Chapter 1 Notes - Craft

7/27/2019 Chapter 1 Notes - Craft

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Chapter 1 Notes Part 2

Recovery Factor(1)UR =HVfH1 - SwcLRF

Where RF is the recovery factor. It depends upon

1. Economic Considerations

2. Environmental and ecological considerations

3. Technical Considerations

Primary, secondary, and tertiary recovery book refers to supplementary recovery

Questions

1. What are some of the economic considerations?

2. What are some of the technical considerations?

3. Why is the recovery factor important?

Isothermal Compressibility

(2)c = -1

V

!V

!pT

Compressibility is a positive number.!V

!p is negative

(3)dV = c VDp

In this case, Dp = pi- p, where p is less than piso the term is positive.

Expansion manifests itself as production.

Questions

1. Why is it isothermal?

2. Why is the equation proceeded by a negative sign?

3. In producing a reservoir, what is the primary goal?

Optimizing ProductionIn an oil zone, it is desirable to tap the oil zone and not the water or gas zones.

During production, the pressure in the reservoir drops. This brings about an expansion in the oil, water,

and gas zones.

(4)dVtotal = Oil Production = dVo + dVw + dVg


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Which can be rewritten as

(5)dVtotal = co Vo Dp + cw Vw Dp + cg Vg Dp

The compressibility for each component is approximately

c0 = 15 10-6

1

psi

cw = 3 10-6

1

psi

cg = 500 10-6

1

psi

The compressibility of gas is much greater than that of oil or water, therefore, oil and water must be

present in large volumes to have the effect of gas.

This is part of the reason that wells are produced on choke. It is not desirable for gas or water to be

produced before most of the producible oil is removed from the reservoir.

Gas Reservoir EngineeringThe relationship between pressure, temperature, and volume for a gas can be expressed as

(6)p V = n R T

The derivation of this equation was presented in ENGR 4303 so we will not rederive it here.

Equation (6) only holds for simple gases over a limited range of temperatures and pressures. There

have been many attempts to find a universal equation for gas behavior and several are in use today.

Probably the simplest is

(7)p V = z n R T

Where z is the compressibility factor or z-factor. It is evaluated empirically and used in situations that do

not involve liquid-vapor equilibrium. The z-factor has been generalized by using reduced temperature

and pressure. Reduced values are derived by dividing the temperature and pressure by the critical

values for the species of interest. In the case of mixtures, the temperature and pressure are divided by

the pseudo critical temperature and pressure as given by

(8)Tpc =ni Ti(9)ppc =ni pi

Where ni is the mole fraction of component i.

The z - factor is plotted as a function of the reduced or pseudo reduced temperature and pressure.

2 Lecture_2.nb


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Tr = 1.05

Tr = 1.5

Tr = 2.0

Tr = 4.0

0 2 4 6 8

0.2

0.4

0.6

0.8

1.0

1.2

Reduced Pressure

z-Factor

z-Factor as a function ofPr and Tr

When the composition is unknown, the pseudo critical temperatures and pressures can be obtained by

using the gas gravity and Figure 1.7 in Dake.

z - factors can be calculated using the method of Hall and Yarborough. A copy of a spreadsheet will be

provided.

Gas Expansion Factor

(10)

E =volume of n moles at standard conditionsHscL

volume of n moles of gas at reservoir conditions

Lecture_2.nb 3


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E =Vsc

V=

p

psc

Tsc

T

zsc

z

Gas Initially In Place(11)G = V fH1 - SwcLEi Ei evaluated at initial reservoir pressure

Gas Density

(12)

p V = z n R T given that n =m

M p

V

m=

z R T

Msincer =

m

Vinverting yields

r =m

v=

M p

z R T

Where m is the mass and M is the molecular weight.

Gas GravityFor gases the specific gravity is given by

(13)

sg = gg =rgas

rair

at the same conditions of T and P

using the definition of density above and the

same conditions of T and P everything cancels Expand

sg =Mgas

Mairwhere Mair = 29

Isothermal Compressibility of a Real GasGiven

(14)V =z n R T

p

!V

!p= -

z n R T

p

1

p-

1

z

!z

!p

Which yields

(15)cg = -1

V

!V

!p=

1

p-

1

z

!z

!p

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Gas Material BalanceMaterial balances for hydrocarbon systems are volume balances (initial volume and current volume).

Two types of gas reservoirs

1. Volumetric depletion reservoirs (little or no influx of water with production)

2. Water drive reservoirs

Volumetric Depletion Reservoirs

(16)Reservoir volume occupied by hydrocarbons = HCPV = V @hH1-SwcL= GEi

G is initial gas in placeEiis the initial gas expansion factor

Gpvolume of production

Dp = pi- p

(17)Gp = G - HHCPVLE = G - GEi

E Gp

G= 1 -

E

Ei

p

z=

pi

zi1 -

Gp

G

Gp

Gis the fractional gas recovery at time i

If E is the expansion factor evaluated at the proposed abandonment pressure,

thenGp

G

is the gas recovery factor

Problems

The above analysis assumes that the connate water does not expand and the rock matrix pore volume

remains constant. Connate water does expand and a decline in pressure results in an increase in grain

pressure which translates into a smaller pore volume.

