Hydrocarbon Reserves Volumetric Method

Prof Attia M. Attia

Contents

Calculating oil and gas in place by Volumetric method

Calculation Unit Recovery from Volumetric Gas Reservoirs

Calculation of Unit Recovery from Gas Reservoirs under Water Drive

Recovery factor

Example

Petroleum Reservoir Engineering (Book) (Craft &Hawkins)

Volumetric Method

The volumetric method is probably the easiest method one can use to estimate the reserves.

This method requires a limited amount of information, and can be used even in the absence of the actual drilling of a well. Obviously, if data can be collected from a well, the volumetric estimates will be subject to much less uncertainty than if no well data were available.

In the absence of a drilled well, most of the parameters are estimated by analogy, i.e., data based on geological and geophysical inferences based on nearby wells.

Volumetric Estimate of Oil Reserves

Original oil in place (OOIP)

Original oil reserves = Npa = N x Rf

Original solution gas in place = N x Rsi

7758 (1 ) /wi oiN A H S B

where Npa is the recoverable standard barrels of oil from the reservoir,

Ah is the reservoir volume (typically product of area times thickness) in acre-foot,

is the fractional porosity of the reservoir,

Swi is the fractional residual water saturation,

Boi is the formation volume factor in bbl/STB, and

RF is the recovery factor which is a fraction.

Estimating Reservoir Bulk Volume

Reservoir volume is very difficult to determine and can be subject to significant errors. If the well is to be drilled on a required spacing, an obvious choice of drainage area is the spacing of a unit. This may not be true, but is often used as an approximation for defining the area.

In addition to area, we also need the payzonethickness, h, to determine the reservoir volume. If a well has already been drilled, the payzone can be defined based on the log data. In the absence of information from a drilled well, information can be inferred from nearby wells by creating contour maps with respect to payzone thickness

• Another method one can use to estimate the reservoir volume is by creating isopach maps. These maps are the maps drawn at constant depth of productive zones. The map shows the area of oil bearing formation. These maps are typically drawn at constant intervals.

• To calculate the reservoir volume, we can use any of the standard numerical methods. For example, if we use a trapezoidal rule, we can calculate the volume, Ah, as,

Ah = 12h A1 + 2A2 + .... + 2An – 1 + An + tavgAn

• where is the isopach map interval, A1 is the area of deepest oil bearing zone, An is the area of the shallowest oil bearing zone, and tavg is the average thickness above the maximum thickness isopach line. Another commonly applied numerical method is the pyramidal rule.

• Here, the volume between two successive isopach contours is given by

Ahi= 1

3h Ai + Ai + 1 + AiAi + 1 .

In this case, we estimate the volume of the reservoir by summing the volumes between all successive isopach contours.

Sometimes the contour maps are not necessarily circular and can exhibit twisted patterns.

A device called planimeter can be used to measure the area at each contour level.

In modern times, computers are used to calculate the areas, and hence volume, after drawing the contour maps.

Example

The following data are read from a contour map. It shows area as a function of the depth of a production zone. Calculate the reservoir volume using the trapezoidal rule.

Depth, Feet Area, Acres

2,900 20

2,925 35

2,950 50

2,975 70

3,000 95

3,025 110

3,050 130

Average thickness above 2900 ft = 0

Solution

• The interval over which area maps are drawn is 25 feet, i.e., h = 25 feet. Using a trapezoidal rule

Ah = 12h A1 + 2A2 + .... + 2An – 1 + An + tavg An

= 1225 20 + 2 35 + 2 50 + 2 70 +2 95 +2 110 + 130

= 10875acre–ft

Example

Geological contouring of a yet to be explored region indicates that the drainage area of a potential hydrocarbon reserves is 150 acres and the payzone is expected to be 45 feet. The porosity of nearby regions in the same zone is 23% and the water saturation is 20%. If the formation volume factor is assumed to be 1.2 bbl/STB and the recovery factor is estimated to be 0.19, calculate the potential reserves from this project.

Solution

NR =7758 Ah 1 – Swi

Boi

RF

=7758 150 45 0.23 1 – 0.2

1.20.19

= 1,525,611 STB

Potential reserves are 1,525,611 standard barrels

Gas Reservoirs

• Gas reservoirs are hydrocarbon reservoirs that contain dry gas (i.e., the methane mole fraction is greater than 95%).

