Chapter 5 - Undersaturated oil reservoirs.pdf

Chapter 5
Undersaturated Oil Reservoirs
© 2022 Dr. Khalid Elwegaa

Introduction
Oil reservoir fluids are mainly complex of hydrocarbons compounds, which frequently contain impurities such as
nitrogen, carbon dioxide, and hydrogen sulfide. The composition in mole percentage of several typical reservoir
liquids is given in the following table (1) together with the tank gravity of the crude oil, the gas-oil ratio of the
reservoir mixture, and other characteristics of the fluids.
The composition of the tank oils obtained from the reservoir fluids are quite different from the composition of the
reservoir fluids, owing to:
• The release of the most of methane and ethane from solution.
• Vaporization of sizeable fractions of the propane, butane, and pentane as pressure is reduced in passing from the
reservoir to the stock oil.

Table (1): Reservoir Fluid Compositions and Properties

1. Calculations of Initial Oil in Place by Volumetric method and Estimation of Oil Recoveries
The volumetric method for estimating oil in place is based on log and core analysis data to determine the bulk
volume, the porosity, and the fluid saturations, and on fluid analysis to determine the oil formation volume factor.
• Under initial conditions:
One acre-foot of bulk oil productive rock contains
𝐼𝑛𝑡𝑒𝑟𝑠𝑡𝑖𝑡𝑖𝑎𝑙 𝑤𝑎𝑡𝑒𝑟 = 7758 × ∅ × 𝑆𝑤
𝑅𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟 𝑜𝑖𝑙 = 7758 × ∅ × (1 − 𝑆𝑤) 𝑏𝑏𝑙
𝑎𝑐. 𝑓𝑡
𝑆𝑡𝑜𝑐𝑘 𝑡𝑎𝑛𝑘 𝑜𝑖𝑙 (𝑁𝑖) =
7758 × ∅ × (1 − 𝑆𝑤)
𝐵𝑜𝑖
𝑆𝑇𝐵
𝑎𝑐. 𝑓𝑡

Example 1:
An undersaturated oil reservoir has the following data:
Average porosity is 20% and average connate water was 20%. Boi at the initial reservoir pressure was calculated
to be 1.24 bbl/stb. Calculate the initial oil in place per acre-foot?
Solution
𝑆𝑡𝑜𝑐𝑘 𝑡𝑎𝑛𝑘 𝑜𝑖𝑙 𝑁𝑖 =
7758 × 0.20 × 1 − 0.20
1.24
= 1000 𝑆𝑇𝐵
𝑎𝑐. 𝑓𝑡

For oil reservoirs under Volumetric control, there is no water influx to replace the produced oil, so it must be
replaced by the gas saturation of which increases as the oil saturation decreases. Sg is the gas saturation and Bo is
the oil volume factor at abandonment.
• At abandonment conditions:
One acre foot of bulk rock contains
𝐼𝑛𝑡𝑒𝑟𝑠𝑡𝑖𝑡𝑖𝑎𝑙 𝑤𝑎𝑡𝑒𝑟 = 7758 × ∅ × 𝑆𝑤
𝑅𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟 𝑔𝑎𝑠 = 7758 × ∅ × 𝑆𝑔
𝑅𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟 𝑜𝑖𝑙 = 7758 × ∅ × 1 − 𝑆𝑤 − 𝑆𝑔
𝑏𝑏𝑙
𝑎𝑐. 𝑓𝑡
𝑆𝑡𝑜𝑐𝑘 𝑡𝑎𝑛𝑘 𝑜𝑖𝑙 (𝑁𝑎) =
7758 × ∅ × (1 − 𝑆𝑤 − 𝑆𝑔)
𝐵𝑜
𝑆𝑇𝐵
𝑎𝑐. 𝑓𝑡

Oil recovery by volumetric control in stock tank barrels per acre foot is
𝑂𝑖𝑙 𝑟𝑒𝑐𝑜𝑣𝑒𝑟𝑦 = 7758 × ∅ ×
(1 − 𝑆𝑤)
𝐵𝑜𝑖
−
(1 − 𝑆𝑤 − 𝑆𝑔
𝐵𝑜
𝑆𝑇𝐵
𝑎𝑐. 𝑓𝑡
And the recovery factor is
𝑅𝑒𝑐𝑜𝑣𝑒𝑟𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 = 1 −
(1 − 𝑆𝑤 − 𝑆𝑔)
(1 − 𝑆𝑤)
×
𝐵𝑜𝑖
𝐵𝑜

