material balance calculations for petroleum reservoirs · 2018-03-17 · 5 3. oil recovery by...

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MATERIAL BALANCE CALCULATIONS FOR

PETROLEUM RESERVOIRS

In 1936, Schilthuis developed a general material balance equation

that can be applied to all reservoir types. Although it is a tank

model equation, it can provide great insight for the practicing

reservoir engineer.

Hassan S. Naji,

Professor,

[email protected]

1

Consider the sketch below:

The MB is written from start of production to any time (t) as follows:

Expansion of oil in the oil zone + Expansion of gas in the gas zone +

Expansion of connate water in both zones +

Compaction of pore volume in both zones +

Water influx + Water injected + Gas injected =

Oil produced + Gas produced + Water produced

(1)

𝑁(𝐵𝑡 − 𝐵𝑡𝑖) + 𝐺(𝐵𝑔 − 𝐵𝑔𝑖) + (𝑁𝐵𝑡𝑖 + 𝐺𝐵𝑔𝑖) (𝑆𝑤𝑐𝑐𝑤1 − 𝑆𝑤𝑐

)∆𝑝𝑡

+ (𝑁𝐵𝑡𝑖 + 𝐺𝐵𝑔𝑖) (𝑐𝑓

1 − 𝑆𝑤𝑐)∆𝑝𝑡 +𝑊𝑒 +𝑊𝐼𝐵𝐼𝑤 + 𝐺𝐼𝐵𝐼𝑔

= 𝑁𝑝𝐵𝑡 + (𝐺𝑝 − 𝑁𝑝𝑅𝑠𝑜𝑖)𝐵𝑔 +𝑊𝑝𝐵𝑤= 𝑁𝑝𝐵𝑡 + (𝑁𝑝𝑅𝑝 − 𝑁𝑝𝑅𝑠𝑜𝑖)𝐵𝑔 +𝑊𝑝𝐵𝑤= 𝑁𝑝[𝐵𝑡 + (𝑅𝑝 − 𝑅𝑠𝑜𝑖)𝐵𝑔] +𝑊𝑝𝐵𝑤

(2)

Rearranging (2) yields:

𝑁(𝐵𝑡 − 𝐵𝑡𝑖) + 𝐺(𝐵𝑔 − 𝐵𝑔𝑖) + (𝑁𝐵𝑡𝑖 + 𝐺𝐵𝑔𝑖) (𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓

1 − 𝑆𝑤𝑐) ∆𝑝𝑡 +𝑊𝑒

+𝑊𝐼𝐵𝐼𝑤 + 𝐺𝐼𝐵𝐼𝑔 = 𝑁𝑝[𝐵𝑡 + (𝑅𝑝 − 𝑅𝑠𝑜𝑖)𝐵𝑔] +𝑊𝑝𝐵𝑤

(3)

Where:

𝑁 original oil in place OOIP, STB

𝑁𝑝 cumulative oil produced, STB

𝐺 original gas in place OGIP, SCF

𝐺𝐼 cumulative gas injected into reservoir, SCF

𝐺𝑝 cumulative gas produced, = 𝑁𝑝𝑅𝑝, SCF

𝑊𝑒 water influx into reservoir, bbl

𝑊𝐼 cumulative water injected into reservoir, STB

𝑊𝑝 cumulative water produced, STB

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2

𝐵𝑡𝑖 initial two-phase formation volume factor = 𝐵𝑜𝑖, bbl/STB

𝐵𝑜𝑖 initial oil formation volume factor, bbl/STB

𝐵𝑔𝑖 initial gas formation volume factor, bbl/SCF

𝐵𝑡 two-phase formation volume factor = 𝐵𝑜 + 𝐵𝑔(𝑅𝑠𝑜𝑖 − 𝑅𝑠𝑜), bbl/STB

𝐵𝑜 oil formation volume factor, bbl/STB

𝐵𝑔 gas formation volume factor, bbl/SCF

𝐵𝑤 water formation volume factor, bbl/STB

𝐵𝐼𝑔 injected gas formation volume factor, bbl/SCF

𝐵𝐼𝑤 injected water formation volume factor, bbl/STB

𝑅𝑠𝑜𝑖 initial solution gas-oil ratio, SCF/STB

𝑅𝑠𝑜 current solution gas-oil ratio, SCF/STB

𝑅𝑝 cumulative produced gas-oil ratio, SCF/STB

𝑐𝑓 formation compressibility, psia-1

𝑐𝑤 water isothermal compressibility, psia-1

𝑆𝑤𝑐 connate water saturation

∆𝑝𝑡 reservoir pressure drop = 𝑝𝑖 − 𝑝(𝑡), psia

𝑝𝑖 initial reservoir pressure, psia

𝑝(𝑡) current reservoir pressure, psia

1. Straight Line Form of the Material Balance Equation

Normally, when using the material balance equation, each pressure and the

corresponding production and PVT data are considered as being a separate point from

other pressure values. From each separate point, a calculation is made and the results of

these calculations are averaged. However, a method is required to make use of all data

points with the requirement that these points must yield solutions to the material balance

equation that behave linearly to obtain values of the independent variable. The straight-

line method begins with the material balance, equation (3), written as:

𝑁(𝐵𝑡 − 𝐵𝑡𝑖) + 𝐺(𝐵𝑔 − 𝐵𝑔𝑖) + (𝑁𝐵𝑡𝑖 + 𝐺𝐵𝑔𝑖) (𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓

1 − 𝑆𝑤𝑐)∆𝑝𝑡 +𝑊𝑒

+𝑊𝐼𝐵𝐼𝑤 + 𝐺𝐼𝐵𝐼𝑔 = 𝑁𝑝[𝐵𝑡 + (𝑅𝑝 − 𝑅𝑠𝑜𝑖)𝐵𝑔] +𝑊𝑝𝐵𝑤

(4)

Defining the ratio of the initial gas cap volume to the initial oil volume as:

𝑚 =𝐺𝐵𝑔𝑖

𝑁𝐵𝑡𝑖 ⇒ 𝐺𝐵𝑔𝑖 = 𝑁𝑚𝐵𝑡𝑖 ⇒ 𝐺 = 𝑁𝑚

𝐵𝑡𝑖𝐵𝑔𝑖

(5)

Replacing 𝐺 and 𝐺𝐵𝑔𝑖 into equation (4) and taking 𝑁 and 𝑁𝑝 as common yields:

𝑁 [(𝐵𝑡 − 𝐵𝑡𝑖) + 𝑚𝐵𝑡𝑖𝐵𝑔𝑖

(𝐵𝑔 − 𝐵𝑔𝑖) + 𝐵𝑡𝑖(1 + 𝑚) (𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓

1 − 𝑆𝑤𝑐)∆𝑝𝑡] +𝑊𝑒

= 𝑁𝑝[𝐵𝑡 + (𝑅𝑝 − 𝑅𝑠𝑜𝑖)𝐵𝑔] +𝑊𝑝𝐵𝑤 −𝑊𝐼𝐵𝐼𝑤 − 𝐺𝐼𝐵𝐼𝑔

(6)

3

Let us define the following cumulative expansions:

𝐸𝑜 = cumulative oil expansion = (𝐵𝑡 − 𝐵𝑡𝑖) (7)

𝐸𝑔 = cumulative gas expansion =𝐵𝑡𝑖𝐵𝑔𝑖

(𝐵𝑔 − 𝐵𝑔𝑖) (8)

𝐸𝑓,𝑤 = cumulative connate water expansion and formation compaction

= 𝐵𝑡𝑖 (𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓

1 − 𝑆𝑤𝑐)∆𝑝𝑡

(9)

𝐹 = cumulative reservoir voidage= 𝑁𝑝[𝐵𝑡 + (𝑅𝑝 − 𝑅𝑠𝑜𝑖)𝐵𝑔] +𝑊𝑝𝐵𝑤 −𝑊𝐼𝐵𝐼𝑤 − 𝐺𝐼𝐵𝐼𝑔

(10)

Thus, we rewrite (6) as follows:

𝑁[𝐸𝑜 +𝑚𝐸𝑔 + (1 +𝑚)𝐸𝑓,𝑤] +𝑊𝑒 = 𝐹 (11)

Let us define the cumulative total expansion as:

𝐸𝑡 = 𝐸𝑜 +𝑚𝐸𝑔 + (1 +𝑚)𝐸𝑓,𝑤 (12)

Plugging (12) into (11) yields:

𝑁𝐸𝑡 +𝑊𝑒 = 𝐹 (13)

Dividing (13) all through by 𝐸𝑡, we get:

𝑁 +𝑊𝑒

𝐸𝑡=

𝐹

𝐸𝑡 (14)

Which is written as 𝑏 + 𝑥 = 𝑦. This would suggest that a plot of 𝐹

𝐸𝑡 as the y coordinate

and 𝑊𝑒

𝐸𝑡 as the x coordinate would yield a straight line with slope equal to 1 and intercept

equal to 𝑁.

4

2. Drive Indexes from the Material Balance Equation

The three major driving mechanisms are:

Depletion Drive (expansion of oil in the oil zone),

Segregation Drive (expansion of gas in the gas zone), and

Water Drive (water influx in the water zone).

To determine the relative magnitude of each of these driving mechanisms, the

compressibility term in the material balance equation is neglected and the equation is

rearranged as follows:

𝑁(𝐵𝑡 − 𝐵𝑡𝑖) + 𝐺(𝐵𝑔 − 𝐵𝑔𝑖) + (𝑊𝑒 −𝑊𝑝𝐵𝑤) = 𝑁𝑝[𝐵𝑡 + (𝑅𝑝 − 𝑅𝑠𝑜𝑖)𝐵𝑔] (15)

Dividing all through by the right hand side of the equation yields:

𝑁(𝐵𝑡 − 𝐵𝑡𝑖)

𝑁𝑝[𝐵𝑡 + (𝑅𝑝 − 𝑅𝑠𝑜𝑖)𝐵𝑔]+

𝐺(𝐵𝑔 − 𝐵𝑔𝑖)

𝑁𝑝[𝐵𝑡 + (𝑅𝑝 − 𝑅𝑠𝑜𝑖)𝐵𝑔]

+(𝑊𝑒 −𝑊𝑝𝐵𝑤)

𝑁𝑝[𝐵𝑡 + (𝑅𝑝 − 𝑅𝑠𝑜𝑖)𝐵𝑔]= 1

(16)

The terms on the left hand side of the above equation represent the depletion drive index

(DDI), the segregation drive index (SDI), and the water drive index (WDI) respectively.

