IPR-Fetkovich

Reservoir Eng Hndbk Ch 07 2001-10-24 16:49 Page 498

498

Reservoir Engineering Handbook

It should be noted that one of the main disadvantages of Standing’s methodology is that it requires reliable permeability information; in addition, it also requires material balance calculations to predict oil saturations at future average reservoir pressures. Fetkovich’s Method

Muskat and Evinger (1942) attempted to account for the observed nonlinear flow behavior (i.e., IPR) of wells by calculating a theoretical productivity index from the pseudosteady-state flow equation. They expressed Darcy’s equation as: 0.00708 kh Qo = È re ˘ Í ln - 0.75 +s˙ Î rw ˚

pr

Ú f(p) dp

(7 - 25)

p wf

where the pressure function f(p) is defined by: f ( p) =

k ro m o bo

(7 - 26)

where kro = oil relative permeability k = absolute permeability, md Bo = oil formation volume factor mo = oil viscosity, cp Fetkovich (1973) suggests that the pressure function f(p) can basically fall into one of the following two regions:

Region 1: Undersaturated Region The pressure function f(p) falls into this region if p > pb. Since oil relative permeability in this region equals unity (i.e., kro = 1), then: Ê 1 ˆ f ( p) = Á ˜ Ë m o Bo ¯ p

(7 - 27)


Oil Well Performance

499

Fetkovich observed that the variation in f(p) is only slight and the pressure function is considered constant as shown in Figure 7-10.

Region 2: Saturated Region In the saturated region where p < pb, Fetkovich shows that the (kro/ moBo) changes linearly with pressure and that the straight line passes through the origin. This linear is shown schematically in Figure 7-10 can be expressed mathematically as: Ê 1 ˆ Ê pˆ f ( p) = Á ˜ Á ˜ Ë m o Bo ¯ p Ë p b ¯

(7 - 28)

b

Where mo and Bo are evaluated at the bubble-point pressure. In the application of the straight-line pressure function, there are three cases that must be considered:

Figure 7-10. Pressure function concept.


500


• –pr and pwf > pb • –pr and pwf < pb • –pr > pb and pwf < pb All three cases are presented below. Case 1: p–r and pwf > pb

This is the case of a well producing from an undersaturated oil reservoir where both pwf and –pr are greater than the bubble-point pressure. The pressure function f(p) in this case is described by Equation 7-27. Substituting Equation 7-27 into Equation 7-25 gives: 0.00708 kh Qo = Ê re ˆ ln Á ˜ - 0.75 + s Ë rw ¯

pr

Ú p wf

Ê 1 ˆ Á ˜ dp Ë m o Bo ¯

Ê 1 ˆ Since Á ˜ is constant, then: Ë m o Bo ¯

Qo =

0.00708 kh ( p r - p wf ) È Ê re ˆ ˘ m o Bo Íln Á ˜ - 0.75 + s˙ ÍÎ Ë rw ¯ ˙˚

(7 - 29)

or Qo = J (p–r - pwf)

(7 -30 )

The productivity index is defined in terms of the reservoir parameters as: J=

0.00708 kh È Ê re ˆ ˘ m o Bo Íln Á ˜ - 0.75 + s˙ ÍÎ Ë rw ¯ ˙˚

– + p )/2. where Bo and mo are evaluated at (p r wf

(7 - 31)



501

Example 7-8 A well is producing from an undersaturated-oil reservoir that exists at an average reservoir pressure of 3000 psi. The bubble-point pressure is recorded as 1500 psi at 150°F. The following additional data are available: • stabilized flow rate = 280 STB/day • stabilized wellbore pressure = 2200 psi • h = 20¢ rw = 0.3¢ re = 660¢ s = -0.5 • k = 65 md • mo at 2600 psi = 2.4 cp • Bo at 2600 psi = 1.4 bbl/STB Calculate the productivity index by using both the reservoir properties (i.e., Equation 7-31) and flow test data (i.e., Equation 7-30).

Solution • From Equation 7-31 J=

0.00708 (65) (20) = 0.42 STB/ day / psi È Ê 660 ˆ ˘ (2.4) (1.4) Íln - 0.75 - 0.5˙ Î Ë 0.3 ¯ ˚

• From production data: J=

280 = 0.35 STB/ day / psi 3000 - 2200

Results show a reasonable match between the two approaches. It should be noted, however, that there are several uncertainties in the values of the parameters used in Equation 7-31 to determine the productivity index. For example, changes in the skin factor k or drainage area would change the calculated value of J. Case 2: p–r and pwf < pb

When the reservoir pressure –pr and bottom-hole flowing pressure pwf are both below the bubble-point pressure pb, the pressure function f(p) is


