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Reservoir Material Balance Review

Reservoir Material Balance Fundamentals

This section will cover the following learning objectives:

Learning Objectives

Describe how the equations can be simplified based on certain assumptions and importance of mechanisms

Relate material balance equations to different types of reservoirs

Discuss the basic material balance equations and the assumptions

Reservoir Material Balance Fundamentals ═════════════════════════════════════════════════════════════════════════

© PetroSkills, LLC. All rights reserved._____________________________________________________________________________________________

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Why is this important?

Material balance technique provides a fundamental building block for reservoir engineering.

Estimating the initial hydrocarbons in place is important to determine what the ultimate recovery from the reservoir would be.

The initial estimates are made using volumetric analysis; hence significant uncertainty exists in those estimations.

• Material balance method is the next logical step in that estimation.


Material balance technique:

Is much more reliable because it is based on available production data.

Provides us with connected volume of hydrocarbons.

Is a fundamental building block for reservoir simulation models.

Provides an anchor point to understand a good estimate of initial oil in place.

• This initial estimate ensures that some of the unrealistic solutions are automatically eliminated from further consideration.



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Learning this module would allow an engineer to evaluate any type of field for estimating initial hydrocarbons in place.

The module covers all the five types of reservoirs: 1. Dry gas

2. Wet gas

3. Retrograde condensate

4. Volatile oil

5. Black oil

The module also covers the influence of water influx and shows how to match the performance of the reservoir in the presence of water influx.

This module would allow an engineer to estimate the initial hydrocarbon in place using a simple Microsoft Excel sheet.

The initial estimate of hydrocarbons in place can be used as stand-alone information for predicting the ultimate recovery of hydrocarbons using recovery factor, or it can be used as an input in reservoir simulation model.



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Method, Equation, Simplification

Material Balance Method

Material Balance is:

A “zero” dimensional equation. This means that it assumes that no gradients of any kind exist in the reservoir

Useful for calculating the initial amount of hydrocarbons in place. It is not useful to predict the performance of the reservoir to time

Application of material balance requires following data as a function of time:

1. Average reservoir pressure

2. Cumulative production of oil, gas and water

3. Fluid and rock properties as a function of pressure and temperature

Average pressure is the most difficult to obtain, followed by cumulative production of water and gas respectively.



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Material Balance Equation

Recall that we covered the generalized material balance equation as follows:

The blue portion of the equation represents the net withdrawal of gas, oil and water respectively. The black portion of the equation represents the five mechanisms which contribute to the production. Those mechanisms are respectively, gas expansion, oil expansion, water expansion, formation compaction and water influx respectively

This material balance equation is applicable to any type of reservoir and for any mechanism

1. Although material balance equation is complex, in many situations, the equation can be simplified depending on the type of reservoir from which the hydrocarbons are produced and the type of mechanisms which are important to consider in producing those hydrocarbons.

2. In practice, it is sometimes difficult to know which terms in material balance are important

and which are not. Depending on the assumptions made, you will get different results of hydrocarbons in place.

3. The biggest uncertainty in material balance calculations comes from prediction of water

influx. Water influx is a function of time, so it varies as the reservoir pressure declines. Further, it is difficult to measure. In addition, aquifer properties are rarely known or measured.

4. In practice, the most common application of material balance is for single phase gas and

black oil reservoirs. Although the material balance equation is applicable to other types of reservoirs including condensate and volatile oil types of reservoirs, compositional simulation is often used to describe and model those types of reservoirs.

5. Under certain situations, material balance equation can be re-cast as a straight-line

equation and by plotting certain graphs, the initial oil in place or gas in place can be easily calculated. Havlena Odeh method is valuable to achieving such an objective.



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Material Balance Simplification

For single phase gas reservoir, we can write material balance equation as:

This simplification is achieved by assuming that there is no volatile oil in gas phase, and there is no free oil initially present in the reservoir, and compressibility’s of the formation and water are independent of pressure

As we will illustrate, this equation can be further simplified and applied

For black oil reservoir, we can simplify the equation as:

This simplification is achieved by assuming that there is no volatile oil in gas phase

As we will illustrate, this equation can be further simplified and applied in certain situations

Similar to single phase gas and black oil reservoirs, we can also write simplified forms for other types of reservoirs

1



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Dry and Wet Gas Reservoirs



Learning Objectives

Describe the application of material balance equation for gas reservoirs

Consider the simplifications of material balance equation for absence or presence of different mechanisms

Evaluate the uncertainties associated with mis-characterization of different mechanisms

Apply various straight line manipulations for determining the gas in place for gas reservoirs



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Single Phase Gas Reservoir

For typical reservoir properties, material balance technique is extremely popular for determining the gas in place for single phase gas reservoirs.

Part of the reason for such popularity is the quick equilibration time required for gas phase. It is easier to obtain representative average pressure for gas reservoirs and hence appropriate application of material balance.

We will not consider the water influx in this section. Water influx will be discussed separately after we cover other mechanisms.



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Material Balance Equation

We can write the generalized material balance equation for gas reservoir as:

1

1

If we neglect water influx, we can simplify this equation as:

1

1

Havlena Odeh formulation allows us to write this equation as:

,,

Where F ;Eg

and Ef,w

Volumetric Expansion

If we assume that gas compressibility is much more significant than either formation or water compressibility, we can neglect the impact of Ef,w and write the material balance equation as:

)

This equation has several alternate straight line formulations:

You can plot F vs. Eg to obtain gas in place which would be slope of the straight line

You can plot F/Eg vs. cumulative gas produced, or average pressure, or time and the intercept of y axis can provide the gas in place

You can ignore water production and re-write the equation as

1⁄

where a plot of Gp vs. p/z would provide a

straight line with intercept on x axis as gas in place



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Example 1

Following data are collected on a gas reservoir. The specific gravity of gas is 0.666 and the reservoir temperature is 112 F [44.4 C]. Assume volumetric expansion only and determine the gas in place:

psia kPa MMSCF MSm3

1,556 10,728 0 01,543 10,639 19 538 1,538 10,604 45 1,274 1,504 10,370 108 3,058 1,443 9,949 214 6,060 1,432 9,873 275 7,787 1,423 9,811 319 9,033 1,360 9,377 427 12,091 1,274 8,784 522 14,781

GpPressure



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z Bg Eg

psia kPa MMSCF MSm3

MMft3

Mm3

MMSCF MSm3

psia kPa1,556 10,728 0 0 0.808 0.008406 0 0 0 1,926 13,278 1,543 10,639 19 538 0.809 0.008487 0.161 4.57 8.13E-05 1,983 56,158 1,907 13,150 1,538 10,604 45 1,274 0.8094 0.008519 0.383 10.86 0.000113 3,389 95,969 1,900 13,101 1,504 10,370 108 3,058 0.8121 0.008741 0.944 26.73 0.000335 2,820 79,851 1,852 12,769 1,443 9,949 214 6,060 0.8173 0.009168 1.962 55.56 0.000763 2,573 72,856 1,766 12,173 1,432 9,873 275 7,787 0.8183 0.00925 2.544 72.03 0.000844 3,013 85,315 1,750 12,066 1,423 9,811 319 9,033 0.8191 0.009318 2.972 84.17 0.000912 3,259 92,298 1,737 11,978 1,360 9,377 427 12,091 0.8249 0.009818 4.192 118.72 0.001413 2,968 84,043 1,649 11,367 1,274 8,784 522 14,781 0.8333 0.010588 5.527 156.51 0.002182 2,533 71,722 1,529 10,541

p/zF/EgPressure Gp F

Solution: Example 1

Knowing the pressure and the temperature and the specific gravity, we can calculate the z factor as well as Bg.

