# METHODS CALCULATE AREA DRAINED BY HORIZONTAL WELLS

S.D. JoshiJoshi Technologies International Inc.

Tulsa

The drainage area of a horizontal well can be calculated by averaging the results obtained from two different methods. For anisotropic reservoirs, nonuniform permeability has to be included in drainage area calculations.

### Well spacing

Horizontal wells, in general, give higher reserves than vertical wells. Two reasons for obtaining higher ultimate reserves are increase in the drainage area and increase in the recovery factor.

The most important reason for obtaining higher ultimate reserves from horizontal wells is the significant enhancement of the drainage area, especially as compared to vertical wells. The other reason for an increase in reserves is improvement in the recovery factors. Recovery factor is the percentage of the original oil in place that can be produced.

At the present time, the field histories tend to indicate that the recovery factors are, in general, 2-5% higher than those for vertical wells. Until more long term histories are available, it is difficult to make an exact judgment on the improvement of recovery factors.

In a given time period, a horizontal well drains a larger reservoir volume than a vertical well. Thus, the spacing used for a horizontal well should be larger than that used for a vertical well.

The drainage area of a horizontal well also depends upon natural fracturing in a fractured reservoir.

In a naturally fractured reservoir, a horizontal well drains more volume in the direction parallel to the natural fractures than in the direction perpendicular to the natural fractures. Therefore, the well spacing requirement along a fracture trend is different than that perpendicular to the fracture trend.

It is important to note that even for a vertical well, the well spacing is based upon the reservoir parameters and economic criteria. Theoretically speaking, one can drill a single well in a large reservoir and it will drain the entire reservoir, although it may take a very long time.

Well spacings are designed so as to maximize the oil recovery and economic benefit of production. To estimate the drainage area of a horizontal well, first one must estimate the drainage area (or well spacing) of a vertical well.

### Pseudosteady state

Dimensionless time, tD, which is used to define various regions, is given as:

[SEE FORMULA (1)]

and area-based dimensionless time

[SEE FORMULA (2)]

Thus

[SEE FORMULA (3)]

where: k = permeability (md), t = time (hr), f = porosity (fraction), viscosity (cp), cti = initial total compressibility (psi-1), A = area sq ft, and rw = well bore radius, ft

For a vertical well located at the center of a drainage circle or a square, the time to reach pseudosteady state is tDA = 0.1. Substituting this in Equation 3, we have

[SEE FORMULA (4)]

[SEE FORMULA (5)]

tpss = time to reach pseudosteady state in hours

[SEE FORMULA (6)]

tpdss is the time to reach pseudosteady state in days.

Generally, oil wells are developed on 40-acre spacing and gas wells are developed on 160-acre spacing. Hence 40 acres 40 x 43,560 sq ft/acre 1.7424 x 106 sq ft (7)

160 acres = 160 x 43,560 sq ft/acre = 6.9696 x 106 sq ft (8)

Substituting these areas into Equation 6 gives for a 40 acre well:

[SEE FORMULA (9)]

and for a 160 acre well:

[SEE FORMULA (10)]

Equations 4 and 5 show that transient time depends on the basic reservoir properties, such as permeability, porosity, and compressibility. Time to reach pseudosteady state does not depend on well stimulation.

In the case of oil wells, normally, the time to reach pseudosteady state in of the order of a few days to months.

In contrast, for gas wells in low-permeability reservoirs, the time to reach pseudosteady state could be very long; in some cases as long as a few years.

### Oil well example

For an oil well drilled on 40-acre spacing, calculate the time to reach pseudosteady state given f = 10%, cti 0.00005 psi 1, k = 35 md, and m = 4.2 cp (shallow well-dead oil). Using Equation 5, tpss = 0.0002274 A = 396 hr = 16.5 days.

### Gas well example

Calculate the time required to reach pseudosteady state for a gas well drilled at either 20 or 160-acre spacing in a reservoir with an initial pressure (pi) of 1,450 psi and the reservoir properties of f = 7%, k = 0.03 md, m = 0.015 cp, and cti = 0.000690 psi-1.

Again using Equation 5 tpss = 0.00915A. For 20 acres, tpss 7,974 hr = 332 days 0.91 years.

For 160 acres, tpss 63,772 hr 2,657 days 7.3 years of infinite-acting period.

As noted in the previous example, gas wells (or oil wells) drilled in a very tight reservoir, especially gas wells in reservoirs with permeability less than 0.1 md, can take years for the transient state to end. In such tight reservoirs, it is very difficult to drain the reservoir economically. In these cases, methods are needed to accelerate reservoir drainage.

