QUESTION 2.1

Given data;

dist. b/w injectors and producers=2000ft porosity=0.18

dist. b/w wells=625ft Bo=1.3rb/stb

Reservoir thickness=40ft Bw=1rb/stb

Flooding rate=750stb/day

Compare the times and displacement efficiencies at breakthrough, 10%, 20%, 50%, 80%, 90%, 99% water cut for the following cases;

Case 1- µo=50cp, µw=0.5cp Case 2- µo=10cp, µw=0.5cp Case 3-µo=0.4cp, µw=1cp

Sw Krw Kro Fw(Case 1) Fw( Case 2)

Fw(Case 3)

0.2 0 0.8 0 0 00.25 0.002 0.61 0.247 0.0615 0.0010.3 0.009 0.470 0.657 0.277 0.008

0.35 0.02 0.370 0.844 0.519 0.0210.4 0.033 0.285 0.921 0.698 0.044

0.45 0.051 0.22 0.959 0.823 0.0850.5 0.075 0.15 0.980 0.909 0.167

0.55 0.1 0.095 0.991 0.955 0.2960.6 0.132 0.06 0.995 0.978 0.468

0.65 0.170 0.03 0.998 0.991 0.6940.7 0.208 0.015 0.999 0.996 0.847

0.75 0.251 0.010 0.9996 0.998 0.9090.8 0.3 0 1 1 1

The values of fractional water flow, Fw for case1, 2 and 3 where obtained using the relation;

Fw= (1+ (Kroµw/Krwµo))

1

and the curves Krw, Kro vs Sw and Fw vs Sw have been plotted for the

several cases.

CASE 1; µo=50cp, µw=0.5cp

At breakthrough;

Swbt=0.29, fwbt=0.56, Swbt=0.35 (from plot)

L =

Qibt= = (1/7.33) = 0.136

tbt= (2000×625×0.18×40×0.136)/ (5.615×750)= 291.56days= 0.7988yr

Displacement efficiency= ED= = 0.1875; ED= 18.75%

Surface water cut at fwbt=0.56;

Fws = ; Bw= 1rb/stb; Bo= 1.3rb/stb

Fws=62.3% surface water-cut

All % water-cut below 62.3% have saturations below shock front saturation and at every time t they travel equal distance.

At 80% water cut;

0.8=

1

1 11 11.3 wef

; fwe=0.756 and corresponding Swe=0.32 (from plot)

15.556

0.38 0.2w

w

f

S

(2000)(625)(0.18)(40)384.65 1.0538

(5.615)(750)(5.556)t days yrs

5.615iwtbt (dfw/dSw)AØ

dfw

dSw

1

Swbt

Swbt-Swi

1-Swi

Swe

(1 )wewe we

w

w

fS S

f

S

1 0.7560.32 0.38

4.076weS

; Displacement efficiency, ED=

0.38 0.20.225

1 1 0.2we wi

wi

S S

S

Displacement efficiency at 80% water cut is 22.5%

At 90% water cut;

10.9

1 11 11.3 wef

; fwe=0.873 and corresponding Swe=0.36, 0.42weS (from

plot)

14.545

0.42 0.2w

w

f

S

(625)(2000)(0.18)(40)470.2 1.289

(5.615)(750)(4.545)t days yrs

Displacement efficiency, 0.42 0.2

0.2751 1 0.2we wi

Dwi

S SE

S

Displacement efficiency at 90% water cut is 27.5%

At 99% water cut;

10.99

1 11 11.3 wef

; Fwe= 0.987 and corresponding Swe=0.52, 0.6weS

12.5

0.6 0.2w

w

f

S

(625)(2000)(40)(0.18)854.9 2.342

(5.615)(750)(2.5)t days yrs

Swe

Swe

Swe

Displacement efficiency=0.6 0.2

0.51 0.2DE

; ED= 50%

CASE 2; µo=10cp, µw=0.5cp

At breakthrough;

Swbt=0-38, fwbt=0.64, Swbt=0.48 (from plot)

L =

Qibt= = (1/3.57) = 0.28

tbt= (2000×625×0.18×40×0.28)/ (5.615×750)= 598.397days= 1.639yr

Displacement efficiency= ED= = 0.35; ED= 35%

Surface water cut at fwbt=0.64;

Fws =

1

11 1W

O wbt

B

B f

; Bw= 1rb/stb; Bo= 1.3rb/stb

Fws=69.8% surface water-cut

All % water-cut below 69.8% have saturations below shock front saturation and at every time t they travel equal distance.