The following table is constructed:



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Solution: Example 1 (Approach 1)

0

20

40

60

80

100

120

140

160

180

0

1

2

3

4

5

6

0 0.001 0.002 0.003

F [

MS

M3 ]

F (

MM

ft3 ]

Eg

The graph on the right shows the first approach where the slope provides gas in place.

The field data exhibits some uncertainty; so all the data points do not fall perfectly on the straight line. This is an indication of some uncertainty in estimating gas in place.

The slope is equal to 2.75 BCF [78 MM Sm3].



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-

57,500

115,000

0

1000

2000

3000

4000

0 200 400 600

F/E

g, M

Sm

3

F/E

g, M

MS

CF

Gp

The second approach shows more variability (more uncertainty) but no particular trend

The uncertainty in data are accentuated in this graph because y axis represents a denominator which is a “difference” between current and initial values

The intercept on y axis is equal to 2.75 BCF [78 MM Sm3]

Difference – in general –always magnifies the measurement errors



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y = -0.7021x + 1930

-

8,500

17,0000 5000 10000 15000 20000

0

500

1000

1500

2000

2500

0 200 400 600

p/z

, kP

a

Gp, MSm3

p/z

, psi

a

Gp, MMSCF

The third approach shows that data points fall nicely on the straight line

The intercept on x axis is equal to 2.75 BCF [78 MM Sm3]

Compared to other two approaches this approach is much more resistant to uncertainties in the measurements

The reason is that the data are plotted as absolute magnitude rather than “difference”; so less sensitive to measurement errors



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Observations

Among the three approaches used to evaluate the production data for gas reservoirs, the most robust approach (which is resistant to errors in measurements) is p/z plot. The flip side of this robustness is lack of sensitivity. If indeed the data indicates potential presence of other mechanisms, it would be difficult to see using p/z graph. A better approach would be to select F/Eg graph to indicate the presence of other mechanisms.

If the reservoir is influenced by gas expansion only, every approach provides the same result.

p/z method is relatively insensitive to other mechanisms and may provide a straight line even though other mechanisms may be important.

F/Eg method is much more sensitive to other mechanisms and as seen in the plot on the right-hand side, provide clues as to importance of other mechanisms.



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Effect of Formation Compressibility

Example

A reservoir with 2,000 psi [13,790 kPa] may have gas compressibility which is approaching 500 times 10 to the minus 6 psi [41 kPa] inverse. And if the raw compressibility is 3 times 10 to the minus 6, we're talking about two order of magnitude difference between gas compressibility and formation compressibility. Under those conditions, whether we include the formation compressibility or not is not going to make any difference.

Exceptions to the generalized rule

If you have a reservoir which is significantly abnormally pressured; that is the pressure exceeds hydrostatic pressure, or if the reservoir is producing from unconsolidated formations, in a lot of those situations the formation compressibility can be significantly higher.

if your reservoir pressure is greater, the gas compressibility decreases. So, one hand the gas compressibility is decreasing; on the other hand, the formation compressibility is increasing.

For gas reservoirs, gas compressibility in most cases is greater than formation compressibility by several orders of magnitude. The effect of formation compressibility on material balance is considered negligible under those conditions.



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Observations

Example

If you had used cf equal to 5x10-6 psi-1 [7.26x10-7 kPa-1], the gas in place using material balance would have been 3.3 TCF [93.5 BSm3]. This value is significantly different than what was estimated by assuming 13.8x10-6 psi-1[2x10-6 kPa-1] formation compressibility value.

For high pressure gas, the gas compressibility is lower; hence the importance of formation compressibility is greater.

In this reservoir, knowledge of formation compressibility is critical to correctly assess the gas in place.

P/z method is not appropriate since it is going to overestimate the gas in place.

Depending on the selection of formation compressibility, the estimation of gas in place can vary significantly.



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Example 2

The water compressibility is assumed to be 3x10-6 psi-1 [4.35x10-7 kPa-1] and the average water saturation in the reservoir is 0.4. The specific gravity of gas is 0.59.

Calculate the gas in place by neglecting the compressibility of the formation and water

Calculate the gas in place by incorporating the compressibility of the formation and water

Explain the percentage difference in the two values

An early production data were collected from abnormally pressured reservoir. The gas reservoir is deep and the initial reservoir pressure is 17,800 psia [122.9 MPa]. The reservoir temperature is 347 F [172 C]. The formation compressibility has some uncertainty and the value ranges from 5x10-6 psi-1 [7.26x10-7 kPa-1] to 13.8x10-6 psi-1 [2x10-6 kPa-1].

psia kPa MMSCF MMSm3

17,819 122,860 0 017,683 121,920 13,727 388.7017,629 121,550 19,020 538.5917,386 119,870 46,015 1303.0017,056 117,600 77,350 2190.32

Pressure Gp



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Example 2: Solution

You can determine the gas in place by first calculating the values of z factor and formation value factor. This table is generated after estimating z factor and Bg. This table ignores the affect of formation compressibility.

z Bg Eg

psia Mpa MMSCF MMSm3 MMft3 MMm3 MMSCF MMSm3 psia Mpa17,819 122.8 - - 1.884 0.002402 0 0 0 9,459 65.2 17,683 121.8 13,727 389 1.875 0.002409 33.072 0.94 6.951E-06 4,758,024 134,836 9,432 65.0 17,629 121.5 19,020 539 1.871 0.002412 45.877 1.30 9.715E-06 4,722,396 133,826 9,421 64.9 17,386 119.8 46,015 1,304 1.855 0.002425 111.578 3.16 2.247E-05 4,965,642 140,719 9,372 64.6 17,056 117.5 77,350 2,192 1.833 0.002443 188.937 5.35 4.026E-05 4,693,094 132,996 9,304 64.1

Pressure Gp F F/Eg p/z


Data points fall on the straight line

The slope is equal to 4.75 TCF [135 B Sm3]

The straight line shows very little uncertainty



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- 2,000 4,000

85,000

95,000

105,000

115,000

125,000

135,000

145,000

155,000

3000000

3500000

4000000

4500000

5000000

5500000

- 50,000 100,000

Gp

F/E

g, M

Mm

3

F/E

g, M

MS

CF

Gp

In this graph, there is more uncertainty in predicting the straight line

The intercept on y axis is equal to 4.75 TCF BCF [135 B Sm3]

The difference between this answer and previous answer is very small


55.0

60.0

65.0

70.0 - 1,000 2,000 3,000

8,000 8,200 8,400 8,600 8,800 9,000 9,200 9,400 9,600 9,800

10,000

- 50,000 100,000

p/z

, Mp

a

Gp, MMSm3

p/z

, psi

a

Gp, MMSCF

The third approach shows that data points fall nicely on the straight line

The intercept on x axis is equal to 4.73BCF [134 B Sm3]

The results are consistent with the first approach although slightly lower

All data points do fall on the straight line indicating very little uncertainty



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Solution: Example 2

If we decide to include the compressibility of the formation and water, we will need to calculate the value of Ef,w. Recall that

1, 1

For example, if we assume the first pressure data in the table, we can calculate the value as

13.8 10 3 10 0.4 17,819 17,683 8.177 10,0.0024021 0.4

13.8 10 3 10 0.4 17,819 17,683 = 8.177 10

Since this is a dimensionless quantity, we will get the same value in SI units.

z Bg Eg Ef,w Eg+Ef,w

psia kPa MMSCF MSm3 MMft3 Mm3 MMSCF MSm3

17,819 122,860 - - 1.884 0.002402 0 0 0 0 017,683 121,920 13,727 389 1.875 0.002409 33.072 0.94 6.951E-06 8.177E-06 1.513E-05 2,186,149 61,952 17,629 121,550 19,020 539 1.871 0.002412 45.877 1.30 9.715E-06 1.14E-05 2.111E-05 2,173,176 61,585 17,386 119,870 46,015 1,304 1.855 0.002425 111.578 3.16 2.247E-05 2.601E-05 4.848E-05 2,301,501 65,221 17,056 117,600 77,350 2,192 1.833 0.002443 188.937 5.35 4.026E-05 4.576E-05 8.602E-05 2,196,526 62,246

Pressure Gp F F/(Eg+Ef,w)

Example 2: Solution

This table can be constructed.