Thus, using Pirson's abbreviations, we write:

DDI + SDI + WDI = 1

𝐹𝐸𝑡

𝑊𝑒𝐸𝑡

Aquifer too small

Aquifer too big

5

3. Oil Recovery by Internal Gas Drive

Oil is produced from volumetric, undersaturated reservoirs by the expansion of reservoir

fluids. Down to the bubble-point pressure, the production is caused by liquid (oil and

connate water) expansion and rock compressibility. Below the bubble-point pressure, the

expansion of the connate water and the rock compressibility are negligible, and as the oil

phase contracts owing to release of gas from solution, production is a result of expansion

of the gas phase. As production proceeds, pressure drops and the gas and oil viscosities

and volume factors continually change. Tracy method for predicting the performance of

internal gas drive reservoir uses the material balance equation for a volumetric

undersaturated oil reservoir, which is written as:

𝑁(𝐵𝑡 − 𝐵𝑡𝑖) + 𝐺(𝐵𝑔 − 𝐵𝑔𝑖) +𝑊𝑒 = 𝑁𝑝𝐵𝑡 + 𝑁𝑝(𝑅𝑝 − 𝑅𝑠𝑜𝑖)𝐵𝑔 +𝑊𝑝𝐵𝑤 (17)

Since:

𝐺 = 𝑁𝑚𝐵𝑡𝑖𝐵𝑔𝑖

𝐺𝑝 = 𝑁𝑝𝑅𝑝

𝐵𝑡𝑖 = 𝐵𝑜𝑖

𝐵𝑡 = 𝐵𝑜 + 𝐵𝑔(𝑅𝑠𝑜𝑖 − 𝑅𝑠𝑜)

(18)

Plugging into the equation, rearranging, and solving for N yields:

𝑁 =𝑁𝑝[𝐵𝑜 − 𝐵𝑔𝑅𝑠𝑜] + 𝐺𝑝𝐵𝑔 + (𝑊𝑝𝐵𝑤 −𝑊𝑒)

(𝐵𝑜 − 𝐵𝑜𝑖) + 𝐵𝑔(𝑅𝑠𝑜𝑖 − 𝑅𝑠𝑜) + 𝑚𝐵𝑜𝑖𝐵𝑔𝑖

(𝐵𝑔 − 𝐵𝑔𝑖) (19)

6

Let:

Φ𝑁 =𝐵𝑜 − 𝐵𝑔𝑅𝑠𝑜


(𝐵𝑔 − 𝐵𝑔𝑖)

(20)

Φ𝐺 =𝐵𝑔



(21)

Φ𝑊 =1



(22)

Plugging into the above equation yields:

𝑁 = 𝑁𝑝Φ𝑁 + 𝐺𝑝Φ𝐺 + (𝑊𝑝𝐵𝑤 −𝑊𝑒)Φ𝑊 (23)

If there is no water influx or water production, the equation is written as:

𝑁 = 𝑁𝑝Φ𝑁 + 𝐺𝑝Φ𝐺 (24)

Φ factors can be calculated at all desired pressures using data from a reservoir fluid

analysis. Then a table or plot of these factors can be used to calculate oil in place or

predict future performance. The process starts as follows:

At time level j, the above equation is written as:

𝑁 = (𝑁𝑝𝑗−1

+ ∆𝑁𝑝)Φ𝑁𝑗+ (𝐺𝑝

𝑗−1+ ∆𝐺𝑝)Φ𝐺

𝑗 (25)

Since:

∆𝐺𝑝 = ∆(𝑁𝑝𝑅𝑝) = 𝑅𝑝𝑎𝑣𝑔

∆𝑁𝑝 = [𝑅𝑝𝑗−1

+ 𝑅𝑝𝑗

2] ∆𝑁𝑝 (26)

Expanding and rearranging yields:

𝑁 = 𝑁𝑝𝑗−1

Φ𝑁𝑗+ ∆𝑁𝑝Φ𝑁

𝑗+ 𝐺𝑝

𝑗−1Φ𝐺

𝑗+ 𝑅𝑝

𝑎𝑣𝑔∆𝑁𝑝Φ𝐺

𝑗 (27)

𝑁 = (𝑁𝑝𝑗−1

Φ𝑁𝑗+ 𝐺𝑝

𝑗−1Φ𝐺

𝑗) + (Φ𝑁

𝑗+ 𝑅𝑝

𝑎𝑣𝑔Φ𝐺

𝑗)∆𝑁𝑝 (28)

Rearranging and solving for ∆𝑁𝑝, we get:

7

∆𝑁𝑝 =𝑁 − 𝑁𝑝

𝑗−1Φ𝑁

𝑗− 𝐺𝑝

𝑗−1Φ𝐺

𝑗

Φ𝑁𝑗+ 𝑅𝑝

𝑎𝑣𝑔Φ𝐺

𝑗 (29)

Where 𝑁𝑝𝑗−1

, 𝐺𝑝𝑗−1

are the cumulative oil and gas production at the old time level 𝑗 − 1

respectively. The Tarner method for predicting reservoir performance by internal gas

drive will be presented in a form proposed by Tracy as follows:

Calculate Φ𝑁 and Φ𝐺 at 𝑝 = 𝑝𝑖 − ∆𝑝 using:

Φ𝑁 =𝐵𝑜 − 𝐵𝑔𝑅𝑠𝑜



(30)

Φ𝐺 =𝐵𝑔



(31)

Assume 𝑅𝑝𝑗 = 𝑅𝑠𝑜

𝑗

Calculate 𝑅𝑝𝑎𝑣𝑔

using:

𝑅𝑝𝑎𝑣𝑔

= [𝑅𝑝𝑗−1

+ 𝑅𝑝𝑗

2] (32)

Calculate ∆𝑁𝑝 using:

∆𝑁𝑝 =𝑁 − 𝑁𝑝

𝑗−1Φ𝑁

𝑗− 𝐺𝑝

𝑗−1Φ𝐺

𝑗

Φ𝑁𝑗+ 𝑅𝑝

𝑎𝑣𝑔Φ𝐺

𝑗 (33)

Calculate:

𝑁𝑝 = 𝑁𝑝𝑗−1

+ ∆𝑁𝑝 (34)

Calculate 𝑆𝑜 and 𝑆𝐿 as follows:

𝑆𝑜 = (1 − 𝑆𝑤) (1 −𝑁𝑝

𝑁)𝐵𝑜𝐵𝑜𝑖

(35)

𝑆𝐿 = 𝑆𝑤 + 𝑆𝑜 (36)

Evaluate 𝑘𝑜 at 𝑆𝑤 and 𝑘𝑔 at 𝑆𝐿

8

Calculate 𝑅𝑝𝑗 using the following equation:

𝑅𝑝𝑗= 𝑅𝑠𝑜 +

𝐵𝑜𝐵𝑔

𝜆𝑔

𝜆𝑜= 𝑅𝑠𝑜 +

𝐵𝑜𝐵𝑔

𝑘𝑔 𝜇𝑔⁄

𝑘𝑜 𝜇𝑜⁄= 𝑅𝑠𝑜 +

𝐵𝑜𝐵𝑔

𝑘𝑔

𝑘𝑜

𝜇𝑜𝜇𝑔

(37)

Calculate the difference between the assumed and the calculated 𝑅𝑝𝑗 value. If these two

values agree within some tolerance, then the calculated ∆𝑁𝑝 is correct, else, the

calculated value should be used as the new guess and steps 3 through 9 are repeated.

4. Oil Reservoirs Above Bubble-Point Pressure

The material balance equation for oil reservoirs above the bubble-point pressure is

written as:

𝑁(𝐵𝑜 − 𝐵𝑜𝑖) + 𝑁𝐵𝑜𝑖 (𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓

1 − 𝑆𝑤𝑐)∆𝑝𝑡 +𝑊𝑒 = 𝑁𝑝𝐵𝑜 +𝑊𝑝𝐵𝑤

= 𝑁𝐵𝑜 − 𝑁𝐵𝑜𝑖 + 𝑁𝐵𝑜𝑖 (𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓


= 𝑁𝑝𝐵𝑜 +𝑊𝑝𝐵𝑤 = 𝑁𝐵𝑜𝑖 [𝐵𝑜𝐵𝑜𝑖

− 1 + (𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓


= 𝑁𝑝𝐵𝑜 +𝑊𝑝𝐵𝑤

(38)

Since:

𝐵𝑜𝐵𝑜𝑖

= 1 + 𝑐𝑜∆𝑝𝑡

(39)


𝑁𝐵𝑜𝑖 (𝑐𝑜∆𝑝𝑡 + (𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓

1 − 𝑆𝑤𝑐) ∆𝑝𝑡) +𝑊𝑒 = 𝑁𝑝𝐵𝑜 +𝑊𝑝𝐵𝑤

= 𝑁𝐵𝑜𝑖 ((1 − 𝑆𝑤𝑐)𝑐𝑜 + 𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓


= 𝑁𝑝𝐵𝑜 +𝑊𝑝𝐵𝑤 −𝑊𝑒 = 𝑁𝐵𝑜𝑖 (𝑆𝑜𝑖𝑐𝑜 + 𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓


= 𝑁𝑝𝐵𝑜 +𝑊𝑝𝐵𝑤 −𝑊𝑒 = 𝑁𝐵𝑜𝑖 (𝑐𝑡


= 𝑁𝑝𝐵𝑜 +𝑊𝑝𝐵𝑤 −𝑊𝑒

(40)

Let:

𝑐𝑒 =𝑐𝑡

1 − 𝑆𝑤𝑐

(41)

9


𝑁 =𝑁𝑝𝐵𝑜 +𝑊𝑝𝐵𝑤 −𝑊𝑒

𝐵𝑜𝑖𝑐𝑒∆𝑝𝑡 ⟹ 𝑁𝐵𝑜𝑖𝑐𝑒∆𝑝𝑡 = 𝑁𝑝𝐵𝑜 +𝑊𝑝𝐵𝑤 −𝑊𝑒 (42)

The Van Everdingin & Hurst model for water influx is given by:

𝑊𝑒(𝑡𝑛+1) = 𝐵 ∑(∆𝑝𝑗) 𝑊𝑒𝐷 (𝑡𝐷

𝑛+1 − 𝑡𝐷𝑗)

𝑛

𝑗=0

(43)

Where:

𝐵 =2𝜋

5.6146∅𝑐𝑡ℎ𝑟𝑤

2𝜃

360= 1.11908∅𝑐𝑡ℎ𝑟𝑤

2𝜃

360 (44)


𝑁𝐵𝑜𝑖𝑐𝑒∆𝑝𝑡 = 𝑁𝑝𝐵𝑜 +𝑊𝑝𝐵𝑤 − 𝐵 ∑(∆𝑝𝑗) 𝑊𝑒𝐷 (𝑡𝐷𝑛+1 − 𝑡𝐷

𝑗)

𝑛

𝑗=0

(45)

Writing the equation in a straight-line form yields:

𝑁𝑝𝐵𝑜 +𝑊𝑝𝐵𝑤

𝐵𝑜𝑖𝑐𝑒∆𝑝𝑡= 𝑁 + 𝐵

∑ (∆𝑝𝑗) 𝑊𝑒𝐷 (𝑡𝐷𝑛+1 − 𝑡𝐷

𝑗)𝑛

𝑗=0

𝐵𝑜𝑖𝑐𝑒∆𝑝𝑡 (46)

This form suggests plotting 𝑁𝑝𝐵𝑜+𝑊𝑝𝐵𝑤

𝐵𝑜𝑖𝑐𝑒∆𝑝𝑡 on the y-axis versus

∑ (∆𝑝𝑗) 𝑊𝑒𝐷 (𝑡𝐷𝑛+1−𝑡𝐷

𝑗)𝑛

𝑗=0

𝐵𝑜𝑖𝑐𝑒∆𝑝𝑡 on

the x-axis and drawing the best-fit line through the set of points. The slope of the best-fit

line would be B and the y-intercept would be N as it can be seen in the next Figure.

