Tarek Ahmed. Reservoir engineering handbook

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imposed on Equation 9-21 with Z = 3.5 to give an expression for calcu-

lating the critical oil flow rate, or

Example 9-11

Calculate the water breakthrough using the Sobocinski-Cornelius

method for a vertical well producing at 250 STB/day. The following

reservoir data are available:

= 250 STB/day h = 50 ft hp = 15 ft r

= 63.76 lb/ft

= 47.5 lb/ft

= 0.73 cp B

= 1.1 bbl/STB k

= 9 md

= 93 md f=13% M = 3

Solution

Step 1. Solve for the dimensionless cone height Z from Equation 9-21 to

give

Step 2. Calculate the dimensionless breakthrough time by using Equation

9-22.

Step 3. Estimate time to breakthrough from Equation 9-23.

20,325 (0.73) (0.13) (50) (0.5481)

(63.76 47.5) (9) (1 + )

= 123 days

- 3

(

)

(4) (0.64866) 1.75

(0.6486)

0.75

(0.6486)

7 2 (0.6486)

= 0.5481

Z = 0.492

(63.76 47.5) (93) (50) (50 15)

(0.73) (1.1) (250)

= 0.6486¥

= 0.141

()

h (h h )

(9 - 25)

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Example 9-12

Using the data given in Example 9-11, approximate the critical oil

flow rate by using Equation 9-25.

Solution

The Bournazel-Jeanson Method

Based on experimental data, Bournazel and Jeanson (1971) developed

a methodology that uses the same dimensionless groups proposed in the

Sobocinski-Cornelius method. The procedure of calculating the time to

breakthrough is given below.

Step 1. Calculate the dimensionless core height Z from Equation 9-21.

Step 2. Calculate the dimensionless breakthrough time by applying the

following expression:

Step 3. Solve for the time to breakthrough t

by substituting the above-cal-

culated dimensionless breakthrough time into Equation 9-23, i.e.,

As pointed out by Joshi (1991), Equation 9-26 indicates that no

breakthrough occurs if Z ≥ 4.286. Imposing this value on Equa-

tion 9-21 gives a relationship for determining Q

woh p

= 0.1148

khh h

()()rr

(9 - 27)

20,325 h (

)

()

(1+

)

(

)

3 0.7 Z-

(9 - 26)

= 0.141

(63.76 47.5) (93) (50) (50 15)

(0.73) (1.1)

= 46.3 STB/day

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Example 9-13

Resolve Example 9-11 by using the Bournazel-Jeanson method.

Step 1. Solve for the dimensionless cone height Z = 0.6486

Step 2. Calculate the dimensionless breakthrough time from Equation 9-26.

Step 3. Calculate the time to breakthrough by apply Equation 9-23 to give

Step 4. From Equation 9-27, the critical oil rate is

= 37.8 STB/day

AFTER BREAKTHROUGH PERFORMANCE

Once the water breakthrough occurs, it is important to predict the per-

formance of water production as a function of time. Normally, using

numerical radial models solves such a problem. Currently, no simple ana-

lytical solution exists to predict the performance of the vertical well after

breakthrough. Kuo and Desbrisay (1983) applied the material balance

equation to predict the rise in the oil-water contact in a homogeneous

reservoir and correlated their numerical results in terms of the following

dimensionless parameters:

• Dimensionless water cut (f

)

• Dimensionless breakthrough time t

DBT

• Dimensionless limiting water cut (WC)

limit

The specific steps of the proposed procedure are given below.

Step 1. Calculate the time to breakthrough t

by using the Sobocinski-

Cornelius method or the Bournazel-Jeanson correlation.

Step 2. Assume any time t after breakthrough.

20,325 (0.73) (0.13) (50) (0.2548)

(63.76 47.5) (9) (1 + )

= 57.2 days

- 3

(

)

0.6486

3 0.7 (.6486)

= 0.2548

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Step 3. Calculate the dimensionless breakthrough time ratio t

DBT

from:

DBT

= t/t

(9 - 28)

Step 4. Compute the dimensionless limiting water cut from:

With the parameters in Equation 9-29 as defined below:

h = H

(1 - R) (9 - 31)

= H

+ H

R (9 - 32)

where (WC)

limit

= current limiting value for water cut

M = mobility ratio

)

sor

= relative permeability for the water and residual oil

saturation (S

)

swc

= relative permeability for the oil at the connate

water saturation (S

)

, m

= oil and water viscosities, cp

= initial oil zone thickness, ft

= initial water zone thickness, ft

h = current oil zone thickness, ft

= current water zone thickness, ft

= cumulative oil production, STB

N = initial oil in place, STB

Step 5. Calculate the dimensionless water cut (f

)

based upon the

dimensionless breakthrough time ratio as given by the following

relationships:

)

= 0 for t

DBT

< 0.5 (9-34)

R=(

/N)

or wc

(9 - 33)

(

)

(

)

sor

swc

(9 - 30)

(WC

)

M(h/h)

limit

(9 - 29)

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)

= 0.29 + 0.94 log (t

DBT

) for 0.5 ≤ t

DBT

≤ 5.7 (9-35)

)

= 1.0 for t

DBT

> 5.7 (9-36)

Step 6. Calculate the actual water cut f

from the expression:

= (f

)

(WC)

limit

(9-37)

Step 7. Calculate water and oil flow rate by using the following expres-

sions:

= (f

) Q

(9-38)

= Q

- Q

(9-39)

where Q

, Q

are the water, oil, and total flow rates, respectively.

It should be pointed out that as oil is recovered, the oil-water contact

will rise and the limiting value for water cut will change. It also should

be noted the limiting water cut value (WC)

limit

lags behind one time step

when calculating future water cut.

Example 9-14

The rock, fluid, and the related reservoir properties of a bottom-water

drive reservoir are given below:

well spacing = 80 acres

initial oil column thickness = 80 ft

= 20¢r

= 47 lb/ft

= 63 lb/ft

= 1053¢

= 0.25¢ M = 3.1 f=14% S

= 0.35

= 0.25 B

= 1.2 bbl/STB m

= 1.6 cp m

= 0.82 cp

= 60 md k

= 6 md

Calculate the water cut behavior of a vertical well in the reservoir

assuming a total production rate of 500, 1000, and 1500 STB/day.

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Solution

Step 1. Calculate the dimensionless cone height Z by using Equation 9-21.

500 1000 1500

Z 0.2362 0.1181 0.0787

Step 2. Calculate the dimensionless breakthrough time by applying Equa-

tion 9-26.

Q 500 1000 1500

Z 0.2362 0.1181 0.0787

)

0.08333 0.04048 0.02672

Step 3. Calculate the time to breakthrough from Equation 9-23.

Q 500 1000 1500

)

0.08333 0.04048 0.02672

106.40 51.58 34.11

Step 4. Calculate initial oil in place N.

N = 7758 Af h ( 1 - S

)/B

N = 7758 (80) (0.14) (80) (1 - 0.25) / 1.2 = 4,344,480 STB

Step 5. Calculate the parameter R by applying Equation 9-33.

R=[

/(4,344, 480)]

1 0.25

1 0.35 0.25

= 4.3158 N

BT D BT

(20,325) (1.6) (0.14) (80)

(63 47) (6) (1+ 3.1

)

(

)

= 1276 76.()

Z = 0.492

(63 47) (60) (80) (80 20)

(1.6) (1.2)

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Step 6. Calculate the limiting water cut at breakthrough.

500 1000 1500

106.4 51.58 34.11

53,200 51,580 51,165

R 0.02296 0.022261 0.022082

h 78.16 78.22 78.23

21.84 21.78 21.77

(WC)

limit

0.464 0.463 0.463

Step 7. The water cut calculations after an assumed elapsed time of 120

days at a fixed total flow rate of 500 STB/days are given below:

• From Equation 9-28, calculate t

DBT

= 120/106.4 = 1.1278

• Apply Equation 9-36 to find (f

)

= 0.29 + 0.96 log (1.1278) = 0.3391

• Solve for the present water cut from Equation 9-37:

= (0.3391) (0.464) = 0.1573

Step 8. Calculate water and oil flow rate:

= (0.1573) (500) = 78.65 STB/day

= 500 - 78.65 = 421.35 STB/day

Step 9. Calculate cumulative oil produced from breakthrough to 120 days:

Step 10. Calculate cumulative oil produced after 120 days:

= 53,200 + 6,265.18 = 59,465.18 STB

Step 11. Find the recovery factory (RF):

RF = 59,465.18/4,344,480 = 0.0137

DN=

500 + 421.35

(120 106.4) = 6,265.18 STB

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Step 12. Assume an elapsed time of 135 days, repeat the above steps at

the same total rate of 500 STB/day:

• R = 4.3158 ¥ 10

-7

(59,465.18) = 0.020715

• h

= 21.66

• h = 78.34

• (W

)

limit

= 0.4615

• (f

)

= 0.29 + 0.94 log (135/106.4) = 0.3872

• f

= (0.3872) (0.4615) = 0.1787

• Q

= (500) (0.1787) = 89.34 STB/day

= 500 - 89.34 = 410.66 STB/day

• N

= 59,465.18 + 6,240.0 = 65,705.22

• RF = 0.0151

Tables 9-1 through 9-3 summarize the calculations for water cut versus

time for total flow rates of 500, 100, and 1500 STB/day, respectively.