Infill drilling and horizontal drilling provide alternatives to drain the reservoir effectively.

The dimensionless time to reach pseudosteady state is tDA = 0.1, as long as the well is centrally located in a drainage plane, i.e., when the well is at the center of a circle or a square (xe/ye = 1).

When the drainage area becomes rectangular, the time to reach pseudosteady state increases. For example, when one side of a drainage rectangle is five times larger than the other side (xe/ye = 5) the dimensionless time to reach pseudosteady state is tDA = 1.0, i.e., ten times longer than a vertical well located centrally in the drainage plane.

Thus, vertical wells are unable to drain effectively rectangular drainage areas in uniform permeability reservoirs.

As shown in Fig. la, a long horizontal well in a given time can drain a larger area than a vertical well.

A 40-acre spacing vertical well reaches pseudosteady state in 16 days. By the same principle (Fig. 1) a 2,000 ft long well would reach pseudosteady state in a 101-acre area in 16 days.

Table 1 tells us that the time to reach pseudosteady state using vertical well draining a rectangle with dimensions xe/ye = 2.5 would 2.5 times longer than that for a 2,000 ft long horizontal well (assumes L is 1 - the quantity ye/xe).

Thus, horizontal wells can be utilized to drain a larger reservoir volume than vertical wells in a given time period. This becomes very important in tight reservoirs when close vertical well spacing is required to drain the reservoir effectively.

Therefore, in a tight reservoir, horizontal wells can be used to enhance drainage volume per well in a given time period.

### Areal anisotrophy

The discussion so far has been restricted to reservoirs with homogeneous areal permeability, namely kx = ky (Fig. 1b). In naturally fractured reservoirs, the permeability along the fracture trend is larger than in a direction perpendicular to fractures. In such cases, a vertical well would drain more length along the fracture trend.

The derivation starting with Equation 11 can be used to estimate each side of a drainage area in an areally anisotropic reservoir. The equation assumes a single phase, steady-state (time independent) flow through an homogeneous formation.

[SEE FORMULA (11)]

Assuming constant values of kx and ky in x and y directions, respectively, Equation 11 is rewritten as

[SEE FORMULA (12)]

Multiplying and dividing throughout by kxky, Equation 12 becomes

[SEE FORMULA (13)]

This can be transformed into

[SEE FORMULA (14)]

where

[SEE FORMULA (15)]

and

[SEE FORMULA (16)]

Thus, an areally anisotropic reservoir would be equivalent to a reservoir with effective permeability of kxky. The drainage length along the high-permeability side is larger by a factor of ky/kx than the drainage length along a low-permeability side.

Thus, if permeability along the fracture trend is 16 times larger than perpendicular to it, then drainage length along the fracture is four times larger than the length perpendicular to the fracture (Fig 1b).

In such areally anisotropy reservoirs, using vertical wells, it is difficult to drain the larger reservoir lengths in the low-permeability direction.

A horizontal well drilled along the low-permeability direction has a potential to drain a significantly larger area than a vertical well and therefore recover more reserves than vertical wells.

Thus, horizontal wells are highly beneficial in areally anisotropic reservoirs.

It is obvious that in naturally fractured formations, horizontal wells drilled in a direction perpendicular to the natural fractures are highly beneficial (Fig 1c). The success of horizontal wells in naturally fractured reservoirs, such as Austin chalk formation in Texas and Bakken formation in North Dakota, illustrates the advantage of horizontal drilling in areally anisotropic formations.

For fractured vertical wells, limited results are available to calculate the time to reach pseudosteady state in square drainage boundaries.1 2 Khan has obtained results for fractured vertical wells in rectangular areas.3

Recently, similar results were also available for horizontal wells.4-6 Mutalik, et al.,4 calculated the time to reach the pseudosteady state for fractured vertical wells and horizontal wells in rectangular drainage areas (Tables 1 and 2).

It is important to note that there is some discrepancy in calculating the time to start pseudosteady state. For a single-phase flow in a homogeneous reservoir, the relationship between the dimensionless pressure and the dimensionless time for a well producing at a constant rate in a bounded reservoir (i.e., reservoir with a fixed drainage area) is a given, as

[SEE FORMULA (17)]

where A' is a constant.

Taking derivative of Equation 17 gives

[SEE FORMULA (18)]

Thus, in a single-phase flow calculation, pseudosteady begins when slope, m, becomes 2p.

Some engineers assume that when m reaches within 10% of 2p value, pseudosteady begins. Others use 5% criteria and a few use 1% criterion.

