Oil Reserves and Reserve Estimate
12.1 INTRODUCTION
This chapter consists of two sections: Section 1 deals with reserves and reserve estimate. Section 2 is devoted to economic evaluation and applications, with solved examples.
Evaluation of an oil property depends on the development of the underground accumulation of hydrocarbons and on the amount of money that will be received from selling the produced hydrocarbons. Such evaluation includes estimate of reserves, estimate of gross income, estimate of net income after all types of taxes and production costs, and calculations of present worth value of the property.
Development of an oil and/or gas reservoir depends on the producible amount of hydrocarbons. This amount is called “reserves”. The proved reserve is that form of reserve which is recoverable by the force of natural energy existing in the reservoir or by secondary processes. The probable reserve is the reserve which has not been proved by production at a commercial flow rate.
Reserve estimation is one of the most essential tasks in the petroleum industry. The total estimated amount of oil in an oil reservoir, including both producible and non-producible oil, is called “oil in place”. Practically speaking, because of reservoir characteristics and limitations in petroleum extraction technologies, only a fraction of this oil can be brought to the surface, and it is only this producible fraction that is considered to be reserve.
An oil evaluation study has as its primary purpose the determination of the value of oil in place. Such evaluation includes estimates of reserves. Methods most commonly used to estimate the reserve of recoverable hydrocarbons are presented first in this chapter. These include: volumetric, material balance, and decline curve methods. The role of economic evaluation for oil properties is illustrated by many case studies.
SECTION 1: RESERVES AND RESERVE ESTIMATE
12.2 OVERVIEW
Many factors impact the demand for and supply of oil and natural gas, influence how and where energy suppliers invest their capital, and determine the manner in which countries compete to attract foreign investment. Globally, we currently consume the equivalent of over 11 billion tonnes of oil from fossil fuels every year. Crude oil reserves are vanishing at a rate of more than 4 billion tonnes per year—so if we carry on, as we are, our known oil deposits could run out in just over 53 years.
World oil supply derives from the following factors:
- • The investment decisions of individual companies
- • The political decisions of countries in regard to licensing
- • The degree of foreign investment and a multitude of other variables that influence system dynamics, including price, inventory levels, geopolitics, market psychology and manipulation, organization of the petroleum exporting countries (OPEC) policy, exchange rates, unexpected events, and resource availability
In its latest Statistical Review of World Energy, BP estimated the world had
I. 7297 trillion barrels of crude oil remaining at the end of 2018. That was up from 1.7275 trillion barrels a year earlier and 1.4938 trillion barrels in 2008 (Nov
II, 2019).
Oil and gas assets represent the majority of value of an E&P company.
The Oil and Gas Financial Journal describes reserves as “a measurable value of a company’s worth and a basic measure of its life span”.
Interest in the determinants of investment in crude oil and natural gas reserves derives from three basic sources:
Oil and gas assets represent the majority of value of an E&P company.
The Oil and Gas Financial Journal describes reserves as “a measurable value of a company’s worth and a basic measure of its life span”.
Interest in the determinants of investment in crude oil and natural gas reserves derives from three basic sources:
- • First, it is always interesting to find a satisfactory explanation of investment behavior in any industry
- • Second, an aspect of the current concern with the “energy crisis” is the domestic crude petroleum industry’s productive capacity, which is an increasing function of the stock of proved oil and gas reserves
- • Third, there is a decades-old controversy over the special provisions of the federal corporation income tax law
The American Society of Appraisers defines the fair market value as reserves as: The price, expressed in terms of cash equivalents, at which property would change hands between a hypothetical willing and able buyer and a hypothetical willing and able seller, acting at arm’s length in an open and unrestricted market, when neither is under compulsion to buy nor sell and when both have reasonable knowledge of the relevant facts.
The American Society of Appraisers recognizes three general approaches to valuation:
- 1. The Cost Approach,
- 2. The Income Approach, and
- 3. The Market Approach.
When valuing acreage rights comparable transactions do provide the best indication of value. However, when valuing reserves, a D.C.F. (discounted cash flow) is often the best way to allocate value to different reserve categories because comparable transactions are very rare as the details needed to compare these specific characteristics of reserves are rarely disclosed.
