Tarek Ahmed. Reservoir engineering handbook

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and n

as the values of the performance coefficient and the flow exponent

at the bubble-point pressure p

, Klins and Clark introduced the following

dimensionless parameters:

• Dimensionless performance coefficient = C/C

• Dimensionless flow exponent = n/n

• Dimensionless average reservoir pressure = p

–

The authors correlated (C/C

) and (n/n

) to the dimensionless pressure

by the following two expressions:

and

where C

= performance coefficient at the bubble-point pressure

= flow exponent at the bubble-point pressure

The procedure of applying the above relationships in adjusting the coef-

ficients C and n with changing average reservoir pressure is detailed below:

Step 1. Using the available flow-test data in conjunction with Fetkovich’s

equation, i.e., Equation 7-34, calculate the present (current) val-

ues of n and C at the present average pressure p

–

Step 2. Using the current values of p

–

, calculate the dimensionless values

of (n/n

) and (C/C

) by applying Equations 7-37 and 7-38,

respectively.

=- -

1 3 5718 1 4 7981 1

2 3066 1

. (7 - 38)

=+ -

1 0 0577 1 0 2459 1

0 503 1

(7 - 37)

508 Reservoir Engineering Handbook

Reservoir Eng Hndbk Ch 07 2001-10-24 16:50 Page 508

Step 3. Solve for the constants n

and C

from:

and

It should be pointed out that if the present reservoir pressure

equals the bubble-point pressure, the values of n and C as calcu-

lated in Step 1 are essentially n

and C

Step 4. Assume future average reservoir pressure p

–

and solve for the cor-

responding future dimensionless parameters (n

) and (C

)

by applying Equations 7-37 and 7-38, respectively.

Step 5. Solve for future values of n

and C

from

= n

(n/n

)

= C

)

Step 6. Use n

and C

in Fetkovich’s equation to generate the well’s future

IPR at the desired average reservoir pressure (p

–

)

. It should be

noted that the maximum oil flow rate (Q

)

max

at (p

–

)

is given by:

)

max

= C

[(p

–

)

]

(7-41)

Example 7-10

Using the data given in Example 7-9, generate the future IPR data

when the reservoir pressure drops to 3200 psi.

Solution

Step 1. Since the reservoir exists at its bubble-point pressure, then:

= 0.854 and C

= 0.00079 at p

= 3600 psi

(/ )

(7 - 40)

(7 - 39)

Oil Well Performance 509

Reservoir Eng Hndbk Ch 07 2001-10-24 16:50 Page 509

Step 2. Calculate the future dimensionless parameters at 3200 psi by

applying Equations 7-37 and 7-38:

Step 3. Solve for n

and C

= (0.854) (1.0041) = 0.8575

= (0.00079) (0.6592) = 0.00052

Therefore, the flow rate is expressed as:

= 0.00052 (32002 - p

)

0.8575

When the maximum oil flow rate, i.e., AOF, occurs at p

= 0, then:

)

max

= 0.00052 (3200

- 0

)

0.8575

= 534 STB/day

Step 4. Construct the following table:

3200 0

2000 349

1500 431

500 523

0 534

Figure 7-14 compares current and future IPRs as calculated in Exam-

ples 7-9 and 7-10.

=- -

1 3 5718 1

3200

3600

4 7981 1

3200

3600

2 3066 1

3200

3600

0 6592

=+ -

1 0 0577 1

3200

3600

0 2459 1

3200

3600

0 5030 1

3200

3600

1 0041

510 Reservoir Engineering Handbook

Reservoir Eng Hndbk Ch 07 2001-10-24 16:50 Page 510

Case 3: p

–

> p

and p

< p

Figure 7-15 shows a schematic illustration of Case 3 in which it is

assumed that p

< p

and p

–

> p

. The integral in Equation 7-25 can be

expanded and written as:

Substituting Equations 7-27 and 7-18 into the above expression gives:

oo b oo

ÚÚ

0 00708

075

11.

ln .

fpdp fpdp

ÚÚ

0 00708

075

ln .

() ()

Oil Well Performance 511

500

1000

1500

2000

2500

3000

3500

4000

100 200 300 400 500 600 700 800 900 1000

Flow Rate (STB/Day)

Pressure (psi)

Current IPR

Future IPR

Figure 7-14. IPR.

Reservoir Eng Hndbk Ch 07 2001-10-24 16:50 Page 511

where m

and B

are evaluated at the bubble-point pressure p

Arranging the above expression gives:

Integrating and introducing the productivity index J into the above

relationship gives:

pp pp

bwf rb

=-+-

()()

pdp dp

ÚÚ

0 00708

075

ln .m

512 Reservoir Engineering Handbook

Figure 7-15. (k

) vs. pressure for Case #3.

Reservoir Eng Hndbk Ch 07 2001-10-24 16:50 Page 512

Example 7-11

The following reservoir and flow-test data are available on an oil well:

• Pressure data: p

–

= 4000 psi p

= 3200 psi

• Flow test data: p

= 3600 psi Q

= 280 STB/day

Generate the IPR data of the well.

Solution

Step 1. Calculate the productivity index from the flow-test data.

