WBS-IARF Model-Manual Log-Log- Analysis-WTA 1

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WBS-IARF Model-Manual Log-Log- Analysis-WTA 2

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WBS-IARF Model-Manual Log-Log- Analysis-WTA 3

Here tw is any time value to be chosen on the unit-

MOnur Sept. 2013

Here tw is any time value to be chosen on the unit-

slope line and ∆pw (or Bourdet derivative ∆p’w) is

the corresponding pressure change value to be

read on the unit-slope line from the vertical axis as

shown above.

WBS-IARF Model-Manual Log-Log- Analysis-WTA 4

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WBS-IARF Model-Manual Log-Log- Analysis-WTA 5

Here tIARF is any time value to be chosen in the time interval when the zero-

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IARF

slope line applies and ∆pIARF and Bourdet derivative ∆p’IARF are the

corresponding pressure change and derivative values to be read for mthe

pressure change and derivative data during IARF period as from the vetical

axis as shown above.

WBS-IARF Model-Manual Log-Log- Analysis-WTA 6

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WBS-IARF Model-Manual Log-Log- Analysis-WTA 7

Given the following data, estimate permeability and skin factor for the diagnostic

MOnur Sept. 2013

plot:

q = 50 STB/D pwf = 2095 psia

h = 15 ft φ = 18.3% rw = 0.25 ft

B = 1.36 RB/STB ct = 17.9 x 10−6 psi−1 µ = 0.563 cp

WBS-IARF Model-Manual Log-Log- Analysis-WTA 8

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WBS-IARF Model-Manual Log-Log- Analysis-WTA 9

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WBS-IARF Model-Manual Log-Log- Analysis-WTA 10

From the horizontal part of the field data derivative curve,

MOnur Sept. 2013

From the horizontal part of the field data derivative curve,

we estimate (∆p’)IARF. to be 14.

We pick a convenient time tIARF, 20, within the horizontal

portion of the derivative, and read the coordinates of the

field data pressure change curve at that time, ∆pIARF to

be 400.

WBS-IARF Model-Manual Log-Log- Analysis-WTA 11

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WBS-IARF Model-Manual Log-Log- Analysis-WTA 12

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WBS-IARF Model-Manual Log-Log- Analysis-WTA 13

Given the following data, analyze the data in the log-log

MOnur Sept. 2013

Given the following data, analyze the data in the log-log

graph above. Answer: C = 2.3x10-3 bbl/psi, k = 41 md, s =

0.

q = 50 STB/D

h = 25 ft φ = 27.6 % rw = 0.36 ft

Bo = 1.099 RB/STB ct = 9.4 x 10−6 psi−1 µο = 5.28 cp

WBS-IARF Model-Manual Log-Log- Analysis-WTA 14

Given the following data, analyze the data in the log-log

MOnur Sept. 2013

Given the following data, analyze the data in the log-log

graph above. Answer: C = 8.3x10-3 bbl/psi, k = 16 md, s =

1.44

qg = 5108 Mscf/D

h = 4.4 ft φ = 10 % rw = 0.33 ft

Bgi = 0.781 RB/Mscf cti = 1.66 x 10−4 psi−1 µ gi = 0.0214 cp