We cannot use p/z method since it is restricted to gas expansion only.



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Solution: Example 2 (Approach 1)(Formation and Water Expansion)

0

1

2

3

4

5

6

0

50

100

150

200

250

0 0.00002 0.00004 0.00006 0.00008 0.0001

F [

MM

SM

3 ]

F (

MM

ft3 ]

Eg + Ef,w

Data points fall on the straight line

The slope is equal to 2.22 TCF [62.9 B Sm3]

The straight line shows very little uncertainty

The difference between with and without formation compressibility is almost 100% indicating importance of formation compressibility in this calculation

Solution: Example 2 (Approach 2)(Formation and Water Expansion)

- 2,000 4,000

25,000

35,000

45,000

55,000

65,000

75,000

85,000

1000000

1400000

1800000

2200000

2600000

3000000

- 50,000 100,000

Gp

F/(

Eg

+ E

f,w),

MM

m3

F/(

Eg

+ E

f,w),

MM

SC

F

Gp

In this graph, there is more uncertainty in predicting the straight line

The intercept on y axis is equal to 2.23TCF BCF [62 B Sm3]

Similar to previous approach, the difference with and without considering formation compressibility is almost 100%



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Observations

Example

If you had used cf equal to 5x10-6 psi-1 [7.26x10-7 kPa-1], the gas in place using material balance would have been 3.3 TCF [93.5 BSm3]. This value is significantly different than what was estimated by assuming 13.8x10-6 psi-1[2x10-6 kPa-1] formation compressibility value.

For high pressure gas, the gas compressibility is lower; hence the importance of formation compressibility is greater.

In this reservoir, knowledge of formation compressibility is critical to correctly assess the gas in place.

P/z method is not appropriate since it is going to overestimate the gas in place.

Depending on the selection of formation compressibility, the estimation of gas in place can vary significantly.



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This section has covered the following learning objectives:

Learning Objectives

Describe the application of material balance equation for gas reservoirs

Consider the simplifications of material balance equation for absence or presence of different mechanisms

Evaluate the uncertainties associated with mis-characterization of different mechanisms

Apply various straight line manipulations for determining the gas in place for gas reservoirs



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Black Oil Reservoirs



Learning Objectives

Recognize the important drive mechanisms for black oil reservoirs

Estimate the oil in place in oil reservoirs when the reservoir is above bubble point

Estimate the oil in place in oil reservoirs when the reservoir is producing below bubble point

Estimate the oil in place in oil reservoirs when the reservoir is influenced by gas cap

Quantify the uncertainties in oil in place based on the assumptions in the strength of drive mechanisms



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Black Oil Reservoir

We start with original material balance and eliminate terms from that equation

For black oil reservoir, the key assumption we make is Rv is zero

Simplifying, we obtain:

Black Oil Reservoir

Substituting, we obtain

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In addition, we can combine the water compressibility and formation compressibility into a single equation where the pore volume can be written as

11

1

We can re-write black oil model by assuming that the ratio of initial gas cap size to initial oil rim size can be written equal to m. That is,



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Black Oil Reservoir Considerations

The equation for black oil model can be further simplified depending on the important mechanisms which are influencing the production of the hydrocarbons. We will consider three important cases:

Reservoir producing above bubble point and without a gas cap

Reservoir producing below bubble point and without a gas cap

Reservoir producing below bubble point and with a gas cap



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If we assume that oil reservoir is producing at a pressure above bubble point with no gas cap, we can write the equation as:

1

Since the production is above bubble point, ; substituting

1

If we assume, no water injection, no gas injection, no water production and no water influx, we can write the equation as:

1

In general, when oil is producing above bubble point, we cannot ignore formation compressibility (why?), so this is the simplest form we can use.

Black Oil Reservoir (Above Bubble Point)



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Havlena and Odeh Method

For single phase oil above bubble point and without water influx, we can write:

1 ,

Where , =

By plotting F vs. Eo Ef,w should yield a straight line with a slope equal to Nfoi

Alternately, by plotting ,

versus any time varying parameter (oil

produced, pressure, time) we should be able to see a horizontal line with an intercept on y axis equal to

Note: The definition of Ef,w is different for oil reservoir than

gas reservoir. For gas reservoir, Boi is replaced by Bgi.



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Example 3

Calculate the oil in place by incorporating the compressibility of the formation and water

Calculate the oil in place by neglecting the compressibility of the formation and water

Which calculation among the above provide a reasonable estimate of oil in place and why?

An initially undersaturated oil reservoir has a connate water saturation of 24% at an initial pressure of 8,001 psia [55,165 kPa]. Bubble point pressure was estimated to be 4,475 psia [30,854 kPa]. The API gravity of oil is measured to be 28, the specific gravity of gas is 0.74, and the reservoir temperature is 290 F [143.3 C]. Assume cw = 3.62 x 10-6 psi-1 [5.25 x 10-7 kPa-1] andcf = 9.20 x 10-6 psi-1[1.33 x 10-6 kPa-1]. Additional production data are provided.

Example 3

psia kPa STB Sm3 STB Sm3 SCF/STB Sm3/Sm3 bbl/STB m3/Sm3 bbl/SCF m3/Sm3 bbl/STB m3/Sm3

8,001 55,165 ‐ ‐ ‐ ‐ 735 130 1.36470 1.36470 0.00077 0.00432 1.05174 1.05174

7,670 52,883 2,074,066 329,750 ‐ ‐ 735 130 1.37013 1.37013 0.00073 0.00415 1.05286 1.05286

7,341 50,614 4,483,248 712,779 ‐ ‐ 735 130 1.37567 1.37567 0.00071 0.00399 1.05399 1.05399

7,012 48,346 6,834,181 1,086,548 ‐ ‐ 735 130 1.38133 1.38133 0.00068 0.00386 1.05513 1.05513

6,683 46,078 9,238,653 1,468,828 ‐ ‐ 735 130 1.38712 1.38712 0.00066 0.00374 1.05628 1.05628

6,354 43,809 10,836,244 1,722,825 ‐ ‐ 735 130 1.39306 1.39306 0.00064 0.00364 1.05744 1.05744

6,025 41,541 12,284,003 1,953,000 150 24 735 130 1.39918 1.39918 0.00063 0.00355 1.05861 1.05861

5,695 39,266 14,529,644 2,310,029 910 145 735 130 1.40549 1.40549 0.00061 0.00346 1.05980 1.05980

5,366 36,997 16,215,173 2,578,007 2,450 390 735 130 1.41202 1.41202 0.00060 0.00339 1.06100 1.06100

5,037 34,729 19,426,843 3,088,621 5,010 797 735 130 1.41880 1.41880 0.00059 0.00332 1.06221 1.06221

4,708 32,461 22,757,625 3,618,173 6,500 1,033 735 130 1.42588 1.42588 0.00080 0.00453 1.06343 1.06343

P Np Wp Rs Bo Bg Bw



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Solution: Example 3 – Sample Calculation

Field Units

F 2,841,738 1.37013 0x1.05286

2,841,738bbl

SI Units

329,750 1.37013 0x1.05286

451,800 3

Field Units

1.37013 1.3647 735 735 0.00073

0.00543

SI Units

1.37013 1.3647 130 130 0.00415

0.00543

At P =7,670 psia,

The underground withdrawal, F can be calculated using the equation

The expansion of oil, Eo

Eo =Bo – Boi +(Rsi – Rs)Bg


Field Units

,

1.36471 0.24

3.62x10−6 0.24 9.2x10−6 (8,001−7,670)=0.00598

SI Units

,

1.36471 0.24

5.25x10−7 0.24 1.33x10−6 (55,165−52,883)=0.005983

3

The expansion of initial water and reduction in the pore volume, Ef,w

Similarly we can calculate the F, Eo and Ef,w at other average reservoir pressures.