10

5. Oil Reservoirs Below Bubble-Point Pressure

The material balance equation for oil reservoirs below the bubble-point pressure is

written as:

𝑁 [(𝐵𝑡 − 𝐵𝑡𝑖) + 𝑚𝐵𝑡𝑖𝐵𝑔𝑖

(𝐵𝑔 − 𝐵𝑔𝑖) + (1 + 𝑚)𝐵𝑡𝑖 (𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓


= 𝑁𝑝[𝐵𝑡 + 𝐵𝑔(𝑅𝑝 − 𝑅𝑠𝑜𝑖)] +𝑊𝑝𝐵𝑤

(47)

Rearranging in a straight-line form yields:

𝑁 +𝑊𝑒

[(𝐵𝑡 − 𝐵𝑡𝑖) + 𝑚𝐵𝑡𝑖𝐵𝑔𝑖

(𝐵𝑔 − 𝐵𝑔𝑖) + (1 + 𝑚)𝐵𝑡𝑖 (𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓1 − 𝑆𝑤𝑐

)∆𝑝𝑡]

=𝑁𝑝[𝐵𝑡 + 𝐵𝑔(𝑅𝑝 − 𝑅𝑠𝑜𝑖)] +𝑊𝑝𝐵𝑤


(𝐵𝑔 − 𝐵𝑔𝑖) + (1 + 𝑚)𝐵𝑡𝑖 (𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓1 − 𝑆𝑤𝑐

)∆𝑝𝑡]

(48)

The Van Everdingin & Hurst model for water influx is given by:



𝑛

𝑗=0

(49)

Where:

𝐵 =2𝜋


2𝜃

360= 1.11908∅𝑐𝑡ℎ𝑟𝑤

2𝜃

360 (50)


𝑁𝑝[𝐵𝑡 + 𝐵𝑔(𝑅𝑝 − 𝑅𝑠𝑜𝑖)] +𝑊𝑝𝐵𝑤


(𝐵𝑔 − 𝐵𝑔𝑖) + 𝐵𝑡𝑖(1 + 𝑚)(𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓1 − 𝑆𝑤𝑐

)∆𝑝𝑡]

= 𝑁 + 𝐵 ∑ (∆𝑝𝑗) 𝑊𝑒𝐷 (𝑡𝐷

𝑛+1 − 𝑡𝐷𝑗)𝑛

𝑗=0


(𝐵𝑔 − 𝐵𝑔𝑖) + 𝐵𝑡𝑖(1 + 𝑚) (𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓1 − 𝑆𝑤𝑐

) ∆𝑝𝑡]

(51)

This form suggests plotting 𝑁𝑝[𝐵𝑡+𝐵𝑔(𝑅𝑝−𝑅𝑠𝑜𝑖)]+𝑊𝑝𝐵𝑤

[(𝐵𝑡−𝐵𝑡𝑖)+𝑚𝐵𝑡𝑖𝐵𝑔𝑖

(𝐵𝑔−𝐵𝑔𝑖)+𝐵𝑡𝑖(1+𝑚)(𝑆𝑤𝑐𝑐𝑤+𝑐𝑓

1−𝑆𝑤𝑐)∆𝑝𝑡]

on the y-axis

versus ∑ (∆𝑝𝑗) 𝑊𝑒𝐷 (𝑡𝐷

𝑛+1−𝑡𝐷𝑗)𝑛

𝑗=0

[(𝐵𝑡−𝐵𝑡𝑖)+𝑚𝐵𝑡𝑖𝐵𝑔𝑖

(𝐵𝑔−𝐵𝑔𝑖)+𝐵𝑡𝑖(1+𝑚)(𝑆𝑤𝑐𝑐𝑤+𝑐𝑓

1−𝑆𝑤𝑐)∆𝑝𝑡]

on the x-axis and drawing the best-

11

fit line through the set of points. The slope of the best-fit line would be B and the y-

intercept would be N as it can be seen in the next Figure.

12

6. Closed or Volumetric Gas Reservoirs with Negligible Water and

Rock Compressibilities

For closed or volumetric gas reservoirs, the material balance equation is written as

follows:

𝐺(𝐵𝑔 − 𝐵𝑔𝑖) = 𝐺𝑝𝐵𝑔 (52)

Since the gas formation volume factor is defined as:

𝐵𝑔 =𝑉𝑔,𝑟𝑒𝑠

𝑉𝑔,𝑠𝑐𝑓⁄ (53)

The Universal Gas Law is written as:

𝑝𝑉 = 𝑧𝑛𝑅𝑇 ⟹ 𝑉 =𝑧𝑛𝑅𝑇

𝑝 (54)

Therefore,

𝐵𝑔 =

(𝑧𝑛𝑅𝑇𝑝 )

𝑟𝑒𝑠

(𝑧𝑛𝑅𝑇𝑝 )

𝑠𝑐𝑓

= (𝑝

𝑇𝑧)𝑠𝑐𝑓

(𝑇𝑧

𝑝)𝑟𝑒𝑠

=(14.696)

(519.67)(5.6146)𝑇𝑧

𝑝

= (0.0050367764503𝑇)𝑧

𝑝= 𝐶

𝑧

𝑝

(55)


𝐺 (𝑧

𝑝−𝑧𝑖𝑝𝑖) =

𝑧

𝑝𝐺𝑝 (56)

Dividing by 𝐺, then multiplying by 𝑝

𝑧, and finally multiplying by

𝑝𝑖

𝑧𝑖 yields:

𝑧

𝑝−𝑧𝑖𝑝𝑖

=𝑧

𝑝

𝐺𝑝

𝐺 ⟹ 1 −

𝑝 𝑧⁄

𝑝𝑖 𝑧𝑖⁄=𝐺𝑝

𝐺 ⟹

𝑝𝑖𝑧𝑖−𝑝

𝑧=𝑝𝑖𝑧𝑖

𝐺𝑝

𝐺 (57)

Which can arranged in a straight-line form, the p over z form, as follows:

𝑝

𝑧=𝑝𝑖𝑧𝑖−

𝑝𝑖𝑧𝑖𝐺

𝐺𝑝 (58)

13

This form suggests plotting 𝑝

𝑧 on the y-axis versus 𝐺𝑝 on the x-axis and drawing the

best-fit line through the set of points. The slope of the best-fit line would be −𝑝𝑖

𝑧𝑖𝐺 and

the y-intercept would be 𝑝𝑖

𝑧𝑖.

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14

7. Gas Reservoirs with Water Influx and Negligible Water and Rock

Compressibilities

For gas reservoirs with water influx, the material balance equation is written as

follows:

𝐺(𝐵𝑔 − 𝐵𝑔𝑖) +𝑊𝑒 = 𝐺𝑝𝐵𝑔 +𝑊𝑝𝐵𝑤 (59)

Rearranging and dividing all through by 𝐵𝑔 yields:

𝐺 (1 −𝐵𝑔𝑖

𝐵𝑔) = 𝐺𝑝 +

𝑊𝑝𝐵𝑤 −𝑊𝑒

𝐵𝑔 (60)

Substituting for 𝐵𝑔 = 𝐶𝑧

𝑝 on the left-hand side of (60) and dividing by 𝐺 yields:

(1 −𝑝 𝑧⁄

𝑝𝑖 𝑧𝑖⁄) =

1

𝐺(𝐺𝑝 +


𝐵𝑔) (61)

Solving for 𝑝

𝑧 yields:

𝑝



(𝐺𝑝 +𝑊𝑝𝐵𝑤 −𝑊𝑒

𝐵𝑔) (62)


𝑧 on the y-axis versus (𝐺𝑝 +

𝑊𝑝𝐵𝑤−𝑊𝑒

𝐵𝑔) on the x-axis and

drawing the best-fit line through the set of points. The slope of the best-fit line would be

−𝑝𝑖

𝑧𝑖𝐺 and the y-intercept would be

𝑝𝑖

𝑧𝑖.

8. Gas Reservoirs with Water Influx and Water and Rock

Compressibilities

For gas reservoirs with water influx and water and rock compressibilities, the material

balance equation is written as follows:

𝐺(𝐵𝑔 − 𝐵𝑔𝑖) + 𝐺𝐵𝑔𝑖 (𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓

1 − 𝑆𝑤𝑐)∆𝑝𝑡 +𝑊𝑒 = 𝐺𝑝𝐵𝑔 +𝑊𝑝𝐵𝑤 (63)

15

Factor out 𝐺 as common and divide all through by 𝐵𝑔 and rearrange yields:

𝐺 [1 −𝐵𝑔𝑖

𝐵𝑔+𝐵𝑔𝑖

𝐵𝑔(𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓

1 − 𝑆𝑤𝑐)∆𝑝𝑡] = 𝐺𝑝 +


𝐵𝑔 (64)

Let us define the effective compressibility as follows:

�̂�𝑒 = (𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓

1 − 𝑆𝑤𝑐) (65)

Plugging (65) into (64) and solving for 𝐺𝑝 yields:

𝐺𝑝 = 𝐺 [1 −𝐵𝑔𝑖

𝐵𝑔(1 + �̂�𝑒∆𝑝𝑡)] −


𝐵𝑔 (66)

Substituting for 𝐵𝑔 = 𝐶𝑧

𝑝 on the left-hand side of (66) and dividing by 𝐺 yields:

𝐺𝑝 = 𝐺 [1 −𝑝 𝑧⁄

𝑝𝑖 𝑧𝑖⁄(1 + �̂�𝑒∆𝑝𝑡)] −


𝐵𝑔 (67)

Solving for 𝑝

𝑧 and rearranging in a straight line form yields:

𝑝

𝑧(1 − �̂�𝑒∆𝑝𝑡) =

𝑝𝑖𝑧𝑖−


[𝐺𝑝 +𝑊𝑝𝐵𝑤 −𝑊𝑒

𝐵𝑔] (68)


𝑧(1 − �̂�𝑒∆𝑝𝑡) on the y-axis versus (𝐺𝑝 +


𝐵𝑔) on the

x-axis and drawing the best-fit line through the set of points. The slope of the best-fit

line would be −𝑝𝑖


𝑝𝑖

𝑧𝑖.

9. Abnormally High-Pressure Gas Reservoirs

In abnormally high-pressure gas reservoirs, 𝑐𝑓 is a strong function of pressure,

therefore we set:

�̂�𝑒∆𝑝𝑡 = ∑�̂�𝑒𝑗∆𝑝𝑗

𝑛

𝑗=0

= ∑�̂�𝑒𝑗(𝑝𝑗+1 − 𝑝𝑗)

𝑛

𝑗=0

(69)

Therefore equation (68) is written as follows:

𝑝

𝑧(1 +∑�̂�𝑒

𝑗(𝑝𝑗+1 − 𝑝𝑗)

𝑛

𝑗=0

) = 𝑝𝑖𝑧𝑖−


[𝐺𝑝 +𝑊𝑝𝐵𝑤 −𝑊𝑒

𝐵𝑔] (70)

16


𝑧(1 − ∑ �̂�𝑒

𝑗(𝑝𝑗+1 − 𝑝𝑗)𝑛

𝑗=0 ) on the y-axis versus (𝐺𝑝 +


𝐵𝑔) on the x-axis and drawing the best-fit line through the set of points. The slope

of the best-fit line would be −𝑝𝑖


𝑝𝑖

𝑧𝑖.

