Over Pressures

OVERPRESSURESOVERPRESSURES

1. PRESSURE CONCEPTS

1.1. DEFINITIONS

A pressure is a force divided by the surface where this force applies.

Pressure Pascal = Force Newton / Surface m2

The official pressure unit is the PascalIt is a very small unit: 1 Pascal = 1 Newton/m2 1 bar = 105 Pascal 1 atm = 1,013 *105 Pascal

A practical unit on the rig is the kgf/cm2

1 kgf/cm2 = 0.981 bar

In API , the unit is the pound per square inch (psi)1 bar = 14.4988 psi

Exercise 1:

Enter a value in the yellow cell, click on result (blue cell) and press F9 to get the answer. Calculation & explanation on the last page…..

15 Bars = psi155 Psi = Bars

1.1.1. HYDROSTATIC PRESSURE : P h

Pressure exerted by the weight of a static column of fluid

It is a function of fluid specific gravity and of vertical height of the fluid

Ph = d * g * H

With Ph = hydrostatic pressure (Pascal)d = Fluid specific gravity (kg/m3)

H = Vertical height of fluid (m)g = gravity acceleration (9.81 m/s2)

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OVERPRESSURES : AN INITIATION


Using well site units, the formula becomes :

Ph = H* d 10

With Ph= hydrostatic pressure (bar) d = Fluid specific gravity (kg/l)H = Vertical height of fluid (m)

NB : the term 10 is approached, for precision, you should use 10.2 with pressure in bars and 9.6 for pressure in bar

In API, the formula is:

Ph = 0.052 * H * d

With Ph= hydrostatic pressure (psi)d = Fluid specific gravity (ppg)H = Vertical height of fluid (ft)

Consequently, in the following sketches, the Ph is always the same:

Pascal was betting he could destroy a barrel with just a pint of water:He fixed a long and thin tube on the barrel and poured the water, the volume of water was small, but the height was enough to make the barrel explode!

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H


‘U tube’ effect: If the mud weight in the pipes and in the annulus are different, the ‘U tube’ effect is the difference of Ph in the 2 branches of the ‘U tube’ formed by annulus and pipes

Exercise 2:Calculate Ph in the following examples

Height of fluid (m):

1000 Fluid SG (kg/l):

1.5 Ph bar:

Height of fluid (ft):

1000 Fluid SG (ppg):

10 Ph psi:

Exercise 3: A heavy mud slug has been pumped, what extra volume do you get in the mud pit ? (‘U tube’ effect)

Pipes internal volume (l/m) :

9.1 Slug volume (l):

2000 Slug SG (kg/l):

1.6

Mud Weight (kg/l):

1.08 Extra volume in pit (l):

Exercise 4: A heavy mud slug has been pumped, what is the air gap height in the pipes ? (‘U tube’ effect)

Pipes internal volume (l/m) :

9.1 Slug volume (l):

3000 Slug SG (kg/l):

1.7

Mud Weight (kg/l):

1.10 Height of air gap in pipes (m)

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1.1.2. OVERBURDEN: S

At a given depth, the overburden is the pressure (applying on fluids) or stress (applying on matrix) exerted by the weight of the overlying sediments.

S = H * b

10

With S = Overburden stress (bar)b = Formation average bulk density (no unit)H = Vertical thickness of overlying sediments (m)

In API, the formula is:

S = H * b * 0.433

With S = Overburden stress (psi)b = Formation average bulk density (no unit)H = Vertical thickness of overlying sediments (ft)

The bulk density of a sediment is a function of the matrix density, the porosity and the density of the fluid in the pores.

b = ( * f) + (1-) * m

With b = Bulk density (no unit)f = Formation fluid density (no unit) = Porosity (from 0 to 1)m = Matrix density (no unit)

With depth, the sediment porosity will decrease under the effect of compaction (proportional to overburden) and of course, the bulk density will increase.

