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PGE 312 Physical and chemical Behavior of Fluids
Home Work 9 Solutions
Problem 92
Given: Cumulative Oil Production,
MSTB
260
455
761
1024
1319
1699
1910
2201
2400
2607
2800
2959
3111
3231
3360
3480
3591
3676
3799
3890
3992
4039
4110
4205
4275
4317
4390
4435
4500
4540
4600 Producing gasoil ratio,
scf/STB
310
322
286
305
317
402
434
515
575
610
661
602
723
769
848
911
974
990
1092
1044
1143
1170
1318
1271
1407
1485
1494
1373
1467
1305
1446 2
Separator Pressure = 30psig = 30+14.7 = 44.7psia
Separator Temperature = 120oF
Separator gas gravity = 0.888
Stock tank oil gravity = 36.1oAPI
Laboratory solution gasoil ratio = 318 scf/STB Solution:
Plotting the production history against cumulative oil production, we can deduce the solution gasoil
ratio as the producing gasoil ratio at bubble point. The plot is shown below Therefore, the solution gasoil ratio is 315scf/STB. 3
Adjusting the solution gasoil ratio for vented stock tank gas by using stock tank gasoil ratio correlation
in Fig. 93, we get
Rsb = 315 + 23 = 338 scf/STB.
The estimated solution gasoil ratio can be compared with the laboratory value as follows: Since the percentage error is less than 10%, we have obtained a very good estimate of the solution gasoil ratio. Problem 93
Given:
Reservoir pressure,
psia
3122
2982
1830
1275
965
1095
933
495 Cumulative oil production,
MSTB
110
370
1050
1400
3060
3160
3360
4620 Bubble point pressure measured in the lab = 1441psig = 1445.7psia. 4
Solution:
A plot of the pressure history against the cumulative oil production is shown below. The estimated bubble point pressure is 1480psia.
The estimated bubble point pressure can be compared with the laboratory value as follows: Since the percentage error is less than 10%, we have obtained a very good estimate of the bubble point
pressure. Problem 111
Stock tank oil gravity 40.3o API 5
Specific Gravity = 0.756
Solution gas oil ratio = 1000
Reservoir Temperature = 205oF
The bubble point pressure at the above stated values is found out from Fig. 111 as 3450psia Problem 115
Given:
Pressure = 14.696psia
Temperature = 60oF
Ideal hydrocarbon liquid of known composition
Solution:
Component Composition
mole fraction,
lb mol
xj Molecular
weight, lb/lb mol
Mj 0.025
0.325
0.650
1.000 44.1
58.1
72.2 Propane
iButane
nPentane
Σ at 60oF and 14.696psia Problem 117
Given:
Bubble point pressure, pb= 3385psia
Temperature = 205oF
Specific gravity of heptanes plus γo = 0.85 = 53.04 Mass, lb
xjMj 1.1025
18.8825
46.9300
66.9150 lb Liquid density
at 60oF and
14.696psia,
lb/cuft
ρoj
31.64
35.12
39.36 Liquid volume
at 60oF and
14.696psia,
cu ft
xjMj/ ρoj
0.0349
0.5377
1.1923
1.7648 cu ft 6
Solution:
We use a first trial value of 43.70lb/cu.ft, which corresponds to 0.70g/cc. The apparent liquid density of
methane = 0.32g/cc (19.98 lb/cu.ft) and apparent liquid density of ethane = 0.47g/cc (29.34 lb/cu.ft)
Component methane
ethane
propane
ibutane
nbutane
ipentane
npentane
Hexanes
Heptanes plus Composition
mole fraction
xj
0.4642
0.0637
0.0561
0.0119
0.0383
0.0168
0.0195
0.0285
0.3010
1.0000 Molecular
weight,
lb /lb mole
Mj
16.043
30.070
44.097
58.123
58.123
72.150
72.150
86.177
202.000 Mass,
lb
xjMj
7.4472
1.9155
2.4738
0.6917
2.2261
1.2121
1.4069
2.4560
60.8020
80.6313 lb Liquid Density
60oF & 14.7 psia
lb/cu.ft
ρoj
19.98
29.34
31.62
35.11
36.42
38.96
39.36
41.40
53.04 Liquid Volume 60oF
& 14.7 psia
cu.ft
xjMj /ρoj
0.3727
0.0653
0.0782
0.0197
0.0611
0.0311