10. Abnormally High-Pressure Gas Reservoirs with Dissolved Gas in

Water (Fetkovitch Model)

For high-pressure gas reservoirs with dissolved gas in water, Fetkovitch et al. expressed

the material balance equation, equation (70), as follows:

𝑝

𝑧(1 − 𝑐�̅�∆𝑝𝑡) =

𝑝𝑖𝑧𝑖−


[(𝐺𝑝 − 𝐺𝐼 +𝑊𝑝𝑅𝑠𝑤) + (𝑊𝑝𝐵𝑤 −𝑊𝑒 −𝑊𝐼𝐵𝐼𝑤

𝐵𝑔)] (71)

Where:

𝑐�̅� =𝑐�̅� + 𝑆𝑤𝑐𝑐�̅�𝑤 +𝑀(𝑐�̅� + 𝑐�̅�𝑤)

1 − 𝑆𝑤𝑐

𝑐�̅� =1

𝑉𝑝𝑖(𝑉𝑝𝑖 − 𝑉𝑝

𝑝𝑖 − 𝑝) =

1

∅𝑖(∅𝑖 − ∅

𝑝𝑖 − 𝑝)

𝑐�̅�𝑤 =1

𝐵𝑡𝑤𝑖(𝐵𝑡𝑤 − 𝐵𝑡𝑤𝑖𝑝𝑖 − 𝑝

)

𝐵𝑡𝑤 = 𝐵𝑤 + 𝐵𝑔(𝑅𝑠𝑤𝑖 − 𝑅𝑠𝑤)

(72)

𝑀 is defined as the associated water volume ratio. The associated water and pore

volumes external to the net pay include Non-Net Pay (NNP) such as inter-bedded shales

and dirty sands plus external water volume found in AQuifers. This volume is expressed

as a ratio relative to the pore volume of the net-pay reservoir; i.e.

𝑀 = 𝑀𝑁𝑁𝑃 +𝑀𝐴𝑄

𝑀𝑁𝑁𝑃 =(𝑉𝑝)𝑁𝑁𝑃(𝑉𝑝)𝑅𝐸𝑆

=ℎ𝑁𝑁𝑃ℎ𝑅𝐸𝑆

=ℎ𝐺𝑅𝑂𝑆𝑆 − ℎ𝑅𝐸𝑆

ℎ𝑅𝐸𝑆

𝑀𝐴𝑄 =(𝑉𝑝)𝐴𝑄

(𝑉𝑝)𝑅𝐸𝑆

≈ (𝑟𝑒2

𝑟𝑤2− 1)

(73)


𝑧(1 − 𝑐�̅�∆𝑝𝑡) on the y-axis versus (𝐺𝑝 − 𝐺𝐼 +𝑊𝑝𝑅𝑠𝑤) +

(𝑊𝑝𝐵𝑤−𝑊𝑒−𝑊𝐼𝐵𝐼𝑤

𝐵𝑔) on the x-axis and drawing the best-fit line through the set of points.

The slope of the best-fit line would be −𝑝𝑖


𝑝𝑖

𝑧𝑖.

17

11. Abnormally High-Pressure Gas Reservoirs with Dissolved Gas in

Water (Kazemi's Model)

We start with Fetkovitch material balance equation as follows:

𝑝

𝑧(1 −∑�̃�𝑒

𝑗∆𝑝𝑗

𝑛

𝑗=0

)

= 𝑝𝑖𝑧𝑖−


[(𝐺𝑝 − 𝐺𝐼 +𝑊𝑝𝑅𝑠𝑤) + (𝑊𝑝𝐵𝑤 −𝑊𝑒 −𝑊𝐼𝐵𝐼𝑤

𝐵𝑔)]

(74)

Where:

�̃�𝑒𝑗=𝑐𝑓𝑗+ 𝑆𝑤𝑐𝑐𝑎𝑤

𝑗+𝑀(𝑐𝑓

𝑗+ 𝑐𝑎𝑤

𝑗)

1 − 𝑆𝑤𝑐

𝑐𝑓𝑗=1

∅(𝜕∅

𝜕𝑝) =

1

∅𝑗(∅𝑗+1 − ∅𝑗

𝑝𝑗+1 − 𝑝𝑗)

𝑐𝑎𝑤𝑗

=−1

𝐵𝑤(𝜕𝐵𝑤𝜕𝑝

− 𝐵𝑔𝜕𝑅𝑠𝑤𝜕𝑝

) =−1

𝐵𝑤𝑗(𝐵𝑤𝑗+1

− 𝐵𝑤𝑗

𝑝𝑗+1 − 𝑝𝑗− 𝐵𝑔

𝑗 𝑅𝑠𝑤𝑗+1

− 𝑅𝑠𝑤𝑗

𝑝𝑗+1 − 𝑝𝑗)

(75)


𝑧(1 − ∑ �̃�𝑒

𝑗∆𝑝𝑗𝑛

𝑗=0 ) on the y-axis versus (𝐺𝑝 − 𝐺𝐼 +

𝑊𝑝𝑅𝑠𝑤) + (𝑊𝑝𝐵𝑤−𝑊𝑒−𝑊𝐼𝐵𝐼𝑤

𝐵𝑔) on the x-axis and drawing the best-fit line through the set

of points. The slope of the best-fit line would be −𝑝𝑖


𝑝𝑖

𝑧𝑖.

12. Graphical Technique for the Abnormally High-Pressure Gas

Reservoirs (A Non-p/z Form of the Gas Material Balance)

The graphical technique for the abnormally high-pressure gas reservoirs is written as:

𝐺(𝐵𝑔 − 𝐵𝑔𝑖) + 𝐺𝐵𝑔𝑖 (𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓


= (𝐺𝑝 − 𝐺𝐼)𝐵𝑔 +𝑊𝑝𝑅𝑠𝑤 + (𝑊𝑝 −𝑊𝐼)𝐵𝑤

(76)

𝐺(𝐵𝑔 − 𝐵𝑔𝑖) + 𝐺𝐵𝑔𝑖∑�̃�𝑒𝑗∆𝑝𝑗

𝑛

𝑗=0

+𝑊𝑒

= (𝐺𝑝 − 𝐺𝐼)𝐵𝑔 +𝑊𝑝𝑅𝑠𝑤 + (𝑊𝑝 −𝑊𝐼)𝐵𝑤

(77)

Let us define the following cumulative expansions:

𝐸𝑔 = cumulative gas expansion = (𝐵𝑔 − 𝐵𝑔𝑖) (78)

18

𝐸𝑓,𝑤 = cumulative connate water expansion and formation compaction

= 𝐵𝑔𝑖 (𝑆𝑤𝑐𝑐𝑤 + 𝑐𝑓

1 − 𝑆𝑤𝑐)∆𝑝𝑡 = 𝐵𝑔𝑖𝑐�̅�∆𝑝𝑡 for Fetkovitch model

= 𝐵𝑔𝑖∑�̃�𝑒𝑗∆𝑝𝑗

𝑛

𝑗=0

for Kazemi′s model

(79)

𝐹 = (𝐺𝑝 − 𝐺𝐼)𝐵𝑔 +𝑊𝑝𝑅𝑠𝑤 + (𝑊𝑝 −𝑊𝐼)𝐵𝑤 (80)

Plugging (78), (79), and (80) into (77) yields:

𝐹 = 𝐺[𝐸𝑔 + 𝐸𝑓,𝑤] +𝑊𝑒 (81)

Let us define the cumulative total expansion as:

𝐸𝑡 = 𝐸𝑔 + 𝐸𝑓,𝑤 (82)


𝐹 = 𝐺𝐸𝑡 +𝑊𝑒 (83)

Dividing all through by 𝐸𝑡, we get:

𝐺 +𝑊𝑒

𝐸𝑡=

𝐹

𝐸𝑡 (84)

Which is written as 𝑏 + 𝑥 = 𝑦. This would suggest that a plot of 𝐹

𝐸𝑡 as the y coordinate

and 𝑊𝑒

𝐸𝑡 as the x coordinate would yield a straight line with slope equal to 1 and intercept

equal to 𝑁 as it can be seen in the next Figure.

19

Since the Van Everdingin & Hurst model for water influx is given by:



𝑛

𝑗=0

(85)

Where:

𝐵 =2𝜋


2𝜃

360= 1.11908∅𝑐𝑡ℎ𝑟𝑤

2𝜃

360 (86)


𝐺 + 𝐵 ∑ (∆𝑝𝑗) 𝑊𝑒𝐷 (𝑡𝐷

𝑛+1 − 𝑡𝐷𝑗)𝑛

𝑗=0

𝐸𝑡=

𝐹

𝐸𝑡 (87)

This form suggests plotting 𝐹

𝐸𝑡 on the y-axis versus

∑ (∆𝑝𝑗) 𝑊𝑒𝐷 (𝑡𝐷𝑛+1−𝑡𝐷

𝑗)𝑛

𝑗=0

𝐸𝑡 on the x-axis

and drawing the best-fit line through the set of points. The slope of the best-fit line

would be B and the y-intercept would be N as it can be seen in the next Figure.

21

13. Wet Gas Reservoirs

In wet gas reservoirs, the stock tank barrels of the Liquid condensate produced must be

accounted for in the cumulative gas production. This is done as follows:

(𝐺𝑝)𝑒𝑓𝑓 = 𝐺𝑝(dry gas) + 𝐺𝑝(wet gas)

(𝐺𝑝)𝑒𝑓𝑓 = 𝐺𝑝 + (𝐺𝐿𝐶𝐾𝐿𝐶)

(88)

Where:

(𝐺𝑝)𝑒𝑓𝑓 is the effective cumulative gas produced, SCF

𝐺𝑝 is the cumulative gas produced, SCF

𝐺𝐿𝐶 is the cumulative liquid condensate produced, STB

𝐾𝐿𝐶 is the liquid condensate conversion factor, SCF/STB, which is given by:

𝐾𝐿𝐶 = 132,790𝑆𝐺𝐿𝐶𝑀𝑊𝐿𝐶

𝑆𝐺𝐿𝐶 is the specific gravity of the liquid condensate which is given by:

𝑆𝐺𝐿𝐶 =141.5

131.5 + °𝐴𝑃𝐼

𝑀𝑊𝐿𝐶 is the molecular weight of the liquid condensate which is given by:

𝑀𝑊𝐿𝐶 =6084

°𝐴𝑃𝐼 − 5.9 Cragoe′sequation

𝑀𝑊𝐿𝐶 =5954

°𝐴𝑃𝐼 − 8.8 modified by McCain

Therefore the effective gas production would be given by:

(𝐺𝑝)𝑒𝑓𝑓 = 𝐺𝑝 + 3155.8255 (°𝐴𝑃𝐼 − 8.8

131.5 + °𝐴𝑃𝐼)𝐺𝐿𝐶

22

14. Closed Form Solutions

Some derivations for G from dynamic calculations:

Ramey (1966) expressed the produced gas at abandonment pressure with the following

equation:

(𝐺𝑝)𝑎𝑏 = 𝐺 {1 − 𝐸𝑣 [𝑆𝑔𝑡

𝑆𝑔+1 − 𝐸𝑣𝐸𝑣

](𝑝 𝑧⁄ )𝑎𝑏(𝑝 𝑧⁄ )𝑖

} (90)

Since:

𝐺(𝐵𝑔 − 𝐵𝑔𝑖) +𝑊𝑒 = 𝐺𝑝𝐵𝑔 +𝑊𝑝𝐵𝑤 (91)

Therefore,

𝐺𝐵𝑔 − 𝐺𝐵𝑔𝑖 = 𝐺𝑝𝐵𝑔 −𝑊𝑒 +𝑊𝑝𝐵𝑤 (92)

Assuming there is no water produced, and considering the trapped gas, we rewrite

equation (92) as follows:

[𝐺𝐵𝑔 − 𝐺𝐵𝑔𝑖𝐸𝑣(𝑡)𝑆𝑔𝑡

𝑆𝑔] − 𝐺𝐵𝑔𝑖 = 𝐺𝑝𝐵𝑔 −𝑊𝑒 (93)

