Reservoir Rock Properties

Oil Reservoir Engineering

Contents (1) Porosity • Geological factors affecting porosity • Experimental porosity measurements • Preparation of samples for measuring porosity • Glass pycnometer • Rusel volumeter • Ruska porosimeter (2) Fluid saturation • fluid saturation measurements • Distillation method( A S T M ) Oil Reservoir Engineering

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• Centrifugal method • Factors affecting fluid saturation • The uses of core determined fluid saturation (3) Permeability • Limitations of Darcy's law applications • Dimensions of permeability • Reservoir flow system applications • Conversion units of Darcy's law • Permeability of combined layers • Linear beds in serried • Linear beds in parallel • Radial beds in series permeability of channels and fractures • Laboratory measurements of permeability • Perm-plug method • Hole-core measurements • Factors affecting permeability measurements • Effective and relative permeability • Characteristics of two-phase relative permeability curves • Three phase relative permeability • Factors affecting relative permeability • Relative permeability ratio • Measurements of relative permeability data • Uses of effective and relative permeability data (4) Capillary forces • Surface and capillary pressures • Adhesion tension • Rise of fluids in capillaries • Calculations of capillary pressure Oil Reservoir Engineering

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• Capillary pressure in unconsolidated sands • Capillary pressure curves • Drainage and imbibition capillary pressure curves • Laboratory determination of capillary pressure curve • Jamin effect • Calculation of wettability • Relationship between gravity and capillary forces • Converting laboratory capillary pressure data • Extending the range of laboratory Pc - Sw data • Calculation of effective and absolute permeability from capillary pressure data • Calculation of relative permeability from capillary pressure data (5) Petrophysics • Properties of clean rocks • Relation between porosity, permeability, tortuosity and mean capillary radius • Specific surface • Kozeny equation • Flow of electric current through clean rocks • Formation resistivity factor • Lithologic factor affecting formation factor • Resistivity of rocks partially saturated with water • Saturation exponents • Tortuosity determination • Effective tortuosity • Hydraulic formation factor and index Oil Reservoir Engineering

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• • • • •

The effective hydraulic index Relative permeability to the wetting-phase Imbibition direction Drainage direction Relative permeability to the non wetting-phase Imbibition phase Drainage direction

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Oil Reservoir Engineering Oil Reservoirs (Definitions) Oil is produced from wells drilled into underground porous rock

formations. The ensemble of wells draining a common oil accumulation or source or surface area defined by the well distribution termed an "Oil

Field" or "Oil Pool". The part of the rock that is oil productive is termed an "Oil Reservoir" by variety of the subsurface location of the reservoir

rock; its entrained fluids are subject to elevated temperature and pressure – the reservoir temperature and reservoir pressure.

Reservoir rocks are mostly sedimentary in origin. They are either

mechanical or chemical deposition of solid–materials or simply the remains of animals or plant life.

Physical properties of reservoir rocks Considered on hand–specimen scale reservoir rocks have defined

ranges of physical properties which are of paramount interest to the reservoir engineer. The three engineering characteristics of the reservoir rock are porosity, oil, gas and water saturation and permeability, Specific, effective and relative.

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Porosity (Φ %) Porosity of a material is defined as that fraction of the bulk volume

of this material that isn't occupied by the solid framework of the material.

In oil reservoirs, the porosity represents the percentage of the total

space that is available for occupancy by either liquids or gases.

One may distinguish two types of porosity, namely: absolute and

effective.

Absolute porosity The percentage of total void space with respect to the bulk volume

regardless of the interconnection of the pore voids. A rock may have considerable absolute porosity and yet have no conductivity to fluid for lack of pore interconnection.

Effective porosity The percentage of interconnected void space with respect to the

bulk volume. It is an indication of conductivity to fluid but not necessarily a measure of it.

Porosity in sediments both treated and descried by natural

geological processes. Geological conditions are responsible for both primary and secondary porosity.

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Primary porosity Primary porosity results from voids which are left between mineral

fragments and grains after their accumulation as sediments.

Secondary porosity Secondary porosity results from geological agents such as leaching,

fracturing, and fissuring which occur after lithification of sediments.

Effective porosity is a function of a great many lithological factors.

Some of the most important of this are heterogeneity of grin size,

packing, clay content, cementation, weathering and leaching, clay types and clay hydration status.

Geological factors affecting porosity 1– Degree of sorting Well–sorted, moderately rounded sand grains settle in water giving

a packing of 30 to 40% porosity. In poorly sorted sediments, the smaller grains fit into the space between the larger ones, and porosity is considerable decreased.

2 – Compaction It's a geological factor which reduces porosity due to overburden

pressure of the overlying sediments. Sandstones, whoever, exhibit very

little compressibility 3x10–6 whereas shales may be reduced to a small fraction of there original sedimentation volume.

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3 – Cementation It's the agent which has the greatest effect on the original porosity

and which affects the size, shape and continuity of pore channels.

4 – Clay content Clay may often act as cementing material. Clay is deposits of the

same time as sand grains and generally it adheres to them so that after deposition considerable porosity still exists and the over–all porosity of sandstone may not be lowered greatly by a small amount of clay.

5 – Granulation and crushing of sand grains Their effect on porosity at great depth under overburden pressure is

of interest. With increasing overburden pressure, quartz grains in sandstone show a progressive change from random packing to closer

packing. Some crushing and plastic deformation of the sand grain occurs.

6 – Mode of packing One may get qualitative picture of the geometrical structure of

sands by consideration of packing of spheres of uniform size. This, too,

is of infinite variety. However, it will suffice to note here two basic and extremely types, namely: the cubic and rhombohedra packing. Unit cells of such packing are shown in fig (1).

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Figure 1

r

denoted the radius of sphere, thus for cubic packing:

Bulk

volume = (2r ) = 8r 3

Solid

4 volum = πr 3 3

3

3 3 4 p.v B.v − S .v 8r − 3 πr ∴ porosity Φ = = = = 47.6% B.v B.v 8r 3

For rhombohedra system:

B.V . = 8r 3 sin 60

S .V . =

4 3 πr 3

8r 3 sin 60 − 4 πr 3 3 ∴φ = = 39.5% 3 8r sin 60 In general we can write:

φ = 1−

π σ (1 − cos θ ) 1 + 2 cos θ

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Of particular interest is the fact that the radii cancel and the porosity

of packing of uniform spheres is a function of packing only.

7– Rock compressibility: Assume that:

1 dvp ( ) = − . c Pore volume compressibility p

Bulk volume compressibility (c B ) = −

vp dp

1 dvB . v B dp

1 dvs ( ) = − . c s Solid volume compressibility v s dp

v B = v p + vs dvB dvp dv s = + dp dp dp ∴ −v B c B = − c p v p − c s v s ∴

∴ cB = cs

vp vs + cp vB vB

∴ c B = c s (1 − φ ) + c pφ ∴φ =

cB− cs c p− cs

Experimental porosity measurements Experimental porosity – determination procedures may be divided in

to two classes, namely, those designed to measure effective porosity and those which measure absolute porosity. Oil Reservoir Engineering

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Preparation of samples for measuring it is porosities: They are selected to be preferably10 to 20 cm3 in bulk volume and

are obtained from the center of the core .their surfaces are cleaned to remove traces of drilling mud. The samples are extracted in a soxhlit

using oil solvents such as benzene, toluene alight hydrocarbon fraction.

During the extraction, the sample should be kept in a paper thimble,

covered with plug of cotton in order to avoid erosion of loosely cemented grains.

After extraction, the samples are dried in an over a 100 to 105 co

and cooled in a desiccators. This operation removes the solvent and moisture from the samples.

Effective – porosity measurements: For approximate work, some methods of obtaining effective porosity

(grain volume methods) may be used. In these methods: the bulk

volume is determined either by the displacement of a liquid which does not penetrate the sample or by saturating the sample and volumetrically

displacing a suitable liquid with the saturated sample. The grain volume may be measured by the volumetric displacement of a gas or a liquid,

while the pore volume may be measured by determining the amount of liquid necessary to saturate the sample.

An alternate method of obtaining the grain volume is to divide the

dry weight of the sample by the average grain density of 2.65 (the average density of most reservoir rock minerals).

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Example: A coated sample has the following data: Weight of dry sample in air (Wa) = 20 gms. Weight of coated sample (with paraffin of density 0.9 gm/cc) = 20.9 gms. Weight of coated sample immersed in water = 10 gms. Calculate the bulk volume of the sample. Solution: Wt. of paraffin = 20.9 – 20 = 0.9 gm. Volume of paraffin = 0.9/0.9 = 1 cm3. Volume of water displaced = (20.9 – 10)/ew = 10.9 cm3. So: B.V. = volume of water displaced – volume of paraffin = 10.9 – 1 = 9.9 cm3. Example: If the sample of Ex.1 has been saturated (100%) by water and; its weight in air become 21.5 gms. When it has been immersed in water, it weights 11.6 gms; calculate B.V. Solution: Wt. of displaced water = 21.5 – 11.6 = 9.9 gms. B.V. = volume of water displaced = 9.9 / 1 = 9.9 cm3.

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Laboratory measurements for porosity measurements: Glass pycnometer: A glass pycnometer with a cap which rests on a ground taper joint

and with a sample hole through the cap is filled with mercury, the cap is

pressed into its seat and the excess mercury which overflows through a

hole in the cap is collected and removed. The pycnometer is opened and the sample is placed in the surface of the mercury and submerged by a set of pointed rods which project from the lower side of the cap, fig (2).

Figure 2

The cap is again pressed into its seat, which causes a certain

amount of mercury equivalent to the bulk volume of the sample to overflow. The rods which submerge the sample should be adjusted so that the sample does not touch the sides of the pycnometer; this avoids trapping air bubbles.

Either the volume of mercury which overflows or the loss of weight

of the mercury in the pycnometer may be measured and the core bulk volume calculated.

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Example: For a core weights in air 20 gms; given: Weight of pycnometer filled with Hg = 350 gms. Wt. of pycnometer filled with Hg and sample = 235.9 gms. Of the mercury density is 13.546 gms/cc. Calculate B.V. Solution: Weight of pycnometer + weight of Hg + weight of sample = 350+20 = 370 gms. Weight of Hg displaced = 370 – 235.9 = 134.1 gms. So: B.V. of sample = 134.1 / 13.546 = 9.9 cm3.

Russel volumeter

Figure 3

As the determination of the bulk volume by glass pycnometer can

not be applied to loosely cemented samples which have a tendency to Oil Reservoir Engineering

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disintegrate when immersed in mercury, and a serious source of error of

the trapping of air bubbles at the surface of the sample, Russle

volumeter provides for direct reading of bulk volume. A saturated sample is placed in a sample bottle after a zero reading is established with fluid in the volumeter. The resulting increase in volume is the bulk volume. Only saturated or coated samples may be used in the device. This

device has the advantage of applicability to loosely cemented sample with irregular surfaces. Since the liquid used is transparent, trapped air bubbles may be seen and steps taken to remove them.

Ruska porosimeter A schematic diagram is given in Fig (4)

Figure 4

A micrometer piston is used to pressure the sample cup, so that the

mercury reaches a given reference on the manometer. Let the reading be Rb in the absence of a core sample in the cup. When the core floats

on the mercury within the cup, the displacement of the micrometer piston gives a reading Rc to reach the same reference mark. The porosity of the sample is then calculated by:

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 

φ = 1001 −

Rc   Rb 

a number of other devices has been designed for measuring pore

volume, grain volume, and porosity including the cope porosimeter, the mercury pump porosimeter, Washburn–Bunting porosimeter, Stevens porosimeter, … etc.

In the determination of absolute porosity, it is required that all

unconnected as well as interconnected pores are accounted for. The procedure required that the sample be crushed. The method is as follows:

Break of the well core, clean the surface of the sample to remove

the drilling mud, measure the bulk volume by any one of the procedure described above, crush the sample to its grains, wash the grains with

suitable solvent to remove oil mud and water, and determine the volume of the grains. It is of course necessary to dry the rock grains before their

volume is determined. The volume of the dry grains may be determined in a pycnometer containing a suitable liquid as kerosene.

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Fluid saturation Cores or underground rock samples, which brought to the surface,

are universally found to have entrained in their pores varying amounts of

liquid. In a typical oil field, water called interstitial or connate–water and frequently free gas pressured in addition to the oil.

The water saturation Sw is defined by the equation:

Sw =

pore volume filled by water total pore volume

Similarly;

So =

pore volume filled by oil total pore volume

If oil and water are the only fluids present, the volume filled by water

plus the volume filled by oil must equal the total pore volume; thus: Sw + S o = 1

In many pools, in addition to oil and connate water, free gas is also

present. The free gas saturation is defined by:

Sw =

pore volume filled by total pore volume

gas

And then: Sw + S o + S g = 1 Three factors should be remembered concerning fluid saturation:

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The saturation will vary from place to place; the water saturation

tending to be higher at the lower part due to gravity.

Water tending to be higher in the less porous section. The saturation will vary with cumulative with drawal.

Figure 5

Fluid saturation measurements Methods for the determination of reservoir fluid saturation in place

consist in analyzing reservoir core simples for water and oil, the

saturation in gas being obtained by difference since the sum of the three fluids is equal to unity.

Distillation method (ASTM): • Take a sample ranging in volume from 50 to 60 cc from the central part of the larger core.

• Place the core in an extraction thimble and weighed. • Put the thimble in the flask containing a liquid solvent such as toluene or a gasoline fraction boiling at about 150 ˚C.

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• A reflux condenser is fitted to the flask to return the condensate to a calibrated glass trap. Fig (6)

• The liquid hydrocarbon is boiled and the water present in the sample vaporized, carried into the flask condenser, and caught in the trap. When the volume in the trap remains

constant under continued extraction, the volume of the water

collected is read and the sample containing the sample is then transferred to a soxhlet for the final extraction.

Figure 6

• The thimble and the sample are then dried and weighted. • The total fluid saturation is obtained by weight difference and includes both oil and water.

• By weight difference again, the weight of oil is obtained, and, by use of an appropriate oil density, its volume is calculated.

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• The saturations on a percentage of pore–volume base are readily calculated for both water and oil

Centrifugal method A solvent is injected into the centrifuge just of center.

Figure 7

Owing to centrifugal force, it is thrown to the outer radii, being

forced to pass through the core sample. The solvent removes the water and oil from the core. The outlet fluid is trapped and the quantity of water in the core is measured.