Depending upon the criterion used, one can estimate different values for the beginning of pseudosteady state. Different criterion can give significantly different values for the beginning of pseudosteady state.7 8

At present, there is no consensus about the criterion, but most engineers accept tDA = 0.1 as a dimensionless time to start a pseudosteady state for a vertical well located centrally in either a circular or square drainage area.

Reference 1 does not include information about criterion that were used to calculate tDA = 0.1, probably because these results were obtained using a numerical simulator.

The results by Mutalik, et al.4 for calculation of pseudosteady state for horizontal wells are probably conservative because they used a slope requirement of 5% within the value of 2p.

The above discussion indicates that before using any dimensionless time to reach pseudosteady state, it is important to critically review the criterion that has been used. This is especially important in determining well spacing in leases that last only for a short time, say less than 10 years.

In these reservoirs, knowing the beginning of pseudosteady state becomes important to drain a reservoir effectively in a limited time period.

### Drainage

Due to longer well length, in a given time period under similar operating conditions, a horizontal well will drain a larger reservoir area than a vertical well. If a vertical well drains a certain reservoir volume of area in a given time, then this information can be used to calculate a horizontal well drainage area.

A horizontal well can be looked upon as a number of vertical wells drilled next to each other and completed in a limited pay-zone thickness. Then as shown in Fig. 1d, each end of a horizontal well would drain half a circular area, with a rectangular drainage area at the center.

This concept implicitly assumes that the reservoir thickness is considerably smaller than the sides of the drainage area. It is possible to calculate the drainage area of a horizontal well by assuming an elliptical drainage area in the horizontal plane, with each end of a well as a foci of drainage ellipse.

The methods to estimate drainage areas of horizontal wells generally give fairly similar results. As a rule of thumb, a 1,000 ft long horizontal well can drain twice the area of a vertical well, while a 2,000 ft long well will drain three times a vertical well, in a given time.

Thus, it is important to use larger well spacing for a horizontal well development than that used for a vertical well development.

The following examples for drainage area calculations are for reservoirs with uniform permeability in the areal plane. In a fractured reservoir, where permeability in one direction is higher than the other, then the well would accordingly drain a larger length in a high-permeability direction by a factor of ky/kx. The ky represents higher permeability and kx represents lower permeability in the horizontal plane (Fig. 1c).

### Horizontal wells needed

A 400-acre lease is to be developed using 10 vertical wells. An engineer suggested drilling either 1,000 or 2,000 ft long horizontal wells. Calculate the possible number of horizontal wells that will drain the lease effectively. Assume that a single vertical well effectively drains 40 acres.

If rev is a drainage radius of a vertical well, then a 40 acre vertical well drains an area of a circle = prev2 = 40 acres x 43,560 sq ft/acre, rev = 745 ft.

Two methods can be employed to calculate horizontal well drainage area on the basis of 40 acre drainage area of a vertical well.

In Method 1 (Fig. 1d), a 1,000 ft long well will drain 74 acres. The drainage area is presented as two half circles at each end and a rectangle in the center. Similarly (Fig. ld) a 2,000 ft long well will drain 108 acres.

In Method 2, if we assume that the horizontal well drainage area is an ellipse in a horizontal plane, then for a 1,000 ft long well:

a = half major axis of an ellipse = (L/2) + rev = (1,000/2) + 745 = 1,245 ft

b = half minor axis of an ellipse = rev - 745 ft

Drainage area = pab/43,560 = 67 acres

Similarly for a 2,000 ft long well, a = (L/2) +745, 1,745 ft

b = 745 ft, and drainage area = pab/43,560 = 94 acres.

The two methods give different answers for drainage area. If average areas are used the 1,000 ft well will drain 71 acres, and a 2,000 ft well will drain 101 acres. Thus, a 400-acre field can be drained by ten vertical wells, six 1,000 ft long wells, or four 2,000 ft long wells.

Horizontal wells are very appropriate for offshore and hostile environment applications where a substantial upfront savings can be obtained by drilling long horizontal wells. Because a large area can be drained with less wells, fewer slots are required on offshore platforms, and therefore, costs are significantly reduced.

### Alternative well lengths

A 600-acre lease is to be developed with ten vertical wells. Another alternative is to drill 500, 1,000, or 2,000 ft long horizontal wells. Table 3 shows the possible number of horizontal wells that will drain the leases effectively. A 60 acre vertical well would drain a circle of radius, rev, of 912 ft. Area of a circle = prev2 = 60 acres x 43,560 sq ft/acre, rve = 912 ft.

Again using the two methods described in the previous example, Table 3 shows that a 600-acre field can be effectively drained either by ten vertical wells, eight 500 ft long horizontal wells, six 1,000 ft long wells, or five 2,000 ft long wells.

