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EXERCISE-11 cw = 3.0 10-6 / psi c f = 8.6 10-6 / psi S wc = .2

p i = 4000 psi B oi = 1.2417 rb/stb

pb = 3330 psi B ob = 1.2511 rb/stb

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To maintain pressure at 2700 psia thetotal underground withdrawal at theproducing end of a reservoir block must equal the water injection rate at

the injection end of the block.The total withdrawal associated withone STB of oil is:

Bo + (R Rs)Bg rb

SOLUTION

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AS, Bo + (R Rs)Bg rbEvaluating at 2700 psia, using the PVTdata

1.2022 + (3000 401) 0.00107 = 4.0 rbThus to produce 10,000 stb/d oil, aninitial injection rate of 40,000 rb/d of water will be required ,

If the injection had been started at, orabove bubble point pressure, a maximuminjection rate of only 12,500 b/d of waterwould have been required.

GO TO SLIDE 23

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The PVT properties of the crude oil system are tabulated below:

Assume that the water injection will commence

when the reservoir pressure declines to 1650 psi;find the following:Pressure that is required to dissolve the trappedgas.The required rate of water injection so as have

the production of 5000 bbl/dayt

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Solution-Part -1Step 1. Calculate initial oil in place N

N = 7758 A h (1 Swi)/Boi (22 MMSTB)

Step 2. Calculate remaining oil saturation at 1650 psi:

Step 3. Calculate gas saturation at 1650 psi:

Step 4. Calculate the trapped gas saturation from Equation-1, to give

Sgt = 12.6%Step 5. Calculate the gas solubility when all the trapped gas is dissolved inthe oil by applying Equation :

(So= 0.619)

(Sg=0.181 )

Step 6. Enter the tabulated PVT data with the new gas solubility of 814scf/STB and find the corresponding pressure of approximately 2140 psi.This pressure is identified as the pressure that is required to dissolve thetrapped gas.

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Factors important in WI pressuremaintenance and water flooding:

1 Reservoir Geometry2.Lithology3.Reservoir depth.4.Porosity5.Permeability (Magnitude and variation)

6.Fluid properties and Relative permeabilityRelationship.

7.Continuity of reservoir rock properties8.Magnitude and distribution of fluid saturations.

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RemarksThese factors influence ultimate recovery,

rate of return and ultimate economics.Therefore the process along with theseFactors must be considered collectivelyTo evaluate economic feasibility .FactorsLike oil price , marketing conditions,operating expenses and availability of wateralso influence the implementation of WI/Pressure maintenance

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EXERCISE-11

Determine the fractional oil recovery, duringdepletion down to bubble point pressurecw = 3.0 10-6 / psi c f = 8.6 10-6 / psi S wc =p i = 4000 psi B oi = 1.2417 rb/stbpb = 3330 psi B ob = 1.2511 rb/stb

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Step-1:average compressibility of the under saturated oilbetween initial and bubble point pressure:

Step-2:The recovery at bubble point pressure can becalculated as:

SOLUTION

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or 1.52% of the original oil in place

REMARKSConsidering that the 670 psi pressure droprepresents about 17% of the initial,absolute pressure, the oil recovery is

extremely low. This is because the effectivecompressibility is small providing thereservoir contains just liquid oil and water.The situation will, however, be quitedifferent once the pressure has fallen belowbubble point.

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EXERCISE-2

Below bubble point pressure (saturated oil)The reservoir described in exercise 1 will beproduced down to an abandonment pressureof 900 psia. Determine1. An expression for the recovery at

abandonment as a function of thecumulative gas oil ratio Rp.

2. Derive an expression for the free gassaturation in the reservoir at abandonmentpressure .

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Assuming m = 0; no initial gas cap, negligiblewater influx and not considering the equivalentcompressibility effect, the recovery factor atabandonment pressure of 900 psia is given as:

SOLUTION

in which all the PVT parameters Bo, Rs and Bg

are evaluated at the abandonment pressure.Using the given data the recovery factor can beexpressed as:

1

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Liberated gas2

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EXERCISE-3WATER INJECTION BELOW BUBBLE POINT PRESSURE

It is planned to initiate a water injection inthe reservoir whose PVT properties aredefined in table of Exercise-1. Theintention is to maintain pressure at the

level of 2700psia (p b = 3330 psia). If thcurrent producing gas oil ratio of the field(R) is 3000 scf/stb, what will be the initialwater injection rate required to produce

10,000 stb/d of oil.

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Exercise-1 It is planned to initiate water injection scheme in the

reservoir as per the PVT properties given inThe table below.

Intention is to maintain pressure at least of 2700 psia(Pb=3330psia)

If the current producing GOR of the field

is3000 scf/stb. Calculate what will be the initial water injection rate

required to produce 10,000stb/d of oil.

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Essentiality

To maintain pressure (say 2700psi)Total underground withdrawal rate from the producing end of areservoir rock must be equal to water injection rate at the injecend.