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The expansion of initial water and reduction in the pore volume, Ef,w

Similarly we can calculate the F, Eo and Ef,w at other average reservoir pressures.

psia kPa bbl m3 bbl/STB m3/Sm3 bbl/STB m3/Sm3

8,001 55,165 ‐ ‐ ‐ ‐ ‐ ‐

7,670 52,883 2,841,738 451,800 0.00543 0.00543 0.00598 0.00598

7,341 50,614 6,167,455 980,547 0.01096 0.01096 0.01193 0.01193

7,012 48,346 9,440,228 1,500,876 0.01662 0.01662 0.01788 0.01788

6,683 46,078 12,815,114 2,037,440 0.02242 0.02242 0.02383 0.02383

6,354 43,809 15,095,586 2,400,006 0.02836 0.02836 0.02978 0.02978

6,025 41,541 17,187,671 2,732,621 0.03448 0.03448 0.03573 0.03573

5,695 39,266 20,422,136 3,246,860 0.04078 0.04078 0.04169 0.04169

5,366 36,997 22,898,533 3,640,576 0.04731 0.04731 0.04764 0.04764

5,037 34,729 27,567,822 4,382,933 0.05410 0.05410 0.05359 0.05359

4,708 32,461 32,456,231 5,160,128 0.06118 0.06118 0.05954 0.05954

P F Eo Ef,w



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The plot of Fvs. Eo Ef,w should yield a straight line with a slope equal to Nfoi

Original oil in place • FieldUnit Nfoi

254,453,022STB

• SIUnit Nfoi 40,454,798 3



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Solution: Example 3 (Without Formation Compressibility)

Original Oil In Place • Nfoi 499,848,751STB• Nfoi 79,469,601Sm3

The difference with and without formation compressibility is almost 100% indicating importance of formation compressibility in this calculation

We cannot neglect the formation compressibility since the order of magnitude of oil compressibility and formation compressibility are similar

Production data alone is unable to differentiate this error



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Original Oil In Place

Nfoi258,716STB

Nfoi41,132,559Sm3

The graph show more variability than Approach 1. This graph is more sensitive to the errors in the measurements

This method is better to understand any signature in the data



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Example 4

Calculate the original oil in place

psia kPa STB Sm3 SCF/STBSm3/Sm3 STB Sm3 SCF/STBSm3/Sm3 bbl/STB m3/Sm3 bbl/SCF m3/Sm3 bbl/STB m3/Sm3

3,516 24,242 ‐ ‐ ‐ ‐ ‐ ‐ 800 142 1.39680 1.39680 0.00072 0.00408 1.00812 1.00812

3,089 21,298 1,994,856 317,157 810 144 11 2 678 120 1.33265 1.33265 0.00077 0.00434 1.00940 1.00940

2,946 20,312 3,159,622 502,340 908 161 120 19 640 113 1.31251 1.31251 0.00083 0.00467 1.00983 1.00983

2,662 18,354 5,246,535 834,132 1,120 198 1,120 178 565 100 1.27407 1.27407 0.00090 0.00510 1.01070 1.01070

2,377 16,389 8,099,726 1,287,754 1,235 219 2,110 335 494 88 1.23807 1.23807 0.00101 0.00567 1.01157 1.01157

2,092 14,424 10,515,312 1,671,801 1,301 231 3,342 531 427 76 1.20451 1.20451 0.00115 0.00646 1.01245 1.01245

P Np Rp Rs Bo BgWp Bw

A reservoir with a connate water saturation of 0.26 is at bubble point at the initial reservoir pressure of 3,516 psia [24,242 kPa] (i.e., no gas cap). API gravity of the oil is measured to be 35 , the specific gravity of gas is 0.74, and the reservoir temperature is 150 F [65.6 C]. Water and formation compressibility are estimated to 3x10-6 psi-1 [4.35x10-7 kPa-1] and 3.5 x 10-6 psi-1 [5.08 x10-7 kPa-1] respectively.

Average isothermal oil compressibility is estimated to 2 x 10-4 psi-1 [2.90 x 10-5 kPa-1].The following table summarizes relevant PVT and production data.



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Solution: Example 4

Field Units

1,994,856 1.33265 810 678 0.00077 11x1.0094 2,860,360bbl

SI Units

317,157 1.33265 144 120 0.00434 2x1.0094 454,761m3

Havlena and Odeh equation for oil reservoir below bubble point

Plot of vs will yield a straight line with the slope equal to

Sample Calculation at P = 3,089 psia [21,298 kPa]

The underground withdrawal, F

W

Solution: Example 4

Field Units

1.33265 1.3980 800 678 0.00077 0.02937bblSTB

SI Units

1.33265 1.3980 142 120 0.00434 0.02937m3

Sm3


Eo Bo – Boi Rsi – Rs Bg

Similarly F and Eo can be calculated for different reservoir pressures



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Solution: Example 4

psia kPa bbl m3 bbl/STB m3/Sm3

3,516 24,242 ‐ ‐ ‐ ‐

3,089 21,298 2,860,360 454,761 0.029375 0.02937

2,946 20,312 4,848,337 770,824 0.048264 0.04826

2,662 18,354 9,314,977 1,480,963 0.089292 0.08929

2,377 16,389 16,061,259 2,553,536 0.148509 0.14851

2,092 14,424 23,197,597 3,688,123 0.235093 0.23509

P F Eo

Field Units

N 100,156,360STB

SI Units

N 15,923,589Sm3

Original oil in place



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Black Oil Reservoir (With Gas Cap)

If we assume gas cap is present and compressibility effects are negligible, we can write:

If we assume no water or gas injection, we can write:

If we assume no water production or water influx, we can write:

Havlena and Odeh Method

For an oil reservoir with gas cap but no water influx, we can write the equation as:

Plot of F vs. should yield a straight line if the value

of m is known If m is not known, a plot of vs. will provide a straight line with an

intercept equal to Nfoi and slope providing a value of m

Intercept will give us the oil in place and the slope will give us the product of m

times the initial oil in place



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Example 5

Estimate the initial oil and initial gas in place

psia kPa STB Sm3SCF/STB Sm3/Sm3

SCF/STB Sm3/Sm3bbl/STB m3/Sm3

bbl/SCF m3/Sm3

3,964 27,331 ‐ ‐ ‐ ‐ 1,200 213 1.67030 1.67030 0.00077 0.00435

2,837 19,560 8,643,070 1,374,138 1,510 268 782 139 1.43650 1.43650 0.00101 0.00570

BgP Np Rp Rs Bo

The following are PVT and production data from an initially saturated oil reservoir at an initial pressure of 3,964 psia [27,331 kPa]. API gravity of oil is measured to be 45, the specific gravity of gas is 0.75, and the reservoir temperature is 210 F [98.9 C]. Geologist estimates that the ratio of the initial free-gas volume to free oil volume is 0.84. Assume no water influx.