The total present worth of future income is then discounted further, a percentage based on market conditions, to determine the fair market value. The costs of any expected additional equipment necessary to realize the profits are included in the annual expense, and the proceeds of any expected salvaged of equipment is included in the appropriate annual income.
12.3 METHODS OF RESERVE ESTIMATE
The methods most commonly used to estimate the reserve of recoverable hydrocarbons are:
- 1. Volumetric methods
- 2. Material balance methods
- 3. Decline curve methods
Each of these methods will be discussed separately.
12.3.1 Volumetric Methods
The estimation of reserve is done on the basis of an equation which is not complicated to use provided the required data are available. The data include the area of the production zone (A), the formation thickness (h), the porosity (ф), and the initial water saturation (S_{wi}). The equation has the form:
where:
N = bbls of initial oil in place at surface temperature and pressure Condition, which is called stock tank
B_{oj} = initial oil formation volume factor, which is defined as bbl at reservoir condition (rb), divided by bbl at surface condition (STB)
Once the recovery factor is known, then the recoverable oil can be known. The bulk volume of the reservoir can be calculated using subsurface and isopachous maps. The isopachous map consists of isopach lines that connect points of formations having equal thickness. The areas lying between the isopach lines of the entire reservoir under consideration are used to calculate the volume contained in it.
Simpson’s rule, trapezoidal rule and pyramidal rule are normally used to determine the reservoir bulk volume (V_{B}). Simpson’s rule provides the following equation:
where:
h = interval between the isopach lines in ft B_{0} = area in acres enclosed by successive isopach lines in acres A„ A_{2}, A_{3}, A_{n} = areas enclosed by successive isopach lines in acres t_{n} = average thickness above the top
Trapezoidal rule provides the following equation:
Pyramidal rule has the form:
This equation calculates the reservoir bulk volume between any two successive areas (AV_{B}), and the total reservoir bulk volume is the summation of all the calculated bulk volumes.
The accuracy of trapezoidal rule and pyramidal rule depends on the ratio of the successive areas. If the ratio of the areas is smaller than 0.5, the pyramidal rule is used; otherwise the trapezoidal rule is used.
The formula as provided in Equation (12.1) can be applied to calculate free gas in a gas reservoir as given below:
where:
G = gas in place
B,, = gas formation volume factor V_{B} = reservoir bulk volume S_{w} = connate water
12.3.2 Material Balance Equation
Material balance equation accounts for the fluids that leave, enter, or accumulate in the reservoir at any time. The oil reservoir is classified as an undersaturated or saturated reservoir based on the reservoir pressure. A reservoir with pressure higher than the bubble point pressure is considered to be an under-saturated reservoir. The material balance for such reservoir, with the assumption that the oil is produced by the fluid expansion only and the reservoir is constant, is derived below:
Assume that the initial production, P_{r} dropped to P due to N STB produced. Then,
Initial volume = NB_{m} bbl at the reservoir condition, rb Final volume = (N- /V_{p})B_{0} bbl at the reservoir condition, rb
Since the reservoir volume is constant, then:
A reservoir with pressure lower than the bubble point pressure will cause gas to form, resulting in a free gas phase. Such a reservoir is called a saturated reservoir. The derivation of material balance equation for this case is given next:
Assume the reservoir volume is constant, then:
where:
/V = oil in place, rb
yV_{P} = oil produced, STB
B_{0} = formation volume factor, rb/STB
B_{m} = initial formation volume factor, rb/STB
B_{?} = gas formation volume factor, rb/STB
R_{si} = initial gas in solution. SCF/STB
R_{s} = gas in solution at a pressure lower than P_{;}
R_{p} = cumulative gas-oil ratio
If the reservoir has a gas cap at the time of discovery, then the material balance equation will have the form:
where:
m = volume of free gas/oil volume ^{=} G_{f}B_{gi}/NB_{oi}
If the reservoir is under water drive, the water influx as well as the water production needs to be added to the material balance. Then, Equations (12.7) and (12.8) become:
All these terms, except N_{p}, R_{p}, W_{c}, and W_{p}, are functions of pressure and also are properties of the fluids. These data should be measured in the laboratory. R_{p} depends on the production history. It is the quotient of both the gas produced (G_{p}) and the oil produced (N_{p}). A water influx can be calculated by using different methods depending on the flowing conditions. The boundary pressure as well as the time are used to calculate the water influx. The value of m is determined from the log data, which provides the gas-oil and oil-water contacts and also from the core data. Therefore, the accuracy of the calculated oil in place depends upon how accrately we take these measurements for such calculations.