This method provides a very rapid method because of the high

forces which can be applied; at the same time that the water content is

determined, the core is cleaned in preparation for the other measurements.

Factors affecting fluid saturation Mud filtration In the case of rotary drilling, the differential pressure across the well

face causes mud and mud filtrate to invade to formation immediately

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adjacent to the well surface, this flushing the formation with mud and this

filtrate displacing some of the oil and perhaps some of the original

interstitial water. The displacement process changes the original fluid contents to fluid saturation.

Pressure gradient Pressure gradient between the surface and the formation permits

the expansion of the entrapped water, oil and gas. Thus the contents of

the core at the surface have been changed from those which existed in the formation.

Uses of core determined fluid saturations to:

The saturation values obtained directly from rock samples are used • Determine the original oil–gas contact, original oil water contact and weather sand is productive of oil or gas.

• Establish a correlation of the water content of cores and permeability from which it can be determined whether a formation will be productive of hydrocarbon.

In summary, it is seen that although fluid–saturation determinations

made on core samples at the surface may not give a direct indication of

the saturation within the reservoir, they are of value and do yield very useful and necessary information.

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Permeability Permeability is a property of the porous medium and is a measure

of the capacity of the medium to transmit fluids. The measurement of

permeability is a measure of the fluid conductivity of the particular material.

By analogy with electrical conductors, the permeability represents

the reciprocal of the resistance which the porous medium offers to fluid flow.

If the reservoir rock system is considered to be a bundle of circular

tubes such that the flow could be represented by a summation of the flow from all the tubes as described by poiseuille's equation:

n π r 4 ∆p Q= 8 µ l Where:

Q

: Flow rate, cm3/sec.

r

: Radius of tubes, cm.

∆p

: Pressure lose over length, dyne/cm2.

µ

: Fluid viscosity, cp.

l

n

: Length over which

∆p is measured, cm.

: Number of tubes.

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Figure 8

As there are numerous tubes and radii involved in each segment of

porous rock, it is an impossible task to use poiseuille's flow equation for porous medium.

In 1856, as a result of experimental studies on the flow of water

through unconsolidated sand filter bed, Darcy formulated a law which bears his name; this law describes, with some limitation, the movement of fluid in porous medium.

Darcy's equation states that the velocity of homogeneous fluid in a

porous medium is proportional to the pressure gradient and inversely proportional to the fluid viscosity;

ν =

Q ∆p ∝ A µ dl

Or

ν =−

∆P µ ∆l

K

ν is the apparent velocity in cm/sec. and A is the apparent or total

across–section area of the rock, cm2.

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In other words, A includes the area of the rock material as well as

dP is in the area of the pore channels. The pressure gradient dl atmosphere per cm, taken in the same direction as ν and Q .

The proportionality constant (K) is permeability of the rock

expressed in Darcy units.

The negative sign indicates that if the flow is taken as positive in the

positive direction, then the pressure decrease in that direction, so that

dP the slope is negative. dl

Limitation of Darcy's law applications 1. Darcy's law applies only in the region of laminar flow, for in

turbulent flow, which occurs at higher velocities, the pressure gradient increase at a greater rate than does the flow rate.

2. It does not apply to flow within individual pore channels, but to

portions of a rock whose dimensions are reasonably large compared with the size of the pore channels.

3. Because actual velocity is in general not measurable, apparent velocity from the basis of Darcy's law. Actual velocity can be related to apparent velocity as following: q = Aapp. . Vapp. = Aact. . Vact.

∴Vapp = Vact ⋅

Aact l P.V . ⋅ = Vact ⋅ = Φ Vact . Aapp l B.V .

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Figure 9

This means that the actual velocity of a fluid will be the apparent velocity divided by the porosity where the fluid completely saturates the rock

4. The fluid flow region is steady–state ,isothermal 5. Fluids used are non compressible fluids. 6. Use non reactive fluids. the unit of permeability is the darcy .where 1 dareg = 1000 md = 1.0133 x 106 dyne/ cm2.atm.

Dimension of permeability The dimension of permeability can be established by substituting

units of the other terms in Darcy's law as:

v=

l t

,

,

P = F = MLT −2 / A = L2 , L = L

∴ Q = AV =

µ = ML−1 T −1

(

)

KA∆P µL

∴ LT 1= = K .ML−1T −2 / ML−1T −1.L ∴ K = L2 Oil Reservoir Engineering

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Reservoir flow system applications: Several simple flow systems are frequently encountered in the measurement and application of permeability.

Figure 10

Linear flow

Figure 11

It is assumed that flow occurs through a constant cross–sectional

area that the ends of the system are parallel planes and that the pressure at either end of the system is constant over the end surface.

If the block is 100% saturated with an incompressible fluid and

steady–state flow of constant rate g;

Darcy's low for a " dl " segment of this system:

q=

KAdp µdL

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Integrating between the limits "0" and "L" and p1 and p2

q = o ∫ L dl = − ∴ q[ L] = − ∴q =

KA

µ

KA

µ

p1

∫ p 2 dp

[ p 2 − p1]

KA[ p1 − p 2] µL

If a compressible fluid is flowing, the quantity of q varies with the

pressure. The usually assumed valuation is that p.q=pmqm = constant where :

pm =

p1 + p 2 2

And Qm is the flow rate at pm the integral is therefore;

q o ∫ L dL =

− KA

µ

p1

∫ p 2 dp

[ qm pm / p]o ∫ L dL = [ − KA / µ ] p1 ∫ p 2 dp ∴o ∫ L dL =

− KA p2 pdp p1 ∫ qm pm µ 2

− KA p 2 2 − p1 ∴L = [ ] qm pm µ 2 =

KA 2  ( p1 − p 2 )( p + p 2 ) . .  qm µ p1 + p 2  2

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∴ qm =

KA( p1 − p 2) µL

Thus it is evident for the linear system that gas flow and liquid flow

can be calculated by the same equation provided the rate is measured of the mean pressure of the system

Radial flow A redial flow system, analogous to flow in to a well bore from a

cylindrical drainage system. When dray's law in differential from is

applied to a (dr) cylinder of the system, the resultant integrated equation of flow is

Figure 12

K (2πrh )∆p µdr dr 2πKh pe ∴ q rw ∫ re = dp pw ∫ µ r 2πKh( p e − p w ) ∴q = µ ln(re / rw) Q=

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in case of compressible flow: qmpm = q.p= constant

∴ q = qm pm / p ∴ qm p m / p ∴ qm pm r rw ∫ ve

re rw ∫

dr + 2πkh pe = dp pw ∫ r µ

dr 2πKh pe = pdp pw ∫ r µ

2πKh( p 2 e − p 2 w ) ∴ qm = 2 µlv(re / rw)

∴ qm =

2πKh( pe − pw ) µen re / rw

3–Spherical flow

Figure 12

such a system might be closely approached where is producing

formation was of a thickness approximately half the distance between

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wells and where the formation was penetrated for only a short distance comparable to the total pay thickness.

A = 4 ⋅π ⋅ r 2 ×

1 = 2 ⋅π ⋅ r 2 2

(

)

k ⋅ 2 ⋅ π ⋅ r 2 ⋅ ∆p q= µ ⋅ dr ∴ rw ∫ re

dr 2 ⋅ π ⋅ k pe = dp pw ∫ 2 r q⋅µ

 1 1  2 ⋅π ⋅ k ∴ −  = [ pe − pw ] q⋅µ  rw re 

q=

2 ⋅ π ⋅ K ( pe − pw ) µ ⋅  1 r − 1 r  e  w

Q rw > 1

re

2 ⋅ π ⋅ k ( pe − pw )rw

µ

For compressible fluids, the same equation can be used as:

qm =

2 ⋅ π ⋅ k ( pe − pw )rw

µ

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Conversion units of dray's 1-linear flow Qincomp =

K ⋅ A ⋅ ∆P k ⋅ A ⋅ Β ⋅ ∆P , Qincomp = µ⋅L µ⋅L

where : Q : cm3 / sec

,

A : cm 2

, ∆p = atm, µ : cp, L : cm , Qincom = 1.127 when

k ⋅ A ⋅ ∆P µ⋅L

K : darey Qincom = 1.127

: Q : bb1 / day

,

A : t2

,

K ⋅ A ⋅ ∆P µ⋅L

∆p : psia

, µ : cp , K : Darcy

Qincomp = 6.323 ⋅

K ⋅ A ⋅ ∆P µ⋅L

Where;

Q : ft 3 / day , A : ft 2 , ∆P : psia, µ : cp, L : ft. and

K : Darcy

2-Radial flow •

Qincomp = 7.08

K ⋅ h( Pe − Pw ) µ ⋅ Ln re r w

Where;

Q : bbl / day , h : ft , r : ft. and

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P : psia,

µ : cp,

K : Darcy

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•

Qincomp = 39.76

K ⋅ h( Pe − Pw ) µ ⋅ Ln re r w

Where;

Q : ft 3 / day , h : ft , r : ft. and •

Qincomp =

P : psia,

µ : cp,

K : Darcy

2 ⋅ π ⋅ K ⋅ h( Pe − Pw ) µ ⋅ Ln re r w

Where;

Q : cm3 / sec, h : cm,

P : atm, µ : cp,

K : Darcy

and

r : cm.

Permeability of combination layers The foregoing flow equations were all derived on the basis of one

continuous value of permeability between the inflow and outflow face. It

is seldom that rocks are so uniform – most porous rocks will have space variations of permeability.

If the rock system is comprised of different layer of fixed

permeability, the average permeability of the flow system can be determined by one of the several averaging procedures:

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Linear beds in series Consider the case of the average permeability Kav, can be

computed as follows:

Figure 13

Qt = Q1 = Q2 = Q3 ∆pt = ∆p1 + ∆p2 + ∆p3 L = L1 + L2 + L3 ∴ Qt =

Q2 =

K av . A.∆Pt , µ L

K 2 A ∆p 2 µL2

Q1 =

and

K1 A∆p1 µL1

Q3 =

K 3 A∆p3 µL3

Solving for pressure and substituting for ∆p ;

∴

Qt µL Q1 µL1 Q2 µL2 Q3 µL3 = + + K av A K1 A K2 A K3 A

∴

L L1 L 2 L3 = + + Kav K 1 K 2 K 3

∴ Kav =

L ∑ n Li / Ki

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i =1 (33)

As the permeability is a property of the rock and note of the fluids

flowing through permeability must be equally applicable to gases.

For example , the average permeability of 10 md , 50 md and 1000

md beds , which are 6 ft , 18 ft and 40 ft respectively in length but of equal cross– section when placed in series is:

Kav =

∑ Li 6 + 18 + 40 = = 64md . 6 18 40 ∑ Li / Ki + + 10 50 1000

1-Linear beds in parallel

Figure 14

cons tan ts : ∆p , µ var iable : k , A and

and

L

q

Consider fig (15). The total flow rate is the sum of the individual flow rates, q =q +q +q t

1

2

3

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k av . At ∆p K1 A1∆p K 2 A2 ∆P K 3 A3 ∆p = + + µL µL µL µL

∴ k av. At = K1 A1 + K 2 A2 + K 3 A3 ∴ K av = ∑ K i Ai / At if all beds are of the same width , their areas are proportional to

their thicknesses.

∴ k av = ∑ K i Ai / ∑ hi for example the average permeability of three beds of 10 md , 50

md and 1000 md and 6ft , 18 ft and 36 ft respectively in thickness but of equal with , when placed in parallel is K av =

∑ K I hi 10 × 6 + 18 × 50 + 36 × 1000 = = 616md . ∑ hi 6 + 18 + 36

2– Radial beds in parallel Constants: ∆p , r and µ variable, q and h Many

producing

formations

are

composed of strata which many vary widely in permeability and thickness,

Figure 15

If these are producing fluid to a

common well bore under the same draw– Oil Reservoir Engineering

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down and from the same drainage radius, then;

q t = q1 + q 2 + q 3 ∴

+

7.08 K av ⋅ ht ⋅ ( Pe − Pw ) 7.08 K1 = re µ Ln µ rw

7.08 K 2

h2

µ Ln

( Pe − Pw ) re rw

++

( Pe − Pw ) r Ln e rw

h1

7.08 K 3

h3

µ Ln

( Pe − Pw ) re rw

∴ K av ⋅ ht = K1 ⋅ h + K 2 ⋅ h2 + K 3 ⋅ h3 ∴ K av =

ΣK i ⋅ hi Σhi

This is the same as for the parallel flow in linear beds with the same

bed width.