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Solution: Example 5

Initial Oil In Place

Solving for , 30

Field Units

8,643,070 1.43650 0.00101 1510 782

1.4365 1.67030 0.00101 1200 782 0.84x1.67030.00077

0.00101 0.00077

Material Balance equation

Solution: Example 5

SI Units

1,374,138 1.43650 0.0057 268 139

1.4365 1.67030 0.0057 213 139 0.84x1.67030.00435

0.0057 0.00435

Field Units

0.84x30x10 x1.67030.00077

= 54.7

Solving for , = 4,769,619 Sm3

Initial Gas In Place

SI Units

0.84x4,769,619x1.6703

0.00435 = 1.54 3



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Learning Objectives

Recognize the important drive mechanisms for black oil reservoirs

Estimate the oil in place in oil reservoirs when the reservoir is above bubble point

Estimate the oil in place in oil reservoirs when the reservoir is producing below bubble point

Estimate the oil in place in oil reservoirs when the reservoir is influenced by gas cap

Quantify the uncertainties in oil place based on the assumptions in the strength of drive mechanisms



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Volatile Oil and Retrograde Condensate Reservoirs



Learning Objectives

Recognize the important drive mechanisms for retrograde condensate and volatile oil reservoirs

Estimate the oil in place in retrograde condensate and volatile oil reservoirs under different mechanisms

Quantify the uncertainties in oil in place based on the assumptions in the strength of drive mechanisms



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Retrograde Condensate Reservoir Calculation

Both retrograde condensate reservoirs and volatile oil reservoirs are some of the most difficult types of reservoirs where material balance is applied. Part of the reason it is difficult to apply material balance to these types of reservoirs, is because there are very few terms we can eliminate in a standard material balance equation. As a result, developing linear approximations can become tedious.

We start with original material balance and eliminate terms from that equation; no original oil in place and no water injection.


Material balance equation for retrograde condensate is different than wet gas because we explicitly account for the oil production rather than assume that condensate yield is constant throughout the life of the reservoir.



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psia kPa STB Sm3SCF/STB Sm3/Sm3

bbl/STB m3/Sm3bbl/SCF m3/Sm3

scf/STB Sm3/Sm3STB/MMSCF Sm3/MMSm3

3,700 25,511 - - - - - - 0.00087 0.00491 - 86.5 488.2 3,650 25,166 40,115 6,411 11,888 2,107 2.417 2.417 0.00088 0.00497 2378 421 81.5 459.9 3,400 23,442 166,438 26,598 12,766 2,262 2.192 2.192 0.00092 0.00519 2010 356 70.5 397.9 3,100 21,374 341,935 54,644 14,410 2,553 1.916 1.916 0.00099 0.00559 1569 278 56.2 317.2 2,800 19,305 460,594 73,606 15,751 2,791 1.736 1.736 0.00108 0.00609 1272 225 46.5 262.4

RvP Np Rp Bo Bg Rs

Example 6

Use material balance to calculate the initial gas in place, neglect the rock and formation compressibility and assume no water influx and no water production.

A retrograde condensate reservoir with an initial reservoir pressure of 3,700 psia [25,511 kPa] is assumed to be at dew point. The initial volatilized Oil-Gas Ratio (Rvi) is 86.5 STB/MMscf [488.2 Sm3/MMSm3].



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Solution: Example 6

Field Units

40,115x11,8880.00088 2.417x

81.510

181.510 x2,378

40,1152.417 0.00088x2,378

181.510 x2,378

420,168

Sample Calculation at P = 3,650 psia [ 25,166 kPa] The underground withdrawal, F

x1 1

1 1Material Balance Equation Simplified

SI Units

6,411x2,1070.00497 2.417x

459.910

1459.910 x421

6,4112.417 0.00497x421

1459.910 x421

67,146

Solution: Example 6

Field Units

0.00088 1 86.510 x2,378 2.41786.510

81.510

1 2,378x 81.510

0.00088

SI Units

0.00497 1 488.210 x421 2.417488.210

459.910

1 421x 459.910

0.00498

The expansion of the gas, Eg

Similarly we can calculate the Btg at initial condition

0.00087 0.00491

11



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Solution: Example 6

psia kPa bbl m3 bbl/SCF m3/Sm3 bbl/SCF m3/Sm3

3700 25511 - - 0.00087 0.00491 0.00000 0.000003650 25166 420,168 67,146 0.00088 0.00498 0.00001 0.000073400 23442 1,961,403 313,446 0.00093 0.00523 0.00006 0.000323100 21374 4,904,031 783,698 0.00100 0.00566 0.00013 0.000752800 19305 7,882,605 1,259,695 0.00110 0.00618 0.00023 0.00127

P F Btg Eg

Field Units

0.00088 0.00087 0.00001

SI Units

0.00498 0.00491 0.00007

Similarly, we could calculate F, Btg and Eg at different reservoir pressures



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Initial gas in place• The plot of Fvs.Eg should yield a straight line with a slope equal to Gfgi

• Field Unit Gfgi 35.3BSCF

• SI Unit Gfgi 998.8MMSm3



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Initial gas in place• Field Unit Gfgi 35.5BSCF

• SI Unit Gfgi 1.00BSm3

The graph show more variability than Approach 1; this graph is more sensitive to errors in measurements

This method allows better understanding of any signature in the data



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Volatile Oil Reservoir Calculation

Start with original material balance and eliminate terms which represent free gas in the reservoir.


Material balance equation for volatile reservoir is perhaps the most complex since no term in the equation can be easily eliminated. It is possible that volatile oil reservoir may contain gas cap as well.

1 1

1 1



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Example 7

• Using material balance calculate the initial oil in place.

• Neglect water and formation compressibility and assume no water influx.

psia kPa STB Sm3 SCF/STB Sm3/Sm3 bbl/STB m3/Sm3 bbl/SCF m3/Sm3 SCF/STB Sm3/Sm3 STB/MMSCF Sm3/MMSm3

5,070 34,956 - - - - 2.704 2.704 - - 2,909 515 - - 4,798 33,081 2,804,775 445,924 2,861 507 2.740 2.740 - - 2,909 515 - - 4,658 32,116 4,933,151 784,308 2,953 523 2.707 2.707 0.000830 0.004684 2,834 502 116 655 4,598 31,702 6,494,156 1,032,488 2,920 517 2.631 2.631 0.000835 0.004712 2,711 480 111 626 4,398 30,323 11,693,845 1,859,173 2,857 506 2.338 2.338 0.000853 0.004814 2,247 398 106 598 4,198 28,944 16,696,347 2,654,507 2,881 510 2.204 2.204 0.000874 0.004932 2,019 358 94 530

RvP Np Rp Bo Bg Rs

An initial under saturated volatile oil reservoir with an initial pressure of 5070 psia [34,956 kPa] and an estimated bubble point pressure of 4,677 psia [32,247 kPa] is producing. The initial solution gas oil ratio is 2,909 scf/STB [515 Sm3/Sm3].



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Solution: Example 7

Field Units

4,933,151x2,9530.00083 2.707x

11610

111610 x2,834

4,933,1512.707 0.00083x2,834

111610 x2,834

13,804,167

1 1Simplified Material Balance Equation

SI Units

784,308x5230.004684 2.707x

65510

165510 x502

784,3082.707 0.004684x502

165510 x502

2,194,687

Sample calculation at P = 4,658 psia [32,116 kPa]

The underground withdrawal, F

x1 1

Solution: Example 7


Field Units

SI Units

2.707 1 515x65510 0.004684 515 502

1 502x65510

2.76465mSm

Similarly, we can calculate Btoi,Btoi 2.704 2.704

11

2.707 1 2,909x11610 0.00083 2,909 2,834

1 2,834x11610

2.76465



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Solution: Example 7

Similarly, we could calculate F, Btg and Eg at different reservoir pressure

psia kPa bbl m3 bbl/STB m3/Sm3 bbl/STB m3/Sm3

5,070 34,956 - - 2.70400 2.70400 0.00000 0.000004,798 33,081 7,685,083 1,221,831 2.74000 2.74000 0.03600 0.036004,658 32,116 13,804,167 2,194,687 2.76465 2.76465 0.06065 0.060654,598 31,702 18,139,214 2,883,905 2.78478 2.78478 0.08078 0.080784,398 30,323 33,003,027 5,247,062 2.86388 2.86388 0.15988 0.159884,198 28,944 48,643,888 7,733,760 2.93649 2.93649 0.23249 0.23249

P F Bto Eo

Field Units

Field Units

2.76465 2.704 0.06065

SI Units

Field Units

2.76465 2.704 0.06065



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Initial oil in place• The plot of Fvs.Eo should yield a straight line with a slope equal to Nfoi

• Field Unit Nfoi 204MMSTB• SI Unit Nfoi 32.4MMSm3



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Initial oil in place• Field Unit Nfoi 216MMSTB• SI Unit Nfoi 34MMSm3

The graph show more variability than Approach 1; this graph is more sensitive to errors in measurements

This method allows better understanding of any signature in the data



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Retrograde Condensate vs. Volatile Oil Reservoirs

Although retrograde condensate and volatile oil reservoir represent different phases in the reservoirs, both reservoirs represent the initial composition which is close to the critical point.