The total withdrawal associated with each(1) rb of oil is given as:

Bo + ( Rp - Rs ) Bg rb

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Exercise-2-PVT PROPERTIES

Pressure[psia]

Bo[[rb/stb]

R s[scf/stb]

Bg[rb/scf]

4000 1.2417 510

3500 1.2480 5103330(P b) 1.2511 510 0.00087

3000 1.2222 450 0.00096

2700 1.2022 401 0.00107

2400 1.1822 352 0.00119

2100 1.1633 304 0.00137

1800 1.1450 257 0.001611500 1.1287 214 0.00196

1200 1.1115 167 0.00249

900 1.0940 122 0.00339

600 1.0763 78 0.00519

300 1.0583 35 0.01065

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Evaluating @2700psi and using the data of PVTprovided earlier.

Bo + ( Rp - Rs ) Bgrb1.2511 +510 x 0.00087 rb

= 4.0 rbThus the total withdrawal associated with

each(1) rb of oil, for a pressure of 2700psia is4rb(four times)

Therefore to produce 10,000 stb/d oil, the initialwater injection rate should be 40000rb/d

Calculations

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(1) If the injection had been initiated at ,or above thebubble point pressure, The requires injection rate for

water would have been 12500b/d.{10000x1.25(B o)}(2) Since the water injection is now at a pressure lower than

Pb a higher/additional injection rate of the order of 27500rb/d which amounts to{(27.5/40)x 100} =68.75%70%of injection rate will be needed to displace theliberated gas.

Observations

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PVT Studies

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PVT StudiesDifferential liberation analysis of BHS @125 F

Pressurepsig

Bobbl/STB

Bgbbl/SCF

SolutionGOR

SCF/STB

B tbbl/ST

3100 1.4235 - 885 1.42352800 1.4290 - 885 1.42902400 1.4370 - 885 1.43702000 1.4446 - 885 1.4446

1725 1.4509 - 885 1.45091700 1.4468 0.00141 876 1.45951600 1.4303 0.00151 842 1.49521500 1.4139 0.00162 807 1.5403

1400 1.3978 0.00174 772 1.5444

Pi

Pb

Produced GOR

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Example

Vital data:(1) Initial reservoir pressure (@4300ftsubsea) 3100(2) Bubble point pressure (@4300ftsubsea) 1725

(3) Average reservoir temperature 125 F(4) Average porosity 7.7%(5) Average connate water 20%(6) Critical gas saturation 10%

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Example- Performance Prediction

Up to bubble point pressure, recovery is by expansion

of oilWhen pressure falls from P i=3112 to p b=1725

i.e. for N=1 STB of oil, Np = 0.0189 STB

0189.04509.14235.14509.1

ooio

p

B B B

N

N

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For P > P b, the produced gas-oil-ratio at the surface

separator can be regarded as 885 SCF/STB, the initialsolution GOR At P

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Average producing GOR between two consecutivepressures P 1(1725) and P 2(1700)

Rave = 880.5 SCF/STB and Npb =0.0189 (as determined earlier) Let N p1 to be determined

221 so so

avg R R R =885+876

2

p

STEP-1 : To calculate R p based on cumulative production up to P = 1700psia

STEP-2 : To calculat e Np1 using Material balance Equation

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To cont. prediction: At P= 1600 psig Material balance envisages:

(Release of dissolved gas)

)()(

()(

121

2211

p p pb p pb

pavg pb pavg pb soi p N N N N N

N R N N R N R R

Where SCF/STB842876

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Where SCF/STB

Therefore

From the material balance

Which meansNp2 = 0.0486 STB

for every 1 STB of initial OIP

8592

8428762avg R

2

28596161.0 p

p p N

N R

0717.0)297.100093.0

1589.0(2

22 N N

N N

p p p

The oil saturation in the reservoir at this

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The oil saturation in the reservoir at thispressure:

Oil Saturation:

Gas Saturation:

.0)1(

)()(22

wcoi

o po po

S B N

B N N

PV

B N N S

765.0oS

03076501 wcg SS

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RapidDeclineDown to Pb

FlaCudu

GaEx

GRAPHICAL REPRESENTATION

The said reservoir has an areal extent of 40 acres and thenet formation thickness is 200 feet :

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net formation thickness is 200 feet :

If the reservoir is produced at 92 BOPD, time to produce11.35%(refer graph) of initial reserves :

MM69.24235.1

)2.01(077.0200775840 N

years08.936592

1135.01069.2 6t

Constant x, Ax xh(1-Swi)

Boi N =

Performance Prediction

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For pressure drop from 3112 psig to 1725 psig:

(up to bubble point)

(compared to 0.0189 without compressibility)