Let:

𝐺𝐵𝑔𝑖𝐸𝑣(𝑡)𝑆𝑔𝑡

𝑆𝑔≡ (𝐺𝐵𝑔𝐸𝑣(𝑡)

𝑆𝑔𝑡

𝑆𝑔)𝐵𝑔𝑖

𝐵𝑔≡ (𝐺𝐵𝑔𝐸𝑣(𝑡)

𝑆𝑔𝑡

𝑆𝑔)(𝑝 𝑧⁄ )

(𝑝 𝑧⁄ )𝑖 (94)

Thus, we write:

[𝐺𝐵𝑔 − 𝐺𝐵𝑔𝐸𝑣(𝑡)𝑆𝑔𝑡

𝑆𝑔

(𝑝 𝑧⁄ )

(𝑝 𝑧⁄ )𝑖] − 𝐺𝐵𝑔𝑖 = 𝐺𝑝𝐵𝑔 −𝑊𝑒

𝐺𝐵𝑔 [1 − 𝐸𝑣(𝑡)𝑆𝑔𝑡

𝑆𝑔

(𝑝 𝑧⁄ )

(𝑝 𝑧⁄ )𝑖] − 𝐺𝐵𝑔𝑖 = 𝐺𝑝𝐵𝑔 −𝑊𝑒

𝐺 [1 − 𝐸𝑣(𝑡)𝑆𝑔𝑡

𝑆𝑔

(𝑝 𝑧⁄ )

(𝑝 𝑧⁄ )𝑖] − 𝐺

(𝑝 𝑧⁄ )

(𝑝 𝑧⁄ )𝑖= 𝐺𝑝 −

𝑊𝑒

𝐵𝑔

[1 − 𝐸𝑣(𝑡)𝑆𝑔𝑡

𝑆𝑔

(𝑝 𝑧⁄ )

(𝑝 𝑧⁄ )𝑖] −

(𝑝 𝑧⁄ )

(𝑝 𝑧⁄ )𝑖=𝐺𝑝

𝐺−

𝑊𝑒

𝐵𝑔𝐺

[1 − 𝐸𝑣(𝑡)𝑆𝑔𝑡

𝑆𝑔

(𝑝 𝑧⁄ )

(𝑝 𝑧⁄ )𝑖] −

(𝑝 𝑧⁄ )

(𝑝 𝑧⁄ )𝑖=𝐺𝑝

𝐺−

𝑊𝑒

𝐵𝑔𝐺

23

𝐺𝑝

𝐺= [1 − 𝐸𝑣(𝑡)

𝑆𝑔𝑡

𝑆𝑔

(𝑝 𝑧⁄ )

(𝑝 𝑧⁄ )𝑖] −

(𝑝 𝑧⁄ )

(𝑝 𝑧⁄ )𝑖+

𝑊𝑒

𝐵𝑔𝐺

𝐺𝑝

𝐺

𝑝𝑖𝑧𝑖= [1 − 𝐸𝑣(𝑡)

𝑆𝑔𝑡

𝑆𝑔

(𝑝 𝑧⁄ )

(𝑝 𝑧⁄ )𝑖]𝑝𝑖𝑧𝑖−𝑝

𝑧+

𝑊𝑒

𝐵𝑔𝐺

𝑝𝑖𝑧𝑖

𝑝

𝑧=𝑝𝑖𝑧𝑖[1 − 𝐸𝑣(𝑡)

𝑆𝑔𝑡

𝑆𝑔

(𝑝 𝑧⁄ )

(𝑝 𝑧⁄ )𝑖−𝐺𝑝

𝐺+

𝑊𝑒

𝐵𝑔𝐺]

𝑝

𝑧=𝑝𝑖𝑧𝑖[1 − 𝐸𝑣(𝑡)

𝑆𝑔𝑡

𝑆𝑔

(𝑝 𝑧⁄ )

(𝑝 𝑧⁄ )𝑖−1

𝐺(𝐺𝑝 −

𝑊𝑒

𝐵𝑔)]

𝑝

𝑧(1 + 𝐸𝑣(𝑡)

𝑆𝑔𝑡

𝑆𝑔) =

𝑝𝑖𝑧𝑖[1 −

1

𝐺(𝐺𝑝 −

𝑊𝑒

𝐵𝑔)] (95)


𝑧(1 + 𝐸𝑣(𝑡)

𝑆𝑔𝑡

𝑆𝑔) on the y-axis versus (𝐺𝑝 −

𝑊𝑒

𝐵𝑔) on the x-

axis and drawing the best-fit line through the set of points. The slope of the best-fit line

would be B and the y-intercept would be N as it can be seen in the next Figure.

𝑝

𝑧(1 + 𝐸𝑣(𝑡)

𝑆𝑔𝑡

𝑆𝑔) =

𝑝𝑖𝑧𝑖[1 − (

𝐺𝑝

𝐺−

𝑊𝑒

𝐺𝐵𝑔𝑖

𝐵𝑔𝑖

𝐵𝑔)]

𝑝

𝑧(1 + 𝐸𝑣(𝑡)

𝑆𝑔𝑡

𝑆𝑔) =


𝐺𝑝

𝐺−

𝑊𝑒

𝐺𝐵𝑔𝑖

(𝑝 𝑧⁄ )

(𝑝 𝑧⁄ )𝑖)]

Since:

𝐸𝑣(𝑡) =𝑊𝑒

𝐺𝐵𝑔𝑖=𝑊𝑒 +𝑊𝑝𝐵𝑤

𝐺𝐵𝑔𝑖

𝑝

𝑧(1 + 𝐸𝑣(𝑡)

𝑆𝑔𝑡

𝑆𝑔) =


𝐺𝑝

𝐺− 𝐸𝑣(𝑡)

(𝑝 𝑧⁄ )

(𝑝 𝑧⁄ )𝑖)]

𝑝

𝑧(1 + 𝐸𝑣(𝑡) (

𝑆𝑔𝑡

𝑆𝑔− 1)) =

𝑝𝑖𝑧𝑖[1 −

𝐺𝑝

𝐺]

We may bring it to the form of Ramey as follows:

𝐺𝑝 = 𝐺 {1 − 𝐸𝑣 [𝑆𝑔𝑡

𝑆𝑔+1 − 𝐸𝑣𝐸𝑣

](𝑝 𝑧⁄ )

(𝑝 𝑧⁄ )𝑖} (96)

24

Example #1: Material Balance Calculations for a Saturated Oil

Reservoir Neglecting Water and Rock

Compressibilities.

Given the following data:

Volume of bulk oil zone 112,000 acre-ft

Volume of bulk gas zone 19,600 acre-ft

Initial reservoir pressure 2710 psia

Initial oil FVF 1.340 bbl/STB

Initial gas FVF 0.006266 ft3/SCF

Initial dissolved GOR 562 SCF/STB

Oil produced during the interval 20 MM STB

Reservoir pressure at the end of the interval 2000 psia

Average produced GOR 700 SCF/STB

Two-phase FVF at 2000 psia 1.4954 bbl/STB

Volume of water encroached 11.58 MM bbl

Volume of water produced 1.05 MM STB

Water FVF 1.028 bbl/STB

Gas FVF at 2000 psia 0.008479 ft3/SCF

A. Calculate the stock tank oil initially in place.

B. Calculate the driving indexes.

C. Discuss your results.

Solution to Example #1:

A. The material balance equation is written as:

𝑁 =𝑁𝑝[𝐵𝑡 + (𝑅𝑝 − 𝑅𝑠𝑜𝑖)𝐵𝑔] − (𝑊𝑒 −𝑊𝑝𝐵𝑤)

(𝐵𝑡 − 𝐵𝑡𝑖) + 𝑚𝐵𝑡𝑖𝐵𝑔𝑖


Since:

Np = 20 x 106 STB

Bt = 1.4954 bbl/STB

Rp = 700 SCF/STB

Rsoi = 562 SCF/STB

Bg = 0.008479 ft3/SCF = 0.008479/5.6146 = 0.001510 bbl/SCF

We = 11.58 x 10 6 bbl

Wp = 1.05 x 106 STB

Bw = 1.028 bbl/STB

Bti = 1.34 bbl/STB

m = 𝐺𝐵𝑔𝑖

𝑁𝐵𝑡𝑖 = 19,600/112,000 = 0.175

Bgi = 0.006266 ft3/SCF = 0.006266/5.6146 = 0.001116 bbl/SCF

25

Thus:

𝑁 =20[1.4954 + (700 − 562)0.001510] − (11.58 − (1.05)(1.028))

(1.4954 − 1.34) + 0.1751.34

0.001116(0.001510 − 0.001116)

106

= 98.97 MMSTB

B. In terms of drive indexes, the material balance equation is written as:



𝐺(𝐵𝑔 − 𝐵𝑔𝑖)


(𝑊𝑒 −𝑊𝑝𝐵𝑤)

𝑁𝑝[𝐵𝑡 + (𝑅𝑝 − 𝑅𝑠𝑜𝑖)𝐵𝑔]= 1

Thus the depletion drive index (DDI) is given by:


𝑁𝑝[𝐵𝑡 + (𝑅𝑝 − 𝑅𝑠𝑜𝑖)𝐵𝑔]=

98.97𝑥106(1.4954 − 1.34)

20𝑥106[1.4954 + (700 − 562)0.001510]= 0.45

The segregation drive index (SDI) is given by:

𝑁𝑚𝐵𝑡𝑖𝐵𝑔𝑖


𝑁𝑝[𝐵𝑡 + (𝑅𝑝 − 𝑅𝑠𝑜𝑖)𝐵𝑔]

=98.97𝑥106𝑥0.175𝑥

1.340.001116

(0.001510 − 0.001116)

20𝑥106[1.4954 + (700 − 562)0.001510]= 0.24

The water drive index (WDI) is given by:

(𝑊𝑒 −𝑊𝑝𝐵𝑤)

𝑁𝑝[𝐵𝑡 + (𝑅𝑝 − 𝑅𝑠𝑜𝑖)𝐵𝑔]=

(11.58𝑥106 − 1.05𝑥1.028𝑥106𝐵𝑤)

20𝑥106[1.4954 + (700 − 562)0.001510]= 0.31

C. The drive mechanisms as calculated in part B indicate that when the reservoir

pressure has declined from 2710 psia to 2000 psia, 45% of the total production was by

oil expansion, 31% was by water drive, and 24% was by gas cap expansion.

26

Example #2: Material Balance Calculations for a Volumetric Gas

Reservoir Neglecting Water and Rock Compressibilities.


Initial pressure 3250 psia

Reservoir temperature 213 °F

Standard pressure 15.025 psia

Standard temperature 60 °F

Cumulative production 1 x 109 SCF

Average reservoir pressure 2864 psia

Gas deviation factor at 3250 psia 0.910



A. Calculate the initial gas in place for a closed gas reservoir with negligible water and

rock compressibilities.

