Shawket Exam Questions Dec 06
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Transcript of Shawket Exam Questions Dec 06
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Possible Possible Exam QuestionsExam Questions
Heriot-Watt UniversityINSTITUTE OF PETROLEUM ENGINEERING
HeriotHeriot--Watt UniversityWatt UniversityINSTITUTE OF PETROLEUM ENGINEERING
Question: Chapter 2 – Section 3Q. Explain briefly what is meant by:
– (a) a normal pressured reservoir, and
– (b) an overpressured reservoir?
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Solution: Chapter 2 – Section 3
Question: Chapter 2 – Section 3
With the aid of a diagram, comment on the fluid pressures gradients in an oil reservoir with a gas cap with a supporting aquifer for a normally pressured reservoir?
Illustrate the gradient for an overpressuredreservoir?
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0.45 /water
dP psi ftdD
⎛ ⎞ =⎜ ⎟⎝ ⎠
0.35 /oil
dP psi ftdD
⎛ ⎞ =⎜ ⎟⎝ ⎠
0.08 /dP gas psi ftdD
⎛ ⎞ =⎜ ⎟⎝ ⎠
Pressure
Dep
th
0
Solution: Chapter 2 – Section 3
0.45 /water
dP psi ftdD
⎛ ⎞ =⎜ ⎟⎝ ⎠
0.35 /oil
dP psi ftdD
⎛ ⎞ =⎜ ⎟⎝ ⎠
0.08 /dP gas psi ftdD
⎛ ⎞ =⎜ ⎟⎝ ⎠
Pressure
Dep
th
0
Question: Chapter 3 – Section 4
Explain briefly what you understand by:– (a) The compositional model description for the
characterization of a reservoir fluid?
– (b) The black oil model description for the characterization of a reservoir fluid?
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Solution: Chapter 3 – Section 4
Question: Chapter 4 – Section 3Methane is a significant component in reservoir fluids.
Using a sketch for a binary of methane and n-decane (C10), illustrate the impact of methane on the critical point loci of C1-C10 binary mixtures?
What is the significant of this diagram?
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Solution: Chapter 4 – Section 3
Mixture of methane and n-octane has a critical point much greater than pure component values.
Methane will tend to make the mixtures to be more volatile, i.e. of lower Tc and higher Pc.
Question: Chapter 4 – Section 5Draw a pressure temperature diagram of retrograde-gas condensate fluid, indicating the following key features:
– (a) Bubble point and dew point lines,– (b) Critical point,– (c) Cricondentherm,– (d) Isovol lines of constant proportion of gas-
liquid,– (e) Region of retrograde condensation?
What is gas cycling and why in some cases is it used?
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Solution: Chapter 4 – Section 5
Solution: Chapter 4 – Section 5
Gas CyclingGas Cycling is the reinjection of separated gas back into the reservoir – in some cases with gases from other near reservoirs.
It is used to keep the reservoir above the dew point line – preventing condensation in the reservoir.
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Question: Chapter 6 – Section 3Define:
– Oil FVF
– Total FVF
– Solution GOR
S 3.2.1-R 4
Solution: Chapter 6 – Section 3Above bubble point as
pressure reduces oil expands due to compressibility.
Below bubble point oil shrinks as a result of
gas coming out of solution.
Oil FVF:
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Total FVF:
( )t o g sb sB B B R R= + −
Above Pb, Bt = Bo
Solution: Chapter 6 – Section 3
Gas Solubility
Above bubble pointAll gas in solution
At bubble pointAll gas in solution
Below bubble pointFree gas and solution gas
At surface conditionsNo gas in solution
Solution: Chapter 6 – Section 3
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S9 –R5
The dew point pressure of a condensate gas field is 6250 psia. The initial reservoir conditions are 240°F and 8500 psia. When the reservoir was initially tested, a condensate to gas ratio of 80 stb per million scf of gas was obtained. The produced gas and condensate composition were as follows:
Question: Chapter 6 – Section 9
S9 –R5
The reservoir pore volume is considered to be 5x1011
ft3 with Swc = 0.17.
Calculate the condensate fluids produced (stb) and the gas produced (scf) in producing the reservoir down to a pressure of 6750 psia.
