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Table of Contents: CHAPTER

PAGE

CHAPTER 1

3

CHAPTER 2

23

CHAPTER 3

24

CHAPTER 4

47

CHAPTER 5

62

CHAPTER 6

84

CHAPTER 7

104

CHAPTER 8

138

CHAPTER 9

156

CHAPTER 10

167

1

2

CHAPTER 1

INTRODUCTION TO RESERVOIR ENGINEERING

PROBLEM 1.1 Calculate the volume 1 lb-mole of ideal gas will occupy at: a) 14.7 psia and 60°F b) 14.1 psia and 32°F c) 14.7 plus 10 oz and 80°F d) 15.025 psia and 60°F

ANSWER: a)

pV = nRT Ÿ V =

3 nRT 1 × 10.73(460 + 60) = = 379.6 ft p 14.7

b)

V=

1 × 10.73 × (460 + 32) = 359.1 ft 3 14.7

c)

1 psia p = 14.7 psia + 100Ζ( ) = 15.325 psia 160Ζ V=

1 × 10.73 × (460 + 80) = 378.7 ft 3 15.325 3

d)

1 × 10.73 × (460 + 60) = 371.4 ft 3 15.025

V=

PROBLEM 1.2 A 500 cu ft tank contains 10 lb of methane and 20 lb of ethane at 90°F. a) How many moles are in the tank? b) What is the pressure of the tank is psia? c) What is the molecular weight of the mixture? d) What is the specific gravity of the mixture?

ANSWER: a)

nt =

m1 m 10 20 + 2 = + = 1.292 mw1 mw 2 16 30

b)

pa =

nRt 1.292(10.73)(550) = = 15.25 psia V 500

c)

Mw =

mt 30 = = 23.22 nt 1.292

d)

γg =

M 23.22 = = 0.80 M air 28.97

4

PROBLEM 1.3 What are the molecular weight and specific gravity of a gas that is one third each of methane, ethane, and propane by volume?

ANSWER: CH 4 = 0.333 C 2 H 6 = 0.333 C3 H 8 = 0.333 M w = 0.333(16) + (0.333)(30) + (0.333)(44) = 30

γg =

Mw 30 = = 1.035 28.97 28.97

p1V1 p2V2 = n1T1 n2T2 10 = 0.227lb − mole 44 pV = nRT Ÿ 14.7(50) = n × 10.73 × 535 → n2 = 0.128lbmol

n1 =

Ÿ

p 2 (50) 14.7(50) = Ÿ p2 = 38.48 psia 0.128(535) (0.128 + 0.227)(505)

PROBLEM 1.4 A 10 lb block of dry ice is placed in a 50 cu ft tank that contains air at atmospheric pressure 14.7 psia and 75°F. What will be the final pressure of the sealed tank when all the dry ice has evaporated and cooled the gas to 45°F?

ANSWER: p1V1 p 2V2 = n1T1 n2T2

5

10 = 0.227lbmole 44 pV = nRT Ÿ 14.7(50) = n × 10.73 × 535 → n2 = 0.128lbmol

n1 =

Ÿ

p2 (50) 14.7(50) = Ÿ p2 = 38.48 psia 0.128(535) (0.128 + 0.227)(505)

PROBLEM 1.5 A welding apparatus for a drilling rig uses acetylene (C2H2), which is purchased in steel cylinders containing 20 lb of gas, costs $4.50 exclusive of the cylinder. If a welder is using 200 cu ft per day measured at 16 oz gauge and 85°F, what is the daily cost of acetylene? What is the cost per MCF at 14.7 psia and 60°F?

ANSWER: oz = 1 psi in 2 p = p gauge + 14.7 = 15.7 psi

16

n=

m 20 = = 0.77 M w 26

PV = ZnRT → V =

ZnRT 1(0.77)(10.73)(545) = = 286.8 ft 3 p 15.7

4.5$ × 200 ft 3 = 3.14$ 286.8 ft 3 ZnRT (1)(0.77)(10.73)(520) V= = = 292.26 ft 3 p 14.7 4.5$ × 1000 ft 3 = 15.40$ 292.26 ft 3

6

PROBLEM 1.6 (a) A 55,000 bbl (nominal) pipeline tank has diameter of 110 ft and a height of 35 ft. It contains 25 ft of oil at the time suction is taken on the oil with pumps that handle 20,000 bbl per day. The breather and safety valves have become clogged so that a vacuum is drawn on the tank. If the root is rated to withstand ¾ oz per sq in. pressure, how long will it be before the roof collapses? Barometric pressure is 29.1 in. Neglect the fact that the roof is peaked and that there may be some leaks. (b) Calculate the total force on the roof at the time of collapse. (c) If the tank had contained more oil, would the collapse time have been greater or less?

ANSWER: The solution of this problem is left to readers.

PROBLEM 1.7 (a) What percentage of methane by weight does a gas of 0.65 specific gravity contain if it is composed only of methane and ethane? What percentage by volume? (b) Explain why the percentage by volume is greater than the percentage by weight.

ANSWER: a)

Mw → M w = 0.65(28.97) = 18.83 28.97 x(16) + (1 − x)(30) = 18.83 → x = 0.798

γ g = 0.65 → γg = c1 % = 79.8% =

16(0.798) = 67.8 18.83

b) Molecular weight of methane is less than ethane. Therefore in a mixture of fixed volume methane’s weight is less than ethane. 7

PROBLEM 1.8 A 50 cu ft tank contains gas at 50 psia and 50°F. It is connected to another tank that contains gas at 25 psia and 50°F. When the valve between the two is opened, the pressure equalizes at 35 psia at 50°F. What is the volume of the second tank?

ANSWER: PV RT PV P1V1 P2V2 = + Ÿ PV = P1V1 + P2V2 RT RT RT Vt = V1 + V2 Ÿ P (V1 + V2 ) = P1V1 + P2V2 z = 1, n =

Ÿ 35(50 + V2 ) = (50)(50) + ( 25)(V2 ) Ÿ V2 = 75 ft 3

PROBLEM 1.9 Gas was contracted at $1.10 per MCF at contract conditions of 14.4 psia and 80°F. What is the equivalent price at a legal temperature of 60°F and pressure of 15.052 psia?

ANSWER: P1V1 = n1 RT1 → n1 =

P1V1 14.4(1000) = = 2.485 ft 3 RT1 10.73(540)

P2V2 = n2 RT2 → n2 = Ÿ 2.697 ft 3 ×

P2V2 15.025(1000) = = 2.697 ft 3 10.73(520) RT2

101$ = 1.19$ 2.485 ft 3

PROBLEM 1.10 A cylinder is fitted with a leakproof piston and calibrated so that the volume within the cylinder can be read from a scale for any position of the piston. The cylinder is 8

immersed in a constant temperature bath maintained at 160°F, which is the reservoir temperature of the Sabine Gas Field. Forty-five thousand cc of the gas, measured at 14.7 psia and 60°F, is charged into the cylinder. The volume is decreased in the steps indicated below, and the corresponding pressures are read with a dead weight tester after temperature equilibrium is reached.

V,cc

2529

964

453

265

180

156.5

142.2

P, psia

300

750

1500

2500

4000

5000

6000

(a) Calculate and place in tabular form the gas deviation factors and the ideal volumes that the initial 45,000 cc occupies at 160°F and at each pressure. (b) Calculate the gas volumes factors at each pressure, in units of ft3/SCF. (c) Plot the deviation factor and the gas volume factors calculated in part (b) versus pressure on the same graph.

ANSWER: The solution of this problem is left to readers.

PROBLEM 1.11 (a) If the Sabine Field gas gravity is 0.65, calculate the deviation factors from zero to 6000 psia at 160°F, in 1000 lb increments, using the gas gravity correlation of Fig. 1.4. (b) Using the critical pressures and temperatures in Table 1.1, calculate and plot the deviation factors for the Sabine gas at several pressures and 160°F.The gas analysis is as follows:

9

Component

C1

C2

C3

iC4

nC4

iC5

Mole

0.875

0.083

0.021

0.006

0.008

0.003

Component

nC5

C6

C7+

Mole

0.002

0.001

0.001

Fraction

Fraction

Use the molecular weight and critical temperature and pressure of octane for the heptanes-plus. Plot the data of Prob. 1.10(a) and Prob. 1.11(a) on the same graph for comparison. (c) Below what pressure at 160°F may the ideal gas law be used for the gas of the Sabine Field if errors are to be kept within 2%? (d) Will a reservoir contains more SCF of a real or of an ideal gas at similar conditions? Explain.

ANSWER: The solution of this problem is left to readers.

PROBLEM 1.12 A high-pressure cell has a volume of 0.330 cu ft and contains gas at 2500 psia and 130°F, at which conditions its deviation factor is 0.75. When 43.6 SCF measured at 14.7 psia and 60°F were bled from the cell through a wet test meter, the pressure dropped to 1000 psia, the temperature remaining at 130°F. What is the gas deviation factor at 1000 psia and 130°F?

10

ANSWER: P1V = Z 1 n1 RT 2500 (0.33) = 0.75 × n1 × (10 .73)(590 ) Ÿ n1 = 0.174 mol PscVsc = n2 RTsc

(14.7)(43.6) = n2 (10.73)(520) → n2 = 0.115mol ntotol = n1 − n2 = 0.174 − 0.115 = 0.059mol Ÿ P2V = Z 2 ntotal RT Ÿ 1000(0.33) = Z 2 (0.059)(10.73)(590) Ÿ Z 2 = 0.885

PROBLEM 1.13 A dry gas reservoir is initially at an average pressure of 6000 psia and temperature of 160°F. The gas has a specific gravity of 0.65. What will the average reservoir pressure be when one-half of the original gas (in SCF) has been produced? Assume the volume occupied by the gas in the reservoir remains constant. If the reservoir originally contained 1 MM ft3 of reservoir gas, how much gas has been produced at a final reservoir pressure of 500 psia?

ANSWER: The solution of this problem is left to readers.

PROBLEM 1.14 A reservoir gas has the following gas deviation factors at 150Ԭ. Plot z versus p and graphically determine the slopes at 1000 psia, 2200 psia, and 4000 psia, Then, using Eq. (1.19), find the gas compressibility at these pressures.

11

P,psia

0

500

1000

2000

3000

4000

5000

z

1.00

0.92

0.86

0.80

0.82

0.89

1.00

ANSWER: The solution of this problem is left to readers.

PROBLEM 1.15 Using Figs. 1.4 and 1.5, find the compressibility of a 70% specific gravity gas at 5000 psia and 203°F.

ANSWER: from fig1 − 4 → T pc = 375° R, p pc = 660 psia T pr =

T 663 = = 1.77 TPc 375

Ppr =

p 5000 = = 7.6 Ppc 660

Ÿ fig 1 − 5 → Z = 1.01

PROBLEM 1.16 Using Eq. (1.22) and the generalized chart for gas deviation factors, Fig. 1.5, find the pseudoreduced compressibility of a gas at a pseudoreduced temperature of 1.30 and a pseudoreduced pressure of 4.00. Check this value on Fig. 1.7.

ANSWER: The solution of this problem is left to readers.

12

PROBLEM 1.17 Estimate the viscosity of a gas condensate fluid at 7000 psia and 220°F. It has a specific gravity of 0.90 and contains 2% nitrogen, 4% carbon dioxide, and 6% hydrogen sulfide.

ANSWER: ­° Ppc = 756.8 − 131γg − 3.6γ 2 g = 636 psia 2 °¯T pc = 169.2 + 349.5γ g − 74γ g = 424° R

γ = 0.90®

from fig1 − 9 → M1 = 0.0117 cp correction for CO 2 = 0.0003 cp correction for N 2 = 0.00024 cp correction H 2s = 0.00032 cp → M 1 = 0.0117 + 0.0003 + 0.00024 + 0.00032 = 0.01256cp P 7000 ­ ° Ppr = P = 636 = 11 M pc ° → fig1 − 10 → =3 ® M1 °T = T = 680 = 1.6 °¯ pr T pc 24 M = 3(0.01256) = 0.0376cp

PROBLEM 1.18 Experiments were made on a bottom-hole sample of the reservoir liquid taken from the LaSalle Oil Field to determine the solution gas and the formation volume factor as functions of pressure. The initial bottom-hole pressure of the reservoir was 3600 psia and bottom-hole temperature was 160°F; so all measurements in the laboratory were made at 160°F. The following data, converted to practical units, were obtained from the measurements:

13

Pressure

Solution gas SCF/STB

Formation Volume

psia

at 14.7 psia and 60°F

Factor, bbl/STB

3600

567

1.310

3200

567

1.317

2800

567

1.325

2500

567

1.333

2400

554

1.310

1800

436

1.263

1200

337

1.210

600

223

1.140

200

143

1.070

a) What factors affect the solubility of gas in crude oil? b) Plot the gas in solution versus pressure. c) Was the reservoir initially saturated or undersaturated? Explain. d) Does the reservoir have an initial gas cap? e) In the region of 200 to 2500 psia, determine the solubility of the gas from your graph in SCF/STB/psi. f) Suppose 1000SCF of gas had accumulated with each stock tank barrel of oil in this reservoir instead of 567 SCF. Estimate how much gas would have been in solution at 3600 psia. Would the reservoir oil then be called saturated or undersaturated?

ANSWER: a) Pressure, API, Specific gravity b) The solution of this problem is left to readers.

14

c) Undersatured. There is no evolved gas in the reservoir from initial reservoir pressure until pressure drops to bubble point pressure. d) No. There is no initial gas cap in undersatured reservoirs. e) The solution of this problem is left to readers. f) The solution of this problem is left to readers.

PROBLEM 1.19 a) From the bottom-hole sample given in Prob. 1.18: b) Plot the formation volume factor versus pressure. c) Explain the break in the curve. d) Why is the slope above the bubble- point pressure negative and smaller than the positive slope below the bubble-point pressure? e) If the reservoir contains 250MM reservoir barrels of oil initially, what is the initial number of STB in place? f) What is the initial volume of dissolved gas in the reservoir? g) What will be the formation volume factor of the oil when the bottom-hole pressure is essentially atmospheric if the coefficient of expansion of the stock tank oil is 0.0006 per °F?

ANSWER: a) The solution of this problem is left to readers. b) In pressures below bubble point pressure gas evolves and comes out of solution. The break of the curve is due to this phenomenon. c) Some fraction of evolved gas expands at above bubble point pressure and this will result in negative and smaller slope.

15

d) B0 =

=

V V0 250 →N= 0 = = 190 .83MMSTB N B0 1.31

e) Vsolution gas = VO ( sc ) × Rs = 190 .83(567 ) = 108 .2 MMMscf

VT = V60 [1 + B (T − 60) ] → V160 = 1 × [1 + 0.0006 (160 − 60) ] V160 = 1.06 bb1

sTB

PROBLEM 1.20 If the gravity of the stock tank oil of the Big Sandy reservoir is 30°API and the gravity of the solution gas is 0.80, estimate the solution gas-oil ratio and the single-phase formation volume factor at 2500 psia and 165°F. The bubble-point pressure is 2800 psia.

ANSWER:

API = 30, γ = 0.8, p = 2500,T = 160°F fig1 − 11 →R s = 567 scf P = 2500psia Ÿ fig 1 - 12 → BO = 1.333bb1

sTB

STB

PROBLEM 1.21 A 1000 cu ft tank contains 85 STB of crude and 20,000 SCF of gas, all at 120°F. When equilibrium is established, (i.e., when as much gas has dissolved in the oil as will), the pressure in the tank is 500 psia. If the solubility of the gas in the crude is 0.25 SCF/STB/psi and the deviation factor for the gas at 500 psia and 120°F is 0.90, find the liquid formation volume factor at 500 psia and 120°F.

16

ANSWER: Rs = 0.25(500) = 125 scf

sTB solution gas at surface = 85(125) = 10625scf free gas at surface = 2000 − 10625 = 9375scf 3 (0.9)(580) ZT = 0.02829 = 0.0295 ft Bg = 0.02829 scf 500 P free gas at reservoir condition = 9375(0.0295) = 276.56 ft 3 1000 − 276.56 = 128.8bb1 solution gas at reservior condition + oil volume = 5.615 128.8 = 1.515 bb1 B0 = sTB 85

PROBLEM 1.22 A crude oil has a compressibility of 20×10-6 psi-1 and a bubble point of 3200 psia. Calculate the relative volume factor at 4400 psia (i.e., the volume relative to its volume at the bubble point), assuming constant compressibility.

ANSWER: Vr =

V0 0.9799 = = 0.9864 Vbp 0.9934

PROBLEM 1.23 (a) Estimate the viscosity of an oil at 3000 psia and 130°F. It has a stock tank gravity of 35°API at 60°F and contain-san estimated 750 SCF/STB of solution gas at the initial bubble point of 3000 psia. (b) Estimate the viscosity at the initial reservoir pressure of 4500 psia.

17

(c) Estimate the viscosity at 1000 psia if there is an estimated 300 SCF/STB of solution gas at that pressure.

ANSWER: a)

μ 0 = Aμ od

B

A = 10.715( Rs0 + 100) −0.515 = 10.715(750 + 100) −0.515 = 0.332 B = 5.44( Rs0 + 150) −0.338 = 5.44(750 + 100) −0.338 = 0.556 ­μ od = 10 x − 1 ° Z −1.163 ° x = y (T − 460) , y = 10 °°Z = 3.0324 − 0.02024(35) = 2.324 ® 2.324 = 210.9 ° y = 10 ° x = 210.9(130) −1.163 = 0.734 ° °¯μ od = 10 0.734 − 1 = 4.42cp

μ 0 = Aμ od = 0.332(4.42) 0.556 = 0.76cp B

b)

P m ) Pb m = 2.6 P1.187 exp[− 11.513 − 8.98(10) −5 P ]P = 4500 → m = 0.377 4500 0.377 μ 0 = 0.76( ) = 0.866cp 3000

μ 0 = Aμ od (

c)

μ 0 = Aμ od

B

A = 10.715( Rso + 100) −0.515 , Rso = 300 Ÿ A = 0.49 B = 5.44( Rso + 150) −0.338 , Rso = 300 Ÿ B = 0.718

18

μ od = 10 x − 1 x = y (T − 460) −1.163 y = 10 Z Z = 3.0324 − 0.02024(35) = 2.324 y = 10 2.324 = 210.9 x = 210.86(130) −1.163 = 0.734 Ÿ μ od = 10 0.734 − 1 = 4.42 cp Ÿ μ o = 0.49 ( 4.42 ) 0.718 = 1.42 cp

PROBLEM 1.24 Given the following laboratory data: Cell

Oil Volume in cell

Gas Volume in

Cell Temperature

Pressure

(cc)

cell (cc)

(°F)

2000

650

0

195

1500 = Pbp

669

0

195

1000

650

150

195

500

615

700

195

14.7

500

44,500

60

(psia)

Evaluate Ro, Bo, and B, at the stated pressures. The gas deviation factors at 1000 psia and 500 psia have been evaluated as 0.91 and 0.95, respectively.

ANSWER: The solution of this problem is left to readers.

19

PROBLEM 1.25 (a) Find the compressibility for a connate water that contains 20.000 parts per million of total solids at a reservoir pressure of 4000 psia and temperature of 150°F. (b) Find the formation volume factor of the formation water of part (a). ANSWER: a) T = 150 ° F , P = 4000 psia cw =

[7.033 p + 541 .5C

1 NaCl

− 537T + 403300 ]

= 3.2 × 10 −6 psi −1

b) ΔVw = −1.00010 ×10 −2 + 1.33391×10 −4 T + 5.050654 × 10 −7 T 2 = 0.02 ΔVwp = −1.95301× 10−9 PT − 1.72834 × 10−13 p 2T − 3.58922 × 10 −7 p − 2.2534 × 10 −10 p 2 = −0.0066 Bw = (1 + Δvwt )(1 + Δvwp ) = (1 + 0.02)(1 − 0.0066) = 1.013 bb1 Bw = 1.013 bb1

stb

stb

PROBLEM 1.26 (a) What is the approximate viscosity of pure water at room temperature and atmospheric pressure? (b) What is the approximate viscosity of pure water at 200°F?

ANSWER: a)

s=0 A = 109.574 − 8.40564S + 0.313314S 2 + 8.72213 × 10−3 S 3 = 109.574 B = −1.12166 + 2.63951× 10− 2 S − 6.79461× 10−4 S 2 − 5.47119 × 10−5 S 3 + 1.55586 × 10−6 S 4

20

B = −1.1216 M w = AT B = 109.574(60)−1.1216 = 1.11cp b)

S = 0.18 A = 108.1 B = −1.1169 Ÿ M w = AT B = 108.1(200) −1.1169 = 0.291cp

PROBLEM 1.27 A container has a volume of 500 cc and is full of pure water at 180°F and 6000 psia. (a) How much water would be expelled if the pressure were reduced to 1000 psia? (b) What would be the volume of water expelled if the salinity were 20,000 ppm and there were no gas in solution? (c) Rework part (b) assuming that the water is initially saturated with gas and that all the gas is evolved during the pressure change. (d) Estimate the viscosity of the water.

ANSWER: a) ΔVwp = −1.95301 × 10 −9 PT − 1.72834 × 10 −13 P 2T − 3.58922 × 10 −7 p − 2.25341 × 10 −10 F T1 = T2 = 180 ° F

at P = 6000 psia Ÿ ΔVwp 1 = −0.013495cc at P = 1000 psia Ÿ ΔVwp 2 = −9.669 × 10 −4 cc ΔVwp = −0.01253 Ÿ V2 = 500 + 0.01253 = 500.01cc

21

b) M w1 = AT B salinity = 20000 ppm → S = 20000 × 10 −4 = 2 Ÿ A = 94.086 B = −1.072 Ÿ M w1 = 94.086(180) −1.072 = 0.36 M w = Mw1× (0.9994 + 4.0295 × 10 −5 P + 3.1062 × 10 −9 P 2 ) , P = 1000 psia M w = 0.375cp

c) The solution of this problem is left to readers. d) The solution of this problem is left to readers.

22

CHAPTER 2

THE GENERAL MATERIAL BALANCE EQUATION

This chapter contains no questions.

23

CHAPTER 3

SINGLE-PHASE GAS RESERVOIRS

PROBLEM 3.1 A volumetric gas field has an initial pressure of 4200 psia, a porosity of 17.2%, and connate water of 23%. The gas volume factor at 4200 psia is 0.003425 cu ft/SCF and at 750 psia is 0.01852 cu ft/SCF. a) Calculate the initial in-place gas in standard cubic feet on a unit basis. b) Calculate the initial gas reserve in standard cubic feet on a unit basis, assuming an abandonment pressure of 750 psia. c) Explain why the calculated initial reserve depends on the abandonment pressure selected. d) Calculate the initial reserve of a 640-acre unit whose average net productive formation thickness is 34 ft, assuming an abandonment pressure of 750 psia. e) Calculate the recovery factor based on an abandonment pressure of 750 psia.