You will note that the porosity shale curve is exponential

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b

Depth

2.31b

Depth

2.31

Sea botttom


Relationship between porosity and depth

Relationship between bulk density and depth

On shore Off shore

1.1.3. FORMATION PRESSURE: P f

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Porosity

Depth

Sand

Shale


Also called Pore pressure :Pp

Pressure of the fluid contained in the pores of the sediment.

1.1.3.1. NORMAL P f: Pf=Ph

The Pf equals the Ph due to the column of fluid in the sediment.It depends on the density of the water(usually from 1.00 to 1.08)

1.1.3.2. NEGATIVE P f ANOMALY : Pf<Ph

In the following example, the outcrop is lower than the point where the well enter the formation. The water does not reach this zone.

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1.1.3.3. POSITIVE P f ANOMALY: Pf>Ph

Artesian well

.In that case Pf = H*d instead of h * d10 10

Hydrocarbon column

Due to the difference of densities between water and hydrocarbons, the pressure at the top of the reservoir is almost the same that at hydrocarbon –water contact

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Pressure

Depth

gas

Water


The formula for the pressure anomaly (excess of pressure respect to normal) is

Phc = H * (dw – dhc)10

With Phc = Pressure anomaly at the top of the hydrocarbon column (bar) H = Height of the hydrocarbon column (m) dw = Water SG (kg/l) dhc = Hydrocarbon SG (kg/l)

Note that this anomaly is proportional to the height of the hydrocarbon column and to the diffeence of SG between water and hydrocarbon.

This anomaly can be very high in case of reservoir in a vertical lense:

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How does it work ?

Exercise 5:

Imagine the reservoir as a U tube: C is the top of the hydrocarbon:

Water SG (kg/l) 1.05Hydrocarbon SG (kg/l) 0.25Point A & C depth (m) 2000Point B depth (Hydrocarbon/water contact)(m) 2590Calculate Pf in A (bar)Calculate Pf in B (bar)Calculate Ph of the column of water AB (bar)Calculate Ph of the hydrocarbon column (bar)Calculate the Pf in C (bar) 257.20Pressure anomaly at top of hydrocarbon (bar) 47.20

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AC

B


1.2. PRESSURE REPRESENTATIONS

1.2.1. EQUILIBRIUM & EQUIVALENT MUD WEIGHT

Equivalent MW is the MW corresponding to a mud column pressure, related to depth.

Equilibrium MW represents the average MW needed to counterbalance Pf.From the Ph formula, we can recover:

MW = P * 10 H

Exercise 6:

Two rigs are drilling a well in the same overpressured formation, calculate the equilibrium MW for each rig:

Distance from A to formation (m) 2000Distance from B to formation (m) 1200Pf (bar) 200Equilibrium MW for rig AEquilibrium MW for rig B

In a well, the mud weight can be different than the equivalent MW.

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The following exercise will show you some examples:Exercise 7:

Well total depth (m) 2500Mud weight in hole (kg/l) 1.22Calculate the equivalent MW in hole in the following occurrences:During a trip, the driller forgot to fill the hole and the mud level is lower than normal , its distance from flow line is (m)

150

During a Leak Off Test, the pressure reached (bar) : 20You start circulating, the pressure losses in the annulus is (bar): 10

You can see that , according to the cases, Equivalent MW can be smaller or bigger than actual MW in hole.

Here are some examples:

Equivalent MW < MW Equivalent MW > MWMud losses Leak Off TestSwabbing Circulating (ECD)Fluid influx in hole (Kick) Surge

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1.2.2. PRESSURE GRADIENT: G

A pressure gradient G is the unit increase in pore pressure for a vertical increase in depth unit, but to get consistency with MW, we will take 10m.

It is used to give a degree of consistency to pressure data: Pressure gradient and MW will be comparable

Exercise 8:

Mud weight in hole (kg/l) 1.20Calculate the hydrostatic pressure gradient (bar/10m):

1.20

As the figures are similar, MW and Pressure gradient may be plotted on the same graph, allowing a comparison between MW, Formation pressure gradient, Fracturation gradient and Overburden gradient.