Question: Chapter 6 – Section 9
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S9 –R5
Solution: Chapter 6 – Section 9
S9 –R5
The
0.9809 = 0.07*11.01+0.21 0.8159 = 9.8013/12.0127
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Possible Exam Questions
S9 –R5
The
Possible Exam Questions
S9 –R5
The
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S9 –R5
Question 2: Chapter 6 – Section 9
S9 –R5
Question 2: Chapter 6 – Section 9
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S9 –R5
Question 3: Chapter 6 – Section 9
S9 –R5
Question 3: Chapter 6 – Section 9
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For a low permeability rock, the measured permeability of the rock, using gas as the fluid is more than it is using a liquid. Comment briefly on this and how the permeability of a rock can be obtained using a gas as the fluid.
Question: Chapter 7 – Section 4
Solution: Chapter 7 – Section 4Measurements on gas compared to liquid give higher values than the liquid for low perm rocks.
The phenomena is attributed to Klinkenberg.
Klinkenberg effect is due to the slippage of gas molecules along grain surfaces.
Occurs when the diameter of the pore throat approaches mean free path of the gas.
Darcy’s law assumes laminar flow and viscous theory specifies zero velocity at the boundary.
Darcy’s law is not valid when mean free path approaches diameter of pore.
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Solution: Chapter 7 – Section 4
Smaller the molecule , the larger the effect
Gas permeabilty is plotted versus reciprocal mean pressure
The system below represents the common arrangement for measuring the permeability of the core plug using a gas.
Derive an equation to calculate the gas permeabilities of a rock using the above system?
Question 2: Chapter 7 – Section 4
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Flow rate, Qb, measured at ambient pressure, PbQ in the core at P,For ideal gas:
b bb b
Q PQP Q P QP
= =
Solution: Chapter 7 – Section 4
kA PQdx∆
= −µ
b bQ P kA PP dx
∆= −
µ
( )( )2 2
2 1b b
P PkAQ P L 02
−− = −
µ
( )2 21 2
bb
kA P PQ
2 LP
−=
µ
( )b b
2 21 2
2 Q P LkA P Pµ
=−
L P 2
b bo P1
kAQ P dx PdP= −µ∫ ∫
Solution: Chapter 7 – Section 4
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Draw a graphical, normalized, representation of the results would expect to see for sandstones with high, medium, and low permeabilityies at ambient conditions?
Question: Chapter 7 – Section 5
Solution: Chapter 7 – Section 5
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Question - Solution: Chapter 7 – Section 7
In the context of Pc define the FWL?
It is the shallowest depth in the reservoir at which Sw=100% and Pc =0.
It lies below the OWC.
Question – Solution : Chapter 7 – Section 7
Explain briefly why, because of Pc, it is possible to have Sw of large values, up to 100%, above the OWC and above layers with lower Sw?
Each rock type has a different Pc curve.
At the same level in the reservoir the Pc curve of a lower quality reservoir rock shows higher Sw than a higher quality reservoir rock.
Then it is possible to have Sw of large values, up to 100%, above the OWC and above layers with lower Sw, when we have reservoir layers of lower rock quality on top of layers of good rock quality.
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Question – Solution : Chapter 7 – Section 8
Explain briefly the reason for significant oil saturation remaining in the water swept zones of a reservoir after natural water drive or water injection?
As water rises capillary forces move ahead of natural level, more in narrow pores than in larger pores. The capillary forces then isolate oil in larger pore which is then held by capillary forces in the swept position of the reservoir.
Question – Solution : Chapter 8 – Section 4
Briefly explain how the permeability sensitivity of rocks to stress can be measured in the lab?
It could be measured in the lab by using permeability measurement tools with simulating stresses.
This is done by using high pressure core holder for biaxial loading.
Use tubes running parallel along confining rubber sleave that are pressured to simulate the stress distribution of interest.
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List the rock properties that should be determined in a rock-mechanical oriented, special core analysis programme?
Special Core AnalysisDetailed mineral description
Relative permeability studiessteady state/ unsteady state
Capillary pressure
Resistivity measurements
Studies under reservoir conditions
Question – Solution : Chapter 8 – Section 6
Core samples have been obtained from a well and air-mercury capillary pressure curves generated for an oil reservoir system.
The lowest limit of 100% Sw was found at the bottom of the well in rock type A as shown in the following figure.