ANSWER:

pi = 4200 psi Bg i = 0.003425 ft 3 / scf @ pi = 4200 psia

φ = 17.2% Bg = 0.0 / 852 ft 3 / scf @ p = 750 psia Swi = 23%

24

a)

G=? G=

43560 × vb × φ × (1 − Swi ) = Bg i

43560 × 1 × 0 .172 × (1 − 0 .23 ) = 1684 .4 MSCF 0 .003425 (b) Initial gas reserve?

43560 × vb × φ × (1 − Swi ) = Bgab 43560 ×1× 0.172 × (1 − 0.23) = 316MSCF 0.01852 Reserve = G − Gab = 1684.4 − 316 = 1368.4MSCF

Gab =

OR

RF =

Bgab − Bgi Bgab

=

0.01852 − 0.003425 = 0.815 0.01852

Reserve = RF × G = 0.815 × 1684.4 = 1372MSCF c) We use Gab to calculate Reserve and Gab is a function of pressure. Therefore reserve is a function of pressure too. (d) G = 1684 .4 × (640 × 34 ) = 36652544 MSCF

G r = (RF )(G ) = (0.812 )(36652544 ) = 29 .8 MMMSCF

(e)

Bgab − Bgi 0.01852 − 0.003425 = = 0.815 0.01852 Bgab RF = 81.5%

RF =

25

PROBLEM 3.2 The discovery well No.1 and wells No.2 and No.4 produce gas in the 7500-ft reservoir of the Echo Lake Field (Fig. 3.14). Wells Nos. 3 and 7 were dry in the 7500-ft reservoir; however, together with their electric logs and the one from well No.1, the fault that seals the northeast side of the reservoir was established. The logs of wells Nos. 1, 2, 4, 5, and 6 were used to construct the map of Fig. 3.14, which was used to locate the gas-water contact and to determine the average net sand thickness. The reservoir had been producing for 18 months when well No.6 was drilled at the gas-water contact. The static wellhead pressures of the production wells showed virtually no decline during the 18month period before drilling wellNo.6 and averaged near 3400 psia. The following data were available from electric logs, core analysis, and the like.

Average well depth = 7500 ft Average static wellhead pressure = 3400 psia Reservoir temperature = 175°F Gas specific gravity =0.700 Average porosity = 27% Average connate water = 22% Standard conditions = 14.7 psia and 60°F Bulk volume of productive reservoir rock at the time No.6 was drilled= 22,500 ac-ft





a) Calculate the reservoir pressure. b) Estimate the gas deviation factor and the gas volume factor. c) Calculate the reserve at the time well No.6 was drilled, assuming a residual gas saturation of 30%. d) Discuss the location of well No.1 with regard to the overall gas recovery. e) Discuss the effect of sand uniformity on overall recovery--for example, a uniform permeable sand versus a sand in two beds of equal thickness, one of which has a permeability of 500 md and the other, 100 md.

ANSWER: (a) § Pwh ·§ Depth · ΔP = 0.25¨ ¸¨ ¸ΔP = Pws − Pwh © 100 ¹© 100 ¹ § 3400 ·§ 7500 · Pws = 3400 + 0.25¨ ¸¨ ¸ = 4038 Psia © 100 ¹© 100 ¹

OR § 0.01875γ g .Depth· ¸¸ Pws = Pwh.Exp¨¨ Zav.Tav © ¹

Zav=?

Pav =

Pws + Pwh = 3718.75 2 ­Tpc = 168 + 325γg − 12.5γ 2 g = 389.4

γg = 0.7®

2 ¯Ppc = 677 + 15γg − 37.5γ g = 659.125

­Tpr = T = 635 = 1.63 389.4 Tpc ° Ÿ Z av = 0.885 ® °Ppr = P Ppc = 371.75659.125 = 5.64 ¯

§ 0.01875(0.7)(7500) · Pws = 3400EXP¨ ¸ = 4038Psia 0.885 × 635 ¹ © 27

b)

Bgi = 0.02827

ZT 0.885(635) = 0.02827 = 0.003934 ft3 sef 4038 P

c) G=

43560 × VB × φ × (1 − Swi ) = Bg i

RF =

G=

1 − Swi − Sgr 1 − 0.22 − 0.30 = = 0.615 1 − Swi 1 − 0.22

43560 × 22500 × 0.27 × (1 − 0.22 ) = 52 .4 × 10 9 SCF 0.003934

GReserve = RF .G = 0.615(52.4 × 10 9 ) = 32.2 × 10 9 SCF

d) The solution of this problem is left to readers. e) The solution of this problem is left to readers.

PROBLEM 3.3 The M Sand is a small gas reservoir with an initial bottom-hole pressure of 3200 psia and bottom-hole temperature of 220°F. It is desired to inventory the gas in place at three production intervals. The pressure-production history and gas volume factors are as follows:

28

Pressure

Cumulative Gas

Gas Volume

psia

Production

Factor

MM SCF

Cu ft/SCF

3200

0

0.0052622

2925

79

0.0057004

2525

221

0.0065311

2125

452

0.0077360

a) Calculate the initial gas in place using production data at the end of each of the production intervals, assuming volumetric behavior. b) Explain why the calculations of part (a) indicate a water drive. c) Show that a water drive exists by plotting the cumulative production versus p/z. d) Based on electric log and core data, volumetric calculations on the M Sand showed that the initial volume of gas in place is 1018 MM SCF. If the sand is under a partial water drive, what is the volume of water encroached at the end of each of the periods? There was no appreciable water production.

ANSWER: a) RF =

Bg − Bg i Gp GP Bg ŸG= & RF = GP = Bg G RF Bg − Bg i

First period Ÿ G =

0.0057004 (79) = 1027MMSCF 0.0057004 − 0.0052622

Second period Ÿ G = Third period Ÿ G =

0.0065311 (221) = 1137MMSCF 0.0065311− 0.0052622

0.0077360 (452) = 1413.5MMSCF 0.0077360 − 0.0052622 29

b) The reservoir is water drive because the amount of G is increasing in the reservoir. c) P

Z=Bg.P/0.02829T

P/Z

GP(MMSCF)

3200

0.875

3657

0

2925

0.867

3374

79

2525

0.857

2946

221

2125

0.855

2488

452

d)

G(Bg − Bgi) + we = GpBg+ WpBwŸ We = GpBg− G(Bg − Bgi) Water influx after the first period: Ÿ We = 79 × 10 6 (0.0057004 ) − 1018 × 10 6 (0.0057004 − 0.0052622 ) = 4244 ft 3 = 755 .8bbl

Water influx after the second period: Ÿ We = 221 × 10 6 (0.0065311 ) − 1018 × 10 6 (0.0065311 − 0.0052622 ) = 151 .6 × 10 3 ft 3 = 27005 bbl

Water influx after the third period:

Ÿ We = 452 × 10 6 (0.0077360) − 1018 × 10 6 (0.0077360 − 0.0052622) = 174237.5bbl

PROBLEM 3.4 When the Sabine Gas Field was brought in, it had a reservoir pressure of 1700 psia, and temperature of 160ºF. After 5.00 MMM SCF was produced, the pressure had fallen to 30

1550 psia. If the reservoir is assumed to be under volumetric control, using the deviation factors of Prob. 1.10, calculate the following: a) The hydrocarbon pore volume of the reservoir. b) The SCF produced when the pressure falls to 1550, 1400, 1100,500, and 200 psia. Plot cumulative recovery in SCF versus p/z. c) The SCF of gas initially in place. d) From your graph, find how much gas can be obtained without the use of compressors for delivery into a pipeline operating at 750 psia. e) What is the approximate pressure drop per MMM SCF of production? f) Calculate the minimum value of the initial reserve if the produced gas measurement is accurate to ± 5% and if the average pressures are accurate to ± 12 psi when 5.00 MMM SCF have been produced and the reservoir pressure has dropped to 1550 psia.

ANSWER: The solution of this problem is left to readers.

PROBLEM 3.5 If, however, during the production of 5.00 MMM SCF of gas in the preceding problem, 4.00 MM bbl of water had encroached into the reservoir and still the pressure had dropped to 1550 psia, calculate the initial in-place gas. How does this compare with Prob. 3.4(c)?

ANSWER: The solution of this problem is left to readers.

31

PROBLEM 3.6 (a) The gas cap of the St. John Oil Field had a bulk volume of 17,000 ac-ft when the reservoir pressure had declined to 634 psig. Core analysis shows an average porosity of 18% and an average interstitial water of 24%. It is desired to increase the recovery of oil from the field by repressuring the gas cap to 1100 psig. Assuming that no additional gas dissolves in the oil during repressuring, calculate the SCF required. The deviation factors for both the reservoir gas and the injected gas are 0.86 at 634 psig and 0.78 at 1100 psig, both at 130°F. (b) If the injected gas has a deviation factor of 0.94 at 634 psig and 0.88 at 1100 psig, and the reservoir gas deviation factors are as above, recalculate the injected gas required. (c) Is the assumption that no additional solution gas will enter the reservoir oil a valid one? (d) Considering the possibility of some additional solution gas and the production of oil during the time of injection, will the figure of part (a) be maximum or minimum? Explain. (e) Explain why the gas deviation factors are higher (closer to unity) for the injected gas in part (b) than for the reservoir gas.

ANSWER: a)

Bg = 0.0282

ZT 0.78(590) = 0.02820 = 0.0117 ft3 Ÿ sef P 1114.7

43560 × 17000 × 0.18 × (1 − 0.24) = 8.6 × 10 9 SCF 0.0117 0.78(590) ZT = 0.02827 = 0.0221 ft 3 / scf Ÿ Bg = 0.0282 648.7 P 43560 × 17000 × 0.18 × (1 − 0.24) Ÿ G2 = = 4.5 × 10 9 scf 0.0221

Ÿ G1 =

32

The amount of injected gas into the reservoir=G2-G1= 4.5×109-8.6×109= - 4.1×109 The negative sign means that gas is injected to the reservoir.

b) P=634 psig=648.7 psia

Z=0.94

P=1100 psig=1114.7 psia

Z=0.88

· § 1 1 ¸¸ Ginj = V g ¨¨ − Bg ( 648 . 7 ) Bg ( 1114 . 7 ) ¹ © Bg 648.7 = 0.02417 ft 3

scf ft 3 Bg1114.7 = 0.01317 scf Vg = 43560 × Vb × φ × (1 − Swi) 1 1 § · 9 − G inj = 43560 × 17000 × 0.18 × (1 − 0.24)¨ ¸ = −3.5 × 10 SCF 0 . 02417 0 . 01317 © ¹

c) No d) It will be maximum. e) Because in this case fewer amount of gas is required for injection.

PROBLEM 3.7 The following production data are available from a gas reservoir produced under volumetric control: Pressure

Cumulative

(psia)

Gas Production (MMM SCF)

5000

200

4000

420

33

The initial reservoir temperature was 237ºF and the reservoir gas gravity is 0.7. a) What will be the cumulative gas production at 2500 psia? b) What fraction of the initial reservoir gas will be produced at 2500 psia? c) What was the initial reservoir pressure?

ANSWER: P p § p· = i − i G p 2500 ¨ ¸ © Z ¹ 2500 Z i GZ i

First of all we have to find the value of Pi/ZiG Tpc = 168 + 325δg − 12 .5δg 2 = 389 .375 Ppc = 677 + 15δg − 37 .5δg 2 = 669 .125

697 ­ T °Tpr = Tpc = 389.375 1.79 P = 2500 ® Ÿ ° Ppr = P = 2500 = 3.74 Ppc 669.125 ¯

( Z)

Ÿ Z 2500 = 0.89 Ÿ P

2500

= 2809

( )

­Tpr = 1.79 P = 5000® Ÿ Z 5000 = 1.01 Ÿ P Z ¯ Ppr = 7.747

( )

­Tpr = 1.79 P = 4000 ® Ÿ Z 4000 = 0.95 Ÿ P Z ¯ Ppr = 6

Slope =

Δy = Δx

(P Z )

5000

( Z)

− P

4000

G p 4000 − G p 5000

=

5000

4000

= 4950.5

= 4210 .5

4950 .5 − 4210 .5 = −3.36 × 10 − 9 200 × 10 9 − 420 × 10 9

Pi = 3.36 × 10 −9 ZiG In the next step, we have to find Pi/Zi:

34

Pi p § p· − i G5000 ¨ ¸ = Z Z GZ © ¹ 5000 i i

4950.5 =

pi − (3.36 × 10 − 9 zi

) × (200 × 10 ) Ÿ 9

pi = 5623 zi

Now, substitute the calculated values into equation *:

2809 = 5623 − (3.36 × 10 −9 )× GP 2500 G p 2500 = 837.56 × 109 SCF b)

RF = P P

Gp Gp2500 = G G

Z

= 0 Ÿ Gp = G

Z

= pi

Zi

− Pi

ZiG

.Gp

0 = 5623 − (3.36 × 10 −9 )Gp Ÿ Gp = G = 1673.5 × 10 9 SCF RF =

837.56 × 10 −9 = 0.500 Ÿ RF = 50% 1673.5 × 10 9

It means that half of the total initial gas in place can be recovered at 2500 psia. c) Using trial and error method and by considering that Pi/Zi=5623, then: First assumption: P=6000 psia

35

Tpr = 1.79 ½ 6000 ° P = 6000 = 5555.5? Pi/Zi = 5623 6000 ¾ → Z = 1.08 → 1.08 Ppr = = 8.96° 669.125 ¿

Second assumption: P=6200 psia Tpr = 1.79 P = 6200 Ppr = 6200

½° ¾ Ÿ Z = 1.1 → 62001.1 = 5636 = 9.26° 669.125 ¿

Since 5636 is almost close to initial Pi/Zi=5623, then we can conclude that Pi=6200 psia.

PROBLEM 3.8 (a) A well drilled into a gas cap for gas recycling purposes is found to be in an isolated fault block. After 50 MM SCF was injected, the pressure increased from 2500 to 3500 psia. Deviation factors for the gas are 0.90 at 3500 and 0.80 at 2500 psia and the bottomhole temperature is 160°F. What is the cubic feet of gas storage space in the fault block? (b) If the average porosity is 16%, average connate water is 24%, and average sand thickness is 12 ft, what is the areal extent of the fault block?

ANSWER: (a) Vg=?

1 1 − ) Bgi Bg 0.8(620) ZT = 0.0282 = 0.0056 ft 3 Bgi = 0.0282 Scf 2500 P

Ginj = Vg (

36

0.9(620) = 0.0045 3500 1 · − 50 × 10 + 6 § 1 − 50 × 10 + 6 = Vg ¨ − ¸ → Vg = − 43.58 © 0.0056 0.0045 ¹ Vg = 1.15MMSCF Bg = 0.0282

(b)

A=? h = 12 ft A=?

φ = 16% Swi = 24%

Vg = 43560× ( A × h) × φ × (1 − Swi) 1.15 × 106 = 43560× A × 40 × 0.16 × (1 − 0.24) Ÿ A = 18.09acres

PROBLEM 3.9 The initial volume of gas in place in the P Sand reservoir of the Holden Field is calculated from electric log and core data to be 200 MMM SCF underlying 2250 productive acres, at an initial pressure of 3500 psia and 140°F. The pressure-production history is

Pressure, psia

Production

Gas Deviation

MMM SCF

Factor at 140ºF

3500 (initial)

0.0

0.85

2500

75.0

0.82

37

(a) What is the initial volume of gas in place as calculated from the pressureproduction history, assuming no water influx? (b) Assuming uniform sand thickness, porosity, and connate water, if the volume of gas in place from pressure-production data is believed to be correct, how many acres of extension to the present limits of the P Sand are predicted? (c) If, on the other hand, the gas in place calculated from the log and core data is believed to be correct, how much water influx must have occurred during the 75 MMM SCF of production to make the two figures agree?

ANSWER: (a) G=? P P Pi 2500 3500 3500 = − i Gp  = − (75) Z Zi ZiG 0.82 0.85 0.85G Ÿ G = 289 MMMSCF

(b) G=

43560 × ( A × h ) × φ × (1 − Swi ) Bgi

ZT = 4.1 × 10 −3 @ p = 3500 P Bg = 5.5 × 10 −3 @ P = 2500 Bgi = 0.0282

200 × 10 9 =

43560 × ( A × h) × φ × (1 − Swi) Ÿ hφ (1 − Swi) = 8.36 4.1 × 10 −3

43560 × A × 8.36 Ÿ A = 3253acre 4.1 × 10 −3 3253 − 2250 = 1003acre 289 × 10 9 =

38

c) We=?

G( Bg − BGi) = GpBg − We Ÿ We = GpBg − G( Bg − Bgi) We = 75(5.5 × 10 −3 ) − 200(5.5 × 10 −3 ) − (4.1 × 10 −3 ) = 0.126 × 10 9 ft 3 ÷ 5.615 We = 22.8MMbbl

PROBLEM 3.10 Explain why initial calculations of gas in place are likely to be in greater error during the early life of depletion reservoirs. Will these factors make the predictions high or low? Explain.

ANSWER: Reservoir pressure drops during production from the reservoir and this phenomenon increases the Bg and consequently the calculated gas in place has some error from the its real value. Water influx into the reservoir maintains the reservoir pressure and this pressure stabilization will not allow Bg to be increased.

PROBLEM 3.11 A gas reservoir under partial water drive produced 12.0 MMM SCF when the average reservoir pressure had dropped from 3000 psia to 2200 psia. During the same interval, an estimated 5.20 MM bbl of water entered the reservoir based on the volume of the invaded area. If the gas deviation factor at 3000 psia and bottom-hole temperature of 170ºF is 0.88 and at 2200 psia is 0.78, what is the initial volume of gas in place measured at 14.7 psia and 60ºF?

39

ANSWER: Gp = 12 × 10 9 SCF Pi = 3000 Psia ( Z = 0.88) P = 2200 Psia ( Z = 0.76) T = 170 $ F = 630 $ R We = 5.2 × 10 6 bbl G =?

G( Bg − Bgi) + We = GpBg 0.88(630) Bg i = 0.0282 = 0.0054 3000 0.78(630) Bg = 0.0282 = 0.00629 2200 G (0.00629 − 0.0054) + (5.2 × 10 6 × 5.615) = 12 × 10 9 (0.000629) G = 42.8MMMSCF

PROBLEM 3.12 A gas-producing formation has uniform thickness of 32 ft, a porosity of 19%, and connate water saturation of 26%. The gas deviation factor is 0.83 at the initial reservoir pressure of 4450 psia and reservoir temperature of 175°F. (a) Calculate the initial in-place gas per acre-foot of bulk reservoir rock. (b) How many years will it take a well to deplete by 50% a 640 ac unit at the rate of 3 MM SCF/day? (c) If the reservoir is under an active water drive so that the decline in reservoir pressure is negligible, and during the production of 50.4 MMM SCF of gas water invades 1280 acres, what is the percentage of recovery by water drive? (d) What is the gas saturation as a percentage of total pore space in the portion of the reservoir invaded by water?

40

ANSWER: (a)

Bgi = 0.0282 G=

3 0.83(635) = 3.35 × 10 −3 ft Scf 4450

43560 × vb × φ × (1 − Swi ) 43560 × 1 × 0.19 × (1 − 0.26 ) = = 1.83 × 10 6 Scf Bgi 3.35 × 10 − 3

(b) G=

43560 × (640 × 32) × 0.19 × (1 − 0.26) = 37.435 × 10 9 SCF 0.00335

(50%) =18.7×109SCF q=

v v 18.7 × 10 9 Ÿt = = = 6233day t q 3 × 10 6

t=

6233 = 17.1year 365

(c) RF=? Active Water drive → Gp = 50.4 × 10 9 SCF

A=1280 acre Sgr=?

RF =

1 − Swi − Sgr 1 − Swi

S Wf =

Connate Water + We 43560 × φ × vb × Swi + We = Vp 43560 × φ × vb

We =?

Bg = Bgi Water Avctive → G ( Bg − ↑ 0 Bgi) + We = GpBg + BW ↑ 0 Wp We = GpBg  50.4 × 10 9 × (0.00335) = 0.169 × 10 9 Sft 3 Swf =

43560 × (1280 × 32) × 0.19 × 0.26 + 0.169 × 10 9 43560 × (1280 × 32)0.19 41

Sgr = 1 − Swf = 1 − 0.7576 = 0.2424 1 − Swi − Sgr 1 − 0.26 − 0.2424 RF = = = 0.67 ⎯ ⎯→ RF = 67% 1 − Swi 1 − 0.26 d) Sg =

Vg (G − Gp ) Bg = GBgi Vp 1 − Swi

Water Active drive → Bg = Bgi Ÿ Sg =

Sg =

G − Gp G 1 − Swi

(74.88 − 50.4) × 10 9 = 0.2424 74.88 × 10 9 1 − 0.26

PROBLEM 3.13 Fifty billion standard cubic feet of gas has been produced from a dry gas reservoir since its discovery. The reservoir pressure during this production has dropped to 3600 psia. Your company, which operates the field, has contracted to use the reservoir as a gas storage reservoir. A gas with a gravity of 0.75 is to be injected until the average pressure reaches 4800 psia. Assume that the reservoir behaves volumetrically, and determine the amount of SCF of gas that must be injected to raise the reservoir pressure from 3600 to 4800 psia. The initial pressure and temperature of the reservoir were 6200 psia and 280°F, respectively, and the specific gravity of the reservoir gas is 0.75.