### Development patterns

A 360 acre lease (Fig. 2) is to be developed using nine vertical wells. How many 1,000 ft long horizontal wells could drain this reservoir effectively? How many 2,000 ft long horizontal wells could drain this effectively? What is the suggested development pattern.

As shown in one of the previous examples, if a vertical well drains 40 acres effectively, 1,000 ft and 2,000 ft long horizontal wells would drain 80 and 120 acres, respectively.

With 1,000 ft long wells, the 360 acre lease could be developed using either four horizontal wells and one vertical well or three horizontal wells and three vertical wells. The possible configurations are shown in Fig. 2.

Because a 2,000 ft long horizontal well could drain 120 acres. A 360 acre lease also can be developed using three 2,000 ft long horizontal wells.

### Anisotropic development

A well, Harris-1, drains approximately 40 acres in a 35 ft thick naturally fractured reservoir.

Pressure tests conducted between the Harris-1 and the well to the east between Harris-1 and the well to the north, indicate permeability differences along the two directions. The permeability along the east-west, kx, is 0.5 md, while the permeability along the north-south direction, ky, is 4.5 md. An engineer proposed to drill a 2,000 ft long horizontal well along the east-west direction. Estimate the drainage area and dimensions of each drainage area side.

Let us assume that the vertical well, Harris-1, drains a rectangle area due to anisotropy.

If the reservoir has a uniform permeability, then the well would drain a 40 acre square with each side being

[SEE FORMULA]

The reservoir has nonuniform permeability in the areal plane with kx = 0.5 md and ky 4.5 md. Hence, ky/kx = 4.5/0.5 = 9 and ky/kx = 3

If the drainage rectangle has sides 2xe and 2ye, and if we assume that Harris-1 drains only 40 acres (Equations 15 and 16):

(2xe) x (2ye) = 40 x 43,560

additionally due to anisotropy, 2ye/2xe = 3.

Solving the above two equations simultaneously,

2xe = 762 ft and 2ye = 2,286 ft

Thus, for a vertical well, the drainage length along the north-south direction, which is a high permeability direction is 2,286 ft.

Hence, vertical well spacing along the north-south direction, 2ye, should be three times as large as along the east-west direction, 2xe.

Assuming that each well tip of a horizontal well drains half of a vertical well, for a 2,000 ft long horizontal well drilled along the east-west direction, the drainage length along this direction is 2xe = 2,000 + 762 = 2,762 ft.

Similarly, drainage length along the north-south direction will be the same as that for a vertical well which is 2ye = 2,286 ft.

Therefore, well spacing should be at least 2,286 ft along the north-south direction and the horizontal well tips should be spaced at least 762 ft apart.

Thus, well spacing requirements for vertical, as well as horizontal wells, are different in isotropic and anisotropic reservoirs.

### References

- Earlougher, R.C. Jr., "Advances in Well Test Analysis," Soc. of Patrol. Eng., 1977.
- Gringarten, A.C., Ramey, H.J., Jr., and Raghavan, R., "Unsteady-State Pressure Distribution Created by a Well with a Single Infinite-Conductivity Vertical Fracture," Soc. of Petrol. Eng. Journal, pp. 347-60, August 1974.
- Khan, A., "Pressure Behavior of a Vertically Fractured Well Located at the Center of a Rectangular Drainage Region," MS Thesis, The University of Tulsa, 1978.
- Mutalik, P.M., Godbole, S.P., and Joshi, S.D., "Effect of Drainage Area Shapes on the Productivity of Horizontal Wells," paper 18301, SPE 63rd Annual Technical Conference Houston, Oct. 2-5, 1988.
- Daviau, F., Mouronval, G., Bourdarot, G., and Crutcher, P., "Pressure Analysis for Horizontal Wells." SPE Formation Evaluation, December 1988, pp. 716-24.
- Goode, P.A., and Thambyanyagam, R.K.M., "Pressure Drawdown and Buildup Analysis of Horizontal Wells in Anisotropic Media," SPE Formation Evaluation, December 1987, pp. 683-97.
- Onur, M. and Reynolds, A.C., "A New Approach for Constructing Derivative Type Curves for Well Test Analysis," SPE Formation Evaluation, March 1988, pp. 197-206.
- Vongvuthipornchai, S., and Raghavan, R., "A Note on the Duration of the Transitional Period of Responses Influenced by Wellbore Storage and Skin," SPE Formation Evaluation, March 1988, pp. 207-14.
- Dake, L.P., "Fundamentals of Reservoir Engineering," Elsevier Scientific Publishing Co., New York, 1978.

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