Shawket Exam Questions Dec 06

Heriot -Watt University Heriot-Watt INSTITUTE OF PETROLEUM ENGINEERING

Possible Exam Questions

Question: Chapter 2 – Section 3 z

Q. 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


z

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 overpressured reservoir?

2

Solution: Chapter 2 – Section 3 Pressure

0

⎛ dP ⎞ = 0.45 psi / ft ⎜ ⎟ ⎝ dD ⎠ water

Depth

⎛ dP ⎞ ⎜ ⎟ gas = 0.08 psi / ft ⎝ dD ⎠

⎛ dP ⎞ ⎜ ⎟ = 0.35 psi / ft ⎝ dD ⎠oil

0

Pressure

⎛ dP ⎞ = 0.45 psi / ft ⎜ ⎟ ⎝ dD ⎠ water

Depth

⎛ dP ⎞ ⎜ ⎟ gas = 0.08 psi / ft ⎝ dD ⎠

⎛ dP ⎞ ⎜ ⎟ = 0.35 psi / ft ⎝ dD ⎠oil


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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z

z

Methane is a significant component in reservoir fluids. Using a sketch for a binary of methane and ndecane (C10), illustrate the impact of methane on the critical point loci of C1-C10 binary mixtures? What is the significant of this diagram?

4

Solution: Chapter 4 – Section 3 z

z

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.


Draw 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 gasliquid, – (e) Region of retrograde condensation?

z

What is gas cycling and why in some cases is it used?

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z

Gas 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.

6


Define: – Oil FVF – Total FVF – Solution GOR

S 3.2.1-R 4

Solution: Chapter 6 – Section 3 Oil FVF:

Above bubble point as pressure reduces oil expands due to compressibility.

Below bubble point oil shrinks as a result of gas coming out of solution.

7

Solution: Chapter 6 – Section 3 Total FVF:

B t = B o + B g ( R sb − R s ) Above Pb, Bt = Bo

Solution: Chapter 6 – Section 3 Gas Solubility At bubble point All gas in solution

Above bubble point All gas in solution

Below bubble point Free gas and solution gas

At surface conditions No gas in solution

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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:

S9 –R5


z

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.

S9 –R5

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S9 –R5

z

The

0.9809 = 0.07*11.01+0.21

0.8159 = 9.8013/12.0127

S9 –R5

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Possible Exam Questions z

The

S9 –R5

Possible Exam Questions z

The

S9 –R5

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Question 2: Chapter 6 – Section 9

S9 –R5


S9 –R5

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S9 –R5


S9 –R5

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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.


z z

z

z

z

Measurements 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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z

Smaller the molecule , the larger the effect Gas permeabilty is plotted versus reciprocal mean pressure

Question 2: Chapter 7 – Section 4 z

z

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?

15


Flow rate, Qb, measured at ambient pressure, Pb Q in the core at P, For ideal gas:

QP = Q b Pb

Q=

Q b Pb P

Solution: Chapter 7 – Section 4 Q=−

kA ∆P µ dx

Q b Pb kA ∆P =− P µ dx

L

Q b Pb ∫ dx = − o

(

kA P2 − P1 Q b Pb ( L − 0 ) = − µ 2

k=

2

2

)

Qb =

2µQ b Pb L

(

A P1 − P2 2

2

P2

kA ∫ PdP µ P1

(

kA P1 − P2 2

2

)

2µLPb

)

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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?


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Question - Solution: Chapter 7 – Section 7 z

z

z

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 z

z z

z

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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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.


z

z

z

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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Question – Solution : Chapter 8 – Section 6 List the rock properties that should be determined in a rock-mechanical oriented, special core analysis programme? z

Special Core Analysis

Ê Detailed mineral description Ê Relative permeability studies Ê steady state/ unsteady state

Ê Capillary pressure Ê Resistivity measurements Ê Studies under reservoir conditions


z

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?

z

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

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S7.2 – R5

Solution 2: Chapter 8 – Section 6

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S7.2 – R5


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Capillary pressure data are obtained from core samples which represent a small part of the reservoir. Leverett derived a “J” Function using the Poiseuille equation for laminar flow:

to relate Pc to permeability and porosity. z

Derive the J Function and comment on one of its limiting assumptions?

Solution 3: Chapter 8 – Section 6 z

Based on flow through a core as represented by a bundle of capillary tubes.

