Pe 04025 Reservoir i

MINISTRY OF SCIENCE AND TECHNOLOGY Department of Technical and Vocational Education B-Tech Year II

Final Examination Petroleum Engineering PE-04025 Reservoir Engineering II

Date:

.

. 2010

Time:

to

.

Answer any FIVE questions. 1. Calculate the water influx at the third and fourth quarter years of production for the

reservoir reservoir shown in following following figure. figure. Use φ = 0.209; k = 275 md (average reservoir   permeability, presumed the same for the aquifer); µ = 0.25 cp; c e = 6 × 10-6 psi-1; h = 19.2 ft; area of reservoir = 1216 ac; estimated area of aquifer = 250,000 ac; θ = 180°. (20-marks)

2.

(a) Explain linear flow of incompressible fluids steady state.

(10-marks)

(b) Given: Pressure differential = 100 psi

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Permeability = 250 md Fluid viscosity = 2.5 cp Length = 450 ft Cross-sectional area = 45 square feet Determine the flow rate.

(10-marks)

3. A sand body is 1500 feet long, 300 feet wide, and 12 feet thick. It has a uniform  permeability of 345 md that had been a gas reservoir with a BHT of 140 ° F. 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 MMSCF/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 fiver times the gas flow? (20-marks)

4.

A sand body is 1500 feet long, 300 feet wide, and 12 feet thick. It has a uniform  permeability of 345 md to oil at 17 per cent, connate water saturation. The porosity is 32  per cent. The oil has a reservoir viscosity of 3.2 cp and an FVF of 1.25 at bubble point. (a) If flow takes place above saturation pressure, what pressure drop will cause 100 reservoir barrels per day to flow through the sand body, assuming the fluid behaves essentially as and incompressible fluid? What for 200 reservoir BPD? (b) What is the apparent velocity of the oil in feet per day at the 100 BPD flow rate? (c) What is the actual average velocity? (d) What time will be required for complete displacement of the oil from the sand? (20-marks)

5.

Two wells are located 2500 ft apart. The static well pressure at the top of perforations ( 9332 ft 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 pressure 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?

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(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? (20-marks) 6.

The following data pertains to volumetric gas reservoir.  Net formation thickness = 15 ft Hydrocarbon porosity = 20 per cent Initial reservoir pressure = 6000 psia Reservoir Temperature = 190° F Gas viscosity = 0.020 cp Casing diameter = 6 inches Assuming ideal gas behavior (z = 1.00). Suppose four wells are drilled on the 640-acre unit and each is produced at 4.00MM CFD. For 6 md and 500 psia minimum flowing well pressure, calculate the recovery. (20-marks) -------------------------------------------- End of the Question------------------------------------

MINISTRY OF SCIENCE AND TECHNOLOGY Department of Technical and Vocational Education B-Tech Year II

Final Examination Petroleum Engineering PE-04025 Reservoir Engineering II Solution

1

Solution:

φ = 0.209; k = 275 md; µ = 0.25 cp; c e = 6 × 10-6 psi-1; h = 19.2 ft; area of reservoir = 1216 ac; estimated area of aquifer = 250,000 ac; θ = 180°.

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Reservoir is against a fault A = r w2 =

1 2

π r w2

1216 × 43,560 0.5 × 3.1416 (4-marks)

For t = 91.3 days (one quarter or one time period) t  D = 6.323 × 10 −3

t  D = 6.323 × 10 −3

kt  φµ c e r w2 275 × 91.3 0.209 × 0.25 × 6 × 10 −6 × (5807 ) 2

 B = 1.119 × φ  × c e × r w2 × h ×  B = 1.119 × 0.209 × 6 × 10

∆ p 3 = ∆ p 4 =

1 2 1 2

( p1 − p 2 ) = ( p 2 − p 4 ) =

1 2 1 2

−6

= 15.0

θ  360 2

× (5807 ) × 19.2 ×

180 360

(4-marks) = 455 bbl/psi

(3788 − 3748) = 20.0 psi (4-marks) (3774 − 3709) = 32.5 psi

Water influx at the end of the third quarter 

Water influx at the end of the fourth quarter 

(4-marks)

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W e (3 rd  quarter ) =  BΣ∆ p × Q(t ) = 455 × 416.0 = 189,300 bbl

(4-marks)

W e (4t h quarter ) =  BΣ∆ p × Q(t ) = 455 × 948.4 = 431,500 bbl

2.

Solution: (a) Following figure represents linear flow through a body of constant cross section, where both ends are entirely open to flow, and where no flow crosses the sides, top or bottom.

(4-marks)

If the fluid is incompressible, or essentially so for all engineering purposes, then the velocity is the same at all points, as is the total flow rate across any cross section, so that, v=

q  A

=

− 1.127

k  dp

(4-marks)

 µ  dx

Separating variables and integrating over the length of the porous body, q

l 

k 

 p 2

 A

∫ dx = −1.127 µ  ∫ dp

q

=

0

 p1

1.127kA( p 1 − p 2 )  µ  L

(4-marks)

In this integration q, µ  , and k  have been removed from the integral sign, assuming they are invariant with pressure. Actually, for flow above the  bubble point, the volume, and hence the rate of flow, will vary with the c ( p − p )  pressure as expressed by; V  = V i e . i

As the net overburden pressure is the gross less the internal fluid  pressure, a variation of permeability with pressure is indicated, particularly in the shallower reservoir, values at the average pressure may be used for  most purposes.

