

4 250.75 2630.1 30 34.3 2673.3

6 167.5 2638.9 36 28.75 2677.1

8 125.875 2645.1 48 21.8125 2683.1

10 100.9 2649.9 60 17.65 2687.7

12 84.25 2653.8 72 14.875 2691.4

15 67.6 2658.6

( ) ( )cyclepsi50

1log10log750,2700,2

t

ttlog

t

ttlog

ppm

1

1p

2

2p

1ws2ws

=

−−=

∆∆+

−

∆∆+

−=

Solutions 18

NExT April 2000

2500

2550

2600

2650

2700

2750

2800

1101001000

Horner Time Ratio

Pre

ssu

re, p

si

p1hr = 2600

p* = 2750

m = 50 psi/cycle


Solution

2. Next, we calculate the effective permeability to oil using the slope of the straight line.

3. The original reservoir pressure is found by extrapolating the semilog straight line to infinite shut-in time, i.e., (tp + ∆t)/ ∆t = 1. The pressure at this point is pi = 2,750 psia. At ∆t = 1 hour, the Horner time ratio is (tp + ∆t)/ ∆t = (999 + 1)/1 = 1000. From the extrapolated semilog straight line, p1hr = 2,600 psia at this value.

4. The skin factor is

( )( )( )( )( ) md30

325095.225.1806.162

mhBq6.162

k ==µ=

( )

( )( )( )( )( )

12

23.325.0101595.215.0

30log

508.1847600,2

151.1

23.3rc

klog

mpp

151.1s

26

2wt

wfhr1

=

+

×−−=

+

µφ−−=

−

Solutions 19

NExT April 2000

Wellbore Storage

Solutions 20

NExT April 2000

Exercise 2


Calculate WBS Coefficient For Single-Phase Liquid

Solution

First, calculate the wellbore volume, using the equation for thevolume of a right circular cylinder:

Next, calculate the WBS coefficient.

( )( ) ( )

bbl5.88

bbl/ft615.5

ft497

ft497

2600122

921.5

hrV

3

3

3

2

2

=

=

=

π=

π=

( ) ( )psi/bbl10X54.3

10X45.88

cVc

4

6

wbwb

−

−

=

=

=

Solutions 21

NExT April 2000

Exercise 3


Calculate WBS Coefficient For Rising Liquid Level

Solution

First, calculate the cross-sectional area of the wellbore, using the equation for the area of a circle:

Next, calculate the WBS coefficient:

2

2

ft191.0

12.2921.5

rA

=

π=

π=

( ) ( )

psi/bbl10x56.7

cm/g

ft/lbm4.62xcm/g04.1

191.065.25

A65.25c

2

3

33

wb

wb

−=

=

ρ=

Solutions 22

NExT April 2000

Exercise 4


Calculate WBS Coefficient for Single-Phase Gas

Solution

First, calculate the wellbore volume using the equation for the volume of a right circular cylinder:

Next, use FLPROP to calculate the gas compressibility

cg = 3.21 x 10-4 psi-1

Finally, calculate the WBS coefficient:

C = Vwb cwb

= (54) (3.21 x 10-4) = 0.017 bbl/psi

( ) ( ) ( ) ( ) ( ) ( )

bbls54

ft306

375122

921.57200

122441.2

hrhr

VVV

3

22

csg2csgtbg

2tbg

csgtbgwb

=

=

+

π=

π+π

+=

Solutions 23

NExT April 2000

Flow Regimes and the Diagnostic Plot

Solutions 24

NExT April 2000

FLOWREGM.WTD (Diagnostic Plot)

0.01

0.1

1

10

100

1000

0.0001 0.001 0.01 0.1 1 10 100 1000 10000

Wellbore storage Linear flow

Adj

uste

d pr

essu

re c

hang

e, p

si

Radial equivalent adjusted time, hr

Exercise 1 Flow Regimes and the Diagnostic Plot

Solution

A2 - Solutions

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