Giovanni Benvegna2014/2015

SommarioState of matter............................................................................................................................................1

PVT properties:..........................................................................................................................................11

Gas.........................................................................................................................................................12

Oil..........................................................................................................................................................18

Water.....................................................................................................................................................22

Experimental test......................................................................................................................................23

Rock petrophysical properties...................................................................................................................27

Darcy law...............................................................................................................................................28

Darcy law for gas...................................................................................................................................32

Production parameters..............................................................................................................................34

Rock fluid interaction properties...............................................................................................................35

Mobility ratio.........................................................................................................................................38

The GOR.................................................................................................................................................38

Surface tension and interfacial tension.................................................................................................39

Imbibition and drainage........................................................................................................................46

Production drive mechanism.....................................................................................................................48

Pressure gradient......................................................................................................................................50

Diffusivity equation...................................................................................................................................52

Steady state solution.............................................................................................................................58

Skin effect..........................................................................................................................................59

Productivity index and completion factor..........................................................................................60

Pseudo steady state solution.................................................................................................................61

Pressure behavior with different flow rate............................................................................................62

Transient state solution.........................................................................................................................64

Skin effect in transient.......................................................................................................................67

Diffusivity equation for gasses...................................................................................................................68

Depletion...................................................................................................................................................74

State of matterWe define these state:

Liquid Vapour Gasses Dew point Bubble point Vapour pressure

When we say standard point we mean: P = 1 atm and T = 15°C. using m3sc we are talking standard

condition if we use m3st we refer to stock condition.

In a single component system:

We define the line AO as the vapour pressure; OB as the sublimation pressure and the line OC as the melting pressure, the slope of this last line is negative for water and positive for hydrocarbons.

Pc represent the minimum pressure necessary for liquefaction of vapor at the critical temperature. Tc is the critical temperature, above which a vapor cannot be liquefied, regardless of the applied pressure. So at the end A represent the critical point.

Suppose that the reservoir condition initial is in the terms of P and V at red dot. During the production the pressure decrease and the volume increase until it reach the bubble point. From bubble point until the complete vaporization at the dew point the pressure remain constant and volume continue to increase. From the dew point oil is in gasses phase, continuing the production pressure slightly decrease and volume increase as well.

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Collecting this behavior for different temperature condition it is possible to define a curve which pass among all the bubble point and the dew point. Which represent the region of the different states.

The fluid in a reservoir is not composed only by HC, we have also CO2, N2, H2S and so on. The composition of fluid oil define the fluid category.

In a multi component system:

The value of the bubble point pressure is different form dew point:

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If we increase the temperature, B will be higher, C as well and the distance between B and C decrease. So if we perform the experiment at different temperature and we plot all the result and so we connect all dew and bubble point we obtain the phase envelop.

In the phase envelop we find a point named Critical point, in which the phase property of a fluid is the same, vapor, liquid and solid.

The general phase diagram for a multicomponent system we have this kind of shape defined by own critical point, and line:

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In this graph we can see:

Bubble point line; Dew point line; Critical point; Cricondetherm: the higher temperature in which liquid can exist; So the fluid at the right side of this line is “Dry Gas”; above the bubble point and to the left of critical vertical we have “unsaturated fluid: liquid”. Among cricondetherm and vertical critical temperature we have “Gas condensate”.

Into a real reservoir we can image that initial pressure is Pi (300 bar) and temperature is Tr (100°C). we assume that temperature in a reservoir doesn’t change due to the lower dimension respect the earth; so it is possible only to reduce the pressure. Anyway, during the rising of the fluid the pressure decrease and temperature as well. If the P-T transformation of the fluid during the rising, remain at the right side of cricondetherm we talk about the Dry gas cos remain gas.

If we fall inside the envelop curve we talk about wet gas because there is gas in the reservoir and liquid gas at surface:

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If the pressure and temperature of a reservoir is in the gas condensate condition. Production will give me only liquid and gas at surface and into the reservoir I will have liquid in the reservoir if the pressure is lower dew point, I have at the end gas condensate not strictly connected to the path

We prefer to produce oil, so it is better to maintain the pressure above the dew point, because if the condition is inside the envelop line we have a lower volume of oil compared to the gas condensate.

The shape of the envelop curves depends on the composition of HC. In the picture below is shown 2 different condition of two reservoir: 1 is heavier compared to 2.

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If we consider the thermodynamics transformation during the rising of the HC from the reservoir (A) to the surface (B) we have different composition:

Another way to describe the multicomponent system is done using multi-diagrams:

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Volatile oil: release a large amount of gas at pressure reference, we don’t know if we have gas with saturated oil or oil with saturated gas.

In case of a gas-condensate reservoir (A) the production path cross the dew line in B, we can see that liquid start to increase and continuing the production after a while it will return in a low level of liquid. This phenomena is called Retrograde condensation.

This phenomena depends to the difference energy of attraction and repulsion of the molecular gasses.

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We make another differentiation of fluid inside the reservoir: volatile oil (to the left) and black oil (to the right)

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Let’s consider:

Pbubble = 250 bar P initial = 300 bar in the reservoir T reference = 15°C

We know that the P reference at surface is 1 bar

After the starting production Pi is in reduction direction. When it reach the Pb the composition is still oil, but after this level oil start to release gas:

Density decrease during the line path.

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PVT properties:PVT properties are the properties of a fluid which depend to the pressure volume and temperature value. The reference condition at which the value of the properties are define are 2:

Normal condition: P = 1 atm = 14.69 psi; T = 0°C Standard condition (sc for gas or stock tank condition st for oil): P = 1 atm = 14.69 psi; T = 15°C Reservoir condition (r): P = 300 bar ; T = 100°C

Sometimes happen to consider some properties not strictly connected to pressure and temperature as a PVT, for example Viscosity.

Viscosity as an intensive propriety, not related to PVT property but it is function of pressure and temperature, so in the reservoir engineering is considered in this way. Viscosity is friction within a fluid that results from the strength of molecule to molecule attractions. In other words it is a measure of the resistance of a fluid which is being deformed by either shear stress or extensional stress.

The unit of measure is Pa s or cP. We refer to Newtonian viscosity, dynamic viscosity, coefficient of viscosity as Viscosity μ.

Water at surface reference condition is 1 cP which correspond 10-3 Pa s. In the reservoir condition is 0.5 – 0.9 cP. For an oil we have at reference condition more or less ??, instead at reservoir condition is 0,5-50 cP (250 for very heavy oils).

The conversion factor among the different unit are:

1 P = 10-1 kg·m−1·s−1 = 10-1 Pa·s

We have to stay tuned to kinematic viscosity ν, which is measured in m2/s or St (stokes).

ν=μ/ ρ

Which the conversion factor is:

1 stokes = 100 centistokes = 1 cm2 s−1 = 0.0001 m2 s−1.

Let’s consider the different fluid in physic: Gas, Oil, Volatile Oil and Water.