Example: Calculate the average permeability permeability data given below: Depth, ft

K, md

5012 – 13

500

5013 – 16

460

5016 – 17

5

5017 – 19

235

Oil Reservoir Engineering

of

the

depth

(36)

5019 – 23

360

5023 – 24

210

5024 – 29

3

Solution: ∴ K av =

ΣK i ⋅ hi Σhi

(500)(1) + (460)(3) + (5)(1) + (235)(2) + (360)(4) + (210)(1) + (3)(5) 11 = 327 md

=

3-Radial beds in series

Figure 16

Constants

:

q, µ and

Variables

:

∆P, K and

Oil Reservoir Engineering

h r

(37)

Consider Fig (17)

Pe − Pwf = (∆P1 + ∆P1 + ∆P1 ) = ( P1 − Pwf ) + ( P2 − P1 ) + ( Pe − P2 ) qt =

2 π

K av

µ q1 =

2 π

K av

µ q2 =

2 π

K av

µ q3 =

2 π

K av

µ q µ ∴

2 π q µ

+

2 π

h ( Pe − Pw ) r Ln e rw

h ( P1 − Pw ) r Ln 1 rw h ( P2 − Pw ) r Ln 2 rw

h ( P3 − Pw ) r Ln 3 rw

re r q µ Ln 1 rw rw = K av h 2 π K1 h Ln

r r2 q µ Ln e r1 r2 + K2 h 2 π K2 h Ln

Oil Reservoir Engineering

(38)

Ln ∴

re rw

K av

Ln =

r1 rw

K1

Ln +

r2 r1

K2

Ln +

re r2

K3

re rw ∴ Kav = r r r Ln 1 Ln 2 Ln e rw r1 r2 + + K1 K2 K3 Ln

Example: What is the average permeability of four beds in series having equal formation thickness under the following conditions? 1. For a linear flow. 2. For a radial flow system if the radius of the penetrating well bore is 6 in, and the radius of effective drainage is 2000 ft. Bed

1

2

3

4

L, ft

250

250

500

1000

K, md

25

50

100

200

Solution: Assume bed (1) adjacent to the well bore; a) Linear flow:

Oil Reservoir Engineering

(39)

K av =

L L Σ i Li

250 + 250 + 500 + 1000 250 250 500 1000 + + + 25 50 100 200 2000 = = 80md 25

=

b) Radial flow: ln re

rw ∑[(ln ri / ri − 1) / Ki ] ln(2000 / .05) = = 30.4md ln(250 / .05) ln (500 / 250) ln(1000 / 500) ln (2000 / 1000) + + + 25 50 100 200 Kav =

Example: A well of 6 bore is drilled in to a pay of 500 md on a spacing of 40 Ares (re = 750 ft). Assume that the mud penetrated for a distance of one foot in to the pay and that experiment indicates that the pay will be reduced in permeability to a value of 10 % of its original. It is desired to know to what average permeability the well system is reduced by the mud penetration. Solution: The reduced permeability =0.1x500=50md

Oil Reservoir Engineering

(40)

∴ Kav =

ln(750 / 0.25) = 478md ln 1.25 / 0.25 ln 750 / 1.25 + 50 500

Example: A well of 40 acres spacing with a 6 ft bore produce 50 bb1/day of fluid from a pay of 50 md permeability before acidizing and 90 bb1/day after acidizing. If the acid had been injected to penetrated 15 ft into the formation, from these data can you calculate the permeability increase which would have had to occur in the acidized section to produce the observed increase in the production rate. Solution:

K after acid q after acidizind 90 = av = = 1.8 q before acidizind K av before acid 50 90 ) = 90md acid 50 750 Ln 0.25 ∴ 90 = 15.25 750 ( ) ( ) 0 . 25 15 . 25 Ln + Ln K1 50

∴ K av after

= 50(

∴ K 1 = 373md

Oil Reservoir Engineering

(41)

Permeability of channels and fractures In some sands and carbonates the formation frequently contains

solution channels and fractures do not change the effective permeability

of the flow network. In order to determine the contribution made by fracture or channel to the total conductivity of the system, it is necessary to express their conductivity in term of the darey.

Channels Considering Poiseulle's equation for fluid conductivity in capillary

tubes,

Q=

π

r4

∆P 8 µ L

The total area available for flow is; A=π

r2

So that the equation reduces to

r 2 ∆P Q = A⋅ ⋅ 8 µ L From Darcy's law, it is also known that;

Q = A⋅ K ⋅

∆P µ L

Equating Darcy's and Poiseuille's equations for fluid flow in a tube,

r2 K= 8 Oil Reservoir Engineering

(42)

Where If

r

K

and

r

is in cm, then

are in consistent units.

K

in Darcy is given by:

r2 6 2 K= = 12 . 5 × 10 r −9 8(9.869) × 10 If

r

is in inches; then

K

in Darcy is given by:

K = 12.5 × 106 (2.54) 2 r 2 = 80 × 106 r 2 Example:

Consider a cube of reservoir rock one foot on the side and having a matrix permeability of 10 md. If a liquid of one cp. viscosity flows linearly through the rock, under a pressure gradient one psi per ft, the rate of flow will be: Solution::

Q = 1.127(0.01) × 1 × 1 = 0.01127bbl / day If a circular opening 0.01 ft diameter traverses the same rock , then the total flow rate can be considered to be the above value (Q1) value plus the rate of flow through the circular opening (Q2) Q2 = 1.127[80 × 10 (.005) ] × 6

2

π (0.005)2 1

144 Qt = 0.01127 + 0.00122 = 0.012491

× = 0.00122 1 bb1 / day

bb1 / day

Therefore the combined rate is 0.012491 or an increase of about 11 %. Fractures

Oil Reservoir Engineering

(43)

For flow through slots or fine clearances and unit width (w) , by analogy to channels relationships , K=w2/12 If the width is in cm and K in darcys, the permeability of fractures is given by K = 84.4 x 105 w2 If w is in inches and K in dareys, K = 54.4 x 106 w2 Example: A core of very low permeability (0.01 md). It contains fracture of (0.005") wide and 1 ft in lateral extent per square foot if the core. Assume that the fracture is in the direction in which flow is desired. Calculate the average permeability of the core: Solution:

Qaw =

∑ KiAi ∑ Ai

0.00001[144 − 12 × 0.005] + 54.4 × 10 6 (0.005) × 12 × 0.005 = = 562md 144 2

Oil Reservoir Engineering

(44)

Laboratory measurements of permeability The permeability of a porous medium can be determines from

samples extracted from the formation or by in place testing. Two methods – will be presented here – are used to evaluate the permeability of cores.

perm– plug method

Figure 17

The tested samples are usually cut with a diamond drill from the

well cores in a direction parallel to the bedding plane of the formations.

Perm plugs are approximately 2 cm in diameter and 2 to 3 cm long. Samples are dried after extraction process "as described before".

The residual oil or fluids are thus: remove and the core sample becomes

100 % saturated with air. The perm plug is then inserted saturated with air. The perm plug is then inserted in a core holder of the permeability device.

Fig18 samples are mounted in such a way that the sides of the

samples are sealed, and a fluid pressure differential can be. Applied

Oil Reservoir Engineering

(45)

across their full length, and the rate of flow of fluid "air" through the plug is observed.

Obtaining data "for conditions of viscous flow" at several flow rates

and plotting results as shown in fig (19) from eq:

Q/ A =

K

µ

[( p1 − p 2) / L]

for viscous flow condition : the data should plot a straight line

passing through the origin. Turbulence is indicated by curvature of the plotted points.

The slope of the straight line portion is equal to (k/μ) from which the

absolute permeability (k) can be computed.

In case of using liquids in stead of air. Data are taken only at one

flow rate .

To assure condition of viscous flow, it is the lowest possible rate

which can be accurately measured.

Figure 18

Oil Reservoir Engineering

(46)

Whole – core measurement

Figure 19

The core must be prepared in the same manner as per–plug

method preparation. The core is then mounted in a special holding device such as shown in fig: 20 the measurements are the same as for perm–plugs but the calculation are slightly different. Measurements of

permeability on long cores generally yield better indication of the

permeability than do the small cores especially for rocks which contain fractures as limestone.

Factors affecting permeability measurements 1-Klinkenberg effect "Gas slipping" Klinkenberg has reported variations in permeability as determined

using gases as the flowing fluids from that obtained when using no reactive liquids.

Oil Reservoir Engineering

(47)

The permeability of a rock for any liquid will be the same but for a

gas will depend upon the individual gas and the mean pressure of flow.

The measured permeability of a porous medium to gas is greater

than that to a liquid for the liquid has a zero velocity as the wall past which it flows but that the gas has a finite velocity at the wall. Fig (21)

Figure 20

This is said due to the gas slippage. The gas slippage is a function

of the mean pressure and the type of rock. The phenomena of gas

slippage occurs when the diameter of the capillary openings approach the mean free path λ.

The mean free path of a gas is a function of the molecular size and

the kinetic energy.

The Klinkenberge effect is a function of the gas for which the

permeability is determined.

The amount of slippage causes a change in permeability that can

be represented by the following equation:

K g = K l (1 +

4Cλ ) r

Where;

Kg

: Gas permeability

Oil Reservoir Engineering

(48)

Kl

: Liquid permeability

r : Mean radius of capillary tubes. C

: Constant nearly equal One.

The last equation can else be written in the following formula:

K g = K l (1 +

b ) Pm

Where;

Pm observed,

b

: Mean flowing pressure of the gas at which

Pm ∝

Kg

1

λ

: Constant for a given gas and rock,

1 b ∝ ,λ r

Figure 21

Oil Reservoir Engineering

(49)

was

Fig (22) is apart of the equation at various mean pressures using

Hydrogen, Nitrogen and Carbon dioxide.

Note that for each gas a straight line is obtained for the observed

permeability as a function of the reciprocal of the mean pressure of the test.

The data obtained with lowest molecular, (H2), weight gas yield the

straight line with greater slope, indicative of a greater slippage effect.

All the lines when extrapolated to infinite mean pressure,

1 =0 P ), intercept the permeability axis at a common point. ( m According to Fig (22); The slope of the straight line, m, is equal The constant

b

To

accurate

Kl ⋅ b

.

depends on the mean free path λ of the gas and

the size of the openings in the porous medium. obtain

permeability

measurements,

requires

approximately 12 flow test under viscous flow conditions from which the permeability to liquid can be graphically determined.

2-Clay content Many clays act as cementing minerals or are as part of the rock

matrix. These minerals or are as part of the rock matrix. These minerals

are usually very complex in molecular structure and posses the ability to

attract and hold +ve ions such as hydrogen , sodium or calcium. These

Oil Reservoir Engineering

(50)

minerals also have the property of hydration, i.e., holding water within their molecular structure. This is normally called "clay swelling"

Reservoir waters generally contain ions such as sodium and

calcium or magnesium which can be transferred to the clay. The greater

the amount of ions in solution the move will be absorbed by the clay and vice verse. If the reservoir water is replaced by fresh water, the clay

must given up some of its +ve ions to the water until a new equilibrium is

established. The result of this ion exchange and the change of ion

concentration in the flowing liquid is an increase in clay volume and consequently reducing the pore volume available for flow.

Figure 22

Fig (32) shows such a decrease in flow rate with fresh water flow

thus permeability determination with fresh water on a core containing clays will be less than that in the natural state.

3-Reactive liquids Reactive liquids after the internal geometry of the porous. Medium,

due to precipitation or corrosion.

Oil Reservoir Engineering

(51)

4-Overburden pressure When the core is removed from the formation, the confining forces

are removed. The rock matrix is permitted to expand in all direction partially changing the shapes of the fluid– flow paths inside the core.

Compaction of the core due to overburden pressure may cause as

mush as 60 % reduction in the permeability of various formations.

5-Grain size It was found that the rate of fluid flow is proportion al to the square

of the grain diameter , hence the finer sand the smaller the permeability.

6-Mode of packing: The effect of packing as a factor which influence permeability can

be introduced as:

K = 10.2 d2 / c Where

d: diameter of spheres, cm

mode of packing

C: packing constant depending on porosity . table:1 Φ

c

Hexagonal

26

52.5

30 52.5 40 20.3 cubic

Oil Reservoir Engineering

45

23.7

(52)

Effective and relative permeability It was note that Darcy's low for flow in porous media. was

predicated upon the condition that the porous media was entirely saturated with the flowing fluid such a circumstance does not often exist

in nature , particular in petroleum reservoir. Gas or oil is usually fauna coexistent with water and frequently gas, oil and water may occupy together the pores of reservoir.

Ability of aporous medium to conduct a fluid when the saturation of

that fluid in the material is less than 100 %of the pore sbace is known as

the "effective permeability of the porous medium to that fluid. The

effective permeability is written by using a subscript to designate the fluid under consideration. For example Effective permeability to oil ,

Ko =

Effective permeability to gas ,

Qo µ o L A∆p

Kg =

Qg µ g L A∆p

And Effective permeability to water ,

Kw =

Qw µ w L A∆p

Of course , it would be expected that for a given system (A , Δp , μ ,

L being constant) , the value of Qx would increase as the pore space of

the porous medium in question contained more fluid (x). it has been found experimentally that at a given value of fluid saturation , the value

of the effective permeability to that fluid is constant. Thus , a change in effective permeability depends only upon a change in saturation.

Oil Reservoir Engineering

(53)

Effective permeability will , of course , vary from zero to the value of

the permeability at 100 % saturation. No two porous bodies will necessarily have the same variation of effective permeability with saturation.

Relative permeability Relative permeability is defined as the ratio of the effective

permeability to a given fluid at a definite saturation to the permeability at 100 % saturation. The terminology most widely used is simply (Kg/ K),

(Ko /K) and (Kw /K), meaning the relative permeability to gas, oil and

water respectively. Since K is constant for a given porous medium, the same fashion as dose the effective permeability. The relative

permeability to a fluid will vary from a value of zero at some low

saturation of that fluid (critical saturation), to a value of 1 at 100 % saturation of that fluid.

Characteristics of two – phase relative permeability curves: Figure 23

a) Rapid

fall

permeability

in

to

wetting – phase (Krw)

wetting

as

saturation

the

phase

Oil Reservoir Engineering

first

(54)

decreases from 100 %. b) Its approach to zero value at saturation greater than zero %. c) Relative permeability to the no wetting phase (krnw) apparently dose not rice.

Above zero until the wetting phase saturation has fallen to

approximately (1– Snwc) where Snwc is the critical non-wetting saturation or equilibrium saturation.

1. Rapid rise of the permeability (Krnw) as the wetting phase saturation decreases.

2. Virtual attainment of 100 %permeability to non-wetting phase before all the wetting phase is completely removed.

Interpretation 1. The rapid decline in Krw indicates that the larger pores or larger flow paths are occupied first by the non-wetting fluid.

Figure 24

Oil Reservoir Engineering

(55)

As the saturation of the non-wetting phase increases, the average

pore size saturated with wetting phase be comes successively smaller. Fig (25)

2. Blocking off of central regions of pores by non-wetting phase will leading to an increase in flow resistance or rapid fall in krw.

This behavior will continue with decreasing (Sw) until (Swe),

where a minute film which would wet the surface of the grains.

This film would decrease the diameter of the larger tube, thus

reducing the flow capacity for the non-wetting phase, and yet the film itself would contribute no flow capacity. Fig (26)

Figure 25

To the wetting phase (Krw = 0). Thus the total fluid

of the tubes would be decreased.

capacity

3. This point represents the equilibrium saturation which is the value at which the non-wetting phase becomes mobile.

4. At any saturation value above this saturation wetting-phase saturation falls into the pendular saturation state, so that it

losses its own mobility and this confirmed by the rapid rise in Krnw.

Oil Reservoir Engineering

(56)

5. The attainment of 100 % Krnw at saturation of non-wetting –

phase less than 100 % indicates that a portion of the available pore space even though interconnected contributes little to the fluid conductive capacity of the porous medium.

Three– phase– relative permeability There are many instances when not two fluids but three fluids exist

in rock simultaneously as oil, water and gas.

