How to Continuously Compute Sum in Influx Database

Water Influx

Tarek Ahmed , Paul D. McKinney , in Advanced Reservoir Engineering, 2005

Example 2.2

Calculate the cumulative water influx that result from a pressure drop of 200 psi at the oil-water contact with an encroachment angle of 80°. The reservoir-aquifer system is characterized by the following properties:

	Reservoir	Aquifer
radius, ft	2600	10 000
porosity	0.18	0.12
c f, psi−1	4 × 10−6	3 × 10−6
c w, psi−1	5 × 10−6	4 × 10−6
h, ft	20	25

Solution

Step 1.

Calculate the initial volume of water in the aquifer from Equation 2.3.2:

$\begin{array}{l}{W}_{\text{i}}=\left[\frac{\pi (r_{\text{a}}^2-r_{\text{e}}^2)h\Phi }{5.615}\right]\\ =\left[\frac{\pi (10{000}^2-{2600}^2)(25)(0.12)}{5.615}\right]=156.5\text{MMbbl}\end{array}$

Step 2.

Determine the cumulative water influx by applying Equation 2.3.3:

$\begin{array}{l}{W}_{\text{e}}\text{undefined}=\text{undefined}(c_{\text{w}}\text{undefined}+\text{undefined}{c}_{\text{f}}\text{undefined})W_{\text{i}}f\text{undefined}(p_{\text{i}}-\text{undefined}p)\\ =(4.0+3.0){10}^{-6}(156.5\times {10}^6)\left(\frac{80}{360}\right)(200)=48689\text{bbl}\end{array}$

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Water Influx

Tarek Ahmed , in Reservoir Engineering Handbook (Fifth Edition), 2019

The Pot Aquifer Model

The simplest model that can be used to estimate the water influx into a gas or oil reservoir is based on the basic definition of compressibility. A drop in the reservoir pressure, due to the production of fluids, causes the aquifer water to expand and flow into the reservoir. The compressibility is defined mathematically as:

(10-3) $\mathrm{\Delta V}=\mathrm{c}\mathrm{V}\mathrm{\Delta }\mathrm{p}$

Applying the above basic compressibility definition to the aquifer gives:

$\text{Water influx}=\left(\text{aquifer compressibility}\right)\left(\text{initial volume of water}\right)\left(\text{pressure drop}\right)$

or

(10-4) ${\mathrm{W}}_{\mathrm{e}}=\left({\mathrm{c}}_{\mathrm{w}}+{\mathrm{c}}_{\mathrm{f}}\right){\mathrm{W}}_{\mathrm{i}}\left({\mathrm{p}}_{\mathrm{i}}-\mathrm{p}\right)$

where

W_e = cumulative water influx, bbl

c_w = aquifer water compressibility, psi^–1

c_f = aquifer rock compressibility, psi^–1

W_i = initial volume of water in the aquifer, bbl

p_i = initial reservoir pressure, psi

p = current reservoir pressure (pressure at oil-water contact), psi

Calculating the initial volume of water in the aquifer requires the knowledge of aquifer dimension and properties. These, however, are seldom measured since wells are not deliberately drilled into the aquifer to obtain such information. For instance, if the aquifer shape is radial, then:

(10-5) ${\mathrm{W}}_{\mathrm{i}}=\left[\frac{\mathrm{\pi }\left({\mathrm{r}}_{\mathrm{a}}^2-{\mathrm{r}}_{\mathrm{e}}^2\right)\mathrm{h}\mathrm{\varphi }}{5.615}\right]$

where

r_a = radius of the aquifer, ft

r_e = radius of the reservoir, ft

h = thickness of the aquifer, ft

ϕ = porosity of the aquifer

Where the effective radius of the reservoir is expressed in terms of the reservoir pore volume "V_P" as given by:

${\mathrm{r}}_{\mathrm{e}}=\sqrt{\frac{360{\mathrm{V}}_{\mathrm{P}}}{\mathrm{\pi }\mathrm{h}\mathrm{\varphi }\mathrm{\theta }}}$

Where reservoir pore volume "V_P" is expressed in ft³.

Equation 10-5 suggests that water is encroaching in a radial form from all directions. Quite often, water does not encroach on all sides of the reservoir, or the reservoir is not circular in nature.

To account for these cases, a modification to Equation 10-4 must be made in order to properly describe the flow mechanism. One of the simplest modifications is to include the fractional encroachment angle f in the equation, as illustrated in Figure 10-2, to give:

Figure 10-2. Radial aquifer geometries.

(10-6) ${\mathrm{W}}_{\mathrm{e}}=\left({\mathrm{c}}_{\mathrm{w}}+{\mathrm{c}}_{\mathrm{f}}\right){\mathrm{W}}_{\mathrm{i}}\mathrm{f}\left({\mathrm{p}}_{\mathrm{i}}-\mathrm{p}\right)$

where the fractional encroachment angle f is defined by:

(10-7) $\mathrm{f}=\frac{{\left(\text{encoachment angle}\right)}^{°}}{{360}^{°}}=\frac{\mathrm{\theta }}{{360}^{°}}$

The above model is only applicable to a small aquifer, i.e., pot aquifer, whose dimensions are of the same order of magnitude as the reservoir itself. Dake (1978) points out that because the aquifer is considered relatively small, a pressure drop in the reservoir is instantaneously transmitted throughout the entire reservoir-aquifer system. Dake suggests that for large aquifers, a mathematical model is required which includes time dependence to account for the fact that it takes a finite time for the aquifer to respond to a pressure change in the reservoir.

Example 10-2

Calculate the cumulative water influx that results from a pressure drop of 200 psi at the oil-water contact with an encroachment angle of 80°. The reservoir-aquifer system is characterized by the following properties:

	Reservoir	Aquifer
radius, ft	2600	10,000
porosity	0.18	0.12
cf, psi–1	4 × 10–6	3 × 10–6
cw, psi–1	5 × 10–6	4 × 10–6
h, ft	20	25

Solution

Step 1.

Calculate the initial volume of water in the aquifer from Equation 10-4.

${\mathrm{W}}_{\mathrm{i}}=\left(\frac{\mathrm{\pi }\left(10,000^2-2600^2\right)\left(25\right)\left(0.12\right)}{5.615}\right)=156.5\text{MMbbl}$

Step 2.

Determine the cumulative water influx by applying Equation 10-5.

${\mathrm{W}}_{\mathrm{e}}=\left(4+3\right)10^{-6}\left(156.5\times 10^6\right)\left(\frac{80}{360}\right)\left(200\right)=\mathrm{48,689}\mathrm{bbl}$

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Water Influx

Tarek Ahmed , in Reservoir Engineering Handbook (Fourth Edition), 2010

Problems

1.

Calculate the cumulative water influx that results from a pressure drop of 200 psi at the oil-water contact with an encroachment angle of 50°. The reservoir-aquifer system is characterized by the following properties:

	Reservoir	Aquifer
radius, ft	6000	20,000
porosity	0.18	0.15
cf, psi−1	4 × 10−6	3 × 10−6
cw, psi−1	5 × 10−6	4 × 10−6
h, ft	25	20

2.

An active water-drive oil reservoir is producing under the steady-state flowing conditions. The following data are available:

p_i = 4000 psi

Q_o = 40,000 STB/day

GOR = 700 scf/STB

z = 0.82

Q_w =0

p = 3000 psi

B_o = 1.3 bbl/STB

R_s = 500 scf/STB

T = 140°F

B_w = 1.0 bbl/STB

Calculate Schilthuis' water influx constant.

3.

The pressure history of a water-drive oil reservoir is given below:

t, days	p, psi
0	4000
120	3950
220	3910
320	3880
420	3840

The aquifer is under a steady-state flowing condition with an estimated water influx constant of 80 bbl/day/psi. Using the steady-state model, calculate and plot the cumulative water influx as a function of time.

4.

A water-drive reservoir has the following boundary pressure history:

Time, months	Boundary pressure, psi
0	2610
6	2600
12	2580
18	2552
24	2515

The aquifer-reservoir system is characterized by the following data:

	Reservoir	Aquifer
radius, ft	2000	∞
h, ft	25	30
k, md	60	80
ϕ, %	17	18
μw, cp	0.55	0.85
cw, psi−1	0.7 × 10−6	0.8 × 10−6
cf, psi−1	0.2 × 10−6	0.3 × 10−6

If the encroachment angle is 360°, calculate the water influx as a function of time by using:

a.

The van Everdingen-Hurst method

b.

The Carter-Tracy method

5.

The following table summarizes the original data available on the West Texas water-drive reservoir:

	Oil Zone	Aquifer
Geometry	Circle	Semi-circle
Area, acres	640	Infinite
Initial reservoir pressure, psia	4000	4000
Initial oil saturation	0.80	0
Porosity, %	22	—
Boi, bbl/STB	1.36	—
Bwi, bbl/STB	1.00	1.05
co, psi−1	6 × 10−6	—
cw, psi−1	3 ×10−6	7 × 10−6

The aquifer geological data estimate the water influx constant at 551 bbl/psi. After 1,120 days of production, the reservoir average pressure has dropped to 3,800 psi and the field has produced 860,000 STB of oil. The field condition after 1,120 days of production is given below:

p = 3800 psi

N_p = 860,000 STB

B_o = 1.34 bbl/STB

B_w = 1.05 bbl/STB

W_e = 991,000 bbl

t_D = 32.99 (dimensionless time after 1120 days)

W_p = 0 bbl

It is expected that the average reservoir pressure will drop to 3,400 psi after 1,520 days (i.e., from the start of production). Calculate the cumulative water influx after 1,520 days.

6.

A wedge reservoir-aquifer system with an encroachment angle of 60° has the following boundary pressure history:

Time, days	Boundary Pressure, psi
0	2850
365	2610
730	2400
1095	2220
1460	2060

Given:

h = 120′

μ_w = 0.7 cp

c_f = 5 × 10⁻⁶ psi⁻¹

k = 60 md

c_w = 4 ×10⁻⁶ psi⁻¹

ϕ = 12%

reservoir area = 40,000 acres

aquifer area = 980,000 acres

T = 140°F

Calculate the cumulative influx as a function of time by using Fetkovich's method.

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Water Influx

Tarek Ahmed , D. Nathan Meehan , in Advanced Reservoir Management and Engineering (Second Edition), 2012

2.3.5 The Carter and Tracy Water Influx Model

The van Everdingen and Hurst methodology provides the exact solution to the radial diffusivity equation and therefore is considered the correct technique for calculating water influx. However, because superposition of solutions is required, their method involves tedious calculations. To reduce the complexity of water influx calculations, Carter and Tracy (1960) proposed a calculation technique that does not require superposition and allows direct calculation of water influx.

The primary difference between the Carter–Tracy technique and the van Everdingen and Hurst technique is that Carter–Tracy technique assumes constant water influx rates over each finite time interval. Using the Carter–Tracy technique, the cumulative water influx at any time, t _n , can be calculated directly from the previous value obtained at t _n−1, or:

(2.33) ${(W_{\mathrm{e}})}_n={(W_{\mathrm{e}})}_{n-1}+[{(t_{\mathrm{D}})}_n-{(t_{\mathrm{D}})}_{n-1}]\times \left[\frac{B\mathrm{\Delta }{p}_n-{(W_{\mathrm{e}})}_{n-1}{(p_{\mathrm{D}}^{\mathrm{\backslash }})}_n}{{(p_{\mathrm{D}})}_n-{(t_{\mathrm{D}})}_{n-1}{(p_{\mathrm{D}}^{\mathrm{\backslash }})}_n}\right]$

where

B=the van Everdingen and Hurst water influx constant as defined by Eq. (2.23)

t _D=the dimensionless time as defined by Eq. (2.17)

n=the current time step

n−1=the previous time step

Δp _n =total pressure drop, p _i−p _n , psi

p _D=dimensionless pressure

$p_{\mathrm{D}}^{\mathrm{\backslash }}$ =dimensionless pressure derivative

Values of the dimensionless pressure p _D as a function of t _D and r _D are tabulated in Chapter 1, Table 1.2. In addition to the curve-fit equations given in Chapter 1 ( Eqs. (1.90) through (1.95) (1.90) (1.91) (1.92) (1.93) (1.94) (1.95) , Edwardson et al. (1962) developed the following approximation of p _D for an infinite-acting aquifer:

(2.34) $p_{\mathrm{D}}=\frac{370.529\sqrt{t_{\mathrm{D}}}+137.582t_{\mathrm{D}}+5.69549{(t_{\mathrm{D}})}^{1.5}}{328.834+265.488\sqrt{t_{\mathrm{D}}}+45.2157t_{\mathrm{D}}+{(t_{\mathrm{D}})}^{1.5}}$

The dimensionless pressure derivative can then be approximated by:

(2.35) $p_{\mathrm{D}}^{\mathrm{\backslash }}=\frac{E}{F}$

where

E=716.441+46.7984(t _D)^0.5+270.038t _D+71.0098(t _D)^1.5

F=1296.86(t _D)^0.5+1204.73t _D+618.618(t _D)^1.5+538.072(t _D)²+142.41(t _D)^2.5

When the dimensionless time t _D>100, the following approximation can be used for p _D:

$p_{\mathrm{D}}=\frac{1}{2}[\ln (t_{\mathrm{D}})+0.80907]$

with the derivative given by:

$p_{\mathrm{D}}^{\mathrm{\backslash }}=\frac{1}{2t_{\mathrm{D}}}$

Fanchi (1985) matched the van Everdingen and Hurst tabulated values of the dimensionless pressure p _D as a function of t _D and r _D in Table 1.2 by using a regression model and proposed the following expression:

$p_{\mathrm{D}}=a_{\mathrm{0}}+a_1{t}_{\mathrm{D}}+a_2\ln (t_{\mathrm{D}})+a_2{[\ln (t_{\mathrm{D}})]}^2$

in which the regression coefficients are given below:

r eD	a 0	a 1	a 2	a 3
1.5	0.10371	1.6665700	−0.04579	−0.01023
2.0	0.30210	0.6817800	−0.01599	−0.01356
3.0	0.51243	0.2931700	0.015340	−0.06732
4.0	0.63656	0.1610100	0.158120	−0.09104
5.0	0.65106	0.1041400	0.309530	−0.11258
6.0	0.63367	0.0694000	0.41750	−0.11137
8.0	0.40132	0.0410400	0.695920	−0.14350
10.0	0.14386	0.0264900	0.896460	−0.15502
∞	0.82092	−0.000368	0.289080	0.028820

It should be noted that the Carter and Tracy method is not an exact solution to the diffusivity equation and should be considered as an approximation.