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GasThe PVT properties for a gas are:

Viscosity Density or gas gravity Compressibility factor Gas formation volume factor Compressibility

The real gas are considered when we have higher pressure and low temperature. In order to consider the real behavior of a gas instead of a perfect gas we introduce the compressibility factor z:

PV=nRT→PV=z nRT

The compressibility factor born by the non-negligible molecular volume and the non-negligible attractive force among the molecule.

We have to consider the unit used:

R = 8.314 kJ kgmol-1 K-1. R=10.73 psia ft3 lbmol-1 R-1.

For the compressibility factor of a gas:

In the reservoir condition z is more or less 0.8 - 0.9

The z coefficient is not known previous, so we have to determine it (z = 1 at standard condition because the ratio is 1 at standard condition for each gas), in a purely theoretical way:

pnV n=n znRT r

pscV sc=nz scRT sc

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zn=pn

psc

V n

V sc

T sc

T r

The law of corresponding states:

All natural gasses, in the same pseudo reduced pressure and pseudo reduced temperature condition shown the same volumetric behavior and so the same value of z.

Let’s consider two different gas C1 and C2.

Each component is characterized by different critical point: Pressure and temperature.

The reduce pressure and temperature are determined as:

Pr=Pmeasured

Pcriticalc i

;T r=T measured

T criticalc i

This quantity is dimensionless so if we plot z as function of Pr and Tr the different curves overlapping together due to the same behavior.

For a given gas with different component inside we have to consider the different behavior defining a pseudo-reduced pressure and temperature:

Ppr=Pmeasured

Ppseudo critical

;T pr=T measured

T pseudocritical

The definition of pseudo-critical pressure and temperature are:

Ppc=∑i=1

n

Pc rc iy i;T pc=∑

i=1

n

T cr ciy i

Where y is the molar fraction.

y i=Molei

∑Mole

For example: given 2 gas in a mixture C1: 75% and C2 25%:

Ppc= yC1Pcrc1

+ yC2Pcrc2

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In this graph the part below the white line is reported to the bottom where it was in the upper part.

Unfortunately in a reservoir there is also H2S gas, so we have to apply some corrections; we use this graph:

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The formation volume factor (FVF of Bg) is the ratio between the volume of free gas, at reservoir condition. and the volume of the same gas at standard condition:

Bg=V ( p ,T )V sc

Which is measured in bbl/m3.

Consider that: (sc is standard condition and r reservoir condition)

pscV sc=nRT sc

prV r=zrnRTr

Bg=zrPsc

Pr

T r

T sc

Where Psc ,T r , T sc are constant, so qualitatively:

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The range of the FVF is between 0.002-0.05 m3/Sm3.

The compressibility in a real gas is

Cg=−1V

δVδP

=1ρδρδP

The minus disappear because an increasing of pressure correspond to and increasing of density.

The compressibility of a gas could be determined approximately as:

Cg=1P

−1zδzδP

≈1P

Just to remind some value

Cgas = 10-4 psi-1. Coil = 10-5 psi-1. Cwater = 10-6 psi-1. Csoil = 10-6 psi-1.

The compressibility of the rock depend to the porosity and consolidation.

The gas density is

ρg=MPzRT

ρgasres con=ρgasSCBg

So the Gas gravity is

G=ρg

ρair=

M g

M air

Where Mair is 28.96 g/mol and ρairSC is 1.225 kg/m3.

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ρga src=G ρairSC

Bg

Viscosity in gas change the behavior according to pressure and temperature in this way

Exercise 1 to 4

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OilCompressibility is the slope of the lines obtained in liquid phase by a CCE experiment (see the next chapter).

co=−1V

∂V∂ p

Which it is in an isothermal process equal to

co=−2

V 1+V 2

V 1−V 2

p1−p2

Formation volume factor for oil is defined as: the volume of oil and dissolved gas at reservoir condition that must be produced to obtain 1 cubic meter of stable oil at stock tank conditions (standard condition)

Bg=V oilr

V oilst

The behavior of this factor is:

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According to our experience it has the range of 1 - 2.2 m3/Sm3, maximum 2.8.

The initial increasing of the Bo in a reducing pressure until reach the bubble point is increasing due to the slightly compressibility of the oil. Of course at left of the bubble point a reduction of pressure cause a reduction of the Bo due to the liberation of gas and the volume of oil at reservoir condition is reduced.

Volume of gas dissolved (Rs) is the volume of gas which is dissolved in the oil at reservoir pressure and temperature, but it is measured at surface condition, referred to 1 m3 of stable oil (which means no release of gas at stock tank condition).

R s=V g (Pr , T r )sc

V oilst

Assuming we have a unsaturated reservoir, up to the bubble curve and we start the production, the pressure decrease and temperature remain the same, until to reach the bubble curve the gas dissolved in the oil is still the same. After the bubble point it start to be delivered into the reservoir, according to a non-necessarily linear law:

V or=V b Soϕ

OOIP=N=V or

Bg

The range of this parameter is among 100-350 m3sc/m3

st.

The volume of residual oil is the volume of oil at stock tank condition at the end of DLE experiment, corrected from the reservoir condition to the stock tank condition.

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We use the API gravity

API= 141.5ρo , rel@st

−131.5

Where

°API Description10 Water

<20 Heavy oil25<API<30 Medium oil30<API<40 Light oil40<API<50 Volatile oil

ρoil (p ,T )=ρoil ,st+ρgas ,sc R s(p ,T )

Bo(p ,T )

If I have the Gas Gravity (GG) and Bg, which is a curve function of pressure

ρga srescond=ρga sstcon

Bg

To oil we have

ρoil ,res .cond=ρoil , stock cond+ ρgas ,stand condR s

Bo

ρoil ,res .cond=

mo

V ostock cond

+mg

V g , standcond

V g (Pr , T r )@sc

V o , stock cond

V o ,r

V o , st

=mo+mg

V o ,r

Viscosity is determined as

Decreasing the pressure the density decrease (?) So the viscosity decrease as well. When the pressure reach the bubble point the oil start to liberate the gas and the density of oil increase and the viscosity increase.

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From the first graph we can see that the first behavior is due to the liquid phase, so it is the viscosity of the oil versus the pressure. Up to the bubble point the behavior is due to gas viscosity.

Exercise 5-6

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WaterSalinity: It affect to: density, compressibility, resistivity and so on.

Gas solubility: it is negligible

Water formation volume factor: it is the volume of water and its dissolved gas at reservoir conditions that must be produced to obtain 1 m3 of water at stock tank condition.

Bw=V ( p ,T )V st

≈1

Viscosity: 0.4 – 0.5 cP

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Compressibility: Cw = 2 – 4 10-6 psi-1.