Figure 26

Fig (27) show the triangular diagram representing the saturation

conditions of rock the coordinates of any point with the triangle represent the different saturations in all three phases. For example;

Point So Sw Sg

1 20 20 60

Oil Reservoir Engineering

2 20 40 40

3 40 40 20

4 60 20 20

5 40 60 0

(57)

The symmetry of water permeability with respect to any axis of the

triangle is due to water is considered as wetting – phase with respect to each of oil and gas phases.

Figure 27

The symmetry of gas relative permeability is due to similar behavior

of gas as anon wetting phase with respect to oil and water.

Figure 28

Oil Reservoir Engineering

(58)

The relative permeability to oil is not symmetrical with respect to any

axis of the triangle. This symmetry is distorted toward the high percentage of gas saturation which indicates a lowering of mobility of oil by the presence of gas.

Figure 29

Fig: 31 shown the regions in which single–phase, two–phase and

three phase fluids flow normally occur. Figure 30

The region of three – phase flow

is externally small it is illustrated by

the hatched area two – phase flow region

the

white

area

,

which

indicates that in most cases two

phase relative permeability curves are quite satisfactory the single –

phase regions are illustrated by the shaded area.

Oil Reservoir Engineering

(59)

Factors effecting relative permeability 1-Rock wettability Figure 31

The

permeability

relative

values

are

affected by the change in fluid distribution brought

about by different wetting characteristics.

Curve (1) indicates

that the system is water wet system while, curve (2) indicates oil wet

system.

2-Structure history (Drainage or imbibition)

Figure 32

Oil Reservoir Engineering

(60)

It is note that the imbibition processes causes the non-wetting

phase (oil), to lose its mobility at higher value of structure than does the drainage process.

The drainage method causes the wetting Phase (water) to lose it'

mobility at higher value of (Sw) than does the imbibition method.

3-Pore configuration and pore size distribution

Figure 33

Curve 1

: Capillary tubes.

Curve 2

: Dolomite.

Curve 3

: Unconsolidated Sand.

Curve 4

: Consolidated sand.

Oil Reservoir Engineering

(61)

Relative permeability ratio

Figure 34

The relative permeability ratio expresses the ability of a reservoir to

permit flow of one fluid as related to its ability to permit flow of another fluid under the same circumstances.

Kg The two most useful permeability ratios are ( permeability to gas with respect to that to oil and (

Ko Kw

). The relative

K o ), the relative

permeability to water with respect to that to oil, it is understand that both quantities in the ratio are determined simultaneously on a given system.

The relative permeability ratio may vary in magnitude from zero to infinity.

Oil Reservoir Engineering

(62)

Consider a system flowing gas and oil. At high oil saturation, the flow of gas is zero,

Kg

Kg and hence,

Ko

is zero.

Kg

increase but

Figure 35

As the gas saturation increases,

Ko

decrease

Kg and therefore

Ko

increase. When the oil saturation becomes

Kg sufficiently low,

Ko

approach zero and the value of

Ko

approach

Kg infinity. Fig (35) is a typical plot of

Ko

versus the oil structure. To

Kg give linearity,

Ko

was plotted against oil structure saturation or

liquid saturation on a semi–log paper, Fig (36). It has become common

Oil Reservoir Engineering

(63)

usage to express the central straight line portion of the relationship in the analytical form:

log( Kg Ko

Kg Ko

) = log a − b ⋅ S o

= a ⋅ e− b⋅S o

Where; and

b

are constants characteristic only of a given reservoir

material and

a

given set of fluids. Slope of the linear plot is denoted b .

a

The use of this analytical expression has been justified in view of

agreement between theoretical relationship which may be deduced from it and actual observed data.

Measurement of relative permeability data There are essentially for means by which relative permeability data

can be obtained:

(1) Laboratory method. (2) Capillary pressure method. (3) Field data. (4) Petrochemical data.

1-Laboratory method:

Oil Reservoir Engineering

(64)

Figure 36

A small core is

choose and prepared for

the

mounted

test.

pressurized

in

It

is

a

rubber

sleeve. The two fluids

are introduced simultaneously at the inlet and through different pipe

systems, at a predetermined flow ratio and are flowed through the core until the produced ratio is equal to the injected ratio.

The saturation of the various fluids is determined in one of different

methods (as measuring the fluid resistively by means of two electrodes).

Once the saturation has been measured, the relative permeability of

the two phase at this saturation conditions can calculated by means of the injection ratio.

The injection ratio is varied and the process continually repeated

until a complete relative–permeability curve is obtained.

2-Field determination Due to Darcy's equation for gas and oil, the relative permeability

ratio can be defined by the following equation:

Qg Qo

A(

Kg

= A(

Ko

Oil Reservoir Engineering

µg

)(

∆Pg

µg

)(

∆Po

∆L ∆L

)

)

(65)

Figure 37

If the

Qo and Qg

are expressed at reservoir condition and if it is

assumed the pressure drops (

∴

∆Pg

and

∆Po ) are the same,

Qg K g ⋅ µo ⋅ β o free gas = = Qo K o ⋅ µ g ⋅ β g oil produced

∴ Rp =

total gas prod free gas gas in sol = + oil prod oil produced oil produced

Or;

∴ Rp =

Qg Qo

+ rs

K g ⋅ µo ⋅ β o ∴ Rp = + rs Ko ⋅ µ g ⋅ β g ∴

Kg Ko

=

µg ⋅ β g ( R p − rs ) µo ⋅ β o

The normal procedure is to use field average GORs which are

normally the most accurate values obtainable. The saturation at which

this particular value of relative permeability ratio applies must be calculated from field production data as follow:

Oil Reservoir Engineering

(66)

So =

oil volume ( N − Np ) β o = N ⋅ Bo i pore volume (1 − S w i )

So = (1 −

Np β o )( )(1 − S wi ) N Boi

Example:

Given the following data: Pressure Rp βo β g x10–

r

6

3000 2900 2800 2700 2600 2500 2400 2300 2200 2100 2000 1900 1800 1700 psi

850 1.443 840 752 920 1.432 875 725 990 1.420 910 695 1020 1.403 970 657 1000 1.393 1010 632 1180 1.382 1062 608 1420 1.371 1122 580 1510 1.364 1162 565 1666 1.354 1230 540 1920 1.340 1330 509 2220 1.326 1453 476 2480 1.313 1590 446 2710 1.301 1758 416 2800 1.298 1795 410 SCF/STR bbl/STB bbl/SCF SCF/STB

µo

Np

µg

30.4 32.1 34.0 36.8 38.4 40.5 42.4 43.6 45.5 48.0 50.8 53.8 57.4 58.2

N 0.0041 0.0150 0.0240 0.0399 0.0517 0.0632 0.0752 0.0847 0.0965 0.1083 0.1202 0.1318 0.1420 0.1453

Calculate constants a and b in the equation if S wi = 28.5%

Kg and K o

= a ⋅ e − bS o .

Solution: Oil Reservoir Engineering

(67)

Kg

For each pressure step calculate So and K using equations; o

µg ⋅ β g ∴ = ( R p − rs ) K o µo ⋅ β o Kg

And,

So = (1 −

Np β o )( )(1 − S wi ) N Boi

Example of calculations at P=2500 psi

Kg Ko

=

1 × 0.001062 (1180 − 608) = 0.01085 40.5 × 1.382

1.382 So = (1 − 0.0632)( )(1 − 0.285) = 0.638 1.443

P

So

Kg Ko

3000

0.710

0.00187

2900

0.696

0.00321

2800

0.684

0.00556

2700

0.664

0.00682

2600

0.652

0.00694

2500

0.638

0.01085

2400

0.625

0.01620

2300

0.615

0.01848

2200

0.609

0.0224

2100

0.589

0.0292

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(68)

2000

0.575

0.0377

1900

0.563

0.0456

1800

0.550

0.0540

1700

0.540

0.0567

Figure 38

Determination of constants a and b ; At So = 0.516 At So = 0.648

Q

Kg Ko

Kg , K = 0.1 o

Kg , K = 0.01 o

= a ⋅ e − bS o

∴ 0.1 = a ⋅ e −0.516b

Oil Reservoir Engineering

(69)

And,

∴ 0.01 = a ⋅ e −0.648b So,

ln10 = 0.132b

∴b = 2.3 ln(

Kg Ko

0.132

= 17.42

) = ln a − b ⋅ So

∴ ln(0.1) = ln a − 8.988 ∴ ln(a ) = 8.988 + (−2.3) = 6.688 ∴ a = 803

Uses of effective and relative permeability data Relative permeability data are essential to all flow work in the field

of reservoir engineering. Just a few of its uses are mentioned engineering. Just a few of its uses are mentioned here.

Determination of free water level: • From the relative permeability curves, it should have become apparent that the point of 100% water flow is not necessarily the point of 100% water saturation. It is recognized that there are two water levels. • Free water level: zero capillary pressure level. • Initial WOC: the level below which fluid production is 100% water, from relative permeability data the engineer can determine what the fluid saturation must be at the point of zero oil permeability. When the fluid saturation determined from well test data and relative Oil Reservoir Engineering

(70)

permeability curves are used, the capillary pressure can be determined and the height above the free water level can be calculated. Other uses of relative permeability data 1- Determination of residual fluid saturation. 2- Fractional flow and frontal advance calculation to determine the fluid distribution. 3- Making future prediction for all types of oil reservoir where two phase flow is involved. 4- Emulation of drill–steam and production tests.

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Capillary forces Surface and capillary pressure Consider two immiscible fluids (water and oil, fluid commonly found

in petroleum reservoirs). A water molecule which is within the body of the water will uniformly attached in all directions, by an attractive force,

by other molecules and thus the resultant force on the molecule will be zero

Figure 39

A water molecule at the interface has force acting upon it from the

oil lying immediately above the interface and water molecules lying

below the interface. The resulting forces are unbalanced and give rise to interfacial tension.

Figure 40

This attractive force tends to attract the surface molecules in to the

liquid and minimize the surface area. A certain amount of work is

required to move a water molecule from within the body in the liquid

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(72)

through the interface; this work is referred to as the free surface energy of the liquid.

Surface tension The force in dynes acting in the surface perpendicular to a line cm

of length and for a distance of one com in order to produce the new until area of the surface, or it is the force per unit length required to create a unit new surface.

Surface tension is the force acting on the surface between a liquid

phase and a gaseous phase while the interfacial tensing is created at the interface between two liquids.

liquid liquid or solid

vapour or air liquid surface tension

int erfacial tension Figure 41

One of the simplest example of the surface tension is the tendency

of free volume to take the minimum possible form as, for example, a sphere in the case of a free drop liquid.

Capillary pressure force The forces that are active at the interface between two immiscible

liquid faces

Adhesion tension, At: Oil Reservoir Engineering

(73)

Figure 42

σ so = σ ws + σ wo cosθ ∴σ wo cosθ = σ so − σ ws = At + (ve) ∴ At = σ wo cosθ A positive adhesion tension indicates that the dense phase (water in

this case) wets the solid surface.

An adhesion tension of zero indicates that both phases have an

equal affinity for the surface.

Figure 43

At = σ wo cosθ

σ wo cos 90 = 0 Thus the adhesion tension determines the ability of the wetting

phase to adhere to the solid and to spread over the surface of the solid.

Oil Reservoir Engineering

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Figure 44

If the contact angle σ

f 90 ,

an outside energy will be required to

cause the dense fluid to spread over the surface.

Adhesion tension or the degree of spreading depends upon the

contact angle of the system that affected by the mineralogy of surface (rock) and the kind of the two immiscible fluids.

Rise of fluids in capillaries

Figure 45

The rise in height, Fig (46) is due to the attractive force (adhesion

tension) between the liquid and tube that tends to pull liquid upward. This total upward force is balanced by the weight of the liquid column

Upward force

Downward force

At and 2 ⋅ π ⋅ r ,

π ⋅r2 ⋅h ⋅ g

Equating this two balanced forces:

Oil Reservoir Engineering

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∴ 2 ⋅ σ cosθ = r ⋅ ρ ⋅ g ⋅ h ∴

2 ⋅ σ cosθ = ρ ⋅ g ⋅ h = ∆P r

(1)

It is noticed from the shape of the interface between the two phases

in the tube that the pressure in the liquid phase beneath the interface (A) is less than above the interface.

This difference in pressure existing across the interface is referred

to as capillary pressure of the system.

Pc = Pnw − Pw = ρ ⋅ g ⋅ h =

2 ⋅ σ cosθ r

(2)

Figure 46

Fig (47) shows the condition of two liquid phases compared with

case of liquid and gaseous phase.

Oil Reservoir Engineering

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Calculation of capillary pressure, Pc 1-For liquid–air system

Figure 47

At the point B' within the capillary the tube pressure is the same as

that. At the point B outside the capillary, which pressure is atmospheric.

At the point A' just under the meniscus with in the capillary, the

pressure is equal to that at B' within the capillary minus the head of water.

The pressure at A' is therefore;

PA' = PB ' − ρ w ⋅ g ⋅ h

(3)

Now, at the point A, just above the meniscus within the capillary, the

pressure is the same as that at B.

This statement can be made because the density can be neglected.

Oil Reservoir Engineering

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The pressure across the meniscus (or the phase boundary) is

therefore;

Pair − Pwater = Pnw − Pw = Pc

(4)

= PA − PA' Where;

Pw and Pnw are

the wetting phase pressure (water) and the non

wetting–face pressure (air), and

ρw

is the water density.

Substituting equation (3) in equation (4);

∴ Pc = PA − PA' = PA − ( PB ' − ρ w ⋅ g ⋅ h) = PA − PB ' + ρ w ⋅ g ⋅ h

(5)

As mentioned before;

PA = PB ' = PB = Patm ∴ Pc = ρ w ⋅ g ⋅ h

(6)

For two immiscible liquids (oil and water) In this case the pressure at A is now equal to that at B minus the

head of oil, because the oil density isn't negligible compared to water density.

Oil Reservoir Engineering

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PA = PB − ρo ⋅ g ⋅ h

(7)

Substituting equation (7) into equation (5);

∴ Pc = ( PB − ρo ⋅ g ⋅ h) − PB ' + ρ w ⋅ g ⋅ h = ( ρ w − ρo ) ⋅ g ⋅ h

(8)

It was noted in equation (1) that the quantity ( h ) could be expressed by the surface tension ( σ ) and the contact angle ( θ );

therefore, the capillary pressure can be expressed as:

Pc =

2 ⋅ σ ⋅ cosθ = ∆ρ ⋅ g ⋅ h r

(9)

It noticed from equation (9) that the capillary pressure is a function

∴ At = σ cosθ

of the adhesion tension (

the radius of the capillary tube.