For oil reservoirs under hydraulic control (active water drive), where there is no decline in reservoir pressure,
water influx either edge water drive or bottom water drive.
• At abandonment conditions:
The oil remaining is:
𝑅𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟 𝑜𝑖𝑙 = 7758 × ∅ × (𝑆𝑜𝑟) 𝑏𝑏𝑙
𝑎𝑐. 𝑓𝑡
𝑆𝑡𝑜𝑐𝑘 𝑡𝑎𝑛𝑘 𝑜𝑖𝑙 (𝑁𝑎) =
7758 × ∅ × 𝑆𝑜𝑟
𝐵𝑜𝑖
𝑆𝑇𝐵
𝑎𝑐. 𝑓𝑡
Where Sor is the residual oil saturation remaining after water displacement, since it was assumed that the
reservoir pressure was maintained at its initial value by water influx, no free gas saturation develops in the oil
zone and the oil volume factor at abandonment remains Boi.

Oil recovery by active water drive in stock tank barrels per acre foot is
𝑂𝑖𝑙 𝑟𝑒𝑐𝑜𝑣𝑒𝑟𝑦 = 7758 × ∅ ×
(1 − 𝑆𝑤 − 𝑆𝑜𝑟
𝐵𝑜𝑖
𝑆𝑇𝐵
𝑎𝑐. 𝑓𝑡
And the recovery factor is
𝑅𝑒𝑐𝑜𝑣𝑒𝑟𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 =
(1 − 𝑆𝑤 − 𝑆𝑜𝑟)
(1 − 𝑆𝑤)
• In statistical study of Craze and Buckley’s water-drive recovery data, Guthrie and Greenberger, using multiple
correlation analysis methods, found the following correlation between water-drive recovery and five variables
that affect recovery in sandstone reservoirs.
𝑅𝐹 = 0.114 + 0.272 𝐿𝑜𝑔 𝐾 + 0.256 𝑆𝑤 − 0.136 log 𝜇𝑜 − 1.538 ∅ − 0.00035 ℎ

Example 2:
An undersaturated oil reservoir has the following properties:
K = 1000 md, 𝑆𝑤 = 0.25, 𝜇𝑜= 2.0 cp, ∅ = 0.20, h = 10 ft.
Use the Guthrie and Greenberger correlation to calculate the oil recovery factor if the reservoir under water drive?
Solution
𝑅𝐹 = 0.114 + 0.272 𝐿𝑜𝑔 1000 + 0.256 × 0.25 − 0.136 log 2 − 1.538 × 0.20 − 0.00035 × 10 = 0.642
𝑅𝐹 = 64.2 %

2. Material Balance Equation (MBE) for undersaturated oil reservoirs
The MBE for undersaturated oil reservoir was developed before in chapter 2 which is as follows:
𝑁 𝐵𝑡 − 𝐵𝑡𝑖 + 𝑁𝐵𝑡𝑖
𝑐𝑤𝑆𝑤𝑖 + 𝑐𝑓
1 − 𝑆𝑤𝑖
∆𝑝 + 𝑊
𝑒 = 𝑁𝑝 𝐵𝑡 + 𝑅𝑝 − 𝑅𝑠𝑜𝑖 𝐵𝑔 + 𝐵𝑤𝑊
𝑝
• Neglecting the change in porosity of rocks with the change of internal fluid pressure, reservoirs with zero or
negligible water influx (𝑊
𝑒 = 0) are constant volume or volumetric reservoirs.
• Water production from volumetric reservoirs is generally small or negligible (𝑊
𝑝= 0).
Above the bubble point pressure (Bp):
From initial reservoir pressure down to the bubble point pressure, the reservoir oil volume remains constant and
oil is produced by liquid expansion. Only liquid will exist in the reservoir and any gas that is produced on the
surface will be gas coming out of solutions as oil moves up through the wellbore and through the surface
facilities.
𝑅𝑝 = 𝑅𝑠𝑜 𝑎𝑛𝑑 𝑅𝑠𝑜 = 𝑅𝑠𝑜𝑖

The MBE becomes:
𝑁 𝐵𝑡 − 𝐵𝑡𝑖 = 𝑁𝑝𝐵𝑡
The fractional recovery (RF) is:
𝑅𝐹 =
𝑁𝑝
𝑁
=
𝐵𝑡 − 𝐵𝑡𝑖
𝐵𝑡
Note:
Bt = Bo + Bg (Rsb − Rs )
And above BP, Bt = Bo

Below the bubble point pressure (Bp):
Below the Bp pressure, a free gas develops and for a volumetric undersaturated oil reservoir with no water
production, the hydrocarbon pore volume remains constant.
The figure below shows schematically the changes that occur between initial reservoir pressure and some pressure
below bubble point.
Diagram showing the formation of a free gas phase in a volumetric reservoir below the bubble point