Example 2.9

Rework Example 2.7 by using the Carter and Tracy method.

Solution Example 2.7 shows the following preliminary results:

•

Water influx constant B=20.4   bbl/psi;

•

t _D=0.9888t.

Step 1.

For each time step n, calculate the total pressure drop Δp _n =p _i−p _n and the corresponding t _D:

n	t 1 (days)	p n	Δp n	t D
0	0	2500	0	0
1	182.5	2490	10	180.5
2	365.0	2472	28	361.0
3	547.5	2444	56	541.5
4	730.0	2408	92	722.0

Step 2.

Since the values of t _D are greater than 100, use Eq. (1.91) to calculate p _D and its derivative $p_{\mathrm{D}}^{\mathrm{\backslash }}$ . That is:

$\begin{array}{l}{p}_{\mathrm{D}}=\frac{1}{2}[\ln (t_{\mathrm{D}})+0.80907]\\ p_{\mathrm{D}}^{\mathrm{\backslash }}=\frac{1}{2t_{\mathrm{D}}}\end{array}$

n	t	t D	p D	$p_{\mathrm{D}}^{\mathrm{\backslash }}$
0	0	0	–	–
1	182.5	180.5	3.002	2.770×10−3
2	365.0	361.0	3.349	1.385×10−3
3	547.5	541.5	3.552	0.923×10−3
4	730.0	722.0	3.696	0.693×10−3

Step 3.

Calculate cumulative water influx by applying Eq. (2.33)

W _e after 182.5 days:

${(W_{\mathrm{e}})}_n={(W_{\mathrm{e}})}_{n-1}+[{(t_{\mathrm{D}})}_n-{(t_{\mathrm{D}})}_{n-1}]\times \left[\frac{B\mathrm{\Delta }{p}_n-{(W_{\mathrm{e}})}_{n-1}{(p_{\mathrm{D}}^{\backslash })}_n}{{(p_{\mathrm{D}})}_n-{(t_{\mathrm{D}})}_{n-1}{(p_{\mathrm{D}}^{\mathrm{\backslash }})}_n}\right]=0+[180.5-0]\times \left[\frac{(20.4)(10)-(0)(2.77\times {10}^{-3})}{3.002-(0)(2.77\times {10}^{-3})}\right]=\mathrm{12,266}\text{bbl}$

W _e after 365 days:

$W_{\mathrm{e}}=\mathrm{12,266}+[361-180.5]\times \left[\frac{(20.4)(28)-(\mathrm{12,266})(1.385\times {10}^{-3})}{3.349-(180.5)(1.385\times {10}^{-3})}\right]=\mathrm{42,545}\text{bbl}$

W _e after 547.5 days:

$W_{\mathrm{e}}=\mathrm{42,546}+[541.5-361]\times \left[\frac{(20.4)(56)-(\mathrm{42,546}\mathrm{)}\mathrm{(}0\text{.923}\times {\text{10}}^{-3})}{3.552-(361)(0.923\times {10}^{-3})}\right]=104,\text{406}\text{bbl}$

W _e after 720 days:

$W_{\mathrm{e}}=\mathrm{104,406}+[722-541.5]\times \left[\frac{(20.4)(92)-(\mathrm{104,406})(0.693\times {10}^{-3})}{3.696-(541.5)(0.693\times {10}^{-3})}\right]=\mathrm{202,477}\text{bbl}$

The following table compares the results of the Carter and Tracy water influx calculations with those of the van Everdingen and Hurst method:

Time (months)	Carter and Tracy, W e (bbl)	van Everdingen and Hurst, W e (bbl)
0	0	0
6	12,266	7085
12	42,546	32,435
18	104,400	85,277
24	202,477	175,522

The above comparison indicates that the Carter and Tracy method considerably overestimates the water influx. However, this is due to the fact that a large time step of 6 months was used in the Carter and Tracy method to determine the water influx. The accuracy of this method can be increased substantially by restricting the time step to 1 month. Recalculating the water influx on a monthly basis produces an excellent match with the van Everdingen and Hurst method as shown above.

Time (months)	Time (days)	p (psi)	Δp (psi)	t D	p D	$p_{\mathrm{D}}^{\mathrm{\backslash }}$	Carter–Tracy W e (bbl)	van Everdingen and Hurst W e (bbl)
0	0	2500.0	0.00	0	0.00	0	0.0	0
1	30	2498.9	1.06	30.0892	2.11	0.01661	308.8
2	61	2497.7	2.31	60.1784	2.45	0.00831	918.3
3	91	2496.2	3.81	90.2676	2.66	0.00554	1860.3
4	122	2494.4	5.56	120.357	2.80	0.00415	3171.7
5	152	2492.4	7.55	150.446	2.91	0.00332	4891.2
6	183	2490.2	9.79	180.535	3.00	0.00277	7057.3	7088.9
7	213	2487.7	12.27	210.624	3.08	0.00237	9709.0
8	243	2485.0	15.00	240.713	3.15	0.00208	12,884.7
9	274	2482.0	17.98	270.802	3.21	0.00185	16,622.8
10	304	2478.8	21.20	300.891	3.26	0.00166	20,961.5
11	335	2475.3	24.67	330.981	3.31	0.00151	25,938.5
12	365	2471.6	28.38	361.070	3.35	0.00139	31,591.5	32,435.0
13	396	2467.7	32.34	391.159	3.39	0.00128	37,957.8
14	426	2463.5	36.55	421.248	3.43	0.00119	45,074.5
15	456	2459.0	41.00	451.337	3.46	0.00111	52,978.6
16	487	2454.3	45.70	481.426	3.49	0.00104	61,706.7
17	517	2449.4	50.64	511.516	3.52	0.00098	71,295.3
18	547	2444.3	55.74	541.071	3.55	0.00092	81,578.8	85,277.0
19	578	2438.8	61.16	571.130	3.58	0.00088	92,968.2
20	608	2433.2	66.84	601.190	3.60	0.00083	105,323.0
21	638	2427.2	72.75	631.249	3.63	0.00079	118,681.0
22	669	2421.1	78.92	661.309	3.65	0.00076	133,076.0
23	699	2414.7	85.32	691.369	3.67	0.00072	148,544.0
24	730	2408.0	91.98	721.428	3.70	0.00069	165,119.0	175,522.0

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Gas Reservoirs

Tarek Ahmed , in Reservoir Engineering Handbook (Fifth Edition), 2019

The material balance method

If enough production-pressure history is available for a gas reservoir, the initial gas in place G, the initial reservoir pressure p_i, and the gas reserves can be calculated without knowing A, h, ϕ, or S_w. This is accomplished by forming a mass or mole balance on the gas as:

(13-5) ${\mathrm{n}}_{\mathrm{p}}={\mathrm{n}}_{\mathrm{i}}-{\mathrm{n}}_{\mathrm{f}}$

where:

n_p = moles of gas produced

n_i = moles of gas initially in the reservoir

n_f = moles of gas remaining in the reservoir

Representing the gas reservoir by an idealized gas container, as shown schematically in Figure 13-1, the gas moles in Equation 13-5 can be replaced by their equivalents using the real gas law to give:

Figure 13-1. An idealized water-drive gas reservoir.

(13-6) $\frac{{\mathrm{p}}_{\mathrm{sc}}{\mathrm{G}}_{\mathrm{p}}}{\mathrm{R}{\mathrm{T}}_{\mathrm{sc}}}=\frac{{\mathrm{p}}_{\mathrm{i}}\mathrm{V}}{{\mathrm{z}}_{\mathrm{i}}\mathrm{RT}}-\frac{\mathrm{p}\left[\mathrm{V}-\left({\mathrm{W}}_{\mathrm{e}}-{\mathrm{W}}_{\mathrm{p}}\right)\right]}{\mathrm{zRT}}$

where:

p_i = initial reservoir pressure

G_p = cumulative gas production, scf

p = current reservoir pressure

V = original gas volume, ft³

z_i = gas deviation factor at p_i

z = gas deviation factor at p

T = temperature, °R

W_e = cumulative water influx, ft ³

W_p = cumulative water production, ft³

Equation 13-6 is essentially the general material balance equation (MBE). Equation 13-6 can be expressed in numerous forms depending on the type of the application and the driving mechanism. In general, dry gas reservoirs can be classified into two categories:

○

Volumetric gas reservoirs

○

Water-drive gas reservoirs

The remainder of this chapter is intended to provide the basic background in natural gas engineering. There are several excellent textbooks that comprehensively address this subject, including the following:

○

Ikoku, C., Natural Gas Reservoir Engineering, 1984

○

Lee, J. and Wattenbarger, R., Gas Reservoir Engineering, SPE, 1996

Volumetric Gas Reservoirs

For a volumetric reservoir and assuming no water production, Equation 13-6 is reduced to:

(13-7) $\frac{{\mathrm{p}}_{\mathrm{sc}}{\mathrm{G}}_{\mathrm{p}}}{{\mathrm{T}}_{\mathrm{sc}}}=\left(\frac{{\mathrm{p}}_{\mathrm{i}}}{{\mathrm{z}}_{\mathrm{i}}\mathrm{T}}\right)\mathrm{V}-\left(\frac{\mathrm{p}}{\mathrm{zT}}\right)\mathrm{V}$

Equation 13-7 is commonly expressed in the following two forms:

Form 1. In terms of p/z

Rearranging Equation 13-7 and solving for p/z gives:

(13-8) $\frac{\mathrm{p}}{\mathrm{z}}=\frac{{\mathrm{p}}_{\mathrm{i}}}{{\mathrm{z}}_{\mathrm{i}}}-\left(\frac{{\mathrm{p}}_{\mathrm{sc}}\mathrm{T}}{{\mathrm{T}}_{\mathrm{sc}}\mathrm{V}}\right){\mathrm{G}}_{\mathrm{p}}$

Equation 13-8 is an equation of a straight line when (p/z) is plotted versus the cumulative gas production G_p, as shown in Figure 13-2. This straight-line relationship is perhaps one of the most widely used relationships in gas-reserves determination. The straight-line relationship provides the engineer with the reservoir characteristics:

Figure 13-2. Gas material balance equation "Tank Model."

○

Slope of the straight line is equal to:

(13-9) $\text{slope}=\frac{{\mathrm{p}}_{\mathrm{sc}}\mathrm{T}}{{\mathrm{T}}_{\mathrm{sc}}\mathrm{V}}$

The original gas volume V can be calculated from the slope and used to determine the areal extend of the reservoir from:

(13-10) $\mathrm{V}=\mathrm{43,560}\mathrm{Ah}\mathrm{\varphi }\left(1-{\mathrm{S}}_{\mathrm{wi}}\right)$

where A is the reservoir area in acres.

○

Intercept at G_p = 0 gives p_i/z_i

○

Intercept at p/z = 0 gives the gas initially in place G in scf

○

Cumulative gas production or gas recovery at any pressure

Example 13-2 ¹

A volumetric gas reservoir has the following production history.

Time, t years	Reservoir pressure, p psia	z	Cumulative production, Gp MMMscf
0.0	1798	0.869	0.00
0.5	1680	0.870	0.96
1.0	1540	0.880	2.12
1.5	1428	0.890	3.21
2.0	1335	0.900	3.92

The following data is also available:

ϕ = 13%

S_wi = 0.52

A = 1060 acres

h = 54 ft.

T = 164°F

Calculate the gas initially in place volumetrically and from the MBE.

Solution

Step 1.

Calculate B_gi from Equation 13-1

${\mathrm{B}}_{\mathrm{gi}}=0.02827\frac{\left(0.869\right)\left(164+460\right)}{1798}=0.00853\mathrm{f}{\mathrm{t}}^3/\mathrm{scf}$

Step 2.

Calculate the gas initially in place volumetrically by applying Equation 13-3.

$\mathrm{G}=\mathrm{43,560}\left(1060\right)\left(54\right)\left(0.13\right)\left(1–0.52\right)/0.00853=18.2\text{MMMscf}$

Step 3.

Plot p/z versus G_p as shown in Figure 13-3 and determine G.

$\mathrm{G}=14.2\text{MMMscf}$

Figure 13-3. Relationship of p/z vs. Gp for Example 13-2.

This checks the volumetric calculations.

The initial reservoir gas volume V can be expressed in terms of the volume of gas at standard conditions by:

$\mathrm{V}={\mathrm{B}}_{\mathrm{g}}\mathrm{G}=\left(\frac{{\mathrm{p}}_{\mathrm{sc}}}{{\mathrm{T}}_{\mathrm{sc}}}\frac{{\mathrm{z}}_{\mathrm{i}}\mathrm{T}}{{\mathrm{p}}_{\mathrm{i}}}\right)\mathrm{G}$

Combining the above relationship with that of Equation 13-8 gives:

(13-11) $\frac{\mathrm{p}}{\mathrm{z}}=\frac{{\mathrm{p}}_{\mathrm{i}}}{{\mathrm{z}}_{\mathrm{i}}}-\left[\left(\frac{{\mathrm{p}}_{\mathrm{i}}}{{\mathrm{z}}_{\mathrm{i}}}\right)\frac{1}{\mathrm{G}}\right]{\mathrm{G}}_{\mathrm{p}}$

This relationship can be expressed in a more simplified form as:

$\frac{\mathrm{p}}{\mathrm{Z}}=\frac{{\mathrm{p}}_{\mathrm{i}}}{{\mathrm{Z}}_{\mathrm{i}}}-\left[\mathrm{m}\right]{\mathrm{G}}_{\mathrm{p}}$

where the coefficient m is essentially constant and represents the resulting straight line when P/Z is plotted against G_P. The slope, m is defined by:

$\mathrm{m}=\left(\frac{{\mathrm{p}}_{\mathrm{i}}}{{\mathrm{Z}}_{\mathrm{i}}}\right)\frac{1}{\mathrm{G}}$

Equivalently, m is defined by Equation 13-9 as:

$\mathrm{m}=\frac{\mathrm{T}{\mathrm{p}}_{\mathrm{sc}}}{{\mathrm{T}}_{\mathrm{sc}}\mathrm{V}}$

where:

G = Original gas in place, scf

V = Original gas in place, ft³

Again, Equation 13-11 shows that for a volumetric reservoir, the relationship between (p/z) and G_p is essentially linear. This popular equation indicates that by extrapolation of the straight line to abscissa, i.e., at p/z = 0, will give the value of the gas initially in place as G = G_p.

