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PGE 312 – Physical and Chemical Behavior of Fluids
Homework 7 – Solutions Problem 84
Given:
API Gravity = 47.3 oAPI
Solution:
The API gravity is given by equation 82
o API = Therefore,
The specific gravity of the stock tank liquid is 0.7914 Problem 86
Given:
Density = 56.4
Solution:
o API = Liquid specific gravity,
Therefore, oAPI = Problem 89
Volume of oil at reservoir conditions = 86.3 res bbl/d
Volume of oil in stock tank = 57.9 STB/d
Total volume of gas at evolved at surface = 43150+7240 = 50390 scf/d 2 Formation volume factor, Bo = Therefore, Bo = = 1.491 Solution gas oil ratio Rs = Therefore, Rs = = 870.29 Since the Solutiongas oil ratio (Rs) is constant at pressures above the bubble point pressure,
the value calculated applies also at bubble point. Problem 810
Pressure
p
psig
5000
4500
4000
3500
3400
3300
3200
3100
3054
2700
2400
2100
1800
1500
1200
900
600 Oil Formation
Volume gas–oil factor
Bo
res bbl/STB
1.498
1.507
1.517
1.527
1.530
1.532
1.534
1.537
1.538
1.484
1.441
1.401
1.361
1.323
1.287
1.252
1.215 Solution gasoil ratio
Rs
scf/STB Gas Formation
volume factor Bg
res bbl/scf 941
941
941
941
941
941
941
941
941
819
732
646
562
481
400
321
240 0.000866
0.000974
0.001090
0.001252
0.001475
0.001795
0.002285
0.003108
0.004760 3 The shapes of the Bo vs p and Rs vs p plots match the shapes in Fig. 81 and Fig. 82,
respectively. Above the bubble point pressure, Bo decreases with an increase in pressure as the
single phase oil is supercompressed. The solution gas oilratio on the other hand is unaffected
by increase in pressure because above the bubble point, the volume of gas dissolved in the oil is
constant. 4 For all pressures below the bubble point, both the formation volume factor and the solution
gasoil ratio decrease with decrease in pressure due to the evolution of gas from the oil. Problem 811
The solution gasoil ratio at bubble point Rsb = 941 scf/STB (from table)
Total formation volume factor
Bt = Bo + (Rsb – Rs)* Bg  (1) At 2400 psig
Bo = 1.441 , Rs = 732 , Bg = 0.001090 Therefore, Bt = 1.441+(941732)*0.001090 = 1.669 Problem 812
Since the pressure 3500 psig is above the bubble point pressure of 3054 psig, the volume of gas
evolved in the reservoir pore space is 0. Therefore, Rs = Rsb. Thus, the total formation volume
factor is equal to the formation volume factor of oil Bo.
Bt = Bo = 1.527 Problem 813
Given:
Temperature: 300oF
Pressure
psig
5000
4500
4000
3500
3000
2500
2200
1972 = pb Oil volume
cc
219.80
220.55
221.33
222.17
223.07
223.99
224.57
225.05 5 Solution:
The coefficient of isothermal compressibility of oil as a function of pressure at constant
reservoir temperature at pressures above the bubble point is given by equation 8 8 on page
232 in the textbook. Therefore, plotting ln V vs. p we can obtain as the negative slope of the curve at specific
pressure, p. The dotted lines show the lines for slope calculation at different pressure. Pressure
psig (p)
5000
4500
4000
3500
3000
2500
2200
1972 = pb Oil volume
cc (V)
219.80
220.55
221.33
222.17
223.07
223.99
224.57
225.05 Oil volume
cuft
0.007763
0.007790
0.007817
0.007847
0.007879
0.007911
0.007932
0.007949 Ln V
4.8583
4.8549
4.8514
4.8476
4.8436
4.8395
4.8369
4.8347 (Slope) 6.00E6
6.00E6
6.00E6
6.50E6
7.50E6
1.25E5 6 Plotting the coefficient of compressibility of the oil as a function of the pressure above the
bubble point we observe that the trend is pretty close to the shape in Fig. 8 5 on page 232 in
the textbook. Problem 816
Given:
Pressure between 3500 – 4000 (Pressure range > Bubble point pressure Pb= 3054 psig)
The coefficient of isothermal compressibility at pressure greater than bubble point pressure of
the oil is given in equation 89. Integrating the above equation, assuming co remains constant with pressure we get 7 results in
 (1)
At p1 = 3500 psig, Bo1 = 1.527
values into equation (1) we get and p2 = 4000psig, Bo2 = 1.517 . Therefore, . Substituting these = 13.14 x 106 psi1 Problem 817
The coefficient of isothermal compressibility at pressure lesser than bubble point pressure of
the oil is given in equation 824 in book.
 (1)
At 2400 psig
Bo = 1.441, Bg = 0.001090, 8 The value of is the slope of the tangent in the above graph and is = 0.06/400 = 1.5x10 4 The value of is the slope of the tangent in the above graph and is = Therefore, substituting the values into equation (1) we get
= 113.0x106 psi1 = 0.2875 ...
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 Spring '08
 Peters

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