Well Test Problems

8/13/2019 Well Test Problems

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Well test exercises.

Dr Jamshidi.

Arash Nasiri Savadkouhi/92250744

An 8856 ft deep oil well completed with a 7.675 inch ID production casing. The well is scheduled for drawdown

test with the following data:

B=1.2 bbl/STB, =0.8 cp, rw=0.2 ft, =0.11, h=116 ft, Co= o=50 lbm/ft3, Ct =6If this well is placed under a single phase, single rate (q=400 STB/D) production through the casing and the

following data recorded test:

a) Classify flow history into four periods (if there is!)

b) Calculate wellbore storage coefficient and its dimensionless value.

c) Determine the formation permeability, skin factor, apparent

wellbore radius and reservoir flow efficiency.

d) Estimate the pore volume of the reservoir

Solution:

Steps:

1. Plot log() log(), log-log paper.2. Plot log(), semi-log paper3. Plot in Cartesian coordinate4. From log-log paper, find the line with tangent 1 cycle per cycle.5. Seek lines with tangent of and cycle per cycle in log-log paper.6. We calculate C and CD.7. If we found #5 we seek for start of transient flow after the last pointlocated on those lines. If not, after finding we go 1.5 cycle forwardand after that time in the semi-log paper seeking the MTR line.

8. In semi-log paper by finding the MTR line we notify the start andend of transient flow time. With the slope and P1hr of the line we

calculate k, s, rw.

9. By using plot of Pwf vs t we gain the pore volume of the reservoir. If we consider the shape of the reservoir as a circle then:

= .... 00264 = ...

. 00264

Other equations: = . ...

=.

.( log

3.230.8694) Ei solution is only valid in MTR period.

= . , = 1.151 ( ... 3.23), =... , = .....

= .. , = , =

Time(hr) Pwf(psi) = 0 4412 -

0.1 4212 200

0.15 4117 295

0.2 4012 400

1.94 3699 713

2.79 3653 759

4.01 3636 776

4.82 3616 796

5.78 3607 805

6.94 3600 812

8.32 3593 819

9.99 3586 826

14.4 3573 839

17.3 3567 845

20.7 3561 851

24.9 3555 857

29.8 3549 863

35.8 3544 868

43 3537 875

51.5 3532 880

61.8 3526 886

74.2 3521 891

89.1 3515 897

107 3509 903

128 3503 909

154 3497 915185 3490 922

222 3481 931

266 3472 940

319 3460 952

383 3446 966

460 3429 983


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Results:

In ETR region:

twbs=6.94 hr, C=0.0107, CD=2967.43 /we also have a fracture near well. Fracture length= rstart-rend=119.6 ft

In MTR region: m=-76 psi/cycle; => k=7.08 mD

=3667 T1=6.94 hr, T2=222 hr => r1=313 ft, r2=1772 ft, hence a substantial amount of formation

has been sampled; thus I can be more

confident that that the permeability is

representative.

S=5.23

rwa=0.053 ft, = 0.869 = 345.4 = . =0.64LTR Region: m=

=-0.2194 psi/hr => =85.3210

Plots:

Log-log

MTR region: Pwf vs t


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LTR region:

Cartesian coordination Pwf vs t


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Problem 2 - Following Table shows the pressure buildup data from an oil well with an estimated drainage

radius of 2640 ft. Before shut-in, the well had produced at a stabilized rate of 4900 STB/day for 310 hours.

Known reservoir data is:

Depth:10476 ft, rw=0.354 ft, Ct =22.6Q=4900 STBD h=482 ft, re=2640 ft

= 0.20 , = 0.09 B=1.55 bbl/STB tp=310 hr;

a) By Horner analysis calculate:The average permeability

The skin factor

The additional pressure drop due to skin.

b) Is this reservoir an infinitive acting reservoir during pressure buildup test (why or why not)?if yes, estimate initial reservoir pressure.

If no(is bounded reservoir):

Estimate the False pressure.Calculate the average reservoir pressure from MBH method (well drainage area is circular).

c) By MDH method calculate (well drainage area is circular and no flow across drainage boundary):The average permeability

The skin factor

The additional pressure drop due to skin

The false pressure

The reservoir initial pressure

Solution:

Steps:

1. Plot log

(log-log paper)(ref:John Lee)

But we do as is mentioned in the class

Plot log 2. Plot Pws vs log((tp+)/) (semi-log paper)(Horner)3. From log-log paper we obtain:ETR region and start of MTR, then we calculate C,Cd.

4. From semi log paper we obtain K,s,, rw,FEMDH method.

5. = . ... & = . 6. 7. =

.. MBH method.

8. Pws= .log+ , & = ... .9. From MBH curves . / ,

. /

RESULTS.

Since the number of points are too much to plot I just pick

them up arbitrarily.

ETR: no line with slope of 1/0.5/0.25 cycle per cycle could

pass through the early points however the end of wellbore

storage can be seen as 0.84 hr.

Hence (Horner results):

Slope=54, = . =9.18

Time

(hr)

Pws

psi = 1

Pws-Pwf

0 2761 0 - -0.1 3057 0.1 3101 296

0.21 3153 0.2099 1472 392

0.31 3234 0.3097 1001 473

0.52 3249 0.5191 597.2 488

0.63 3256 0.6287 493.1 495

0.73 3260 0.7283 425.7 499

0.84 3263 0.8377 370 502

0.94 3266 0.9372 330.8 505

1.05 3267 1.0465 296.2 506

1.15 3268 1.1457 270.6 507

1.36 3271 1.3541 228.9 510

1.68 3274 1.6709 185.5 513

1.99 3276 1.9773 156.8 515

2.51 3280 2.4898 124.5 519

3.04 3283 3.0105 103 522

3.46 3286 3.4218 90.6 525

4.08 3289 4.0270 77 528

5.03 3293 4.9497 62.6 532

5.97 3297 5.8572 52.9 536

6.07 3297 5.9534 52.1 536

7.01 3300 6.8550 45.2 539

8.06 3303 7.8558 39.5 542

9 3305 8.7461 35.4 544

10.05 3306 9.7344 31.8 545

13.09 3310 12.5597 24.7 549


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= 1.151 ( ... 3.23)=4.93() =0.869 =231.3 psib) This reservoir does not act as an infinitive reservoir since

the the Horner line doesnt converge to Pi. So:

=3367 Using MBH method: tDA=0.04038

Then: 2.09= .(). =3358.25

Utilizing MDH method:

We generate:

m=43.49, =3309 psi, K=11.78 mD = 1.151 ( ... 3.23)=8.59() =0.869 =324.64 psiFor r(drainage radius)=2640 ft, t=2978 hr then, A=27861164.18 ft2

= . =0.106, = =(for t=3.04 hr and Pws=3283 psi)=0.00001Hence 4.36= = (). => Pi=3447.7 psi

16.02 3313 15.2328 20.4 552

20 3317 18.7879 16.5 556

26.07 3320 24.0477 12.9 559

31.03 3322 28.2066 11 561

34.98 3323 31.4331 9.9 562

37.54 3324 33.4851 9.3 563

40 3325 35.4286 8.8 564

42 3326 36.9886 8.4 565

44.3 3328 38.7609 8 567

48 3330 41.5642 7.5 569

53.5 3338 45.6259 6.8 577

55 3340 46.7123 6.6 579


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Well Test Problems

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