At 80% water cut;

0.8=

1

1 11 11.3 wef

; fwe=0.756 and corresponding Swe=0.42, 0.52weS (from

plot)

5.615iwtbt (dfw/dSw)AØ

dfw

dSw

1

Swbt

Swbt-Swi

1-Swi

13.125

0.52 0.2w

w

f

S

(2000)(625)(0.18)(40)683.88 1.874

(5.615)(750)(3.125)t days yrs

Displacement efficiency, ED=0.52 0.2

0.41 1 0.2we wi

wi

S S

S

Displacement efficiency at 80% water cut is 40%

At 90% water cut;

10.9

1 11 11.3 wef

; fwe=0.873 and corresponding Swe=0.47, 0.55weS (from

plot)

12.857

0.55 0.2w

w

f

S

(625)(2000)(0.18)(40)748.03 2.05

(5.615)(750)(2.857)t days yrs

Displacement efficiency, 0.55 0.2

0.43751 1 0.2we wi

Dwi

S SE

S

Displacement efficiency at 90% water cut is 43.75%

At 99% water cut;

10.99

1 11 11.3 wef

; Fwe= 0.987 and corresponding Swe=0.62, 0.7weS

12

0.7 0.2w

w

f

S

Swe

Swe

Swe

(625)(2000)(40)(0.18)1068.57 2.93

(5.615)(750)(2)t days yrs

Displacement efficiency=0.7 0.2

0.6251 0.2DE

; ED= 62.5%

CASE 3; µo=0.4cp, µw=1.0cp

At breakthrough;

Swbt=0.7, fwbt=0.86, Swbt=0.79 (from plot)

L =

Qibt= = (1/1.695) = 0.589

tbt= (2000×625×0.18×40×0.589)/ (5.615×750)= 1258.77days= 3.449yr

Displacement efficiency= ED= = 0.7375; ED= 73.75%

Surface water cut at fwbt=0.86;

Fws =

1

11 1W

O wbt

B

B f

; Bw= 1rb/stb; Bo= 1.3rb/stb

Fws=88.9% surface water-cut

All % water-cut below 88.9% water cut have saturations below shock front saturation and at every time t they travel equal distance.

5.615iwtbt (dfw/dSw)AØ

dfw

dSw

1

Swbt

Swbt-Swi

1-Swi

If we compare the times and displacement efficiencies of Case1, 2 and 3, we will be can safe concluding that case 3 is the most effective. Because Case 3 takes a longer time before breakthrough at the producer well and at the same times delivers the highest displacement efficiency.

QUESTION 2.2

Calculation of displacement efficiency for a vertical Gas flood in a high permeability reservoir a 200ft oil zone in a high permeability light oil reservoir is overlaid by a large gas cap gas injected at the cost of the gas cap to maintain reservoir pressure above the bubble point

The displacement of oil by gas is essential a 1-d displacement (downward)

Reservoir fluid properties

Ø= 0.35

Kv =500md

Bo=1.8RB/STb

µo= 0.5cp

µg=0.013cp

ℓo= 0.75gm/cm3

ℓg=0.12gm/cm3

Relative permeability curves are given as follows in the table

assumption made:

1.The displacement is esssntially a 1-d downward displacement

2.The capillary pressure is neglected

3.The effect of gravity is also neglected

Assuming: Fg= qg/(qg + qo)

Where fg is fractional flow of gas

4.Initial water saturation assumed to be zero

Sg kro Krg Kro/krg fg

0 1 0 ∞ 00.05 0.77 0 ∞ 00.08 0.66 0 ∞ 00.1 0.59 0.0003 1966.7 0.001950.15 0.44 0.0013 338.46 0.11360.2 0.32 .0062 51.61 0.4270.25 0.22 0.0172 12.79 0.750.3 .15 0.0359 4.18 0.9020.35 .1 0.0639 1.56 0.9610.4 .06 .10 0.60 0.9850.45 .037 .15 .25 0.9940.5 .02 .213 .094 0.9970.55 .01 .29 .034 0.9990.6 .004 .37 .011 0.9990.65 .0012 .47 .0026 10.7 .0004 .57 .0007 10.75 .00002 .68 .000029 10.8 0 .80 0 1

ED= displacement efficiency

ED=( soi –so)/(soi-sor)

Where

Sor=1 from the question given

So= 1-sg

ED= sg/(1-sor)

From the graph

Sg= 0.32

1-sor= 0.80

ED= So/(1-sor) = 0.32/.80

ED= 40%

Displacement efficiency is 40%