pee 562 assignment 2
TRANSCRIPT
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%