– (a) determine the FWL and indicate it on the well diagram?
– (b) Construct the Sw profile in the space provided?
– (c) Calculate the oil-in-place per unit cross-section over the thickness of the reservoir?
Data:– Specific gravity of Water = 1.03
– Specific gravity of oil = 0.80
– Density of water = 62.4 lb/ft3
– PcAir-Hg = 10 PcW-O
– Oil FVF = 1.22 bbl/stb
Question 2: Chapter 8 – Section 6
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S7.2 – R5
Question 2: Chapter 8 – Section 6
Solution 2: Chapter 8 – Section 6
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Solution 2: Chapter 8 – Section 6
S7.2 – R5
Solution 2: Chapter 8 – Section 6
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Question 3: Chapter 8 – Section 6
Capillary pressure data are obtained from core samples which represent a small part of the reservoir. Leverettderived a “J” Function using the Poiseuille equation for laminar flow:
to relate Pc to permeability and porosity.
Derive the J Function and comment on one of its limiting assumptions?
Based on flow through a core as represented by a bundle of capillary tubes.
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cap
r PPoiseuille's equation q8 Lπ ∆
=µ
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ncap
n r PFor n tubes q8 Lπ ∆
=µ
Solution 3: Chapter 8 – Section 6
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Solution 3: Chapter 8 – Section 6
2n rPorosity of bundle of tubes Aπ
φ = A
coreq LPermeability kA Pµ
=∆
Lcore
Combining equations gives:
cap2
core
L8krL
=φ
cap
core
L is the 'tortuosity' of the bundle of tubes
L
If in the reservoir rock the tortuosity remains constant then
Solution 3: Chapter 8 – Section 6cap2
core
L8krL
=φ kr constant=
φ
Substituting in capillary pressure equation c2 CosP
rσ θ
=
c2 CosP
kconstant
σ θ=
φ ckP
1 Jconstant Cos
φ= =σ θ
Sometimes J function written without σCosθ term
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Question: Chapter 9 – Section 2
Derive an equation for the average permeability, resulting from radial circular flow into a well from layers of different permeabilities and thicknesses?
From where would such an average be obtained?
Solution: Chapter 9 – Section 2
Case of several layers flowing simultaneously in a well
( )i i e wi
e
w
2 h k P PQ rln
r
π −=
µ
( ) ( ) ( )T e w e wi 1 1 2 2 3 3
e e
w w
2 h k P P 2 P PQ Q k h k h k hr rln ln
r r
π − π −= = = + +
µ µ∑
i i
T
h kk
h= ∑ The value to be compared
through well flow tests
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Explain briefly the importance of characterizing the permeability variations in a reservoir in relation to the prediction of the behaviour of natural and injected water drive systems.
The answer should be limited to the behaviour in the vertical plane rather than the areal plane.
Question: Chapter 9 – Section 3
Solution: Chapter 9 – Section 3
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Briefly explain the need for the development of transient flow solutions to the diffusivity equation in reservoir engineering?
Question: Chapter 10 – Section 3
Solution: Chapter 10 – Section 3
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Question: Chapter 10 – Section 3
Briefly
Solution: Chapter 10 – Section 3
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S7.2 – R5
Briefly
Solution: Chapter 10 – Section 3
Briefly
Solution: Chapter 10 – Section 3
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Briefly
Solution: Chapter 10 – Section 3
Solution: Chapter 10 – Section 3
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Describe the method by which the line source solution may be adapted to accommodate a zone of reduced permeability around a wellbore (a skin)
Question: Chapter 10 – Section 3
Solution: Chapter 10 – Section 3
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Solution: Chapter 10 – Section 3
Question: Chapter 10 – Section 4
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Solution: Chapter 10 – Section 4
Solution: Chapter 10 – Section 4
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Solution: Chapter 10 – Section 4
Question 2: Chapter 10 – Section 4
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Possible Exam Questions
S7.2 – R5
Solution 2: Chapter 10 – Section 4
S7.2 – R5
Solution 2: Chapter 10 – Section 4
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Solution 2: Chapter 10 – Section 4
Solution 2: Chapter 10 – Section 4
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Describe briefly the drive mechanisms associated with producing an under-saturated oil reservoir, without a supporting aquifer, down to a pressure well below the bubble point.