), and inverse proportional to

Thus, across a fluid boundary which is within a larger vessel, the

capillary pressure will be zero, or substantially so, because infinitely large.

r

becomes

Capillary pressure in unconsolidated sands Reservoir rocks are varying in complexity of pore structure, and a

simple or ideal porous structure is required as a starting point to explain their capillary behavior. The ideal pore configuration usually chosen is that made up of spherically uniform particles of definite size, i.e., unconsolidated sand.

Oil Reservoir Engineering

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Figure 48

Consider to spherical grains in contact as shown in Fig (10) with a

wetting fluid at the point of contact. A contact angle of zero will be assumed in order to have the condition of a continuous film of the wetting phase a round the sand grain. In this system, the capillary pressure is given by;

Pc = σ {

1 1 + } R1 R2

(10)

Where;

R1

and

R2

are the radii of the curvature of the interface and

the interfacial tension between the two fluids. The values of

R1

σ

is

and

R2 expresses the amount of fluid that is contained at the contact, or the

saturation of that fluid within the porous body if the number of such contacts are considered. It is practically impossible to measure the

Oil Reservoir Engineering

(80)

values of radius (

R1

and

R2

so they are generally referred to by the mean

Rm ); where;

1 1 1 = + Rm R1 R2

(11)

The mean radius is empirically determined from other measurement

on a porous medium.

Referring to equation (10), it is seen that if

R1

and

R2 both

decreased (i.e. the quantity of the wetting–face decreases); the

magnitude of the capillary pressure would in turn have to increase in size. It is therefore possible to express the capillary pressure as a function of rock saturation when two immiscible fluids are used within the

porous medium. In other words, smaller water saturation gives a greater capillary pressure.

Equation (8) and Equation (10) together demand that: at a given

height within a reservoir, the amount of water that is held by capillary pressures will increase as the permeability decreases. Also the capillary pressure increase with height in the reservoir.

Capillary pressure curves The relationship between water saturation at any point in a porous

body and the capillary pressure at that point is known as the capillary

pressure. Since the capillary pressure must vary with height above the free water level in a porous section, the capillary pressure curve

expresses also the relationship of water saturation to height above the Oil Reservoir Engineering

(81)

free water level. Fig (50) shows a typical curve, any point in this curve represents an equilibrium condition.

First, it is assumed that the reservoir rock was originally filled with

water.

Figure 49

Second, this water was displayed by the oil which accumulates in

the reservoir. The capillary pressure can be measured by finding out

how much pressure must be applied on oil (non wetting fluid) in order to reach certain saturation in that fluid (wetting–phase fluid). If the largest

capillary opening be considered as circular of radius r , the pressure needed for forcing the oil will be

∆P =

2 ⋅ σ ⋅ cos θ r

.

It is the minimum pressure at which the non-wetting fluid starts to

enter the core because any capillary of small radius will require a higher

Oil Reservoir Engineering

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pressure application. This minimum pressure is called the "displacement pressure" of the core.

As the driving pressure upon the non-wetting fluid is increased,

capillaries of smaller and smaller radii are penetrated by the non–

wetting fluid. Should the capillaries of the specimen be highly uniform in

size , no excess pressure would be required to saturate them in nonwetting fluid and the plot of pressure applied vs. fluid saturation would be very flat nearly until the irreducible saturation (Swi) is reached. This is illustrated incurve "1" fig (51).

Figure 50

On the contrary, should the capillaries be of very heterogeneous

size, the capillary – pressure curve would be very step such as curve (3)

curve (2) is the capillary pressure for capillary size distribution of medium heterogeneity.

It may be summarized that the capillary pressure of a reservoir rock

as a function of fluid saturation is a measure of capillary size distribution,

Oil Reservoir Engineering

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which in turn is a measure of rock texture. Fine textured rocks made up of small cemented grains, closely packed exhibit a higher capillary

pressure at a given saturation then coarse textured rocks made up of large grains, poorly cemented and loosely packed.

There is no sharp line between oil and water level, Fig (51). The

depth internal within which the saturation (Sw) changes from 100% to the irreducible saturation (Swi) is known as the "transition zone", thus, the

water–oil contact con not be said to exist at a definite depth, but rather within a range of depth.

In general, less permeable sands are expected to have a greater

transition zone, Fig (52).

Figure 51

A distinction must be made between the displacement pressures

and "threshold pressure", the former refers to the entrance pressure of

the non-wetting fluid into the porous medium fully saturated with the wetting fluid.

Oil Reservoir Engineering

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However, if the medium is partially

saturated with non-wetting fluid, the

entrance pressure is reduced to a point

below

the

displacement

pressure. If the saturation in the nonwetting

fluid

is

the

equilibrium

saturation to this phase, the threshold pressure is now zero. Fig (53). Figure 52

Drainage and imbibition capillary pressure curves To obtain the drainage curve,

the core was saturated with water

and then let a non-wetting fluid (oil or air), be forced into the core. In this

desaturation direction, the water is

displaced from the larger toward the smaller capillaries. Fig (54) Figure 53

By contrast, when the water

saturation changes are toward larger

saturation values, "in the imbibition direction", which is also the direction

of displacement of oil or air by water. The oil saturation is reduced until it reaches to the ultimate oil saturation "residual oil saturation".

Oil Reservoir Engineering

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A higher value of water saturation for a given capillary pressure

value would be obtained if the porous system was being desaturated (drainage), then was being resaturated (imbibition) with wetting fluid.

We can summarize factors affecting capillary pressure curve as:

• The pore size distribution. • The saturation history "drainage or imbibition direction". • Rock homogeneity or heterogeneity. • Rock permeability. • The kind of fluid and solids that are involved. Thus in order to use capillary pressure data, these factors must be

taken into consideration before the actual application of the data.

Laboratory determination of capillary pressure curve A typical apparatus which is used for the determination of capillary

pressure curve is shown in Fig (16). Many versions of the apparatus are used but the basic principles are the same.

Oil Reservoir Engineering

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The porous diaphragm is chosen so that it will permit water to flow

through it, but will not permit air to flow through it under the pressure necessary. This means that the diaphragm must have very small pores and must be water wet.

To determine the capillary pressure curve to quantities must be

measured:

1-Sw in the core at any time. 2-Pc at the same time By definition, Pc is difference in pressure between the gas (air) and

water phases in the core. The water saturation (Sw) can be determined either by wetting the core from time to time or by measuring the volume of water that has been removed from the core from time to time.

Procedure • The porous diaphragm must first be completely saturated with water.

• On the diaphragm is placed a thin layer of finely powder, the

purpose of which is to ensure good contact between the diaphragm and the core. This layer is completely saturated with water.

• The core itself is also completely saturated with water and placed on the layer of powder.

• The apparatus is then closed so that an air pressure can be applied. This pressure is set at some definite and contact value (Pair).

Oil Reservoir Engineering

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• Because the diaphragm has been chosen with such a fine pore structure that air can not penetrate it, there will be no air

flow except into the core. Air flow into the core will displace water which itself can path through the diaphragm. However air will displace water from the core through the diaphragm for a while but eventually all flow will stop and the system will be static.

• When the system has reached this point of no flow, the pressure in both the water and air phases within the core is known.

A. Pair = the applied pressure. B. Pw = Patm (the pressure below the diaphragm) – the head, ΔP from the bottom of the diaphragm to the center of the core.

∴ Pc = Papplied − ( Patm −

hcore

2

⋅ g ⋅ ρ ⋅ w)

• At this point, the core can be removed from the holder and the wide determined, weight of the core minus the dry weight gives the weight of water present and consequently the saturation.

• The core is replaced in the holder and the process is repeated,

a new air pressure higher than the first beginning used to determine a new capillary pressure and a new saturation.

• By making a succession of such determinations, the entire capillary pressure curve is obtained and plotted.

Oil Reservoir Engineering

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Example:

A dry core weights 8.59 gms and then 100% saturated with water it weights 9.74 gms. It is placed in a closed container diaphragm. A constant pressure is applied and when equilibrium is attended the core reweighed for various applied pressures, the following data were obtained: Pressure

Weight

10 20 30 40 50 60 80 100 150 200 300 400

9.74 9.74 9.68 9.57 9.41 9.29 9.50 8.96 8.88 8.85 8.82 8.82

mmHg

gms

a. Plot Pc versus Sw. b. Find Pd and S wi . Solution:

Oil Reservoir Engineering

(89)

Figure 54

P.V . = =

wt . of

100% sat . sample − wt

of

dry

sample

ρ fluid

9.74 − 8.59 = 1.15cm 2 1 Sw =

volume of P.V .

fluid

And,

Pc = Papplied Pc

Oil Reservoir Engineering

Pc

Sw

10

100

20

100

30

94.7

40

85.2

50

71.3 (90)

60

60.8

80

40.0

100

32.2

150

25.2

200

22.6

300

20.0

400

20.0

From Fig

Pd = 20 mmHg

S wi = 20%

The Jamin effect The idea of the capillary pressure is that a pressure existing across

an interface within a capillary system.

Figure 55

When more than one interface is present in a given channel–

condition may be such that the resistance to flow is markedly increase,

or may become great enough to prohibit flow. This effect is named "Jamin effect".

Consider at first a straight pore cylindrical capillary. The capillary

pressure which is equavelent to the displacement is given by:

Oil Reservoir Engineering

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Pc =

2 ⋅ σ cosθ r

This the pressure difference between point A and B, it is also the

pressure necessary to keep the interface from moving to the right at point B within the capillary, Fig(57)

Figure 56

∴ J effect = PB − PA =

2 ⋅ σ ⋅ cosθ r

from   → to   → A B

Consider a discrete global of one fluid within another fluid with

which it is immiscible, Fig (20).

Figure 57

There are now two interface, the pressure drop across each

interface is the same but opposite in the direction to the other, there is

no net pressure necessary to prevent motion, the total pressure drop between the point A and B is zero as seen by;

J = PB − PA = (

2 ⋅ σ ⋅ cosθ 2 ⋅ σ ⋅ cosθ )A − ( )B = 0 r r

Now if either term of this equation were modified, the net pressure

drop between point A and B would not be zero.

Oil Reservoir Engineering

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This condition gives the Jamin condition, i.e. the resistance to flow.

The difference may not be zero due to a change in any one of the three terms σ ,

cosθ

or r .

Variation in " r ": The difference in pressure between point A and B is:

PB − PA = 2 ⋅ σ ⋅ cosθ (

1 1 − ) rA rB

∴ rB pp rA ∴ PA f PB A positive pressure is required at point A to retain the bubble in

position shown. If flow were to the right, a bubble of oil in the water stream could block such a channel until the pressure drop between point A and B was sufficiently great to push the bubble through the smallest construction.

Variation in contact angle θ Consider Fig (61). The resultant pressure between point A and B:

Figure

PB − PA =

58

2 ⋅σ (cosθ A − cosθ B ) r

Oil Reservoir Engineering

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As θ A f θ B

∴ cosθ A p cosθ B ∴ PA f PB

Figure 59

A pressure drop between A and B is necessary to initial flow toward

the right in the figure shown.

Variation of interfacial tension σ

Figure 60

If a bubble of gas is bounded on one side by oil and on the other by

water. The net effect between A and P is then:

PB − PA =

2 (σ A ⋅ cosθ A − σ B ⋅ cosθ B ) r

If σ B ⋅ cos θ B f σ A ⋅ cos θ A

∴ PA f PB A positive pressure drop from A to B would be necessary to initial

flow to the right.

The overall Jamin effect is of course increased in direction

proportional to the total number of bubbles that exit in a given channel.

Oil Reservoir Engineering

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Calculation of wettability There are some means expressing the degree of wettability:

1-contact angle θ A contact angle of zero would indicate complete wetting by the

dense phase.

An angle of 90o indicated that either phase wets the solid. An angle of 180 o complete wetting by the less dense phase.

2-the sessile drop ratio

Figure 61

It is the ratio of the height of droplet in the surface to the breads of

the droplet.

r=

h b

Sessile drop ratio of one indicates complete non-wetting the solid

surface by the dense phase.

A ratio of zero indicates complete wetting the solid surface by the

dense fluid.

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3-the wettability number Wettability number (W.N.) =

σ o.a PT ⋅ σ w.o PT

w. o

o.a

Where;

PTw.o

= the threshold pressure of core for oil to enter when core

PTo . a

= threshold pressure of core for air to enter when core initially

initially saturated with water.

saturated with oil.

A wettability number of one would indicate a complete wetting by

water.

A wettability number of zero would indicate complete wetting by oil.

Relationship between gravity and capillary forces Gravity force: the force tends to expel water from the rock. It is also the force that causes oil to force water out of the rock

pores and it is opposite by the capillary force.

The water saturation at any point in the reservoir is the result of a

balance between the capillary and gravity force.

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As the vertical distance above the free water level (Pc = 0)

increases, the gravity forces in the reservoir increases and water saturation tends to be lower.

The gravity pressure in the reservoir is analogous to the capillary

pressure in the laboratory. Gravity pressure

≡

gravity pressure

≡ ∆ρ ⋅ g ⋅ h

The reservoir point that can generally be determined from electric

logs is the depth of the free water level.

The laboratory capillary pressure test starts out with 100% water

saturation in the core and zero capillary.

However, the starting laboratory point corresponds to the reservoir

free water level not the original water oil contact (OWC). This is directly

used in laboratory to convert laboratory data to field data, when need to calculate the depth of the free water level in the reservoir.

Fig (63) shows the distinction between the free water level and the

original water oil contact.

Oil Reservoir Engineering

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Figure 62

Converting laboratory capillary pressure data To use laboratory data of capillary pressure it is necessary to

convert to the reservoir condition.

( Pc )lab =

2 ⋅ σ w.a cosθ L 2 ⋅ σ L cos L = r r

( Pc ) Re s =

2 ⋅ σ o.w cosθ R 2 ⋅ σ R cos R = r r

If

θ L = θ R = 180o

∴ Pc R = Pc L ⋅

σR σL

Example:

Calculate the reservoir capillary pressure from the following laboratory data: Oil Reservoir Engineering

(98)

Pc L = 18 psi ,

S w = 35% ,

(σ o.w ) R = 24dyne / cm ,

(σ a.w ) L = 72dyne / cm ,

ρ w = 68lb / ft 3 and

ρo = 53ib / ft 3 .