ANSWER:

γg = 0.75 γg inj = 0.75 T = 220$ F = 740$ R

42

Ginj = ? Gp = 50 × 109 Scf P1 = 3600Psia P2 = 4800Psia Pi = 6200Psia ª º 1 1 G inj = Vg « − »∗ Bg ( 4800 ) Bg ( 3600 ) ¬ ¼

­Tpc = 168 + 32.5(0.75) − 12.5(0.75) 2 = 404.72

γ g = 0.75®

2 ¯Ppc = 677 + 15(0.75) − 37.5(0.75) = 667.15

­Tpr = T = 340 = 1.83 Tpc 404.72 ° for p1 = 3600® Ÿ Z 1 = 0.933 ° Ppr = P Ppc = 3600 667.15 = 5.4 ¯

forp 2 = 4800

Tpr = 1.83 Ppr = 7.2

Ÿ Z2 = 1

­Tpr = 1.83 forpi = 6200® Ÿ Z i = 1.1 ¯ Ppr = 9.3

Bg1 = 0.0282 ZT Bg 2 = 0.0282 ZT

3

P

= 0.005425 ft = 0.004361 ft

3

P

Bg i = 0.003714 ft

3

Scf

scf

@ (P1 = 3600 )

Scf

@ (P2 = 4800 )

@ (Pi = 6200 )

G ( Bg 1 − Bg i ) = GpBg 1 Ÿ G =

Gp.Bg 1 50 × (0.005425) = ( Bg 1 − Bg i ) (0.005425 − 0.003714 )

G = 158.53 × 10 9 (0.003714 ) = 588.8 × 10 6 Scf V g = GB gi = 158 .53 × 10 9 × (.003714 ) = 588 .8 × 10 6 SCF

1 1 ª º ∗ Ÿ G inj = 588.8 × 10 6 « − = 26.48 × 10 9 Scf ¬ 0.004361 0.005425 »¼

43

PROBLEM 3.14 The production data for a gas field are given below. Assume volumetric behavior and calculate the following: (a) Determine the initial gas in place. (b) What percentage of the initial gas in place will be recovered at a p/z of 1000? (c) The field is to be used as a gas storage reservoir into which gas is injected during summer months and produced during the peak demand months of the winter. What is the minimum p/z value that the reservoir needs to be brought back up to if a supply of 50 MMM SCF of gas is required and the abandonment p/z is 1000?

p/z (psia)

Gp (MMM SCF)

6553

0.393

6468

1.642

6393

3.226

6329

4.260

6246

5.504

6136

7.538

6080

8.749

ANSWER: a) P P Pi = − i Gp Z Zi ZiG − Pi Δy 6553 − 6080 = Slope = = = 56 × 10 −9 Z iG Δx 8.749 − 0.393

44

P for ( P )6553 Ÿ 6553 = i − (56 × 10 9 )(0.393 × 10 9 ) Z Zi Pi = 6575.008 Zi

P = 0 Ÿ G = Gp Ÿ 0 = 6575.008 − (56 × 10 −9 )Gp Z Gp = G = 117.41× 10 9 Scf (b) if

P = 100 Ÿ Z

∗ Ÿ 1000 = 6575.008 − 56 × 10 −9 Gp Ÿ Gp = 99.55 × 10 9 Scf RF =

Gp 99.55 = = 0.8478 Ÿ RF = 85% G 117.41

(c)

G − Gp = 117.41 × 10 9 − 99.55 × 10 9 = 17.86 × 10 9 Scf 17.86 × 10 9 + 50 × 10 9 = 67.86 × 10 9 scf P P m = i = 56 × 10 9 → i = 6575 ZiG Zi P P Pi = − i Gp → Z Zi ZiG P = 6575 − (56 × 10 −9 )(50 × 10 9 = 3775 Z

PROBLEM 3.15 Calculate the daily gas production including the condensate and water gas equivalents for a reservoir with the following daily production:

45

Separator gas production = 6 MM SCF Condensate production = 100 STB Stock tank gas production = 21 M SCF Fresh water production = 10 bbl Initial reservoir pressure = 6000 psia Current reservoir pressure = 2000 psia Reservoir temperature = 225ºF Water vapor content of 6000 psia and 225°F = 0.86 bbl/MM SCF Condensate gravity = 50°API

ANSWER:

G EO = 133000

γο M WO

= 133000

Scf 0.78 = 751.7 138 STB

141.5 5954 γο = = 0.78 , Mwo = = 138 ο 131.5 + API API − 8.811

GEO = 75174Scf / STB GEO × NP = 751.7(100) = 75174Scf GEW = 7390(6 × 0.86) = 38132Scf

Gp = Sep Gas Produced = S.T Gas Produced + G Eo + GEW = 21000 + 6 × 10 6 + 75174 + 38132 = = 6.134 × 10 6 Scf Gp = 6.134 × 10 6 Scf

46

CHAPTER 4

GAS-CONDENSATE RESERVOIRS

PROBLEM 4.1 A gas-condensate reservoir initially contains 1300M SCF of residue (dry or sales gas) per acre-foot and 115 STB of condensate. Gas recovery is calculated to be 85% and condensate recovery 58% by depletion performance. Calculate the value of the initial gas and condensate reserves per acre-foot if the condensate sells for $20.00/bbl, and the gas sells for $1.90 per 1000 std cu ft.

ANSWER:

gas = 1300Mscf × 0.85(

20 cent ) = 22100 cent 1Mscf

1$ 22100 cent( ) = 221$ 100 cent 2.5$ liquid = 115bb1× 0.85( ) = 166.75$ 1bb1

PROBLEM 4.2 A well produces 45.3 STB of condensate and 742 M SCF of sales gas daily. The condensate has a molecular weight of 121.2 and a gravity of 52.0°API at 60°F. (a) What is the gas-oil ratio on a dry gas basis? 47

(b) What is the liquid content expressed in barrels per million standard cubic feet on a dry gas basis? (c) What is the liquid content expressed in GPM on a dry gas basis? (d) Repeat parts (a), (b), and (c) expressing the figures on a wet, or gross, gas basis.

ANSWER: a)

GOR =

742Mscf 742 ×1000 scf = = 16380 45.3bb1 45.3 bb1

b)

45.3bb1 bb1 = 0.0611 742Mscf Mscf c)

45.3 × 42 = 2.57 gpm 742 d) i)

γo =

141.5 141.5 = = 0.7711 API + 131.5 52 + 131.5

GE ° =

133000 133000 × 0.7711 ×n = × 45.3 = 38.33Mscf Mw° 121.2

G = 38.33 + 742 = 780.33mscf GOR =

780.33Mscf scf = 17225 45.3bb1 bb1

ii)

bb1 45.3 = 0.0581 780.33 Mscf

48

iii)

45.3 × 42 bb1 = 0.0244 780.33 Mscf

PROBLEM 4.3 The initial daily production from a gas-condensate reservoir is 186 STB of condensate, 3750 M SCF of high-pressure gas, and 95 M SCF of stock tank gas. The tank oil has a gravity of 5l.2°API at 60°F. The specific gravity of the separator gas is 0.712 and of the stock tank gas, 1.30. The initial reservoir pressure is 3480 psia, and reservoir temperature is 220°F. Average hydrocarbon porosity is 17.2%. Assume standard conditions of 14.7 psia and 60°F. (a) What is the average gravity of the produced gases? (b) What is the initial gas-oil ratio? (c) Estimate the molecular weight of the condensate. (d) Calculate the specific gravity (air =1.00) of the total well production. (e) Calculate the gas deviation factor of the initial reservoir fluid (vapor) at initial reservoir pressure. (f) Calculate the initial moles in place per acre-foot. (g) Calculate the mole fraction that is gas in the initial reservoir fluid. (h) Calculate the initial (sales) gas and condensate in place per acre-foot.

ANSWER: a) g=

g sep Rs sep + g st Rs st Rs sep + Rs st

=

0.712 × 3750 + 1.3 × 95 = 0.727 3759 + 95

49

b)

GOR =

3750 + 95 scf × 1000 = 20672 186 bb1

c)

M wo =

5954 5954 = = 140.5 ApI − 8.811 51.2 − 8.811

d)

γ well =

R1 + 4602 + R3 133000 R1 + + R3 Mw°

γo =

141.5 = 0.7745 131.5 + API

R1 =

3750 ×10 3 scf = 20161.3 186 stb

R3 =

95 × 10 3 scf = 510.752 186 stb

Ÿ γ well =

20161.3 × 0.712 + 4602 × 0.7745 + 510.752 × 103 = 0.868 133000 × 0.7745 20161.3 + + 510.752 140.5

e) well Ÿ

Tpc = 420 ppc = 640

Ÿ= 0.865

f)

n=

pV 43560 × 3480 × 0.172 = = 4131 RT 0.865 ×10.73 × (580)

g)

ng = fg = ng + n o

3750000 95000 R1 R + 3 + 379.4 379.4 379.4 379.4 = = 0.99 R3 350 3750000 95000 350 × 0.7745 R1 + + + + 379.4 379.4 140.5 379.4 379.4 M wo 50

h) pV = nRT → G =

G=

379.4 pV 379.4 × p × 43560 × φ × (1 − sw ) = RT RT

379.4 × 3480 × 43560 × 0.172 × (1 − 0) Mscf = 1567.066 0.865(10.732)(68) 1ac − ft

G × f g = (1567.066)(0.964) = 1510.65

Mscf 1511 × 103 ,N = = 72 STB ac − ft R1 + R2

PROBLEM 4.4 (a) Calculate the .gas deviation factor for the gas-condensate fluid the composition of which is given in Table 4.1 at 5820 psia and 265°F. Use the critical values of C8 for the C7+ fraction. (b) If half the butanes and all the pentanes and heavier gases are recovered as liquids, calculate the gas-oil ratio of the initial production. Compare with the measured gas-oil ratio.

ANSWER: a) 1.072 b) 15300

PROBLEM 4.5 Calculate the composition of the reservoir retrograde liquid at 2500 psia for the data of Tables 4.4 and 4.5 and Ex. 4.3. Assume the molecular weight of the heptanes-plus fraction to be the same as for the initial reservoir fluid.

51

ANSWER:

Δnl = 0.25(0.028) + (0.50)(0.019) + (0.75)(0.016) + 0.034 = 0.0625 nc4 =

0.25(0.028) = 0.112 0.0625

nc5 =

0.50(0.019) = 0.152 0.0625

nc6 =

0.75(0.016) = 0.192 0.0625

+

nc7 =

1(0.034) = 0.544 0.0625

PROBLEM 4.6 Estimate the gas and condensate recovery for the reservoir of Ex. 4.3 under partial water drive if reservoir pressure stabilizes at 2500 psia. Assume a residual hydrocarbon saturation of 20% and F = 52.5%.

ANSWER:

p = 2500 psi,Vi = 7623Gp =

379.4 pV  1000RT

from table 4 .4 Ÿ 0 .066 × 7623 = 503 .118

V at 2500psi= 7623− 503.118 = 7119.882 and at 2500psi = 0.794 T = 655 ° R R = 10 .73

Ÿ Gp =

379.4(2500)(7119.882) = 1210.18Mscf / ft 3 1000(0.794)(10.73)(655)

volume under partial water drive at 2500 psi = 283.7 ft 3 / ac − ft

52

1210.18 − 283.7 = 0.765 1210.18

f = 0.525 → (0.765)(0.525) = 0.402 (1210.18)(0.402) = 486.38Mscf / ac. ft 486.38 ×

225.1 1000 Mscf bb1 = 456 → 456 × = 31.02 240.1 14700 ac. ft ac. ft

from table4.5 → depletion to2500psi(condensate) = 15.3bb1/ac.ft depletion to 2500psi (Gas) = 225.1 Mscf / ac. ft

condensate recovery = 15.3 + 31.02 = 46.32

bb1 ac. ft

Gas recovery= 225.1 + 456 = 681.1Mscf / ac. ft

PROBLEM 4.7 Calculate the recovery factor by cycling in a condensate reservoir if the displacement efficiency is 85%, the sweep efficiency is 65%, and the permeability stratification factor is 60%.

ANSWER:

Recovery Factor = 0.85(0.65)(0.60) = 0.332

PROBLEM 4.8 The following data are taken from a study on a recombined sample of separator gas and separator condensate in a PVT cell with an initial hydrocarbon volume of 3958.14 cu cm. The wet gas GPM and the residue gas-oil ratios were calculated using equilibrium ratios for production through a separator operating at 300 psia and 70°F. The initial

53

reservoir pressure was 4000 psia, which was also close to the dew-point pressure, and reservoir temperature was 186°F. (a) On the basis of an initial reservoir content of 1.00 MM SCF of wet gas, calculate the wet gas, residue gas, and condensate recovery by pressure depletion for each pressure interval. (b) Calculate the dry gas and condensate initially in place in 1.00 MM SCF of wet gas. (c) Calculate the cumulative recovery and the percentage of recovery of wet gas, residue gas, and condensate by depletion performance at each pressure. (d) Place the recoveries at an abandonment pressure of 605 psia on an acre-foot basis for a porosity of 10% and a connate water of 20.

54

Composition in Mole Percentages Pressure, psia

4000

3500

2900

2100

1300

605

CO2

0.18

0.18

0.18

0.18

0.19

0.21

N2

0.13

0.13

0.14

0.15

0.15

0.14

C1

67.62 63.10 65.21

69.79

70.77

66.59

C2

14.10 14.27 14.10

14.12

14.63

16.06

C3

8.37

8.25

8.10

7.57

7.73

9.11

i-C4

0.98

0.91

0.95

0.81

0.79

1.01

n-C4

3.45

3.40

3.16

2.71

2.59

3.31

i-C5

0.91

0.86

0.84

0.67

0.55

0.68

n-C5

1.52

1.40

1.39

0.97

0.81

1.02

C6

1.79

1.60

1.52

1.03

0.73

0.80

C7+

6.85

5.90

4.41

2.00

1.06

1.07

Mol. Wt C7+

143

138

128

116

111

110

0.762

0.819

0.902

224.0 474.0

1303

2600

5198

Wet gas GPM (calculated)

5.254 4.578 3.347

1.553

0.835

0.895

Residue gas-oil ratio

7,127 8,283 11,621 26,051

Gas deviation factor for wet 0.867 0.799 0.748 gas at 186Ԭ Wet gas production, cu cm at 0 cell P and T

Retrograde liquid, percentage 0

3.32

of cell volume

ANSWER: a)

3500− 4000psia 53.71 Mscf,48.46 Mscf,5.85 bb1 55

19.36

23.91

49,312 45,875 22.46

18.07

b) 891.6 Mscf 125.1 bb1

c)

767.5 Mscf, 739.2 Mscf,29.7bb1 and 76.74%,82.90%,23.74% d) 675 Mscf 650 Mscf 26.14 bbl

PROBLEM 4.9 If the retrograde liquid for the reservoir of Prob. 4.8 becomes mobile at 15% retrograde liquid saturation, what effect will this have on the condensate recovery?

ANSWER: It will increase the recovery factor.

PROBLEM 4.10 If the initial pressure of the reservoir of Prob. 4.8 had been 5713 psia with the dew point at 4000 psia, calculate the additional recovery of wet gas, residue gas, and condensate per acre-foot. The gas deviation factor at 5713 psia is 1.107, and the GPM and GOR between 5713 and 4000 psia are the same as at 4000 psia.

56

ANSWER: 675 Mscf 650 Mscf 26.13 bbl

PROBLEM 4.11 Calculate the value of the products by each mechanism in Table 4.9 assuming (a) $17.50 per STB for condensate and $1.50 per M SCF for gas; (b) $20.00 per STB and $1.90 per M SCF; and (c) $20.00 per STB and $2.50 per M SCF.

ANSWER: a)

Depletion = (20)(71.6) + (1.90)(1200) = 3712$ / ac. ft water drive = (20)(81.8)+ (1.90)(823) = 3199.7$/ac.ft partial water drive = (20)(59.6)+ (1.90)(1010) = 3111$/ac.ft b)

depletion = (25)(71.6) + (3)(1200) = 5390$ / ac. ft

waterdrive= (25)(81.8)+ (3)(823)= 45144/ac.ft partial water drive = (25)(59.6)+ (3)(1010)= 4520$/ac.ft

PROBLEM 4.12 In a PVT study of a gas-condensate fluid, 17.5 cu cm of wet gas (vapor), measured at cell pressure of 2500 psia and temperature of 195Ԭ, was displaced into an evacuated low-pressure receiver of 5000 cu cm volume that was maintained at 250Ԭto ensure that no liquid phase developed in the expansion. If the pressure of the receiver rises to 620 57

mm Hg, what will be the deviation factor of the gas in the cell at 2500 psia and 195Ԭ, assuming the gas in the receiver behave ideally? ANSWER: pV = nRT → n = n1 = n2 →

pV RT

p1V1 p2V2 = R1T1 R2T2

V1 = 17.5cm 3 p1 = 2500 psia T1 = 655° R =?

V2 = 5000cm 3 p2 = 620mmHg = 12 psia T2 = 250° F = 710° R 2 = 1(ideal behavior ) Ÿ=

p1V1T2 2500 (17.5)(710) = = 0.79 p2V2T1 12(5000 )(655)

PROBLEM 4.13 Using the assumptions of Ex. 4.3 and the data of Table 4.4, show that the condensate recovery between 2000 and 1500 psia is 13.3 STB/ac-ft and the residue gas-oil ratio is 19,010 SCF/bbl.

ANSWER: The solution of this problem is left to readers.

58

PROBLEM 4.14 A stock tank barrel of condensate has a gravity of 55°API. Estimate the volume in ft3 occupied by this condensate as a single- phase gas in a reservoir at 2740 psia and 215°F. The reservoir wet gas has a gravity of 0.76.

ANSWER:

Mw =

5954 5954 = = 128.9 AIP − 8.811 55 − 8.8.11

mo = 350 yc = 350(0.76) = 266 no =

mo 266 = = 2.06 lb − mol Mwo 128.9

g = 0.79 → V=

Tpc = 392.1°R,

Tpr = 1.72

Ppc = 655.2psia,

PPr = 4.18 →= 0.874

nRT = 4.77 ft 3 P

PROBLEM 4.15 A gas-condensate reservoir has an areal extent of 200 acres, an average thickness of 15 ft, an average porosity of 0.18, and an initial water saturation of 0.23. A PVT cell is used to simulate the production from the reservoir, and the following data are collected:

59

Pressure

Wet Gas

z

Condensate

(psia)

Produced

Wet Gas

Produced from

(cc)

Separator (moles)

4000(dew point)

0

0.75

0

3700

400

0.77

0.0003

3300

450

0.81

0.0002

The initial cell volume was 1850 cc, and the initial gas contained 0.002 moles of condensate. The initial pressure is 4000 psia, and the reservoir temperature is 200°F. Calculate the amount of dry gas (SCF) and condensate (STB) recovered at 3300 psia from the reservoir. The molecular weight and specific gravity of the condensate are 145 and 0.8, respectively.

ANSWER: The solution of this problem is left to readers.

PROBLEM 4.16 Production from a gas-condensate reservoir is listed below. The molecular weight and the specific gravity of the condensate are 150 and 0.8, respectively. The initial wet gas in place was 35 MMM SCP, and the initial condensate was 2 MM STB. Assume a volumetric reservoir and that the recoveries of condensate and water are identical, and determine the following: (a) What is the percentage of recovery of residue gas at 3300 psia? (b) Can a PYT cell experiment be used to simulate the production from this reservoir? Why or why not?

60

Pressure, psia

4000

3500

3300

Compressibility of wet gas,z

0.85

0.80

0.83

0

2.4 MMM

2.2 MMM

0

80,000

70,000

0

5000

4375

Wet gas produced during pressure increment. SCF Liquid condensate produced during pressure increment, STB Water produced during pressure increment, STB ANSWER:

liquid recovery =

80000 70000 + = 0.075 20000000 2000000

Residue Gas = (1 − liquid recovery) × GIIP = (1 − 0.075) × 35 × 10 9 = 32.375 × 10 9 scf

PROBLEM 4.17 A PVT cell is used to simulate a gas-condensate reservoir. The initial cell volume is 1500 cc, and the initial reservoir temperature is 175°F. Show by calculations that the PVT cell will or will not adequately simulate the reservoir behaviour. The data generated by the PVT experiments as well as the actual production history are as follows: Pressure, psia

4000

3600

3000

Wet gas produced in pressure increment,

0

300

700

Compressibility of produced gas

0.70

0.73

0.77

Actual Production history, M SCF

0

1000

2300

cc

ANSWER: The solution of this problem is left to readers. 61

CHAPTER 5

UNDERSATURATED OIL RESERVOIRS

PROBLEM 5.1 Using the letter symbols for reservoir engineering, write expressions for the following terms for a volumetric, undersaturated reservoir: a) The initial reservoir oil in place in stock tank barrels. b) The fractional recovery after producing Np STB. c) The volume occupied by the remaining oil (liquid) after producing Np STB. d) The SCF of gas produced. e) The SCF of initial gas. f) The SCF of gas in solution in the remaining oil. g) By difference, the SCF of escaped or free gas in the reservoir after producing Np STB. h) The volume occupied by the escaped, or free, gas.

ANSWER: a) NBoi b) RF=Np/N c) (N-Np)Bo d) GBg e) NRsio f) NRsio-NpRp 62

g) NRsio-(N-Np) h) GfBg

PROBLEM 5.2 The physical characteristics of the 3-A-2 reservoir are given in Fig. 5.2: (a) Calculate the percentage of recovery, assuming this reservoir could be produced at a constant cumulative produced gas-oil ratio of 1100 SCF/STB, when the pressure falls to 3550, 2800, 2000, 1200, and 800 psia. Plot the percentage of recovery versus pressure. (b) To demonstrate the effect of increased GOR on recovery, recalculate the recoveries, assuming that the cumulative produced GOR is 3300 SCF/STB. Plot the percentage of recovery versus pressure on the same graph used for the previous problem. (c) To a first approximation, what does tripling the produced GOR do to the percentage of recovery? (d) Does this make it appear reasonable that to improve recovery high-ratio (GOR) wells should be worked over or shut in when feasible?

ANSWER: The solution of this problem is left to readers.

PROBLEM 5.3 If 1 million STB of oil have been produced from the 3-A-2 reservoir at a cumulative produced GOR of 2700 SCF/STB, causing the reservoir pressure to drop from the initial reservoir pressure of 4400 psia to 2800 psia, what is the initial stock tank oil in place?

63

ANSWER: Pi=4400 psia (Bti=1.57 bbl/STB and Rsoi=1100 scf/STB) P=2800 psia (Bt=1.72 bbl/STB , Bg=1.017×10-3 /5.615=1.811×10-4 bbl/scf) NP[ Bt + ( Rp − Rsi) Bg ] Bt − Bti 1× 10 6 [1.72 + (2700 − 1100)(1.811× 10 − 4 ) N= 1.72 − 1.57 N = 22MMSTB N=

PROBLEM 5.4 The following data are taken from an oil field that had no original gas cap and no water drive Oil pore volume of reservoir = 75 MM cu ft Solubility of gas in crude = 0.42 SCF/STB/psi Initial bottom-hole pressure = 3500 psia Bottom-hole temperature = 140°F Bubble-point pressure of the reservoir = 2400 psia Formation volume factor at 3500 psia = 1.333 bbl/STB Compressibility factor of the gas at 1500 psia and 140°F= 0.95 Oil produced when pressure is 1500 psia = 1.0 MM STB Net cumulative produced GOR = 2800 SCF/STB

(a) Calculate the initial STB of oil in the reservoir. (b) Calculate the initial SCF of gas in the reservoir. (c) Calculate the initial dissolved GOR of the reservoir. (d) Calculate the SCF of gas remaining in the reservoir at 1500 psia. (e) Calculate the SCF of free gas in the reservoir at 1500 psia.

64

(f) Calculate the gas volume factor of the escaped gas at 1500 psia at standard conditions of 14.7 psia and 60ºF. (g) Calculate the reservoir volume of the free gas at 1500 psia. (h) Calculate the total reservoir GOR at 1500 psia. (i) Calculate the dissolved GOR at 1500 psia. (j) Calculate the liquid volume factor of the oil at 1500 psia. (k) Calculate the total, or two-phase, oil volume factor of the oil and its initial complement of dissolved gas at 1500 psia.