B. If abandonment pressure is 500 psia, calculate the initial gas reserve.

C. Based on a 500-psia abandonment pressure, calculate the remaining gas.


A. Since

𝑝



𝐺𝑝

Rearranging and solving for 𝐺 yields:

𝐺 =

𝑝𝑖𝑧𝑖

(𝑝𝑖𝑧𝑖−𝑝𝑧)

𝐺𝑝 =

32500.910

(32500.910 −

28640.888)

109 = 10.316 MMM SCF

B. Rearranging the above equation for 𝐺𝑝 we obtain:

𝐺𝑝 = 𝐺(𝑝𝑖𝑧𝑖−𝑝𝑧)

𝑝𝑖𝑧𝑖

=(32500.910 −

28640.888)

32500.910

(10.316)109 = 8.797 MMM SCF

C. The remaining gas is given by:

𝐺 − 𝐺𝑝 = (10.316 − 8.797)109 = 1.519 MMM SCF

27


Reservoir Neglecting Water and Rock Compressibilities:

p/z plot


Initial reservoir pressure 4200 psia

Reservoir temperature 180 °F


p/z 𝑮𝒑 MMM SCF

4600

3700

2800

0

1

2

A. What will be the cumulative gas produced when the average reservoir pressure has

dropped to 2000?

B. Assuming the reservoir rock has a porosity of 12%, water saturation of 30%, and

reservoir thickness is 15 ft. How many acres does the reservoir cover?


A. Preparing the 𝑝

𝑧 plot and drawing the best-fit line through the data points, we get:

y = -900x + 4600

0

500

1000

1500

2000

2500

3000

3500

4000

4500

5000

0 0.5 1 1.5 2 2.5

p/z

Gp MMM SCF

28

The equation for the best fit is:

𝑝



𝐺𝑝 = 4600 − 900𝐺𝑝 ⟹ 𝐺𝑝 =(4600 −

𝑝𝑧)

900

Thus at 𝑝

𝑧=

2000

0.8= 2500 ⟹ 𝐺𝑝 =

(4600− 2500)

900= 2.33 MMMSCF

B. Setting 𝑝

𝑧= 0 in the best-fit equation yields:

𝐺𝑝 =(4600 − 0)

900= 5.11 MMM SCF

Since:

𝐺 =43,560(𝐴)(ℎ)(∅)(1 − 𝑆𝑤)

𝐵𝑔𝑖=43,560(𝐴)(15)(0.12)(1 − 0.3)

𝐵𝑔𝑖= 5.11 MMM SCF

Moreover:

𝐵𝑔𝑖 = (0.02828)𝑇 (𝑧

𝑝)𝑖

=(0.02828)(640)

𝑝𝑖𝑧𝑖

=18.0992

4600= 0.0039346

𝑓𝑡3

𝑆𝐶𝐹

Therefore, solving for the area yields:

𝐴 =(5.11)(0.0039346)

43,560(15)(0.12)(1 − 0.3)= 366.4 acres

29


Reservoir Neglecting Water and Rock Compressibilities:

p/z plot


Reservoir

Pressure, p

psia

Gas Deviation Factor, z

dimensionless

Cumulative 𝑮𝒑

MMM SCF

2080

1885

1620

1205

888

645

0.759

0.767

0.787

0.828

0.866

0.900

0.000

6.873

14.002

23.687

31.009

36.207

A. Estimate the amount of gas initially in place.

B. Discard the first point and estimate the amount of initial gas in place.

C. Discuss your results.


A. Preparing the 𝑝

𝑧 plot and drawing the best-fit line through the data points, we get:

y = -57.137x + 2806.8

y = -59.702x + 2877.4

0

500

1000

1500

2000

2500

3000

0 5 10 15 20 25 30 35 40

p/z

Gp MMM SCF

p/z Plot

30


𝑝



𝐺𝑝 = 2806.8 − 57.137𝐺𝑝 ⟹ 𝐺𝑝 =(2806.8 −

𝑝𝑧)

57.137

A. Setting 𝑝


𝐺𝑝 =(2806.8 − 0)

57.137= 49.124 MMM SCF

B. Discarding the first point and drawing the best fit through the rest of points, (red line

in the above Figure) yields the following best-fit equation:

𝐺𝑝 =(2877.4 −

𝑝𝑧)

59.702=2877.4

59.702= 48.2 MMM SCF

31

Example #5: Material Balance Calculations for a Gas Reservoir with

Water Influx and Neglecting Water and Rock

Compressibilities


Initial bulk reservoir volume 415.3 MM ft3

Average porosity 0.172

Average connate water saturation 0.25

Initial pressure 3200 psia

Initial gas FVF, 𝐵𝑔𝑖 0.005262 ft3/SCF

Final pressure 2925 psia

Gas FVF at final pressure 0.0057 ft3/SCF

Cumulative water production 15,200 STB

Water FVF, 𝐵𝑤 1.03 bbl/STB

Cumulative gas produced, 𝐺𝑝 935.4 MM SCF

Bulk volume invaded by water at final pressure 13.04 MM ft3

A. Calculate initial gas in place.

B. Calculate water influx, 𝑊𝑒.

C. Calculate residual gas saturation, 𝑆𝑔𝑟.


A. The amount of gas in place by the volumetric method is given by:

𝐺 =43,560 𝐴 ℎ ∅ (1 − 𝑆𝑤)

𝐵𝑔𝑖=415.3𝑥106𝑥0.172𝑥(1 − 0.25)

0.005262= 10.18 MMM SCF

B. Since:

𝐺(𝐵𝑔 − 𝐵𝑔𝑖) +𝑊𝑒 = 𝐺𝑝𝐵𝑔 +𝑊𝑝𝐵𝑤

Rearranging and solving for 𝑊𝑒, we get:

𝑊𝑒 = 𝐺𝑝𝐵𝑔 +𝑊𝑝𝐵𝑤 − 𝐺(𝐵𝑔 − 𝐵𝑔𝑖)

= 935.4𝑥106𝑥0.0057 + 15,200𝑥5.6146𝑥1.03− 10.18𝑥106𝑥(0.0057 − 0.005262) = 960,400 CF

C.

𝑆𝑤 =Water Volume

Pore Volume=𝑉𝑤𝑉𝑝

=connate water + water influx − water produced

𝑝𝑜𝑟𝑒 𝑣𝑜𝑙𝑢𝑚𝑒

=13.04x106x0.172x0.25 + 960,400 − 15,200x5.6146x1.03

13.04x106x0.172= 0.64

32

Therefore, 𝑆𝑔𝑟 = 1 − 0.64 = 0.36 or 36%


Water Influx and Neglecting Water and Rock

Compressibilities: p/z plot


Areal extent, A = 8870 acres

Formation thickness, h = 32.5 ft

Porosity, φ = 30.8 %

Initial water saturation, Swi = 42.5 %

Gas deviation factor at standard conditions, zb = 1.0

Performance History

Time

years

𝐺𝑝

MMM SCF

p

psia

z 𝐵𝑔

bbl/SCF

0

2

4

6

8

10

12

14

16

18

20

0

2.305

20.257

49.719

80.134

105.930

135.350

157.110

178.300

192.089

205.744

2333

2321

2203

2028

1854

1711

1531

1418

1306

1227

1153

0.882

0.883

0.884

0.888

0.894

0.899

0.907

0.912

0.921

0.922

0.928

0.001,172

0.001,180

0.001,244

0.001,358

0.001,496

0.001,630

0.001,820

0.001,995

0.002,187

0.002,330

0.002,495

1. Estimate the initial gas in place by:

A. Volumetric calculation

B. MB calculation neglecting water influx

C. MB calculation with water influx

2. Discuss the results


A. The volumetric calculation of the initial gas in place yields:

𝐺 =43,560 𝐴 ℎ ∅ (1 − 𝑆𝑤)

𝐵𝑔𝑖=43,560(8870)(32.5)(0.308)(1 − 0.425)

(0.001172)(5.6146)

= 338 MMM SCF

33

B. Prepare a 𝑝

𝑧 plot versus 𝐺𝑝 assuming no water influx:


𝑝



𝐺𝑝 = 2632.1 − 6.82595𝐺𝑝 ⟹ 𝐺𝑝 =(2632.1 −

𝑝𝑧)

6.82595

Setting 𝑝


𝐺𝑝 =(2632.1 − 0)

6.82595= 385.6 MMM SCF

This value is much higher than the volumetric calculated value. A close look to the plot

indicates an upward curvature of the line connecting the data points indicating water

encroachment. Thus, the 𝑝

𝑧 plot versus 𝐺𝑝 cannot be used to estimate gas in place.

C. Including water influx, 𝑊𝑒, in the gas material balance equation yields:

𝐺(𝐵𝑔 − 𝐵𝑔𝑖) +𝑊𝑒 = 𝐺𝑝𝐵𝑔

Preparing the following table:

y = -6.8259x + 2632.1

0

500

1000

1500

2000

2500

3000

0 50 100 150 200 250

p/z

Gp MMM SCF

34

𝑝

𝑧 𝑊𝑒, bbl

𝑊𝑒 = 𝐺𝑝𝐵𝑔 − 𝐺(𝐵𝑔 − 𝐵𝑔𝑖) 𝐺𝑝 −

𝑊𝑒

𝐵𝑔

2645.12

2628.54

2492.08

2283.78

2073.83

1903.23

1687.98

1554.82

1418.02

1330.80

1242.46

0.00E+00

1.62E+04

8.66E+05

4.66E+06

1.04E+07

1.79E+07

2.73E+07

3.53E+07

4.69E+07

5.62E+07

6.62E+07

0.00E+00

2.29E+09

1.96E+10

4.63E+10

7.32E+10

9.50E+10

1.20E+11

1.39E+11

1.57E+11

1.68E+11

1.79E+11

Plot 𝑝

𝑧 on the y-axis versus 𝐺𝑝 −

𝑊𝑒

𝐵𝑔 on the x-axis and draw the best-fit line through the

set of points, see next Fig. The equation for the best fit is:

𝑝



(𝐺𝑝 −𝑊𝑒

𝐵𝑔) = 2646.5 − 7.8674 (𝐺𝑝 −

𝑊𝑒

𝐵𝑔)


𝐵𝑔) =

(2646.5 − 𝑝𝑧)

7.8674

Setting 𝑝


𝐺𝑝 =(2646.5 −

𝑝𝑧)

7.8674= 336.39 MMM SCF

This value is much closer to the volumetric calculated value.

D. Discussion of results - The amount of water influx was calculated on the assumption

that the volumetric calculation of the gas in place is valid. However, a more elaborate

technique for water influx calculation should be used. Such techniques will be

covered in the "Water Influx" chapter.