As a result, as the reservoirs go through depletion phase and cross either bubble point or dew point, the behavior of the two reservoirs becomes very similar.

If early sampling is not done for these reservoirs, it is very easy to mischaracterize these reservoirs. That is, identify retrograde condensate as volatile oil and vice-versa.

Depending on the assumptions we make, we either will calculate the initial gas in place or initial oil place.

This mischaracterization can be important if we want to apply different secondary recovery processes.



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Learning Objectives

Recognize the important drive mechanisms for retrograde condensate and volatile oil reservoirs

Estimate the oil in place in retrograde condensate and volatile oil reservoirs under different mechanisms

Quantify the uncertainties in oil place based on the assumptions in the strength of drive mechanisms



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Water Influx



Learning Objectives

Recognize the importance of water influx in the material balance calculations

Learn how to estimate the water influx using pot aquifer as well as pseudo-steady state methods

Describe trial and error procedure required to estimate the aquifer influx

Recognize the uncertainties associated with the estimation of aquifer size and the strength



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Aquifer in Oil and Gas Reservoir

The estimation of aquifer influx is the most difficult part of material balance calculation because:

The water influx is not a constant but changes with time; we will need to calculate the water influx as a function of time.

The aquifer properties are rarely measured – permeability, areal extent, the shape, etc., so estimation becomes more difficult.

The data collected in material balance data may not have sufficient resolution to uniquely identify the aquifer properties. As a result, multiple solutions of aquifer strength and size are possible.

Importance of Aquifer – Gas Reservoir

Considered a disadvantage. The water influx can trap mobile gas, and this can result in lower gas recovery thus adversely impacting the performance of gas reservoirs. Because water moves lot slower than gas phase, once water reaches producing well, it is difficult to produce from that well and well will likely have to be abandoned. In many gas wells, water breakthrough happens without warning and the solutions to mitigate this are often impractical.

Importance of Aquifer – Oil Reservoir

The presence of aquifer, in general, for oil reservoirs is considered an advantage. Solution gas drive is very inefficient, and the recovery factors range between 5 to 25%. Presence of water influx can improve the recovery factor by providing additional pressure support. An efficient bottom water drive can obviate the need for water flooding; thus, reducing the cost of developing an oil field. Even in oil wells, the water coning can be a problem and needs to be avoided.



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In general, the shape of the aquifer is assumed to be radial and consistent with the shape of the reservoir. θ is the angle of partial circle over which aquifer is connected to the reservoir.

It is also possible that linear aquifer could be assumed if the reservoir can be approximated with linear shape.

In general, we assume that the influx is caused by expansion of the water from the aquifer.



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Modeling of Aquifer

The pot aquifer method is too simplistic and the transient method requires computer calculations

The analytical solution is difficult and either require graphical solution or approximation using simplified formulations [Klinset al. (SPERE, 1988)]

Instantaneous equilibrium with reservoir pressure (pot aquifer)

Pseudo-steady state calculation [Fetkovich(JPT, July 1971)]

Transient state calculation [Hurst and Van Everdingen, (AIME, 1949)]

We will consider Fetkovichpseudo-steady state method as a reasonable compromise

It works well in majority of the situations

The calculation of aquifer influx can

be considered with three possibilities



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Assuming expansion is the only mechanism by which water influx can happen, the maximum amount of water influx possible is where ca is the total compressibility of aquifer (cf +cw).

Fetkovich Method

We will assume that before any production, the reservoir and aquifer are in equilibrium. That is, the pressure in aquifer as well as in the reservoir is the same and is equal to pi.

The initial amount of water present in the aquifer is:

where

Is it possible that the aquifer could be recharged by surface water as the depletion is happening?

As the reservoir depletes, the aquifer and the reservoir are no longer in equilibrium and the pressure in the reservoir is always assumed to be less than the pressure in the aquifer.

Fetkovich Method

We can determine maximum amount of water influx when aquifer pressure drops to as

The rate at which water influx occurs under pseudo-steady state conditions can be written as

The productivity index is calculated as

Where:

qa = Rate of influxJ = Productivity index = Average pressure

Where C is dependent on whether we use field or SI units.

3434



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Fetkovich Method

Integrating the equation for the rate and with some additional manipulation, we obtain the incremental water influx in a given step as

∆ 1∆

Where ∆ is the amount of water influx in time step j. is the average pressure in the aquifer in the previous time step, and is the average pressure in the reservoir at the current time step.

Where ∆ is the amount of water influx in time step j. is the average pressure in the aquifer in the previous time step, and is the average pressure in the reservoir at the current time step.

The calculation of water influx requires the knowledge of the size of the aquifer as well as the knowledge of the productivity index – both values being unknownThe calculation of water influx requires the knowledge of the size of the aquifer as well as the knowledge of the productivity index – both values being unknown

The size of the aquifer - - determines how much is the total amount of water influx is possible; the value of is important in determining how quickly the water influx will happen. The size of the aquifer - - determines how much is the total amount of water influx is possible; the value of is important in determining how quickly the water influx will happen.



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Fetkovich Method Overview

Fetkovich method assumes that pseudo-steady condition prevails in the aquifer. This is true if . That is the radius of the aquifer

should not be greater than 8 times the radius of the reservoir.

If the aquifer has to be assumed larger than this value, we may need to use transient method such as Hurst and Van Everdingan.

In applying Fetkovich equation, the two unknown parameters are the size of the aquifer [Wei] and the productivity index [J]. The actual performance of the reservoir will need to be matched by adjusting these two parameters.

It is possible that multiple solutions can provide us with satisfactory results.



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Consider material balance equation for gas reservoir

Or , Or ,

We can write this equation as

Material Balance Calculations in the Presence of Aquifer

That is a plot of ( ) vs. , ) should provide a straight line and the slope would provide a value of gas in place.

If the aquifer influx is too small, the line will concave downwards. If the water influx is too large, the line will curve concave upwards. Whatever value of aquifer influx which provides with the best straight line,

we will use it.

Water Influx Calculations

Starting with time, t = 0, assume the first time step where the average reservoir pressure is measured (which is less than initial pressure).

Assume or calculate the value of Wei and J for aquifer.

The amount of water influx for the first time step is calculated as:

∆ 1∆

This assumes that the initial pressure in aquifer is which is equal to initial reservoir pressure.

For the next time step, we calculate the new average pressure in the aquifer as

1 where We represents cumulative influx so far.

We then calculate the water influx in the next time step as:

∆ 1∆

The same calculation can be repeated at other time steps.



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Water Influx Calculation

Different combinations of Wei and J can be assumed, and water influx values can be calculated till we obtain a straight-line plot.

Once a straight-line plot is obtained, gas in place can be calculated.