The graphical representation of Equation 13-11 can be used to detect the presence of water influx, as shown graphically in Figure 13-4. When the plot of (p/z) versus G_p deviates from the linear relationship, it indicates the presence of water encroachment.

Figure 13-4. Impact of water drive on the P/z straight-line plot.

Many other graphical methods have been proposed for solving the gas MBE that are useful in detecting the presence of water influx. One such graphical technique is called the energy plot, which is based on arranging Equation 13-11 and taking the logarithm of both sides to give:

(13-12) $log\left[1-\frac{{\mathrm{z}}_{\mathrm{i}}\mathrm{p}}{{\mathrm{p}}_{\mathrm{i}}\mathrm{z}}\right]=log{\mathrm{G}}_{\mathrm{p}}-\text{logG}$

Figure 13-5 shows a schematic illustration of the plot.

Figure 13-5. The Energy-plot.

From Equation 13-12, it is obvious that a plot of [1 – (z_i p)/(p_i z)] versus G_p on log-log coordinates will yield a straight line with a slope of one (45° angle). An extrapolation to one on the vertical axis (p = 0) yields a value for initial gas in place, G. The graphs obtained from this type of analysis have been referred to as energy plots. They have been found to be useful in detecting water influx early in the life of a reservoir. If W_e is not zero, the slope of the plot will be less than one, and will also decrease with time, since W_e increases with time. An increasing slope can only occur as a result of either gas leaking from the reservoir or bad data, since the increasing slope would imply that the gas-occupied pore volume was increasing with time.

It should be pointed out that the average field, (p/Z)_Field, can be estimated from the individual wells' p/Z versus G_P performance by applying the following relationship:

${\left(\frac{\mathrm{p}}{\mathrm{Z}}\right)}_{\mathrm{Field}}=\frac{{\mathrm{p}}_{\mathrm{i}}}{{\mathrm{Z}}_{\mathrm{i}}}-\frac{\sum _{\mathrm{j}=1}^{\mathrm{n}}{\left({\mathrm{G}}_{\mathrm{P}}\right)}_{\mathrm{j}}}{\sum _{\mathrm{j}=1}^{\mathrm{n}}{\left[\frac{{\mathrm{G}}_{\mathrm{P}}}{\frac{{\mathrm{p}}_{\mathrm{i}}}{{\mathrm{Z}}_{\mathrm{i}}}-\frac{\mathrm{p}}{\mathrm{Z}}}\right]}_{\mathrm{j}}}$

The summation Σ is taking over the total number n of the field gas wells, that is, j = 1, 2, ... n. The total field performance in terms of (p/Z)_Field versus (G_P)_Field can then be constructed from the estimated values of the field p/Z and actual total field production, that is, (p/Z)_Field versus ΣG_P. The above equation is applicable as long as all wells are producing with defined static boundaries, that is, under pseudo-steady-state conditions. However, when using the MBE for reserve analysis for an entire reservoir that is characterized by a distinct lack of pressure equilibrium throughout, the following average reservoir pressure decline, (p/Z)_Field, can be used:

${\left(\frac{\mathrm{p}}{\mathrm{Z}}\right)}_{\mathrm{Field}}=\frac{\sum _{\mathrm{j}=1}^{\mathrm{n}}{\left(\frac{\mathrm{p}\mathrm{\Delta }{\mathrm{G}}_{\mathrm{P}}}{\mathrm{\Delta }\mathrm{p}}\right)}_{\mathrm{j}}}{\sum _{\mathrm{j}=1}^{\mathrm{n}}{\left(\frac{\mathrm{\Delta }{\mathrm{G}}_{\mathrm{P}}}{\mathrm{\Delta }\mathrm{p}/\mathrm{Z}}\right)}_{\mathrm{j}}}$

where Δp and ΔG_P are the incremental pressure difference and cumulative production, respectively.

The gas recovery factor (RF) at any depletion pressure is defined as the cumulative gas produced "G_P" at this pressure divided by the gas initially in place "G" i.e.:

$\mathrm{RF}=\frac{{\mathrm{G}}_{\mathrm{P}}}{\mathrm{G}}$

Introducing the gas RF to Equation 8-60 gives

$\frac{\mathrm{p}}{\mathrm{Z}}=\frac{{\mathrm{p}}_{\mathrm{i}}}{{\mathrm{Z}}_{\mathrm{i}}}\left[1-\frac{{\mathrm{G}}_{\mathrm{P}}}{\mathrm{G}}\right]$

or

$\frac{\mathrm{p}}{\mathrm{Z}}=\frac{{\mathrm{p}}_{\mathrm{i}}}{{\mathrm{Z}}_{\mathrm{i}}}\left[1-\mathrm{RF}\right]$

Solving for the recovery factor at any depletion pressure gives:

$\mathrm{RF}=1-\left[\frac{{\mathrm{Z}}_{\mathrm{i}}}{\mathrm{Z}}\frac{\mathrm{p}}{{\mathrm{p}}_{\mathrm{i}}}\right]$

Form 2. In terms of B_g

From the definition of the gas formation volume factor, it can be expressed as:

${\mathrm{B}}_{\mathrm{gi}}=\frac{\mathrm{V}}{\mathrm{G}}$

Combining the above expression with Equation 13-1 gives:

(13-13) $\frac{{\mathrm{p}}_{\mathrm{sc}}}{{\mathrm{T}}_{\mathrm{sc}}}\frac{{\mathrm{z}}_{\mathrm{i}}\mathrm{T}}{{\mathrm{p}}_{\mathrm{i}}}=\frac{\mathrm{V}}{\mathrm{G}}$

where:

V = volume of gas originally in place, ft³

G = volume of gas originally in place, scf

p_i = original reservoir pressure

z_i = gas compressibility factor at p_i

Equation 13-13 can be combined with Equation 13-7, to give:

(13-14) $\mathrm{G}=\frac{{\mathrm{G}}_{\mathrm{p}}{\mathrm{B}}_{\mathrm{g}}}{{\mathrm{B}}_{\mathrm{g}}-{\mathrm{B}}_{\mathrm{gi}}}$

Equation 13-14 suggests that to calculate the initial gas volume, the only information required is production data, pressure data, gas specific gravity for obtaining z-factors, and reservoir temperature. Early in the producing life of a reservoir, however, the denominator of the right-hand side of the material balance equation is very small, while the numerator is relatively large. A small change in the denominator will result in a large discrepancy in the calculated value of initial gas in place. Therefore, the material balance equation should not be relied on early in the producing life of the reservoir.

Material balances on volumetric gas reservoirs are simple. Initial gas in place may be computed from Equation 13-14 by substituting cumulative gas produced and appropriate gas formation volume factors at corresponding reservoir pressures during the history period. If successive calculations at various times during the history give consistent values for initial gas in place, the reservoir is operating under volumetric control and computed G is reliable, as shown in Figure 13-6. Once G has been determined and the absence of water influx established in this fashion, the same equation can be used to make future predictions of cumulative gas production function of reservoir pressure.

Figure 13-6. Graphical determination of the gas initially in place G.

Ikoku (1984) points out that successive application of Equation 13-14 will normally result in increasing values of the gas initially in place G with time if water influx is occurring. If there is gas leakage to another zone due to bad cement jobs or casing leaks, however, the computed value of G may decrease with time.

Example 13-3

After producing 360 MMscf of gas from a volumetric gas reservoir, the pressure has declined from 3200 psi to 3000 psi, given:

${\mathrm{B}}_{\mathrm{gi}}=0.005278{\text{ft}}^3/{\text{scfB}}_{\mathrm{g}}=0.005390{\text{ft}}^3/\mathrm{scf}$

a.

Calculate the gas initially in place.

b.

Recalculate the gas initially in place assuming that the pressure measurements were incorrect and the true average pressure is 2900 psi. The gas formation volume factor at this pressure is 0.00558 ft³/scf.

Solution

a.

Using Equation 13-14, calculate the gas-in-place "G":

$\mathrm{G}=\frac{{\mathrm{G}}_{\mathrm{p}}{\mathrm{B}}_{\mathrm{g}}}{{\mathrm{B}}_{\mathrm{g}}-{\mathrm{B}}_{\mathrm{gi}}}\mathrm{G}=\frac{360\times 10^6\left(0.00539\right)}{0.00539-0.005278}=17.325\text{MMMscf}$

b.

Recalculate G by using the correct value of B_g.

$\mathrm{G}=\frac{360\times 10^6\left(0.00668\right)}{0.00558-0.005278}=6.652\text{MMMscf}$

Thus, an error of 100 psia, which is only 3.5% of the total reservoir pressure, resulted in an increase in calculated gas in place of approximately 160%, a 2½-fold increase. Note that a similar error in reservoir pressure later in the producing life of the reservoir will not result in an error as large as that calculated early in the producing life of the reservoir.

Water-Drive Gas Reservoirs

If the gas reservoir has a water drive, then there will be two unknowns in the material balance equation, even though production data, pressure, temperature, and gas gravity are known. These two unknowns are initial gas in place and cumulative water influx. In order to use the material balance equation to calculate initial gas in place, some independent method of estimating W _e, the cumulative water influx, must be developed as discussed in Chapter 11.

Equation 13-14 can be modified to include the cumulative water influx and water production to give:

(13-15) $\mathrm{G}=\frac{{\mathrm{G}}_{\mathrm{p}}{\mathrm{B}}_{\mathrm{g}}-\left({\mathrm{W}}_{\mathrm{e}}-{\mathrm{W}}_{\mathrm{p}}{\mathrm{B}}_{\mathrm{W}}\right)}{{\mathrm{B}}_{\mathrm{g}}-{\mathrm{B}}_{\mathrm{gi}}}$

The above equation can be arranged and expressed as:

(13-16) $\mathrm{G}+\frac{{\mathrm{W}}_{\mathrm{e}}}{{\mathrm{B}}_{\mathrm{g}}-{\mathrm{B}}_{\mathrm{gi}}}=\frac{{\mathrm{G}}_{\mathrm{p}}{\mathrm{B}}_{\mathrm{g}}-{\mathrm{W}}_{\mathrm{p}}{\mathrm{B}}_{\mathrm{W}}}{{\mathrm{B}}_{\mathrm{g}}-{\mathrm{B}}_{\mathrm{gi}}}$

Equation 13-16 reveals that for a volumetric reservoir, i.e., W_e = 0, the right-hand side of the equation will be constant regardless of the amount of gas G_p that has been produced. For a water-drive reservoir, the values of the left-hand side of Equation 13-16 will continue to increase because of the W_e/(B_g – B_gi) term. A plot of several of these values at successive time intervals is illustrated in Figure 13-7. Extrapolation of the line formed by these points back to the point where G_p = 0 shows the true value of G, because when G_p = 0, then W_e/(B_g – B_gi) is also zero.

Figure 13-7. Impact of water influx on estimating Original-Gas-In-Place.

This graphical technique can be used to estimate the value of We, because at any time the difference between the horizontal line (i.e., true value of G) and the sloping line [G + (W_e)/(B_g – B_gi) will give the value of We/(Bg – Bgi).

Because gas often is bypassed and trapped by encroaching water, recovery factors for gas reservoirs with water drive can be significantly lower than for volumetric reservoirs produced by simple gas expansion. In addition, the presence of reservoir heterogeneities, such as low-permeability stringers or layering, may reduce gas recovery further. As noted previously, ultimate recoveries of 80% to 90% are common in volumetric gas reservoirs, while typical recovery factors in water-drive gas reservoirs can range from 50% to 70%. As illustrated schematically in Figure (13-7A), the amount of gas that is trapped in a region that has been flooded by water encroachment can be estimated by defining the following characteristic reservoir parameters and taking the steps outlined below (Figure 13-7A):

Figure 13-7A. Accounting for the trapped-gas by the water influx.

(P.V)   = reservoir pore volume, ft³

(P.V)_wiz  = pore volume of the water-invaded zone, ft³

S_grw  = residual gas saturation to water displacement

S_wi  = initial water saturation

G   = gas initially in place, scf

G_P  = cumulative gas production at depletion pressure p, scf

B_gi  = initial gas formation volume factor, ft³/scf

B_g  = gas formation volume factor at depletion pressure p, ft³/scf

Z   = gas deviation factor at depletion pressure p

Step 1.

Express the reservoir pore volume, (P.V), in terms of the initial gas in place, G, as follows:

$\mathrm{G}{\mathrm{B}}_{\mathrm{gi}}=\left(\mathrm{P}.\mathrm{V}\right)\left(1–{\mathrm{S}}_{\mathrm{wi}}\right)$

Solving for the reservoir pore volume gives:

$\left(\mathrm{P}.\mathrm{V}\right)=\frac{\mathrm{G}{\mathrm{B}}_{\mathrm{gi}}}{1-{\mathrm{S}}_{\mathrm{wi}}}$

Step 2.

Calculate the pore volume in the water-invaded zone:

${\mathrm{W}}_{\mathrm{e}}–{\mathrm{W}}_{\mathrm{p}}{\mathrm{B}}_{\mathrm{w}}={\left(\mathrm{P}.\mathrm{V}\right)}_{\mathrm{wiz}}\left(1–{\mathrm{S}}_{\mathrm{wi}}–{\mathrm{S}}_{\mathrm{grw}}\right)$

Solving for the pore volume of the water-invaded zone, (P.V)_wiz, gives:

${\left(\mathrm{P}.\mathrm{V}\right)}_{\mathrm{wiz}}=\frac{{\mathrm{W}}_{\mathrm{e}}-{\mathrm{W}}_{\mathrm{p}}{\mathrm{B}}_{\mathrm{w}}}{1-{\mathrm{S}}_{\mathrm{wi}}-{\mathrm{S}}_{\mathrm{grw}}}$

Step 3.