dHHCPVL = - dVw + dVfVw = initial connate water volume

Vf = initial pore volume

This leads to

cf = -1

Vf

!Vf

!HGPLGP = grain pressure

dHFPL = - dHGPL

Lecture_2.nb 5


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\ cf =1

Vf

!Vf

!HFPL =1

Vf

!Vf

!p

Looking at the equation

(18)dHHCPVL = - dVw + dVfIt can be rewritten as

(19)dHHCPVL = -Hcw Vw + cf VfL DpUsing

(20)Vf = PV =HCPV

H1 - SwcL =

G

EiH1 - SwcL(21)Vw = PV Swc =

G Swc

EiH1 - SwcLTotal reduction in HC pore volume becomes

(22)Gp

G

= 1 -1 - Hcw Swc + cfLDp

1 - Swc

E

Ei

Your book points out that this results in approximately a 1.3% decrease in pore volume for a Dp of

1000 psi

Most depleetion type gas reservoirs can be described by

p

z=

pi

zi1 -

Gp

G

Which is the equation of a straight line

RF = 0.58

0.0 0.2 0.4 0.6 0.8

0

500

1000

1500

2000

2500

GpG

Pz

Explain the paragraph below Figure 1.10

KpzOab

value at abandonment

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Gas Contracts delivered at some rate and pressure (usually pipeline pressure)

Dp sources loss in the formation, flowing pressure into the wellbore, flow in the wellbore, Dp

processing, and drops between the processing point and the pipeline

RF can be increased by compression (capital and operational costs must be accounted for)

History Matching

This is a technique for analyzing production data. Easy way is to fit to a simple model (linear)

G

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0

Gp

Pz

This analysis only works for very small water influx

Water Drive Reservoirs

Pressure decrease can result in a water influx. This changes the material balance equation

Production = GIIP - Unproduced Gas

(23)Gp = G -G

Ei- We E

We cumulative amount of water influx resulting from the pressure drop

If some water is produced the equation can be expressed as

(24)p

z=

pi

zi1 -

Gp

G 1 - We Ei

G

We Ei

Gis the fraction of hydrocarbonpore volume floodedby water less than one

Theequation is also nonlinear.

Lecture_2.nb 7


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C

B

A

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0

GpG

Pz

RF = 0.58

0.0 0.2 0.4 0.6 0.8

0

500

1000

1500

2000

2500

GpG

Pz

C

B

0.0 0.2 0.4 0.6 0.8

0.0

0.2

0.4

0.6

0.8

1.0

GpG

Pz

Impact of water influx provides pressure support i.e. slows pressure decline

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In all cases, extrapolation leads to over estimates of G

The rate of gas withdrawal is directly proportional to the ability for water to encroach

High withdrawal rate coupled with a strong aquifer could lead to early coning and pockets of trapped

gas

The size and properties of the aquifer coupled with the withdrawal rate, residual gas saturation, and

volumetric sweep effeciency impact the ultimate gas recovery

See .

Water influx reduces gas volume

Rewrite the material balance equations

Original VolumeHrcfL = remaining volumeHrcfL + net water influxHrcfL

Assuming no injection and rock and water compressibility changes are small.

Also gas solubility in water is negligible

(25)p

z=

pi

zi1 -

Gp

G+

1

G BgIWe - 5.615 Wp BWM

We = cumulative gas influxHrcfLWp = cumulative water productionHstbLBw = Water formation volume factorHrbbl stbLBg = Gas formation volume factorHrcf scfLBgi = Initial gas formation volume factorHrcf scfLMaking substitutions and rearranging yields

Lecture_2.nb 9


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(26)G =Gp Bg - IWe - 5.615 Bw WpM

Bg - Bgi

Looking at the denominator, at short times the difference is small and errors in G will be large (use

over long periods of time)

Recovery Efficiency in a water - drive gas reservoir

(27)rfwd = 1 -Bgi Sgr

Bga Sgi= 1 -

pa

za

zi

pi

Sgr

Sgi

Example: The reservoir is the same as described in the previous example except that pressure is

fully maintained at its original value by a strong water drive. Assume that the entire gas reservoir is

swept by water. What is the recovery factor? Given: Sgi = 75% and Sgr = 35%, respectively.

(28)rfwd = 1 -Ev

Bgi

Bga

Sgr

Sgi+

1 - Ev

Ev

Sincethe pressure isconstant for the life of the reservoir, a simplified form of Eq. H27L becomes,

rfwd = 1 -Sgr

Sgi= 1 -

0.35

0.75= 53%

If Ev = 60 %

rfwd = 1 - 0.60.35

0.75+

1 - 0.6

0.6= 32 %

Note: partial water drive does not maintain pressure completely some gas is produced by pressure

depletion recovery effeciency improves.

Pressure gradients can exist in the reservoir

For large aquifers, the pressure responce may not be instaneous and the early time data can appear

linear extrapolation then results in the estimate of G being too large.

Method to estimate GIIP using production history

(29)Ga =Gp

1 - E EiA slight modification yields

(30)G =Gp - We E

1 - E EiSubtraction yields

(31)Ga = G +We E

1 - E EiPlot GAvs

We E

1-EEiwill yield a straight line if the aquifer model is selected

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Ultimate gas recovery is a function of the aquifer and pa

Hydrocarbon Phase Behavior

Time to move to Chapter 2

Lecture_2.nb 11

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Chapter 1 Notes - Craft