12.3.3 Material Balance Equation for Gas Reservoir
a. No water drive: If the reservoir volume stays constant and G_{p}, gas produced during a time t, and B_{gi} drop to then material balance is given by Equation (12.11) as follows:
b. With water drive: The material balance:
If the measured data are accurate, the calculated gas in place will always be accurate. In Equation (12.12), the water influx can be found using the pressure drop during the production history with other parameters.
12.3.4 Material Balance Equation, Straight-line Concept
The material balance equation given in Equation (12.10) may be expressed as a straight-line equation which will have the form:
where:
F represents the total underground withdrawal while E_{0} denotes the oil expansion and the expansion of associated gas, while E,, represents the gas cap expansion.
Equation (12.13) includes all the drive mechanisms. If any one of these mechanisms is not acting in the reservoir, then the term representing such a mechanism must be deleted from the equation.
a. No water drive, no original gas cap:
A plot of F versus £_{c} gives a straight line passing through the origin with a slope of N (initial oil in place).
b. No water drive (W_{c} = 0). Equation (12.13) will be reduced to:
Again plotting F versus (£„ + mEj yields a straight line passing through the origin with a slope of N.
c. No water drive and m is not known. Equation (12.15) can be written differently:
A plot of F/E_{a} versus EJE_{a} should result in a straight line with the intercept of N with Y-axis. The value of m can be known from the slope.
d. For water drive reservoir, m = 0, Equation (12.13) will have the form:
Divide by E_{0}:
A plot of F/E_{a} versus WJE_{a} should give a straight line witli N being the Y intercept provide the calculated water influx is correct.
The same concept can be applied to the gas reservoir to express the gas material balance equation as a straight line. Equation (12.12) becomes:
Plotting G_{p}j3„ versus £_{g} should give a straight line with G being the slope. If the reservoir is under water drive, Equation (12.12) can be written as:
Divide by £_{g}:
A plot of G_{p}B„ + W_{p}/Е versus WJE,, should result in a straight line with G being the Y intercept.
Using the straight-line technique to estimate oil or gas reserves will minimize the error in the calculated reserve because a number of data will be used for the reserve estimation and the error in the data will be averaged.
The gas in place can be estimated by another approach which requires plotting P/z versus cumulative gas production for a volumetric reservoir. Such a plot results in a straight line with G being the X-axis intercept. Estimation of gas reserve using early production data may result in error by as much as a factor of 2. Therefore, this method should be used only when the cumulative gas production reaches a stage of about 20% of the gas in place.
12.3.5 Decline Curve Methods
Predicting the reserve using decline curve methods requires production rate of all the wells. The production rate generally declines with time, reaching an end point, which is referred to as the economic limit. The economic limit is a production rate at which the income will just meet the direct operating cost of a well or a certain field. Typical decline curve analysis consists of plotting production rate versus time and trying to fit the obtained data into a straight line or other forms which can be extrapolated up to the economic limit to estimate the reserve on the assumption that all the factors affecting the well performance have exactly the same effect in the future as they had in the past.
The commonly used decline curve methods are:
- 1. Constant percentage decline
- 2. Hyperbolic decline
- 3. Harmonic decline
- 12.3.5.1 Constant Percentage Decline
The constant percentage decline is known as the exponential decline and is used widely more than the other forms of decline due to its simplicity. In this case, the decline rate is assumed to be constant during the production time. The decline rate in production rate with time is:
where:
D = decline rate
Aq = q_{i}- q q_{t} is initial production rate and q is production at a time (t)
At = time t required for q_{t} to decline to q
Integrating Equation (12.20) to get rate-time relation:
Integrating Equation (12.21) with respect to time: or
From Equation (12.22):
Substitute in Equation (12.22); then:
Equation (12.23) can be rearranged as follows:
A plot of q versus N_{p} will result in a straight line. The slope of the line is D and q, is the intercept of the Y-axis. Equation (12.21) also yields a straight line if q is plotted against t on semilog paper. The slope of such a plot is D and the intercept is q,. The A_{p} is the cumulative production between any two production rates.