Question: Chapter 11 – Section 2
Two stages of drive:Above the bubble point: Fluid production is driven by compressibility drive.
Effective compressibility of the system:Oil
Water
Pore space
Compressibility drive
Below the bubble point: Fluid production is caused by solution gas drive or dissolved gas drive:
– Expanding gas provides force to drive oil.
Solution: Chapter 11 – Section 2
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Derive two equations in terms of composition and equilibrium ratios to determine the dew point and bubble pint pressure of a reservoir fluids.
Explain briefly how the equations are used, when the temperature of the reservoir and composition of the fluid are known.
Question: Chapter 12 – Section 4
Solution: Chapter 12 – Section 4
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Solution: Chapter 12 – Section 4
Solution: Chapter 12 – Section 4
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Derive an equation in terms of equilibrium ratios and composition to predict liquid and vapour ratios and compositions resulting from the flash separation of a reservoir fluid.
Explain briefly the application of the equation when the reservoir fluid composition, temperature and the pressure of separation are known?
Question 2: Chapter 12 – Section 4
Explain briefly why surface samples from a wet or condensate reservoir can be unrepresentative if collected too early after a shut down or major well disturbance. What suggestions would you give to get more representative samples?
Question: Chapter 14 – Section 3
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Surface samples from a wet or condensate reservoir can be unrepresentative if collected too early after a shut down or major well disturbance, because in these incidents:
– Well acts as a separator
– Liquid rains down & accumulates at bottom of well.
– Pressure builds up in the well & disturbed formation
– Some gas goes back into solution.
– Large variations in compositions of produced fluids.
– Early period lean gas produced. High GOR
– When fluids produced from bottom of well. Liquids much lower GOR.
– Then fluids from disturbed reservoir zone
– Eventually fluids from undisturbed reservoir
Solution: Chapter 14 – Section 3
Explain briefly the three following tests carried out on reservoir fluid samples in relation to a PVT study, and comment on their application.
– Relative Volume (Flash Vaporization Test)
– Separator Test
– Differential Test
Question: Chapter 14 – Section 6
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Solution: Chapter 14 – Section 6
Solution: Chapter 14 – Section 6
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Solution: Chapter 14 – Section 6
Solution: Chapter 14 – Section 6
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Table below gives the results for a volume/ pressure investigation of a reservoir fluid at reservoir temperature. The system composition remained constant throughout the test
Question: Chapter 14 – Section 12
In another test on the fluid a sample of oil at its bubble pointpressure and reservoir temperature in a PVT cell were passed through a two stage spearator at 100 psig and 75°F and 0 psig and 60°F. 34 cc of oil were displaced from the PVT cell and 27.4 cc of oil were collected from the last separator stage. 4976 cc of gas were collected at standard conditions during the test.
In a further test the pressure in a PVT cell at reservoir temperature was reduced in stages and the gas produced at each stage removed and the remaining oil volume measured. The totoal gas produced at standard conditions was recorded and is presented in the table below:
Question: Chapter 14 – Section 12
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(a) Determine the bubble point pressure of the reservoir fluid at reservoir temperature.
(b) the oil FVF at 3650 psig
(c) the solution GOR at 3650 psig and 2700 psig
(d) the solution GOR at 1200 psig
(e) the total FVF factors at 3650 psig and 1200 psig.
Question: Chapter 14 – Section 12
Solution: Chapter 14 – Section 12
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Solution: Chapter 14 – Section 12
Solution: Chapter 14 – Section 12
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Solution: Chapter 14 – Section 12
Solution: Chapter 14 – Section 12
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Solution: Chapter 14 – Section 12
Question 2: Chapter 14 – Section 12Similar to Question 1, above
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Question 2: Chapter 14 – Section 12
Question 2: Chapter 14 – Section 12
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Question 2: Chapter 14 – Section 12
A laboratory cell, contained 290 cc of reservoir liquid at its bubble point of 2100 psia at 145°F. 21cc of Hg were removed from the cell and the pressure dropped to 1700 psia. Mercury was then re-injected at constant temperature and pressure and 0.153 scf of gas was removed leaving 270 cc of liquid in the cell. The process was repeated reducing the pressure to 14.7 psia and the temperature to 60°F. Then 0.45 scf of gas was removed and 207.5 cc of liquid remained in the cell.