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Experimental testThe constant composition oil expansion (CCE) experiment is used to determine bubble point pressure, undersaturated oil density, isothermal oil compressibility for oil it is a non-destructive test:

Procedure:

Initially: oil sample at reservoir pressure and temperature equal to reservoir condition isothermal fluid expansion by reducing pressure in steps measurement of mixture volume @ equilibrium

We determine the bubble point pressure as the intersection of the 2 behavior of the point fitting.

The differential liberation experiment (DLE) experiment is designed to approximate the depletion process of an oil reservoir and thereby provide suitable PVT data to calculate reservoir performance in terms of gas liberation. It is a destructive test on a sample fluid (200 – 250 cm3)

Procedure:

Initially: oil sample at reservoir pressure and temperature in the laboratory PVT cell are the reservoir condition

isothermal fluid expansion by reducing pressure in steps measurement of mixture volume @ equilibrium removal of the liberated gas at each pressure step, maintaining the same oil volume Reach the atmospheric pressure last step: flash expansion to reach atmospheric temperature (stock tank conditions)

Rsd and Bod are relative to residual oil volume, Residual oil volume: is the oil volume at the end of the DLE experiment, corrected from reservoir to standard temperature

Ro=Roend DLE

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We start with a PVT cell at bubble point which P2 is equal to Psat. So we reduce the pressure down to the saturation and we wait for the equilibrium. Maintaining the same pressure we remove the gas until have only oil in the cell.

We can measure only the removed gas.

The final condition change if it is followed a different path. DLE is not the transformation which oil do in the plant. DLE follow this transformation path.

At stock condition we want avoid the release of gas, this is the reason of the last step.

Multistage separator test (MST) is performed on an oil sample primarily to provide a basis for converting differential liberation data from a residual oil to a stock tank oil basis. It is a destructive test on a sample fluid of 250 cm3.

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Bod=V ores

V oℜsSC

Bo=V ores

V osto .con .

Procedure (in the laboratory PVT cell):

Initially: oil sample is at saturation conditions. (Measured parameter: oil volume) The sample is then brought to the pressure and temperature of the first stage separator. All the

gas is removed. (Measured parameters: oil volume, volume and number of moles of removed gas are measured)

Oil remaining after gas removal is brought to the conditions of the next separator stage…etc… Last step: stock tank conditions are reached

We do it in order to: design the level of pressure and temperature which obtain the maximum gas extraction from the oil and evaluate the real stable oil volume, which is higher than residual oil obtained by DLE. The green line represent the Bo during the MST experiment.

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In the real life, when we are on the field, the volume of the oil that we have at surface is the stock tank oil. Residual is obtained only in laboratory.

Converting from differential to stock tank basis is done using these relationship:

Bo=Bod ( Bof

Bod)pb

R s=(R sf )pb−( (Rsd ) pb−Rsd )( Bof

Bod)pb

The way to describe in thermodynamic graph the experiment is not possible because, per each step we have a new composition of the oil so we changing the envelop line. We can imagine that the thermodynamic path of the MST is a multy DLE path which reach the same end point with a different composition with more oil.

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Rock petrophysical propertiesThe rock analysis is done in order to obtain the permeability, porosity and water saturation; they enter to the routine core analysis (RCAL). The properties which define the interaction between fluid and rock is the special core analysis (SCAL) and measure the wettability, capillary pressure and relative permeability.

The plug from the core is more little in order to reduce the time used to analyze. The shape factor (length/diameter) must be higher than 1.

For reservoir engineering the porosity used is always the effective porosity, which is the ratio of the interconnected void space in the rock to the bulk volume of the rock.

The fluid saturation, which is defined as the ratio of the volume of a fluid and the volume of the pore.

S f=V f

V p

;∑ Si=1

V f=V bϕ S f=V b

V p

V b

V o

V p

Rock compressibility which is about 4 10-6 psi-1, is higher in shallow and slightly in consolidated formations.

C r=+1V

∂V∂P

We have + because we considering the system from the pore point of view which is trying to compress the rock from inside.

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Darcy lawApplied only if:

steady state condition: same flow in and outlet uncompressible fluid single fluid laminar flow No interaction between rock and fluid

So the flow is constant. The time transient of the experiment depend on permeability.

v= qA

=−Kh2−h1

L

Where L is the length of the sample porous medium, h is the height of above standard datum and K is the hydraulic conductivity [m/s].

If we want to change the fluid we move from K to k: using instead of hydraulic conductivity the intrinsic permeability defined as:

k= Kμρw g

Generalizing the Darcy law we obtain

v=−kμ

(∇ p−gρ∇ z )

Using k as the absolute permeability (even if we are using small k for intrinsic permeability), which is the permeability of the rock fully saturated by fluid.

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Reservoir

Well

In mono-direction (x) horizontal we can neglect gravity, and considering a constant cross section, we obtain (L is the length of the sampling):

v=−kμ

δpδx

→=−kμ

∆ PL

In mono-direction (z) vertical we have this assumption Pinlet=Poutlet and ∇ z=1 in atmospheric environment modify the Darcy equation:

v=−kμ

( ρg ∙1 )

The z direction in an arrow directed in down-direction.

In a three dimension Cartesian system:

In reservoir engineering Darcy law is used in radial geometry:

The hypothesis are:

negligible gravitational effect constant producing thickness

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ReRw R

P

Pi Pe

well perforate across the entire producing thickness

The flow is positive when we produce and negative if we inject. Darcy law is:

qA

= kμdpdr

Integrating this equation we obtain the Muskat equation:

q=2πhkμ

∆ p

ln ( rerw )So

pu=pe−qμ

2πhkln

rerw

So the pressure outlet is function of geometrical terms and property of fluid, not function of time. If we plot p versus radius we have a square scratch.

This graph means that we have reach the steady state, where the pressure at the external boundary give us a pressure constant equal to Pe.

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ReRw R

P

Pi Pe

Pw

The true scratch is

Pw change according to the flow rate and the slope of the depletion line change as well.

If we plot in a semi-logarithmic scale we have a straight line where slope is qμ

2πhk the coefficient of the

logarithm.

Permeability k has a tensor behavior, so the component of velocity has to take into account the different value of k function of the direction. In most case we consider that kx = ky and kz is lower.

We can consider different kind of material according to the value of K in different direction:

Isotropy: the prop. Are independent by direction Anisotropy: the prop. Are not independent by direction Heterogeneous: value of k change with space Homogeneous: value of k don’t depend to the location

All the properties which are scalar (mass, density, etcetera) are homogeneous.

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Darcy law for gasAssuming that we have this porous element (plug) where we inject gas:

if { q¿=qout uncompressibleq¿≠qout compressible gas

So for gas we know that is compressible, so we have to correct the Darcy law. The assumption are:

Negligible gravitational force Constant cross section Compressible fluid Z=1 Laminar flow

The gas law say:

p¿V ¿=poutV out

If we divide by time:

p¿q¿=pout qout

Remember that q=−Akμdpdx

pout qout=−pAkμdpdx

Integrating it in the space we obtain

pout qout∫0

L

dx=−Akμ∫p¿

pout

p dp

So

qout=Akμp¿

2−pout2

2 pout L

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According to the lower viscosity of gas compared to water or fluid, the laboratory test are done with gas in order to have a little waste time used for transient depleted.