502


represented by the straight line relationship as expressed by Equation 7-28. Combining Equation 7-28 with Equation 7-25 gives: ˘ È ˙ Í 0.00708 kh ˙ Qo = Í ˙ Í Êr ˆ Í ln Á e ˜ - 0.75 + s ˙ ˙˚ ÍÎ Ë rw ¯

pr

Ú p wf

Ê pˆ 1 Á ˜ dp (m o Bo ) p b Ë p b ¯

ÈÊ 1 ˆ Ê 1 ˆ ˘ Since the term ÍÁ ˜ Á ˜ ˙ is constant, then: ÍË m o Bo ¯ p Ë p b ¯ ˙ b ˚ Î ˘ È ˙ Í Ê 1ˆ 0.00708 kh ˙ 1 Qo = Í ˙ (m o Bo ) p ÁË p b ˜¯ Í Êr ˆ b Í ln Á e ˜ - 0.75 + s ˙ ˙˚ ÍÎ Ë rw ¯

pr

Ú p dp p wf

Integrating gives:

Qo =

Ê 1 ˆ -2 0.00708 kh 2 Á ˜ ( p r - p wf ) È Ê re ˆ ˘ Ë 2 pb ¯ (m o Bo ) p b Íln Á ˜ - 0.75 + s˙ ÍÎ Ë rw ¯ ˙˚

(7 - 32)

Introducing the productivity index into the above equation gives: Ê 1 ˆ -2 2 Qo = J Á ˜ ( p r - p wf ) Ë 2 pb ¯

(7 - 33)

Ê J ˆ The term Á ˜ is commonly referred to as the performance coeffi Ë 2 pb ¯ cient C, or: Q o = C ( p r2 - p 2wf )

(7 - 34)



503

To account for the possibility of non-Darcy flow (turbulent flow) in oil wells, Fetkovich introduced the exponent n in Equation 7-35 to yield: Q o = C ( p r2 - p 2wf ) n

(7 - 35)

The value of n ranges from 1.000 for a complete laminar flow to 0.5 for highly turbulent flow. There are two unknowns in Equation 7-35, the performance coefficient C and the exponent n. At least two tests are required to evaluate these two parameters, assuming –pr is known: By taking the log of both sides of Equation 7-35 and solving for log (p2r - p2wf), the expression can be written as: log ( p r2 - p 2wf ) =

1 1 log Q o - log C n n

A plot of –p2r - p2wf versus Qo on log-log scales will result in a straight line having a slope of 1/n and an intercept of C at –p2r - p2wf = 1. The value of C can also be calculated using any point on the linear plot once n has been determined to give: C=

( p r2

Qo - p 2wf ) n

Once the values of C and n are determined from test data, Equation 7-35 can be used to generate a complete IPR. To construct the future IPR when the average reservoir pressure – ) , Fetkovich assumes that the performance coefficient C is declines to (p r f a linear function of the average reservoir pressure and, therefore, the value of C can be adjusted as: – ) /(p –) ] (C)f = (C)p [(p r f r p

(7-36)

where the subscripts f and p represent the future and present conditions. Fetkovich assumes that the value of the exponent n would not change as the reservoir pressure declines. Beggs (1991) presented an excellent and comprehensive discussion of the different methodologies used in constructing the IPR curves for oil and gas wells. The following example was used by Beggs (1991) to illustrate Fetkovich’s method for generating the current and future IPR.


504


Example 7-9 A four-point stabilized flow test was conducted on a well producing from a saturated reservoir that exists at an average pressure of 3600 psi. Qo, STB/day

pwf, psi

263 383 497 640

3170 2890 2440 2150

a. Construct a complete IPR by using Fetkovich’s method. b. Construct the IPR when the reservoir pressure declines to 2000 psi.

Solution Part A.

Step 1. Construct the following table: Qo, STB/day

Pwf, psi

– 2 - p2 ) ¥ 10-6, psi2 (p r wf

263 383 497 640

3170 2890 2440 2150

2.911 4.567 7.006 8.338

Step 2. Plot (–p2r - p2wf) verses Qo on log-log paper as shown in Figure 7-11 and determine the exponent n, or: n=

log(750) - log(105) = 0.854 log (10 7 ) - log (10 6 )

Step 3. Solve for the performance coefficient C: C = 0.00079



505

Figure 7-11. Flow-after-flow data for example 7-9 (After Beggs, D., “Production Optimization Using Nodal Analysis,” permission to publish by the OGCI, copyright OGCI, 1991.)

Step 4. Generate the IPR by assuming various values for pwf and calculating the corresponding flow rate from Equation 7-25: Qo = 0.00079 (36002 - p2wf)0.854


506


Pwf

Qo, STB/day

3600 3000 2500 2000 1500 1000 500 0

0 340 503 684 796 875 922 937

The IPR curve is shown in Figure 7-12. Notice that the AOF, i.e., (Qo)max, is 937 STB/day. Part B.