ANSWER: No original gas cap and no water drive

pb = 2400 psia Z = 0.95( p = 1500 psia, T = 140$ F ) N P = 1MMSTB @ p = 1500 Rp = 2800 scf

STB

@ p = 1500

Voi = 75 MMft 3 dRs scf / sTB = 0.42 dP psi Pi = 3500 Psia T = 140 $ F Boi = 1.333 bbl

STB

(a) Voi = Nboi Ÿ N =

voi 75 × 10 6 ft 3 5.615bbl = = 10.02 × 10 6 STB Boi 1.333 bbl STB

(b) G = Initial Gas Reservoir( scf) = ? G = NRsοi = (10.02)(0.42 × 2400 ) = 10.1MMMSCF

65

(c)

GORi =

G 10.1× 10 9 Scf = = 1008 scf STB N 10.02 × 10 6 STB

OR

GORi = 0.42 × 2400 = 1008

Scf STB

(d) G 1500 = G − Gp = 10.1 × 10 9 − ( NpRp ) = 10.1 × 10 9 − (1 × 10 6 × 2800) = 7.3 × 10 9 scf

(e)

Gf = NRsi − NpRp − ( N − Np) Rs =

= (10.1×109 ) − (1×106 × 2800) − [(10.2 ×106 − 1×106 )0.42 ×1500] = 1.62 ×109 Scf

(f)

Bg = 0.0282

ZT 0.95(600) = 0.0282 = 0.0107 ft 3 scf P 1500

(g)

VFreegas = Gf .Bg = 1.6 ×109 × (0.0107) = 17.33 ft 3 (h) GOR 1500 =

7.3 × 10 9 Scf 7.3 × 10 9 scf = = 809 N − NpSTB (10.02 − 1) × 10 6 STB

(i) Dissolved GOR at Psia=0.42×1500=630

scf STB

(j) Voi=Vo+Vg

NBoi = ( N − Np ) Bo + G f Bg  Bo =

NBoi − GfBg N − Np

(10.02 × 10 6 × 1.333) − (1.62 × 109 × 0.0107 / 5.615) 10.0 × 10 6 − 1 × 10 6 Bo = 1.14 bbl STB Bo =

66

(k)

Bt = Bo + ( Rsi − Rs) Bg = 1.14 + (0.42 × 2400 − 0.42 × 1500) × Bt = 1.86 bbl

0.0107 5.615

STB

PROBLEM 5.5 (a) Continuing the calculations of the Kelly-Snyder Field, calculate the fractional recovery and the gas saturation at 1400 psig. (b) What is the deviation factor for the gas at 1600 psig and BHT 125°F?

ANSWER: The solution of this problem is left to readers.

PROBLEM 5.6 The R Sand is a volumetric oil reservoir whose PVT properties are shown in Fig. 5.9. When the reservoir pressure dropped from an initial pressure of 2500 psia to an average pressure of 1600 psia, a total of 26.0 MM STB of oil had been produced. The cumulative GOR at 1600 psia was 954 SCF/STB, and the current GOR was 2250 SCF/STB. The average porosity for the field is 18% and average connate water 18%. No appreciable amount of water was produced, and standard conditions were 14.7 psia and 60°F.

67

a) Calculate the initial oil in place. b) Calculate the SCF of evolved gas remaining in the reservoir at 1600 psia. c) Calculate the average gas saturation in the reservoir at 1600 psia. d) Calculate the barrels of oil that would have been recovered at 1600 psia if all the produced gas had been returned to the reservoir. e) Calculate the two-phase volume factor at 1600 psia. f) Assuming no free gas flow, calculate the recovery expected by depletion drive performance down to 2000 psia. g) Calculate the initial SCF of free gas in the reservoir at 2500 psia.

ANSWER:

φ = 18% Swi = 18% Wp = 0 Psc = 14.7 psia Tsc = 60 $ F pb = 2200 psi Tres = 150 $ F = 610 $ R 





Pi = 2500 psia p = 1600 psia Np = 26 × 10 6 Rp 1600 = 954 scf

STB

CurrentGoR = 2250 scf STB =

qg q0

(a)

RF =

Np Bt − Bti = N Bt + ( Rp − Rsoi) Bg

N=? Boi @ 2500 = 1.29, Rsoi = 575 scf STB Bo @1600 = 1.215, Rso = 383 scf STB , Z = 0.82

ZT (0.82)(610) = 0.00502 = 1.57 × 10 −3 bbl scf 1600 P −3 Bt = Bo + Bg ( Rsoi − Rs ) → Bt = 1.215 + 1.57 × 10 (575 − 383)

Bg = 0.00502

Bt = 1.516 bbl

STB 1.516 − 1.29 RF = = 0.106 1.516 + (954 − 575)1.57 × 10 −3 Np 26 × 10 6 = = 246 MMSTB 0.106 RF N = 246 MMSTB

N=

(b) Gf = NRsi – NpRp - (N-Np) Rs =246 (575) - (26×954) - (246-26) 383 = 31.9×109 SCF (c)

1.215 So = (1 − RF )(1 − Sw)(Bo / Boi) = (1 − 0.106)(1 − 0.18)( ) = 0.69 1.29

Sg = 1− So − Swc = 1 − 0.69 − 0.18 Ÿ Sg = 13%

69

(d) RF =

NP Bt − Bti = N Bt + ( Rp ↓ 0 − Rsi ) Bg if Rp=0 ŸNp=?

1.516 − 1.29 = 0.365 1.516 + (0.575)1.57 × 10 −3 NP = N .RF = 0.365(246) = 89.89MMSTB

RF =

(e)

Bt = Bo + ( Rsoi − Rso) Bg = 1.215 + 1.57 ×10 −3 (575 − 383)1.57 ×10 −3 = 1.516 bbl

STB

(f) Assume: No free gas flows

Bg = 0.00502

0.82(610) = 1.25 ×10−3 bbl scf 2000

Rso = 508.3scf / stb ……

Bo@2000= 1.272bbl / stb

Bt = Bo + ( Rsi − Rs ) Bg = 1.272 + 1 .25 × 10 −3 (575 − 508 .3) = 1.355 bbl / stb Bt − Bti 1 .355 − 1 .29 = = 0 .0479 1 .355 Bt Np = N .RF = ( 246 )( 0.0479 ) = 11 .8 MMSTB

RF =

(g) 2500psia=O@ free gas

PROBLEM 5.7 If the reservoir of Prob. 5.6 had been a water-drive reservoir, in which 25×106 bbl of water had encroached into the reservoir when the pressure had fallen to 1600 psia, calculate the initial oil in place. Use the same current and cumulative GORs, the same PVT data, and assume no water production.

70

ANSWER: Np[Bt + ( Rp − Rsi ) Bg ] − We Bt − Bti 26[1.516 + (954 − 575)1.57 × 10 −3 ] − 25 N= 1.516 − 1.29 6 N = 134 × 10 STB N=

PROBLEM 5.8 The following production and gas injection data pertain to a reservoir. (a) Calculate the average producing GOR during the production interval from 6 MM to 8 MM STB. (b) What is the cumulative produced GOR when 8 MM STB has been produced? (c) Calculate the net average producing GOR during the production interval from 6 MM to 8 MM STB. (d) Calculate the net cumulative produced GOR when 8 MM STB has been produced? (e) Plot on the same graph the average daily GOR, the cumulative produced gas, the net cumulative produced gas, and the cumulative injected gas versus cumulative oil production.

71

Cumulative Oil

Average Daily Gas-

Cumulative Gas

Production, Np

Oil Ration, R

Injected, GI

MM STB

SCF/STB

MM SCF

0

300

0

1

280

0

2

280

0

3

340

0

4

560

0

5

850

0

6

1120

520

7

1420

930

8

1640

1440

9

1700

2104

10

1640

2743

ANSWER: (a)

Rav = =

( NpRp) 6 + ( NpRp) 7 + ( NpRp)8 Np6 + Np7 + Np8

scf (6 × 106 × 1120) + (7 × 106 × 1420) + (8 × 106 × 1640) = 1420 STB 6 × 106 + 7 × 106 + 8 × 106

(b)

Rp =

³

Np

RdNp

= NP R8−7 ( Np8 − Np7 + R7 −6 (1) + R6−5 (1) R5− 4 (1) + R4−3 (1) + R3−2 (1) + R2−1 + R1−0 8 0

1530(1) + 1270(1) + 958(1) + 705(1) + 450(1) + 310(1) + 280(1) + 290(1) 8 RP = 728 scf STB 72

=

(c) Ravg.net = =

³

Np 2

Np1

RdNp

Np 2 − Np1

=

R8−7(Np8 − Np7 ) + R7−8(Np 7 − Np 6 ) − ( 1440 − 520 ) Np8 − Np 6

1530(1) + 1270(1) − 1440 + 520 = 940 Scf STB 8−6

(d)

( ³0 RdNp) − Ginj Np

Rp =

Np R (1) + R7−6 (1) + R6−5 (1) + R5−4 (1) + R4−3 (1) + R3− 2 (1) + R2−1 (1) + R1−0 (1) − 1440 = 8 −7 8 1530 + 1270 + 985 + 705 + 450 + 310 + 280 + 290 − 1440 = 8 Rp = 548 Sf STB

PROBLEM 5.9 An undersaturated reservoir producing above the bubble point had an initial pressure of 5000 psia, at which pressure the oil volume factor was 1.510 bbl/STB. When the pressure dropped to 4600 psia, owing to the production of 100,000 STB of oil, the oil volume factor was 1.520 bbl/STB. The connate water saturation was 25%, water compressibility 3.2 x 10-6 psi-1, and, based on an average porosity of 16%, the rock compressibility was 4.0 x 10-6 psi-1. The average compressibility of the oil between 5000 and 4600 psia relative to the volume at 5000 psia was 17.00 x 10-6 psi-1. a) Geologic evidence and the absence of water production indicated a volumetric reservoir. Assuming this was so, what was the calculated initial oil in place? b) It was desired to inventory the initial stock tank barrels in place at a second production interval. When the pressure had dropped to 4200 psia, formation volume factor 1.531 bbl/STB, 205 M STB had been produced. If the average oil compressibility was 17.65 x 10-6 psi-1, what was the initial oil in place?

73

c) When all cores and logs had been analyzed, the volumetric estimate of the initial oil in place was 7.5 MM STB. If this figure is correct, how much water entered the reservoir when the pressure declined to 4600 psia?

ANSWER: Pi = 5000 psia

Swc = 25% −6

Cw = 3.2 × 10 psi φ = 16%

−1

Bo i = 1 . 510 b Np = 100 / 00 STB

Cf = 4 ×10 −6 psi −1

Bo = 1 . 520 @ p = 4800

Co = 17 × 10 −6 psi −1

p = 4600 psia

(a) a) N=? We = 0,Wp = 0 So=1-0.25 =0.75 Ce =

C 0 S 0 + C w S w + cf 1 − S wi

NBtiCeΔp = NpBt → N =

Np Bt . CeΔp Bti

17 × 10 −6 (0.75) + 3.2 × 10 −6 (0.25) + 4 × 10 −6 = 2.34 × 10 −5 psi −1 1 − 0.25 100000 1.520 N= × = 10.75MMSTB −5 2.34 × 10 (5000 − 4600) 1.51

Ce =

(b) p = 4200

Np=205MSTB

N=?

B$ = 1.531bbl

C$ = 17.65 × 10−6 psi−1

STB

17 ×10 (0.75) + 3.2 × 10 (0.25) + 4 ×10 = 2.405 ×10 −5 psi −1 1 − 0.25 205 1.531 N= × = 10.8MMSTB 2.405 ×10 −5 (5000 − 4200) 1.510

Ce =

−6

−6

−6

74

(c) N=7.5 MMSTB

P=4600 psia

we=?

NBtiC e Δp = N p Bt − we → we = N p Bt − NBtiC e Δp

we = 100000(1.320) − (7.5 × 106 )(1.510)(2.34 × 10−5 )(400) we = 46000 bbl

PROBLEM 5.10 Estimate the fraction recovery from a sandstone reservoir by water drive if the permeability is 1500 md, the connate water is 20%, the reservoir oil viscosity is 1.5 cp, the porosity is 25%, and the average formation thickness is 50 ft.

ANSWER: Fraction Recovery (RF)=? k = 1500 md Swc = 20%

μ = 1.5cp φ = 25% h = 50 ft

RF = 0.114 + 0.272 "gk + 0.256 Sw − 0.136 "gμo − 1.538φ − 0.00035 h = 0.114 + 0.272 "g1500 + 0.256 ( 20 ) − 0.136 ("g1.5) − 1.538 (0.25 ) − 0.00035 (50 ) = 0.603 × 100 = 60 .3%

PROBLEM 5.11 The following PVT data are available for a reservoir, which from volumetric reserve estimation is considered to have 275 MM STB of oil initially in place. The original pressure was 3600 psia. The current pressure is 3400 psia, and 732,800 STB have been

75

produced. How much oil will have been produced by the time the reservoir pressure is 2700 psia? Pressure

Solution

Formation

(psia)

Gas Oil Ratio

Volume Factor

(SCF/STB)

(bbl/STB)

3600

567

1.310

3200

567

1.317

2800

567

1.325

2500

567

1.333

2400

554

1.310

1800

434

1.263

1200

337

1.210

600

223

1.140

200

143

1.070

ANSWER: N=

Np Bt ce Δp Bt i

3600 1.31 3400 Bo 3200 1.310 275 × 10 6 =

3600 − 3400 3600 − 3200 = → Bo = 1.3135 1.31 − Bo 1.31 − 1.317 732800 1.3135 × Ÿ C e = 1.336 × 10 −5 ce × 200 1.310

2800 1.325 2700 Bo 2500 1.333

2800 − 2700 2800 − 2500 = → Bo = 1.327 1.325 − Bo 1.325 − 1.333

NP = 275×106 (1.336 ×10−6 )(3600 − 2700)

1.310 = 3.27MMSTB 1.327

76

PROBLEM 5.12 Production data, along with reservoir and fluid data, for an undersaturated reservoir follow. There was no measurable water produced, and it can be assumed that there was no free gas flow in the reservoir. Determine the following: a) Saturations of oil, gas, and water at a reservoir pressure of 2258. b) Has water encroachment occurred and, if so, what is the volume?

Gas gravity = 0.78 Reservoir temperature = 160°F Initial water saturation = 25% Original oil in place = 180 MM STB Bubble-point pressure = 2819 psia

The following expressions for Bo and Rso as functions of pressure were determined from laboratory data:

Bo= 1.00 + 0.00015p, bbl/STB Rso = 50 +0.42p, SCF/STB Pressure

Cumulative

Cumulative

Instantaneous

(psia)

Oil Produced

Gas Produced

GOR

(STB)

(SCF)

(SCF/STB)

2819

0

0

1000

2742

4.38 MM

4.38 MM

1280

2639

10.16 MM

10.36 MM

1480

2506

20.09 MM

21.295 MM

2000

2403

27.02 MM

30.26 MM

2500

2258

34.29 MM

41.15 MM

3300

77

ANSWER: a) ­° Ppc = 756 .8 − 131γ g − 3.6γ g 2 = 625 .5 Psia γ g == 0.78 ® 2 ° °¯Tpc = 169 .2 − 349 .5γ g − 74γ g = 311 .5 R

2258 P = = 3.46 Ppc 652.5 620 T = = 1.99 Tr = 311 .5 T  Z = .94 PC pr =

Bg = 0.00504

(0.94)(620) zT = 0.00504 = 1.3 ×10 −3 bbl scf 2258 p

­ Bo = 1.3387 P = 2258 ® ¯ Rs = 998 .36

So =

So =

Sg =

V $ ( N − Np ) Bo = NBoi Vp 1 − swi

­ Boi = 1.42285 pi = 2819 ® ¯ Rsi = 1234

180 − 34.29 195.0619 = = 0.571 → So = 0.571 180(1.42285) 341.484 1 − 0.25 Vg Vp

Gp

 º ª Vg = « NRsi − ( N − Np ) Rs − NpRp » Bg «¬ »¼

V g = [180(1234 ) − (180 − 34.29)998.36 − 41.15 × 10 3 ]× 1.3 × 10 −3 V g = 46.15 × 10 6 scf

Sg =

46.15 = 0.135 → Sg = 13.5% 341.484

Sw = 1 − So − Sg = 1 − 0.571− 0.135→ Sw = 0.294

78

b)

N=

Np[Bo + ( Rp − Rsi) Bg ] − We Bo − Boi + ( Rsi − Rs) Bg

RP =

G P 41.15 × 10 9 = 1200 SCF / STB = N P 34.29 × 10 6

Ÿ We = Np[Bo + ( Rp − Rsi) Bg ] − N [Bo − Boi + ( Rsi − Rs) Bg ] We = 34.29 × 10 6 [1.3387 + (1200 − 1234)1.3 × 10 −3 ]

− 180 × 10 6 [1.3387 − 1.42285 + (1234 − 998.36)1.3 × 10 −3 ]

We = 4390000bbl

PROBLEM 5.13 The following table provides fluid property data for an initially undersaturated lense type of oil reservoir. The initial connate water saturation was 25%. Initial reservoir temperature and pressure were 97°F and 2110 psia, respectively. The bubble-point pressure was 1700 psia. Average compressibility factors between the initial and bubblepoint pressures were 4.0 (10)-6 psia-1 and 3.1 (10)-6 psia-1 for the formation and water, respectively. The initial oil formation volume factor was 1.256 bbl/STB. The critical gas saturation is estimated to be 10%. Determine the recovery versus pressure curve for this reservoir.

79

Pressure

Oil Formation

Solution

Gas Formation

(psia)

Volume Factor

GOR

Volume Factor

(bbl/STB)

(SCF/STB)

(ft3/SCF)

1700

1.265

540

0.007412

1500

1.241

490

0.008423

1300

1.214

440

0.009826

1100

1.191

387

0.011792

900

1.161

334

0.014711

700

1.147

278

0.019316

500

1.117

220

0.027794

ANSWER:

swi = 0.25 Pi = 2110 psia Pb = 1700 Sgc = 10%

Boi = 1.256 bbl

STB

Cw = 3.1 × 10 −6 T = 97 $ F C f = 4 × 10 −6 ª cusi + cf º Bt − Bti + Bti « Δp NP 1 − Swi »¼ ¬ (1) P = Pb = rb = N Bt 1.255 − 1.256 + 1.256(6.36 ×10 −3 ) + 10 = = rb = 0.00970 1.265

(2) P < Pb

P = 1500 psia

Bt + B0 + ( Rsi − Rs) Bg = 1.241(540 − 490)

0.008423 = 1.316 5.615 80

540 + 490 = 515 2 ªs c + cf Bt − Bti + Bti « wi w ¬ 1 − swi r1 = Bt + ( RP − Rsi ) Bg Ravg1 =

Rp =

rb Rswi + (r1 − rb ) Ravg1 r1

º » ΔP 1.316 − 1.256 + 1.256(6.36 × 10−6 )610 ¼ = 0.008423 1.316 + ( Rp − 540) 5.615

=

0.0097(540) + (r1 − 0.0097)515 0.2425 + 515r1 = = 0.05042 r1 r1 P=1300 psia

Bt=1.389

Ravg2=465

1.389 − 1.256 + 1.256(6.36 × 10 −6 )810 0.009826 1.389 + ( Rp − 540) 5.615 0.0097(540) + (0.05042 − 0.0097)515 + ( r2 − 0.05042) 465 Rp = r2

r2 =

Rp =

2.76 + 405r2 r2

r2 = 0.1071

P=1100 psia Bt=1.512

, Ravg3=413.5

1.512 − 1.256 + 1.256(6.36 × 10 −6 )1010 0.011792 1.512 + ( Rp − 540) 5.615 0.0097(540) + (0.05042 − 0.0097)515 + Rp = r3

r3 =

(0.1071 − 0.05042)405 + (r

3

− 0.1071)413.5

r3 r3 = 0.2372 The value of r at P=500, 700, and 900 psia can be calculated using the same method and procedure as stated above.

81

PROBLEM 5.14 The Wildcat reservoir was discovered in 1970. The reservoir had an initial pressure of 3000 psia, and laboratory data indicated a bubble point pressure of 2500 psia. The connate water saturation was 22%. Calculate the fractional recovery, Np/N, from initial conditions down to a pressure of 2300 psia. State any assumptions which you make relative to the calculations.

Porosity = 0.165 Formation compressibility = 2.5 (10)-6 psia-1 Reservoir temperature =150ºF

z

Pressure

Bo

Rso

(psia)

(bbl/STB)

(SCF/STB)

Bg

Viscosity

(bbl/SCF)

Ratio µ0/µg

3000

1.315

650

0.745

0.000726

53.91

2500

1.325

650

0.680

0.000796

56.60

2300

1.311

618

0.663

0.000843

61.46

ANSWER:

pi = 3000 psia

φ = 0.165

Pb = 2500 psia

C f = 2.5 × 10 −6 psia −1

Swc = 22%

T = 150 F = 610 R

Np = ?@ 2300 psia N

82

ª Cf º Δp Bt − Bti + Bti « ¬1 − Swi »¼ (1) P > Pb Bt ª 2.5 × 10 −6 º 1.325 − 1.315 + 1.315« »500 Np ¬ 1 − 0.22 ¼ = = 0.00914 N 1.325 NP = N

(2) P < Pb

NP = N

ª Cf º Bt − Bti + Bti « Δp ¬1 − Swi »¼ Bt + ( Rp − Rsoi ) Bg

⊗

Bt = Bo + ( Rsoi − Rso) Bg Ÿ Bt = 1.311 + (650 − 618)0.000843 = 1.38 bbl Bg = 0 .00503

Rp =

STB

ZT = 0 .000843 bbl scf P

0.00914(650) + (r − 0.00914)634 r

NP =r N

ª 2.5 × 10 −6 º 1.338 − 1.315 + 1.315« 700 1 − 0.22 »¼ ¬ ⊗→r = ª 0.00914(650) + ( r − 0.00914)634 º 1.338 + « − 650» 0.000843 r ¬ ¼

r=

Np = 0.0195 N

83

CHAPTER 6

SATURATED OIL RESERVOIRS

PROBLEM 6.1 Calculate the values for the second and fourth periods through the fourteenth step of Table 6.1 for Conroe Field.

ANSWER: DDI =

N (Bt − Bti ) 17 × 12 = N P [Bt + (RP − Rsi )B g ] 14

ªB º NmBti « g − 1» 17 × 11 ¬ B gi ¼ = SDI = N P [Bt + (R P − Rsi )B g ] 14 WDI =

We − WP Bw 15 = N P [Bt + (RP − Rsi )B g ] 14

PROBLEM 6.2 Calculate the drive indexes at Conroe for the second and fourth periods.