The MBE for pressure below bubble point becomes:
𝑁 =
𝑁𝑝 𝐵𝑡 + 𝑅𝑝 − 𝑅𝑠𝑜𝑖 𝐵𝑔
And the fractional recovery is:
𝑅𝐹 =
𝑁𝑝
𝑁
=
𝐵𝑡 + 𝑅𝑝 − 𝑅𝑠𝑜𝑖 𝐵𝑔

Example 3:
Calculation to show the effect of the produced gas-oil ratio Rp on fractional recovery in volumetric undersaturated
reservoirs and estimation of the original oil in place? Given PVT data from 3-A-2 reservoir (figure 5-2)
Cumulative GOR at 2800 psia = 3300 SCF/STB,
Reservoir temperature = 190 °F = 650 °R,
Standard conditions = 14.7 psia and 60 °F
Solution:

• From figure 5-2:
Rsoi = 1100 SCF/STB, Boi = 1.572 bbl/STB, Rso at 2800 psia = 900 SCF/STB, Bo at 2800 psia = 1.52 bbl/STB
Rp = 3300 SCF/STB
• 𝐵𝑔 = 0.005035 × 0.87 ×
650
2800
= 0.00102 𝑏𝑏𝑙/𝑆𝐶𝐹
• 𝐵𝑡 = 1.52 + 0.00102 1100 − 900 = 1.724 𝑏𝑏𝑙/𝑆𝑇𝐵
1. Determine RF at 2800 psia when Rp = 3300 SCF/STB
𝑅𝐹 =
𝑁𝑝
𝑁
=
=
1.724 − 1.572
1.724 + 0.00102(3300 − 1100)
= 0.0383 = 3.83%
Bt = Bo + Bg (Rsb − Rs )

2. Determine RF at 2800 psia when Rp = 1100 SCF/STB
If two-third of the produced gas had been returned to the reservoir at the same pressure of 2800 psia:
𝑅𝐹 =
𝑁𝑝
𝑁
=
=
1.724 − 1.572
1.724 + 0.00102(1100 − 1100)
= 0.088 = 8.8%
3. Determine Ni if Np = 1.486 MM STB at Rp = 3300 SCF/STB
𝑁 =
𝑁𝑝 𝐵𝑡 + 𝑅𝑝 − 𝑅𝑠𝑜𝑖 𝐵𝑔
=
1.486 × 106
1.724 + 0.00102 3300 − 1100
1.724 − 1.572
= 38.8 𝑀𝑀𝑆𝑇𝐵

3. Material Balance Equation (MBE) for undersaturated oil reservoirs including formation and water
compressibilties:
The effect of compressibilties above the bubble point on calculations for N are examined before using the
following equation with 𝑅𝑝 − 𝑅𝑠𝑜𝑖
𝑁 𝐵𝑡 − 𝐵𝑡𝑖 + 𝑁𝐵𝑡𝑖
1 − 𝑆𝑤𝑖
∆𝑝 + 𝑊
𝑒 = 𝑁𝑝𝐵𝑡 + 𝐵𝑤𝑊
𝑝
This equation is rearranged to solve for N
𝑁 =
𝑁𝑝𝐵𝑡 − 𝑊
𝑒 + 𝐵𝑤𝑊
𝑝
𝐵𝑡 − 𝐵𝑡𝑖 + 𝐵𝑡𝑖
1 − 𝑆𝑤𝑖
∆𝑝

An oil compressibility , co is introduced with the following relationship:
𝑐𝑜 =
𝑣𝑜 − 𝑣𝑜𝑖
𝑣𝑜𝑖 𝑃𝑖 − 𝑃
=
𝐵𝑜 − 𝐵𝑜𝑖
𝐵𝑜𝑖∆𝑃
Or 𝐵𝑜 = 𝐵𝑜𝑖 + 𝐵𝑜𝑖𝑐𝑜∆𝑃
The definition of co uses the single-phase formation volume factor, but it should be apparent that as long as the
calculations are being conducted above the bubble point, 𝐵𝑜 = 𝐵𝑡
Now Substitute the Bo equation in the MBE results in the following equation:
𝑁 𝐵𝑡𝑖 − 𝐵𝑡𝑖 + 𝑁𝐵𝑡𝑖𝑐𝑜∆𝑃 + 𝑁𝐵𝑡𝑖
1 − 𝑆𝑤𝑖
∆𝑝 + 𝑊
𝑝