Step 1. Calculate future C by applying Equation 7-36 2000 ˆ (C)f = 0.00079 Ê = 0.000439 Ë 3600 ¯

4000

3500

Pressure (psi)

3000

2500

2000

1500

1000

500

0 0

100

200

300

400

500

600

700

Flow Rate (STB/Day)

Figure 7-12. IPR using Fetkovich method.

800

900

1000


507


Step 2. Construct the new IPR curve at 2000 psi by using the new calculated C and applying the inflow equation. Qo = 0.000439 (20002 - p2wf)0.854 pwf

Qo

2000 1500 1000 500 0

0 94 150 181 191

Both the present time and future IPRs are plotted in Figure 7-13. Klins and Clark (1993) developed empirical correlations that correlate the changes in Fetkovich’s performance coefficient C and the flow exponent n with the decline in the reservoir pressure. The authors observed the exponent n changes considerably with reservoir pressure. Klins and – ) are Clark concluded the “future” values of (n)f and (C) at pressure (p r f related to the values of n and C at the bubble-point pressure. Denoting Cb

4000

3500

Pressure (psi)

3000

2500

2000

1500

1000

500

0 0

100

200

300

400

500

600

700

Flow Rate (STB/Day)

Figure 7-13. Future IPR at 2000 psi.

800

900

1000


508


and nb as the values of the performance coefficient and the flow exponent at the bubble-point pressure pb, Klins and Clark introduced the following dimensionless parameters: • Dimensionless performance coefficient = C/Cb • Dimensionless flow exponent = n/nb • Dimensionless average reservoir pressure = –pr /pb The authors correlated (C/Cb) and (n/nb) to the dimensionless pressure by the following two expressions: Ê nˆ Ê pr ˆ Ê pr ˆ Á ˜ = 1 + 0.0577 Á1 - ˜ - 0.2459 Á1 - ˜ Ë nb ¯ Ë pb ¯ Ë pb ¯

2

pr ˆ 3 Ê + 0.503 Á1 - ˜ Ë pb ¯

(7 - 37)

and Ê Cˆ Ê Ê pr ˆ pr ˆ Á ˜ = 1 - 3.5718 Á1 ˜ + 4.7981 Á1 ˜ pb ¯ pb ¯ Ë Cb ¯ Ë Ë Ê p ˆ - 2.3066 Á1 - r ˜ Ë pb ¯

2

3

(7 - 38)

where Cb = performance coefficient at the bubble-point pressure nb = flow exponent at the bubble-point pressure The procedure of applying the above relationships in adjusting the coefficients C and n with changing average reservoir pressure is detailed below: Step 1. Using the available flow-test data in conjunction with Fetkovich’s equation, i.e., Equation 7-34, calculate the present (current) values of n and C at the present average pressure –pr. Step 2. Using the current values of p–r, calculate the dimensionless values of (n/n b ) and (C/C b ) by applying Equations 7-37 and 7-38, respectively.


509


Step 3. Solve for the constants nb and Cb from: nb =

n n / nb

(7 - 39)

C (C / C b )

(7 - 40)

and Cb =

It should be pointed out that if the present reservoir pressure equals the bubble-point pressure, the values of n and C as calculated in Step 1 are essentially nb and Cb. Step 4. Assume future average reservoir pressure –pr and solve for the corresponding future dimensionless parameters (nf /nb) and (Cf/Cb) by applying Equations 7-37 and 7-38, respectively. Step 5. Solve for future values of nf and Cf from nf = nb (n/nb) Cf = Cb (Cf /Cb) Step 6. Use nf and Cf in Fetkovich’s equation to generate the well’s future – ) . It should be IPR at the desired average reservoir pressure (p r f – ) is given by: noted that the maximum oil flow rate (Qo)max at (p r f – )2]nf (Qo)max = Cf[(p r

(7-41)

Example 7-10 Using the data given in Example 7-9, generate the future IPR data when the reservoir pressure drops to 3200 psi.

Solution Step 1. Since the reservoir exists at its bubble-point pressure, then: nb = 0.854

and Cb = 0.00079

at pb = 3600 psi


510


Step 2. Calculate the future dimensionless parameters at 3200 psi by applying Equations 7-37 and 7-38: 2 Ê nˆ Ê1 - 3200 ˆ - 0.2459 Ê1 - 3200 ˆ = 1 + 0 0577 . Á ˜ Ë 3600 ¯ Ë 3600 ¯ Ë nb ¯ 3

3200 ˆ + 0.5030 Ê1 = 1.0041 Ë 3600 ¯ 2 Ê Cˆ Ê1 - 3200 ˆ + 4.7981 Ê1 - 3200 ˆ = 1 3 5718 . Á ˜ Ë 3600 ¯ Ë 3600 ¯ Ë Cb ¯