ANSWER:

DDI =

N ( Bt − Bti)

Np[( Bt + ( Rp − Rsoi) Bg ]

=

534(0.14) = 0.2911 256.983 84

NmBti [Bg − Bgi] 534(0.1295) Bgi = = 0.2694 SDI = Np[( Bt + ( Rp − Rsoi) Bg ] 256.7983 WDI =

we − Bwwp 112.8 = = 0.4395 Np[( Bt + ( Rp − Rsoi) Bg ] 256.7983

PROBLEM 6.3 If the recovery by water drive at Conroe is 70%, by segregation drive 50%, and by depletion drive 25%, using the drive indexes for the fifth period calculate the ultimate oil recovery expected at Conroe.

ANSWER:

Recovery= 70(0.646) + 50(0.174) + 25(0.180) = 58.42%

PROBLEM 6.4 Explain why the first material balance calculation at Conroe gives a low value for the initial oil in place.

ANSWER: It is due to fact that increasing water influx into the reservoir decreases the initial oil in place.

85

PROBLEM 6.5 (a) Calculate the single-phase formation volume factor on a stock tank basis, From the PVT data giveninTables6.3 and 6.4, at a reservoir pressure of 1702 psig, for separator conditions of 100 psig and 76°F. (b) Calculate the solution GOR at 1702 psig on a stock tank basis for the same separator conditions. (c) Calculate the two-phase formation volume factor by flash separation at 1550 psig for separator conditions of 100 psig and 76°F.

ANSWER:

1720psia< pb a)

Bo = Bod (

Bofb 1.335 ) = 1.245( ) Ÿ B$ = 1.2428bbl STB Bodb 1.341

b) B $ fb · § Rso = Rsofb − ¨ [Rsodb − Rsod )]× ¸ $ db ¹ B © 1.335 º ª = 349.6 = (505 + 49) − «(638 − 425) 1.391 »¼ ¬

c)

Bt =

v Vb × = 1.2691×1.335 = 1.6942bbl STB )c Vb vr

PROBLEM 6.6 From the core data that follow, calculate the initial volume of oil and free gas in place by the volumetric method. Then, using the material balance equation, calculate the cubic

86

feet of water that have encroached into the reservoir at the end of the four periods for which production data are given.

Pressure

Bt

Bg

Np

RP

Wp

(psia)

(bbl/STB)

(ft3/SCF)

(STB)

(SCF/STB)

(STB)

3480

1.4765

0.0048844

0

0

0

3190

1.5092

0.0052380

11.17 MM

885

224.5 M

3139

1.5159

0.0053086

13.80 MM

884

534.2 M

3093

1.5223

0.0053747

16.41 MM

884

1005.0 M

3060

1.5270

0.0054237

18.59 MM

896

1554. 0 M

Average porosity = 16.8% Connate water saturation = 27% Productive oil zone volume = 346,000 ac-ft Productive gas zone volume> 73,700 ac-ft Bw = 1.025 bbl/STB Reservoir temperature =207°F Initial reservoir pressure =3480 psia

ANSWER:

φ = 16.8% Swi = 27% Bw = 1.025 bbl

Stb

T = 207 $ F Pi = 3480 psia

a) m=

Vg VO

=

73700 = 0.213 34600

87

N=

7758 × vb × φ × (1 − Swi ) 7758 × 346000 × 0.168 × (1 − 0.27) = = 39 × 10 6 STB Bti 1.4765 × 5.615

Vo = NBoi = 43560 × 346000 × 0.168 × (1 − 0.27) = 58.6 × 10 6 ft 3 m=

GBgi mNBoi 0.213(223)(1.4765 × 5.615) ŸG= = = 14.09 × 109 SCF 0.0048844 NBoi Bgi

Vg = GBgi = 43560 × 73700 × 0.168 × (1 − 0.27) = 68.38 × 10 6 ft 3 b) We=?

F = NEo +

NmBti Eg + We Bgi

­F = Np[Bt + ( Rp − Rsi) Bg ] + BwWp ° ®Eo = Bt − Bti °Eg = Bg − Bgi ¯ Rsoi =

G 80.6 × 10 9 Scf = = 365 N 223 × 10 6 STB

F = 11.17[1.5092 + (885 − 358)

0.005238 ] + (1.025)(0.2245) 5.615

= 22.5MMbbl

223 × 0.21× 1.4765 (0.005238 − 0.0048844) + We 0.0048844 We = 10.2MMbbbl For P = 3190 22.5 = 223(1.5092 − 1.4765) +

we = Np[Bt + ( Rp − Rsi ) Bg ] + BwWp − N ( Bt + − Bti ) −

for : P = 3139psi

88

NmBti ( Bg − B gi ) Bgi

0.0053086º ª we = 13.8«1.5159 + (884 − 358) + 1.025 × (0.5342) − 223(1.5159 − 1.4765) 5.615 »¼ ¬

−

223 × 0.213 × 1.4765 (0.0053086 − 0.0048844) 0.0048844

we = 13 .45 MMbbl

For P = 3693psi 0.0053747 º ª + 1.025 × (1.005) − 223(1.5223 − 1.4765) we = 16.41«1.5223 + (884 − 358) 5.615 »¼ ¬

−

223× 0.213× 1.4765 (0.0053086 − 0.0048844) 0.0048844

we = 17 Mbbl

For : P = 3660psi 0.0054237 º ª + 1.025 × (1.554) − 223(1.5270 − 1.4705) we = 18.59 «1.5270 + (896 − 358) 5.615 »¼ ¬

−

223 × 0.213 × 1.4765 (0.0053086 − 0.0048844) 0.0048844

we = 20 .6 MMbbl

89

PROBLEM 6.7 The following PVT data are for the Aneth Field in Utah: Pressure

Bo

Rso

Bg

(psia)

(bbl/STB)

(SCF/STB)

(bbl/SCF)

µ0/ µg

2200

1.383

727

-

-

1850

1.388

727

0.00130

35

1600

1.358

654

0.00150

39

1300

1.321

563

0.00182

47

1000

1.280

469

0.00250

56

700

1.241

374

0.00375

68

400

1.199

277

0.00691

85

100

1.139

143

0.02495

130

40

1.100

78

0.05430

420

The initial reservoir temperature was 133°F. The initial pressure was 2200 psia, and the bubble-point pressure was 1850 psia. There was no active water drive. From 1850 psia to 1300 psia a total of 720 MM STB of oil were produced and 590.6 MMM SCF of gas. a) How many reservoir barrels of oil were in place at 1850 psia? b) The average porosity was 10%, and connate water saturation was 28%. The field covered 50.000 acres. What is the average formation thickness in feet?

ANSWER:

T = 133$ F Pi = 2200 psia Pb = 1850 psia Np = 720MMSTB ½ ¾ p = 1850 → 1300 psia Gp = 590.6 MMMscf ¿

90

(a) N=?

at Pb=1850psia

Rp =

Gp = 820 scf STB Np

There is no gas cap in the reservoir, therefore: m=0 N=

N=

Np[Bt + ( Rp − Rsi ) Bg ] − ( we − wpBw) − WinjBaing SwiC w + Cf Bg )Δp Bt − Bti + mBti ( − 1) + (1 + m) Bti ( 1 − Swi Bgi Np[Bt + ( Rp − Rsoi ) Bg ] Bt − Bti

Bt = Bo + ( Rsoi − Rso) Bg Bt = 1.324 + (727 − 563)0.00182 = 1.6195 N=

720 × 10 6 [1.6195 + (820 − 727)0.00182] = 5446 × 10 6 STB 1.6195 − 1.383

(b)

φ = 10% swi = 0.28

N=

A = 5000 ac h=?

43560 × A × h × φ × (1 − swi ) Boi

5446 × 10 6 =

43560 × 50000 × h × 0.10 × (1 − 0.28) 5.615 ft 3 1.383 bbl × STB 1bbl

h = 270 ft

91

PROBLEM 6.8 You have been asked to review the performance of a combination solution gas, gas-cap drive reservoir. Well test and log information show that the reservoir initially had a gas cap half the size of the initial oil volume. Initial reservoir pressure and solution gas-oil ratio were 2500 psia and 721 SCF/STB, respectively. Using the volumetric approach, initial oil in place was found to be 56 MM STB. As you proceed with the analysis, you discover that your boss has not given you all the data you need to make the analysis. The missing information is that at some point in the life of the project a pressure maintenance program was initiated using gas injection. The time of the gas injection and the total amount of gas injected are not known. There was no active water drive or water production. PVT and production data are in the following table:

Pressure Bt

Bg

Np

RP

(psia)

(bbl/STB)

(bbl/SCF)

(STB)

(SCF/STB)

2500

0.001048

1.498

0

0

2300

0.001155

1.523

3.741 MM

716

2100

0.001280

1.562

6.849 MM

966

1900

0.001440

1.620

9.173 MM

1297

1700

0.001634

1.701

10.99 MM

1623

1500

0.001884

1.817

12.42 MM

1953

1300

0.002206

1.967

14.39 MM

2551

1100

0.002654

2.251

16.14 MM

3214

900

0.003300

2.597

17.38 MM

3765

700

0.004315

3.209

18.50 MM

4317

500

0.006163

4.361

19.59 MM

4839

92

a) At what point (i.e., pressure) did the pressure maintenance program begin? b) How much gas in SCF had been injected when the reservoir pressure is 500 psia? Assume that the reservoir gas and the injected gas have the same compressibility factor.

ANSWER: Initial Gas cap=50% OF initial oil volume Pi = 2500 psia Rsi = 721 Scf

STB N = 56 MMSTB

N=

Np[Bt + ( Rp − Rsi ) Bg ] − GinjBginj ª Bg º − 1» Bt − Bti + mBti « Bgi ¼ ¬

º ª Bg Ginj B g inj = Np [Bt + ( Rp − Rsi ) B g ] − N « Bt − Bti + mBti ( − 1) » Bgi ¼ ¬ P = 2300 Ÿ

Ginj Bg inj = 3.741[1.523 + (716 − 721)0.001155] 0.001155 º ª − 56 «(1.523 − 1.498) + 0.5 × 1.498( −1 = 0 0.001.48 »¼ ¬ P = 2100 Ÿ

Ginj Bg inj = 6.849[1.562 + (966 − 721)0.00128] 0.001440 º ª − 56 «(1.620 − 1.498) + 0.5 × 1.498( −1 = 0 0.001048 »¼ ¬ Ginj Bg inj = 9.173[1.620 + (1297 − 721)0.001440] 0.001440 º ª − 56 «(1.620 − 1.498) + 0.5 × 1.498( −1 = 0 0.001048 »¼ ¬

93

Ginj Bg inj = 12.42[1.817(1953 − 721)0.001884] 0.001884 º ª − 56 «(1.701 − 1.4981) + 0.5 × 1.498( −1 = 0 0.001048 »¼ ¬ Ginj Bg inj = 14.39[1.967 + ( 2551 − 721)0.002206] 0.002206 º ª − 56 «(967 − 1.498) + 0.5 × 1.498( −1 = 0 0.001048 »¼ ¬

Ginj Bg inj = 13.8 ×106 bbl Analyzing the calculated results show that we have gas injection at P=1300 psia.

b) Ginj Bg inj = 19.59[4.361 + (4839 − 721)0.0006163] 0.006163 ª º − 56«(4.361 − 1.498) + 0.5 × 1.498( − 1)» 0.0001048 ¬ ¼ 6 Ginj Bg inj = 217.5 × 10 bbl → Ginj =

217.5 × 10 6 217.5 × 10 6 = = 35.3 × 10 9 scf Bginj 0.006163

PROBLEM 6.9 An oil reservoir initially contains 4 MM STB of oil at its bubble-point pressure of 3150 psia with 600 SCF/STB of gas in solution. When the average reservoir pressure has dropped to 2900 psia, the gas in solution is 550 SCF/STB. Boi was 1.34 bbl/STB and Bo at a pressure of 2900 psia is 1.32 bbl/STB. Other data: Rp = 600 SCF/STB at 2900 psia Swi = 0.25 Bg = 0.0011 bbl/SCF at 2900 psia

94

Volumetric reservoir No original gas cap

a) How many STB of oil will be produced when the pressure has decreased to 2900 psia? b) Calculate the free gas saturation that exists at 2900 psia,

ANSWER: p = 2900 Psia Rp = 600ςp = 2900 scf swi = 0.25 Pb = 3150 psia Rs = 550 STB scf Rsi = 600 Bo = 1.32 bbl Bg = 0.0011bbl STB Scf STB bbl We = 0 m=0 Boi = 1.34 STB N = 4MMSTB

a)

Np @ 2900 psia = ? Bt = 1.32 + (600 − 550)0.0011 = 1.775 Np =

N ( Bt − Bti) 4(1.375 − 1.34) = Bt + ( Rp − Rsoi) Bg 1.375 + (600 − 600)0.0011

Np = 101.818 × 10 3

b) RF = 1 −

(1 − Sw − Sg ) Boi × (1 − Sw) Bo

Np 101.818 × 10 3 = = 0.0254 N 4 × 10 6 1 − 0.25 − Sg 1.34 0.0254 = 1 − × → Sg = 0.02996 1 − 0.25 1.32 Sg = 2.99% ≅ 3% RF =

95

G free = [NRsi − NpRp − ( N − Np ) Rs ]Bg

= [4 × 103 (600) − (101.818 × 600) − (4 × 103 − 101.818)550)]0.0011

G free = 194.9 × 10 6 scf × 0.00111 = 214400bbl

OR Np Bo )(1 − swi )( ) N Boi 101.818 × 10 3 1.32 = (1 − (1 − 0.25)( ) = 0.72 4 × 10 6 1.34 So = (1 −

Sg = 1 − So − Swi = 1 − 0.72 − 0.25 ≅ 0.03 → Sg = 3%

OR Sg =

Vg NBoi /(1 − S wi )

=

214400 = 0.03 Ÿ Sg = 3% ( 4 × 10 6 )1.34 / 0.75

PROBLEM 6.10 Given the following data from laboratory core tests, production data, and logging information: Well spacing = 320 ac Net pay thickness = 50 ft with the gas/oil contact 10 ft from the top Porosity: 0.17 Initial water saturation = 0.26 Initial gas saturation =0.15 Bubble-point pressure =3600 psia Initial reservoir pressure =3000 psia Reservoir temperature = 120°'F 96

Boi = 1.26 bbl/STB Bo = 1.37 bbllSTB at the bubble-point pressure Bo = 1.19 bbllSTB at 2000 psia Np = 2.0 MM STB at 2000 psia Gp = 2.4 MMM SCF at 2000 psia Gas compressibility factor, z = 1.0 – 0.001p Solution gas-oil ratio, Rso = O.2p

Calculate the amount of water that has influxed and the drive indexes at 2000 psia.

ANSWER:

A = 320ac ­hoil = 40 ft ® ¯hgas = 10 ft φ = 0.17 S wi = 0.26

Boi = 1.26 bbl

STB

Bo = 1.37 bbl

STB

Bo = 1.19 bbl

S gi = 0.15

@ Pb

STB @ p = 2000

Pb = 3600

N p = 2 MMSTB @ p = 2000

P = 3000 psi

G p = 2.4MMMScf @ p = 2000

T = 120 F pi = 3000

Z = 1 − 0.0001P

P = 3000 psia

& zi = 0.7 

Bg i = 0.0282

We = ? WDI = ? Rp =

Rso = 0.2 P

Gp scf = 1200 Np STB

3 0.7(120 + 460) z = 0.0282 = 3.816 × 10 −3 ft scf 3000 p

P = 2000 psia

, z = 0.8 

3 0.8(120 + 460) zT = 0.0282 = 6.54 × 10 −3 ft Bg = 0.0282 scf 2000 p

97

Voi = 43560 × Vb × φ × (1 − swi ) = 43560 × ( 40 × 320 ) × 0.17 × (1 − 0.26) = 70142054 ft 3

Vgi = 43560 × Vb × φ × ( sgi ) = 43560 × (10 × 320) × 0.17 × 0.15 = 3554496 ft 3

m=

Vg 3554496 = = 0.05 Vo 70142054

N=

Voi 70142054 = = 9.91×106 Boi 1.26 × 5.615

Bt = Bo + ( Rsi − Rs ) Bg

Bt = 1.19 + (0.2 × 3000 − 0.2 × 2000)

bbl 6.54 × 10 −3 = 1.423 STB 5.615

we = N p [Bt + ( Rp − Rsi ) Bg ] − N ( Bt − Bti ) −

NmBti ( Bg − B gi ) Bg i

6.54 × 10 6 º ª we = 2 × 10 6 «1.423 + (1200 − 600) 5.615 »¼ ¬ 9.91× 10 6 × 0.05 × 1.26 (6.54 × 10 −3 − 3.816 × 10 −3 ) − 9.91× 10 6 (1.423 − 1.26) − 3.816 × 10 −3 we = 2.18 × 10 6 bbl WDI =

2.18 × 10 6 we − BwWP = = 0.514 Np [Bt + ( Rp − Rsi ) B g ] 4.243 × 10 6

WDI = 51 .4 %

PROBLEM 6.11 From the following information determine: a) Cumulative water influx at pressures 3625, 3530, and 3200 psia. b) Water-drive index for the pressures in (a).

98

Pressure

Np

Gp

Wp

Bg

Rso

Bt

(psia)

(STB)

(SCF)

(STB)

(bbl/SCF)

3640

0

0

0

0.000892

888

1.464

3625

0.06

0.49 MM

0

0.000895

884

1.466

2.31 MM

0.001

0.000899

880

1.468

4.12 MM 0.08 MM

0.000905

874

1.469

5.68 MM 0.26 MM

0.000918

860

1.476

7.00 MM 0.41 MM

0.000936

846

1.482

8.41 MM 0.60 MM

0.000957

825

1.491

9.71 MM 0.92 MM

0.000982

804

1.501

0.001014

779

1.519

(SCF/STB) (bbl/STB)

MM 3610

0.36 MM

3585

0.79

MM

MM 3530

1.21 MM

3460

1.54 MM

3385

2.08 MM

3300

2.58 MM

3200

3.40

11.62

MM

MM

1.38 MM

ANSWER: The solution of this problem is left to readers.

PROBLEM 6.12 The cumulative oil production, NP, and cumulative gas oil ratio, RP , as functions of the average reservoir pressure over the first 10 years of production for a gas cap reservoir

99

follow. Use the Havlena-Odeh approach to solve for the initial oil and gas (both free and solution) in place.

Pressure

Np

RP

Bo

Rso

Bg

(psia)

(STB)

(SCF/STB)

(bbl/STB)

(SCF/STB)

(bbl/SCF)

3300

0

0

1.2511

510

0.00087

3150

3.295 MM

1050

1.2353

477

0.00092

3000

5.903 MM

1060

1.2222

450

0.00096

2850

8.852 MM

1160

1.2122

425

0.00101

2700

11.503

1235

1.2022

401

0.00107

1265

1.1922

375

0.00113

1300

1.1822

352

0.00120

MM 2550

14.513 MM

2400

17.730 MM

ANSWER: Calculated the value of F, Eo, and Eg and prepare the following table. For example F at P=3150 psia can be calculated as below:

F = NE $ + Nm

Bti Eg Bgi

p = 3150 → F = 3.295[1.256 + (1050− 510)]0.00092= 5.80

100

NO.

Bt

Eo=Bt-Bti

F (MM)

F/Eo

Eg=Bg-Bgi

Bti.Eg/Bgi.Eo

3300

1.2511=Bti

0

0

0

0

0

3150

1.2656

0.0145

5.80

400

5×10-5

4.95

3000

1.2798

0.0287

10.67

371.7

9×10-5

4.5

2850

1.2980

0.0469

17.30

368.8

1.4×10-4

4.29

2700

1.3188

0.0677

24.09

355.8

2×10-4

4.24

2550

1.3447

0.0936

31.89

340.7

2.6×10-4

3.99

2400

1.3718

0.1207

41.13

370.76

3.3×10-4

3.93

Plot F/Eo versus Bti.Eg/Bgi.Eo and draw the best straight line. The y intercept equals N and slope of figure is mN. 450 400 350 F/Eo

300 250 200 150 100 50 0 0

1

2

3

4

5

Bti*Eg/Bgi*Eo

N = 100 MMSTB

mN = 55.5 → m =

55.5 55.5 = = 0.55 100 N

Scf Initial gas(free gas + dissolved gas) → G =

G=

NmBoi + NRsoi Bgi

100(0.55)(1.2511) + (100 × 510) = 130MMMScf 0.00087

G = 130MMMScf 101

6

PROBLEM 6.13 Using the following data, determine the original oil in place by the Havlena Odeh method. Assume there is no water influx and no initial gas cap. The bubble-point pressure is 1800 psia. Pressure

Np

RP

Bt

Rso

Bg

(psia)

(STB)

(SCF/STB)

(bbl/STB)

(SCF/STB)

(bbl/SCF)

1800

0

0

1.268

577

0.00097

1482

2.233 MM

634

1.335

491

0.00119

1367

2.981 MM

707

1.372

460

0.00130

1053

5.787 MM

1034

1.540

375

0.00175

ANSWER: ª mBti º F = N « E $ + (1 + m ) BtiE fiw + E g » + we Bgi ¬ ¼

we = 0, m = 0 →

F = NE $

F = Np [Bt + ( Rp − Rsoi ) B g ] Eο = Bt − Bti

Water and formation compressibilities can be ignored since the pressure is below bubble point. Then, calculate the values of F and Eo and prepare the following table.

P

E$

F

1800

0

0

1482

0.067

3.2

1367

0.104

4.59

1053

0.272

13.520

Plot F versus Eo and draw the best straight line. The slope of figure is N.

