Class Exercises

Type Curve Analysis

Advanced

Reservoir Engineering

Class Exercises- Chapter 1

Sem 2- 2014

1

Class Exercises

Type Curve Analysis

Example 1- Buildup Type Curve Analysis

2

Class Exercises

Type Curve Analysis

Example 1A DST has been conducted on an appraisal well in which a transient

pressure response was observed during the test. The initial average

reservoir pressure is estimated to be 3460 psia.

During the test, the well was first opened up for the main flow period

and flowed for 25 hrs, producing a total of 5320 stb of oil to surface.

The final flow rate was 3500 bopd. During the production period the

flowing bottom hole pressure (pwf) declined but reached a stabilised

value of 2970 psia by the end of the flow period.

The well was then shut-in downhole and the bottom-hole pressure was

recorded versus time using a pressure gauge. The validated data from

the gauge is listed in Table 1.

The reservoir and fluid parameters are as follows:

Analyse the DST test data using both MDH and Gringarten Type Curve

techniques and calculate reservoir parameters of k and S.

3

Shut-in Time pws

(hours) (psia)

0.050 3284.1

0.117 3310.6

0.183 3321.9

0.250 3329.1

0.317 3333.6

0.383 3337.1

0.450 3340.3

0.650 3347.5

0.850 3352.8

1.050 3357.1

1.250 3360.6

1.517 3364.2

2.050 3370.0

2.517 3374.1

3.050 3377.9

4.050 3383.8

Table 1pi = 3460 psia φ = 0.25pwf = 2970 psia (at end of 25 hours) ct = 17.0 x 10-6 psi-1

h = 25 ft (fully perforated) µo = 1.0 cprw = 0.510 ft (12 1/4” hole) Boi = 1.30 rb/stb

Class Exercises

Type Curve Analysis

MDH Technique

4

Shut-in Time pws MDH(hours) (psia) Log(∆t) p'ws

0.050 3284.1 -1.300.117 3310.6 -0.93 71.7740.183 3321.9 -0.74 58.1680.250 3329.1 -0.60 53.1410.317 3333.6 -0.50 43.6390.383 3337.1 -0.42 42.6100.450 3340.3 -0.35 45.7050.650 3347.5 -0.19 45.0840.850 3352.8 -0.07 45.4911.050 3357.1 0.02 46.8561.250 3360.6 0.10 46.2221.517 3364.2 0.18 42.8192.050 3370.0 0.31 44.3532.517 3374.1 0.40 46.0013.050 3377.9 0.48 45.5544.050 3383.8 0.61 47.907

�� ���

�� 24 �

5320

3500� 24 � 36.48���

m

Bqkh oµ

6.162= � � 652.2��

+−

−= 23.3log

)1(151.1 2

w

wfwsl

cr

k

m

phrpS

φµ � � 3.41( ) psiaphrp wswsL 33561 ==

Class Exercises

Type Curve Analysis

Type Curve Technique

5

Shut-in Time pws ∆p ∆te [d(∆p)/d(∆t)] [d(∆p)/d(∆t)] x ∆te

(hours) (psia) psi hrs psi/hr psi

0.000 2970.0 0.000 0.000 0.000

0.050 3284.1 314.100 0.050 6282.000 313.670

0.117 3310.6 340.600 0.117 395.522 46.128

0.183 3321.9 351.900 0.182 171.212 31.175

0.250 3329.1 359.100 0.248 107.463 26.683

0.317 3333.6 363.600 0.314 67.164 21.108

0.383 3337.1 367.100 0.379 53.030 20.100

0.450 3340.3 370.300 0.445 47.761 21.231

0.650 3347.5 377.500 0.639 36.000 22.990

0.850 3352.8 382.800 0.831 26.500 22.012

1.050 3357.1 387.100 1.021 21.500 21.943

1.250 3360.6 390.600 1.209 17.500 21.150

1.517 3364.2 394.200 1.456 13.483 19.637

2.050 3370.0 400.000 1.941 10.882 21.121

2.517 3374.1 404.100 2.355 8.779 20.672

3.050 3377.9 407.900 2.815 7.129 20.067

4.050 3383.8 413.800 3.645 5.900 21.507

∆���∆�

1 + ∆� ��⁄

∆� = ��� − ��

• Since this is a buildup test and Gringarten type curves are pr esentedfor drawdown, we need to calculate Agarwal’s equivalent tim e first:

!"#$%&'$%" = ∆'"!(∆))

!(∆')

Class Exercises

Type Curve Analysis

1

10

100

1000

0.01 0.10 1.00 10.00

Δp

(p

si)

a

nd

∆' "

( !(∆)

)/!