The main difference between water and gas are:

Turbulences Slip flow: permeability measured with gas is higher than liquids, so permeability change with

fluid, and permeability change also with different inlet pressure: is higher when pressure is lower.

Assuming we have a flow injection in the pore: the molecule of the flow collide each other in elastic way at the boundaries statistically more often compared to the center of the flow, where they collide in an anaelastic way, this phenomena reduce the friction inside the gas molecule and it is traduced in an increase of velocity. Higher is the pressure lower it is this effect.

k g=kL(1+bpm

)Where p is the pore pressure and b is a coefficient named Klinkemberg slip factor which is function of temperature, mean free path and porous radius. If pressure tends to infinity the permeability tends to kL. this is the reason to perform laboratory analysis with gas.

Where

pm=p¿+ pout

2

We look for kL.

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Production parametersThe production parameter for oil reservoir are (measured):

Flow rate

Qoil=Qo ;Qgas=Q g;Qwater=Qw

Gas oil ratio

GOR=Q gsc

Qost

Water cut

WCT=Qw

QL

=Qw

Qw+Q o

We measure pressure:

WBP: well bottom pressure WBHP: well bottom hole pressure SWBHP: static well bottom hole pressure (stop production) FWBHP: flowing well bottom hole pressure (during production) THP: Tubing head pressure WHP: well head pressure

The minimum THP depend to the facilities of the plant, chosen to transport the oil to all separators and reach the stock condition.

In case of gas reservoir we have

OGR: oil/gas ratio Qg Qw Qo WGR=Qw/Qg Gravity

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Rock fluid interaction propertiesThe rock fluid interaction properties are estimated with SCAR test, we would like to maintain the Darcy law but we some assumption are not so performable, so we must correct the formula with phenomena which are not considered in Darcy kingdom.

Darcy equation for multiphase flow

The effective permeability keff which is the permeability for each phase at a specific saturation. It provide to take the presence of movement of more than a single fluid phase within the pore space.

0<S≤1

0≤keff ≤k|¿|¿

(k o , kg , kw )∨(k o+k g+kw )<k

According to Darcy law applied for each phase

qoil=Akoilμoil

dpdr

qwater=Akwaterμwater

dpdr

qgas=Ak gas

μgas

dpdr

We call the ratio between the effective permeability and the viscosity of the fluid, the mobility λ.

λw=kw

μw

; λg=k g

μg

The mobility of a gas is ten time faster than water, this is the reason to start production in oil phase.

k rw is the permeability relative to the water. Which is

k rw=kwk|¿|¿

We have

(k ro , krg , krw )∨ (kro+krg+krw )<1

Let’s consider an oil-water system in the reservoir.

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The Saturation of oil on oil zone is 1−Sw ir where Sw ir

is the saturation of the irreducible water, which is

the irreducible water due to: unmovable due to the electrostatic forces; water in the isolated pore and water in the shale. This saturation is related to a permeability value of 0.

If he ask to draw the relative permeability curve we must to write the limit, 0 to 1; the name of the axis; the vertical limit Swi and Sor (the residual oil saturation for a water-oil displacement due to capillary forces); Kro which start below 1 due to irreducible water present in the oil layer (kro,iw) [think to the analogy between area and permeability (m2 is the unit of measure)]; the oil is reducing during production and water replace the void freedom, so ko reduce and kw increase. Relative permeability of oil goes to 0 and water goes to the value below 1 due to the irreducible oil which remain in the layer.

Sor is more or less 20-30%. In gas-water system we have the same effect:

Sgr is more or less 40-50%. Statistically the krw,gr is lower than krg,iw but it is not compulsory.

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P

z

gas

oil

Water

Pb

A reservoir with gas cap shown below is a saturated oil reservoir in less word we have a under-saturated oil reservoir.

At GOC we have the bubble point at pressure of saturation.

If this reservoir is closed and we have in the oil layer the production line. The initial saturation is So=1−S wir so I have (kro,iw). When the production start the pressure reduction goes below the bubble point and the oil start to liberate gas so we have a three phase system. The Sgc is the critical gas saturation which means the minimum saturation needed to the gas to start moving. From Sgc gas permeability start to increase and oil continue to reduce. The limit is the oil irreducible saturation plus the irreducible water saturation. Plotting this problem in gas saturation vs relative permeability we have:

In case of a reservoir with gas which fall to dew point and start to condensate I have the same behavior but different name (change the g with o).

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Mobility ratioIt is

M=λwλo

=kw

k o

μoμw

= kk

kr , wk r , o

μoμw

The GORIt is an indicator of the reservoir condition.

GOR=Q gsc

Qost

Value Detail0-50 Heavy

50-200 Medium200-350 Light350-600 Volatile700-800 Gas condensate>35000 Dry or wet gas

GOR=Q g , sc , free+Qg , sc , dissolved

Qo , st

Q g ,sc ,free is the flow rate of gas at surface condition free in to the reservoir.Q g ,sc ,dissolved correspond to the residual.

GOR=Q g , sc , free

Qo , st

+RsQ o , st

Q o , st

GOR=Rs+Qg , rc

Qo ,rc

Bo

Bg

GOR=Rs+krgkro

μo

μg

Bo

Bg

The behavior of GOR is represented in the picture below

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Where Np is cumulative oil production, N is the original oil in place, the ratio of these two element give us the recovery factor RF.

At the beginning of life of unsaturated reservoir (only oil) we have gas only at surface so initially GOR is equal to Rs. When we touch the bubble point (remember the Rs function of P) we start to liberate gas and Rs decrease but we have to wait the Sgcritical so GOR decrease too. After the overpass the Sgc, Rs continue to decrease, but GOR increase due to second term of the equation. At the end of the life of the reservoir we must remember the behavior of the Bg (the gas density change slowly and Bg remain almost constant) so GOR decrease again.

Surface tension and interfacial tensionWe have a fluid, the molecular at surface is not surrounded to all boundary by other liquid molecular, so the sum of all force applied is named interfacial tension

The surface tension is

σ=FL

=[ Nm ]

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The Interfacial tension (IFT) is the tension between two different liquids in a solid.

The fluid can be:

Immiscible Miscible

We’ll talk ever about immiscible fluid. If in the interface between oil and water exist a difference in density or pressure these fluid are miscible.

Wetting phase: when two immiscible fluids are placed in contact with a solid surface one phase usually is attracted to the solid more strongly than the other.

If a drop of water fall to a surface, it is possible to have two shape of the drop on this surface: the first to the left means that the plane is water wet, the last to the right in case of non-water wet.