0.0357
0.0593
1.1463
1.8696 cu.ft The first calculated value of pseudoliquid density =
For a second trial value we use 46.82 lb/cu.ft, which corresponds to 0.75g/cc. Component Methane
Ethane
Propane plus Mass,
lb
xjMj Liquid Density
60oF & 14.7 psia
lb/cu.ft
ρoj
7.4472 21.21
1.9155 30.24
71.2686
80.6313 lb The second calculated value of pseudoliquid density = Liquid Volume 60oF
& 14.7 psia
cu.ft
xjMj /ρoj
0.3511
0.0633
1.4316
1.8460 cu.ft 7 The pseudoliquid density from the above plot ρpo = 43 lb/cu. ft at 14.7 psia and 60oF
Compressibility adjustment = + 1.3 lb/cu ft at 60oF and 3385psia, Fig. 113
ρo = 43 + 1.3 = 44.3 lb/cu. ft at 60oF and 3385psia
Thermal expansion adjustment = 4.1lb/cu ft at 205oF, Fig. 114
ρo = 44.3 – 4.5 = 39.80lb/cu. ft at 205oF and 3385psia. Problem 118
We use the table from the previous problem
Density of propane plus, ρc3+ = = 49.78 lb/cu.ft Weight percent of methane in the mixture W1, percent = = Weight percent of ethane in the ethane and heavier components, W2 % = = 9.24%
= = 2.62%
Using W1 and W2 values in Fig. 116 we get,
at 14.7 psia and 60oF. = 0.858. Therefore, ρpo = 0.858 *49.78 = 42.71 lb/cu.ft 8
Compressibility adjustment = +1.3 lb/cu.ft. Therefore, ρpo = 42.71+1.3 = 44.01 lb/cu.ft at 60oF and
3385psia
Thermal Expansion adjustment = 4.1lb/cu.ft. Therefore, ρpo = 44.014.5=39.51 lb/cu.ft at 205oF and
3385psia
Error % with laboratory measurement = = 1.18% Problem 1113
Given:
Producing gasoil ratio, Rp = 589
Specific gravity of gas, γg = 0.95
Stock tank oil gravity = 39 oAPI
Bubble point pressure, pb = 1763psia
Temperature = 250oF
Weight percent of H2S = 3%
Solution:
Apparent molecular weight of the gas, Mg = 29 γg =29*0.95=27.55
From Fig. 118 using γg = 0.95 and 39 oAPI we get apparent density of dissolved gas ρg = 27.58 lb/cu.ft
Mass of gas = = 42.62 Specific gravity of stock tank oil Mass of stock tank oil = = 290.81 Mass of surface gas and stock tank oil = 290.81+42.62 = 333.43
Liquid volume = 5.615 + = 7.16 9 Pseudoliquid density, ρpo =
Compressibility adjustment = +0.52 lb/cu.ft. Therefore, ρpo = 46.56+0.52 = 47.08 lb/cu.ft at 60oF and
1763psia
Thermal Expansion adjustment = 5.5lb/cu.ft. Therefore, ρpo = 47.085.5=41.58 lb/cu.ft at 250oF and
1763psia
Density adjustment for hydrogen sulphide ρpo = 41.580.26=41.32 lb/cu.ft at 250oF and 1763psia Problem 1118
Solution:
Liquid density, ρpo = 41.32 lb/cu.ft at 250oF and 1763psia = 232.05 lb/res bbl
Formation volume factor of oil Bo =
Therefore, Bo = = 1.437 Problem 1124
Given:
Formation volume factor of oil at bubble point Bob = 1.255
Solution gasoil ratio at bubble point, Rsb = 410
Bubble point pressure, pb =2265psig = 2279.7psia
Temperature = 214oF
Gas specific gravity, γg = 0.811
Stock tank oil gravity = 23.7oAPI
Coefficient of isothermal compressibility of oil, co = 9 * 106 psi1 10
Solution:
Formation volume factor of oil, Bo = for p≥pb Therefore, Bo at 4070 psig = = 1.235 Problem 1131
Formation Volume factor of oil at bubble point Bob = 1.255
Solution gasoil ratio Rsb = 410
Reservoir Pressure Pb = 2265 psia
Reservoir Temperature = 214oF
Specific gravity of gas γg = 0.811
Stock tank gravity = 23.7oAPI
µoD from Fig. 1113 = 8cp; µob = 1.8cp; from Fig. 1115 µo = 2.0 cp ...
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This note was uploaded on 11/27/2011 for the course PGE 312 taught by Professor Peters during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Peters

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