(4-marks)

(b) Solution:

Pressure differential = 100 psi

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Permeability = 250 md Fluid viscosity = 2.5 cp Length = 450 ft Cross-sectional area = 45 square feet q=

1.127kA( p 1 − p 2 )

 µ  L

=

1.127 × 0.25 × 45 × 100 2.5 × 450

(4-marks) = 1.127 bpd 3.

Solution:

L = 1500 ft

A = 300 × 12 = 1600 ft 2

K = 345 md

S w = 17%

φ = 32%

µ = 0.023 cp

B = 1.25

p 1= 2500 psia

z = 0.88

T = 140 ° F

q = 5 MMSCF/day

(a)

Downstream pressure, p2 = ?

(4-marks) (4-marks) P2= 2365 psia (b)

Downstream pressure at 25 MMSCF /day, P 2 = ? (4-marks)

(4-marks) P2= 1724 psia (c)

Gas flow is directly proportional in pressure drop. (4-marks)

4.

Solution:

L = 1500 ft

φ = 32%

A = 300 × 12 = 1600 ft 2

µ = 3.2 cp

K = 345 md

S w = 17%

B = 1.25

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(a) q =

1.127kA∆ p

 µ  L

=

1.127 × 345 × 10 −3 × 3600 × ∆ p 3.2 × 1500

(if q = 100 bpd)

100= 0.2916 ∆ p

∆ p = 342.92 ≅ 343 psi q=

1.127kA∆ p

 µ  L

=

1.127 × 345 × 10 −3 × 3600 × ∆ p 3.2 × 1500

(if q = 200 bpd)

200= 0.2916 ∆ p (5-marks)

∆ p = 685.8 ≅ 686 psi (b) Apparent velocity, v

100

q =

=

 A

3600

= 0.0278 bpd/ft 2 × 5.615 (5-marks) = 0.156 ft/day

(c) Actual average velocity =

apparentvelocity

φ  × S o

=

0.156 0.32 × (1 − 0.17)

= 0.587 ft/day (d) Time (displacement of oil) =

5.

1500 0.587

(5-marks) (5-marks)

= 2555 days

Solution:

Reservoir fluid gradient = 0.25 psi/ft

Well A

2500 ft

Well B

Reservoir permeability = 245 md Fluid viscosity = 0.63 cp Datum Level = 9100 ft

9332 ft Datum 9100 ft

9672 ft

(a)

pA at 9100 ft datum = 4365 – (232 × 0.25) = 4307 psia  pB at 9100 ft datum = 4372 – (572 × 0.25) = 4229 psia

(4-marks)

(b) The difference of 78 psia indicates that fluid is moving down dip, from Well A to Well B.

(4-marks)

(c) Average effective pressure gradient, ∆ p/∆s = 78/2523 = 0.0309 psi/ft

(4-marks)

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(d)

Fluid velocity, v =

q  A

= 1.127

k  dp

= 1.127 ×

245 × 10 −3

× 0.0309 0.63 = 0.01354 bpd/sq-ft × 5.615 = 0.076 ft/day

 µ  ds

(e) Take the positive direction from Well A to Well B. Then, θ = 97° 44′ , and cos θ = cos 97° 44′ = -0.01347 6.

(4-marks)

(4-marks)

Solution:

h = 15 ft

φ = 20 per cent

Pi = 6000 psia

T = 190° F

µ = 0.020 cp

d casing = 6 inches

k = 6 md

z = 1.00

qsc = 4.00 MM CFD

P w = 500 psia

A = 640 acres

Suppose 4 wells are drilled on the 640 acre unit.

πr e2 = (640/4) × 43560 r e = 1489 ft

(4-marks)

 pi

= π r e2 hφ  Hc

G

= π  × (1489) 2 × 15 × 0.2 ×

 p sc

×

T  sc

G

T r  6000 14.7

×

520 650

= 6.83 × 109 SCF (initial gas in place)

  q t     p avg  = 1 −  sc   pi   G  

(4-marks)

  4 × 10 6 t      = 1 − 9   6 . 83 10 ×    

= 6000 – 3.514t

703kh

 µ T  ln(0.472 r e r w )

q sc

=

=

703 × 6 × 10 −3 × 15 0.02 × 650 ln(0.472 × 1489 (3 12))

= 0.6129 (4-marks)

  q sc t   2 2 2  703kh 1 −   ( pi −  p w ) G     

4 × 10 6

 µ T  ln(0.472 r e r w )

= 0.6129 ( 6000 − 3.514t ) 2 − (500) 2

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t = 967 days = 2.65 years (4-marks) Recovery for 2.65 years =

q sc t  G

=

4 × 10

× 967 = 0.5663 = 56.63 % 6.83 × 10 9 6

(4-marks)

-------------------------------------------- End of the Solutions ------------------------------------

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