CONING IN HORIZONTAL WELLS

The applications of horizontal well technology in developing hydro-

carbon reservoirs have been widely used in recent years. One of the main

objectives of using this technology is to improve hydrocarbon recovery

from water and/or gas-cap drive reservoirs. The advantages of using a

horizontal well over a conventional vertical well are their larger capacity

to produce oil at the same drawdown and a longer breakthrough time at a

given production rate.

Many correlations to predict coning behavior in horizontal wells are

available in the literature. Joshi (1991) provides a detailed treatment of

the coning problem in horizontal wells. As in vertical wells, the coning

problem in horizontal wells involves the following calculations:

• Determination of the critical flow rate

• Breakthrough time predictions

• Well performance calculations after breakthrough

DN=

410.66 + 421.34

(135 120) = 6, 240.0 STB

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Table 9-1

Results of Example 9-14

Total Production Rate is 500 STB/day

Time Oil Rate Water Rate Water Cut Cum. Oil Oil Rec.

days STB/day STB/day Fraction MSTB %

120. 406.5 93.5 0.187 59.999 1.38

135. 379.9 120.1 0.240 65.897 1.52

150. 355.8 144.2 0.288 71.415 1.64

165. 333.8 166.2 0.332 76.587 1.76

180. 313.5 186.5 0.373 81.442 1.87

195. 294.5 205.5 0.411 86.002 1.98

210. 276.9 223.1 0.446 90.287 2.08

765. 239.3 260.7 0.521 177.329 4.08

1020. 226.4 273.6 0.547 236.676 5.45

1035. 225.6 274.4 0.549 240.066 5.53

1575. 202.4 297.6 0.595 355.425 8.18

2145. 182.4 317.6 0.635 464.927 10.70

2415. 174.2 325.8 0.652 513.062 11.81

2430. 173.8 326.2 0.652 515.672 11.87

2445. 173.4 326.6 0.653 518.276 11.93

3300. 151.5 348.5 0.697 656.768 15.12

3615. 144.6 355.4 0.711 703.397 16.19

3630. 144.3 355.7 0.711 705.564 16.24

3645. 144.0 356.0 0.712 707.727 16.29

Horizontal Well Critical Rate Correlations

The following four correlations for estimating critical flow rate in hor-

izontal wells are discussed:

• Chaperson’s Method

• Efros’ Method

• Karcher’s Method

• Joshi’s Method

Chaperson’s Method

Chaperson (1986) provides a simple and practical estimate or the criti-

cal rate under steady-state or pseudosteady-state flowing conditions for

an isotropic formation. The author proposes the following two relation-

ships for predicting water and gas coning:

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Table 9-2

Results of Example 9-14

Total Production Rate is 1000 STB/day

Time Oil Rate Water Rate Water Cut Cum. Oil Oil Rec.

days STB/day STB/day Fraction MSTB %

80. 674.7 325.3 0.325 64.14 1.48

95. 594.2 405.8 0.406 73.66 1.70

110. 524.8 475.2 0.475 82.05 1.89

125. 463.2 536.8 0.537 89.46 2.06

140. 407.8 592.2 0.592 95.99 2.21

335. 494.8 505.2 0.505 145.68 3.35

905. 390.8 609.2 0.609 395.80 9.11

1115. 362.4 637.6 0.638 474.81 10.93

1130. 360.5 639.5 0.639 480.23 11.05

1475. 321.8 678.2 0.678 597.70 13.76

1835. 288.8 711.2 0.711 707.42 16.28

1850. 287.5 712.5 0.712 711.74 16.38

1865. 286.3 713.7 0.714 716.04 16.48

1880. 285.1 714.9 0.715 720.33 16.58

1895. 283.8 716.2 0.716 724.59 16.68

2615. 234.4 765.6 0.766 910.20 20.95

2630. 233.5 766.5 0.766 913.71 21.03

3065. 210.4 789.6 0.790 1010.11 23.25

3080. 209.6 790.4 0.790 1013.26 23.32

3620. 185.8 814.2 0.814 1119.83 25.78

3635. 185.2 814.8 0.815 1122.61 25.84

3650. 184.6 815.4 0.815 1125.39 25.90

Water coning

Gas coning

The above two equations are applicable under the following constraint:

1 < 70 and 2y < 4L (9 - 42)

£≤a

Q = 0.0783

()

k [h (h

)]

(9 - 41)

Q = 0.0783

()

k h (h

)

[]

(9 - 40)

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