(∆')

) (p

si)

Δte (hrs)

Engineering plot: ∆) versus ∆'" & ∆'"!(∆))

!(∆')versus ∆'"

6

Class Exercises

Type Curve Analysis

7

Gringarten Type Curve

Class Exercises

Type Curve Analysis

8

Type Curve Match• First make sure the type curves are of the same sca le as the

engineering plot.

Class Exercises

Type Curve Analysis

9

Type Curve Match

Class Exercises

Type Curve Analysis

• For the Match Point:

� Engineering Plot: , Type Curve:

� Also:

10

Type Curve Match Point

Δp= 351.9 psi

Δte= 0.182 hrs

PD= 8

tD/CD= 370

+,"-. = / � 012

Class Exercises

Type Curve Analysis

• Substituting the ordinates of the Match Point from the two graphs along with the values provided for various parameters int o the above left hand side equation:

3 = 452. 26,

• Substituting the abscissa of the Match Point along with the value for k and the rest of parameters into the above right han d side equation:

• Then:

• Since: then:

11

• For buildup:

Calculate k and S:

( )wfwso

D ppBq

khpp −×=∆= −

µσ 31008.7

C

tkh

C

t e

D

D ∆=µ

000295.0

559.688937.0

2==

wtD hrc

CC

φ

+ = 1. 11-0-778

)9$

+,"-. = / � 012 . = :. 2;

Class Exercises

Type Curve Analysis

12

• Using MDH semi-log technique:

• Using log-log Type Curve:

Comparison

3 = ;4-. -6, . = :. 20

3 = 452. 26, . = :. 2;

Class Exercises

Type Curve Analysis

13

Example 2- Drawdown Type Curve Analysis

Class Exercises

Type Curve Analysis

Example 2A welltest has been conducted on an appraisal

well. The well flowed for 12 hours at a constant

rate of 4500 stb/d. Then the well was shut in for

24 hours for a pressure build-up test.

The drawdown test data are provided in Table 1.

The reservoir and fluid parameters are as follows:

Analyse the DST test data using both semi-log

and Gringarten Type Curve techniques and

calculate reservoir parameters of k and S.

14

Table 1

pwf = 4796 psia (end of flow) rw = 0.354 ft h = 50 ft µo = 0.48 cP φ = 0.23 Bo = 1.26 rb/stb c = 10 x 10-6/psi-1 Swc = 0.15

Time Pressure(hours) (psia)

0 49710.001 4947.71

0.0012 4943.230.0015 4938.350.0018 4933.10.0022 4927.570.0026 4921.850.0032 4916.080.0038 4910.40.0046 4904.970.0056 4899.920.0068 4895.370.0082 4891.36

0.01 4887.90.0121 4884.960.0146 4882.450.0177 4880.290.0215 4878.380.026 4876.67

0.0315 4875.080.0382 4873.580.0462 4872.140.056 4870.75

0.0678 4869.390.0822 4868.060.0995 4866.750.1206 4865.45

Time Pressure

(hours) (psia)

0.146 4864.16

0.1769 4862.87

0.2143 4861.56

0.2595 4860.23

0.3144 4858.85

0.3808 4857.41

0.4613 4855.89

0.5587 4854.3

0.6768 4852.63

0.8198 4850.86

0.993 4848.99

1.2028 4847.01

1.457 4844.89

1.7648 4842.58

2.1377 4840.04

2.5894 4837.23

3.1365 4834.09

3.7992 4830.57

4.6019 4826.62

5.5743 4822.16

6.7521 4817.09

8.1787 4811.24

9.9068 4804.4

12 4796.3

Table 1-cnt.