102

16 14 12 10 F

8 6 4 2 0 -2 0

0.05

0.1

0.15 Eo

Slope =

12 − 8 = 50MMSTB 0.25 − 0.17

N=50 MMSTB

103

0.2

0.25

0.3

CHAPTER 7

SINGLE-PHASE FLUID FLOW IN RESERVOIRS

PROBLEM 7.1 Two wells are located 2500 ft apart. The static well pressure at the top of perforations (9332ft subsea) in well A is 4365 psia and at the top of perforations (9672 ft subsea) in well B is 4372 psia. The reservoir fluid gradient is 0.25 psi/ft, reservoir permeability is 245 md, and reservoir fluid viscosity is 0.63 cp. a) Correct the two static pressures to a datum level of 9100 ft subsea b) In what direction is the fluid flowing between the wells? c) What is the average effective pressure gradient between the wells? d) What is the fluid velocity? e) Is this the total velocity or only the component of the velocity in the direction between the two wells? f) Show that the same fluid velocity is obtained using Eq. (7.1). ANSWER: a)

φ a = p a − 0.25ha ha = 9332 − 9100 = 232 ft

φ a = 4365 − (0.25 × 232) = 4307 psia φb = pb − 0.25hb hb = 9672 − 9100 = 572 ft

φb = 4372 − (0.25 × 572) = 4229 psia 104

b)

φa ²φ Ÿ Therefore, fluid direction is from well A toward well B.b c)

L = (2500) + (340) 2 = 2523 ft 2

average pressure gradient =

Δφ 4307 − 4229 = = 0.0309 psi / ft Δl 2523

d)

vapparent = vactual =

q a

q aφ (1 − swi )

V = −0.001127 → 0.0135

k Δφ 245 78 bb1 )( ) = 0.0135 = −0.001127 ( 0.63 2523 day. ft 2 μ L

5.615 ft 3 bb1 × = 0.0758 ft 2 day day. ft bbl

e) This is the velocity in the direction between the two wells. f) V = −0.001127

tanθ =

k ª δp º − 0.433γ o cos θ » μ «¬δL ¼

2500 = 7.353 → θ = tan −1 (7.535) = 82.26° 340

V = −0.001127

V = 0.0135

245 ª 4372 − 4365 º − 0.433(0.58) cos(82.26)» 0.63 «¬ 2523 ¼

bb1 = 0.0758 ft day day. ft 2

105

PROBLEM 7.2 A sand body is 1500 ft long, 300 ft wide, and 12 ft thick. It has a uniform permeability of 345 md to oil at 17% connate water saturation. The porosity is 32%. The oil has a reservoir viscosity of 3.2 cp and Bo of 1.25 bbl/STB at the bubble point. a) If flow takes place above the bubble-point pressure, what pressure drop will cause 100 reservoir bbl/day to flow through the sand body, assuming the fluid behaves essentially as an incompressible fluid? What for 200 reservoir bbl/day? b) What is the apparent velocity of the oil in feet per day at the 100 bbl/day flow rate? c) What is the actual average velocity? d) What time will be required for complete displacement of the oil from the sand? e) What pressure gradient exists in the sand? f) What will be the effect of raising both the upstream and downstream pressures by, say, 1000 psi? g) Considering the oil as a fluid with a very high compressibility of 65(10)-6 psi-1, how much greater is the flow rate at the downstream end than the upstream end at 100 bbl/day? h) What pressure drop will be required to flow 100 bbl/day, measured at the upstream pressure, through the sand if the compressibility of the oil is 65(10)-6 psi1

? Consider the oil to be a slightly compressible fluid.

i) What will be the downstream flow rate? j) What conclusion can be drawn from these calculations concerning the use of the incompressible flow equation for the flow of slightly compressible liquids, even with high compressibilities? ANSWER: a)

q=

− 0.001127kA Δp − qμl → Δp = μ L 0.001127kA 106

Δp =

100(3.2)(1500) = 343 psia 0.001127(345)(12 × 300)

for : q = 200 bb1

day

Ÿ Δp = 686 psia

b)

Vapparent =

q(5.615) 100(5.615) = = 0.156 ft day 3600 A

c) Vactual =

Vapparent

φ (1 − swi )

=

0.156 = 0.587 ft day 0.32(1 − 0.17)

d)

t=

N q

N = Vbφ (1 − S wi ) N = (12 × 300 × 1500)(0.32)(1 − 0.17) = 1434240 ft 3 ÷ 5.615 = 255430.1bb1

t=

25543.1 1year = 2554.3days × ≅ 7 year 100 365days

e)

Δp 343 = = 0.229 psia ft Δx 1500 f) It has no effect. g)

q = qref [1 + c°(Pref − p)] = 100[1 + 65 ×10 −6 (343)] = 102.23 bb1 Δq = 102.23 − 100 = 2.23 bb1

day

h)

Slightly compressib le, q ref = q inlet = 100 bb1

day

107

day

q = 0.001127

kA

μLc °

100 = 0.001127

ln[1 + c° ( p1 − p2 )]

345(3600) ln[1 + (65 × 10 −6 )( p 1 − p 2 )] 3.2(1500)(65 × 10 −6 )

Δp = p1 − p2 = 346.8 psia i) q2 = 0.001127

º kA ª 1 ln « μLc° ¬1 + c° ( p2 − p1 ) »¼

q 2 = 0.001127

º ª 345 (3600 ) 1 ln (3.2)(1500 )( 65 × 10 − 6 ) «¬1 + (65 × 10 − 6 )( −346 .8) »¼

q2 = 102.29 bb1

day

j) Using incompressible flow equation for the flow of slightly compressible liquids will have insignificant error.

PROBLEM 7.3 If the sand body of Prob. 7.2 had been a gas reservoir with a bottom-hole temperature of 140 °F but with the same connate water and permeability to gas, calculate the following: a) With an upstream pressure of 2500 psia, what downstream pressure will cause 5.00 MM SCF/day to flow through the sand? Assume an average gas viscosity of 0.023 cp and an average gas deviation factor of 0.88. b) What downstream pressure will cause 25 MM SCF/day to flow if the gas viscosity and deviation factors remain the same? c) Explain why it takes more than five times the pressure drop to cause five times the gas flow. d) What is the pressure at the midpoint of the sand when 25 MM SCF/day is flowing? 108

e) What is the mean pressure at 25 MM SCF/day? f) Why is there a greater pressure drop in the downstream half of the sand body than in the upstream half? g) From the gas law calculate the rate of flow at the mean pressure pm, and show that the equation in terms of qm, is valid by numerical substitution.

ANSWER: a) 0.111924 kA( p1 − p 2 2 ) TL μz 2

qsc =

0.111924 (345)(3600)(625 × 10 4 − p 2 2 ) Ÿ p2 = poutlet = 2365.33 psia 600(1500)(0.023)(0.88)

5 × 10 6 =

b) 25 × 10 6 =

0.111924 (345 )(3600 )( 625 × 10 4 − p 2 2 ) Ÿ p 2 = 1724 .5 psia 600 (1500 )( 0.23)( 0.88)

c) Flow rate equals square pressure drop. Therefore, pressure drop increases. d) p1 + p 2 2 Ÿ p= 2 2

p=

( 2500 ) 2 + (1725 ) 2 = 2147 .34 psia 2

e) Pm =

p1 + p2 2500 + 1724.5 = = 2112.3 psia 2 2

f) Due to friction pressure drop in the downstream half of the sand body is higher than upstream half. g) qm = (

psc t q 14.7 0.88 × 600 25 × 10 6 )( )( sc ) = ( )( )( ) t sc Pm 5.615 52. 2112.3 5.615

qm = 31461 bb1

day 109

PROBLEM 7.4 (a) Plot pressure versus distance through the sand of the previous problem at the 25 MM SCF/day flow rate. (b) Plot the pressure gradient versus distance through the sand body.

ANSWER: The solution of this problem is left to readers.

PROBLEM 7.5 A rectangular sand body is flowing gas at 10 MM SCF/day under a downstream pressure of 1000 psia. Standard conditions are 14.4 psia and 80°F. The average deviation factor is 0.80. The sand body is 1000 ft long, 100 ft wide, and 10 ft thick. Porosity is 22%, and average permeability to gas at 17% connate water is 125 md. Bottom-hole temperature is 160°F, and gas viscosity is 0.029 cp. a) What is the upstream pressure? b) What is the pressure gradient at the midpoint of the sand? c) What is the average pressure gradient throughout the sand? d) Where does the mean pressure occur?

ANSWER: a)

0.11865kA( p1 − p2 ) Tlzμ g 2

q sc =

2

0.11865 × 103 × 125( p1 − 106 ) Ÿ p1 = 3271 psia 620 × 103 × 0.029 × 0.80 2

10 × 106 =

110

b) p1 + p 2 2 = 2418 .2 psia 2 2

p=

(

psc ZT qsc KA dp )( )( ) = 0.001127 Tsc P 5.615 μ dx

14.4 0.8 × 62. 10 7 125(10 3 ) dp dp ( )( )( ) = 0.001127 Ÿ = 2.01 psi ft 54. 2418.2 5.615 0.029 dx dx c)

dp p1 − p2 3271− 100 = = = 2.271 psi ft dx dx 1000 d)

p1 + p2 = 2135.5 psia 2 2 2 0.11865kA( p1 − p 2 ) q sc = Tlzμ g p=

10 × 10 6 =

0.11865(103 )(125)(2135.5 2 − 10 6 ) Ÿ l = 367 ft 620l (0.80)(0.029)

PROBLEM 7.6 A horizontal pipe 10 cm in diameter (I.D.) and 3000 cm long is filled with a sand of 20% porosity. It has a connate water saturation of 30% and, at that water saturation, a permeability of oil of 200 md. The viscosity of the oil is 0.65 cp and the water is immobile. a) What is the apparent velocity of the oil under a 100 psi pressure differential? b) What is the flow rate? c) Calculate the oil contained in the pipe and the time needed to displace it at the rate of 0.055 cu cm/sec. 111

d) From this actual time and the length of the pipe, calculate the actual average velocity. e) Calculate the actual average velocity from the apparent velocity, porosity, and connate water. f) Which velocity is used to calculate flow rates, and which is used to calculate displacement times? g) If the oil is displaced with water so that 20% unrecoverable (or residual) oil saturation is left behind the water flood front, what are the apparent and actual average velocities in the watered zone behind the flood front if the oil production rate is maintained at 0.055 cu cm/sec? Assume piston-like-displacement of the oil by the water. h) What is the rate of advance of the flood front? i) How long will it take to obtain all the recoverable oil, and how much will be recovered? j) How much pressure drop will be required to produce oil at the rate of 0.055 cu cm/sec when the water flood front is at the midpoint of the pipe?

ANSWER: a)

Vapparent

100 ) (0.2)( q kΔ p 14 .7 → V = = = = 0.0007 cm apparent sec μl (0.65)(3000 ) A

b)

q=

kA Δp , A = πr 2 = π (5) 2 = 78.54cm2 μ L

100 3 (0.2)(78.54) 14.7 q= → q = 0.055 cm sec (0.65) 3000 112

c)

t=

N , N = vbφ (1 − swi), vb = πr 2 l q

N = (π × (5) 2 × (3000))(0.2)(1 − 0.3) = 32986.8

t=

599760 32986.8 = 599760sec → t = 7day 0.055 86400

d)

V=

L 3000 = = 0.005 cm sec t 599760

e) Vactual =

Vapparent 0.0007 = = 0.005 cm sec φ (1 − swi ) 0.2(1 − 0.30)

f) Apparent velocity is used to calculate flow rates and actual velocity is used to calculate displacement times. g)

Vapparent = Vactual =

3 q 0.055 = = 0.0007 cm sec A 78.54

Vapp

φ (1 − swi − sor )

=

3 0.0007 = 0.007 cm sec 0.2(1 − 0.5 − 0.2)

h) V = Vactual = 0.007 cm

3

sec

i) N = vbφS ° = (π × (5) × 3000)(0.2)(0.5) = 23562 cm 3 2

t=

3 N , qact = Avact = (78.54)(0.007) = 0.55 cm sec q

t=

23562 = 42840sec.t − 0.5days 0.55

113

j)

Δp =

qμl (0.055)(0.65)(1500) = = 3.414atm KA (0.2)(78.54)

Δp = 50.2 psia Δpt = Δpthreshold + Δp = 100 + 50.2 = 150.2 psia

PROBLEM 7.7 (a) Three beds of equal cross section have permeabilities of 50, 200, and 500 md and lengths of 40, 10, and 75 ft, respectively. What is the average permeability of the beds placed in series? (b) What are the ratios of the pressure drops across the individual beds for liquid flow? (c) For gas flow will the overall pressure drop through beds in series be the same for flow in either direction? Will the individual pressure drops be the same? (d) The gas flow constant for a given linear system is 900, so that p12- p22 = 900 L/k. If the upstream pressure is 500 psia, calculate the pressure drops in each of two beds for series flow in both directions. The one bed is 10 ft long and 100 md; the second is 70 ft and 900 md. (e) A producing formation from top to bottom consists of 10 ft of 350 md sand, 4 in. of 0.5 md shale, 4 ft of 1230 md sand, 2 in. of 2.4 md shale, and 8 ft of 520 md sand. What is the average vertical permeability? (f) If the 8 ft of 520 md sand is in the lower part of the formation and carries water, what well completion technique will you use to keep the water-oil ratio low for the well? Discuss the effect of the magnitude of the lateral extent of the shale breaks on the well production.

114

ANSWER: a)

k avg =

ΣLi 40 + 10 + 75 = → k avg = 125md Li 40 10 75 Σ + + ki 50 200 500

b)

L1 40 p1 − p2 k1 i =1→ = = 50 = 0.8 125 Lt p1 − p4 k avg 125 L2 10 p 2 − p3 k2 200 = 0.05 = = i=2→ 125 p1 − p4 Lt 125 k avg L3 75 p3 − p 2 k3 500 = 0.15 i =3→ = = 125 p1 − p4 Lt 125 k avg

c)

k avg

4 2 10 + + 4 + + 8 ¦ hi 12 12 = = = 28.72 4 2 h 10 4 8 ¦ i + 12 + + 12 + ki 350 0.5 1230 2.4 520

d) The solution of this problem is left to readers. e) The solution of this problem is left to readers. f) The solution of this problem is left to readers.

115

PROBLEM 7.8 (a) Three beds of 40, 100, and 800 md, and 4, 6, and 10 ft thick, respectively, are conducting fluid in parallel flow. If all are of equal length and width, what is the average permeability? (b) In what ratio are the separate flows in the three beds?

ANSWER: a) k avg =

¦ k i hi ( 40 × 4) + (100 × 6) + (800 × 10) = = 438md ¦ hi 4 + 6 + 10

b) qi kh k hi 40 × 4 q = i i i =1→ 1 = i = = 0.0182 qt k avg ht qt k avg ht 438 × 20 i=2→

i =3→

q

2

qt

=

100 × 6 k 2 h2 = = 0.0685 k avg ht 438 × 20

q3 kh 800 × 10 = 3 3 = = 0.913 qt k avg ht 438 × 20

PROBLEM 7.9 As project supervisor for an in situ uranium leaching project, you have observed that to maintain a constant injection rate in well A, the pump pressure has had to be increased so that pe–pw has increased by a factor of 20 from the value at startup. An average permeability of 100 md was measured from plugs cored before the injection of leachant. You suspect buildup of a calcium carbonate precipitate has damaged the formation near the injection well. If the permeability of the damaged section can be assumed to be 1 md, find the extent of the damage. The wellbore radius is 0.5 ft, and the distance to the outer boundary of the uranium deposit is estimated to be 1000 ft. 116

ANSWER: The solution of this problem is left to readers.

PROBLEM 7.10 A well was given a large fracture treatment, creating a fracture that extends to a radius of about 150 ft. The effective permeability of the fracture area was estimated to be 200 md. The permeability of the area beyond the fracture is 15 md. Assume that the flow is steady-state, single-phase, incompressible flow. The outer boundary at r = re = 1500 ft has a pressure of 2200 psia and the wellbore pressure is 100 psia (rw = 0.5 ft). The reservoir thickness is 20 ft and the porosity is 18%. The flowing fluid has a formation volume factor of 1.12 bbl/STB and a viscosity of 1.5 cp. a) Calculate the flow rate in STB/day. b) Calculate the pressure in the reservoir at a distance of 300 ft from the center of the wellbore.

ANSWER:

q=

k avg

q=

0.00708kh( pe − pw ) stb day r μ° B° ln( e ) rw r 1500 k a ke ln( e ) 200(15) ln( ) rw 0.5 = = = 44md r 1500 150 r ) + 15 ln( ) k a ln( e ) + ke ln( a ) 200 ln( 150 0.5 ra rw

0.00708(44)(20)(2200 − 100) = 972.396 stb day 1500 105(1.12) ln( ) 0.5

117

b)

q=

0.00708kh pr − pw 0.00708(44)(20) pr − 100 → 972.396 = r r 105(1.12) μo ho ln ln( ) rw 0.5

P at r = 300ft = 1778psia

PROBLEM 7.11 (a) A limestone formation has a matrix (primary or intergranular) permeability of less than 1 md. However, it contains 10 solution channels per square foot, each 0.02 in. in diameter. If the channels lie in the direction of fluid flow, what is the permeability of the rock? (b) If the porosity of the matrix rock is 10%, what percentage of the fluid is stored in the primary pores, and what in the secondary pores (vugs, fractures, etc.)? (c) If the secondary pore system is well connected throughout a reservoir, what conclusions must be drawn concerning the probable result of gas or water drive on the recovery of either oil, gas, or gas-condensate? What then are the means of recovering the hydrocarbons from the primary pores?

ANSWER: a)

kavg = 175.5md b) Vvug Vp

= 0.022 %

118

PROBLEM 7.12 During a gravel rock operation the 6 in. I.D. liner became filled with gravel, and a layer of mill scale and dirt accumulated to a thickness of 1 in. on top of the gravel within the pipe. If the permeability of the accumulation is 1000 md, what additional pressure drop is placed on the system when pumping a 1 cp fluid at the rate of 100 bbl/hr?

ANSWER:

q = 100 bbl A=

π 4

= 2400 bb1

hr

day

( ID) 2 = 0.196 ft 2

q = 0.001127

2400 (1)( 1 ) kA Δp qμL 12 → Δp = = μ L 0.001127 KA 0.001127 (1000)(0.196)

Δp = 903psia

PROBLEM 7.13 One hundred capillary tubes of 0.02 in. ID and 50 capillary tubes of 0.04 in. ID, all of equal length, are placed inside a pipe of 2 in. inside diameter. The space between the tubes is filled with wax so that flow is only through the capillary tubes. What is the permeability of this ‘rock’?

ANSWER:

ID = 0.02in(number= 100) K (0.02in ) = 20 × 10 6 × (0.02) 2 = 8000 darcy

119

A(0.02in ) = 100 ×

π 4

× (0.02) 2 = 3.142 × 10 − 2 in 2

ID = 0.04in(number= 50) K (0.04in ) = 20 × 10 6 × (0.04) 2 = 32000darcy

A(0.04in) = 50 ×

k avg =

π 4

× (0.04) 2 = 6.283 × 10 −2 in 2

(8000 × 3.142 × 10 −2 ) + (32000 × 6.283 × 10−2 )

π

4

× (2) 2

= darcy

PROBLEM 7.14 Suppose, after cementing, an opening 0.01 in. wide is left between the cement and an 8 in. diameter hole. If this circular fracture extends from the producing formation through an impermeable shale 20 ft thick to an underlying water sand, at what rate will water enter the producing formation (well) under a 100 psi pressure drawdown? The water contains 60,000 ppm salt and the bottom-hole temperature is 150°F.

ANSWER:

q = 8.7 × 109

w2 Ac ( p1 − p2 ) μBL (

q = 8.7 × 10

9

8 º 0.01 2 ª 8.01 2 ) − ( ) 2 » × 100 ) × π × «( 12 ¼ 12 ¬ 12 = 215BPD (0.49)(1)(20)

PROBLEM 7.15 A high water-oil ratio is being produced from a well. It is thought that the water is coming from an underlying aquifer 20 ft from the oil producing zone. In between the aquifer and the producing zone is an impermeable shale zone. Assume that the water is

120

coming up through an incomplete cementing job that left an opening 0.01 in. wide between the cement and the 8 in. hole. The water has a viscosity of 0.5 cp. Determine the rate at which water is entering the well at the producing formation level if the pressure in the aquifer is 150 psi greater than the pressure in the well at the producing formation level.

ANSWER: K = 7.7 × 1012 w2 = 7.7 × 1012 × (

0.01 ) = 5347222 md = 5347 darcy 12

8 7.99 2 Ac = 3.14(( ) 2 − ( ) ) = 3.5 × 10 −3 12 12 w 2 Ac ( p1 − p 2 ) q = 8.7 × 10 = 8.7 × 10 9 μL 9

(

0.01 2 ) × 3.5 × 10 −3 × 150 12 0.5 × 20

q = 317BPD

PROBLEM 7.16

Derive the equation for the steady-state, semispherical flow of an incompressible fluid.

ANSWER:

q sc =

7.08kre rw ( pe − p w ) μ o Bo (rw − re )

121

PROBLEM 7.17 A well has a shut-in bottom-hole pressure of 2300 psia and flows 215 bbl/day of oil under a drawdown of 500 psi. The well produces from a formation of 36 ft net productive thickness. Use rw = 6 in.; re = 660 feet; µ = 0.88 cp; Bo = 1.32 bbl/STB. a) What is the productivity index of the well? b) What is the average permeability of the formation? c) What is the capacity of the formation?

ANSWER: a)

PI =

q 215 bb1 = = 0.43 Δp 500 day.sia

b)

r qμ ln( e ) kh( pe − pw ) rw →K = q = 0.00708 re 0 . 00708 hΔp μ ln( ) rw 660 ) 0.5 = 10.7 md K= 0.00708 (36)(500 ) ( 215)(0.88) ln(

c)

r 660 qμ ln( e ) (215)(0.88) ln( ) rw 0.5 capacity of formation = kh = = 0.00708( pe − pw ) 0.00708(500)

kh = 384md − ft

122

PROBLEM 7.18 A producing formation consists of two strata: one 15 ft thick and 150 md in permeability; the other 10 ft thick and 400 md in permeability. a) What is the average permeability? b) What is the capacity of the formation? c) If during a well workover the 150 md stratum permeability is reduced to 25 md out to a radius of 4 ft, and the 400 md stratum is reduced to 40 md out to an 8 ft radius, what is the average permeability after the workover, assuming no crossflow between beds? Use re = 500 ft and rw = 0.5 ft. d) To what percentage of the original productivity index will the well be reduced? e) What is the capacity of the damaged formation?

ANSWER: a) k avg =

¦ k i hi (150 × 15) + ( 400 × 10) = = 250 md ¦ hi 15 + 10

b)

Capacity = ¦ ki hi = (150 × 15) + (400 × 10) = 6250md − ft c)

k avag1

r k a ke ln( e ) 25 × 150 ln(500 ) rw 0.05 = 60md = = r 500 ) + 150 ln( 4 r ) k a ln( e ) + ke ln( e ) 25 ln( 4 0.05 rw rw

k avg 2 =

40 × 400 ln(500

) 0.05 = 86.8md ) 40 ln(500 ) + 400 ln( 8 8 0.05

k avg ( total ) =

¦ k i hi (60 × 15) + (86.8 × 10) = = 70.72md ¦ hi 15 + 10

123

d) q )2 k 70.72 Δp Productivi ty Index Ÿ = 2 = = 0.28 or 28% q 250 ( )1 k1 Δp (

e)

Totalcapacityafterdamage= ¦kavg hi = (70.72× 25) = 1768md − ft

PROBLEM 7.19 (a) Plot pressure versus radius on both linear and semilog paper at 0.1, 1.0, 10, and 100 days for pe = 2500 psia, q = 300 STB/day; Bo= 1.32 bbl/STB; µ = 0.44 cp; k = 25 md; h = 43 ft; ct = 18 x 10-6 psi ; ࢥ = 0.16. (b) Assuming that a pressure drop of 5 psi can be easily detected with a pressure gauge, how long must the well be flowed to produce this drop in a well located 1200 ft away? (c) Suppose the flowing well is located 200 ft due east of a north-south fault. What pressure drop will occur after 10 days of flow, in a shut-in well located 600 ft due north of the flowing well? (d) What will the pressure drop be in a shut-in well 500 ft from the flowing well when the flowing well has been shut in for one day following a flow period of 5 days at 300 STB/day?