Multiplying both the numerator and the denominator of the term containing 𝑐𝑜 by 𝑆𝑜 and realizing that above the
bubble point there is no gas saturation, 𝑆𝑜 = 1 − 𝑆𝑤𝑖
𝑁𝐵𝑡𝑖
𝑐𝑜𝑆𝑜∆𝑃
1 − 𝑆𝑤𝑖
+ 𝑁𝐵𝑡𝑖
1 − 𝑆𝑤𝑖
∆𝑝 + 𝑊
𝑝
Or
𝑁𝐵𝑡𝑖
𝑐𝑜𝑆𝑜 + 𝑐𝑤𝑆𝑤𝑖 + 𝑐𝑓
1 − 𝑆𝑤𝑖
∆𝑃 = 𝑁𝑝𝐵𝑡 − 𝑊
𝑒 + 𝐵𝑤𝑊
𝑝
The expression in brackets is called the effective fluid compressibility, 𝑐𝑒 which include the compressibilties of the oil, the
connate water and the formation
𝑐𝑒 =
1 − 𝑆𝑤𝑖

Substituting 𝑐𝑒 in MBE becomes
𝑁𝐵𝑡𝑖𝑐𝑒∆𝑃 = 𝑁𝑝𝐵𝑡 − 𝑊
𝑒 + 𝐵𝑤𝑊
𝑝
And for volumetric reservoirs, We = 0, and Wp is generally negligible, So the MBE will be as follows:
𝑁 =
𝑁𝑝
𝑐𝑒∆𝑃
𝐵𝑡
𝐵𝑡𝑖

Example 4:
Calculations of initial oil in place in volumetric, under-saturated reservoir.
Given:
Bti = 1.35469 bbl/STB
Bt at 3600 psig = 1.375 bbl/STB
Connate water = 0.20
Cw = 3.6*10-6 psi-1
Bw at 3600 psig = 1.04 bbl/STB
Cf = 5.0*10-6 psi-1
Pi = 5000 psig
Np = 1.25 MMSTB
∆𝑝 at 3600 psig = 1400 psi
Wp = 32,000 STB
We = 0

1. Using the following equation to calculate the OOIP without compressibility of reservoir oil:
𝑁 =
𝑁𝑝𝐵𝑡 − 𝑊
𝑒 + 𝐵𝑤𝑊
𝑝
𝐵𝑡 − 𝐵𝑡𝑖 + 𝐵𝑡𝑖
1 − 𝑆𝑤𝑖
∆𝑝
𝑁 =
1,250000 × 1.375 + 32,000 × 1.04
1.375 − 1.35469 + 1.35469
3.6 × 10−6 × 0.20 + 5 × 10−6
1 − 0.2
× 1400
= 51.73 𝑀𝑀𝑆𝑇𝐵
• The average compressibility of reservoir oil is:
𝑐𝑜 =
𝐵𝑜 − 𝐵𝑜𝑖
𝐵𝑜𝑖∆𝑃
=
1.375 − 1.35469
1.35469(5000 − 3600)
= 10.71 × 10−6𝑝𝑠𝑖−1

• Calculating the effective fluid compressibility:
𝑐𝑒 =
1 − 𝑆𝑤𝑖
=
0.8 × 10.71 + 0.2 × 3.6 + 5.0 10−6
0.8
= 17.86 × 10−6𝑝𝑠𝑖−1
2. Calculating OOIP including the compressibility of reservoir oil:
𝑁 =
𝑁𝑝 × 𝐵𝑡 + 𝐵𝑤𝑊
𝑝
𝑐𝑒∆𝑃𝐵𝑡𝑖
=
1,250,000 × 1.375 + 32,000 × 1.04
17.86 × 10−6 × 1400 × 1.35469
= 51.73 𝑀𝑀𝑆𝑇𝐵
3. Calculating OOIP if the water and formation compressibilties are neglected (𝑐𝑒 = 𝑐𝑜)
𝑁 =
𝑁𝑝 × 𝐵𝑡 + 𝐵𝑤𝑊
𝑝
𝑐𝑜∆𝑃𝐵𝑡𝑖
=
1,250,000 × 1.375 + 32,000 × 1.04
10.71 × 10−6 × 1400 × 1.35469
= 86.25 𝑀𝑀𝑆𝑇𝐵

• Note:
As can be seen from the example calculations. The inclusion of the compressibilties terms significantly affects the
value of N (N increased from 51.73 𝑀𝑀𝑆𝑇𝐵 to 86.25 𝑀𝑀𝑆𝑇𝐵)
This is true above the bubble point where the oil producing mechanism is depletion, or the swelling of reservoir
fluids.
After the bubble point is reached, the compressibility have a much smaller effect on the calculations.

Chapter 5 - Undersaturated oil reservoirs.pdf

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