Calculate trapped gas volume in the water-invaded zone, or:

$\begin{array}{l}{\left(\mathrm{V}\right)}_{\text{trapped}-\mathrm{gas}}={\left(\mathrm{P}.\mathrm{V}\right)}_{\mathrm{wiz}}{\mathrm{S}}_{\mathrm{grw}}\\ {\left(\mathrm{V}\right)}_{\text{trapped}-\mathrm{gas}}=\left[\frac{{\mathrm{W}}_{\mathrm{e}}-{\mathrm{W}}_{\mathrm{p}}{\mathrm{B}}_{\mathrm{w}}}{1-{\mathrm{S}}_{\mathrm{wi}}-{\mathrm{S}}_{\mathrm{grw}}}\right]{\mathrm{S}}_{\mathrm{grw}}\end{array}$

Step 4.

Calculate the number n of moles of gas trapped in the water-invaded zone by using the equation of state, or:

$\mathrm{P}{\left(\mathrm{V}\right)}_{\text{trapped}-\mathrm{gas}}=\mathrm{Z}\mathrm{n}\mathrm{R}\mathrm{T}$

Solving for n gives:

$\mathrm{n}=\frac{\mathrm{p}\left[\frac{{\mathrm{W}}_{\mathrm{e}}-{\mathrm{W}}_{\mathrm{p}}{\mathrm{B}}_{\mathrm{w}}}{1-{\mathrm{S}}_{\mathrm{wi}}-{\mathrm{S}}_{\mathrm{grw}}}\right]{\mathrm{S}}_{\mathrm{grw}}}{\mathrm{Z}\mathrm{R}\mathrm{T}}$

The above expression indicates that the higher the pressure, the greater the quantity of trapped gas. Dake (1994) points out that if the pressure is reduced by rapid gas withdrawal, the volume of gas trapped in each individual pore space, that is, S_grw, will remain unaltered, but its quantity, n, will be reduced.

Step 5.

The gas saturation at any pressure can be adjusted to account for the trapped gas, as follows:

$\begin{array}{l}{\mathrm{S}}_{\mathrm{g}}=\frac{\text{remaining}\mathrm{gas}\text{volume}-\text{trapped}\mathrm{gas}\text{volume}}{\text{reservoir pore volume}-\text{pore volume of water invaded zone}}\\ {\mathrm{S}}_{\mathrm{g}}=\frac{\left(\mathrm{G}-{\mathrm{G}}_{\mathrm{p}}\right){\mathrm{B}}_{\mathrm{g}}-\left[\frac{{\mathrm{W}}_{\mathrm{e}}-{\mathrm{W}}_{\mathrm{p}}{\mathrm{B}}_{\mathrm{w}}}{1-{\mathrm{S}}_{\mathrm{wi}}-{\mathrm{S}}_{\mathrm{grw}}}\right]{\mathrm{S}}_{\mathrm{grw}}}{\left(\frac{\mathrm{G}{\mathrm{B}}_{\mathrm{gi}}}{1-{\mathrm{S}}_{\mathrm{wi}}}\right)-\left[\frac{{\mathrm{W}}_{\mathrm{e}}-{\mathrm{W}}_{\mathrm{p}}{\mathrm{B}}_{\mathrm{w}}}{1-{\mathrm{S}}_{\mathrm{wi}}-{\mathrm{S}}_{\mathrm{grw}}}\right]}\end{array}$

As illustrated in Figure (13-7A), the pore volume of the water invaded zone "(P.V)_wiz", in ft³, is given by:

${\left(\mathrm{P}.\mathrm{V}\right)}_{\mathrm{wiz}}=5.615\left[\frac{{\mathrm{W}}_{\mathrm{e}}-{\mathrm{W}}_{\mathrm{p}}{\mathrm{B}}_{\mathrm{w}}}{1-{\mathrm{S}}_{\mathrm{wi}}-{\mathrm{S}}_{\mathrm{grw}}}\right]$

To modify the gas material balance equation to account for the trapped gas a in the water Influx zone; the MBE is expressed in terms of gas moles to give in the following generalized form:

${\mathrm{n}}_{\mathrm{P}}={\mathrm{n}}_{\mathrm{i}}–{\mathrm{n}}_{\mathrm{rfg}}–{\mathrm{n}}_{\mathrm{trapped}}$

Where:

n_p = number of moles of gas produced

n_i = initial number of moles

n_rfg = number of moles of remaining as free gas

n_tapped = number of moles of the trapped gas in the water influx zone

Real gas equation of state and the gas-expansion factor "E_g" as defined by Equation (13-2)

$\mathrm{n}=\frac{\mathrm{V}\mathrm{P}}{\mathrm{Z}\mathrm{R}\mathrm{T}}$

${\mathrm{E}}_{\mathrm{g}}=\frac{{\mathrm{T}}_{\mathrm{es}}}{{\mathrm{P}}_{\mathrm{sc}}}\frac{\mathrm{P}}{\mathrm{Z}\mathrm{T}}$

Replacing moles of gas in the generalized molal material balance gives:

${\mathrm{G}}_{\mathrm{P}}=\left[\frac{{\mathrm{T}}_{\mathrm{sc}}}{{\mathrm{P}}_{\mathrm{sc}}}\frac{{\mathrm{P}}_{\mathrm{i}}}{{\mathrm{Z}}_{\mathrm{i}}\mathrm{T}}\right]\mathrm{V}-\left[\frac{{\mathrm{T}}_{\mathrm{sc}}}{{\mathrm{P}}_{\mathrm{sc}}}\frac{\mathrm{P}}{\mathrm{Z}\mathrm{T}}\right]\left[\left(\mathrm{P}.\mathrm{V}\right){\mathrm{S}}_{\mathrm{gi}}-{\left(\mathrm{P}.\mathrm{V}\right)}_{\mathit{wiz}}\left({\mathrm{S}}_{\mathrm{gi}}-{\mathrm{S}}_{\mathrm{grw}}\right)\right]$

Introducing the gas expansion factor in the above equation and rearranging, gives:

${\mathrm{G}}_{\mathrm{P}}={\mathrm{E}}_{\mathrm{gi}}\mathrm{V}-{\mathrm{E}}_{\mathrm{g}}\left[\frac{\mathrm{G}}{{\mathrm{E}}_{\mathrm{gi}}}-{\left(\mathrm{P}.\mathrm{V}\right)}_{\mathrm{wiz}}\left({\mathrm{S}}_{\mathrm{gi}}-{\mathrm{S}}_{\mathrm{grw}}\right)\right]$

${\mathrm{G}}_{\mathrm{P}}=\mathrm{G}\left(1-\frac{{\mathrm{E}}_{\mathrm{g}}}{{\mathrm{E}}_{\mathrm{gi}}}\right)+{\mathrm{E}}_{\mathrm{g}}{\left(\mathrm{P}.\mathrm{V}\right)}_{\mathrm{wiz}}\left({\mathrm{S}}_{\mathrm{gi}}-{\mathrm{S}}_{\mathrm{grw}}\right)$

The above expression in the linearized form:

$\frac{{\mathrm{G}}_{\mathrm{P}}}{1-\frac{{\mathrm{E}}_{\mathrm{g}}}{{\mathrm{E}}_{\mathrm{gi}}}}=\mathrm{G}+\left({\mathrm{S}}_{\mathrm{gi}}-{\mathrm{S}}_{\mathrm{grw}}\right)\left[\frac{{\mathrm{E}}_{\mathrm{g}}{\left(\mathrm{P}.\mathrm{V}\right)}_{\mathrm{wiz}}}{1-\frac{{\mathrm{E}}_{\mathrm{g}}}{{\mathrm{E}}_{\mathrm{gi}}}}\right]$

The above generalized form of the material balance equation suggests that plotting [G_p/(1–E_g/E_gi)] versus [E_g (P.V)_wiz/(1–E_g/E_gi)] would produce a straight-line with an intercept that corresponds to the gas-initially in-place "G." The main advantage of the modified form is that it reflects the gas loss in the water invaded zone.

Or a more convenient form is to express the modified MBE in terms of p/z, to give:

$\frac{{\mathrm{G}}_{\mathrm{P}}}{1-\frac{\mathrm{p}}{{\mathrm{p}}_{\mathrm{i}}}\frac{{\mathrm{z}}_{\mathrm{i}}}{\mathrm{z}}}=\mathrm{G}+\left({\mathrm{S}}_{\mathrm{gi}}-{\mathrm{S}}_{\mathrm{grw}}\right)\left[\frac{{\mathrm{E}}_{\mathrm{g}}{\left(\mathrm{P}.\mathrm{V}\right)}_{\mathrm{wiz}}}{1-\frac{\mathrm{p}}{{\mathrm{p}}_{\mathrm{i}}}\frac{{\mathrm{z}}_{\mathrm{i}}}{\mathrm{z}}}\right]$

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Oil Recovery Mechanisms and The Material Balance Equation

Tarek Ahmed , in Reservoir Engineering Handbook (Fifth Edition), 2019

The MBE as an Equation of a Straight Line

An insight into the general MBE, i.e., Equation 11-15, may be gained by considering the physical significance of the following groups of terms of which it is comprised:

○

N_p [B_o + (R_p – R_s) B_g]   Represents the reservoir volume of cumulative oil and gas produced.

○

[W_e – W_p B_w]   Refers to the net water influx that is retained in the reservoir.

○

[G_inj B_ginj + W_inj B_w]   This pressure maintenance term represents cumulative fluid injection in the reservoir.

○

[m B_oi (B_g/B_gi – 1)]   Represents the net expansion of the gas cap that occurs with the production of N_p stocktank barrels of oil (as expressed in bbl/STB of original oil in place).

There are essentially three unknowns in Equation 11-15:

a.

The original oil in place N

b.

The cumulative water influx W _e

c.

The original size of the gas cap as compared to the oil zone size m

In developing a methodology for determining the above three unknowns, Havlena and Odeh (1963) expressed Equation 11-15 in the following form:

(11-24) $\begin{array}{l}{\mathrm{N}}_{\mathrm{p}}\left[{\mathrm{B}}_{\mathrm{o}}+\left({\mathrm{R}}_{\mathrm{p}}-{\mathrm{R}}_{\mathrm{s}}\right){\mathrm{B}}_{\mathrm{g}}\right]+{\mathrm{W}}_{\mathrm{p}}{\mathrm{B}}_{\mathrm{w}}=\mathrm{N}\left[\left({\mathrm{B}}_{\mathrm{o}}-{\mathrm{B}}_{\mathrm{oi}}\right)+\left({\mathrm{R}}_{\mathrm{si}}-{\mathrm{R}}_{\mathrm{s}}\right){\mathrm{B}}_{\mathrm{g}}\right]\\ +\mathrm{m}\mathrm{N}{\mathrm{B}}_{\mathrm{oi}}\left(\frac{{\mathrm{B}}_{\mathrm{g}}}{{\mathrm{B}}_{\mathrm{gi}}}-1\right)+\mathrm{N}\left(1+\mathrm{m}\right){\mathrm{B}}_{\mathrm{oi}}\left[\frac{{\mathrm{c}}_{\mathrm{w}}{\mathrm{S}}_{\mathrm{wi}}+{\mathrm{c}}_{\mathrm{f}}}{1+{\mathrm{S}}_{\mathrm{wi}}}\right]\mathrm{\Delta p}\\ +{\mathrm{W}}_{\mathrm{e}}+{\mathrm{W}}_{\mathrm{inj}}{\mathrm{B}}_{\mathrm{w}}+{\mathrm{G}}_{\mathrm{inj}}{\mathrm{B}}_{\text{ginj}}\end{array}$

Havlena and Odeh further expressed Equation 11-24 in a more condensed form as:

$\mathrm{F}=\mathrm{N}\left[{\mathrm{E}}_{\mathrm{o}}+\mathrm{m}{\mathrm{E}}_{\mathrm{g}}+{\mathrm{E}}_{\mathrm{f},\mathrm{w}}\right]+\left({\mathrm{W}}_{\mathrm{e}}+{\mathrm{W}}_{\mathrm{inj}}{\mathrm{B}}_{\mathrm{w}}+{\mathrm{G}}_{\mathrm{inj}}{\mathrm{B}}_{\text{ginj}}\right)$

Assuming, for the purpose of simplicity, that no pressure maintenance by gas or water injection is being considered, the above relationship can be further simplified and written as:

(11-25) $\mathrm{F}=\mathrm{N}\left[{\mathrm{E}}_{\mathrm{o}}+\mathrm{m}{\mathrm{E}}_{\mathrm{g}}+{\mathrm{E}}_{\mathrm{f},\mathrm{w}}\right]+{\mathrm{W}}_{\mathrm{e}}$

In which the terms F, E_o, E_g, and E_f,w are defined by the following relationships:

○

F represents the underground withdrawal and given by:

(11-26) $\mathrm{F}={\mathrm{N}}_{\mathrm{p}}\left[{\mathrm{B}}_{\mathrm{o}}+\left({\mathrm{R}}_{\mathrm{p}}-{\mathrm{R}}_{\mathrm{s}}\right){\mathrm{B}}_{\mathrm{g}}\right]+{\mathrm{W}}_{\mathrm{p}}{\mathrm{B}}_{\mathrm{w}}$

In terms of the two-phase formation volume factor B_t, the underground withdrawal F can be written as:

(11-27) $\mathrm{F}={\mathrm{N}}_{\mathrm{p}}\left[{\mathrm{B}}_{\mathrm{t}}+\left({\mathrm{R}}_{\mathrm{p}}-{\mathrm{R}}_{\mathrm{si}}\right){\mathrm{B}}_{\mathrm{g}}\right]+{\mathrm{W}}_{\mathrm{p}}{\mathrm{B}}_{\mathrm{w}}$

○

E_o describes the expansion of oil and its originally dissolved gas and is expressed in terms of the oil formation volume factor as:

(11-28) ${\mathrm{E}}_{\mathrm{o}}=\left({\mathrm{B}}_{\mathrm{o}}-{\mathrm{B}}_{\mathrm{oi}}\right)+\left({\mathrm{R}}_{\mathrm{si}}-{\mathrm{R}}_{\mathrm{s}}\right){\mathrm{B}}_{\mathrm{g}}$

Or equivalently, in terms of B_t:

(11-29) ${\mathrm{E}}_{\mathrm{o}}={\mathrm{B}}_{\mathrm{t}}-{\mathrm{B}}_{\mathrm{ti}}$

○

E_g is the term describing the expansion of the gas-cap gas and is defined by the following expression:

(11-30) ${\mathrm{E}}_{\mathrm{g}}={\mathrm{B}}_{\mathrm{oi}}\left[\left({\mathrm{B}}_{\mathrm{g}}/{\mathrm{B}}_{\mathrm{gi}}\right)-1\right]$

In terms of the two-phase formation volume factor B_t, essentially B_ti = B_oi or:

${\mathrm{E}}_{\mathrm{g}}={\mathrm{B}}_{\mathrm{ti}}\left[\left({\mathrm{B}}_{\mathrm{g}}/{\mathrm{B}}_{\mathrm{gi}}\right)-1\right]$

○

E_f,w represents the expansion of the initial water and the reduction in the pore volume and is given by:

(11-31) ${\mathrm{E}}_{\mathrm{f},\mathrm{w}}=\left(1+\mathrm{m}\right){\mathrm{B}}_{\mathrm{oi}}\left[\frac{{\mathrm{c}}_{\mathrm{w}}{\mathrm{S}}_{\mathrm{wi}}+{\mathrm{c}}_{\mathrm{f}}}{1-{\mathrm{S}}_{\mathrm{wi}}}\right]\mathrm{\Delta p}$

Havlena and Odeh examined several cases of varying reservoir types with Equation 11-25 and pointed out that the relationship can be rearranged into the form of a straight line. For example, in the case of a reservoir which has no initial gas cap (i.e., m = 0) or water influx (i.e., W_e = 0), and negligible formation and water compressibilities (i.e., c_f and c_w = 0); Equation 11-25 reduces to:

$\mathrm{F}=\mathrm{N}{\mathrm{E}}_{\mathrm{o}}$

The above expression suggests that a plot of the parameter F as a function of the oil expansion parameter E_o would yield a straight line with a slope N and intercept equal to zero.