Although we wrote the procedure for gas reservoirs, similar procedure can be used for oil reservoirs as well.

The equations for oil reservoir are also straight forward.



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Example 8

Estimate the gas in place

Calculate the water influx as a function of time

Are there non-unique solutions to the problem?

Production data are available from a wet gas reservoir suspected of being supported by a strong aquifer [Sills, SPE Reservoir Engineering, May 1996]. The reservoir temperature is 256 F [124.4 C], the cw is 3.5 x 10-6

psi-1[5.08 x 10-7 kPa-1], cf is 2.9x10-6 psi-1 [4.21x10-7 kPa-1]. The drainage radius of the reservoir is estimated to be 3,300 feet [1,006 m] and the reservoir thickness is estimated to be 100 feet [30.5 m]. Water saturation in the reservoir is 0.21, the specific gravity of oil is 0.78 and assume that the reservoir is circular. The radius of the aquifer can be anywhere between 1 times to 25 times larger than the gas reservoir. Permeability of formation is 22 md and viscosity of water is 0.46 cp[0.00046 Pa.s].

Example 8

time (yrs) psia Kpa MMSCF MMSm3MSTB MSm3

MSTB MSm3Bg Bw

0 8490 58496 0 0.0 0 0.0 0 0.0 0.003034 1.0518

0.5 8330 57394 1758 49.8 2 0.3 0 0.0 0.003065 1.052

1 8323 57345 5852 165.7 3 0.5 1 0.2 0.003066 1.052

1.5 8166 56264 10410 294.8 66 10.5 3 0.5 0.003097 1.0522

2 8100 55809 14828 419.9 98 15.6 4 0.6 0.003111 1.0522

2.5 7905 54465 21097 597.4 138 21.9 7 1.1 0.003152 1.0524

3 7854 54114 26399 747.5 180 28.6 9 1.4 0.003163 1.0525

3.5 7858 54142 30042 850.7 215 34.2 10 1.6 0.003162 1.0525

4 7900 54431 32766 927.8 237 37.7 11 1.7 0.003153 1.0524

4.5 7971 54920 34548 978.3 257 40.9 11 1.7 0.003138 1.0524

5 7883 54314 37590 1064.4 282 44.8 12 1.9 0.003157 1.0525

5.5 7728 53246 42446 1201.9 314 49.9 16 2.5 0.003192 1.0526

6 7550 52020 51117 1447.5 375 59.6 54 8.6 0.003234 1.0528

6.5 7446 51303 57697 1633.8 420 66.8 153 24.3 0.003259 1.0529

7 7400 50986 63678 1803.2 465 73.9 433 68.8 0.003271 1.053

7.5 7600 52364 65432 1852.8 475 75.5 715 113.7 0.003221 1.0528

8 7675 52881 65613 1858.0 475 75.5 753 119.7 0.003204 1.0527

8.5 7600 52364 67593 1914.0 477 75.8 1042 165.7 0.003221 1.0528

9 7600 52364 70688 2001.7 484 77.0 1237 196.7 0.003221 1.0528

9.5 7615 52467 72226 2045.2 488 77.6 1575 250.4 0.003218 1.0527

10 7623 52522 72943 2065.5 489 77.8 2383 378.9 0.003216 1.0527

Pressure Gp Np Wp



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Solution: Example 8 (Sample Calculations)

Other values can be calculated similarly

Equivalent Gas. .. .

= 145.2

133,316 ..

= 716.4 SCF/STB, or

23,743 ..= 127.6Sm3/Sm3

At time = 1 year

F 5852 3x716.4/1000 /5.615x0.003066 1/1000x1.052 3.2MMrb FIELD

F 165.7 0.5x127.6/1000 x0.003066 0.2/1000x1.052 0.51MMm3 SI

Eg Bg – Bgi 0.003066– 0.003034 3.14x10‐5Eg Bg – Bgi 0.003066– 0.003034 3.14x10‐5

Ef,w 0.003034/ 1‐0.21 x 3.5x10‐6 0.21x2.9x10‐6 x 8490– 8323 2.64x10‐6Ef,w 0.003034/ 1‐0.21 x 3.5x10‐6 0.21x2.9x10‐6 x 8490– 8323 2.64x10‐6

Solution: Example 8

MMRb MMm3 Eg Ef,w

0.00 0.00 0 0

0.96 0.15 3.03E‐05 2.53E‐06

3.20 0.51 3.14E‐05 2.64E‐06

5.77 0.92 6.29E‐05 5.11E‐06

8.26 1.31 7.64E‐05 6.16E‐06

11.91 1.89 0.000118 9.23E‐06

14.96 2.38 0.000129 1E‐05

17.02 2.71 0.000128 9.97E‐06

18.51 2.94 0.000119 9.31E‐06

19.42 3.09 0.000103 8.19E‐06

21.26 3.38 0.000122 9.58E‐06

24.27 3.86 0.000157 1.2E‐05

29.65 4.71 0.000199 1.48E‐05

33.82 5.38 0.000225 1.65E‐05

37.74 6.00 0.000236 1.72E‐05

38.49 6.12 0.000187 1.4E‐05

38.43 6.11 0.00017 1.29E‐05

40.07 6.37 0.000187 1.4E‐05

42.05 6.69 0.000187 1.4E‐05

43.25 6.88 0.000184 1.38E‐05

44.49 7.07 0.000182 1.37E‐05

F



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0

50000

100000

150000

200000

250000

0

5000

10000

15000

20000

25000

30000

35000

40000

0 20000 40000 60000 80000

F/(E

g+ E f,w), SMMm

3

F/(E

g+ E f,w), M

MSTB

Gp, MMSCF

0

50000

100000

150000

200000

250000

0

5000

10000

15000

20000

25000

30000

35000

40000

0 20000 40000 60000 80000

F/(E

g+ E f,w), SMMm

3

F/(E

g+ E f,w), M

MSTB

Gp, MMSCF

Solution: Example 8

The traditional plots clearly show influence of water influx; no clear straight line on the first plot and increasing trend on the second plot

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

50.0

0 0.00005 0.0001 0.00015 0.0002 0.00025 0.0003

F, M

Mm

3

F, M

Mrb

Eg + Ef,w

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

50.0

0 0.00005 0.0001 0.00015 0.0002 0.00025 0.0003

F, M

Mm

3

F, M

Mrb

Eg + Ef,w

0

50000

100000

150000

200000

250000

0

5000

10000

15000

20000

25000

30000

35000

40000

0 20000 40000 60000 80000

F/(E

g+ E f,w), SMMm

3

F/(E

g+ E f,w), M

MSTB

Gp, MMSCF

0

50000

100000

150000

200000

250000

0

5000

10000

15000

20000

25000

30000

35000

40000

0 20000 40000 60000 80000

F/(E

g+ E f,w), SMMm

3

F/(E

g+ E f,w), M

MSTB

Gp, MMSCF

Solution: Example 8

The two unknowns we need to estimate the water influx is:

Wei [the maximum amount of water influx possible based on expansion of water]

J [productivity index of water influx into the reservoir]

Both these values can be estimated based on physical size of the aquifer and the reservoir properties.

Consider the influence of water influx

For example if we assume that aquifer radius is 5 times greater than gas reservoir and the porosity, thickness and other properties are the same as the reservoir, we can calculate Wei as

∅

16,500 3,300 1000.23 6.4 10

,

.182.7

5029 1006 30.50.23 9.3 10 58,49629.1

Field UnitsField Units

SI UnitsSI Units



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Solution: Example 8

Both these values (Wei and J) can be changed to improve the results of water influx

34

Similarly, to calculate J or productivity index

7.08 1022 100

0.46165003300

34

39.4 /

Field Units

5.3562 1022 30.5

0.0004650291006

34

0.91 /

SI Units



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∆ 8,490 8,330 1 978,041bbls∆182.7 10

84908,490 8,330 1

40 8490 0.5 365182.7 10

978,041bbls

Field Units

Water Influx Calculations

Values for other pressures can be calculated in similar fashion

∆ 1∆ 1∆

The amount of water influx at any time step can be calculated as:

∆ 58,496 57,394 1∆29.1 1058,496

58,496 57,394 10.91 58,496 0.5 365

29.1 10=155,507 m3

SI Units



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Trial and Error Procedure

Once the initial estimates of Wei and J are made, trial and error procedure is needed to determine the best values of those parameters.

