بنك الاسئلة EPS 1ST 2011 441

Problem #2:  Given the following data:

Volume of bulk oil zone = 112,000 acre-ft

Volume of bulk gas zone =  19,600 acre-ft

Initial reservoir pressure = 2710 psia

Initial oil FVF = 1.340 bbl/STB

Initial gas FVF = 0.006266 ft³/SCF

Initial dissolved GOR = 562 SCF/STB

Oil produced during the interval = 20 MM STB

Reservoir pressure at the end of the interval = 2000 psia

Average produced GOR = 700 SCF/STB

Two-phase FVF at 2000 psia = 1.4954 bbl/STB

Volume of water encroached = 11.58 MM bbl

Volume of water produced = 1.05 MM STB

Water FVF = 1.028 bbl/STB

Gas FVF at 2000 psia = 0.008479 ft³/SCF

a)  Calculate the stock tank oil initially in place.

b)  Calculate the driving indexes.

c)  Discuss your results.

Solution:

a)  The material balance equation is written as:

Define the ratio of the initial gas cap volume to the initial oil volume as:

we get:

and solve for N, we get:

Since:

N_p        = 20 x 10⁶ STB

B_t         = 1.4954 bbl/STB

R_p        = 700 SCF/STB

R_soi       = 562 SCF/STB

B_g        = 0.008479 ft³/SCF = 0.008479/5.6146 = 0.001510 bbl/SCF

W_e       = 11.58 x 10 ⁶ bbl

W_p       = 1.05 x 10⁶ STB

B_w        = 1.028 bbl/STB

B_ti        = 1.34 bbl/STB

m         = GB_gi/NB_ti = 19,600/112,000 = 0.175

B_gi        = 0.006266 ft³/SCF = 0.006266/5.6146 = 0.001116 bbl/SCF

Thus:

b)  In terms of drive indexes, the material balance equation is written as:

Thus the depletion drive index (DDI) is given by:

The segregation drive index (SDI) is given by:

The water drive index (WDI) is given by:

c)  The drive mechanisms as calculated in part (b) indicate that when the reservoir pressure has declined from 2710 psia to 2000 psia, 45% of the total production was by oil expansion, 31% was by water drive, and 24% was by gas cap expansion.

Problem #3: A plug, which has a length of 3.0 inches, a diameter of 0.75 of an inch, was 100% saturated with water and placed in a liquid permeameter. The water has a viscosity of 0.5 cp. The linear flow rate was measured at 2.75 cc/sec with a pressure drop of 2.4 atm:

A. Calculate the absolute permeability.

B. If the plug was 100% saturated with a 5 cp oil, calculate the volume flow rate when other parameters are kept the same.

Problem #4:  Given the following data:

Initial bulk reservoir volume = 415.3 MM ft³

Average porosity = 0.172

Average connate water saturation = 0.25

Initial pressure = 3200 psia

Initial gas FVF, B_gi = 0.005262 ft³/SCF

Final pressure = 2925 psia

Gas FVF at final pressure = 0.0057 ft³/SCF

Cumulative water production = 15,200 STB

Water FVF, B_w = 1.03 bbl/STB

Cumulative gas produced, G_p = 935.4 MM SCF

Bulk volume invaded by water at final pressure = 13.04 MM ft³

A)  Calculate initial gas in place.

B)  Calculate water influx, W_e.

C)  Calculate residual gas saturation, S_gr.

Solution:

A)  G = 415.3 x 10⁶ x 0.172 x (1 - 0.25) / 0.005262 = 10.18 MMM SCF

B)  Since:

Rearranging and solving for W_e, we get:

which yields:

W_e = 935.4x10⁶x0.0057 + 15,200x5.6146x1.03 - 10.18x10⁹x(0.0057-0.005262)

     = 960,400 ft³

C)  S_w = water volume/pore volume = V_w/V_p =

         (connate water + water influx - water produced)/pore volume =

         (13.04x10⁶x0.172x0.25 + 960,400 - 15,200x5.6146x1.03)/

         13.04x10⁶x0.172 = 0.64

         Therefore, S_gr = 1 - 0.64 = 0.36 or 36%.