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Performance of Oil Reservoirs

Tarek Ahmed , Paul D. McKinney , in Advanced Reservoir Engineering, 2005

Injected water drive index

The relative efficiency of the water injection pressure maintenance operations is expressed by:

$\text{WII}=\frac{W_{\text{inj}}{B}_{\text{winj}}}{A}$

The magnitude of WII indicates the importance of the injected water as an improved recovery agent.

Injected gas drive index

Similar to the injected water drive index, the magnitude of its value indicates the relative importance this drive mechanism as compared to the other indices, as given by:

$\text{GII}=\frac{G_{\text{inj}}{B}_{\text{ginj}}}{A}$

Note that for a depletion drive reservoir under pressure maintenance operations by gas injection, Equation 4.3.20 is reduced to:

$\text{DDI}\text{+}\text{EDI + GII = 1}\text{.0}$

Since the recovery by depletion drive and expansion of the fluid and rock are usually poor, it is essential to maintain a high injected gas drive index. If the reservoir pressure can be maintained constant or declining at a slow rate, the values of DDI and EDI will be minimized because the changes in the numerators of both terms will essentially approach zeros. Theoretically, the highest recovery would occur at constant reservoir pressure; however, economic factors and feasibility of operation may dictate some pressure reduction.

In the absence of gas or water injection, Cole (1969) pointed out that since the sum of the remaining four driving indexes is equal to 1, it follows that if the magnitude of one of the index terms is reduced, then one or both of the remaining terms must be correspondingly increased. An effective water drive will usually result in maximum recovery from the reservoir. Therefore, if possible, the reservoir should be operated to yield a maximum water drive index and minimum values for the depletion drive index and the gas cap drive index. Maximum advantage should be taken of the most efficient drive available, and where the water drive is too weak to provide an effective displacing force, it may be possible to utilize the displacing energy of the gas cap. In any event, the depletion drive index should be maintained as low as possible at all times, as this is normally the most inefficient driving force available.

Equation 4.3.20 can be solved at any time to determine the magnitude of the various driving indexes. The forces displacing the oil and gas from the reservoir are subject to change from time to time and for this reason Equation 4.3.20 should be solved periodically to determine whether there has been any change in the driving indexes. Changes in fluid withdrawal rates are primarily responsible for changes in the driving indexes. For example, reducing the oil producing rate could result in an increased water drive index and a correspondingly reduced depletion drive index in a reservoir containing a weak water drive. Also, by shutting in wells producing large quantities of water, the water drive index could be increased, as the net water influx (gross water influx minus water production) is the important factor.

When the reservoir has a very weak water drive, but has a fairly large gas cap, the most efficient reservoir producing mechanism may be the gas cap, in which case a large gas cap drive index is desirable. Theoretically, recovery by gas cap drive is independent of producing rate, as the gas is readily expansible. Low vertical permeability could limit the rate of expansion of the gas cap, in which case the gas cap drive index would be rate sensitive. Also, gas coning into producing wells will reduce the effectiveness of the gas cap expansion due to the production of free gas. Gas coning is usually a rate-sensitive phenomenon: the higher the producing rates, the greater the amount of coning.

An important factor in determining the effectiveness of a gas cap drive is the degree of conservation of the gas cap gas. As a practical mater, it will often be impossible, because of royalty owners or lease agreements, to completely eliminate gas cap gas production. Where free gas is being produced, the gas cap drive index can often be markedly increased by shutting in high gas-oil-ratio wells, and, if possible, transferring their allowables to other low gas-oil-ratio wells.

Figure 4.15 shows a set of plots that represents various driving indexes for a combination drive reservoir. At point A some of the structurally low wells are reworked to reduce water production. This results in an effective increase in the water drive index. At point B workover operations are complete, water, gas, and oil producing rates are relatively stable, and the driving indexes show no change. At point C some of the wells which have been producing relatively large, but constant, volumes of water are shut in, which results in an increase in the water drive index. At the same time some of the upstructure, high gas-oil-ratio wells have been shut in and their allowables transferred to wells lower on the structure producing with normal gas-oil ratios. At point D gas is being returned to the reservoir, and the gas cap drive index is exhibiting a decided increase. The water drive index is relatively constant, although it is decreasing somewhat, and the depletion drive index is showing a marked decline. This is indicative of a more efficient reservoir operation, and if the depletion drive index can be reduced to zero, relatively good recovery can be expected from the reservoir. Of course, to achieve a zero depletion drive index would require the complete maintenance of reservoir pressure, which is often difficult to accomplish. It can be noted from Figure 4.15 that the sum of the various drive indexes is always equal to 1.

Figure 4.15. Driving indexes in a combination drive reservoir.

(After Clark N.J., Elements of Petroleum Reservoirs, SPE, 1969)

Example 4.2

A combination drive reservoir contains 10 MMSTB of oil initially in place. The ratio of the original gas cap volume to the original oil volume, i.e., m, is estimated as 0.25. The initial reservoir pressure is 3000 psia at 150°F. The reservoir produced 1MMSTB of oil, 1100 MMscf of gas of 0.8 specific gravity, and 50000 STB of water by the time the reservoir pressure dropped to 2800 psi. The following PVT data is available:

	3000 psi	2800 psi
B o, bbl/STB	1.58	1.48
R s, scf/STB	1040	850
B g, bbl/scf	0.00080	0.00092
B t, bbl/STB	1.58	1.655
B w, bbl/STB	1.000	1.000

The following data is also available:

$S_{\text{wi}}=0.20,c_{\text{w}}=1.5\times {10}^{-6}{\text{psi}}^{-1},c_{\text{f}}=1\times {10}^{-6}{\text{psi}}^{-1}$

Calculate:

(a)

the cumulative water influx;

(b)

the net water influx;

(c)

the primary driving indices at 2800 psi.

Solution Because the reservoir contains a gas cap, the rock and fluid expansion can be neglected, i.e., set c _f and c _w = 0. However, for illustration purposes, the rock and fluid expansion term will be included in the calculations.

(a)

The cumulative water influx:

Step 1.

Calculate the cumulative gas-oil ratio R _p:

$R_{\text{p}}=\frac{G_{\text{p}}}{N_{\text{p}}}=\frac{1100\times {10}^6}{1\times {10}^6}=1100\text{scf/STB}$

Step 2.

Arrange Equation 4.3.17 to solve for W _e:

$\begin{array}{l}{W}_{\text{e}}=N_{\text{p}}[B_{\text{t}}+(R_{\text{p}}-R_{\text{si}})B_{\text{g}}]\\ -N\left[(B_{\text{t}}-B_{\text{ti}})+m{B}_{\text{ti}}\left(\frac{B_{\text{g}}}{B_{\text{gi}}}-1\right)+B_{\text{ti}}(1+m)\left(\frac{S_{\text{wi}}{c}_{\text{w}}+c_{\text{f}}}{1-S_{\text{wi}}}\right)\Delta p\right]+W_{\text{p}}{B}_{\text{wp}}\\ ={10}^6[1.655+(1100-1040)0.00092]-{10}^7\times \left[(1.655-1.58)+0.25(1.58)\left(\frac{0.00092}{0.00080}-1\right)+1.58(1+0.25)\left(\frac{0.2(1.5\times {10}^{-6})}{1-0.2}\right)\times (3000-2800)\right]+50000=411281\text{bbl}\end{array}$

Neglecting the rock and fluid expansion term, the cumulative water influx is 417 700 bbl.

(b)

The net water influx:

$\begin{array}{l}\text{Net water influx =}{W}_{\text{e}}-W_{\text{p}}{B}_{\text{w}}=411281-50000\\ =361281\text{bbl}\end{array}$

(c)

The primary recovery indices:

Step 1.

Calculate the parameter A by using Equation 4.3.19:

$\begin{array}{l}A=N_{\text{p}}[B_{\text{t}}+(R_{\text{p}}-R_{\text{si}})B_{\text{g}}]\\ =\text{(1}{\text{. 0 × 10}}^6\text{)}[1\text{.655 + (1100 - 1040)0}\text{.00092}]\\ =\text{ 1 710 000}\end{array}$

Step 2.

Calculate DDI, SDI, and WDI by applying Equations 4.3.21 through 4.3.23, respectively:

$\begin{array}{l}\text{DDI}\text{=}N(B_{\text{t}}-B_{\text{ti}})/A\\ =\frac{{\text{10 × 10}}^6\text{(1}\text{.655}-1\text{.58)}}{\text{1 710 000}}=\text{ 0}\text{.4385}\\ \text{SDI=}[Nm{B}_{\text{ti}}\text{ (}{B}_{\text{g}}-B_{\text{gi}}\text{)/}{B}_{\text{gi}}]/A\\ =\frac{{\text{10×10}}^6\text{(0}\text{.25)(1}\text{.58)(0}\text{.00092}-0\text{.0008)/0}\text{.0008}}{1710000}\\ =0.3465\\ \text{WDI}=(W_{\text{e}}-W_{\text{p}}{B}_{\text{w}})/A\\ =\frac{411281-50000}{1710000}=0.2112\end{array}$

Since:

$\text{DDI}\text{+}\text{SDI}\text{+WDI}\text{+}\text{EDI}\text{=}1\text{.0}$

then:

$\text{EDI}\text{=}1-0.4385-0.3465-0.2112=0.0038$

The above calculations show that 43.85% of the recovery was obtained by depletion drive, 34.65% by gas cap drive, 21.12% by water drive, and only 0.38% by connate water and rock expansion. The results suggest that the expansion drive index term can be neglected in the presence of a gas cap or when the reservoir pressure drops below the bubble point pressure. However, in high-PV compressibility reservoirs such as chalks and unconsolidated sands, the energy contribution of the rock and water expansion cannot be ignored even at high gas saturations.

A source of error is often introduced in the MBE calculations when determining the average reservoir pressure and the associated problem of correctly weighting or averaging the individual well pressures. An example of such a problem is when the producing formations are comprised of two or more zones of different permeabilities. In this case, the pressures are generally higher in the zone of low permeability and because the measured pressures are nearer to those in high-permeability zones, the measured static pressures tend to be lower and the reservoir behaves as if it contained less oil. Schilthuis explained this phenomenon by referring to the oil in the more permeable zones as active oil and by observing that the calculated active oil usually increases with time because the oil and gas in low-permeability zones slowly expand to offset the pressure decline. This is also true for fields that are not fully developed, because the average pressure can be that of the developed portion only, whereas the pressure is higher in the undeveloped portions. Craft et al. (1991) pointed out that the effect of pressure errors on the calculated values of initial oil and water influx depends on the size of the errors in relation to the reservoir pressure decline. Notice that the pressure enters the MBE mainly when determining the PVT differences in terms of:

(B _o - B _oi)

(B _g - B _gi)

(R _si - R _s)

Because water influx and gas cap expansion tend to offset pressure decline, the pressure errors are more serious than for the undersaturated reservoirs. In the case of very active water drives or gas caps that are large compared to the oil zone, the MBE usually produces considerable errors when determining the initial oil-in-place because of the very small pressure decline.

Dake (1994) pointed out that there are two "necessary" conditions that must be satisfied for a meaningful application of the MBE to a reservoir:

(1)

There should be adequate data collection in terms of production pressure, and PVT, in both frequency and quality for proper use of the MBE.

(2)

It must be possible to define an average reservoir pressure trend as a function of time or production for the field.