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Pressure bar

P

H

Depthm

G = P in bar/10mH

Pressure (bar)

Depth(m)

GPf MW FRAC S


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1.2.3. HYDRODYNAMIC LEVEL

We have seen that the formula P = H * d/10 allows to calculate the pressure or the equivalent MW, you can also calculate H, which represents the hydrodynamic level (ie: the height reached by the water in an artesian well)

H = P * 10 + Z d

With: H = Hydrodynamic level (m)P= Formation pressure at depth Z (bar)d = Formation fluid specific gravity (kg/l)Z = Subsea depth (m)

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Sea level

H

P*10 d

Z


1.2.4. PLOT OF PRESSURE VS DEPTH

May be used with an overlay of pressure gradient

Another purpose of this type of plot is to determine the contacts between fluids in a reservoir by tracing the trend lines of the RFT pressure data:

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1.03 1.20 1.30

Pf & Ph (due to MW) in bar

Depth (m)

Depth(m)

Pressure from RFT (bar)

Gas/oil contact

(m)

Oil/water contact

(m)


1.3. STRESS CONCEPT

Unlike liquids, solids can withstand different loads in various directions:Imagine a cube of porous rock somewhere in the deep….We can divide the stresses in 3 resulting forces according to the 3 directions of space: S1 can be considered as the Overburden , S2 and S3 the tectonic forces (open hole ovalization can give an idea of the difference between S2 and S3).

In a porous rock, the fluid may support part of the stress (due to undercompaction) and the total stress S will have 2 components:

S = Pp + (Terzaghi equation)

With S = Total stress (bar)Pp = Pore pressure (or formation pressure) Effective stress (on the grains of the rock)

Consequently:

S1 = Pp +

S2 = Pp +

S3 = Pp +

So, in theory, the formation pressure is limited by the overburden !

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S3

S1 (overburden)

S2


2. ORIGINS OF NON HYDROSTATIC ABNORMAL PRESSURE

2.1. UNDERCOMPACTION (OVERBURDEN EFFECT)

Normally, the compaction increases with depth and the formation water is expelled as the porosity decreases.In some cases, the water cannot be eliminated in time and remains trapped in the sediment: the main cause of overpressure is due to what is called undercompacted shale.

Water elimination from shale depends on 3 factors:

Clay permeability : very low

Sedimentation and burial rate : if the sedimentation rate is very high, the shale is brought very deep before the water has time to go and it remains trapped in the sediment (ie: deltaic areas)

Drainage efficiency : sand layers act as a drain and helps water elimination, less than 15% of sand content in a shale will cause a lack of drainage and an overpressured zone.

Terzaghi experiment

The springs represent the matrix, and the load on the upper plate represents the overburden

A: the lower tap is closed (no drainage) and S is only supported by the fluid:

S = Pf

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SS S

A CB

Pf


B: The lower tap is open, water escapes and the spring/matrix bears part of the load : At that stage, if you close the tap, you get something similar to an undercompacted shale: the fluid is trapped in the sediment and supports part of the overburden, causing an overpressure.

S = Pf +

C: The springs/matrix fully support the load: this is the case of a normally compacted sediment.

Pf = PhS =

One of the main cause of overpressure.

2.2. AQUATHERMAL EXPANSION

The volume of water increases with temperature, if it is in a sealed environment, its pressure increases.

Actual effect is controversial.

2.3. CLAY DIAGENESIS

With depth, the smectites (as Montmorillonite) will lose its adsorbed water and transform into Illite + free water.

Not really a cause of overpressure, but acts as a contributory factor in case of undercompacted shale.

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2.4. OSMOSIS

Osmosis is the spontaneous movement of water through a semi-permeable membrane separating two solutions of different concentrations, until the concentration of each solution becomes equal.

A clay bed can act as a semi-permeable membrane between two reservoir containing water with different salinity.

Not proven in nature and anyway minor effect if exists.