When the decline rate is not constant, then the hyperbolic decline can be assumed and the decline rate varies according to the following equation:
where:
n = decline constant between zero and 1 Д = initial decline rate
The general equation for hyperbolic rate decline can be obtained by substituting Equation (12.17) into Equation (12.21) and then integrating the resulting equation. The equation thus finally derived will have the form:
The cumulative production rate obtained from the hyperbolic decline can be derived as follows:
Equation (12.20) can be written as:
Substitute D value from Equation (12.25) in the above equation, and then substitute q value in the equation to calculate (V_{p}:
The values of q_{v} D_{t}, and n are assumed to be known and are constant, and thereafter Equation (12.27) can be used without any difficulty. The values of q„ D_{{}, and n can be obtained by comparing the actual decline data w'ith a series of curves of hyperbolic type. A plot of q/q_{i} versus time may fit in one of the curves which gives the values of q- I)_{r} and n.
12.3.5.2 Harmonic Decline
In this curve, when the decline rate is not constant, it decreases as the production rate increases. Such a varying rate in decline is called a harmonic decline. It also occurs if the decline constant n of Equation (12.27) is 1. An equation derived for such decline is:
This type of decline may take place in reservoirs where gravity drainage controls the production. Gravity drainage exists in tilted reservoirs where oil production is affected by drainage of oil from upstructure to downstructure which causes segregation of gas and oil in the reservoir. The cumulative production can be obtained by integrating Equation (12.24) with respect to time:
But from Equation (12.25):
Substitute in Equation (12.25)
A graphic harmonic decline analysis can be obtained by writing Equation (12.24) as:
Plotting lq versus t on Cartesian coordinates should result in a straight line, with ajq, being the slope and 1 lq_{{} the intercept with 1/g-axis. From the slope a_{l} can be known. Also Equation (12.30) can be rewritten in a different form:
A plot of q versus N_{p} on a semilog paper will result in a straight line with slope being ajq_{x} and intercept q, This straight line can be extrapolated into the economic limit to calculate the reserve.
12.4 COMPARISON OF THE METHODS
Comparison of all the predictive methods depends on the data available and on the accuracy of these data. Volumetric methods are usually used in the early life of the reservoir while the material balance equations or the decline curve methods can be used when enough data are collected. However, material balance equation techniques depend on many measurements, such as B„, #_{g}, /?_{s}, R_{p}, and total production; hence more error is anticipated in the calculated reserves. The error in the calculated reserve by the decline curve is less than with other methods.
SECTION 2: ECONOMIC EVALUATION AND APPLICATIONS
Evaluation of an oil property is concerned with its money value, which measures the profitability of such an oil property. The profitability depends on the development of underground accumulations of hydrocarbons and on the sale value of the hydrocarbons, which helps to estimate the present worth value of such property at any time under certain specified conditions. The gross income of hydrocarbon sales depends on the current prices of oil and gas and on the predicted economic conditions. The net profit is related to all the expenses that are deducted from the gross income, such as operating cost, which includes the expenses required to produce the hydrocarbon and to maintain the reservoir, taxes, and royalty when applicable.
The following applications and case studies illustrate the role of economic evaluation for an oil property.
Example 12.1
Given the following data:
a. Calculate the oil in place.
b. Calculate the total gas in solution.
SOLUTION
Part (a):
Then, oil in place = 24.274,000 STB Part (b):
Example 12.2
An oil reservoir has a gas cap at the time of discovery. The size of this gas cap is not known. The production data and the fluid properties are given as a function of pressure in Table 12.1.
a. Calculate the oil in place using the material balance equation as a straight line.
b. Use the material balance equation itself.
SOLUTION
Since the production was due to gas cap expansion and the gas cap size is not known, the following equation can be used:
All the calculations are given in Table 12.2.