Establishing an average pressure decline trend can be possible even if there are large pressure differentials across the field under normal conditions. Averaging individual well pressure declines can possibly be used to determine a uniform trend in the entire reservoir. The concept of average well pressure and its use in determining the reservoir volumetric average pressure was introduced in Chapter 1 as illustrated by Figure 1.24. This figure shows that if ( $\overline{p}$ )_j and V _j represents the pressure and volume drained by the jth well, the volumetric average pressure of the entire reservoir can be estimated from:

${\overline{p}}_{\text{r}}=\frac{{\sum }_j{(\overline{p}V)}_j}{{\sum }_j{V}_j}$

in which:

V _j = the PV of the jth well drainage volume

( $\overline{p}$ )_j = volumetric average pressure within the jth drainage volume

In practice, the V _j are difficult to determine and, therefore, it is common to use individual well flow rates q _i in determining the average reservoir pressure from individual well average drainage pressure. From the definition of the isothermal compressibility coefficient:

$c=\frac{1}{V}\frac{\partial V}{\partial P}$

differentiating with time gives:

$\frac{\partial p}{\partial t}=\frac{1}{cV}\frac{\partial V}{\partial t}$

or:

$\frac{\partial p}{\partial t}=\frac{1}{cV}(q)$

This expression suggests that for a reasonably constant c at the time of measurement:

$V\propto \frac{q}{\partial p/\partial t}$

Since the flow rates are measured on a routine basis throughout the lifetime of the field, the average reservoir pressure can be alternatively expressed in terms of the individual well average drainage pressure decline rates and fluid flow rates by:

${\overline{p}}_{\text{r}}=\frac{{\sum }_j\left[{{(\overline{p}q)}_j/(\partial \overline{p}/\partial t)}_j\right]}{{\sum }_j\left[{q_j/(\partial \overline{p}/\partial t)}_j\right]}$

However, since the MBE is usually applied at regular intervals of 3-6 months, i.e., Δt = 3-6 months, throughout the lifetime of the field, the average field pressure can be expressed in terms of the incremental net change in underground fluid withdrawal, Δ(F), as:

${\overline{p}}_{\text{r}}=\frac{{\sum }_j{\overline{p}}_j\Delta {(F)}_j/\Delta {\overline{p}}_j}{{\sum }_j\Delta {(F)}_j/\Delta {\overline{p}}_j}$

where the total underground fluid withdrawal at time t and t + Δt are given by:

$\begin{array}{l}{F}_{\text{t}}={\int }_0^{\text{t}}[Q_{\text{o}}{B}_{\text{o}}+Q_{\text{w}}{B}_{\text{w}}+(Q_{\text{g}}-Q_{\text{o}}{R}_{\text{s}}-Q_{\text{w}}{R}_{\text{sw}})B_{\text{g}}]\text{d}t\\ F_{t\text{+}\Delta t}={\int }_0^{t+\Delta t}[Q_{\text{o}}{B}_{\text{o}}+Q_{\text{w}}{B}_{\text{w}}+(Q_{\text{g}}-Q_{\text{o}}{R}_{\text{s}}-Q_{\text{w}}{R}_{\text{sw}})B_{\text{g}}]\text{d}t\end{array}$

with:

$\Delta (F)=F_{t+\Delta t}-F_{\text{t}}$

where:

R _s = gas solubility, scf/STB

R _sw = gas solubility in the water, scf/STB

B _g = gas formation volume factor, bbl/scf

Q _o = oil low rate, STB/day

Q _w = water low rate, STB/day

Q _g = gas low rate, scf/day

For a volumetric reservoir with total fluid production and initial reservoir pressure as the only available data, the average pressure can be roughly approximated by using the following expression:

${\overline{p}}_{\text{r}}=p_{\text{i}}-\left[\frac{5.371\times {10}^{-6}{F}_{\text{t}}}{c_{\text{t}}(Ah\varphi )}\right]$

with the total fluid production F _t as defined above by:

$F_{\text{t}}={\int }_0^t[Q_{\text{o}}{B}_{\text{o}}+Q_{\text{w}}{B}_{\text{w}}+(Q_{\text{g}}-Q_{\text{o}}{R}_{\text{s}}-Q_{\text{w}}{R}_{\text{sw}})B_{\text{g}}]\text{d}t$

where:

A = well or reservoir drainage area, acres

h = thickness, ft

c _t = total compressibility coefficient, psi⁻¹

φ = porosity

p _i = initial reservoir pressure, psi

The above expression can be employed in a incremental manner, i.e., from time t to t + Δt, by:

${({\overline{p}}_{\text{r}})}_{t+\Delta t}={({\overline{p}}_{\text{r}})}_{\text{t}}-\left[\frac{5.371\times {10}^{-6}\Delta F}{c_{\text{t}}(Ah\varphi )}\right]$

with:

$\Delta (F)=F_{t+\Delta t}-F_{\text{t}}$

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B978075067733250006X

Performance of Oil Reservoirs

Tarek Ahmed , D. Nathan Meehan , in Advanced Reservoir Management and Engineering (Second Edition), 2012

Developing the MBE

Before deriving the material balance, it is convenient to denote certain terms by symbols for brevity. The symbols used conform, where possible, to the standard nomenclature adopted by the Society of Petroleum Engineers.

p i	Initial reservoir pressure, psi
p	Volumetric average reservoir pressure
Δp	Change in reservoir pressure=p i−p, psi
p b	Bubble point pressure, psi
N	Initial (original) oil-in-place, STB
N p	Cumulative oil produced, STB
G p	Cumulative gas produced, scf
W p	Cumulative water produced
R p	Cumulative gas–oil ratio, scf/STB
GOR	Instantaneous gas–oil ratio, scf/STB
R si	Initial gas solubility, scf/STB
R s	Gas solubility, scf/STB
B oi	Initial oil formation volume factor, bbl/STB
B o	Oil formation volume factor, bbl/STB
B gi	Initial gas formation volume factor, bbl/scf
B g	Gas formation volume factor, bbl/scf
W inj	Cumulative water injected, STB
G inj	Cumulative gas injected, scf
W e	Cumulative water influx, bbl
m	Ratio of initial gas cap gas reservoir volume to initial reservoir oil volume, bbl/bbl
G	Initial gas cap gas, scf
PV	Pore volume, bbl
c w	Water compressibility, psi−1
c f	Formation (rock) compressibility, psi−1

Several of the material balance calculations require the total pore volume (PV) as expressed in terms of the initial oil volume N and the volume of the gas cap. The expression for the total PV can be derived by conveniently introducing the parameter m into the relationship as follows.

Define the ratio m as:

$m=\frac{\text{Initial}\text{volume}\text{of}\text{gas}\text{cap}\text{in}\text{bbl}}{\text{Volume}\text{of}\text{oil}\text{initially}\text{in}\text{place}\text{in}\text{bbl}}=\frac{G{B}_{\text{gi}}}{N{B}_{\text{oi}}}$

Solving for the volume of the gas cap gives:

$\text{Initial}\text{volume}\text{of}\text{the}\text{gas}\text{cap}\text{,}G{B}_{\text{gi}}=mN{B}_{\text{oi}}\text{,}\text{bbl}$

The total initial volume of the hydrocarbon system is then given by:

$\begin{array}{c}\text{Initial}\text{oil}\text{volume}+\text{Initial}\text{gas}\text{cap}\text{volume}=\mathrm{(}\mathrm{PV}\mathrm{)}(1-S_{\text{wi}})\\ N{B}_{\text{oi}}+mN{B}_{\text{oi}}=(\text{PV})(1-S_{\text{wi}})\end{array}$

Solving for PV gives:

(4.1) $\text{PV}=\frac{N{B}_{\text{oi}}(1+m)}{1-S_{\text{wi}}}$

where

S _wi=initial water saturation

N=initial oil-in-place, STB

PV=total pore volume, bbl

m=ratio of initial gas cap gas reservoir volume to initial reservoir oil volume, bbl/bbl

Treating the reservoir PV as an idealized container as illustrated in Figure 4.14, volumetric balance expressions can be derived to account for all volumetric changes that occur during the natural productive life of the reservoir. The MBE can be written in a generalized form as follows:

Figure 4.14. Tank-model concept.

(4.2) $\text{PV}\text{occupied}\text{by}\text{the}\text{oil}\text{initially}\text{in}\text{place}\text{at}{p}_{\mathrm{i}}+\text{PV}\text{occupied}\text{by}\text{the}\text{gas}\text{in}\text{the}\text{gas}\text{cap}\text{at}{p}_{\mathrm{i}}=\text{PV}\text{occupied}\text{by}\text{the}\text{remaining}\text{oil}\text{at}p+\text{PV}\text{occupied}\text{by}\text{the}\text{gas}\text{in}\text{the}\text{gas}\text{cap}\text{at}p+\text{PV}\text{occupied}\text{by}\text{the}\text{evolved}\text{solution}\text{gas}\text{at}p+\text{PV}\text{occupied}\text{by}\text{the}\text{net}\text{water}\text{influx}\text{at}p+\text{change}\text{in}\text{PV}\text{due}\text{to}\text{connate}\text{water}\text{expansion}\text{and}+\text{pore}\text{volume}\text{reduction}\text{due}\text{to}\text{rock}\text{expansion}+\text{PV}\text{occupied}\text{by}\text{the}\text{injected}\text{gas}\text{at}p+\text{PV}\text{occupied}\text{by}\text{the}\text{injected}\text{water}\text{at}p$

The above eight terms composing the MBE can be determined separately from the hydrocarbon PVT and rock properties as follows.

Hydrocarbon PV occupied by the oil initially in place:

(4.3) $\text{Volume}\text{occupied}\text{by}\text{initial}\mathrm{oil}\mathrm{-}\mathrm{in}\mathrm{-}\mathrm{place}=N{B}_{\text{oi}}\mathrm{,}\text{bbl}$

where

N=oil initially in place, STB

B _oi=oil formation volume factor at initial reservoir pressure p _i, bbl/STB

Hydrocarbon PV occupied by the gas in the gas cap:

(4.4) $\text{Volume}\text{of}\text{gas}\text{cap}=mN{B}_{\text{oi}}\mathrm{,}\text{bbl}$

where m is a dimensionless parameter and defined as the ratio of gas cap volume to the oil zone volume.

Hydrocarbon PV occupied by the remaining oil:

(4.5) $\text{Volume}\text{of}\text{the}\text{remaining}\text{oil}=\mathrm{(}N-N_{\mathrm{p}}\mathrm{)}{B}_{\mathrm{o}}\mathrm{,}\text{bbl}$

where

N _p=cumulative oil production, STB

B _o=oil formation volume factor at reservoir pressure p, bbl/STB

Hydrocarbon PV occupied by the gas cap at reservoir pressure p: As the reservoir pressure drops to a new level p, the gas in the gas cap expands and occupies a larger volume. Assuming no gas is produced from the gas cap during the pressure declines, the new volume of the gas cap can be determined as:

(4.6) $\text{Volume}\text{of}\text{the}\text{gas}\text{cap}\text{at}p=\left[\frac{mN{B}_{\text{oi}}}{B_{\text{gi}}}\right]B_{\mathrm{g}}\mathrm{,}\text{bbl}$

where

B _gi=gas formation volume factor at initial reservoir pressure, bbl/scf

B _g=current gas formation volume factor, bbl/scf

Hydrocarbon PV occupied by the evolved solution gas: Some of the solution gas that has been evolved from the oil will remain in the pore space and occupies a certain volume that can be determined by applying the following material balance on the solution gas:

(4.7) $\left[\begin{array}{c}\begin{array}{l}\text{Volume}\text{of}\text{the}\text{evolved}\text{gas}\\ \text{that}\text{remains}\text{in}\text{the}\text{PV}\end{array}\end{array}\right]=\left[\begin{array}{l}\text{Volume}\text{of}\text{gas}\text{initially}\\ \text{in}\text{solution}\end{array}\right]-\left[\text{Volume}\text{of}\text{gas}\text{produced}\right]-\left[\begin{array}{l}\text{Volume}\text{of}\text{gas}\text{remaining}\\ \text{in}\text{solution}\end{array}\right]$

where

N _p=cumulative oil produced, STB

R _p=net cumulative produced gas–oil ratio, scf/STB

R _s=current gas solubility factor, scf/STB

B _g=current gas formation volume factor, bbl/scf

R _si=gas solubility at initial reservoir pressure, scf/STB

PV occupied by the net water influx:

(4.8) $\text{Net}\text{water}\text{influx}=W_{\mathrm{e}}-W_{\mathrm{p}}{B}_{\mathrm{w}}$

where

W _e=cumulative water influx, bbl

W _p=cumulative water produced, STB

B _w=water formation volume factor, bbl/STB

Change in PV due to initial water and rock expansion: The component describing the reduction in the hydrocarbon PV due to the expansion of initial (connate) water and the reservoir rock cannot be neglected for an undersaturated oil reservoir. The water compressibility c _w and rock compressibility c _f are generally of the same order of magnitude as the compressibility of the oil. However, the effect of these two components can generally be neglected for gas cap drive reservoirs or when the reservoir pressure drops below the bubble point pressure.

The compressibility coefficient c, which describes the changes in the volume (expansion) of the fluid or material with changing pressure is given by:

$c=\frac{-1}{V}\frac{\partial V}{\partial p}$

or

$\mathrm{\Delta }V=Vc\mathrm{\Delta }p$

where ΔV represents the net changes or expansion of the material as a result of changes in the pressure. Therefore, the reduction in the PV due to the expansion of the connate water in the oil zone and the gas cap is given by:

$\text{Connate}\text{water}\text{expansion}=\mathrm{[}\mathrm{(}\mathrm{PV}\mathrm{)}{S}_{\text{wi}}]c_{\mathrm{w}}\mathrm{\Delta }p$

Substituting for PV with Eq. (4.1) gives:

(4.9) $\text{Expansion}\text{of}\text{connate}\mathrm{water}\mathrm{=}\left[\frac{N{B}_{\text{oi}}(1+m)}{1-S_{\text{wi}}}{S}_{\text{wi}}\right]c_{\mathrm{w}}\mathrm{\Delta }p$

where

Δp=change in reservoir pressure, p _i–p

c _w=water compressibility coefficient, psi⁻¹

m=ratio of the volume of the gas cap gas to the reservoir oil volume, bbl/bbl

Similarly, as fluids are produced and pressure declines, the entire reservoir PV is reduced (compaction), and this negative change in PV expels an equal volume of fluid as production. The reduction in the PV due to the expansion of the reservoir rock is given by:

(4.10) $\text{Change}\text{in}\text{PV}=\frac{N{B}_{\text{oi}}(1+m)}{1-S_{\text{wi}}}{c}_{\mathrm{f}}\mathrm{\Delta }p$

Combining the expansions of the connate water and formation as represented by Eqs. (4.9) and (4.10) gives:

(4.11) $\text{Total}\text{changes}\text{in}\text{the}\text{PV}=N{B}_{\text{oi}}(1+m)\left(\frac{S_{\text{wi}}{c}_{\mathrm{w}}+c_{\mathrm{f}}}{1-S_{\text{wi}}}\right)\mathrm{\Delta }p$

The connate water and formation compressibilities are generally small in comparison to the compressibility of oil and gas. However, values of c _w and c _f are significant for undersaturated oil reservoirs, and they account for an appreciable fraction of the production above the bubble point. Ranges of compressibilities are given below:

Undersaturated oil	5–50×10−6  psi−1
Water	2–4×10−6  psi−1
Formation	3–10×10−6  psi−1
Gas at 1000   psi	500–1000×10−6  psi−1
Gas at 5000   psi	50–200×10−6  psi−1

PV occupied by the injection gas and water: Assuming that G _inj (volumes of gas) and W _inj volumes of water have been injected for pressure maintenance, the total PV occupied by the two injected fluids is given by:

(4.12) $\text{Total}\text{volume}=G_{\text{inj}}{B}_{\text{ginj}}+W_{\text{inj}}{B}_{\mathrm{w}}$

where

G _inj=cumulative gas injected, scf

B _ginj=injected gas formation volume factor, bbl/scf

W _inj=cumulative water injected, STB

B _w=water formation volume factor, bbl/STB

Combining Eqs. (4.3)–(4.12) with Eq. (4.2) and rearranging gives:

(4.13) $N=\frac{N_{\mathrm{p}}{B}_{\mathrm{o}}+(G_{\mathrm{p}}-N_{\mathrm{p}}{R}_{\mathrm{s}})B_{\mathrm{g}}-(W_{\mathrm{e}}-W_{\mathrm{p}}{B}_{\mathrm{w}})-G_{\text{inj}}{B}_{\text{ginj}}-W_{\text{inj}}{B}_{\mathrm{w}}}{(B_{\mathrm{o}}-B_{\text{oi}})+(R_{\text{si}}-R_{\mathrm{s}})B_{\mathrm{g}}+m{B}_{\text{oi}}[(B_{\mathrm{g}}\mathrm{/}{B}_{\text{gi}})-1]+B_{\text{oi}}(1+m)[(S_{\text{wi}}{c}_{\mathrm{w}}+c_{\mathrm{f}})/(1-S_{\text{wi}})]\mathrm{\Delta }p}$

where

N=initial oil-in-place, STB

G _p=cumulative gas produced, scf

N _p=cumulative oil produced, STB

R _si=gas solubility at initial pressure, scf/STB

m=ratio of gas cap gas volume to oil volume, bbl/bbl

B _gi=gas formation volume factor at p _i, bbl/scf

B _ginj=gas formation volume factor of the injected gas, bbl/scf

Recognizing that the cumulative gas produced G _p can be expressed in terms of the cumulative gas–oil ratio R _p and cumulative oil produced, then gives:

(4.14) $G_{\mathrm{p}}=R_{\mathrm{p}}{N}_{\mathrm{p}}$

Combining Eq. (4.14) with Eq. (4.13) gives:

(4.15) $N=\frac{N_{\mathrm{p}}[B_{\mathrm{o}}+(R_{\mathrm{p}}-R_{\mathrm{s}})B_{\mathrm{g}}]-(W_{\mathrm{e}}-W_{\mathrm{p}}{B}_{\mathrm{w}})-G_{\text{inj}}{B}_{\text{ginj}}-W_{\text{inj}}{B}_{\text{wi}}}{(B_{\mathrm{o}}-B_{\text{oi}})+(R_{\text{si}}-R_{\mathrm{s}})B_{\mathrm{g}}+m{B}_{\text{oi}}[(B_{\mathrm{g}}\mathrm{/}{B}_{\text{gi}})-1]+B_{\text{oi}}(1+m)\times [(S_{\text{wi}}{c}_{\mathrm{w}}+c_{\mathrm{f}})/(1-S_{\text{wi}})]\mathrm{\Delta }p}$

This relationship is referred to as the generalized MBE. A more convenient form of the MBE can be arrived at, by introducing the concept of the total (two-phase) formation volume factor B _t into the equation. This oil PVT property is defined as:

(4.16) $B_{\mathrm{t}}=B_{\mathrm{o}}+(R_{\text{si}}-R_{\mathrm{s}})B_{\mathrm{g}}$

Introducing B _t into Eq. (4.15) and assuming, for the sake of simplicity, that there is no water or gas injection gives:

(4.17) $N=\frac{N_{\mathrm{p}}[B_{\mathrm{t}}+(R_{\mathrm{p}}-R_{\text{si}})B_{\mathrm{g}}]-(W_{\mathrm{e}}-W_{\mathrm{p}}{B}_{\mathrm{w}})}{(B_{\mathrm{t}}-B_{\text{ti}})+m{B}_{\text{ti}}[B_{\mathrm{g}}\mathrm{/}{B}_{\text{gi}}/-1]+B_{\text{ti}}(1+m)[(S_{\text{wi}}{c}_{\mathrm{w}}+c_{\mathrm{f}})/(1-S_{\text{wi}})]\mathrm{\Delta }p}$

(note that B _ti=B _oi) where

S _wi=initial water saturation

R _p=cumulative produced gas–oil ratio, scf/STB

Δp=change in the volumetric average reservoir pressure, psi

B _g=gas formation volume factor, bbl/scf

Example 4.1

The Anadarko Field is a combination drive reservoir. The current reservoir pressure is estimated at 2500   psi. The reservoir production data and PVT information are given below:

	Initial Reservoir Condition	Current Reservoir Condition
p, psi	3000	2500
B o, bbl/STB	1.35	1.33
R s, scf/STB	600	500
N p, MMSTB	0	5
G p, MMMscf	0	5.5
B w, bbl/STB	1.00	1.00
W e, MMbbl	0	3
W p, MMbbl	0	0.2
B g, bbl/scf	0.0011	0.0015
c f, c w	0	0

The following additional information is available:

$\text{Volume}\text{of}\text{bulk}\text{oil}\text{zone}=100\text{,}000\text{acres}\mathrm{-}\text{ft}\text{Volume}\text{of}\text{bulk}\text{gas}\text{zone}=20\mathrm{,}000\text{acres}\mathrm{-}\text{ft}$

Calculate the initial oil-in-place.

Solution

Step 1.

Assuming the same porosity and connate water for the oil and gas zones, calculate m:

$m=\frac{7758\varphi (1-S_{\text{wi}}){(Ah)}_{\text{gas}\text{cap}}}{7758\varphi (1-S_{\text{wi}}){(Ah)}_{\text{oil}\text{zone}}}=\frac{7758\varphi (1-S_{\text{wi}})20\text{,}000}{7758\varphi (1-S_{\text{wi}})100\text{,}000}=\frac{20\text{,}000}{100\text{,}000}=0.2$

Step 2.

Calculate the cumulative gas–oil ratio R _p:

$R_{\mathrm{p}}=\frac{G_{\mathrm{p}}}{N_{\mathrm{p}}}=\frac{5.5\times {10}^{\mathrm{9}}}{5\times {10}^{\mathrm{6}}}=1100\text{scf}\mathrm{/}\text{STB}$

Step 3.

Solve for the initial oil-in-place by applying Eq. (4.15):

$N=\frac{N_{\mathrm{p}}[B_{\mathrm{o}}+(R_{\mathrm{p}}-R_{\mathrm{s}})B_{\mathrm{g}}]-(W_{\mathrm{e}}-W_{\mathrm{p}}{B}_{\mathrm{w}})}{(B_{\mathrm{o}}-B_{\text{oi}})+(R_{\text{si}}-R_{\mathrm{s}})B_{\mathrm{g}}+m{B}_{\text{oi}}[(B_{\mathrm{g}}\mathrm{/}{B}_{\text{gi}})-1]+B_{\text{oi}}(1+m)[(S_{\text{wi}}{c}_{\mathrm{w}}+c_{\mathrm{f}})(1-S_{\text{wi}})]\mathrm{\Delta }p}=\frac{5\times {10}^{\mathrm{6}}[1.33+(1100-500)0.0015]-(3\times {10}^6-0.2\times {10}^6)}{(1.35-1.33)+(600-500)0.0015+(0.2)(1.35)\times [(0.0015\mathrm{/}0.0011)-1]}=31.14\text{MMSTB}$

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Gas Reservoirs

Tarek Ahmed , in Reservoir Engineering Handbook (Fourth Edition), 2010

The Material Balance Method

If enough production-pressure history is available for a gas reservoir, the initial gas-in-place G, the initial reservoir pressure p_i, and the gas reserves can be calculated without knowing A, h, ϕ, or S_w. This is accomplished by forming a mass or mole balance on the gas as:

(13-5) ${\text{n}}_{\text{p}}={\text{n}}_{\text{i}}-{\text{n}}_{\text{f}}$

where n_p = moles of gas produced

n_i = moles of gas initially in the reservoir

n_f = moles of gas remaining in the reservoir

Representing the gas reservoir by an idealized gas container, as shown schematically in Figure 13-1, the gas moles in Equation 13-5 can be replaced by their equivalents using the real gas law to give:

Figure 13-1. Idealized water-drive gas reservoir.

(13-6) $\frac{{\text{p}}_{\text{sc}}{\text{G}}_{\text{p}}}{{\text{R T}}_{\text{sc}}}=\frac{{\text{p}}_{\text{i}}\text{V}}{z_{\text{i}}\text{RT}}-\frac{\text{p}\left[\text{V}-\left({\text{W}}_{\text{e}}-{\text{W}}_{\text{p}}\right)\right]}{\text{zRT}}$

where p_i = initial reservoir pressure

G_p = cumulative gas production, scf

p = current reservoir pressure

V = original gas volume, ft³

z_i = gas deviation factor at p_i

z = gas deviation factor at p

T = temperature, °R

W_e = cumulative water influx, ft ³

W_p = cumulative water production, ft³

Equation 13-6 is essentially the general material balance equation (MBE). Equation 13-6 can be expressed in numerous forms depending on the type of the application and the driving mechanism. In general, dry gas reservoirs can be classified into two categories:

•

Volumetric gas reservoirs

•

Water-drive gas reservoirs

The remainder of this chapter is intended to provide the basic background in natural gas engineering. There are several excellent textbooks that comprehensively address this subject, including the following:

•

Ikoku, C., Natural Gas Reservoir Engineering, 1984

•

Lee, J. and Wattenbarger, R., Gas Reservoir Engineering, SPE, 1996

Volumetric Gas Reservoirs

For a volumetric reservoir and assuming no water production, Equation 13-6 is reduced to:

(13-7) $\frac{{\text{p}}_{\text{sc}}{\text{G}}_{\text{p}}}{{\text{T}}_{\text{sc}}}=\left(\frac{{\text{p}}_{\text{i}}}{{\text{z}}_{\text{i}}\text{T}}\right)\text{V}-\left(\frac{\text{p}}{\text{zT}}\right)\text{V}$

Equation 13-7 is commonly expressed in the following two forms:

Form 1. In terms of p/z

Rearranging Equation 13-7 and solving for p/z gives:

(13-8) $\frac{\text{p}}{\text{z}}=\frac{{\text{p}}_{\text{i}}}{{\text{z}}_{\text{i}}}-\left(\frac{{\text{p}}_{\text{sc}}\text{T}}{{\text{T}}_{\text{sc}}\text{V}}\right){\text{G}}_{\text{p}}$

Equation 13-8 is an equation of a straight line when (p/z) is plotted versus the cumulative gas production G_p, as shown in Figure 13-2. This straight-line relationship is perhaps one of the most widely used relationships in gas-reserve determination.

Figure 13-2. Gas material balance equation.

The straight-line relationship provides the engineer with the reservoir characteristics:

•

Slope of the straight line is equal to:

(13-9) $\text{slope}=\frac{{\text{p}}_{\text{sc}}\text{T}}{{\text{T}}_{\text{sc}}\text{V}}$

The original gas volume V can be calculated from the slope and used to determine the areal extent of the reservoir from:

(13-10) $\text{V}=43.560\text{Ah}\varphi (1-{\text{S}}_{\text{wi}})$

where A is the reservoir area in acres.

•

Intercept at G_p = 0 gives p_i/z_i

•

Intercept at p/z = 0 gives the gas initially in place G in scf

•

Cumulative gas production or gas recovery at any pressure

Example 13-2 ¹

A volumetric gas reservoir has the following production history.

Time, t years	Reservoir Pressure, p psia	z	Cumulative Production, Gp MMMscf
0.0	1798	0.869	0.00
0.5	1680	0.870	0.96
1.0	1540	0.880	2.12
1.5	1428	0.890	3.21
2.0	1335	0.900	3.92

The following data are also available:

ϕ = 13%

S_wi = 0.52

A = 1060 acres

h = 54 ft

T = 164°F

Calculate the gas initially in place volumetrically and from the MBE.

Solution

Step 1.

Calculate B_gi from Equation 13-1.

${\text{B}}_{\text{gi}}=0.02827\frac{(0.869)(164+460)}{1798}=0.00853{\text{ft}}^3/\text{scf}$

Step 2.

Calculate the gas initially in place volumetrically by applying Equation 13-3.

G = 43,560 (1060) (54) (0.13) (1 − 0.52)/0.00853= 18.2 MMMscf

Step 3.

Plot p/z versus G_p as shown in Figure 13-3 and determine G.

G = 14.2 MMMscf

This checks the volumetric calculations.

Figure 13-3. Relationship of p/zvs. Gp for Example 13-2.

The initial reservoir gas volume V can be expressed in terms of the volume of gas at standard conditions by:

$\text{V}={\text{B}}_{\text{g}}\text{G}=\left(\frac{{\text{p}}_{\text{sc}}}{{\text{T}}_{\text{sc}}}\frac{{\text{z}}_{\text{i}}\text{T}}{{\text{p}}_{\text{i}}}\right)\text{G}$

Combining the above relationship with that of Equation 13-8 gives:

(13-11) $\frac{\text{p}}{\text{z}}=\frac{{\text{p}}_{\text{i}}}{{\text{z}}_{\text{i}}}-\left[\left(\frac{{\text{p}}_{\text{i}}}{{\text{z}}_{\text{i}}}\right)\frac{1}{\text{G}}\right]{\text{G}}_{\text{p}}$

This relationship can be expressed in a more simplified form as:

$\frac{\text{p}}{\text{Z}}=\frac{{\text{p}}_{\text{i}}}{{\text{Z}}_{\text{i}}}-\left[\text{m}\right]{\text{G}}_{\text{p}}$

where the coefficient m is essentially constant and represents the resulting straight line when P/Z is plotted against G_P. The slope, m is defined by:

$\text{m}=\left(\frac{{\text{p}}_{\text{i}}}{{\text{Z}}_{\text{i}}}\right)\frac{1}{\text{G}}$

Equivalently, m is defined by Equation 13-9 as:

$\text{m}=\frac{{\text{Tp}}_{\text{sc}}}{{\text{T}}_{\text{sc}}\text{V}}$

where

G = Original gas-in-place, scf

V = Original gas-in-place, ft³

Again, Equation 13-11 shows that for a volumetric reservoir, the relationship between (p/z) and G_p is essentially linear. This popular equation indicates that by extrapolation of the straight line to abscissa, i.e., at p/z = 0, will give the value of the gas initially in place as G = G_p.