One approach is to determine the value of those two parameters such that a best straight line is obtained when a plot of (F–We) versus [Eg + Ef,w] is made.

We use this procedure to maximize the correlation coefficient of that straight line and estimate those values

One possible solution we obtain is Wei = 18.9 MMRB [3 MMm3] and J = 128.4 bbl/d/psi [2.96 m3/d/kPa].



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Solution: Example 8

rb m3Mrb Mm3

MMRb MMm3

0 0

354,853 56,421 355 56 0.61 0.10

370,023 58,833 725 115 2.47 0.39

717,852 114,138 1,443 229 4.33 0.69

863,511 137,297 2,306 367 5.95 0.95

1,295,124 205,924 3,601 573 8.31 1.32

1,406,938 223,702 5,008 796 9.95 1.58

1,396,660 222,068 6,405 1,018 10.61 1.69

1,302,115 207,035 7,707 1,225 10.80 1.72

1,143,347 181,791 8,850 1,407 10.57 1.68

1,337,372 212,641 10,188 1,620 11.07 1.76

1,679,799 267,086 11,868 1,887 12.40 1.97

2,072,892 329,588 13,940 2,217 15.71 2.50

2,301,474 365,932 16,242 2,582 17.58 2.80

2,401,193 381,787 18,643 2,964 19.10 3.04

1,955,226 310,879 20,598 3,275 17.89 2.84

1,786,933 284,121 22,385 3,559 16.04 2.55

1,951,484 310,284 24,337 3,870 15.73 2.50

1,949,532 309,974 26,286 4,180 15.77 2.51

1,914,315 304,374 28,201 4,484 15.05 2.39

1,894,658 301,249 30,095 4,785 14.40 2.29

We F ‐ WeWe

0

0.5

1

1.5

2

2.5

3

3.5

0

5

10

15

20

25

0 0.00005 0.0001 0.00015 0.0002 0.00025 0.0003

F‐We, M

Mm

3

F‐We, M

Mrb

Eg + Ef,w

Solution: Example 8

0

0.5

1

1.5

2

2.5

3

3.5

0

5

10

15

20

25

0 0.00005 0.0001 0.00015 0.0002 0.00025 0.0003

F‐We, M

Mm

3

F‐We, M

Mrb

Eg + Ef,w



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Solution: Example 8

After introducing the water influx, the correlation coefficient of the straight line is 0.97 indicating much better fit. Further, the upward trend in F/(Eg + Ef,w) plot is eliminated.

The gas in place is 70 BCF [1.9 BSm3].

Fortunately, the gas in place calculation is relatively robust and does not change significantly.

If we use Wei = 722 MMrb [115 MMm3] and J equal to 12.5 bbl/d/psi [0.28 m3/d/kPa], we obtain equally good solution.



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Learning Objectives

Recognize the importance of water influx in the material balance calculations

Learn how to estimate the water influx using pot aquifer as well as pseudo-steady state methods

Describe the trial and error procedure required to estimate the aquifer influx

Recognize the uncertainties associated with the estimation of aquifer size and the strength

Applied Reservoir Engineering

This is Reservoir Engineering Core

Reservoir Rock Properties Core

Reservoir Rock Properties Fundamentals

Reservoir Fluid Core

Reservoir Fluid Fundamentals

Reservoir Flow Properties Core

Reservoir Flow Properties Fundamentals

Reservoir Fluid Displacement Core

Reservoir Fluid Displacement Fundamentals

Reservoir Material Balance Core


Decline Curve Analysis and Empirical Approaches Core

Decline Curve Analysis and Empirical Approaches Fundamentals

Pressure Transient Analysis Core

Rate Transient Analysis Core

Enhanced Oil Recovery Core

Enhanced Oil Recovery Fundamentals

Reservoir Simulation Core

Reserves and Resources Core

Reservoir Surveillance Core

Reservoir Surveillance Fundamentals

Reservoir Management Core

Reservoir Management Fundamentals

Properties Analysis Management




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Back to Work Suggestions

Reservoir Material Balance:

Can be applied to any reservoir

Is a good pre-cursor to reservoir simulation

Provides an upper bound compared to decline curve and volumetric analysis


Reservoir Material

Balance Fundamentals

Leverage the skills you’ve

learned by discussing the

skill module objectives with

your supervisor to develop a

personalized plan to

implement on the job. Some

suggestions are provided.

Do you have reliable production data?

Do you have good reservoir pressure data?

Do you have average reservoir pressure data as a function of time which can be used as a proxy to determine the hydrocarbons in place?

Do you have good fluid properties measurements?

Do you have good correlations which we can use to predict the fluid properties?

Do you have reliable production data?

Do you have good reservoir pressure data?

Do you have average reservoir pressure data as a function of time which can be used as a proxy to determine the hydrocarbons in place?

Do you have good fluid properties measurements?

Do you have good correlations which we can use to predict the fluid properties?



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Reservoir Material









When applying the material balance technique:

Is it a gas? Is it a black oil? Is it volatile oil? Is it a condensate or red gas?

Do you know the important mechanisms which are impacting the production?

If you are producing a large amount of water, perhaps water influx needs to be included in your calculations.

If there is enough volatile oil in the gas phase, then perhaps you are producing from red gas, condensate, or volatile oil.

When applying the material balance technique:

Is it a gas? Is it a black oil? Is it volatile oil? Is it a condensate or red gas?

Do you know the important mechanisms which are impacting the production?

If you are producing a large amount of water, perhaps water influx needs to be included in your calculations.

If there is enough volatile oil in the gas phase, then perhaps you are producing from red gas, condensate, or volatile oil.


Reservoir Material









It's important that when you are solving the material balance equation, you formulate the problem in a way that a linear graph can be used to calculate either initial oil or initial gas in place.

Are there any ways by which you can reformulate the equation such that the linear relationship can be established?

It's important that when you are solving the material balance equation, you formulate the problem in a way that a linear graph can be used to calculate either initial oil or initial gas in place.

Are there any ways by which you can reformulate the equation such that the linear relationship can be established?



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Reservoir Material









After you calculate the initial oil and gas in place:

Validate the amount of oil and gas you calculated using the methods.

Compare your initial oil and gas in place with other methods.

Did you use decline curve analysis to calculate how much oil or gas can be recovered?

Is the amount of oil and gas you calculated using material balance exceed what you have already obtained from the decline curve?

Are there mechanisms which we simply did not include?

How does material balance compare to volumetric analysis?

After you calculate the initial oil and gas in place:

Validate the amount of oil and gas you calculated using the methods.

Compare your initial oil and gas in place with other methods.

Did you use decline curve analysis to calculate how much oil or gas can be recovered?

Is the amount of oil and gas you calculated using material balance exceed what you have already obtained from the decline curve?

Are there mechanisms which we simply did not include?

How does material balance compare to volumetric analysis?


Reservoir Material









Can we use this information as a pre-cursor for reservoir simulation?

Reservoir simulation doesn't really have any better ability to calculate the initial oil and gas in place than the material balance technique.

You can obtain the oil and gas base using the material balance technique, then it provides you with an initial starting point which can be utilized in a reservoir simulation method.



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