ANSWER: The solution of this problem is left to readers.

124

PROBLEM 7.20 A shut-in well is located 500 ft from one well and 1000 ft from a second well. The first well flows for 3 days at 250 STB/day, at which time the second well begins to flow at 400 STB/day. What is the pressure drop in the shut-in well when the second well has been flowing for 5 days (i.e., the first has been flowing a total of 8 days)? Use the reservoir constants of Prob. 7.19.

ANSWER: Δp =

φμct r 2 º 70.6qμB ª − Ei ( ) « 0.00105kt »¼ kh ¬

Δp1 =

− 0.16 × 0.44 × 18 × 10 −6 × 500 2 º 70.6( 250)(0.44)(1.32) ª − Ei ( )» « 24( 43) 0.00105( 25)(8 × 24) ¬ ¼

Δp 2 =

− 0.16 × 0.44 × 18 × 10 −6 × 1000 2 º 70.6( 400)(0.44)(1.32) ª − E ( )» i « 24 × 43 0.00105( 25)(5 × 24) ¬ ¼

Δp = Δp1 + Δp 2 Ÿ Δp = 9.536[− E i ( −0.0628)] + 152575 [− E i ( −0.402) ] − E i ( − x ) = − ln( x) − 0.5772 Ÿ − Ei ( −0.06286) = − ln(0.06286) − 0.5772 = 2.19 from tabe l7.1 Ÿ x = 0.402 Ÿ − E i ( − x) = 0.699 Ÿ Δp = (9.536)( 2.19) + (15.2575)(0.699) = 31.54 psi

PROBLEM 7.21 A well is opened to flow at 200 STB/day for 1 day. The second day its flow is increased to 400 STB/day and the third to 600 STB/day. What is the pressure drop caused in a shut-in well 500 ft away after the third day? Use the reservoir constants of Prob. 7.19. 125

ANSWER:

Δpt = Δp1 + Δp2 + Δp3 Δp1 =

70.6(200)(0.44)(1.32) ª − 0.16 × 0.44 × 18 × 10 −6 × 500 2 º − E ( )» = i « 25(43) 0.00105 × 25 × (3 × 24) ¬ ¼

= 7.629[− Ei ( −0.1676)] = 10.44 Δp 2 =

− 0.16 × 0.44 × 18 × 10 −6 × 500 2 × 2 º 70.6( 400 − 200)(0.44)(1.32) ª − Ei ( )» = « 0.00105 × 25 × ( 2 × 24) 25( 43) ¬ ¼

= 7.629[− E ( −0.2514)] = 7.93 Δ p3 =

− 0.16 × 0.44 × 18 × 10 −6 × 500 2 × 2 º 70.6(600 − 400)(0.44)(1.32) ª − Ei ( )» = « 0.00105 × 25 × (2 × 24) 25( 43) ¬ ¼

= 7.629[− Ei ( −0.50286)] = 4.24

Ÿ Δpt = Δp1 + Δp2 + Δp3 = 10.44 + 7.93 + 4.24 = 22.61psi

PROBLEM 7.22 The following data pertain to a volumetric gas reservoir: Net formation thickness = 15 ft Hydrocarbon porosity = 20% Initial reservoir pressure = 6000 psia Reservoir temperature = 190°F Gas viscosity = 0.020 cp Casing diameter = 6 in. Average formation permeability = 6 md

a) Assuming ideal gas behavior and uniform permeability, calculate the percentage of recovery from a 640 ac unit for a producing rate of 4.00 MM SCF/day when the flowing well pressure reaches 500 psia.

126

b) If the average reservoir permeability had been 60 md instead of 6 md, what recovery would be obtained at 4.00 MM SCF/day and a flowing well pressure of 500 psia? c) Recalculate part (a) for a production rate of 2.00 MM SCF/day. d) Suppose four wells are drilled on the 640 ac unit, and each is produced at 4.00 MM SCF/day. For 6 md and 500 psia minimum flowing well pressure, calculate the recovery. ANSWER: (a) 54.8% and 10.25 years (b) 83.6% and 15.64 years (c) 67.6% and 25.6 years (d) 56.7% and 2.65 years

PROBLEM 7.23 A sandstone reservoir, producing well above its bubble-point pressure, contains only one producing well, which is flowing only oil at a constant rate of 175 STB/day. Ten weeks after this well began producing, another well was completed 660 ft away in the same formation. On the basis of the reservoir properties that follow, estimate the initial formation pressure that should be encountered by the second well at the time of completion. ࢥ = 15%

h = 30ft

co = 18(10)-6 psi-1

µ = 2.9 cp

cw = 3(10)-6 psi-1

k = 35 md

cf = 4.3(10)- 6 psi-1

rw = 0.33 ft

Sw = 33%

pi = 4300 psia

Bo = 1.25 bbl/STB

127

ANSWER: The solution of this problem is left to readers.

PROBLEM 7.24 Develop an equation to calculate and then calculate the pressure at well 1, illustrated in Fig. 7.23, if the well has flowed for 5 days at a flow rate of 200 STB/day. ࢥ = 25%

h = 30ft

cr = 30(10)-6 psi-1

µ = 0.5 cp

k = 50 md

Bo = 1.32 bbl/STB

rw = 0.33 ft

pi = 4000 psia

ANSWER: The solution of this problem is left to readers.

128

PROBLEM 7.25 A pressure drawdown test was conducted on the discovery well in a new reservoir to estimate the drainage volume of the reservoir. The well was flowed at a constant rate of 125 STB/day. The bottom-hole pressure data, as well as other rock and fluid property data, follow. What are the drainage volume of the well and the average permeability of the drainage volume? The initial reservoir pressure was 3900 psia. Bo = 1.1 bbl/STB

µo = 0.80 cp

ࢥ = 20%

h = 22 ft

So = 80%

Sw = 20%

co = 10(10)-6 psi-1

cw = 3(10)-6 psi-1

cf = 4(10)- 6 psi-1

rw = 0.33 ft





Time in Hours

Pwf, psi

0.5

3657

1.0

3639

1.5

3629

2.0

3620

3.0

3612

5.0

3598

7.0

3591

10.0

3583

20.0

3565

30.0

3551

40.0

3548

50.0

3544

60.0

3541

70.0

3537

80.0

3533

90.0

3529

100.0

3525

120.0

3518

150.0

3505

ANSWER:

Draw Pwf versus log t in a semi log paper and then find the slope of figure.

m = −56 psi / cycle Plhr = 3639 psi

130

k =−

162.6qμB 162.6(125)(0.80)(1.1) =− = 14.5md mh − 56(22)

ª p − p wf ( Δt = 0) º K − log( ) + 3.23» s = 1.151« lhr 2 m φμct rw ¬ ¼

where : ct =

c° s° + c w s w + c f = 15.75 × 10 − 6 psi −1 1 − sw

ª 3639 − 3900 º 14.5 ) + 3.23» = − log( s = 1.151« −6 2 − 56 0.2(0.80)(15.75 × 10 )(0.33) ¬ ¼ s = 0.1925

Draw Pwf versus time in a Cartesian paper and then find slope of figure. ͵͸ͺͲ ͵͸͸Ͳ ͵͸ͶͲ ͵͸ʹͲ ͵͸ͲͲ

Pwf ͵ͷͺͲ (psi) ͵ͷ͸Ͳ ͵ͷͶͲ ͵ͷʹͲ ͵ͷͲͲ ͵ͶͺͲ Ͳ

ͷͲ

ͳͲͲ

ͳͷͲ

Time (hr)

m / = −0.383 A=−

0.2339 qB 0.2339 (125 )(101) =− = 1211716 ft 2 m / ct h φ ( −0.383)(15 .751 × 10 − 6 )( 22 )(0.20 )

A= 27.8 acres V = Ah = 27.8 × 22 = 611.6ac − ft

131

ʹͲͲ

PROBLEM 7.26 The initial average reservoir pressure in the vicinity of a new well was 4150 psia. A pressure drawdown test was conducted while the well was flowed at a constant oil flow rate of 550 STB/day. The oil had a viscosity of 3.3 cp and a formation volume factor of 1.55 bbl /STB. Other data, along with the bottom-hole pressure data recorded during the drawdown test, follow. Assume that wellbore storage considerations may be neglected, and determine the following: a) The permeability of the formation around the well. b) Any damage to the well. c) The drainage volume of the reservoir communicating to the well. ࢥ = 34.3%

h = 93ft

cr = 1(10)-5 psi-1

rw = 0.5 ft

Time in Hours

Pwf, psi

1

4025

2

4006

3

3999

4

3996

6

3993

8

3990

10

3989

20

3982

30

3979

40

3979

50

3978

60

3977

70

3976

80

3975 132

ASNWER: a) Draw Pwf versus time in a semi log paper. The slope of figure can be found as m= -27 psi/cycle.

k=−

162.6qμB 162.6(550)(3.3)(1.55) =− = 182md mh − 27(93)

b) ª p − p wf ( Δt = 0) º k − log( ) + 3.23» = 0.383 s = 1.151« 1hr 2 m φμct rw ¬ ¼

c) Draw Pwf versus time in a Cartesian paper. The slope of figure (m) can be found as m=-0.1167 psi/cycle.

m = −0.1167psi/cycle A=−

0.2339qB 0.2339(550)(1.55) =− = 5356445 ft 2 = 123ac −5 mct hφ − (0.1167)(1 × 10 )(93)(0.343)

V = Ah = (123)(93) = 11439ac − ft

PROBLEM 7.27 The first oil well in a new reservoir was flowed at a constant flow rate of 195 STB/day until a cumulative volume of 361 STB had been produced. After this production period, the well was shut in and the bottom-hole pressure monitored for several hours. The flowing pressure just as the well was being shut in was 1790 psia. For the data that follow, calculate the formation permeability and the initial reservoir pressure. Bo = 2.5 bbl/STB

µo = 0.85 cp

ࢥ = 11.5%

h = 23 ft

cr = 1(10)-5 psi-1

rw = 0.33 ft

133

ǻt in Hours

Pws, psi

0.5

2425

1.0

2880

2.0

3300

3.0

3315

4.0

3320

5.0

3324

6.0

3330

8.0

3337

10.0

3343

12.0

3347

14.0

3352

16.0

3353

18.0

3356

ANSWER:

tp =

Np 361 = × 24 = 44.43hr q 195

Then calculate tp+ǻt/ǻt for each ǻt and prepare the following table.

134

ǻt (hr)

Pws (psi)

tp+ǻt/ǻt

0.5

2425

89.86

1.0

2880

45.43

2.0

3300

23.21

3.0

3315

15.81

4.0

3320

12.1

5.0

3324

9.88

6.0

3330

8.4

8.0

3337

6.55

10.0

3343

5.44

12.0

3347

4.7

14.0

3352

4.17

16.0

3353

3.77

18.0

3356

3.46

Draw Horner figure which is Pws versus log tp+ǻt/ǻt in a semi log paper and then find the slope of figure: m=-70 psi/cycle.

K =−

162.6qμB 162.6(195)(0.85)(2.15) =− = 36md mh − 70(23)

P1hr?

t p + Δt Δt

=

44.43 + 1 = 45.43 Ÿ P1hr = 3400 psia 1

ª p ( Δ t = 0 ) − p1 hr º k s = 1 .151 « wf − log( ) + 3 .23 » 2 m φμ c t rw ¬ ¼

135

ª1790 − 3400 º 36 s = 1.151« ) + 3.23» = − log( −5 2 (0.115)(0.85)(1 × 10 )(0.33) − 70 ¬ ¼ s = 20.37

PROBLEM 7.28 A well located in the center of several other wells in a consolidated sandstone reservoir was chosen for a pressure buildup test. The well had been put on production at the same time as the other wells and had been produced for 80 hr at a constant oil flow rate of 375 STB/day. The wells were drilled on 80 ac spacing. For the pressure buildup data and other rock and fluid property data that follow, estimate a value for the formation permeability and determine if the well is damaged. The flowing pressure at shut-in was 3470 psia. Bo = 1.31 bbl/STB

µo = 0.87 cp

ࢥ = 25.3%

h = 22 ft

So = 80%

Sw = 20%

co = 17(10)-6 psi-1

cw = 3(10)-6 psi-1

cf = 4(10)- 6 psi-1

rw = 0.33 ft

tp=80 hr ǻt (hr)

pws, (psi)

0.114

3701

0.201

3705

0.432

3711

0.808

3715

2.051

3722

4.000

3726

8.000

3728

17.780

3730 136

ANSWER: Calculate tp+ǻt/ǻt for each ǻt and prepare the following table. ǻt (hr)

Pws (Psi)

tp+ǻt/ǻt

0.114

3701

71.17

0.201

3705

40.8

0.432

3711

19.51

0.808

3715

10.9

2.051

3722

4.9

4.000

3726

3

8.000

3728

2

17.780

3730

1.45

Draw Pws versus log tp+ǻt/ǻt in a semi log paper and then find the slope of figure: m= -53 psi/cycle.

k =−

162qμB 162.6(375)(0.87)(1.31) =− = 60md mh (−53)(22)

P1hr?

tp + Δt 80 + 1 = = 81 Ÿ p1hr = 3780 psia Δt 1 ª p ( Δ t = 0 ) − p1 hr º k s = 1 .151 « wf − log( ) + 3 .23 » 2 m φμ c r t w ¬ ¼

ª 3470 − 3780 º 60 s = 1.151« ) + 3.23» = 1.08 − log( −5 2 (0.253)( 0.87 )(1.82 × 10 )( 0.33) − 53 ¬ ¼

ct = c ° s ° + c w s w + cf → ct = 1.82 × 10 −5 psi −1

137

CHAPTER 8

WATER INFLUX

PROBLEM 8.1 Assuming the Schilthuis steady-state water influx model, use the pressure drop history for the Conroe Field given in Fig. 8.15, and a water influx constant, k ', of 2170ft3/day/psi, to find the cumulative water encroachment at the end of the second and fourth periods by graphical integration for Table 6.1.





ANSWER:

k ′ = 2170

ft 3 / psi ÷ 5.615 day

k ′ = 386.5 bbl

day / psi

For 12 month (365day): t 130 we = k ′³ ( pi − p)dt = 385.5( )365 = 9169712.5bbl 2 o

For 20 month (600day):

130 + 170 ª130 º we = 386.5« (365) + (20 − 12) × 30» = 24.5 × 10 6 bbl 2 ¬ 2 ¼ For 30 month (900day): 170 + 190 ª130 º we = 386 .5 « (365 ) + 36000 + (30 − 20 ) × 30 » 2 ¬ 2 ¼ = 45 .36 × 10 6 bbl

PROBLEM 8.2 The pressure history for the Peoria Field is given in Fig. 8.16. Between 36 and 48 months, production in the Peoria Field remained substantially constant at 8450 STB/day, at a daily gas-oil ratio of 1052 SCF/STB, and 2550 STB of water per day. The initial solution GOR was 720 SCF/STB. The cumulative produced GOR at 36 months was 830SCF/STB, and at 48 months it was 920 SCF/STB. The two-phase formation volume factor at 2500 psia was 9.050 ft3/STB, and the gas volume factor at the same pressure, 0.00490 ft3/SCF. Calculate the cumulative water influx during the first 36 months.

139

ANSWER: ­ dNP STB ° dt = 8450 day ° ° scf 36 − 48month®GOR = 1052 STB ° dwp ° = 2250 STB day °¯ dt

scf ­ ° Rsi = 720 STB ° scf ° (36month) ®Rs = 830 STB ° scf ° °Rs = 920 STB (48month) ¯ ft 3 bbl ­ °° Bt = 9.050 STB ÷ 5.615 = 1.611 STB ® 3 ° B g = 0.0490 ft ÷ 5615 = 8.72 × 10 − 4 bbl scf scf ¯°

ew =

dwe dNp dNp dwp Bg + Bw = Bt + ( R − Rsi ) dt dt dt dt

= 1.611(84501+ (1052 − 720)(8450)8.72 × 10−4 + 2250(1) = 18311bbl 



PROBLEM 8.3 During a period of production from a certain reservoir, the average reservoir pressure remained constant at 3200 psia. During the stabilized pressure, the oil and water producing rates were 30,000 STB/day and 5000 STB/day, respectively. Calculate the incremental water influx for a later period when the pressure drops from 3000 to 2800 psia. Assume the following relationship for pressure and time holds:

dp = −0.003 p, psia / month dt Other data are the following: pi = 3500 psia Rsot = 750 SCF/STB Bi = 1.45 bbl/STB at 3200 psia Bg = 0.002 bbl/STB at 3200 psia R= 800 SCF/STB at 3200 psia Bw = 1.04 bbl/STB at 3200 psia

ANSWER:

STB ­ dNp ° dt = 30000 day ° P = 3200® ° dwp = 5000 STB °¯ dt day Pi = 3500 psia Rsoi = 750 scf

STB

Bt = 1.45 bbl

@ 3200 psia STB B g = 0.002 bbl @ 3200 psia Scf R = 800 scf STB @ 3200 psia Bw = 1.04 bbl STB

141

dwe dNp dNp dwp = Bt + ( R − Rsoi ) B g + Bw dt dt dt dt = 1.45(300000) + (800 − 750)(30000)0.002 + 5000(1.04) = 51700bbl day dwe = C ( pi − p) → dt 51700 = C (3500 − 3200 ) → C = 172.3 bbl

day / psi

dp = −0.003 p ( psi / month) → dt − 1 2800 dp dt = ³ ³ → 0.003 3000 p

−1 −1 (ln p 2 − ln p1 ) = (ln 2800 − ln 3000) 0.003 0.003 Δt = 23 month

Δt =

Δt = 23month Δwe = CΔpΔt = 172.3(300 − 2800 )( 23 × 30) = 23.8 × 10 6 bbl Δwe = 23.8 × 10 6 bbl

PROBLEM 8.4 The pressure decline in a reservoir from the initial pressure down to a certain pressure, p, was approximately linear at -0.500 psi/day. Assuming the Schilthuis steady-state water influx model and a water influx constant of k', in ft3/day-psia, determine an expression for the water influx as a function of time in bbl.

ANSWER:

dp = −0.500 → dp = −0.500dt → pi − p = −0.5dt → dt dp = −0.5t 142

dwe dwe dp = k ( pi − p) → × = k ( pi − p) → dt dp dt dwe ( −0.5) = k ( −0.5dt ) dp

dwe = k .dt → ³ dwe = k .³ dp.dt → ³ dwe = +0.5³ t.dt dp

we = +

0.5 2 0.25 2 kt = +0.25kt 2 pts → we = kt bbl 2 5.615

we = 0.0445kt 2 bbl

PROBLEM 8.5 An aquifer of 28,850 ac includes a reservoir of 451 ac. The formation has a porosity of 22%, thickness of 60 ft, a compressibility of 4(10)-(, psi'", and a permeability of 100 md. The water has a viscosity of 0.30 cp and a compressibility of 3(10)-6 psi-1. The connate water saturation of the reservoir is 26%, and the reservoir is approximately centered in this dosed aquifer. It is exposed to water influx on all of its periphery. a) Calculate the effective radii ofthe aquifer and the reservoir, and their ratio. b) Calculate the volume of water the aquifer can supply to the reservoir by rock compaction and water expansion per psi of pressure drop throughout the aquifer. c) Calculate the volume of the initial hydrocarbon contents of the reservoir. d) Calculate the pressure drop throughout the aquifer required to supply water equivalent to the initial hydrocarbon contents of the reservoir. e) Calculate the theoretical time conversion constant for the aquifer. f) Calculate the theoretical value of B' for the aquifer. g) Calculate the water influx at 100, 200, 400, and 800 days if the reservoir boundary pressure is lowered and maintained at 3450 psia from an initial pressure of 3500 psia. 143

h) If the boundary pressure were changed from 3450 psia to 3460 psia after 100 days and maintained there, what would the influx be at 200, 400, and 800 days as measured from the first pressure decrement at time zero? i) Calculate the cumulative water influx at 500 days from the following boundary pressure history:

t (days)

0

100

p (psia)

3500 3490

200

300

400

500

3476

3458

3444

3420

j) Repeat part (i) assuming an infinite aquifer, and again assuming re / rR = 5.0 k) At what time in days do the aquifer limits begin to affect the influx? l) From the limiting value of WeD for re / rR =8.0, find the maximum water influx available per psi drop. Compare this result with that calculated in part (b).

ANSWER: The solution of this problem is left to readers.

PROBLEM 8.6 Find the cumulative water influx for the fifth and sixth periods in Ex. 8.3 and Table 8.3.

ANSWER: The solution of this problem is left to readers.

144

PROBLEM 8.7 The actual pressure history of a reservoir is simulated by the following data which assume that the pressure at the original oil-water contact is changed instantaneously by a finite amount, ¨P. Use the van Everdingen and Hurst method to calculate the total cumulative water influx. How much of this water influx occurred in the first two years?

Other reservoir properties include the following: Reservoir area = 19,600,000 ft2 Aquifer area =686,900,000 ft2 k = 10.4 md

ࢥ =25%

µw = 1.098 cP

ct =7.01(10)-6 psi -1

h= 10 ft

Time, Years

¨P, psia

0

40

0.5

60

1.0

94

1.5

186

2.0

110

2.5

120

3.0

-

ANSWER: a) ra =

(686900000 ) = 14786 ft 

145

re =

19600000 = 2498 ft 

rD =

ra = 6(Finite reservoir) re kh = φμctre 2

t D = 6.328 × 10 −3

10.4(10) = t D = 0.0548t 0.25(1.098)(7.01× 10 −6 )(2498) 2

= 6.328 × 10 −3

B = 1.119φCt hre f = 1.119(0.25)(7.01× 10 − 6 )(10)(2498) 2 Ÿ B = 122.4 2

we = B ¦ ΔpweD = 1224 × (7686) = 9.409 × 106 bbl

Time/year

tD

weD

¨P

¨P weD

3

60

16.56

40

662.4

2.5

50

16.05

60

963

2

40

14.93

94

1403.42

1.5

30

13.74

186

2555.64

1

20

11.16

110

1227.6

0.5

10

7.293

120

875.16

™¨PweD =7687 b)

we = B ¦ ΔpweD = 1224(3827) = 4.684 ×106 bbl Time/year

tD

weD

¨P

¨P weD

3

60

14.93

40

597.2

2.5

50

13.74

60

824.4

2

40

11.16

94

1049

1.5

30

7.293

186

1356.5

™¨PweD=3827

146

PROBLEM 8.8 An oil reservoir is located between two intersecting faults as shown in the areal view in Fig. 8.17. The reservoir shown is bounded by an aquifer estimated by geologists to have an area of 26,400 ac. Other aquifer data are the following: ࢥ =0.21

k = 275 md

h = 30 ft

ct = 7(10)-6 psi-1

µw = 0.92 cp The average reservoir pressure, measured at three-month intervals is as follow: Time, Years

P, psia

0

2987

91.3

2962

182.6

2927

273.9

2882

365.2

2837

456.5

2793





Use both the van Everdingen and Hurst and the Fetkovich methods to calculate the influx that occurred during each of the three month intervals. Assume that the average reservoir pressure history approximates the oil reservoir aquifer boundary pressure history.