Step 2. Generate the IPR data by applying Equation 7-30 when the

assumed p

> p

and using Equation 7-42 when p

< p

Equation Q

4000 (7-30) 0

3800 (7-30) 140

3600 (7-30) 280

3200 (7-30) 560

3000 (7-42) 696

2600 (7-42) 941

2200 (7-42) 1151

2000 (7-42) 1243

1000 (7-42) 1571

500 (7-42) 1653

0 (7-42) 1680

Results of the calculations are shown graphically in Figure 7-16.

It should be pointed out Fetkovich’s method has the advantage over

Standing’s methodology in that it does not require the tedious material

J STB day psi=

280

4000 3600

07.//

QJpp

orb

bwf

=-+ -() ( )

(7 - 42)

Oil Well Performance 513

Reservoir Eng Hndbk Ch 07 2001-10-24 16:50 Page 513

balance calculations to predict oil saturations at future average reservoir

pressures.

The Klins-Clark Method

Klins and Clark (1993) proposed an inflow expression similar in form

to that of Vogel’s and can be used to estimate future IPR data. To

improve the predictive capability of Vogel’s equation, the authors intro-

duced a new exponent d to Vogel’s expression. The authors proposed the

following relationships:

where

(

)

0 28 0 72 1 24 0 001. . . . (7 - 44)

(

)

max

..1 0 295 0 705 (7 - 43)

514 Reservoir Engineering Handbook

500

1000

1500

2000

2500

3000

3500

4000

4500

200 400 600 800 1000 1200 1400 1600 1800

, STB/day

Pressure (psi)

Figure 7-16. IPR using the Fetkovich method.

Reservoir Eng Hndbk Ch 07 2001-10-24 16:50 Page 514

The computational steps of the Klins and Clark are summarized

below:

Step 1. Knowing the bubble-point pressure and the current reservoir pres-

sure, calculate the exponent d from Equation 7-44.

Step 2. From the available stabilized flow data, i.e., Q

at p

, solve

Equation 7-43 for (Q

)

max

Step 3. Construct the current IPR by assuming several values of p

Equation 7-43 and solving for Q

HORIZONTAL OIL WELL PERFORMANCE

Since 1980, horizontal wells began capturing an ever-increasing share

of hydrocarbon production. Horizontal wells offer the following advan-

tages over those of vertical wells:

• Large volume of the reservoir can be drained by each horizontal well.

• Higher productions from thin pay zones.

• Horizontal wells minimize water and gas zoning problems.

• In high permeability reservoirs, where near-wellbore gas velocities are

high in vertical wells, horizontal wells can be used to reduce near-well-

bore velocities and turbulence.

• In secondary and enhanced oil recovery applications, long horizontal

injection wells provide higher injectivity rates.

• The length of the horizontal well can provide contact with multiple

fractures and greatly improve productivity.

The actual production mechanism and reservoir flow regimes around

the horizontal well are considered more complicated than those for the

vertical well, especially if the horizontal section of the well is of a con-

siderable length. Some combination of both linear and radial flow actual-

ly exists, and the well may behave in a manner similar to that of a well

that has been extensively fractured. Several authors reported that the

shape of measured IPRs for horizontal wells is similar to those predicted

by the Vogel or Fetkovich methods. The authors pointed out that the pro-

ductivity gain from drilling 1,500-foot-long horizontal wells is two to

four times that of vertical wells.

Oil Well Performance 515

Reservoir Eng Hndbk Ch 07 2001-10-24 16:50 Page 515

A horizontal well can be looked upon as a number of vertical wells

drilling next to each other and completed in a limited pay zone thickness.

Figure 7-17 shows the drainage area of a horizontal well of length L in a

reservoir with a pay zone thickness of h. Each end of the horizontal well

would drain a half-circular area of radius b, with a rectangular drainage

shape of the horizontal well.

Assuming that each end of the horizontal well is represented by a ver-

tical well that drains an area of a half circle with a radius of b, Joshi

(1991) proposed the following two methods for calculating the drainage

area of a horizontal well.

Method I

Joshi proposed that the drainage area is represented by two half circles

of radius b (equivalent to a radius of a vertical well r

) at each end and a

rectangle, of dimensions L(2b), in the center. The drainage area of the

horizontal well is given then by:

516 Reservoir Engineering Handbook

Figure 7-17. Horizontal well drainage area.

Reservoir Eng Hndbk Ch 07 2001-10-24 16:50 Page 516

where A = drainage area, acres

L = length of the horizontal well, ft

b = half minor axis of an ellipse, ft

Method II

Joshi assumed that the horizontal well drainage area is an ellipse and

given by:

with

where a is the half major axis of an ellipse.

Joshi noted that the two methods give different values for the drainage

area A and suggested assigning the average value for the drainage of the

horizontal well. Most of the production rate equations require the value

of the drainage radius of the horizontal well, which is given by:

where r

= drainage radius of the horizontal well, ft

A = drainage area of the horizontal well, acres

Example 7-11

A 480-acre lease is to be developed by using 12 vertical wells. Assum-

ing that each vertical well would effectively drain 40 acres, calculate the

possible number of either 1,000- or 2,000-ft-long horizontal wells that

will drain the lease effectively.

43 560,

b=+

(7 - 47)

43 560,

(7 - 46)

Lb b

(

)

43 560

(7 - 45)

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Reservoir Eng Hndbk Ch 07 2001-10-24 16:50 Page 517