Determine:
(i) Bo and Rs at the bubble point?
(ii) Bo, Bt, Bg, Rs and z at 1,700 psia and 145°F.
(iii) Bt at 2100 psia and 145°F.
Question 3: Chapter 14 – Section 12
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(i) Bo and Rs at the bubble point?
stbbblBo 4.1
5.207290
==
stbscf
ftstb
inft
ccincc
scfRsi 462
615.5*
12*
54.2*5.207
)45.0153.0(
333
3
3
3 =+
=
Solution 3: Chapter 14 – Section 12
(ii) Bo, Bt, Bg, Rs and z at 1,700 psia and 145°F.
No data to determine z, so assumed to be equal 1.
stbbblBo 3.1
5.207270
==
stbscf
ftstb
inft
ccincc
scfRs 345
615.5*
12*
54.2*5.207
)45.0(
333
3
3
3 ==
stbbbl
PzTBg 0017936.0
1700)460145(*100504.0 =
+==
stbbbl
scfbbl
stbscf
stbbblBgRsRsiBoBt 509.100179.0*)345462(3.1)( =−+=−+=
Solution 3: Chapter 14 – Section 12
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(iii) Bt at 2100 psia and 145°F.
stbbblBoBt 4.1
5.207290
===
Solution 3: Chapter 14 – Section 12
Question: Chapter 15 – Section 5
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Solution: Chapter 15 – Section 5
Solution: Chapter 16 – Section 5
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Solution: Chapter 16 – Section 5
Solution gas drive above bubble point.MB equation above bubble point simplifies to:-
( ) ( )o oi w wc fp o oi
oi wc
B B c S cN B NB p
B 1 S− +⎡ ⎤
= + ∆⎢ ⎥−⎣ ⎦
No gas capNo gas cap
Aquifer small in volume We = Aquifer small in volume We = WpWp =0=0
RsRs==RsiRsi==RpRp all gas at surface dissolved in oil in reservoirall gas at surface dissolved in oil in reservoir
Solution: Chapter 16 – Section 5
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Oil compressibility - ( )o oio
oi
B Bc
B p−
=∆
Replacing oil term in MB equation givesReplacing oil term in MB equation gives
( )o oio
oi
B Bc
B p−
=∆( )w wc f
p o oi owc
c S cN B NB c p
1 S+⎡ ⎤
= + ∆⎢ ⎥−⎣ ⎦
( )w wc fp o oi o
wc
c S cN B NB c p
1 S+⎡ ⎤
= + ∆⎢ ⎥−⎣ ⎦
So + Swc = 1 o o w wc fp o oi
wc
p o oi e
c S c S cN B NB p1 S
orN B NB c p
⎡ ⎤+ += ∆⎢ ⎥−⎣ ⎦
= ∆
Solution: Chapter 16 – Section 5
o o w wc fp o oi
wc
p o oi e
c S c S cN B NB p1 S
orN B NB c p
⎡ ⎤+ += ∆⎢ ⎥−⎣ ⎦
= ∆
( )e o o w wc fwc
1c c S c S c1 S
= + +− ( )e o o w wc f
wc
1c c S c S c1 S
= + +−
ce is the effective saturation weighted compressibility of the reservoir system
Recovery at bubble point p oie
ob
N B c pN B
= ∆
Solution: Chapter 16 – Section 5
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Question 2: Chapter 16 – Section 5
TableSolution 2: Chapter 16 – Section 5
C
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TableSolution 2: Chapter 16 – Section 5
Instantaneous Gas- Oil Ratio
o e g os
g e o g
B kR R
B kµ
= +µ
1. Above Pb, no free gas. Keg is zero, R=Rs=Rsi.
2. Short time when gas saturation below critical value, keg still zero but R=Rs<Rsi
Solution 2: Chapter 16 – Section 5
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Instantaneous Gas- Oil Ratio
o e g os
g e o g
B kR R
B kµ
= +µ
2-3. Gas reached critical saturation, keg increases as keo decreases. Gas very mobile compared to oil. Free gas produced from oil still in reservoir.3. Maximum GOR value4. Bg is increasing with decreasing pressure.