ANSWER: Van Everdingen & Hurst method: B = 1.119φct re hf 2

re =

43560 (1350 ) = 10600 ft )  (60 360

ra =

43560 ( 26400 ) = 46864 ft π (60 360)

B = 1.119(0.21)(7 × 10 −6 )(10600) 2 (30) t D = 6.328 × 10 −3

60 Ÿ B = 924.2 360

kt Ÿ t D = 0.01146t φμ w ct re2

Time/day

tD

¨P

weD

we

273.9

3.138

12.5

3.42

39509

182.6

2.092

30

2.63

72919.38

91.3

1.046

40

1.712

43326.5

After the third period → ¦ we = 155755 bbl

148

Time/day

tD

¨P

weD

we

365.2

4.18

12.5

4.15

47943

273.9

3.138

30

3.42

94823

182.6

2.092

40

2.63

97226

91.3

1.046

45

1.712

71200

After thefourth period → ¦ we = 311192 bbl

Time/day

tD

¨P

weD

we

456.5

5.23

12.5

4.8

55452

365.2

4.18

30

4.15

115063

273.9

3.138

40

3.42

126430

182.6

2.092

45

2.63

109379

91.3

1.046

44.5

1.712

70409

After thefifth period → ¦ we = 476733bbl The Fetkovich method: Time

Pr

(Pr)n

(Pa)n-1

(Pa)n-1-(Pr)n

(¨we)n

we

0

2987

2987

2987

0

0

0

1

2962

2974.5

2987

12.5

15209

15209

2

2927

2944.5

2985.2

40.7

49550

46759

3

2882

2904.5

2979.4

75

91157

155916

4

2837

2859.5

2968.8

109.3

132974

288890

5

2793

2815

2953.4

138.4

168366

457256

149

rD =

ra = 4.42 ra

ra = 46864 ft re = 10600 ft Step1:

wi =

π

(ra2 − re2 )hφ =

5.615 = 7345345374

π 5615

(468642 − 106002 )30(0.21)

Step2:

wei = qwi pi f = 7 × 10−6 wi (2987)(

60 ) = 25.6 × 106 bbl 360

Step3:

60 ) 0.00708(275)(30)( 0.00708khf 360 = 14.37 J= = μw[LnrD − 0.75] 0.92[Ln(4.42) − 0.75] Jpi 14.37(2987) = = 1.677 × 10 −3 25.6 × 10 6 wei º ª Jp i 1 − EXP « − Δ t » = 1 − Exp [− 1 .677 × 10 − 3 × 91 .3] = 0 .142 we ¬ ¼ i

Step4: ( Δ we ) n =

ª − Jp i º we i = [( p a ) n −1 − ( p r ) n ]«1 − Exp ( Δt » we i pi ¬ ¼

(Δwe) n =

25.6 × 10 6 [( pa) n−1 − ( pr ) n ]0.142 2987

(Δwe) n = 1216.7[( pa) n−1 − ( pr ) n ] ( pa) n−1 = ( p) n (1 −

(we)n ) wei 150

PROBLEM 8.9 For the oil reservoir-aquifer boundary pressure relationship that follows, use the van Everdingen and Hurst method to calculate the cumulative water influx at each quarter (see Fig. 8.18): ࢥ =0.20

k = 200 md

h = 40 ft

ct = 7(10)-6 psi-1

µw = 0.80 cp Area of oil reservoir = 1000 ac Area of aquifer = 15,000 ac

ANSWER: The Van Evedingen and Hurst method

re = 3723 ft  ra = 14421ft 

B = 1.119φct hre2 f = 1.119(0.2)(7 × 10−6 )(40)(3723) 2 = 868.5





t D = 6.328 × 10 −3

kt

= 6.328 × 10 − 3

φμc t re2

(200 )t → 0.2(0.8)(7 × 10 − 6 )(3723 ) 2

t D = 0.0815 t

Step1:

Δp1 = 16

t D = 7.44 → weD = 6.02

we = BΔpweD = 868.5(

4020 − 3988 )6.02 Ÿ we = 83654bbl 2

After 91.3 days

Step2: Δp 2 =

pi − p 2 = 44 t = 14.88 → weD = 9062 D 2

we = ( we ) Δp1 + ( we ) Δp 2 = (868 .5 × 16 × 9062 ) + (868 .5 × 44 × 6.02 ) = 363728 bbl

After 182.6 days

Step3: water influx after 273.9 days: Time/day

tD

¨P

weD

B¨P weD

273.9

22.3

16

13.333

185275

182.6

14.88

44

9.62

367618

91.3

7.44

65

6.02

339844

We=89273 bbl

152

Step4: water influx after 365.2 days: Time/day

tD

¨P

weD

B¨P weD

365.2

29.8

16

16.5

229284

273.9

22.3

44

13.333

509507

182.6

14.88

65

9.62

543073

91.3

7.44

80

6.02

418269

We=1700133 bbl

Step5: water influx after 456.5 days: Time/day

tD

¨P

weD

B¨P weD

456.5

37.2

16

19.7

273751

365.2

29.8

44

16.5

630531

273.9

22.3

65

13.33

752511

182.6

14.88

80

9.62

668379

91.3

7.44

90

6.02

47553

we = 2795743bbl

PROBLEM 8.10 Repeat Prob. 8.9 using the Fetkovich method, and compare the results with the results of Prob. 8.9.

153

ANSWER: Time

Pr

(Pr)n

(Pa)n-1

(Pa)n-1-(Pr)n

(¨we)n

we

0

4020

4020

4020

0

0

0

1

3988

4004

4020

16

80600

80600

2

3932

4960

4006

64

235493

316093

3

3858

3895

3967

72

364794

680887

4

3772

3815

3908

93

467542

1148429

5

3678

3725

3833

108

545409

1693838

Fetkovich method: ra =

43560(150000)

re =

43560(1000)

π π

= 14421 ft

= 3723 ft

ra = 3.87 re

rD =

Step 1:

π

π

(ra2 − re2 )hφ = (144212 − 37232 )40(0.20) 5.615 6.615 = 868811838.4

wi =

Step 2:

wei = ct wi p i f = (7 × 10 −6 ) wi (4020)

360 = 24.45 × 106 bbl 360

Step 3:

J=

0.00708khf 0.00708(200)(40) = = 117.4 μw[Ln rD − 0.75] 0.8[Ln (3.87) − 0.75]

154

Jp i 117 .4( 4020 ) = = 0.0192 24.45 × 10 6 we i ª Jp i º 1 − Exp « − Δ t » = 1 − Exp [− 0 .0192 (91 .3) ]0 .828 ¬ we i ¼

Step 4: ( Δ we ) n =

=

ª º we i [( p a ) n −1 − ( p r ) n ]«1 − Exp ( − Jp i ) Δ t » pi we ¬ ¼ i

24.45 × 106 [( pa) n−1 − ( pr) n ]0.828 4020

(Δwe) n = 5037.5[( pa) n−1 − ( pr ) n ] _

( p a ) n −1 = ( p a ) n (1 −

(Δwe) n ) wei

155

CHAPTER 9

THE DISPLACEMENT OF OIL AND GAS

PROBLEM 9.1 (a) A rock 10 cm long and 2 sq cm in cross section flows 0.0080 cu cm/sec of a 2.5 cp oil under a 1.5 atm pressure drop. If the oil saturates the rock 100%, what is its absolute permeability? (b) What will be the rate of 0.75 cp brine in the same core under a 2.5 atm pressure drop if the brine saturates the rock 100%? (c) Is the rock more permeable to the oil at 100% oil saturation or to the brine at 100% brine saturation? (d) The same core is maintained at 40% water saturation and 60% oil saturation. Under a 2.0 atm pressure drop, the oil flow is 0.0030 cu cm/sec and the water flow is 0.004 cu cm/sec. What are the effective permeabilities to water and to oil at these saturations? (e) Explain why the sum of the two effective permeabilities is less than the absolute permeability. (f) What are the relative permeabilities to oil and water at 40% water saturation? (g) What is the relative permeability ratio ko/kw at 40% water saturation? (h) Show that the effective permeability ratio is equal to the relative permeability ratio.

156

ANSWER: a)

k=

qμl 0.008(2.5)(10) = = 0.067darcy == 67md AΔp 2(1.5)

b) qw =

3 kAΔp 0.067 ( 2)( 2.5) = = 0.0447 cm sec μl 0.75(10)

c) Absolute permeability is one of the properties of rock and it is not related to the fluid type. d)

kw =

q w μl 0.0040(0.750)(10) = = 0.0075 darcy = 7.5 md AΔp 2(2)

ko =

qo μ o l 0.0030(2.5)(10) = = 0.01875darcy = 18.75md AΔp 2(2)

e) Effective permeability is the ability of rock to transmit one specific fluid. When fluid saturation is less than 100% of pore volume, then the two effective permeabilities is less than the absolute permeability. f)

k rw =

k w 0.0075 k 0.01875 = = 0.112, kr° = ° = = 0.28 k k 0.067 0.067

g)

kro 0.28 = = 2.5 krw 0.112 h) ko 0.01875 = = 2.5 k w 0.0075

157

PROBLEM 9.2 The following permeability data were measured on a sandstone as a function of its water saturation: Sw 0

10

20

30*

40

50

60

70

kro 1.0

1.0

1.0

0.94 0.80 0.44 0.16 0.045 0

krw 0

0

0

0

0.04 0.11 0.20 0.30

75*

80

90

100

0

0

0

0.36 0.44 0.68 1.0

*Critical saturations for oil and water a) Plot the relative permeabilities to oil and water versus water saturation on Cartesian coordinate paper. b) Plot the relative permeability ratio versus water saturation on semilog paper. c) Find the constants a and b in Eq. (9.3) from the slope and intercept of your graph. Also find a and b by substituting two sets of data in Eq. (9.3) and solving simultaneous equations. d) If µo = 3.4 cp, µw = 0.68 cp, Bo = 1.50 bbl/STB, and Bw = 1.05 bbl/STB, what is the surface water cut of a well completed in the transition zone where the water saturation is 50%? e) What is the reservoir cut in part (d)? f) What percentage of recovery will be realized from this sandstone under highpressure water drive from that portion of the reservoir above the transition zone invaded by water? The initial water saturation above the transition zone is 30%. g) If water drive occurs at a pressure below saturation pressure such that the average gas saturation is 15% in the invaded portion, what percentage of recovery will be realized? The average oil volume factor at the lower pressure is 1.35 bbl/STB and the initial oil volume factor is 1.50 bbl/STB. h) What fraction of the absolute permeability of this sandstone is due to the least permeable pore channels that make up 20% of the pore volume? What fraction is due to the most permeable pore channels that make up 25% of the pore volume?

158

ANSWER: a) The solution of this problem is left to readers. b) The solution of this problem is left to readers. c)

kro = ae−bsw krw

20 = ae−0.4 Sw → a = 13629.6 0.15 = ae−0.7 sw → b = 16.3 d)

at Sw = 50% →

Ÿ fw =

k ro 0.44 = =4 k rw 0.11

1 1 qw = = = 0.64 qw + qo 1 + ko × μ w Bw 1 + 4( 0.68 )(1.05 ) k w μo B 3.4 1.50

e) fw =

1 1 = = 0.56 ko μ w 0.68 1+ 1 + 4( ) kw μo 3 .4

f)

Recovery =

0.85 − 0.30 = 0.643 0.85

g) S g = 0.15

B = 1.50 bb1

sTB

Assume: Sw = 0.75 1 − s w − s g Boi 1 − 0.75 − 0.15 1.50 Recovery = 1 − ( × ) =1− ( ) Bo 1 − sw 1 − 0.75 1.35

Recovery= 0.55or recovery= 55% 159

h) 0% and 46%.

PROBLEM 9.3 Given the following reservoir data:

Throughput rate = 1000bbl/day Average porosity = 18% Initial water saturation = 20% Cross-sectional area = 50,000 ft2 Water viscosity = 0.62 cp Oil viscosity = 2.48 cp ko/kw versus Sw data in Fig. 9.1 and 9.2

Assume zero transition zone: a) Calculate fw and plot versus Sw. b) Graphically determine ˜fw/˜Sw at a number of points, and plot versus Sw. c) Calculate ˜fw/˜Sw at several values of Sw using Eq. (9.17), and compare with the graphical values of part (b). d) Calculate the distances of advance of the constant saturation fronts at 100, 200, and 400 days. Plot on Cartesian coordinate paper versus Sw. Equalize the areas within and without the flood front lines to locate the position of the flood fronts. e) Draw a secant line from Sw = 0.20 tangent to the fw versus Sw, curve in part (b), and show that the value of Sw at the point of tangency is also the point at which the flood front lines are drawn. f) Calculate the fractional recovery when the flood front first intercepts a well, using the areas of the graph of part (d). Express the recovery in terms of (1) the initial

160

oil in place and (2) the recoverable oil in place (i.e., recoverable after infinite throughput). g) To what surface water cut will a well rather suddenly rise when it is just enveloped by the flood fronts? Use Bo = 1.50 bbl/STB and Bw = 1.05 bbl/STB. h) Do the answers to parts (f) and (g) depend on how far the front has travelled? Explain.

ANSWER: fw =

a)

1 1 for example → for S w = 50% Ÿ f w = = 0.686 ko ȝw 0.22 0.62 1+ 1+ ( )( ) kw ȝo 0.12 2.48

Sw

0.3

0.4

0.5

0.6

0.7

0.8

0.85

Ko/Kw

25

8.8

1.84

0.59

0.14

0.03

0

fw

0.138

0.312

0.685

0.871

0.966

0.99

1

Then draw Sw versus fw. b) for example S w = 50% Ÿ

f − f w1 δf w = w2 = 2 .8 δS w s w 2 − s w1

c)

δfw ( μw / μo)bae − bsw = δsw [1 + ( μw / μo)ae −bsw ]2

& a, b from figure 9.2

Sw

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

įfw/ įSw

0.526

1.56

3.04

2.8

1.284

0.414

0.118

0.0327

d) x=

5.615q′t δf w ( ) sw φAc δS w

161

for t = 100 days Ÿ x100 =

δf 5.615(1000)(100) δf w ( ) sw = 62.4( w ) sw δS w δS w 0.18(5000)

for t = 200 day Ÿ x200 = 124 .8( for t = 400 day Ÿ x400 = 249(

δf w ) δS w sw

δf w ) δS w sw

e) The solution of this problem is left to readers. f) The answers are 0.69 and 0.56. g)

fw =

1 1 = = 0.851 0.62 1.05 ko μ w Bw 1+ 1+ ( )( ) 2.48 1.50 k w μo Bo

PROBLEM 9.4 Show that for radial displacement where rw << r

r =[

5.615q δf w 1 / 2 ( )] πφh δS w

where r is the distance a constant saturation front has travelled.

ANSWER: The solution of this problem is left to readers.

162

PROBLEM 9.5 Given the following reservoir data: Sg

10*

15

20

25

30

35

40

45

50

kgko

0

0.08

0.20

0.40

0.85

1.60

3.00

5.50

10.0

kro

0.70

0.52

0.38

0.28

0.20

0.14

0.11

0.07

0.04

62*

0

*Critical saturations for gas and oil.

Absolute permeability = 400 md Hydrocarbon porosity = 15% Connate water = 28% Dip angle = 20º Cross-sectional area = 750,000 ft2 Oil viscosity = 1.42 cp Gas viscosity = 0.015 cp Reservoir oil specific gravity = 0.75 Reservoir gas specific gravity = 0.15 (water = 1) Reservoir throughput at constant pressure =10,000 bbl/day

a) Calculate and plot the fraction of gas, fg, versus gas saturation similar to Fig. 9.13 both with and without the gravity segregation term. b) Plot the gas saturation versus distance after 100 days of gas injection both with and without the gravity segregation term. c) Using the areas of part (b), calculate the recoveries behind the flood front with and without gravity segregation in terms of both initial oil and recoverable oil.

163

ANSWER: a) fg =

1 − 2.115 k ro 1 + 0.0106 k o / k g

With gravity drainage:

Sg

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.5

0.62

fg

0

0.0881

0.186

0.397

0.569

0.669

0.764

0.85

0.914

Without gravity drainage: Sg

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.5

0.62

fg

1

0.883

0.949

0.974

0.987

0.993

0.996

0.998

0.998

b)

x=

5.615q′t δf g ( ) φAc δS g sg

x100 =

δf 5.615(10000)(100) δf g ( ) sg = 49.9( g ) sg 0.15(750000) δS g δS g įfg/įSg

X(without gravity)

įfg/įSg

X(with gravity)

0.188

9.38

2.02

101.14

0.096

4.8

1.63

81.4

0.055

2.74

1.29

64.5

0.03

1.5

0.99

49.88

0.018

0.9

0.78

38.94

0.011

0.55

0.676

33.73

0.0072

0.36

0.534

26.66

c) The answers are 80.8, 50.1. 28.6, and 17.7. 164

PROBLEM 9.6 Derive an equation including a gravity term similar to Eq. (9.23) for water displacing oil.

ANSWER: qo = fw =

− Kk ro

μo

Aρ o

1

δP ρ o g sin α δφ o − Kk ro = ) Aρ o ( o + δx 1.0133 × 10 6 δx μo

ª Kk ro A δPc Δρg sin α º 1+ ( − ) q ′μ o δx 1.0133 × 10 6 »¼

μ k « 1 + w ro ¬ μ o k rw

Field units: fw =

1

ª º Kk ro A 1 + 1.127 ( −0.433γ sin α ) » q ′μ o ¼

μ k « 1 + w ro ¬ μ o k rw

Equation for water displacing oil with gravity drainage: fw =

1 − 0.488KA( ρ w − ρ o )(sin α )k ro /( μ o q ′) 1 + (k o / k w )(

μw ) μo

PROBLEM 9.7 Rework the water displacement calculation of Table 9.1, and include a gravity segregation term. Assume an absolute permeability of 500 md, a dip angle of 45º, a density difference of 20% between the reservoir oil and water and an oil viscosity of 1.6 cp. Plot water saturation versus distance after 240 days, and compare with Figure 9.11.

165

Sw 20

30

40

50

60

70

80

90

kro 0.93

0.60

0.35

0.22

0.12

0.05

0.01

0

ANSWER: Displacement with gravity drainage=69% Displacement without gravity drainage=59%

PROBLEM 9.8 Continue the calculations of Ex. Prob. 9.1 down to a reservoir pressure of 100 psia using: a) Muskat method b) Schilthuis method c) Tamer method

ANSWER: The solution of this problem is left to readers.

166

CHAPTER 10

HISTORY MATCHING

PROBLEM 10.1 The following data are taken from a volumetric, undersaturated reservoir. Calculate the relative permeability ratio kg/ko at each pressure, and plot versus total liquid saturation. Connate water, Sw = 25% Initial oil in place = 150 MMSTB Boi = 1.552 bbl/STB

P psia

R

Np

SCF/STB MMSTB

Bo

Bg

Rso

bbl/STB

bbl/SCF

SCF/STB

µ o/µ g

4000

903

3.75

1.500

0.000796

820

31.1

3500

1410

13.50

1.430

0.000857

660

37.1

3000

2230

20.70

1.385

0.000930

580

42.5

2500

3162

27.00

1.348

0.00115

520

50.8

2000

3620

32.30

1.310

0.00145

450

61.2

1500

3990

37.50

1.272

0.00216

380

77.3

ANSWER:

R = Rso +

k g μo Bo ko μ g Bg

167

for p = 4000 psia Ÿ 903 = 820 +

kg k 1. 5 (31.1) → g = 1.42 × 10 −3 0.000796 ko ko

for p = 3500 psia Ÿ

kg = 1.21 × 10 −2 ko

for p = 3000 psia Ÿ

kg = 2.60 × 10 −2 ko

for p = 2500 psia Ÿ

kg = 4.43 × 10 −2 ko

for p = 2000 psia Ÿ

kg = 5.73 × 10 −2 ko

for p = 1500 psia Ÿ

kg = 0.079 ko

S L = S w + (1 − S w )(1 −

N p Bo ) N Boi

for p = 4000 psia Ÿ S L = 0.25 + (1 − 0.25)(1 −

3.75 1.5 ) = 0.957 150 1.552

for p = 3500 psia Ÿ S L = 0.879 for p = 3000 psia Ÿ S L = 0.827 for p = 2500 psia Ÿ S L = 0.784 for p = 2000 psia Ÿ S L = 0.747

for p = 1500 psia Ÿ S L = 0.711

P(psia)

4000

3500

3000

2500

2000

1500

Kg/Ko

0.00142

0.0121

0.026

0.0443

0.0573

0.079

SL (%)

95.7

87.9

82.7

78.4

74.7

71.1

168

PROBLEM 10.2 Discuss the effect of the following on the relative permeability ratios calculated from production data: a) Error in the calculated value of initial oil in place. b) Error in the value of the connate water. c) Effect of a small but unaccounted for water drive. d) Effect of gravitational segregation both where the high gas-oil ratio wells are shut in and where they are not. e) Unequal reservoir depletion. f) Presence of a gas cap.

ANSWER: The solution of this problem is left to readers.

PROBLEM 10.3 For the data that follow and that are given in Figs. 10.15 to 10.17 and the fluid property data presented in the chapter, perform a history match on the production data in Figures 10.18 to 10.21, using the computer program in Table 10.1. Use the new permeability ratio data plotted in Figures 10.8 and 10.13 to finetune the match.

Laboratory core permeability measurements Average Absolute Permeability to Air (md) Well

Zone 1

Zone 2

5-6

5.1

4.0

8-16

8.3

6.8

9-13

11.1

6.0

14-12

8.1

7.6

169









PROBLEM 10.4 Write a computer program that uses the Muskat method discussed in Chapter 9 in place of the Schilthuis method used in Chapter 10 to perform the history match on the data in Chapter 10.

ANSWER: The solution of this problem is left to readers.





PROBLEM 10.5 Write a computer program that uses the Tamer method discussed in Chapter 9 in place of the Schilthuis method used in Chapter 10 to perform the history match on the data in Chapter 10.

ANSWER: The solution of this problem is left to readers.

173

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www.get-morebooks.com Kaufen Sie Ihre Bücher schnell und unkompliziert online – auf einer der am schnellsten wachsenden Buchhandelsplattformen weltweit! Dank Print-On-Demand umwelt- und ressourcenschonend produziert.

Bücher schneller online kaufen

www.morebooks.de VDM Verlagsservicegesellschaft mbH Heinrich-Böcking-Str. 6-8 D - 66121 Saarbrücken

Telefon: +49 681 3720 174 Telefax: +49 681 3720 1749

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