ENCH 629 – Fall 2006.
Final Examination – Closed Book Closed Notes
Maximum Marks = 40
Time allowed = 90 minutes
Surname: _________________ Given Name(s):______________________ ID: ______________
Question 1 (8 Marks)
The mobility of a polymer solution through a porous medium can be described by Eq. 5.24 as:
λ p = λ*p u (1−n ) , c
where u is the Darcy velocity. Laboratory core flow tests were used to establish the following
correlations for a specific polymer solution in a target reservoir
k wp = 0.4k w1.2 (Where kw is permeability to water in m2)
λ*p = 0.2k wp
0.6
nc = 0.60
This polymer solution was then injected into a 0.30 m long core of 0.05 m diameter at constant
pressure drop across the core. Given that the core permeability to polymer-free water was 1.0 x
10-13 m2, calculate the flow rate of polymer solution through the core corresponding to 300 kPa
steady-state pressure drop across the core. Use SI units in your calculations.
Solution:
Darcy’s law applied to the flow situation gives
ΔP 300,000
u = λp = λp x = λ p x106 (1)
L 0.30
λ*p = 0.2k wp
0.6
= 0.2(0.4k w1.2 )0.6 = 0.2 × [0.4 × (10−13 )1.2 ]0.6 = 5.038 × 10−11
λ p = λ*pu (1− n ) = 5.038 × 10−11 u (1− 0.6 ) or
c
λ p × 1011
u 0.4
= or
5.038
u = (1.985 × 1010 × λ p )1 / 0.4 or
u = (1.985 × 1010 × λ p ) 2.5 (2)
Combining Eq. 1 and 2
Use the other side if you run out of space
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u = (1.985 × 1010 × λ p ) 2.5 = λ p x 106
⎛ 106 ⎞
(λ p )1.5 = ⎜⎜ ⎟ = 1.8 × 10− 20
10 2.5 ⎟
⎝ (1 . 985 × 10 ) ⎠
λ p = 6.76 × 10 −14
ΔP
From Eq. 1, u = λ p = λ p x106 = 6.76 × 10−14 × 106 = 6.76 × 10− 8 m / s
L
Cross sectional area of the core = πr 2 = π (0.025) 2 = 1.96 × 10−3 m 2
Flow rate is given by Q = u.A = 6.76 x 10-8 x 1.96 x 10-3 = 1.32 x 10-10 m3/s = 0.48 cm3/hr
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Question 2 (20 Marks)
Equal volumes (100 ml of each) of dry gas and oil compositions shown in the following phase
diagram (which has compositions shown in volume fractions) were mixed in a phase behavior cell.
After reaching equilibrium the liquid phase was withdrawn from the cell and the remaining gas
was mixed with equal volume (equal to the gas phase volume in the cell) of fresh oil. At
equilibrium this mixture had certain volume of liquid oil and some gas.
(a) Calculate the compositions and volumes of liquid and gas in the cell.
(b) If this process of withdrawing the liquid and mixing the remaining gas with fresh oil is
repeated many times, what will be the gas composition in the cell after a large number of
contacts with the fresh oil?
Solution
The mixture composition will lie on the line joining the oil composition and the dry gas
composition. When the volumes mixed are equal, the mixture composition is mid-way between
the two mixed compositions. The individual phase compositions will lie on the ends of the tie-
line passing through the mixture composition.
The relative phase volumes can be calculated using the phase rule as:
Gas A 40 mm 2
= = = , where A is the length along the tie line from the mixture point to
Total A + B 60 mm 3
the bubble point curve and B is the length from mixture point to the dew point curve.
Total volume = 200 ml, therefore gas volume = 200 x (2/3) = 133 ml.
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This gas will now be mixed with 133 ml of fresh oil. The mixture composition will be located at
the mid-point of the line joining this gas composition with the original oil composition. The
relative volumes are again given by the lever rule as:
Gas C 22 mm
= = = 0.595
Total C + D 37 mm
Here C is the length along the tie line from the new mixture point to the bubble point curve and
D is the length from this mixture point to the dew point curve.
Total volume = 133 + 133 = 266 ml.
Gas volume = 0.595 x 266 = 158 ml
Liquid volume = 266 – 158 = 108 ml.
Gas composition: C1 = 57%, C2-6 = 33% and C7+ = 10%
Oil composition: C1 = 20%, C2-6 = 30% and C7+ = 50%
(c) The final gas composition will be miscible with the oil as a result of multiple contacts. It
is given by the point of tangency on the dew-point curve for a tangent drawn from the oil
composition.
The actual composition would be: C1 = 43%, C2-6 = 37% and C7+ = 20%
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Question 3 (20 Marks)
(a) Explain why surfactants adsorb at the interface between oil and water and why this reduces
the oil-water interfacial tension.
(b) Explain why surfactants form micelles at higher concentrations
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(c) List three important differences between micro-emulsions and ordinary emulsions.
(d) Describe the effect of salinity on microemulsion phase behaviour
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(e) Explain how the salinity gradient concept can be used to improve microemulsion flooding
performance.
(f) List three mechanisms that result in loss of surfactant from a propagating chemical slug.
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(g) List three factors that increase surfactant adsorption