From the shape of contact is possible to define different elements:

σ ws=IFT solid−water

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σ os=IFT solid−oil

σ ow=IFT oil−water

θ=contact angle

σ os−σ ws=σowcos θ

When

θ={0°→70 ° water wet70→110Neutral110→180Oil wet

The contact angle in a mixture is measured from the fluid with higher density.

Looking to a pipe the distribution of the fluid is:

Which distribution is governed by gravity. If we reduce the size of the pipe the fluid will forget about the gravity and will be distributed according to the capillary force

If the system is:

Water wet Oil wet

opposite

The shape depend to the interaction with the pipe.

If we put pressure sensor at the two boundary I measure the pressure of the oil at the left and the pressure of the water at the right. If the system is in static equilibrium the difference of these two pressure is 0.

So also the contact interface among the two fluid is to equilibrium:

44

∑ F=0

po (π r2 )−pw (π r2 )+σ ws (2πr )−σos (2πr )=0

po−pw=2 (σos−σws)

r

po−pw=2 (σow cosθ )

r

The capillary pressure ( pc , ow) is determine as the difference between the pressure of the non-wetting

phase and the pressure of the wetting phase. As shown in the equation above. The cosine is the absolute value.

If σ os and σ ws have the same length the σ ow will have a 90° contact angle, so the interface is a straight line. The capillary pressure can’t be negative because the definition (it is given by the difference among fluid wetting and non-wetting pressure, and non-wetting push the wetting). Some company fix the pressure of the water as a reference value in oil industries and so they can have negative pressure.

The capillary pressure influence the:

Drainage of the rock The fluid contact

If two fluids are miscible the capillary force tends to have a gradual changing on the composition from one to the other in a certain area.

Let’s consider a system in which we have a glass of water and oil, I put a capillary tube in the middle. If the system is water wet will happens:

45

P

Blue line is outside the capillary tube

Red line in the capillary tube

In water rise:

pA−pD=g ρw ( z A−z B )+g ρo(zB−zD)

pA−pD=g ρw ( z A−zC )+g ρo ( zC−zD )−2πr σowcos θ

π r2

We have obtained the same formulation of the capillary pressure. Now we are interested to calculate the capillary rise or capillary fringe:

g ρw ( zB−zC )−g ρo ( zB−zC )=2σ owcosθ

r

gh (ρw− ρo )=2σow cosθ

r=pc , ow

h=pc, ow

g (ρw−ρo )

Analyzing the capillary pressure and the capillary rise we can see that if we reduce the dimension the pressure increase and so the height increase.

The behavior of the water saturation versus the capillary fringe is represented in the graph below:

46

The capillary fringe give us the depth at which is better to work (up and far from the fringe).

Lighter is the HC smaller will be the capillary fringe, cos smaller is the capillary pressure. On other hand we can say that smaller is the radius of the medium higher will be the capillary rise.

We can see also the variation of the irreducible water changing the medium, due to the specific surface of the pore.

We can define as well a relations between capillary pressure and saturation of water:

47

We use an interpolation from the reservoir close to the new field in order to forecast the reservoir behavior:

J (Sw )=pc

σLcos θL √ kϕImbibition and drainageImbibition is a process in which a wetting fluid displaces the non-wetting fluid (water displace oil). Drainage is a process in which a non-wetting fluid displace a wetting fluid (oil displace water).

48

In a reservoir we have drainage during the migration. We have imbibition when we have a strong aquifer or in water injection.

In the system we have two capillary tubes with different radius, fully saturated with water (1). We inject the oil from the top, we denote that the system is water wet, the oil distribution is this (2) and at the end (3).

If we inject water the oil will be displaced by water and we can see that the water from the tube with smaller radius rise the tube and fall to the tube with larger radius. Oil remain traps and make a bubble in the water.

The bubble of oil will be part of the residue oil in the medium.

The behavior of the saturation in the medium during the imbibition and drainage curve is this:

49

Usually when we start the production the water table raise (not always).

50

Production drive mechanismThe drive mechanisms refer to primary production.

Improved oil recovery: it is the secondary recovery

Enhanced oil recovery: it is the tertiary recovery.

Drive mechanisms are: Depletion drive oil and gas, Gas Cap, and water drive mechanism into primary production drive.

Depletion drive in gas reservoir: it is the mechanism of producing oil thanks to the itself energy. In a CPVR (constant pore volume reservoir) pressure decrease and volume is constant, Bg increase (this is the effect changing in the equation of state) the difference in pressure give us a flow and we produce. The recovery factor expected from literature from 80 to 90 % of GOIP in case of gas reservoir.

Depletion drive in oil undersaturated: In case of an oil reservoir we denote the depletion drive only for the production until the bubble point, the source of energy is the compressibility of the gas dissolved into the oil, it is from 2 to 5 % of OIP.

When we touch the bubble point, according to economical evaluation we chose to do something:

o Inject water to maintain the pressure in the reservoir equal to the bubble point or slightly up to the bubble point because we have the lower viscosity value and the highest mobility

o Dissolved gas drive: we go down the to bubble point so the gas is starting to liberate and the expansion until reach the movable quantity become a driver mechanism (GOR decreases, we are producing slowly). The recovery factor is from 15 to 20 %.

51

Gas cap drive: when the under-pressure sync, near to the well, touch the gas layer due to production, the gas will start to push the oil due to expansion. This phenomena is controlled in horizontal well. Instead, if I use the vertical well the under-pressure near to the well doesn’t be sufficiently large in order to have a drive gas cap mechanism. The energy source is the expansion of the gas cap and solution gas. The rate of oil recovery is from 25 to 30% of OOIP.

Water drive mechanism: water is connected with an aquifer which is huge; due to production water table rise (1 bar per 10 meter in 10 years… it is nothing :D ). The behavior of GOR is constant and horizontal. The oil recovery is 40% the best value. We have also depletion.

Residual gas saturation for water displacement is from 40 to 50%, it is very high and if we have a gas reservoir with water table we can recover from 50 to 60%. It is bad compared . The solution in this case is to produce faster than water rising (the propagation of the pressure sync is function of the rocks and fluid) it mean before the activation of the aquifer.

52

Pressure gradientThis is the pressure versus depth on a reservoir field

Geostatic gradient is 17 kPa/m; for water is 10,2 kPa/m. For gas we have 1,8 kPa/m and oil is 7,5 kPa/m. We can consider different situations:

Hydraulic communication

53

Regional aquifer

Lenticular reservoir

Multilayer reservoir

54

Diffusivity equationIn fluid mechanics of porous media we use the mono-phase flow in order to solve the diffusion and diffusivity equation, gas it is not movable in order to solve the problem analytically.

From the real life I apply a theoretical model (make some hypothesis assumption and simplifying behavior) than I apply the mathematical model combining the system to some mathematical equation. The solution of a mathematical model could be determined analytically or numerically, depending if the equation are implicit or explicit.