Poiseuille's equation q = For n tubes q n =

πr 4 ∆P 8µLcap

nπr 4 ∆P 8µLcap

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Solution 3: Chapter 8 – Section 6 nπr 2 Porosity of bundle of tubes φ = A

Permeability k =

A

q µ L core A∆ P

Lcore

Combining equations gives:

r2 =

8k Lcap φ Lcore

Lcap Lcore

is the 'tortuosity' of the bundle of tubes

If in the reservoir rock the tortuosity remains constant then

Solution 3: Chapter 8 – Section 6 r2 =

8k Lcap φ Lcore

r = constant

k φ

Pc =

Substituting in capillary pressure equation

Pc =

2σCosθ constant

k φ

2σCosθ r

k φ 1 = =J constant σCosθ Pc

Sometimes J function written without σCosθ term

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z

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?


Case of several layers flowing simultaneously in a well

Qi =

Q = ∑ Qi =

2πh i k i ( Pe − Pw ) r µ ln e rw 2πh T k ( Pe − Pw ) 2π ( Pe − Pw ) = ( k 1h 1 + k 2 h 2 + k 3 h 3 ) re re µ ln µ ln rw rw

k=

∑h k i

hT

i

The value to be compared through well flow tests

25


z

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.


26


Briefly explain the need for the development of transient flow solutions to the diffusivity equation in reservoir engineering?


27

Question: Chapter 10 – Section 3


Briefly

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Briefly

S7.2 – R5


Briefly

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Briefly


30


Describe the method by which the line source solution may be adapted to accommodate a zone of reduced permeability around a wellbore (a skin)


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Solution 2: Chapter 10 – Section 4 Possible Exam Questions

S7.2 – R5


S7.2 – R5

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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.


z

Two stages of drive: Above the bubble point: Fluid production is driven by compressibility drive. Effective compressibility of the system:

Ê

Oil

Ê

Water

Ê

Pore space

Ê

Compressibility drive

z

Below the bubble point: Fluid production is caused by solution gas drive or dissolved gas drive: – Expanding gas provides force to drive oil.

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z

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.


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z

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: Chapter 14 – Section 3 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?

40

Solution: Chapter 14 – Section 3 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

Question: Chapter 14 – Section 6 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

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Question: Chapter 14 – Section 12 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 point pressure 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:

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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.


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Question 2: Chapter 14 – Section 12 Similar to Question 1, above

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Question 3: 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: z

(i) Bo and Rs at the bubble point?

z

(ii) Bo, Bt, Bg, Rs and z at 1,700 psia and 145°F.

z

(iii) Bt at 2100 psia and 145°F.

50

Solution 3: Chapter 14 – Section 12 (i) Bo and Rs at the bubble point?

Bo = Rsi =

bbl 290 = 1 .4 stb 207 .5

(0.153 + 0.45) scf scf 462 = stb in 3 ft 3 stb 207.5cc * * * 2.54 3 cc 12 3 in 3 5.615 ft 3

Solution 3: Chapter 14 – Section 12 (ii) Bo, Bt, Bg, Rs and z at 1,700 psia and 145°F.

Bo =

Rs =

270 bbl = 1 .3 207 .5 stb

(0.45) scf scf = 345 3 3 stb in ft stb 207.5cc * * 3 3* 3 3 2.54 cc 12 in 5.615 ft

No data to determine z, so assumed to be equal 1. Bg =

zT 1 * (145 + 460 ) bbl = 0 .0017936 = 0 .00504 P 1700 stb

Bt = Bo + ( Rsi − Rs ) Bg = 1 .3

bbl scf bbl bbl + ( 462 − 345 ) * 0 .00179 = 1 .509 stb stb scf stb

51

Solution 3: Chapter 14 – Section 12 (iii) Bt at 2100 psia and 145°F.