Performance Predictionwith compressibility effects

Additional data required/given

cw = 3 x 10 -6 psi-1

cf = 5 x 10 -6 psi-1

5 11.39 10o oiooi

B Bc psi

B p STEP-1:CalculateC o

028.0)1( owc

oiwcw f oo p

BS

p BS ccS c

N

N STEP-2: ApplyMBE

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When pressure drops to 1700 psig, we havealready seen that :

Substituting in material balance relationship

Recovery@1700psig

1

15.880085.0

p

p p N

N R

p

S

S cc

B

B R R B B

B N B R R B N

wc

wcw f

oi

s sioio

oi g s po p

)1(

)()(

))((1

Applying MBE

3112 1700i p p p =1412

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Substituting numbers :

Which means Np1 = 0.034 STB for every STB of initial oil in place

(compared to 0.024 STB withoutcompressibility effects)

0.0)2415.100012.0

21164.0( 1

1

1 N N

N

N p

p

p

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The flow rate for a crude oil system is customarilyexpressed in surface units, i.e., stock-tank barrels(STB), rather than reservoir units.

Using the symbol Q o to represent the oil flow asexpressed in STB/day, then:

The flow rate in Darcys equation can be expressed in

STB/day to give:q = Bo Q o where B o is the oil formation volume factor bbl/STB

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Integrating the above equation between two radii, rand r 2, when the pressures are p 1 and p 2 yields:

For incompressible system in a uniform formation :

On integration it yields :

If the two radii of interest are the wellbore radius r

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If the two radii of interest are the wellbore radius r wand the external or drainage radius r e .

As the external (drainage) radius r e is usuallydetermined from the well spacing by equating thearea of the well spacing with that of a circle, i.e. r 2= 43,560 A

Then

where A is the well spacing in acres

Neither the external radius nor the wellbore radius is generally known with precision.Fortunately, they enter the equation as a logarithm, so that the error in the equation

will be less than the errors in the radii.

l l

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Pressure Oil Flow Rate The equation:

May be given as:

The equation is used calculate the pressure profile(distribution) and list the pressure drop across variousintervals

Where:

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solutionSTEP -1

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AsIn the present case:

Rearrange and solve for pressure p at radius r

STEP -2

Calculate the pressure at the specified

P fil d th llb

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Pressure profile around the wellbore.

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OIL SATURATION

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or

If the reservoir has produced N p stock-tank barrels of oil, theremaining oil volume is given by:

remaining oil volume = (N N p) Bo

or

Therefore

GAS SATURATION

Sg = 1 So Swi

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STEP-1Calculate reservoir gas pore volume V i (using cum gas

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Calculate reservoir gas pore volume V i (using cum gas production and pressure details)

{( 15.025 x 1.00 x 10 9 x )/520}= {(3250 x V i )/(0.910 x673)} {(2864 x V i)/(0.888 x673)}= 56.17 MM Cu ft

STEP-2Calculate the initial gas in place G (at initial pressure)G= (p i Vi / z i T ) x (T sc / p sc) = (p i Vi Tsc / z i T p sc)=(3250 x 56.17 x 10 6 x 520) / (0.910 x 673 x 15.025)= 10.32 MMM SCF

STEP 3Calculate gas remaining at500psia (Aband Pressure)

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Calculate gas remaining at500psia (Aband. Pressure)G a = (p a Vi / z a T ) x (T sc / p sc)

=(500 x 56.17 x 10 6 x 520) / (0.951 x 673 x 15.025)= 1.52 MMM SCF

As initial gas reserves based on a 500 psia(Aband. Pressure) is the difference betweenthe initial gas in place and the gas remaining at500 psia . Therefore gas reserve may be given

as

G r = G G a = (10.32 1 52) x 10 9

= 8.80 MMM SCF

Examplel l h l d l d l d h

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Calculate the total daily gas production including thewater and condensate using the given data:Daily separator Gas Production =3.25 MM SCF

Daily Stock Tank Condensate =53.2 STBDaily Stock Tank Gas =10 M SCFDaily Fresh Water Production =5.5 bbl*Initial reservoir Pressure =4000 psiaCurrent Reservoir Pressure =1500 psia

Reservoir Temperature =220

FCondensate Gravity = 55 API(0.759 sp.gr.)

Use the Mol. Wt.(M o) as 124 also Consider only 40%water as Produced water.(only 60% water need to be converted to gas equivalent

solution

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1

Gas Equivalent = 53.2 x 133000 x 0.759/124= 43,000 SCF

2

Gas Equivalent (water)= 3.25 x 7390= 24000 SCF

3 Daily Stock tank gas = 10 M SCF

The Daily Gas Production ( G P ) GP = 3250.0 M +10.0 M + 43.0 M + 24.0 M

= 3.327 MM SCF

If the standard conditions are 14.7 psia and 520R and the gasconstant,R,10.73:then the gas equivalent for water produced ascondensate is: 7390 scf/stb of water

Daily Stock Tank Condensate =53.2 STB(Given)