taller ejercicio 1 dd

8/12/2019 Taller Ejercicio 1 Dd

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1. Problem No 1.

A new well, in a small bounded reservoir, North

of Hobbs, New Mexico, was produced at a constant

rate of 360 STB/D.

The initial pressure throughout the reservoir priorto the flow test was 4620 psia.

Part 1. Conventional Techniques

A) Skin Effect1. Calculate reservoir transmissibility and

permeability.

2. Compute the skin factor.3. Calculate flow efficiency, damage ratio and

damage factor. Is the wellbore damaged or

stimulated? Why?

4. Calculate the radial distance of the skin, ifthe formation permeability from lab.measurements is expected to be affected 410

percent by the skin. Compare with theapparent radius, rw.

5. What is the flow rate without the skineffect?

B) Wellbore Storage6. Find the wellbore storage coefficient.7. Estimate the starting time of the semilog

straight line.

C) System Geometry8. Calculate the drainage volume and drainage

area (in acres) of this well.

9. Find the shape factor of this drainage area.10.Determine the system geometry.11.Estimate the drawdown stabilization time.12.Compute the radius of drainage created

during this drawdown test.

Table 1.1. Drawdown Test Data for Problem No. 1

t

[hrs]

Pwf

[psia]

t

[hrs]]

Pwf

[psia]

0 4,620 4.000 4,360

0.020 4,588 7.000 4,325

0.030 4,576 10.000 4,300

0.044 4,552 20.000 4,256

0.060 4,528 30.000 4,230

0.100 4,505 50.000 4,195

0.200 4,486 70.000 4,172

0.400 4,478 80.000 4,160

0.700 4,450 90.000 4,150

1.000 4,435 100.000 4,140

2.000 4,400 125.000 4,120

3.000 4,380

Solution

1. Conventional Techniques

Skin effect

Using data given on Table 1.1, Pwf vs. time isgraphed on Cartesian (Fig. 1.1) and semi log (Fig

1.2). Cartesian plot shows clearly the pseudo stable

state of flow characterized by a straight line

Because the infinite active line is before PSS, thestraight line in semilog is plotted avoiding the PSS.

4000

4100

4200

4300

4400

4500

4600

4700

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140

Time (hours)

Pwf(Psia)

Pint=4245

m*=1

Fig. 1.1 Cartesian plot Pwf vs. time

4000

4100

4200

4300

4400

4500

4600

4700

0.01 0.1 1 10 100 1000

Time(hours)

Pwf(Psia)

m=140

P1hr=4440

Fig. 1.2 Semilog plot Pwf vs. time

From figure 1.2

cyclepsiam /1402

44404160

Permeability is computed from

mh

qk

6.162 (1.1)

mdk 56.538*140

84.0*22.1*360*6.162


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[Homework Number 1] Dora Restrepo 2

Transmissibility

cp

ftmdkh *10.510

84.0

8*56.53

Skin factor is computed from

23.3log1513.1

2

1

wt

hri

rc

k

m

pps

(1.2)p1hris read from Fig. 1.2 on the infinite acting line.

P1h = 4440 psia. Then,

23.3

29.0*10*42*84.0*10.0

56.53log

140

444046201513.1

25s

16.3s

Eq. 1.3 allows find psvalue:

skh

qps

2.141 (1.3)

psips 6.383)16.3(*8*56.53

22.1*84.0*3602.141

Flow efficiency, damage radio and damage factor

are estimated from:

wf

swf

pp

pppFE

(1.4)

Pwf is the last value of pressure at the end of thetest and average pressure will be considered equal

to pi, which implies the assumption that the well is

new or steady state.

%17777.141204620

)3.383(41204620

FE

FEDR

1 (1.5)

%5757.077.1

1DR

FEDF 1 (1.6)

%7777.077.11 DF

These values show that the area next to the wellbore

is stimulated because S < 0, FE > 1, DR < 1, and

DF < 0. According to Tiab (2004) this value of s isinto the range for hydraulically fractured wells (-3

to -5)

Radial distance of the skin:

w

s

s r

r

k

kS ln1

(1.7)

In terms of rs:

1

* sk

k

S

ws err (1.8)

ks = 4.10*k from laboratory measurement

expectations. So, ks= 219.6 md

Then,

fters 8.18*29.01

6.219

56.53

16.3

Apparent radius:

s

ww err ' (1.9)

fterw 8.6*29.0')16..3(

The flow rate without effect skin is calculated using

the next set of equations:

ideal

actual

J

JFE (1.9)

In terms of Jideal:

FE

JJ actual

ideal (1.10)


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and,

skinwf

idealppp

qJ

(1.11)

Finding Jactual:

wf

actualpp

qJ

(1.12)

72.041204620

360

actualJ

Substituting this value on Eq. 1.10,

4.077.1

72.0

idealJ

Using Eq. 1.4 and leaving Eq. 1.11 in terms of q:

FE

qq actual

s 0 (1.13)

DSTBqs /20377.1

3600

Wellbore Storage

Wellbore storage coefficient is estimate using Eq.

1.14 and taking (t/p)i data from Fig. 1.3 at any

convenient point on the unit slope.

10

100

1000

0.01 0.1 1 10 100 1000

Time (hours)

P, t*P'

(t*p')i =70 psia

(p)i

ti = 0.042

TSSL = 4 hr

pr = 390 psia

tr = 30 hr

trpi = 64 hr

Fig. 1.3. Pressure derivative.

ip

tqC

24

(1.14)

psia

bblC 01.0

70

043.0

24

22.1*360

Reading tSSL (starting time of semilog straight

line) from Fig. 1.3; tSSL = 4 hr. Using Eq 1.15 we

find for tSSL:

Ckh

stSSL

/

)*000,12000,200( (1.15)

hrtSSL 6.401.084.0/8*56.53

)]16.3(*000,12000,200[

System geometry

From Fig. 1.1, pressure at the intercept is 4245

psia, and m*is calculated as the slope of the straight

line:

12

12*

tt

ppm

(1.16)

19080

41504160* m

But m* in terms of A is:

Ahc

qm

t

23395.0* (1.17)

Then,

hmc

qA

t )(

23395.0

*

(1.18)

2

536.941,295

8*.)1(*10*1042.0

22.1*360*23395.0ftA

AcresA 79.6

AhV (1.19)309.753,2368*36.941,295*10.0 ftV


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Shape factor

mpp

A

hr

em

mC

)(303.2

*

int1

456.5 ( 1.20)

8.301

140456.5 140)42454454(303.2

eCA

System geometry

pssDA tm

mt

*

1833.0 (1.21)

Where tpss is the starting time of pseudosteady

state flow, and is obtained from Fig. 1.1. tpss= 42hours. Then,

055.042)140(

)1(1833.0

DAt

With CA= 30.8 and tDA= 0.055, the reservoir hasthe next geometry with less than 1% error for tDA:

Where the shape is taken fromAppendix D, page 265 (Pressure

Transient Testing):

CA = 30.8828

tDA= 0.05

Drawdown stabilization time

k

Act ts

)43560(380

(1.22)

1.7856.53

)79.6*43560(10*42*10.0*84.0380

5

st

12. Radius of drainage created during the test

t

s

dc

ktr

029.0 (1.23)

1.30810*42*10.0*84.0

1.78*89.50029.0

5

dr

taller ejercicio 1 dd

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