The characteristic of the problem could be organized in this status:

Number of phase:o Single phaseo Multiphase

Nature of fluidso Compressible (liquid)o Very compressible (gas)

Geometryo Mono-dimensionalo Bi-dimensional (radial flow)o Three-dimensional

Hydraulic regime:o Steady state flowo Pseudo-steady state flowo Transient flow

We have first of all to remark some concept:

Mass continuity equation

∇ ∙ ( ρv )=−∂ (ϕρ )∂t

Flow equation

v=−kμ∇ p

55

State equation

Considering a control volume:

We define the mass flow rate along x the quantity:

m x=δm /δ Ax

δt [ kgsm2 ]

Where Ax=∆ y ∆ z.

m¿=mxx A ∆ t

mout= ˙mxx+∆x A ∆ t

So the balance of the mass is: mass in less mass out equal to accumulation term.

(m xx− ˙mx x+∆ x ) A ∆ t= δδt

( ρϕ∆V )∆ t

Where V is the bulk volume of the control volume. ϕ ∆V is the real volume which consider only the pore volume. The terms ( ρϕ ∆V ) is under the hypothesis of mono-phase flow. In case of multiple phase I must to use a single equation per each phase.

If we dividing by ∆V ∆ t where ∆V=A ∆ x we obtain:

mxx−mxx+∆x

∆ x=δ ( ρϕ )δt

δ mx

δx=

−δ ( ρϕ )δt

We are assuming that medium is undeformable, so bulk volume is constant, the geo-mechanical problem is separated to the fluid addressing problem. Crushproof solid. The minus because is the opposite of the incremental ratio.

Knowing that m x= ρ vx

δρ vxδx

=−δ ( ρϕ )

δt

56

Which in 3D form is

∇ ( ρv )=−δ (ϕρ )δt

That in radial coordinates is:

1rδδr

( rρ vr )=−δ (ϕρ )

δt

Now we can substitute the v with the Darcy law vr=−kμ

δpδr

in order to obtain the diffusion equation.

1rδδr (rρ(−k

μδpδr ))=−δ (ϕρ )

δt

We can do these hypothesis: the medium is homogeneous and isotropic and low gradient pressure, so Permeability constant, Viscosity in case of small pressure gradient doesn’t vary, so viscosity constant and it can leave the derivate.

1rδδr (rρ(−δp

δr ))=−μk

δ (ϕρ )δt

We have obtained the diffusivity equation, which is a partial differential equation:

1rδδr (rρ δpδr )= μ

kδ (ϕρ )δt

The density is function of pressure and porosity change as well. So in order to solve it we have to linearize this function:

δδt

(ϕρ )→ρδϕδt

+ϕ δρδt

→ ρδϕδp

δpδt

+ϕ δρδp

δpδt

Remember the compressibility factor:

cr=1V p

δ V p

δp= 1ϕV p

δ (ϕV b )δp

undeformable→

1ϕδϕδp

So

δϕδp

=crϕ

Considering the hypothesis of a slightly compressible fluid, given the definition of compressibility of a fluid we have:

57

δρδp

=c f ρ

Hence the linearization of diffusion equation will be.

δδt

( ρϕ )=ρ crϕδpδt

+ϕ c f ρδpδt

δδt

( ρϕ )=ρϕ (cr+c f ) δpδt

We also go ahead saying that the sum of the two compressibility is the total compressibility:

CT=c f+cr

That in case of more than one phase

CT=cw Sw+co So+cr

Substituting linearization to the equation we obtain:

1rδδr (rρ δpδr )= μCT ϕρ

kδpδt

Focusing on the first member of the equation, on the derivate, we can write

δδr (rρ δpδr )=r

δpδr

δρδr

+ρ δδr (r δpδr )

δδr (ρr δpδr )=r c f ρ( δρδr )

2

+ρ δδr (r δpδr )

Neglecting the derivative of second order:

δδr (ρr δpδr )=ρ

δδr (r δpδr )

At the end we can move the density out of the derivative and obtain:

ρrδδr (r δpδr )= μCT ϕρ

kδpδt

We simplify density to the equation obtaining the final diffusivity equation:

1rδδr (r δpδr )= μCT ϕ

kδpδt

58

In Cartesian coordinate we have only x and we have a second derivative, where the ¿ (grad (p ) )=∇ p

∇2 p=μCT ϕ

kδpδt

Another way to write the equation is using the diffusivity constant or coefficient:

∇2 p=1ηδpδt

η= kμC tϕ

The diffusivity equation for a mono-phase fluid of a slightly compressible fluid in homogenous and isotropic porous medium and under the Darcy assumption, so the other hypothesis of the diffusivity equation are:

No interaction between rock and fluid (no source of species) Laminar flow Negligible gravitational force Constant producing layer thickness Well perforated across the entire production layer

The diffusivity equation describe how the pressure disturbance evolves within the reservoir.

If the system is homogeneous and isotropic, consider now a cylinder reservoir, the pressure sync will move gradually to the boundaries and it will touch all the boundaries of the system at the same time.

We enter in the topic of the Infinite acting radial flow (IARF) if these hypothesis are satisfied:

Entire thickness is a productive layer Fluid flow horizontal All the hypothesis done before to reach this point

59

r

p

t

q

t

pre

Pe

Pe

rw

We can define an average radial pressure which will give us the reference value to evaluate the parameter of the reservoir:

p= 1V∫V

❑

pdV

We consider mainly three flow regime:

Transient: when we start the production Steady state: when we have an aquifer Pseudo steady state: when we haven’t no aquifer

Starting from the point of view of production we have a initially a transient flow and it will remain until reach the boundaries (red line):

The initial conditions is (blue line)

Pressure in the reservoir equal to the initial pressure pe.

When I touch the boundary of the reservoir my flow change, we can have two situation:

1. Something pushing and pressure remain constant (steady state phase); 2. No flow boundary where pressure decrease in all reservoir zone (pseudo-steady state).

60

61

Steady state solutionThe initial and continue condition are the end condition of the transient regime, so we have:

{p=pe at r=reδpδt

=0∀ t

The flow regime in this state is:

δ2 pδ r2 + 1

rδpδr

=0

Fixed this boundaries (re external radius and rw the internal radius) condition we have this solution:

p (r )=pw+qμ

2πkhln

rrw

pw=p i−qμ

2 πkhlnrerw

62

r

P

t

q

t

p

re

Start of steady state

Transient period

Pi

Pi Pe

Constant flow is given by the effect of the aquifer which maintain a constant delta pressure even if I’m depleting the reservoir

Stabilization time: Time at which we touch the

boundary of the reservoir.

Skin effectOne of the problem which effect the pressure is the skin effect: some fine particles of the mud remain in the pore of the invaded zone creating a damaged zone which lead a loss of pressure near the wellbore environment.