Class Exercises

Type Curve Analysis

Semi-log Technique

15

m

Bqkh oµ

6.162=

� = 565��

+−

−= 23.3log

)1(151.1

2w

wfi

cr

k

m

hrppS

φµ

� = 1.6

Drawdown Semi-log plot

4700

4800

4900

5000

0.0010 0.0100 0.1000 1.0000 10.0000

Time (dt), hours

Flo

win

g P

ress

ure

, Pw

f, P

sia

Wellbore storage

Boundary effectsIARF confirmed by log-log plotData valid for interpretation

y = -16x + 4851

IARF flow

4849)1( =hrpwfpsi

Class Exercises

Type Curve Analysis

Type Curve Technique

16

Time Pressure ∆p d(∆p)/dt t.d(∆p)/dt(hours) (psia) psi psi/hr psi0.0000 4971.0 0.000.0010 4947.7 23.29 23290.0 23.290.0012 4943.2 27.77 22400.0 26.880.0015 4938.4 32.65 16266.7 24.400.0018 4933.1 37.90 17500.0 31.500.0022 4927.6 43.43 13825.0 30.420.0026 4921.9 49.15 14300.0 37.180.0032 4916.1 54.92 9616.7 30.770.0038 4910.4 60.60 9466.7 35.970.0046 4905.0 66.03 6787.5 31.220.0056 4899.9 71.08 5050.0 28.280.0068 4895.4 75.63 3791.7 25.780.0082 4891.4 79.64 2864.3 23.490.0100 4887.9 83.10 1922.2 19.220.0121 4885.0 86.04 1400.0 16.940.0146 4882.5 88.55 1004.0 14.660.0177 4880.3 90.71 696.8 12.330.0215 4878.4 92.62 502.6 10.810.0260 4876.7 94.33 380.0 9.880.0315 4875.1 95.92 289.1 9.110.0382 4873.6 97.42 223.9 8.550.0462 4872.1 98.86 180.0 8.320.0560 4870.8 100.25 141.8 7.940.0678 4869.4 101.61 115.3 7.810.0822 4868.1 102.94 92.4 7.590.0995 4866.8 104.25 75.72 7.530.1206 4865.5 105.55 61.61 7.430.1460 4864.2 106.84 50.79 7.410.1769 4862.9 108.13 41.75 7.39

Time Pressure ∆p d(∆p)/dt t.d(∆p)/dt(hours) (psia) psi psi/hr psi0.2143 4861.56 109.44 35.03 7.51

0.2595 4860.23 110.77 29.42 7.64

0.3144 4858.85 112.15 25.14 7.90

0.3808 4857.41 113.59 21.69 8.26

0.4613 4855.89 115.11 18.882 8.71

0.5587 4854.3 116.7 16.324 9.12

0.6768 4852.63 118.37 14.141 9.57

0.8198 4850.86 120.14 12.378 10.15

0.993 4848.99 122.01 10.797 10.72

1.2028 4847.01 123.99 9.438 11.35

1.457 4844.89 126.11 8.340 12.15

1.7648 4842.58 128.42 7.505 13.24

2.1377 4840.04 130.96 6.811 14.56

2.5894 4837.23 133.77 6.221 16.11

3.1365 4834.09 136.91 5.739 18.00

3.7992 4830.57 140.43 5.312 20.18

4.6019 4826.62 144.38 4.921 22.65

5.5743 4822.16 148.84 4.587 25.57

6.7521 4817.09 153.91 4.305 29.07

8.1787 4811.24 159.76 4.101 33.54

9.9068 4804.4 166.6 3.958 39.21

12 4796.3 174.7 3.870 46.44

∆� = �< − ��

tdt

pdDerivative

)(∆=

Class Exercises

Type Curve Analysis

Engineering plot: ∆) versus ' and '. !(∆))!'

versus '

17

1

10

100

1000

0.001 0.010 0.100 1.000 10.000 100.000

Pw

f(p

si)

an

d

t.d

(Δp

)/d

t(p

si)

Time (hrs)

Class Exercises

Type Curve Analysis

18

Gringarten Type Curve

Class Exercises

Type Curve Analysis

1

10

100

1000

0.001 0.010 0.100 1.000 10.000 100.000Pw

f(p

si)

an

d

t.d

(Δp

)/d

t(p

si)

Time (hrs)

19

Type Curve Match• First make sure the type curves are of the same sca le as the

engineering plot.

Class Exercises

Type Curve Analysis

20

Class Exercises

Type Curve Analysis

21

Type Curve Match Point

• For the Match Point:

� Engineering Plot: , Type Curve:

� Also:

Δp= 113.59

t= 0.3808

PD= 8

tD/CD= 1000

+,"-. = 0 � 012

Class Exercises

Type Curve Analysis

• Substituting the ordinates of the Match Point from the two graphs along with the values provided for various parameters int o the above left hand side equation:

3 = 420. 46,

• Substituting the abscissa of the Match Point along with the value for k and the rest of parameters into the above right han d side equation:

• Then:

• Since: then:

22

• For drawdown:

Calculate k and S:

92.3928937.0

2==

wtD hrc

CC

φ

+ = 1. 11;::778

)9$

+,"-. = 0 � 012 . = 0. ;-

�=

∆�= 7.08 � 10?@

��

�ABC

(�= D=)⁄

�= 0.000295

��

AD

Class Exercises

Type Curve Analysis

23

• Using semi-log technique:

• Using log-log Type Curve:

Comparison

3 = 4;46, . = 0. ;

3 = 420. 46, . = 0. ;-

Class Exercises

Type Curve Analysis

Advanced

Reservoir Engineering

Class Exercises- Chapter 1

Sem 2- 2014

24