35


Water Influx and Water and Rock Compressibilities: p/z

plot


Water compressibility, 𝑐𝑤 3 x 10-6

Formation compressibility, 𝑐𝑓 4 x 10-6

Areal extent, A 8870 acres

Formation thickness, h 32.5 ft

Porosity, φ 30.8 %

Initial water saturation, 𝑆𝑤𝑖 42.5 %

Gas deviation factor at standard conditions, z 1.0

Performance History

Time

years

Gp

MMM SCF

p

psia

z Bg

bbl/SCF

0

2

4

6

8

10

12

14

16

18

20

0

2.305

20.257

49.719

80.134

105.930

135.350

157.110

178.300

192.089

205.744

2333

2321

2203

2028

1854

1711

1531

1418

1306

1227

1153

0.882

0.883

0.884

0.888

0.894

0.899

0.907

0.912

0.921

0.922

0.928

0.001172

0.001180

0.001244

0.001358

0.001496

0.001630

0.001820

0.001995

0.002187

0.002330

0.002495

y = -7.8674x + 2646.5

0

500

1000

1500

2000

2500

3000

0 50 100 150 200

p/z

Gp - We/Bg, MMM SCF

36

1. Estimate the initial gas in place by the MB calculation with water influx and water

and rock compressibilities

2. Discuss the results


The volumetric calculation of the initial gas in place yields:

𝐺 =43,560 𝐴 ℎ ∅ (1 − 𝑆𝑤)

𝐵𝑔𝑖=43,560(8870)(32.5)(0.308)(1 − 0.425)

(0.001172)(5.6146)

= 338 MMM SCF

1. Prepare the following table:

𝑝

𝑧 𝑊𝑒 = 𝐺𝑝𝐵𝑔 − 𝐺(𝐵𝑔

− 𝐵𝑔𝑖), 𝑏𝑏𝑙

𝐺𝑝 −𝑊𝑒/𝐵𝑔 𝑝

𝑧(1 − �̂�𝑒∆𝑝𝑡)

2645.12

2628.54

2492.08

2283.78

2073.83

1903.23

1687.98

1554.82

1418.02

1330.80

1242.46

0.00E+00

1.62E+04

8.66E+05

4.66E+06

1.04E+07

1.79E+07

2.73E+07

3.53E+07

4.69E+07

5.62E+07

6.62E+07

0.00E+00

2.29E+09

1.96E+10

4.63E+10

7.32E+10

9.50E+10

1.20E+11

1.39E+11

1.57E+11

1.68E+11

1.79E+11

2645.12

2628.25

2489.11

2277.39

2064.71

1892.37

1675.56

1541.77

1404.66

1317.30

1229.01

Plot 𝑝

𝑧(1 − �̂�𝑒∆𝑝𝑡) on the y-axis versus 𝐺𝑝 −𝑊𝑒/𝐵𝑔 on the x-axis and draw the best-fit

line through the set of points, see next Fig. The equation for the best fit is:

𝑝

𝑧(1 − �̂�𝑒∆𝑝𝑡) =

𝑝𝑖𝑧𝑖−



𝐵𝑔) = 2643.8 − 7.9153 (𝐺𝑝 −

𝑊𝑒

𝐵𝑔)


𝐵𝑔) =

(2643.8 − 𝑝𝑧)

7.9153

Setting 𝑝


𝐺𝑝 =2643.8

7.9153= 334.0 MMM SCF

This value is much closer to the volumetric calculated value.

37

2. Discussion of results:

The amount of water influx was calculated on the assumption that the volumetric

calculation of the gas in place is valid. However, a more elaborate technique for water

influx calculation should be used. Such techniques will be covered in the "Water Influx"

chapter.

y = -7.9153x + 2643.8

0

500

1000

1500

2000

2500

3000

0 50 100 150 200

p/z

(1 -

Ce

Dp

)

Gp - We/Bg, MMM SCF

38

Example #8: Material Balance Calculations for the Anderson "L"

Abnormally High-Pressure Gas Reservoir (South Texas)

with Water Influx and Water and Rock



water compressibility, 𝑐𝑤 3 x 10-6

formation compressibility, 𝑐𝑓 15 x 10-6

porosity, φ 20 %

connate water saturation, 𝑆𝑤𝑐 35 %

reservoir temperature 267 °F

initial pressure 9507 psia

initial gas formation volume factor 0.5537 RB/MCF

depth -11167 ft

original gas in place 69 BCF by volumetric method

Performance History: Table 1 SPE Paper-16484

Time

Days

p

psia

z Gp

BCF

GLp = GLC

MSTB

0.0

69.0

182.0

280.0

340.0

372.0

455.0

507.0

583.0

628.0

663.0

804.0

987.0

1183.0

1373.0

1556.0

9507

9292

8970

8595

8332

8009

7603

7406

7002

6721

6535

5764

4766

4295

3750

3247

1.4398

1.4196

1.3895

1.3545

1.3300

1.2999

1.2624

1.2443

1.2074

1.1819

1.1652

1.0969

1.0135

0.9778

0.9391

0.9100

0.000

0.393

1.642

3.226

4.260

5.504

7.538

8.749

10.509

11.758

12.789

17.262

22.890

28.144

32.567

36.820

0.00

29.90

122.90

240.90

317.10

406.90

561.20

650.80

776.70

864.30

939.50

1255.30

1615.80

1913.40

2136.00

2307.80

39

PVT Data: Table 3A - SPE Paper-22921

p

psia

Bw

bbl/STB

Rsw

scf/STB

Zg

Bg

ft3/SCF

Btw

bbl/STB

Ctw

10-6 psi-1

9510

9000

8000

7000

6000

5000

4000

3000

2000

1500

1000

750

500

250

100

14.7

1.0560

1.0569

1.0586

1.0604

1.0621

1.0638

1.0654

1.0669

1.0681

1.0686

1.0691

1.0692

1.0693

1.0694

1.0694

1.0694

31.8

31.0

29.2

27.2

25.0

22.5

19.6

16.1

11.8

9.3

6.5

5.0

3.3

1.6

0.5

0.0

1.4401

1.3923

1.2991

1.2072

1.1176

1.0325

0.9562

0.8977

0.8744

0.8832

0.9078

0.9258

0.9472

0.9708

0.9835

1.0000

0.00282

0.00288

0.00303

0.00322

0.00347

0.00385

0.00446

0.00558

0.00815

0.01098

0.01693

0.02302

0.03533

0.07242

0.18341

1.26860

1.056

1.057

1.060

1.063

1.066

1.070

1.075

1.083

1.097

1.113

1.145

1.179

1.249

1.459

2.092

8.254

2.40

2.43

2.51

2.65

2.78

2.98

3.28

3.86

5.19

6.69

9.95

13.30

20.24

41.20

104.23

717.86

40

Example #9: Material Balance Calculations for the Anderson "L"

Abnormally High-Pressure Gas Reservoir (South Texas)

with Water Influx and Water and Rock



water compressibility, 𝑐𝑤 3.1 x 10-6

formation compressibility, 𝑐𝑓 3.2 x 10-6

water viscosity 0.33 cp

porosity, φ 30.6 %

connate water saturation, 𝑆𝑤𝑐 13.9 %




gas radius 13000 ft

aquifer permeability 571 md

water encroachment angle 180 ° original gas in place 806 BCF by volumetric method

theoretical dimensionless time coefficient 12.78 year-1

theoretical radial aquifer constant 10740 RB/psi


Time

Days

p

psia

z Gp

BCF

GLp = GLC

MSTB

Wp

MSTB

0

283

452

799

1296

2172

2607

2966

3298

3663

4028

4390

4767

5156

5414

5504

5579

5809

5940

6293

4243

4165

4108

3935

3760

3538

3433

3408

3331

3178

3138

3100

3054

3078

2933

2917

2897

2875

2807

2496

0.9718

0.9676

0.9651

0.9566

0.9486

0.9395

0.9357

0.9348

0.9322

0.9275

0.9264

0.9253

0.9242

0.9248

0.9214

0.9211

0.9207

0.9202

0.9190

0.9146

0.000

23.451

29.324

57.343

102.318

165.984

190.122

226.208

263.241

297.909

327.622

352.843

370.186

387.236

417.052

424.861

433.398

449.844

465.379

521.950

0.00

129.58

150.52

214.96

338.86

534.18

603.17

698.68

808.73

902.81

1001.46

1070.02

1114.92

1165.00

1251.62

1272.07

1291.01

1330.60

1359.84

1457.16

0.00

0.00

0.72

9.42

27.54

56.95

62.65

78.22

93.05

118.55

173.58

201.13

207.46

240.77

307.01

317.04

343.42

439.21

457.99

667.32

41

6501

6807

6875

2355

2300

2291

0.9142

0.9143

0.9143

549.323

575.786

581.104

1493.04

1524.69

1531.62

861.46

1153.86

1245.00


Infinite Linear Water Influx: p/z plot


water compressibility, 𝑐𝑤 3 x 10-6


porosity, φ 30.8 %

connate water saturation, 𝑆𝑤𝑐 42.5 %





aquifer thickness 32.5

original gas in place 337.7 BCF by volumetric method


Time

Days

p

psia

z Gp

BCF

0

2

4

6

8

10

12

14

16

18

20

2333

2321

2203

2028

1854

1711

1531

1418

1306

1227

1153

0.8829

0.8843

0.8849

0.8893

0.8956

0.9005

0.8997

0.9134

0.9223

0.9231

0.9289

0.000

2.305

20.257

49.719

80.134

105.930

135.350

157.110

178.300

192.089

205.744

42

Example #11: Material Balance Calculations for a Hypothetical

Water Drive Reservoir with a High-Dip Angle: p/z plot



water viscosity 0.25 cp

porosity, φ 20 %

connate water saturation, 𝑆𝑤𝑐 25 %

residual gas saturation, 𝑆𝑔𝑟 25 %


gas gravity 0.6

gas radius 8786 ft


effective permeability to water at 𝑆𝑔𝑟 125 md

aquifer thickness 32.5

water encroachment angle 30 ° dip angle 30 ° original gas in place 20 BCF by volumetric method

aquifer radius/gas radius ratio, ra/rg 5

theoretical dimensionless time coefficient 230.1 year-1

theoretical radial aquifer constant 187.2 RB/psi


Time

Days

Gp

BCF

0.0

182.5

365.0

547.5

730.0

912.5

1095.0

1277.5

1460.0

1642.5

1825.0

2007.5

2190.0

2372.5

2555.0

2737.5

2920.0

3102.5

3285.0

3467.5

0.000

0.730

1.460

2.190

2.920

3.650

4.380

5.110

5.840

6.570

7.300

8.030

8.760

9.490

10.220

10.950

11.680

12.410

13.140

13.870

43

Example #12: Tarner Term Project

Reservoir Data

Initial Reservoir Pressure 4500 Psia

Bubble Point Pressure 3000.0 Psia

Average Reservoir Porosity 0.2000

Average Reservoir Permeability 300.00 Md

Reservoir Thickness 40.0 Feet

Well Spacing 80.0 Acres

Formation Compressibility 0.30000e-05 1/Psi

Initial Water Saturation 0.2500

Abandonment Pressure 50.0 Psia

Laboratory Relative Permeability Data

Sg Krg

0.46

0.45

0.44

0.43

0.42

0.41

0.4

0.39

0.38

0.37

0.36

0.35

0.34

0.33

0.32

0.31

0.3

0.29

0.28

0.27

0.26

0.25

0.24

0.23

0.22

0.21

0.2

0.19

0.443236

0.239504

0.171666

0.137734

0.117306

0.103579

0.093636

0.086018

0.079912

0.074832

0.070467

0.066611

0.063121

0.059896

0.056863

0.053967

0.05117

0.04844

0.045756

0.043103

0.04047

0.03785

0.03524

0.032639

0.030049

0.027474

0.024922

44

0.18

0.17

0.16

0.15

0.14

0.13

0.12

0.11

0.1

0.09

0.08

0.022401

0.01992

0.017493

0.015133

0.012855

0.010677

0.008617

0.006696

0.004936

0.00336

0.001994

0.000864

So Kro

0.29

0.3

0.31

0.32

0.33

0.34

0.35

0.36

0.37

0.38

0.39

0.4

0.41

0.42

0.43

0.44

0.45

0.46

0.47

0.48

0.49

0.5

0.51

0.52

0.53

0.54

0.55

0.56

0.57

0.58

0.59

0.6

0.61

0.000638

0.001449

0.002458

0.003693

0.005184

0.006963

0.009064

0.011525

0.014385

0.017687

0.021475

0.025798

0.030707

0.036254

0.042497

0.049495

0.057313

0.066016

0.075675

0.086362

0.098155

0.111135

0.125385

0.140994

0.158054

0.17666

0.196914

0.218919

0.242784

0.268621

0.296548

0.326686

0.359162

0.394106

45

0.62

0.63

0.64

0.65

0.431654

0.471946

0.515128

Properties Used to Generate the PVT Tables from Correlations

Bubble Point Pressure 3000.0 Psia

Reservoir Temperature 190.0 °F

Oil Gravity 35.00 °API

Gas Gravity 0.7000 (Air=1)