63

We have k '<k which cause an additional pressure drop, from pw to pw' . this delta pressure is called ∆ ps

.

pw=p (r ' )− qμ2 πkh

lnr 'rw

pw' =p (r ' )− qμ

2 π k ' hlnr 'rw

So

∆ ps=pw−pw '

∆ ps=qμ

2πh ( 1k '

−1k ) ln

r '

rw∨ qμ

2πkh ( kk '−1) lnr '

rw

The dimensionless term in the above equation, ( kk ' −1) lnr '

rw, it is called S as the mechanical skin or skin

parameter. It is characterized by well test. The skin parameter could be positive (means damaged zone), negative (improving of the wellbore properties due to fracturing, acidification), null (when nothing affected the formation).

∆ ps=qμ

2πhS

pw' =pe−

qμ2 πkh ( ln

rerw

+s)The normal value goes from 1 to 10, if we have 15 there is something strange and not only damaged zone. When it is from -1 to -5 means we have a fracture

Productivity index and completion factorThis is the rate of oil at stock tank condition divided by the difference of pressure from the reservoir infinite pressure (Pi) and the pressure at borehole (Pw).

PI=qost∆ p

∆ p=p i−pw

pw=p i−qμ

2 πkh (ln rerw

+S)

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∆ p= qμ2 πkh (ln re

rw+S)

So

PI=qo , stq

2πkhμ

1

lnr erw

+S

Which has to be correct by the volumetric factor:

PI=2 πkhμBo

1

lnrerw

+S

The productivity index is one of the main parameter because it give us the information about the magnitude of the amount of oil that we can produce from a given delta pressure.

The tubing head pressure (THP, pressure at the surface in the well) has a minimum which is defined by the pressure of the facilities applied. The difference between this and the bottom hole pressure (BHP) is related by losses and other factor (ideally the minimum THP is the same of BHP). If we have for example 300 bar into reservoir and 220 bar of BHP we can impose 80 bar maximum.

Never use the PI for gas reservoir because it is valid only for oil. We use the gas well liberability.

Completion factor is a ratio between thee real PI and the theoretical PI.

CF=P I real

P I theoretical

CF=∆ p theoretical

∆ preal

=∆ preal−∆ pskin

∆ preal=1−

∆ ps∆ preal

PI is estimated with well test, in order to improve the PI is performed a skin cleaning or other kind of operation like reduce viscosity of the oil, but anyway it is a parameter (characteristic) of the system not a variable.

Pseudo steady state solutionWe consider a pseudo steady state when we have not an aquifer which sustain the pressure in the reservoir, so the pressure is decreasing during this flow regime.

In PSS pseudo steady state, initial condition are the same of the end of the transient period, so we have that:

t=0→p i=prat r=re

65

r

p

t

q

t

Pw

re

Boundaries touched

Transient period

Pi

Pe

PSS regime:Production after 1 h of steady state

SS

PSS

Pw  

Flow rate remain constant because the delta pressure remain the same

The boundary condition in this case is a no flow from the boundaries and a constant flow rate, so:

t>0 {δpδr=0at r=re

δpδt

=c ∀ r

In order to express the pressure variation, we have to consider the non linear compressibility factor of the oil:

Co=−1V

δVδp

= −1Soϕ V b

∆V∆ p

Due to a constant flow we can express ∆V as the cumulative oil product, so ∆V=q t , than:

∆ p= −qSoϕV bCo

t

66

t

t

Build up

Draw down

q

pw

∆ p is a linear function of time, so in PSS we have a line. For a given rate the bigger is the reservoir the smaller will be ∆ p.

The flow regime in this state is:

δ2 pδ r2 + 1

rδpδr

=μCT ϕ

kδpδt

Pressure behavior with different flow rateLet’s consider a period of time characterized by a constant rate (called flow period) in transient regime, if the rate is positive and constant the pressure behavior is called draw down. Let’s suppose to close the well valve until reach the steady state or pseudo steady state. The effect on the pressure at well is an increasing of itself due to re-equilibrium of the pressure in the reservoir; the pressure rise is called build up.

Grow down is only when flow rate is positive. Build up only if the flow rate is null.

When we have injection we have a negative flow rate, in this case the pressure at well wall is increasing, this profile is called Injection; when I close the injection phase the pressure at well wall start to reduce and stabilized at initial reservoir pressure, this profile is called Fall off.

67

t

t

q

pw Fall offInjection

t

t

q

pw

It is a Draw down not a build up due to positive flow rate

t

t

q

pw

 

Let’s consider a different flow rate by step; the effect on the well wall pressure is represented below:

68

P

Pi

rrerw

t

Transient state solutionThe initial condition of this state is the initial condition of the reservoir, so:

t=0→p=pi∀ r

The boundaries condition are: pressure at external radius equal to external pressure (initial pressure); constant flow rate at well wall:

t>0→{ p=p iat r=r eδpδr

=c at r=rw

δpδr

=cmeans q=const

This is called constant terminal rate (Noiman condition)

The flow regime in this state is:

δ2 pδ r2 + 1

rδpδr

=μCT ϕ

kδpδt

The solution of the equation is

p (r ,t )=pi−qμ

2 πkhP( rrw ,t D)

Where P is a function which convert the solution given for dimensionless radius and time and apply to our case, where tD is the dimensionless time which is:

tD=tt c

= kt

μCT ϕ rw2

Where tc is the characteristic time.

69

q

t

Pw

Pi

t

P

Pi

r

rerw

Different time curve

Rd: drainage radius at a certain time, is the radius in which the pressure is still the initial (or investigated radius). When we touch the boundary start the SS

In some case P( rrw , tt c ) is approximated by a function of exponential integral (Ei).

Ei=∫ e x

xdx

P( rrw , tt c )=−12

Ei(−r2

4 ηt )So the last expression is:

pw ( t )=p i−qμ

4 πkhln (2.25 tD)

This equation is very similar to the steady state equation.

According to the superposition of effect:

A transient phenomena can be approximated to a sequence of steady state phenomena that are characterized by an external radius which change in time.

This means that, if we take a picture of the system at 1 h (just to put number) this depletion point can be calculated by using the steady state equation using a radius which is the proper to have a steady state condition:

70

P

rd

Pw

t

Drainage radius doesn’t depends to flow rate!

200 bbl/day

100 bbl/day

pw=p i−qμ

2 πkhlnrdrw

Comparing the two equation at a specific time we have to impose this condition to have an equity:

lnrdrw

=12

ln (2.25 tD )

So we can calculate the dimension of drainage radius as:

rdrw

=√2.25t D

rd=1.5 rw √ ktμCT ϕrw

2 →1.5√ ktμCT ϕ

Drainage radius doesn’t depend to the dimension of the reservoir, and neither to the flow rate, is a characteristic of the fluid and medium permeability and compressibility. Assuming that rate of oil is 100 barrel per day and the flow rate period is 10 h and the permeability of the formation is 100 mD.