Solution:

( Pc ) R = ( Pc ) L ⋅

σR 24 = 18 × = 6 psi σL 72

To convert the capillary pressure saturation data to height saturation it is only necessary to rearrange the equation:

Pc = ∆ρ ⋅ g ⋅ h 144( Pc ) R = ( ρ w − ρo ) g ⋅ h ∴h =

144( Pc ) R 144( Pc ) R = ( ρ w − ρo ) g ∆ρ ⋅ g

Where;

ρ w , ρ o : lb / ft 3 ,

h : ft,

Pc : psi

Example:

Calculate the height of the saturation plane for the last example.

Solution:

∴h =

144( Pc ) R 144 × 6 = = 58 ft. ( ρ w − ρo ) (68 − 53)

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∴ The water saturation (Sw = 35%) exists at a height of 58 ft above the free water level. Example:

Calculate the water saturation profile (depth versus Sw), for the following laboratory capillary pressure data, considering that the initial WOC is located at depth of 3788 ft. Sw

Pc

100 100 85 72 58 52 45 39 35 32 30 30

0 1.9 2.02 2.34 2.74 3.25 3.60 4.57 5.44 7.20 10.00 10.00

%

Psi

Also given;

ρ w = 63.8lb / ft 3 ρ o = 54.7lb / ft 3 σ a.w = 70dyne / cm Oil Reservoir Engineering

(100)

σ w.o = 28dyne / cm cos θ = 1 Solution:

( Pc ) R = ( Pc ) L ⋅

σ w.o σ w. a

∴ ( Pc ) R = ( Pc ) L ⋅

Qh =

28 = 0.4( Pc ) L 70

(1)

144 ⋅ ( Pc ) R 144 = ⋅ ( Pc ) R ρ w − ρo 63.8 − 54.7

Q h = 15.82( Pc ) R = 6.328( Pc ) L

(2)

As the initial OWC that is correspond to the displacement pressure is at depth of 3788 ft, thus the free water level is at a depth of: 6.328 x 1.9=12 ft below the depth of the 1.WOC. The depth of the free water level = 12 + 3788 = 3800 ft. In genera, the relation between the deoth of the free water level at any height value is; D = F.D.L – h

(3)

Using equation (1) and equation (2) and equation (3), we can calculate the (D vs. Sw profile) as follows:

Sw

( Pc ) R

h

D

100 (F.W.L.) 100 (1.WOC) 85

0 0.76 0.81

0 12 12.8

3800 3788 3787

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72 58 52 45 39 35 32 30 30

0.94 1.10 1.30 1.44 1.83 2.18 2.88 4.00 4.00

14.9 17.4 20.6 22.8 29 34.5 45.6 63.6 94.9

3785 3783 3779 3777 3771 3765 3754 3737 3705

%

Psi

ft

ft

In the above example it was assumed that the reservoir was

homogeneous and that the capillary pressure test on a single core simple was sufficient for estimating the entire water saturation profile.

Most of reservoirs are stratified and several simples must be

selected for capillary pressure test, so the actual stratification can be taken into account.

Example:

Calculate the water saturation profile at a well drilled into a reservoir has he following capillary pressure data for this cores of different perm abilities;

Sw

( Pc )core:1

Oil Reservoir Engineering

1000 md

( Pc )core:2

100 md

( Pc )core:3

(102)

10 md

100 80 60 40 20

0.2 0.45 1.0 2.5 6

1.8 2.0 3.2 3.7

3.0 4.9 7.5 13

%

Psi

psi

psi

Also given the following data; Sub sea depth

K

Suitable Pc – S w core

3743 3744 3745 3746 3747 3748

93 970 12 8 1020 108

Core : 2 Core : 1 Core : 3 Core : 3 Core : 1 Core : 2

ft

md

Solution:

After deciding the suitable capillary–saturation curve: Calculate ( Pc ) R for each foot of depth as the previous example.

( Pc ) R =

h 3800 − D = 15.82 15.82

Calculate ( Pc ) L .

( Pc ) L =

( Pc ) R 3800 − D 3800 − D = = 0.4 0.4(15.82) 6.33

Oil Reservoir Engineering

(103)

Or;

( Pc ) L = Read 3)

Sw

h 3800 − D = 6.328 6.328 for each ( Pc ) L from curves Pc – S w for each core (1,2and

Depth

H

( Pc ) L

Curve

Sw

3743

57

8.99

Core : 2

30

3744

56

8.83

Core : 1

20

3745

55

8.68

Core : 3

54

3746

54

8.52

Core : 3

55

3747

53

8.36

Core : 2

20

3748

52

8.20

Core : 2

13

ft

psi

%

Extending the range of laboratory Pc – S w data It is noticed that the permeability at a point n the reservoir often has

a great influence on the water saturation than does its structural location.

The permeability values in the above example were chosen so that the three cores could be used to evaluate

Sw

for each foot of the sand.

In many cases, the permeability of portions of the formation may not be closed to that of the core simples tested. On this case additional

Oil Reservoir Engineering

(104)

Pc

–

Sw

curves can be determined by plotting the laboratory data as in the

proceeding example.

Example:

For the above example, estimate a capillary pressure curve for 40 md core. Solution:

Plot Pc vs. K for the three cores on log–log paper for each saturation level given straight lines. For any K as 40 md, draw a horizontal line and record Pc vs. S w that gives the new curve, Fig (64)

Figure 63

Pc

Sw

1.7 2.5 4

100 80 60

Oil Reservoir Engineering

(105)

6.2 8.4 11.5

50 40 30

Calculation of effective and absolute permeabilities from capillary pressure data: Purcell and Burdines method: Purcell and Burdines have reported on computation of permeability

from capillary pressure data obtained by the mercury– injection. Method, utilized the concept of pore– size distribution as follow:

The minimum capillary pressure required for displacing of a wetting

fluid or injecting a non-wetting fluid into a capillary tube of radius r is given by:

Pc =

2σ cos θ r

∴r =

2σ cos θ pc

(1)

The flow rate through a single tube of radius (r) is given by

poiseuille equation:

πr 4 ∆p Q= 8µL Since the volume of the capillary is :

Oil Reservoir Engineering

(106)

Vi = π ⋅ r 2 ⋅ L

(2)

vi ⋅ r 2 ⋅ ∆P ∴Q = 8 ⋅ µ ⋅ L2 From equation (1) and equation (2);

( σ cosθ )2 vi ∆p ∴Q = 2 ⋅ µ ⋅ L2 ( pc ) 2

(3)

For a porous medium of (n) capillary tubes of equal length L and

different radii.

∴ Qt

( σ ⋅ cos θ )2 ∆p  = ∑n 2 ⋅ µ ⋅ L2

 

vi   ( pc )2 

(4)

Due Darcy's law;

∴ Qt =

K ⋅ A ⋅ ∆P µ⋅L

(5)

From equation (4) and equation (5);

(σ cos θ ) 2 n vi ∴K = ⋅∑ 2 ⋅ A ⋅ L i =1 ( Pc ) 2 As the volume (

vi

of each capillary can expressed as a function of

Si ) of the total void volume vT ∴

(6)

of the system.

vi = Si vT ,

Oil Reservoir Engineering

(107)

Also;

φ=

vT A⋅ L

∴ vi = Si ⋅ vT = S i ⋅ A ⋅ L ⋅ φ

(σ ⋅ cos θ ) 2 ⋅ φ n S i ∴K = ⋅∑ 2 2 i =1 Pci

(7)

Purcell introduced a lithology factor ( λ ) and a conversion factor,

generalizing equation (7) as;

dS nw 2 S =0 Pc 100

∴ K = 10.24(σ ⋅ cos θ ) 2 λ ⋅ φ ⋅ ∫

(8)

Where; :

S

Fractional total pore space occupied by liquid injected

or forced out of sample (

S nw ).

K

:

md.

φ

:

Fraction.

Pc

:

psi.

σ

:

Dyne/cm.

θ

:

Contact angle.

Purcell assumed that the contact angle for mercury was ( θ = 140o)

and that the interfacial tension of Hg = 480 dyne/cm. Oil Reservoir Engineering

(108)

1

∴ K = 14260 ⋅ φ ⋅ λ ⋅

dS ∫ 2 S =0 Pc

The integral is found by reading values of

(

various saturations, calculating value of

as a function of corresponding saturation.

Pc

from (

Pc S – ) curve at

1 ) 2 Pc and plotting these values

An average value of the lithology factor can be taken as

λ ≈ 0.216

for sedimentary rocks.

The value of

{(

the integral is the area under the curve

1 ) vs. S w } 2 Pc .

Calculation of relative permeability from capillary data Generalizing equation (8) and considering capillary pressure data

for displacement of the wetting phase,

∴ K w = 10.24(σ cosθ ) 2 ⋅ λ ⋅ φ ⋅ ∫

Sw

0

dS 2 Pc

(9)

Where;

Kw

: the effective permeability to the wetting phase at any

saturation value (

Sw

).

Oil Reservoir Engineering

(109)

The related permeability to the wetting phase is given by the

following relationship;

Sw

K rw

dS ∫0 Pc 2 Kw = = K 100 dS ∫0 Pc 2

(10)

Where the lithology factor ( λ ) is assumed to be constant for the

porous medium.

The effective permeability to the non-wetting phase (

K nw

) can be

calculated in a similar fashion as equation (9) by assuming that the nonwetting phase is contained in tubes or pores, free of the wetting phases; 100

K nwt = 10.24(σ ⋅ cosθ ) 2 ⋅ λ ⋅ φ ⋅ ∫

S =Sw

by;

dS 2 Pc

The relative permeability to the non-wetting phase (

(11)

K nwt

) is given

100

K nwt

K = nw = K

∫

S =Sw 100

dS Pc

∫ dS P 0

2

2

(12)

c

The following example illustrate how to use the above relationship to calculate

K ab

and

Oil Reservoir Engineering

Kr

from

Pc

data

(110)

Example:

For mercury injection method given the following capillary pressure saturation data; SHg, % 0

10

20

30

40

50

60

70

80

Patm

2.8

3.1

3.5

4.2

5.5

12.2

20

30

1.6

σ = 480dyne / cm , φ = 23%

If λ = 0.216 ,

θ = 140o

and

Calculate;

K and ( K rw vs. S w ). Solution:

dS ∫0 Pc 2 K rw = 100 dS ∫0 Pc 2 Sw

S Hg

Pc

1 2 Pc

0

1.6

0.391

10

2.8

0.128

2.595

0.578

20

3.1

0.104

1.160

0.392

30

3.5

0.082

0.930

0.239

40

4.2

0.067

0.745

0.118

50

5.5

0.029

0.480

0.0405

60

12.2

0.007

0.180

0.0113

70

20

0.003

0.05

0.0032

80

30

0.001

0.020

0

dS ∫ Pc 2

1

Σ = 6.16 Oil Reservoir Engineering

(111)

100

K = 10.24(σ ⋅ cosθ ) ⋅ ∫ 2

0

dS 2 Pc

Where;

K

:

md.

S

:

Fraction.

φ

:

Fraction.

Pc

:

psi.

Or; 100

K = 4.75 × 10− 6 (σ ⋅ cosθ ) 2 ⋅ φ ⋅ λ ⋅ ∫

0

dS 2 Pc

Where;

K

:

md.

S

:

Percent.

φ

:

Percent.

Pc

:

atm.

σ

:

Dyne/cm.

∴ K = 4.75 × 10−6 (480 × 0.766) 2 × 0.216 × 6.16 = 19.65md

Figure 64

Oil Reservoir Engineering

(112)

Petrophysics Petrophysics is the study of the relationship that exist between and

textural rock properties, in other words, it is the structure interpretation of physical rock properties.

Although, the reservoir engineer is mostly interested in porosity,

permeability and fluid saturation of reservoir rocks, there are certain physical properties such as the formation resistively factor (F), the

resistively index (I), and the hydraul formation factor (Fh) which provide

a link between reservoir engineering and logging and from which derivation may be made leading to the possible determination of relative permeability from electric logs. This chapter shows how the properties of

clean rocks mostly common used in fluid–flow mechanics and electric– log interpretation are interrelated.

Petrophysics of clean rocks 1) Permeability Poiseuille (1846) derived an equation which relates the rate of flow of an incompressible fluid of known viscosity through a horizontal straight

capillary of length (L) and radius (r) under the influence of a pressure differential ( ∆P ) as follows;

π ⋅ r 4 ⋅ ∆P Q= 8⋅ µ ⋅ L

(–1)

Where;

Oil Reservoir Engineering

(113)

Q

:

cc/sec.

r

:

cm.

µ

:

Poise.

∆P ( P1 − P2 ) :

Dyne/cm2.

Figure 65

If we considered a linear porous medium of physical length (L) and cross sectional area (A) as made up of a bundle of capillary of average radius (r') and of average length (L'=Lt). Where; (t) is a tortuosity coefficient. Poiseuille's law is written as;

n ⋅ π ⋅ r −4 ⋅ ∆P Q= 8⋅ µ ⋅t ⋅ L

(–2)

Figure 66

The tortuosity coefficient (t) is a dimensionless number

representing the departure of a porous system from being made up by a bundle of straight pore capillaries. It is also a measure of the tortuous path length which a practical fluid must travel,

Oil Reservoir Engineering

(114)

expressed in terms of the shortest distance between two points in that path, i.e. t=L'/L

Comparing Poiseuille's law with Darcy's law expressed in the same system of units;

1.013 A ⋅ K ⋅ ∆P n ⋅ π ⋅ r −4 ⋅ ∆P −8 Q= × 10 = Q = µ⋅L 8⋅ µ ⋅t ⋅ L

n ⋅ π ⋅ r −4 ∴K ≈ × 108 8⋅ A⋅t

(-3)

Where;

∆P

:

Dyne/cm2.

n

:

Number of capillary tubes.

K

:

Darcy.

2) Porosity • Volume porosity (

φv )

of this bundle of capillary tubes is

expressed as the pore-space volume per bulk volume. It is given by;

n ⋅ π ⋅ r −2 ⋅ L ⋅ t n ⋅ π ⋅ r −2 = = t φv = Vβ A⋅ L A Vp

(-4)

φ • Surface porosity porosity ( s ) is the cross sectional area of all

pores that are intersected by a plane surface and

expressed as a fraction of the total cross- sectional area (A) of the rock.