3200 ˆ 3 - 2.3066 Ê1 = 0.6592 Ë 3600 ¯ Step 3. Solve for nf and Cf: nf = (0.854) (1.0041) = 0.8575 Cf = (0.00079) (0.6592) = 0.00052 Therefore, the flow rate is expressed as: Qo = 0.00052 (32002 - p2wf )0.8575 When the maximum oil flow rate, i.e., AOF, occurs at pwf = 0, then: (Qo)max = 0.00052 (32002 - 02)0.8575 = 534 STB/day Step 4. Construct the following table: pwf

Qo

3200 2000 1500 500 0

0 349 431 523 534

Figure 7-14 compares current and future IPRs as calculated in Examples 7-9 and 7-10.


511


Case 3: –pr > pb and pwf < pb

Figure 7-15 shows a schematic illustration of Case 3 in which it is assumed that pwf < pb and p–r > pb. The integral in Equation 7-25 can be expanded and written as: pr È pb ˘ 0.00708 kh Í f ( p) dp + f ( p) dp˙ Qo = Í ˙ Ê re ˆ pb ln Á ˜ - 0.75 + s Îp wf ˚ Ë rw ¯

Ú

Ú

Substituting Equations 7-27 and 7-18 into the above expression gives: È pb Ê 1 ˆÊ p ˆ 0.00708 kh Í Qo = Á ˜ Á ˜ dp + Í Ê re ˆ Ë m o Bo ¯ Ë p b ¯ p ln Á ˜ - 0.75 + s Î wf Ë rw ¯

Ú

˘ Ê 1 ˆ ˙ Á ˜ dp Ë m o Bo ¯ ˙ pb ˚

pr

Ú

4000

3500

Pressure (psi)

3000 Current IPR Future IPR

2500

2000

1500

1000

500

0 0

100

200

300

400

500

600

Flow Rate (STB/Day)

Figure 7-14. IPR.

700

800

900

1000


512


Figure 7-15. (kro/moBo) vs. pressure for Case #3.

where mo and Bo are evaluated at the bubble-point pressure pb. Arranging the above expression gives: È 0.00708 kh Í1 Qo = È Êr ˆ ˘ Í pb m o Bo Íln Á e ˜ - 0.75 + s˙ Î ÍÎ Ë rw ¯ ˙˚

pb

Ú p wf

˘ p dp + dp˙ ˙ pb ˚ pr

Ú

Integrating and introducing the productivity index J into the above relationship gives: È 1 ˘ Qo = J Í ( p 2b - p 2wf ) + ( p r - p b ) ˙ Î 2 pb ˚


513


or Qo = J ( pr - p b ) +

J ( p 2b - p 2wf ) 2 pb

(7 - 42)

Example 7-11 The following reservoir and flow-test data are available on an oil well: –p = 4000 psi r pwf = 3600 psi

• Pressure data: • Flow test data:

pb = 3200 psi Qo = 280 STB/day

Generate the IPR data of the well.

Solution Step 1. Calculate the productivity index from the flow-test data. J=

280 = 0.7 STB / day / psi 4000 - 3600

Step 2. Generate the IPR data by applying Equation 7-30 when the assumed pwf > pb and using Equation 7-42 when pwf < pb. pwf

Equation

Qo

4000 3800 3600 3200 3000 2600 2200 2000 1000 500 0

(7-30) (7-30) (7-30) (7-30) (7-42) (7-42) (7-42) (7-42) (7-42) (7-42) (7-42)

0 140 280 560 696 941 1151 1243 1571 1653 1680

Results of the calculations are shown graphically in Figure 7-16. It should be pointed out Fetkovich’s method has the advantage over Standing’s methodology in that it does not require the tedious material


514


4500 4000 3500

Pressure (psi)

3000 2500 2000 1500 1000 500

0 0

200

400

600

800

1000

1200

1400

1600

1800

Qo, STB/day

Figure 7-16. IPR using the Fetkovich method.

balance calculations to predict oil saturations at future average reservoir pressures. The Klins-Clark Method

Klins and Clark (1993) proposed an inflow expression similar in form to that of Vogel’s and can be used to estimate future IPR data. To improve the predictive capability of Vogel’s equation, the authors introduced a new exponent d to Vogel’s expression. The authors proposed the following relationships: Êp ˆ Êp ˆ Qo = 1 - 0.295 Á wf ˜ - 0.705 Á wf ˜ (Q o )max Ë pr ¯ Ë pr ¯

d

(7 - 43)

where È Ê p ˆ˘ d = Í0.28 + 0.72 Á r ˜ ˙ (1.24 + 0.001 p b ) Ë p b ¯ ˙˚ ÍÎ

(7 - 44)

IPR-Fetkovich

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