If I have 2 different level of permeability

71

Pw

t

k

P

r

ln t

PwP

Pi

t

Skin effect in transientCan I calculate the permeability from the pressure in transient?

pw ( t )=p i−qμ

4 hkπln (2.25 tD)

Can we take into account the skin effect on transient?

pw ( t )=p i−qμ

4 hkπ ( ln (2.25 tD )+2S )

And in steady state?

pw (t )=p i−qμ

2hkπ ( lnrerw

+S)Can I determine the skin effect having the P vs time curve?

pw=cost+m(x+2S)

The pressure sync change in space and time so we make steady state changing the domain of the reservoir using

72

rd=1,5√ ktμC tϕ

In terms of two fluid (oil and water) we use the relative permeability using the value at infinite and:

CT=CoSo+Cw Sw+Cr

To summarize we determine:

Productivity index Skin effect Drainage radius Permeability

73

v

dP/dr

Darcy behavior

Forchheimer behavior

Diffusivity equation for gassesIn case of gas we have:

Low viscosity: means very high velocity High compressibility

These two consideration are in contrast with the previous assumption done to obtain the diffusivity equation. So we need to correct the equation considering new phenomena

The velocity is the ratio of flow rate and area, so when area is lower velocity is higher, so we have the maximum velocity at borehole wall. High velocity means turbulence which means pressure drop. We have to correct the darcy law. We use the Forchheimer equation.

∂ p∂r

=μkv+β ρ v2

Where v is the velocity which is determined by darcy law using the same assumption.

The turbulence factor β (ϕ , kass) this coefficient is purely empirical and there are different correlation to

the physical parameters which lead a really different range of magnitude. β ρ v2 give the turbulence.

Turbulence occurs only in gas reservoir with high probability close to the well, where the area is little.

If I increase porosity velocity decrease and pressure drop decrease and the turbulence.

74

To obtain the solution of the diffusivity equation for liquid we used the mass balance equation, where we change the velocity on it with Darcy velocity. Now I can’t do the same cos the Forchheimer quadratic equation.

From literature we use this number to understand the velocity regime:

Q g{ ¿50000Sm3

d→laminar

¿100000Sm3

d→Turbolent

There are two approach:

1. Neglect Forchheimer equation and use only Darcy and use density of the fluid as gas

We going to underestimate the delta-pressure so the method is not valid close to the wellbore:

ρ= MpzRT

1r∂∂r ( Mp

zRTrkμ∂ p∂ r )= ∂

∂t ( ϕMpzRT )

After some algebraic application we obtain

1r∂∂r ( pzμ r ∂ p∂ r )=CT ϕ

kpz∂ p∂ t

The analytical solution is:

{pzμ

=const i f p>3000 psi→1r∂∂ r (r ∂ p∂r )= μCT ϕ

k∂ P∂ t

zμ=const if p<2000 psi→1r∂∂ r (r p ∂ p∂r )= μCT ϕ

kp∂ P∂ t

→1r∂∂ (r ∂ p

2

∂r )=1η∂ p2

∂t

m ( p )always valid→ 1r∂∂ r (r ∂m( p)

∂ r )=1η∂m (p )∂ t

The first solution is the same of the fluid. The second solution is obtained remember that

∂ p2

∂r=2 p

∂ p∂ r

And the final step represent the Darcy equation for gas.

75

t

M(p)

qout=Akμp¿

2−pout2

2 pout L

The third analytical solution is considered in case of gas the non-linearity is achieved using the pseudo pressure function, in this case become

m (p )=2∫p0

ppzμ

dp

We can take into account the variability of the compressibility and viscosity of the gas variation (which were constant for fluid). It is a function of pressure temperature and has composition.

Considering that ∂m ( p )∂ r

=2Pzμ

∂P∂r

we obtain the so called rigorous solution.

We don’t have turbulence.

2. Forchheimer equation solving

We integrate the Forchheimer equation using a steady state condition. As done previously, we solve the steady state regime of Darcy (result the Muskat equation) and using the same approach we will obtain an equation in radial flow. A transient phenomena can be approximate, again, to a series of steady state, using the drainage radius.

Because gas is a very compressible fluid, the volumetric rate q changes as a function of pressure (q increase when the pressure decreases). On the opposite, the mass rate ρq is constant.

Permeability and turbulence factor are constant in all the reservoir environment and not close to wellbore. As we did previously, I can take into account the skin effect, so for gas I apply the skin effect which take into account the turbulence phenomena as well.

76

77

qsc

t

t

Pwf

Gas well liberability is the index used in gas reservoir

m(pw)

Ln t

m

In steady state condition I have this solution

m (pw )=m ( pi )−Tπkh

Psc

T sc

qsc( lnr erw

+S+Dqsc)pw=p i−

qμ2 πkh (ln re

rw+S ')

D is called the non-Darcy flow coefficient, which describe the difference on the behavior from Darcy

law. S'=S+Dqsc is called the appearance skin which is the skin effect plus the turbulence skin. S’ is the

only parameter that I can measure during well test, but I cannot divide the two terms.

In transient condition the solution become:

m ( pw ( t ) )=m ( p i )−T

2πkh

Psc

T sc

qsc ( ln2,25 tD+2S+2Dqsc )

The problem now is determine the S’, and I do it in a well test following:

tD=kt

μCT ϕ rw2=carat .time t

78

S’

Qsc

D

S

Isochronal flow period

qsc

t

t

Pwf

We try to have the same flow period in order to have the same radius of investigation we call it isochrones period

I cannot separate the 2 terms in the appearance skin effect. So what I can do?

And than

If we perform a stimulation on the formation we will have: (increasing of k -> beta decrease) (increasing phi -> decrease the beta) (what is change is the velocity at microscale)

79

S’

qsc

D

SSnew

Dnew

80

Reservoir

Pressure sync

Reservoir

Pressure sync

t

Pwf

T stabilization

Close well

Pss

ss

Depletion effectPbar

Build up

DepletionSome time we can have the expression of PI as

PI=p−pw

When I start the production pressure decrease, when I reach the boundary I can have steady state (horizontal line blue) when I close the production pressure will increase and tends to the initial pressure. If I fall in PSS the pressure will drop linearly during time. When I close the production the pressure will increase to a value which is not the same of the initial pressure (due to the pressure drop that we have seen on the graph pressure versus radius in pss) so to determine the final pressure I draw a line having the same inclination starting from stable time of pss and when I reach the time of closing well I have the horizontal pressure which represent the pressure p.

This phenomena is called depletion effect. And the effect on the productivity index is constant in steady state and pseudo steady state.

81

PI

tT stabilization

transientPSS or SS

Time = t1 Time = t2 we touch 1 boundary

Late transient

boundaries Pressure sync

   

Time = t3 we touch 4 boundary

PSS

Early transient

Pwf

t

ET LT

PSS

Considering now this kind of structure

82

PSS only

      

SS

   

SS directly

The interaction between reservoir and aquifer are represented below:

83