Engineering Experts only........No plagairism plz

profileJim_Ak
hw_27.docx

Petr 3520 HW 27

Gas MBE Average P Calculation

Avg P Determination Reservoir Map Four Wells 1b

Avg P Determination BHP vs

A gas reservoir is being drained by four wells (A, B, C, D), with the following reservoir properties:

Well

Area [acres]

Height [ft]

Porosity []

A

20

50

.15

B

18

60

.16

C

24

70

.17

D

26

65

.14

For material balance calculations, an average pressure is needed. All four producing wells are shut in and the pressures are allowed to buildup for 60 hours. Cartesian plots of bottom hole pressure (BHP) vs. shut-in time are given above.

1. Read the approximate stabilized pressure at 60 hrs for each of the four wells. Consider these pressures to be the average pressure in the individual areas drained by the wells. (Note: In Pet 341 Well Testing you will learn about more sophisticated ways of determining average pressure in a well’s drainage area from pressure transient tests, but this is an easy and quick and (in many cases) acceptable method).

Well

Stabilized Pressure [psi]

A

B

C

D

2. From these four pressures, determine the arithmetic average pressure for the entire reservoir. [p = 3193.8 psi] (depends on how you read the stabilized pressures)

3. From these four pressures, determine the pore volume weighted average pressure for the entire reservoir.

[p = 3198.2 psi] (depends on how you read the stabilized pressures)

(Note: This is an especially easy calculation in Excel. If you use Excel, paste your calculations table below).

, where Vp = Pore Volume = A·h· (Units cancel, so OK to express Vp in acre-ft)

Sample Excel Calculation:

Gas Material Balance Equations

GBgi = (G-Gp)Bg

4. A gas reservoir, discovered with an initial pressure of 7000 psia, has an average pressure of 6450 psia when 1,170 MCF of gas has been produced. Please determine G. [Answer: G = 47.72 MMSCF ]

P

Z

Bg

Cumulative Gp

(psia)

(RVB/MCF)

(MCF)

7000

1.214

0.5241

0

6450

1.147

0.5373

1,170

5. A gas reservoir, discovered with an initial pressure of 5800 psia, has the following values of average pressure vs. Gp. Please do the following:

(a) Using Excel, calculate p/Z and plot p/Z vs. Gp. Print the plot only and attach to your HW.

(b) Determine G. [ Answer: G = 41.3 MMSCF ]

(c) If the reservoir is produced to an abandonment pressure of 500 psia (Z = 0.943), what will be Gp? [36.2 MMCF] What will be the recovery (Gp/G)? [87.59%]

P

Z

Bg

Cumulative Gp

(psia)

(RVB/MCF)

(MCF)

5800

1.03

0.574

0

5400

0.98

0.593

1.17

5000

0.972

0.631

3.4

4600

0.93

0.652

5.2

6. Derive Bg = 5.035zT/p.

7. (a) Explain this equation in your own words: GBgi = (G-Gp)Bg . (b) What assumptions are made in writing this? (c) What is a “volumetric” gas reservoir? (d) What is the theoretical recovery from a volumetric gas reservoir?

8. Starting with GBgi = (G-Gp)Bg , derive the p/Z vs. Gp form of the gas material balance.

9. (a) Draw a typical p/Z vs. Gp plot. (b) Explain how to get G. (c) What is “abandonment pressure”? (d) How can recovery be increased by adding compression at the surface?

238

Reservoir Average Pressure

(for Gas Material Balance)

WellArea (A)Height (h)

f

PVAvg PPV*P

[acres][acre-ft]in area

A

14300.12

50.42100105,840

B

22400.14

123.22140263,648

C

26500.16

2082120440,960

D

30550.10

1652160356,400

Total PV=546.61,166,848

[acre-ft]

Avg P =2130 (psi)

PV weighted avg Pressure:2134.74 (psi)

ú

ú

û

ù

ê

ê

ë

é

-

=

g

gi

p

B

B

G

G

1

p

i

i

i

i

G

G

Z

p

Z

p

Z

p

ú

û

ù

ê

ë

é

-

=

(

)

(

)

(

)

å

å

=

p

i

p

V

p

V

p

v