Oil Reservoir Engineering

(115)

(-5)

n ⋅ π ⋅ r −2 φs = To aid better in the understanding of fluid flow in A rocks, correlations among porosity, permeability, Surface area, pore size and other variables have been made.

Relation between porosity permeability, toruosity and mean capillary radius Equation (-3) can be written as;

n ⋅ π ⋅ r −2 r −2 K= ⋅ × 108 A 8⋅t By combining this equation with equation (-4);

1 φ ⋅ r −2 ∴ K = ⋅ 2 × 108 8 t By solving for the mean pore radius (r'); 2 ∴ r'= 8 ⋅ K ⋅ t

φ

× 10 − 4

(-6)

Specific surface The specific surface of a porous material is the total area exposed with in the pores pace per unit volume.

Unit volume may be the solid volume in which case the specific

surface is represented by (Ss). The unit value may also be pore space, in which case the specific surface is represented by (Sp). For a packing of capillary tube;

Oil Reservoir Engineering

(116)

S=

int ernal surface area unit volume

SP =

I .S . A. n( 2 ⋅ π ⋅ r ) L ⋅ t 2 = = P.V . n(π ⋅ r 2 ) L ⋅ t r

Ss =

I .S . A. S .V .

S

=

B

(-7)

I .S . A . P .V .

 1 .5 ⋅ A    P.V . S B  Vs  P.V . BV = φ ∴ = = = Ps .  1.5 ⋅ A  Vs SP 1−φ   BV  P.V .   φ  2φ ∴ S s = S P   =  1 − φ  r (1 − φ )

(-9)

1 .5 ⋅ A ) SB ( BV = S .V = (1 − φ ) = 1 .5 ⋅ A SS B.V S .V (-10)

∴ S B = S S (1 − φ ) For a packing of spheres

n ⋅ (4π ⋅ r 2 ) L ⋅ t 3 SS = = n ⋅ (4 π ⋅ r 3 )L ⋅ t r 3

(-11)

And;

n ⋅ (4π ⋅ r 2 ) L ⋅ t 3 1−φ SP = = = r φ n ⋅ (4 π ⋅ r 3 )L ⋅ t ⋅ φ 3 1−φ

Oil Reservoir Engineering

(-12)

(117)

Kozeny equation A useful expression can be derived by combining equation (-6) with equation (-7); 2 r = 8⋅ K ⋅t

SP =

And

× 10− 4

φ 2 r

(1) For capillary tubes or consolidated sands Equating values of 7);

r obtained from equation (-6) and equation (-

2 8K ⋅ t 2 r= = ×10−4 SP φ

∴K =

φ 2

2 ⋅ t ⋅ SP

2

×108

darcy

(-8)

darcy

(-9)

Or;

K=

φ3 2

2

2 ⋅ t ⋅ S S (1 − φ )

2

×108

the above derivations assume that the capillaries of mean radius ( r ) have no roughness, and have constant cross section. It is

assumed that the roughness factor is included in the tortuosity coefficient. The coefficient ( 2t ) in equation (-8) and equation (2

Oil Reservoir Engineering

(118)

9) is called the (Kozeny constant) for consolidated rocks for capillary tubes.

(2) For unconsolidated sands In the case of a packing of spheres, the mean hydraulic radius of

the capillaries is unknown. Therefore the following simple derivation will be used.

Using Poiseuille's law, the internal velocity inside a circular pipe is

given by;

r 2 ⋅ ∆P vi = 8µ ⋅ L

(-10)

Due to the internal roughness of the pipe, equation (-10) becomes;

r 2 ⋅ ∆P vi = ts µ ⋅ L

(-11)

Where;

ts

is the shape factor, which has an average value of 2.5

(t s ≈ 2.5)

.

Applying Darcy's law for the case of a porous medium;

viupp =

K ⋅ ∆P × 10−8 µ⋅L

Oil Reservoir Engineering

(-12)

(119)

The apparent velocity (

viupp

related to the actual velocity (

) from the bases of Darcy's law can be

viact

follow;

) obtained by Poiseuille's law as

Q = Vapp ⋅ Aapp = Vact ⋅ Aact ∴Vapp = Vact ⋅

Aact L ⋅ t PV φ ⋅ = Vact ⋅ = Vact ⋅ Aapp L ⋅ t BV ⋅ t t

∴VDarcy = V pois . ⋅

φ t

K ⋅ ∆P r 2 ⋅ ∆P φ −8 ∴ × 10 = ⋅ ts ⋅ µ ⋅ L t µ⋅L ∴K =

φ ⋅r2 ts ⋅ t

× 108

As;

t s ≈ 2 .5 ∴K =

φ ⋅r2 2.5 ⋅ t

× 108

(-13)

As the mean hydraulic radius of a porous medium can be defined

as the ratio of the pore volume per unit bulk volume, divided by the wetted surface per unit bulk volume;

Oil Reservoir Engineering

(120)

PV r = BV

r=

I .S . A. BV

=

PV 1 = I .S . A. S P

1 SP

Therefore, equation (-13) becomes;

∴K =

φ 2.5 ⋅ t ⋅ S P

2

× 108

(-14)

Or;

∴K =

φ3 2

2.5 ⋅ t ⋅ S S ⋅ (1 − φ ) 2

× 108

(-15)

The value (2.5 t) is called the Kozeny constant for unconsolidated

porous medium.

Equating Kozeny constant for consolidated and unconsolidated

sands;

2.5t = 2t 2 ∴t = 1.25

(-16)

This value of tortuosity (t) agrees very closely with experimental

determinations for unconsolidated sands.

Oil Reservoir Engineering

(121)

When the grains are non spherical, a shape factor (

ts

) must be

introduced which had been determined experimentally. The values of these coefficients are given as:

Grain shape

ts

Spherical Well-rounded Worn Sharp (sub rounded) Angular

1.00 1.02 1.07 1.17 1.27

Introducing the shape factor (

ts

) into Kozeny's equation we

obtained for unconsolidated granular material.

K=

K=

φ 2 .5 ⋅ t s ⋅ t ⋅ S P

2

× 108

φ3 2

2.5 ⋅ t s ⋅ t ⋅ S S (1 − φ ) 2

Where

SP

and

SS

× 108

darcy

(-17)

darcy

(-18)

are respectively the specific surface based on

the pore volume and the soil volume, obtained by the relation (

SP =

1 r ), previously defined.

Oil Reservoir Engineering

(122)

Flow of electric current through clean reservoir rocks The solid framework of the sedimentary reservoir rocks is made up

of minerals, for the most part, non conductive of electricity. Sedimentary

rocks are conductive of electricity only if their interconnected pore space contain electrically conductive fluids, namely formation water, connate water, interstitial water, ground water and the like.

Consider the box like container, Fig (68) is completely filled with

salty water and resistivity Rw ohm-meters.

Figure 67

Let the length of the box be L meters, and its cross sectional area

be A meters2.

The resistance of the base to the flow of current will be in ohms;

R = Rw

L A

(1)

When a voltage (E), in volts, is applied between the sides A, a

current I in amperes will flow, thus, Due to ohm's law;

E = I ⋅ R = I ⋅ Rw Oil Reservoir Engineering

L A (123)

∴I = A⋅

1 E ⋅ Rw L

(2)

This expression, equation (2) is analogous to Darcy's law for the

horizontal flow of the fluids have a unit viscosity.

q = A⋅ K ⋅

∆P L

(3)

Now consider that the box is completely filled with clean sand and

saturated to 100% with salt water of the same resistivity as before, Fig

(69). The resistance of the box will be increased by a factor called the resistivity formation factor "F" which is always larger than one.

Figure 68

A new and smaller current (I') will now flow such that;

E = I '⋅Ro ⋅ As

I 'p I

L A

, then

(4)

Ro f Rw

, where

Ro

is the resistivity of a unit

volume of the box that is completely filled with the porous medium and fully saturated with a conductive fluid of a resistivity

Rw

.

Comparing equation (2) and equation (4);

Oil Reservoir Engineering

(124)

E = I '⋅Ro ⋅

L L = I ⋅ Rw A A

It is shown that the resistivity ( the brine (

Rw

Ro

) is proportional to the resistivity of

). The constant of proportionality is called formation

resistivity factor.

∴F =

Ro Rw

(5)

This relation is an important in log-interpretation because of by knowing;

Rw

and F, it is possible to calculate the resistivity

Ro

which it

has when fully saturated with formation water. Such a condition precludes any possibility of oil production.

The formation factor is a dimensionless quantity by which the

resistivity of the formation water is to be multiplied in order to obtain the resistivity of the rock when 100% saturated in formation water.

Lithologic factors effecting formation factor • Rock porosity is the mean factor that controls the passage of current, i.e. the value of the formation factor. • The salty waters that rock contains in its pores. • Rock cementation and grin size distribution control the size of the interconnected pore of their tortuosity.

Oil Reservoir Engineering

(125)

Various formulas have been proposed to relate the formation factor ( F ) to the lithological factors of porosity ( ( m ).

φ)

and cementation factor

Archie proposed the following formula;

F=

a

φm

(6)

The constant ( a ) is determined empirically. Satisfactory results are usually obtained with;

F=

0.81

φ2

in sands, and

F = φ −2

in compacted sands.

These two formulas differ little from (Humble formula):

F=

0.62

φ 2.15

(7)

The Humble formula and the Archie formula for several vales of

( m ) are represented graphically in Fig (3).

Resistivity of rocks partially saturated with formation waters When oil and gas, which are non conductors of electricity, are

present, with in a porous medium together with a certain amount of salty

Oil Reservoir Engineering

(126)

formation water, the resistivity (

RT )

will be larger than (

Ro

). Since

there is a less available volume for the flow of electric current this

available water volume is represented by the water saturation in the pore space

Sw

.

Resistivity of a partially saturated porous medium (

RT )

depends

not only on the value of ( S w ) but also on its distribution within the pore

space.

The fluid distribution within the porous medium depends on; • The wetting properties of the rock. • The direction in which it was established (drainage or imbibition).

• Porosity type. Fig (69) shows how the ratio

RT Ro

varies as a function of saturation.

Figure 69

Oil Reservoir Engineering

(127)

Curve (1) and (2) are for sands, the slope of which is 2 for the first

and 1.8 for the second. These slopes are called Saturation exponents

(n). (n)

Curve (3) is for oil-wet sand in which case the value of (n) is

available with saturation and the degree of wetting.

The general formula which relates connate water saturation ( and the true resistivity (

the following forms;

Sw = n

RT )

Sw

)

is Archie formula which may be written in

Ro FRw n =n = Rw ⋅ φ − m / RL Rt Rt

(8)

( n ) is the saturation exponent, the value of which is most generally

assumed to be 2 for water wet reservoir rocks.

( n ) can be measured in laboratory by measuring the electric

conductivity of the core at different partial fluids saturations.

Tortuosity determination The simplest method by which tortuosity of rock capillaries can be

determined is made by computation from the value of formation factor F and porosity

φ.

For consolidated rocks where the porosity

φ

may be represented by

tortuous capillaries of actual length (Lt) and ending in number (n) in a

Oil Reservoir Engineering

(128)

cross sectional area (A), a theoretical expression for the formation factor (F), can be derived as follow:

R = Ro ⋅

= Rw ⋅

L A

(9)

L ⋅t n ⋅ (π ⋅ r 2 )

(10)

Using equation (10) and equation (-4);

L ⋅t2 ∴ R = Rw ⋅ A ⋅φ From equation (9) and equation (11);

L L ⋅ t2 ∴ Ro ⋅ = Rw ⋅ A A ⋅φ

∴ Ro = Rw ⋅ ∴

Ro

Rw

=

t2

φ

t2

φ

=F

∴t 2 = F ⋅φ

(12)

Formula (12) provides a ready laboratory means of determining (t) since both (

φ ) and (F) are easily measured.

Oil Reservoir Engineering

(129)

Effective tortuosity In a completely water saturated rock, the passage of electric current

is not expected to take place effectively through the full volume of water.

This analogous to fluid flow in porous media at 100% saturation where all the fluid is not moveable. The non movable water is the irreducible water saturation, (

S wi ). The irreducible water saturation, which occupies

capillaries through which there is neither pressure differential nor potential drop, looks like non conductive mineral framework.

Thus, tortuosity of irreducible water saturation can be written as:

t 2 = F ⋅ φ (1 − S wi )

(13)

The concept of effective tortuosity ( when partial water saturation (

S wi

te

) takes in another aspect

) prevails, for in this situation the non-

wetting phase is non conductive of electricity. An effective tortuosity can then be written as; 2

te = Fe ⋅ φ ( S w − S wi ) Where;

Fe

:

The ratio

Effective formation value

F Fe

is written as:

Oil Reservoir Engineering

(130)

 Ro  F  Rw  Ro 1 = = = Fe  Rt  Rt I  R  w  When

I

is the resistivity index at the partial water saturation (

also the following relationship can be written;

F Ro  t  = =   Fe Rt  te 

2

 S w − S wi     1 − S wi 

Sw

),

(14)

According to this relation, equation (-6) must be changed to; 2 ∴r = 8⋅ K ⋅t

φ ⋅ (1 − S wi )

× 10− 4

(15)

Similar adjustment should be made in Kozeny equation.

Hydraulic formation factor and index

Figure 70

By analogy with the formation resistivity factor, the hydraulic formation factor (

Fh =

Fh

) may be written as;

Darcy flow rate R ≈ o poiseuille flow rate Rw

Oil Reservoir Engineering

(131)

K (π ⋅ R 2 ) ∆P

=

× 10 −8

µ⋅L

4

π ⋅ R ⋅ ∆P

∴ Fh =

8⋅µ ⋅ L

8K × 10 − 8 2 R

(16)

Where;

R

is the radius of the tested core.

Considered the movable water volume;

π ⋅ R 2 ⋅ φ (1 − S wi ) = n(π ⋅ r 2 ) L ⋅ t R 2 ⋅ φ (1 − S wi ) r = n ⋅t 2

(17)

But;

8K ⋅ t 2 r = × 10 −8 φ (1 − S wi ) 2

R 2 ⋅ φ (1 − S wi )

(15)

2  × 10 − 8 = 8 ⋅ K ⋅ t n ⋅ t  φ ⋅ (1 − S wi )

φ 2 ⋅ (1 − S wi ) 2 n ⋅t 2 φ ⋅ (1 − S wi ) 2 ∴ Fh =

3

[

= 8⋅ K

n ⋅ t3

] × 10 R 2

−8

= Fh (18)

Where; Oil Reservoir Engineering

(132)

C:

is the total number of capillaries occupied by the phase fluid.