The graphical representation of Equation 13-11 can be used to detect the presence of water influx, as shown graphically in Figure 13-4. When the plot of (p/z) versus G_p deviates from the linear relationship, it indicates the presence of water encroachment.

Figure 13-4. Effect of water drive on p/z vs. G_p relationship.

Many other graphical methods have been proposed for solving the gas MBE that are useful in detecting the presence of water influx. One such graphical technique is called the energy plot, which is based on arranging Equation 13-11 and taking the logarithm of both sides to give:

(13-12) $log\left[1-\frac{{\text{z}}_{\text{i}}\text{p}}{{\text{p}}_{\text{i}}\text{z}}\right]={\text{log G}}_{\text{p}}-\text{logG}$

Figure 13-5 shows a schematic illustration of the plot.

Figure 13-5. An energy plot.

From Equation 13-12, it is obvious that a plot of [1 − (z_i p)/(p_i z)] versus G_p on log-log coordinates will yield a straight line with a slope of one (45° angle). An extrapolation to one on the vertical axis (p = 0) yields a value for initial gas-in-place, G. The graphs obtained from this type of analysis have been referred to as energy plots. They have been found to be useful in detecting water influx early in the life of a reservoir. If W_e is not zero, the slope of the plot will be less than one, and will also decrease with time, since W_e increases with time. An increasing slope can only occur as a result of either gas leaking from the reservoir or bad data, since the increasing slope would imply that the gas-occupied pore volume was increasing with time.

It should be pointed out that the average field, (p/Z)_Field, can be estimated from the individual wells' p/Z versus G_P performance by applying the following relationship:

${\left(\frac{\text{p}}{\text{Z}}\right)}_{\text{Field}}=\frac{{\text{p}}_{\text{i}}}{{\text{Z}}_{\text{i}}}-\frac{\sum _{\text{j}=1}^{\text{n}}{\left({\text{G}}_{\text{P}}\right)}_{\text{j}}}{\sum _{\text{n}}^{\text{j}=1}{\left[\frac{{\text{G}}_{\text{P}}}{\frac{{\text{p}}_{\text{i}}}{{\text{Z}}_{\text{i}}}-\frac{\text{p}}{\text{Z}}}\right]}_{\text{j}}}$

The summation ∑is taking over the total number n of the field gas wells, that is, j = 1, 2, … n. The total field performance in terms of (p/Z)_Field versus (G_P)_Field can then be constructed from the estimated values of the field p/Z and actual total field production, that is, (p/Z)_Field versus ∑G_P. The above equation is applicable as long as all wells are producing with defined static boundaries, that is, under pseudosteady-state conditions.

When using the MBE for reserve analysis for an entire reservoir that is characterized by a distinct lack of pressure equilibrium throughout, the following average reservoir pressure decline, (p/Z)_Field, can be used:

${\left(\frac{\text{p}}{\text{Z}}\right)}_{\text{Field}}=\frac{{\sum _{\text{j}=\text{l}}^{\text{n}}\left(\frac{\text{P}\Delta {\text{G}}_{\text{p}}}{\Delta \text{p}}\right)}_{\text{j}}}{{\sum _{\text{j}=\text{l}}^{\text{n}}\left(\frac{\Delta {\text{G}}_{\text{P}}}{\Delta \text{p}/\text{Z}}\right)}_{\text{j}}}$

where Δp and ΔG_P are the incremental pressure difference and cumulative production, respectively.

The gas recovery factor (RF) at any depletion pressure is defined as the cumulative gas produced, G_P, at this pressure divided by the gas initially in place, G:

$\text{RF}=\frac{{\text{G}}_{\text{P}}}{\text{G}}$

Introducing the gas RF to Equation 8-60 gives

$\frac{\text{p}}{\text{Z}}=\frac{{\text{p}}_{\text{i}}}{{\text{Z}}_{\text{i}}}\left[1-\frac{{\text{G}}_{\text{P}}}{\text{G}}\right]$

or

Solving for the recovery factor at any depletion pressure gives:

$\text{RF}=1-\left[\frac{{\text{Z}}_{\text{i}}}{\text{Z}}\frac{\text{p}}{{\text{p}}_{\text{i}}}\right]$

Form 2. In terms of B_g

From the definition of the gas formation volume factor, it can be expressed as:

${\text{B}}_{\text{gi}}=\frac{\text{V}}{\text{G}}$

Combining the above expression with Equation 13-1 gives:

(13-13) $\frac{{\text{p}}_{\text{sc}}}{{\text{T}}_{\text{sc}}}\frac{{\text{z}}_{\text{i}}\text{T}}{{\text{p}}_{\text{i}}}=\frac{\text{V}}{\text{G}}$

where V = volume of gas originally in place, ft³

G = volume of gas originally in place, scf

p_i = original reservoir pressure

z_i = gas compressibility factor at p_i

Equation 13-13 can be combined with Equation 13-7, to give:

(13-14) $\text{G}=\frac{{\text{G}}_{\text{p}}{\text{B}}_{\text{g}}}{{\text{B}}_{\text{g}}-{\text{B}}_{\text{gi}}}$

Equation 13-14 suggests that to calculate the initial gas volume, the only information required is production data, pressure data, gas specific gravity for obtaining z-factors, and reservoir temperature. Early in the producing life of a reservoir, however, the denominator of the right-hand side of the material balance equation is very small, while the numerator is relatively large. A small change in the denominator will result in a large discrepancy in the calculated value of initial gas-in-place. Therefore, the material balance equation should not be relied on early in the producing life of the reservoir.

Material balances on volumetric gas reservoirs are simple. Initial gas-in-place may be computed from Equation 13-14 by substituting cumulative gas produced and appropriate gas formation volume factors at corresponding reservoir pressures during the history period. If successive calculations at various times during the history give consistent values for initial gas-in-place, the reservoir is operating under volumetric control and computed G is reliable, as shown in Figure 13-6. Once G has been determined and the absence of water influx established in this fashion, the same equation can be used to make future predictions of cumulative gas production function of reservoir pressure.

Figure 13-6. Graphical determination of the gas initially in place G.

Ikoku (1984) points out that successive application of Equation 13-14 will normally result in increasing values of the gas initially in place G with time if water influx is occurring. If there is gas leakage to another zone due to bad cement jobs or casing leaks, however, the computed value of G may decrease with time.

Example 13-3

After producing 360 MMscf of gas from a volumetric gas reservoir, the pressure has declined from 3,200 psi to 3,000 psi, given:

B_gi = 0.005278 ft³/scf

B_g = 0.005390 ft³/scf

a.

Calculate the gas initially in place.

b.

Recalculate the gas initially in place assuming that the pressure measurements were incorrect and the true average pressure is 2,900 psi. The gas formation volume factor at this pressure is 0.00558 ft³/scf.

Solution

a.

Using Equation 13-14, calculate G.

$\text{G}=\frac{360\times {10}^6(0.00539)}{0.00539-0.005278}=17.325\text{MMMscf}$

b.

Recalculate G by using the correct value of B_g.

$\text{G}=\frac{360\times {10}^6(0.00668)}{0.00558-0.005278}=6.652\text{MMMscf}$

Thus, an error of 100 psia, which is only 3.5% of the total reservoir pressure, resulted in an increase in calculated gas-in-place of approximately 160%, a 2½-fold increase. Note that a similar error in reservoir pressure later in the producing life of the reservoir will not result in an error as large as that calculated early in the producing life of the reservoir.

Water-Drive Gas Reservoirs

If the gas reservoir has a water drive, then there will be two unknowns in the material balance equation, even though production data, pressure, temperature, and gas gravity are known. These two unknowns are initial gas-in-place and cumulative water influx. In order to use the material balance equation to calculate initial gas-in-place, some independent method of estimating W _e, the cumulative water influx, must be developed as discussed in Chapter 11.

Equation 13-14 can be modified to include the cumulative water influx and water production to give:

(13-15) $\text{G}=\frac{{\text{G}}_{\text{p}}{\text{B}}_{\text{g}}-({\text{W}}_{\text{e}}-{\text{W}}_{\text{p}}{\text{B}}_{\text{w}})}{{\text{B}}_{\text{g}}-{\text{B}}_{\text{gi}}}$

The above equation can be arranged and expressed as:

(13-16) $\text{G}+\frac{{\text{W}}_{\text{e}}}{{\text{B}}_{\text{g}}-{\text{B}}_{\text{gi}}}=\frac{{\text{G}}_{\text{p}}{\text{B}}_{\text{g}}+{\text{W}}_{\text{p}}{\text{B}}_{\text{w}}}{{\text{B}}_{\text{g}}-{\text{B}}_{\text{gi}}}$

Equation 13-16 reveals that for a volumetric reservoir, i.e., W_e = 0, the right-hand side of the equation will be constant regardless of the amount of gas G_p that has been produced. For a water-drive reservoir, the values of the right-hand side of Equation 13-16 will continue to increase because of the W_e/(B_g − B_gi) term. A plot of several of these values at successive time intervals is illustrated in Figure 13-7. Extrapolation of the line formed by these points back to the point where G_p = 0 shows the true value of G, because when G_p = 0, then W_e/(B_g − B_gi) is also zero.

Figure 13-7. Effect of water influx on calculating the gas initially in place.

This graphical technique can be used to estimate the value of W_e, because at any time the difference between the horizontal line (i.e., true value of G) and the sloping line [G + (W_e)/(B_g − B_gi) will give the value of W_e/(B_g − B_gi).

Because gas often is bypassed and trapped by the encroaching water, recovery factors for gas reservoirs with water drive can be significantly lower than for volumetric reservoirs produced by simple gas expansion. In addition, the presence of reservoir heterogeneities, such as low-permeability stringers or layering, may reduce gas recovery further. As noted previously, ultimate recoveries of 80% to 90% are common in volumetric gas reservoirs, while typical recovery factors in water-drive gas reservoirs can range from 50% to 70%.

Because gas often is bypassed and trapped by encroaching water, recovery factors for gas reservoirs with water drive can be significantly lower than for volumetric reservoirs produced by simple gas expansion. In addition, the presence of reservoir heterogeneities, such as low-permeability stringers or layering, may reduce gas recovery further. As noted previously, ultimate recoveries of 80% to 90% are common in volumetric gas reservoirs, while typical recovery factors in water-drive gas reservoirs can range from 50% to 70%. The amount of gas that is trapped in a region that has been flooded by water encroachment can be estimated by defining the following characteristic reservoir parameters and taking the steps outlined below:

(P.V) = reservoir pore volume, ft³

(P.V)_water = pore volume of the water-invaded zone, ft³

S_grw = residual gas saturation to water displacement

S_wi = initial water saturation

G = gas initially in place, scf

G_P = cumulative gas production at depletion pressure p, scf

B_gi = initial gas formation volume factor, ft³/scf

B_g = gas formation volume factor at depletion pressure p, ft³/scf

Z = gas deviation factor at depletion pressure p

Step 1.

Express the reservoir pore volume, (P.V), in terms of the initial gas-in-place, G, as follows:

G B_gi = (P.V) (1 − S_wi)

Solving for the reservoir pore volume gives:

$\left(\text{P}.\text{V}\right)=\frac{{\text{GB}}_{\text{gi}}}{1-{\text{S}}_{\text{wi}}}$

Step 2.

Calculate the pore volume in the water-invaded zone:

${\text{W}}_{\text{e}}-{\text{W}}_{\text{p}}{\text{B}}_{\text{w}}={\left(\text{P}.\text{V}\right)}_{\text{water}}\left(1-{\text{S}}_{\text{wi}}-{\text{S}}_{\text{grw}}\right)$

Solving for the pore volume of the water-invaded zone, (P.V)_water, gives:

${\left(\text{P}.\text{V}\right)}_{\text{water}}=\frac{{\text{W}}_{\text{e}}-{\text{W}}_{\text{p}}{\text{B}}_{\text{w}}}{1-{\text{S}}_{\text{wi}}-{\text{S}}_{\text{grw}}}$

Step 3.

Calculate trapped gas volume in the water-invaded zone, or:

Trapped gas volume = (P.V)_water S_grw

Trapped gas volume =

$\left[\frac{{\text{W}}_{\text{e}}-{\text{W}}_{\text{p}}{\text{B}}_{\text{w}}}{1-{\text{S}}_{\text{wi}}-{\text{S}}_{\text{grw}}}\right]{\text{S}}_{\text{grw}}$

Step 4.

Calculate the number n of moles of gas trapped in the water-invaded zone by using the equation of state, or:

p (Trapped gas volume) = Z n R T

Solving for n gives:

$\text{n}=\frac{\text{p}\left[\frac{{\text{W}}_{\text{e}}-{\text{W}}_{\text{p}}{\text{B}}_{\text{w}}}{1-{\text{S}}_{\text{wi}}-{\text{S}}_{\text{grw}}}\right]{\text{S}}_{\text{grw}}}{\text{Z R T}}$

which indicates that the higher the pressure, the greater the quantity of trapped gas. Dake (1994) points out that if the pressure is reduced by rapid gas withdrawal, the volume of gas trapped in each individual pore space, that is, S_grw, will remain unaltered, but its quantity, n, will be reduced.

Step 5.

The gas saturation at any pressure can be adjusted to account for the trapped gas, as follows:

${\text{S}}_{\text{g}}=\frac{\text{remaining gas volume}-\text{trapped gas volume}}{\text{reservoir pore volume}-\text{pore volume of water invaded zone}}$

${\text{S}}_{\text{g}}=\frac{\left(\text{G}-{\text{G}}_{\text{p}}\right){\text{B}}_{\text{g}}-\left[\frac{{\text{W}}_{\text{e}}-{\text{W}}_{\text{p}}{\text{B}}_{\text{w}}}{1-{\text{S}}_{\text{wi}}-{\text{S}}_{\text{grw}}}\right]{\text{S}}_{\text{grw}}}{\left(\frac{\text{G}{\text{B}}_{\text{gi}}}{1-{\text{S}}_{\text{wi}}}\right)-\left[\frac{{\text{W}}_{\text{e}}-{\text{W}}_{\text{p}}{\text{B}}_{\text{w}}}{1-{\text{S}}_{\text{wi}}-{\text{S}}_{\text{grw}}}\right]}$

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