Bt = Bo =

290 bbl = 1 .4 207 .5 stb


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Solution: Chapter 16 – Section 5 Solution gas drive above bubble point. z

MB equation above bubble point simplifies to:-

⎡ ( B − B oi ) ( c w S wc + c f ) ⎤ N p B o = NB oi ⎢ o + ∆p ⎥ B 1 − S oi wc ⎣ ⎦ z

No gas cap

z

Aquifer small in volume We = Wp =0

z

Rs=Rsi=Rp all gas at surface dissolved in oil in reservoir

54


z

Oil compressibility -

co =

( B o − B oi ) B oi ∆ p

Replacing oil term in MB equation gives

⎡) ( c w S wc + c f ) ⎤ ∆ p ( ( ) N p B o = NB ⎥ oi ⎢ c o + 1 − S wc ⎣ ⎦ ⎡ B −cBS + c f ⎤ c ooi =⎢ c o +o w oiwc N p B o = NB ⎥ ∆p ⎣ B oi ∆ p1 − S wc ⎦

So + Swc = 1

⎡ c S + c w S wc + c f N p B o = NB oi ⎢ o o 1 − S wc ⎣ or N p B o = NB oi c e ∆ p

⎤ ⎥ ∆p ⎦

Solution: Chapter 16 – Section 5 ⎡ c S + c w S wc + c f ⎤ N p B o = NB oi ⎢ o o ⎥ ∆p 1 − S wc ⎣ ⎦ or N p B o = NB oi c e ∆ p

ce =

ce =

1 ( c o So + c w S wc + c f ) 1 − S wc

1 ( coSo + c wSwc + cf ) 1 − Swc

ce is the effective saturation weighted compressibility of the reservoir system Recovery at bubble point

Np N

=

B oi c e ∆p B ob

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Solution 2: Chapter 16 – Section 5 Table C

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Solution 2: Chapter 16 – Section 5 Table

Solution 2: Chapter 16 – Section 5 Instantaneous Gas- Oil Ratio B o k eg µ o R = + Rs B g k eo µ g z

z

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
57

Solution 2: Chapter 16 – Section 5 Instantaneous Gas- Oil Ratio

R = z

z z

B o k eg µ o B g k eo µ g

+ R

s

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 value 4. Bg is increasing with decreasing pressure.


58

Solution 3: Chapter 16 – Section 5 Solution Gas Drive-Tarner’s Method z

Similar approach to Schilthuis procedure

z

Above Pb use effective compressibility equation

Np N z

=

B oi c e ∆p B ob

Below bubble point pressure use MB, Instantaneous GOR and Oil Saturation equations


z

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.

NpR p =

N ( Bo + ( R si − R s ) Bg − Bob ) − N p ( Bo − R s Bg ) Bg

z

Step1

z

1. Start at bubble point pressure

z

z

= Gp

2. Select a future pressure and assume a value of Np at that pressure. Sometimes express Np as a function of N. 3. Solve MB eqn. eqn. For NpRp, ie. ie. Gp.

59


4. Using assumed Np solve oil saturation equation for So. This enables keg/keo to be determined.

⎛ N ⎞B So = ⎜ 1 − p ⎟ o (1 − Swc ) N ⎠ Bob ⎝ 5. Calculate instantaneous GOR.

z

z

R=

B o k eg µ o B g k eo µ g

6. Calculate gas produced during pressure drop over period. R i + R i +1

2 Ri = instantaneous GOR at start of period Ri+1 = instantaneous GOR at end of period Np1= cumulative oil produced at end of period

+ Rs

N p1

Assumption R vs Np linear Therefore use small pressure drops


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.

z

Step 2

z

1 Second pressure selected and new Np assumed.

z

2. Solve MB for Np2. This is cumulative gas at end of second pressure.

G 2 = N p 2 R p 2 − N p1R p1 = z

N ( B o + ( R si − R s ) B g − B ob ) − N p 2 ( B o − R s B g ) Bg

− N p1R p1

3. Calculate gas produced during 2nd step by removing from cumulative gas from step 1.

z

4. With assumed value of Np2 from sat’ sat’n eqn. eqn. determine So.

z

5. Calculate IGOR

60

Solution 3: Chapter 16 – Section 5 6. Calculate gas produced during second step

z

( R i+1 + R i+ 2 ) 2 z

z

z

(N

p2

− N p1 ) = G 2

7. G2 from MB compared with G2 from IGOR and new assumed value of Np2 until convergence achieved. By plotting these two values vs Np a convergence point can be determined. Further steps as for step 2.

Question: Chapter 17 – Section 1 Water drive reservoirs are said to be rate sensitive. Explain briefly this statement with respect to different aquifer characteristics.

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Question: Chapter 17 – Section 3 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.


63


Question 2: Chapter 17 – Section 3 Explain

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Solution 2: Chapter 17 – Section 3 Explain


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Shawket Exam Questions Dec 06

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