At partial saturation in the wetting-phase fluid ( hydraulic formation factor (

Fhe w =

Fhe

Sw

), an effective

), may be expressed as;

φ 2 ( S w − S wi ) 2 ne w ⋅ (te w )3

(19)

Where;

ne w

:

is the number of capillaries occupied by the wetting-phase.

te w

:

is their tortuosity.

Similarly for the non wetting-phase;

Fhe nw =

φ 2 (1 − S w ) 2 nn w ⋅ (tn w ) 3

(20)

Where;

nn w phase.

tn w

:

is the number of capillaries occupied by the non wetting-

:

is their tortuosity.

The effective hydraulic index ( I eh ) for the wetting-phase is the ratio of (

equation (19).

Fhe )

Oil Reservoir Engineering

to (

Fh )

that are obtained by equation (18) and

(133)

Fhe n  t ∴ I eh = = Fn new  te w

   

3

 S w − S wi     1 − S wi 

2

(21)

By definition this ratio, by analogy, is also the relative permeability to the wetting phase (

K rw

n  t  ∴ K rw = new  tew 

3

).

 S w − S wi     1 − S wi 

2

(22)

Similarly, the relative permeability to the non-wetting phase ( K rnw )

is obtained by dividing equation (20) by equation (18);

∴ K rnw

n  t = nnw  tew

The ratio (

S wf

   

3

 1 − Sw     1 − S wi 

 S w − S wi    is 1 − S wi  

2

(23)

called the free wetting-phase saturation

), and when substituting in equation (23) and equation (23) we

have;

 n ∴ K rw =   ne  w

 t   te  w

3

  (S wf )2  

(24)

And

Oil Reservoir Engineering

(134)

3

∴ K rnw

n  t  2 ( = 1 − S wf ) nnw  tew 

(25)

The formulas (24) and (25) are fully general and are valid whether

the saturation changes are by imbibition or by drainage. They are not useful without elimination of the number of capillaries involved ( n , and

nnw

) and the tortuosity coefficients ( t ,

tew

and

tnw

order to obtain practical relative permeability formulas.

new

) from them in

Relative permeability to the wetting-phase ( K rw ). * Imbibition direction When the wetting phase is increased from the irreducible saturation (

S wi

), all the capillary acquire simultaneously a movable wetting-phase

saturation. Hence, we have;

n = new

=

nnw

(26)

Substituting equation (26) in equation (24);

∴ K rw = (

t tew

)3 ⋅ Swf

2

(27)

Another formula can be obtained by using equation (14) and

equation (24);

Oil Reservoir Engineering

(135)

∴ K rw(imbibition)

1 Ro 3 2 = ( ) ⋅ S wf 2 Rt

(28)

*Drainage direction When The wetting-phase saturation is decreased because of the

injection of a non-wetting phase starting with too per cent wetting-phase

saturation, the largest capillaries are originally occupied by non wetting fluid, then the next largest, and so on

At any intermediate saturation, the distribution of fluids may be

visualized as being in two bundles of capillaries, one occupied by non wetting and the other by wetting fluid. Hence we have the relation.

∴ n = new

=

nnw

(29)

It is no longer possible to eliminate ( n and and after substitution of (

∴ K rw( drainage) = (

∴ K rw( drainage)

t

new

) from equation (24)

tew ) from equation (14);

n t 3 2 )( ) ⋅ S wf new tew

1 n Ro 3 2 = ( )( ) ⋅ S wf 2 new Rt

(24)

(30)

From laboratory investigation, it has been shown that ( and

K rw(imbibition)

K rw( drainage)

) show very little deviation from one another and that

such deviation may be considered to be due to errors of measurements.

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Hence, in the drainage direction

n ≈ new

and a single relative

permeability formula (27) or (27) may be used for the wetting-phase.

Relative permeability to the non wetting-phase (

K rnw

).

(1) Imbibition direction In

the

imbibition

direction,

the

wetting-phase

fluid

occurs

simultaneously in all the capillaries because the small capillaries are all ready saturated at the start of imbibition with the wetting-phase, i.e. (

S wi

). Because of the large degree of interconnection between the

capillaries

of

all

size,

the

wetting-phase

saturation

increases

simultaneously in all of them and the non wetting-phase becomes constricted in all the pores simultaneously, given rise to coaxial flow of

both phases within a certain range of saturation changes. The non wetting-phase saturation distribution is considered to be a succession of inflations and constriction connected along the capillaries axis.

As the wetting-phase saturation increases, the constriction become

very narrow and eventually break down, leaving an insular non wetting bubble in each pore. When all the filaments are broken in their

continuity, permeability to the non wetting-phase ceases, although a large residual saturation to the non wetting-phase may be present. It will be represented by (

Snwt

saturation saturation.

Oil Reservoir Engineering

) and is called the trapped non wettingwetting-phase

(137)

According to the above physical concept of the non wetting-phase

saturation distribution in a porous medium during imbibition;

n = new

(31)

In addition, tortuosity of the non wetting filaments, which have the

same axis as that of the capillaries them selves, is equal to the tortuosity at 100% saturation;

t = tew

(32)

Therefore;

K rnw(imbibition) = (1 − S wf ) 2 The expression (

1 − S wf

(33)

) represents the free non wettingwetting-phase

saturation. saturation Considering the trapped non wetting-phase saturation, (

Snwt

), the term (

S wf =

S wf

) may be expressed as;

S w − S wi 1 − S wi − S nwt

Hence, the formula for (

K rnw( imbibition)

(34)

K rnw

) should have the form;

 S w − S wi    = 1 −  1 − S wi − S nwt 

Oil Reservoir Engineering

2

(35)

(138)

(2) Drainage direction In the drainage direction, the desaturation in the wetting-Phase

occurs gradually from large capillaries toward smaller ones. At any one

condition of liquid saturation. The non wetting-phase is found in the largest capillaries and the wetting phase in the smaller capillaries. Let

S w Snw ,

and

S wi

represent the specific surface of wetting, non

wetting and irreducible wetting-phase on a pore-volume basis. The volume of the pores having a surface reciprocal of

S w Snw ,

and

S wi

Sw

,

Snw

and

S wi

are the

.

The total pore space volume not occupied by irreducible water is

the sum of the two preceding volumes;

1 1 1 = + S wi S w S nw

(35)

But according to equation (-7);

∴

2 = rwi S wi

and

2 = rw Sw

and

2 = rnw S nw

(36)

From equation (35) and equation (36);

rwi = rw + rnw

(37)

According to equation (15);

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2

8K wtew rw = × 10−4 φ ( S w − S wi )

(38)

2

8K nwtnw ×10−4 rnw = φ (1 − S w ) 8K ⋅ t 2 rwi = ×10−4 φ (1 − S wi )

(38)

(38)

Combining relations (38);

∴

rw tew K rw = S wf rwi t

(39)

And;

rnw t = nw rwi t

K rnw

(1 − S wf )

(39)

Substituting equation (39) into equation (37);

∴

tew K rw t K rnw + nw =1 S wf (1 − S wf ) t t

−1 1 1 −1 tnw t  2 ∴ (K rnw ) ⋅ (1 − S wf ) 2 = 1 −  ew (K rw ) 2 ⋅ S wf 2 t  t 

∴ (K rnw )

1

2

1  t  1 −1   t  =   ⋅ (1 − S wf ) 2 ⋅ 1 −  ew (K rw ) 2 ⋅ S wf 2    t    tnw 

Oil Reservoir Engineering

(140)

Using equation (27);

∴ (K rnw )

1

2

1   3 2   1  t    − 1 t t     2 =   ⋅ (1 − S wf ) 2 ⋅ 1 −  ew   (S wf )  S wf 2    t  t  ew     tnw      

 t  ∴ (K rnw ) =    tnw 

2

  t  −12 1  ⋅ (1 − S wf ) ⋅ 1 −  ew  S wf 2    t  

2

Substituting equation (14) in equation (40);

∴ (K rnw )

1

2

 t  =    tnw 

2

   14  1 R o ⋅ (1 − S wf ) ⋅ 1 −   ⋅ S wf 4    Rt    

2

As the tortuosity of the bundle of largest capillary size controls to a

large extent the value of the tortuosity at full saturation, it is reasonable to postulate that;

t ≈ tnw

(41)

This because the capillary tubes act as conducting circuits in

parallel. When adding, in parallel, circuits of low conductance (small capillaries) to highly conductive circuits (large capillaries), the former change the over-all conductance relatively little.

Hence, the formula for the relative permeability to the non wetting

phase in the drainage direction is;

Oil Reservoir Engineering

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 1 Ro 14  4 = (1 − S wf )1 − S wf ( )  Rt  

K rnw( drainage)

2

(43)

It appears that there are three equations useful in calculating the

relative permeability characteristics from petrophysical consideration,

namely equation (28) for the wetting-phase valid regardless of the

direction of saturation changes, equation (34) for the non wetting-phase in the imbibition direction and equation (43) for the non wetting phase in the drainage direction.

More general formulas for relative permeability in clean water-wet

rock can be written after substitution of Archie's relationship (8) using the value of the saturation exponent (n) = 2

∴ Sw =

Ro Rt 3

∴ K rw = S w ⋅ S wf

1

2

(Drainage or imbibition)

K rnw( drainage) = (1 − S wf )(1 − S wf

K rnw( imbibition)

1

4

(44)

1

⋅ Sw 2 )2

 S w − S wi  = 1 −  1 − S − S wi nwt  

(45)

2

(46)

Example:

Given the following data:

Sw

20

30

Oil Reservoir Engineering

40

50

60

70

80

(142)

90

Ro

Rt

0

0.75

0.165 0.275 0.400 0.535 0.685 0.840

If S wi = 20% and S nwt = 10%, Calculate the wetting and non wetting-phase relative permeabilities in each of drainage and imbibition direction. Solution:

1) For imbibition direction According to equation (28) and equation (34); 1 Ro 3 2 K rw = ( ) ⋅ S wf 2 , Rt

and

K rnw

 S w − S wi   = 1 −  1 − S wi − S nwt 

Sw

Ro

20 30 40 50 60 70 80 90

0.000 0.075 0.165 0.275 0.400 0.535 0.685 0.840

Rt

S wf 0.000 0.125 0.250 0.375 0.500 0.625 0.750 0.875

Oil Reservoir Engineering

 Ro   R t 

2

3

0.0000 0.0203 0.0670 0.1440 0.2520 0.3910 0.5670 0.7700

2

S wf

1

2

0.000 0.354 0.500 0.612 0.707 0.790 0. 866 0.935

K rw = S wf

1

2

Ro 3 2 ( ) Rt

0.0000 0.0072 0.0335 0.0880 0. 1780 0. 3090 0.4910 0.7200

(143)

And;

Sw

S wf

1 − S wf

K rw (1 − S wf ) 2

20 30 40 50 60 70 80 90

0.0000 0.1428 0.2857 0.4286 0.5714 0.7143 0.8571 1.0000

1.0000 0.8572 0.7143 0.5714 0.4286 0.2857 0.1428 0.000

1.0000 0.7348 0.5102 0.3265 0.1837 0.0816 0.0204 0.0000

2) Drainage direction

K rw = K rw(imbibition) K rnw

1  4 1 R   = (1 − S wf )1 − S wf 4  o     Rt   

S w 1 − S wf

(S )

20 30 40 50 60 70 80

0.000 0.595 0.707 0.782 0.840 0.890 0.930

1.000 0.875 0.750 0.625 0.500 0.375 0.250

1

wf

Oil Reservoir Engineering

4

 Ro     Rt 

1

0.000 0.525 0.637 0.724 0.795 0.856 0.910

4

K rnw

2

1  4 1 R   o = (1 − S wf )1 − S wf 4      Rt   

1.000 0.417 0.226 0.119 0.056 0.021 0.006 (144)

2

90

0.125

0.966

0.949

0.001

Example:

For the last example calculate K rw and K rnw for imbibition and drainage direction using the following equations; 3

K rw(imb dra ) = S w ⋅ S wf

1

2

K rnw( imb) = (1 − Sef ) 2 K rnw( drainage) = (1 − Sef )(1 − S wf

1

4

1

⋅ Sw 2 )2

Solution:

(1) Imbibition direction 3

Sw

Sw

20 30 40 50 60 70 80 90

0.008 0.027 0.064 0.125 0.216 0.343 0.512 0.729

100

1

1

K rw( imb dra )

S wf

K rw(imb) = (1 − S wf ) 2

0.000 0. 354 0.500 0.612 0.707 0.791 0.866 0.935

0.000 0.009 0.032 0.076 0.153 0.271 0.443 0.682

0.000 0.143 0.285 0.428 0.571 0.714 0.857 1.000

1.000 0.735 0.510 0.326 0.184 0.082 0.020 0.000

1

1

S wf

2

• S w =100% only for drainage direction where K rw = 1

Oil Reservoir Engineering

(145)

• S w =90% for imbibition process represents the final (Maximum) water saturation value, water there is a trapped non wetting-phase saturation S nwt = 0.10 . (2) Drainage direction

(1 − S

)

2

Sw

(S )

20 30 40 50 60 70 80 90

0.000 0.595 0.707 0.782 0.841 0.889 0.931 0.967

0.447 0.548 0.632 0.707 0.775 0.837 0.894 0.948

1.000 0.454 0.306 0.199 0.121 0.065 0.028 0.018

1.000 0. 397 0.229 0.124 0.061 0.024 0.007 0.002

100

1.00

1.00

0

0

1

wf

4

Sw

1

2

1 wf

4

⋅ Sw

1

2

K rnw

Fig (71) shows the above K rw and the K rnw obtained for the example.

Oil Reservoir Engineering

(146)

Figure 71

Oil Reservoir Engineering

(147)

Referenes 1. Amyx J.D., Bass D.M., Whiting R.L.: "Petroleum reservoir engineering – Physical properties". McGraw Hill Book Company, New York, 1960. 2. Pirson S.J.: "Oil Reservoir engineering". McGraw Hill Book Company, New York, 1958. 3. Craft B.C., Hawkins M.F.: "Applied petroleum reservoir engineering". Prentice–Hall Inc. New Jersey, 1959. 4. Dake L.P.: "Fundamentals of reservoir engineering". Elsevier Scientific publishing company, New York, 1978.

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