Solution 2: Chapter 16 – Section 5
Question 3: Chapter 16 – Section 5
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Solution Gas Drive-Tarner’s Method
Similar approach to Schilthuis procedure
Above Pb use effective compressibility equation
p oie
ob
N B c pN B
= ∆
Below bubble point pressure use MB, Instantaneous Below bubble point pressure use MB, Instantaneous GOR and Oil Saturation equationsGOR and Oil Saturation equations
Solution 3: Chapter 16 – Section 5
Tarner’s method uses MB equation rearranged to calculate gas production Gp.
Procedure is a trial & error approach using independently MB and Instantaneous GOR eqns.
Step1 Step1
1. Start at bubble point pressure1. Start at bubble point pressure
2. Select a future pressure and assume a value of 2. Select a future pressure and assume a value of NpNp at that at that pressure. Sometimes express pressure. Sometimes express NNpp as a function of N.as a function of N.
3. Solve MB 3. Solve MB eqneqn. For . For NNppRRpp, , ieie. . GGpp..
( )( ) ( )o si s g ob p o s gp p p
g
N B R R B B N B R BN R G
B+ − − − −
= =
Solution 3: Chapter 16 – Section 5
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( )p oo wc
ob
N BS 1 1 SN B
⎛ ⎞= − −⎜ ⎟⎝ ⎠
4. Using assumed Np solve oil saturation equation for So. This enables keg/keo to be determined.
5. Calculate instantaneous GOR.5. Calculate instantaneous GOR.o eg o
sg eo g
B kR R
B kµ
= +µ
6. Calculate gas produced during pressure drop over 6. Calculate gas produced during pressure drop over period.period.
i i 1p1
R R N2
++
Ri = instantaneous GOR at start of periodRi+1 = instantaneous GOR at end of periodNp1= cumulative oil produced at end of period
Assumption R vs Np linearTherefore use small pressure
drops
Solution 3: Chapter 16 – Section 5
6. Total gas produced from MB eqn. and IGOR eqn. Compared and assumed value of Np adjusted and steps 2 to 6 repeated until MB and IGOR values for Gp match.
Step 2
1 Second pressure selected and new Np assumed.
2. Solve MB for Np2. This is cumulative gas at end of second pressure.
( )( ) ( )o si s g ob p 2 o s g2 p2 p 2 p1 p1 p1 p1
g
N B R R B B N B R BG N R N R N R
B+ − − − −
= − = −
3. Calculate gas produced during 23. Calculate gas produced during 2ndnd step by removing from step by removing from cumulative gas from step 1.cumulative gas from step 1.
4. With assumed value of Np2 from 4. With assumed value of Np2 from satsat’’nn eqneqn. determine So.. determine So.
5. Calculate IGOR5. Calculate IGOR
Solution 3: Chapter 16 – Section 5
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6. Calculate gas produced during second step
( ) ( )i 1 i 2p2 p1 2
R RN N G
2+ ++
− =
7. G7. G22 from MB compared with Gfrom MB compared with G2 2 from IGOR and new assumed from IGOR and new assumed value of Nvalue of Np2p2 until convergence achieved.until convergence achieved.
By plotting these two values By plotting these two values vsvs NpNp a convergence point can be a convergence point can be determined.determined.
Further steps as for step 2.Further steps as for step 2.
Solution 3: Chapter 16 – Section 5
Water drive reservoirs are said to be rate sensitive. Explain briefly this statement with respect to different aquifer characteristics.
Question: Chapter 17 – Section 1
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Solution: Chapter 17 – Section 1
Solution: Chapter 17 – Section 1
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Explain briefly how the constant terminal pressure solution of the Hurst and van Everdingen unsteady state theory can be used to predict water influx into an oil reservoir with a declining reservoir pressure.
Question: Chapter 17 – Section 3
Solution: Chapter 17 – Section 3
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Solution: Chapter 17 – Section 3
ExplainQuestion 2: Chapter 17 – Section 3
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Question 2: Chapter 17 – Section 3
Solution 2: Chapter 17 – Section 3
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ExplainSolution 2: Chapter 17 – Section 3
Solution 2: Chapter 17 – Section 3
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Question 3: Chapter 17 – Section 3
Question 3: Chapter 17 – Section 3
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Question 3: Chapter 17 – Section 3