TABLE 12.1
Data for Example 12.2
P, psi |
N,„ STB |
fi,„ rb/STB |
R„ SCF/STB |
B_{s}, rb/SCF |
ff_{p}, SCF/STB |
3,200 |
0 |
1.35 |
520 |
0.000932 |
0 |
2,950 |
2.50 x 10^{8} |
1.345 |
444 |
0.00095 |
950 |
1,800 |
3.37 x 10^{8} |
1.34 |
435 |
0.000995 |
1,000 |
2,765 |
4.95 x10^{8} |
1.32 |
410 |
0.0011 |
1,150 |
2,500 |
6.62 x 10^{8} |
1.308 |
395 |
0.00123 |
1,280 |
TABLE 12.2
Solution for Example 12.2
P, psi |
F |
£/£„ |
Vfo |
||
2,950 |
4.57 x 10^{8} |
0.0672 |
0.0255 |
6.8 x 10° |
0.379 |
2,800 |
6 407 x 10^{8} |
0.0745 |
0.09125 |
8 6x10° |
1.22 |
2,650 |
10.56 x 10^{8} |
0.091 |
0.238 |
11.6x 10° |
2.615 |
2,500 |
15.87 x 10^{8} |
0.1118 |
0.4182 |
14.1 x 10° |
3.743 |
FIGURE 12.1 Solution of Example 12.2.
Plotting £/£_{0} against £_{g}/£_{0} as shown in Figure 12.1 yields a straight line. The values of the intercept and the slope are given as follows:
From Equation (12.19), the Y intercept is N and the slope is mN, then:
Now, m is known, the material balance equation can be used to calculate initial oil in place. Equation (12.10) will be used:
Since m is known, Equation (12.15) can be used to determine the oil in place N. The calculation is shown in Table 12.3.
TABLE 12.3
Data to Determine Oil in Place, N
p |
F |
E_{a} + Eg |
2,950 |
4.57 x 10^{8} |
0.07675 |
2,800 |
6.407 x 10^{s} |
0.1087 |
2,650 |
10.56 x 10^{s} |
0.180 |
2,500 |
15.87 x 10^{s} |
0.2684 |
Example 12.3
For application of the constant decline curve, the following production history for a well is given:
Year |
B/day |
1 |
9,600 |
2 |
7,200 |
3 |
6,700 |
4 |
5,700 |
5 |
5,200 |
6 |
4,650 |
7 |
4,300 |
8 |
3,800 |
a. Estimate the remaining life of this field if the economic limit is 800 B/D.
b. What is the recoverable oil as of year 8?
c. What is the net income if the price of oil is assumed to be $85/bbl?
SOLUTION
Since the decline rate follows the constant percentage decline, then a plot of q versus time on semi-log is recommended gives a straight line. The slope of the line represents the decline rate, D.
a. Using Equation (12.21), the revising number of years can be calculated as follows:
or
Recoverable oil = q_{t} - q?/D
b. _ 3,800-800
0.02086 = 143,816 bbl
c. Total income = 143,816 x 85 = $12,224,360
If the operating expenses is taken to be $38/bbl, then the gross income =$6,759,352 If this gross income is to be taxed at 46%, the net profit =$3,650,000
Example 12.4
Use the calculated oil in place in Example 12.1 assuming the following values:
5ale value of the oil = $85/bbl
Operating costs = $47/bbl
The calculated oil in place = 24,274,000 5TB
Cross income = (Oil in place) x Price
= 24,274,000 bbl x 85 $/bbl = $20.6 x 10^{8}
Production taxes = 20.6 x 10^{8} x 0.046 = $0.9476 x 10^{s}
Operating Costs = 24,274,000 x 47 = $11.4 x 10^{8}
Net income = Gross income - (Production costs + Operating costs)
= 20.6x 10^{8}-(0.9476 + 11.4) x 10^{8 }= 58.2524 xIO^{8}
This calculation excludes any capital expenditure that may be justified in the future. Also, the calculation is based on today's oil price, which may change in the future.
Example 12.5
A similar calculation can be done for Example 12.2 assuming the oil price, operating cost, and production taxes are the same as used in the previous calculations.
SOLUTION
Gross income = oil in place x price = 5.9 x 10^{9} x 85 = $500 x 10^{9}
Production taxes = $500 x 10^{9} x 0.046 = $23 x 10^{9}
Operating costs = 5.9 x 10^{9} x 47 = $277 x 10^{9}
Net Income = Gross income - (Production taxes + Operating costs)
= $500 x 10^{9} - 300 x 10^{9 }= S200 x10^{9}
Again, this net income excludes any capital expenditures that may be needed in the future. Other taxes that may be applicable are not combined.