Separator Conditions Pressure = 114.7 psi, Temperature = 75.0 °F

Standard Conditions Pressure = 14.70 psi, Temperature = 60.0 °F

Percent of other gases in the gas mixture CO2 = 0.0%, H2S = 0.0%, N2 = 0.0%

Is the gas a condensate gas No

Salinity Of The Water 20.0000 G/L

PVT DATA P

psi

Bo

RB/STB

Rso

RB/STB

μo

cp

Bg

Ft3/SCF

μg

cp

Cw

1/psi

4600

4550

4500

4450

4400

4350

4300

4250

4200

4150

4100

4050

4000

3950

3900

3850

3800

3750

3700

3650

3600

3550

3500

3450

3400

3350

1.35766

1.35771

1.35775

1.3578

1.35784

1.35789

1.35794

1.35799

1.35804

1.35809

1.35815

1.3582

1.35826

1.35832

1.35837

1.35844

1.3585

1.35856

1.35863

1.3587

1.35876

1.35884

1.35891

1.35899

1.35906

1.35914

606.024

606.024

606.024

606.024

606.024

606.024

606.024

606.024

606.024

606.024

606.024

606.024

606.024

606.024

606.024

606.024

606.024

606.024

606.024

606.024

606.024

606.024

606.024

606.024

606.024

606.024

0.63678

0.63326

0.62978

0.62632

0.62289

0.6195

0.61613

0.61279

0.60948

0.60621

0.60297

0.59976

0.59658

0.59343

0.59032

0.58724

0.5842

0.58119

0.57822

0.57528

0.57238

0.56952

0.56669

0.5639

0.56115

0.55844

0.003748

0.003774

0.003801

0.003829

0.003858

0.003887

0.003917

0.003948

0.00398

0.004014

0.004048

0.004083

0.004119

0.004157

0.004196

0.004236

0.004278

0.004321

0.004365

0.004411

0.004459

0.004509

0.00456

0.004614

0.004669

0.004726

0.079969

0.079442

0.078912

0.078379

0.077844

0.077307

0.076767

0.076225

0.07568

0.075133

0.074584

0.074032

0.073479

0.072923

0.072366

0.071806

0.071245

0.070682

0.070117

0.069551

0.068984

0.068416

0.067846

0.067276

0.066704

0.066132

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

46

3300

3250

3200

3150

3100

3050

3000

2950

2900

2850

2800

2750

2700

2650

2600

2550

2500

2450

2400

2350

2300

2250

2200

2150

2100

2050

2000

1950

1900

1850

1800

1750

1700

1650

1600

1550

1500

1450

1400

1350

1300

1250

1200

1150

1100

1050

1000

1.35923

1.35931

1.3594

1.35949

1.35958

1.35968

1.35978

1.35409

1.34841

1.34275

1.33711

1.33149

1.32589

1.32031

1.31475

1.3092

1.30368

1.29818

1.2927

1.28724

1.2818

1.27639

1.27099

1.26562

1.26028

1.25495

1.24965

1.24438

1.23913

1.23391

1.22871

1.22354

1.2184

1.21328

1.2082

1.20314

1.19812

1.19312

1.18816

1.18323

1.17834

1.17348

1.16865

1.16387

1.15912

1.15441

1.14975

606.024

606.024

606.024

606.024

606.024

606.024

606.024

594.053

582.121

570.227

558.372

546.556

534.78

523.046

511.352

499.701

488.092

476.526

465.005

453.528

442.097

430.712

419.375

408.085

396.845

385.655

374.515

363.428

352.393

341.413

330.488

319.619

308.809

298.058

287.367

276.739

266.175

255.676

245.245

234.884

224.594

214.377

204.237

194.176

184.196

174.3

164.493

0.55577

0.55314

0.55055

0.548

0.54549

0.54303

0.54061

0.54655

0.55264

0.5589

0.56532

0.57192

0.57871

0.58568

0.59286

0.60024

0.60784

0.61567

0.62374

0.63206

0.64064

0.6495

0.65865

0.6681

0.67786

0.68797

0.69843

0.70926

0.72048

0.73212

0.74419

0.75673

0.76977

0.78332

0.79743

0.81212

0.82745

0.84343

0.86013

0.87758

0.89585

0.91498

0.93504

0.9561

0.97823

1.00151

1.02603

0.004786

0.004848

0.004913

0.00498

0.005051

0.005124

0.0052

0.005279

0.005363

0.005449

0.00554

0.005635

0.005734

0.005838

0.005947

0.006061

0.006181

0.006306

0.006438

0.006577

0.006723

0.006877

0.00704

0.007211

0.007391

0.007582

0.007784

0.007997

0.008224

0.008464

0.008719

0.00899

0.009279

0.009587

0.009916

0.010267

0.010644

0.011049

0.011485

0.011954

0.012461

0.013011

0.013608

0.014259

0.014971

0.015753

0.016615

0.06556

0.064988

0.064415

0.063843

0.063271

0.0627

0.062129

0.06156

0.060992

0.060426

0.059862

0.0593

0.058741

0.058185

0.057632

0.057082

0.056536

0.055994

0.055457

0.054925

0.054398

0.053876

0.05336

0.052851

0.052348

0.051852

0.051363

0.050881

0.050407

0.049941

0.049483

0.049034

0.048594

0.048162

0.04774

0.047327

0.046923

0.046529

0.046144

0.045769

0.045404

0.045049

0.044704

0.044369

0.044044

0.043729

0.043424

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

47

950

900

850

800

750

700

650

600

550

500

450

400

350

300

250

200

150

100

50

10

1.14513

1.14055

1.13602

1.13154

1.12711

1.12274

1.11842

1.11417

1.10998

1.10587

1.10183

1.09787

1.09401

1.09024

1.08659

1.08308

1.07973

1.07659

1.07373

1.07183

154.776

145.155

135.633

126.216

116.908

107.715

98.645

89.704

80.902

72.248

63.754

55.436

47.31

39.399

31.732

24.348

17.305

10.694

4.697

0.695

1.05189

1.07919

1.10806

1.13862

1.17101

1.20539

1.24192

1.28079

1.32219

1.36634

1.41344

1.46372

1.51738

1.57459

1.63546

1.69992

1.76762

1.83758

1.90729

1.95758

0.017569

0.018631

0.01982

0.02116

0.02268

0.02442

0.026429

0.028775

0.03155

0.034882

0.038957

0.044053

0.050607

0.05935

0.071593

0.089961

0.12058

0.181824

0.365568

1.835563

0.043129

0.042844

0.042569

0.042304

0.042049

0.041804

0.041568

0.041344

0.041129

0.040924

0.04073

0.040546

0.040373

0.040212

0.040062

0.039924

0.039801

0.039693

0.039604

0.039555

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

3.00E-06

Production Data Pressure

psia Np

STB Gp

SCF Wp

STB Rp

SCF/STB

4500

4480

4460

4440

4420

4400

4380

4360

4340

4320

4300

4280

4260

4240

4220

4200

4180

4160

4140

4120

4100

4080

0

329

658.2

987.62

1317.24

1647.07

1977.12

2307.38

2637.86

2968.57

3299.49

3630.64

3962.02

4293.63

4625.48

4957.56

5289.87

5622.43

5955.23

6288.27

6621.57

6955.11

0

0.199

0.399

0.599

0.798

0.998

1.198

1.398

1.599

1.799

2

2.2

2.401

2.602

2.803

3.004

3.206

3.407

3.609

3.811

4.013

4.215

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

48

4060

4040

4020

4000

3980

3960

3940

3920

3900

3880

3860

3840

3820

3800

3780

3760

3740

3720

3700

3680

3660

3640

3620

3600

3580

3560

3540

3520

3500

3480

3460

3440

3420

3400

3380

3360

3340

3320

3300

3280

3260

3240

3220

3200

3180

3160

3140

7288.91

7622.96

7957.28

8291.85

8626.7

8961.81

9297.19

9632.85

9968.78

10304.99

10641.49

10978.28

11315.35

11652.72

11990.39

12328.36

12666.63

13005.21

13344.1

13683.3

14022.83

14362.68

14702.86

15043.36

15384.2

15725.38

16066.9

16408.77

16750.99

17093.57

17436.51

17779.81

18123.48

18467.52

18811.95

19156.75

19501.95

19847.54

20193.53

20539.92

20886.72

21233.94

21581.58

21929.64

22278.13

22627.06

22976.44

4.417

4.62

4.822

5.025

5.228

5.431

5.634

5.838

6.041

6.245

6.449

6.653

6.857

7.062

7.266

7.471

7.676

7.881

8.087

8.292

8.498

8.704

8.91

9.117

9.323

9.53

9.737

9.944

10.151

10.359

10.567

10.775

10.983

11.192

11.4

11.609

11.819

12.028

12.238

12.448

12.658

12.868

13.079

13.29

13.501

13.713

13.924

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

606

49

3120

3100

3080

3060

3040

3020

3000

2980

2960

2940

2920

2900

2880

2860

2840

2820

2800

2780

2760

2740

2720

2700

2680

2660

2640

2620

2600

2580

2560

2540

2520

2500

2480

2460

2440

2420

2400

2380

2360

2340

2320

2300

2280

2260

2240

2220

2200

23326.27

23676.54

24027.29

24378.5

24730.18

25082.35

25435

29798.63

34271

38848.85

43534.92

48333.86

53245.72

58274.88

63424.3

68697.21

74092.09

79623.14

85280.72

91075.74

97012.09

103079.8

109311.7

115680

122205.9

128888.4

135694.3

142666.5

149749.2

156967.5

164318.1

171759.9

179364.1

187045.2

194838.2

202735.2

210679.1

218763.1

226872

235051.2

243289.1

251513.1

259839.4

268126

276429.3

284736

292963.5

14.136

14.349

14.561

14.774

14.987

15.2

15.414

18.048

20.726

23.446

26.207

29.012

31.86

34.752

37.688

40.67

43.695

46.77

49.889

53.057

56.273

59.532

62.85

66.211

69.625

73.095

76.62

80.235

83.917

87.687

91.55

95.493

99.563

103.724

108.006

112.415

116.932

121.623

126.434

131.407

136.548

141.826

147.331

152.985

158.841

164.903

171.126

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

606

606

606

606

606

606

606

601.2

596.4

591.7

586.9

582.1

577.4

572.6

567.9

563.1

558.4

553.6

548.9

544.2

539.5

534.8

530.1

525.4

520.7

517.9

518

519

520.9

523.7

527.4

532.2

538.2

545.3

553.6

563.2

574.1

586.5

600.3

615.6

632.5

651

671.3

693.3

717.1

742.6

770

50

2180

2160

2140

2120

2100

2080

2060

2040

2020

301248.8

309430.2

317575.6

325673.5

333635.5

341616.8

349444.2

357196.1

364863.6

177.627

184.294

191.193

198.328

205.63

213.252

221.039

229.075

237.359

0

0

0

0

0

0

0

0

0

799.3

830.5

863.6

898.6

935.5

974.5

1015.3

1058.1

1102.8

material balance calculations for petroleum reservoirs · 2018-03-17 · 5 3. oil recovery by...

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