• Behavior of these reservoirs is governed by the gas equation of state and the material balance equation.

• Three quantities—pressure, volume, and temperature—define the state of a gas. As we mentioned, in most hydrocarbon reservoirs the temperature is considered to be constant.

Volumetric Estimate of Gas Reserves

Original gas in place (OGIP)

Original gas reserves = OGIP x R.F.

giwi B/)S1(Ah7758OGIP

Calculation Unit Recovery from Volumetric Gas Reservoirs

One Acre Foot of Gas Reservoir at Initial Conditions Contains:

Initial gas-in-place per unit volume:

G = 43,560 ( 1 - Swi ) / Bgi

Rock Volume = 43,560 ( 1 - )

Water Volume = 43,560 Swi

Gas Volume = 43,560 ( 1 - Swi )

Unit Recovery from Volumetric Gas Reservoirs (continued)

Gas-in-place per unit volume at abandonment:

Ga = 43,560 ( 1 - Swi ) / Bga

Unit Recovery G - Ga = Gr

= 43,560 ( 1 - Swi ) / Bgi

- 43,560 ( 1 - Swi ) / Bga

= 43,560 ( 1 - Swi ) ( 1/ Bgi - 1/ Bga)

Unit Recovery from Volumetric Gas Reservoirs (continued)

Recovery Factor Gr / G

43,560 ( 1 - Swi ) ( 1/ Bgi - 1/ Bga)

43,560 ( 1 - Swi ) / Bgi

=

= 1 - Bgi / Bga

Calculation of Unit Recovery from Gas Reservoirs under Water Drive

Initial gas-in-place per unit volume:

G = 43,560 ( 1 - Swi ) / Bgi

Gas-in-place per unit volume at abandonment:

Ga = 43,560 Sgr / BgaNote that this assumes the entire volume has been swept with

water and that the term Sgr is the residual gas saturation.Sgr is generally accepted to not change with pressure.

Connate water per unit volume:

43,560 ( Swi ) / Bgi

Unit Recovery from Water Drive Gas Reservoirs (continued)

Recovery Factor Gr / G

1 - Swi Sgr

Bgi Bga

1 - Swi

Bgi

-

=

Example

Calculate the initial gas reserve of a 160-acre unit of Xgas Field by volumetric depletion under partial and complete water drive

Given : Area=160 acre h=40 ftAverage porosity= 22% - Connate water=0.23Residual gas saturation afetr water displacement

= 0.34Big @Pi 3250 psia = 188.0 SCF/CuftBg @P 2500 psia = 150.0 SCF/CuftBgi @P 500 psia = 27.6 SCF/Cuft

Solution

Pore volume

= 43560x0.22x160x40= 61.33x10^6 cuft

Initial gas in place

G1 = 61.33x10^6x(1-0.23)x188=8878MMSCF

Gas in place after volumetric depletion to 2500 psia

G2 =61.33x10^6x(1-0.23)x150=7084MMSCF

Gas in place after volumetric depletion to 500 psia

G3 =61.33x10^6x(1-0.23)x27.6=1303MMSCF

Gas in place after water invasion at 3250psia

G4 =61.33x10^6x0.34x188=3920MMSCF

Gas in place after water invasion at 2500psia

G5 =61.33x10^6x0.34x150=3128MMSCF

Initial reserve by depletion to 500 psia=

G1-G5= (8878-1303)x10^6= 7575 MMSCF

Initial reserve by water drive at 3250 psia

G1-G4= (8878-3920)x10^6= 4958 MMSCF

Initial reserve by water drive at 2500 psia=

(G1-G2)+(G2-G5)=(G1-G5)=

(8878- 3128)x10^6= 5650 MMSCF

Summary

OOIP = 7758 Ah (1 - Swi)/Boi

Original solution gas in place = N x Rsi

Oil reserves = OOIP x R.F.

OGIP = 7758 Ah (1 - Swi)/Bgi

Gas reserves = OGIP x R.F.

Remaining reserves = Reserves at original conditions - cumulative production