2.5. EVAPORITES

2.5.1. SEALING ROLE

Evaporites are impermeable and can make a good seal that will block water expelled from underlying sediment, creating overpressure by overburden effect.

Major role in overpressure generation, specially if interlayed with shale.

2.5.2. SULFATE DIAGENESIS

Gypsum is the precipitated form of CaSO4, transformation to Anhydrite may occur early in the burial process:

CaSO4,2H2O (Gypsum) CaSO4 (Anhydrite) + 2H2O

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Clay layer

Pure water

Salt water

Osmotic flowPf decreases

Pf increases


The water amounts to 38% of the original volume, if it cannot be expelled, overpressure develops. Similar increase of volume is created by rehydration of Anhydrite .

Minor effect as the diagenesis of gypsum to anhydrite often occurs at shallow depth, this allows the water to escape. Rehydration of Anhydrite is not proven on a scale that would be enough to generate overpressure.

2.6. ORGANIC MATTER TRANSFORMATION

At shallow depth, bacteria will transform organic matter in biogenic methane.From a depth of 250m, thermo chemical cracking will transform heavy hydrocarbons to lighter ones, with increase of volume. If these processes occur in a close environment, they create overpressure .

Important role in overpressure generation in confined series of shaly sands or carbonate.

2.7. TECTONICS

2.7.1. RELIEF & STRUCTURING

As already seen in chapter 1.1.3, relief can be the cause of pressure anomalies (ie: artesian well).

2.7.2. REORGANISATION OF STRESS FIELD

Sediments are subjected to overburden and to horizontal tectonic stresses.

2.7.3. FAULTS

Faults can create a seal and stop the water or on the contrary, bring an overpressured zone in front of a permeable zone, allowing the water to escape.

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2.8. MISCELLANEOUS

2.8.1. CARBONATE COMPACTION

Normally, carbonates do not have problems of under compaction. Chalk is the exception. Chalk is due to the deposit of tiny discs called coccoliths (calcareous plates protecting some phytoplankton) and Chalk structure looks like Clay structure, with the same problem of low vertical permeability.

2.8.2. PERMAFROST

Typical of the artic zones, due to the fact that the sediments are. The problem is due to unfrozen pockets (called taliks) inside the permafrost. If a talik freeze, its volume tends to increase (remember that ice is bigger than original volume of water), but the permafrost impedes expansion, thus creating overpressure.

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Phytoplankton with its « shield » of coccoliths

When phytoplankton dies, the coccoliths deposit with a « clay like » structure.


3. PREDICTION & DETECTION OF OVERPRESSURE

3.1. NORMAL COMPACTION TREND

To be able to detect an overpressure linked to compaction anomaly; it is useful to define what the normal compaction trend is.

3.2. CHARACTERISTICS OF UNDERCOMPACTED ZONES

If two zones of different potential are separated by a seal, there will be an abrupt change in pressure.

But if the seal is not perfect we get a gradual transition zone which is easier to detect.

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Impervious barrier

Hydrostatic pressure

Overpressure

Depth

Pressure

Hydrostatic pressure

Overpressure

Depth

Pressure

Transition zone


A plot of shale density versus depth on a semi-log scale will show the top of the undercompacted/overpressure zone:

3.3. PREDICTION

By prediction, we intend methods before drilling…

3.3.1. REGIONAL GEOLOGY

Study of the lithostratigraphy (do we have shale and evaporite, deltaic formation…?), tectonics (do we have faults, diapirs…?), etc may give indications of the possibility of overpressure.

3.3.2. GEOPHYSICAL METHODS

Seismic data may give some indications of overpressure.

3.3.3. GRAVIMETRY

A negative anomaly in gravimetry may be due to undercompacted shales.

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Normal compaction

Undercompaction/Overpressure

Depth

Log of shale density

Top of transition zone


3.4. METHODS WHILE DRILLING

3.4.1. REAL TIME METHODS

3.4.1.1. RATE OF PENETRATION : ROP

As compaction of the sediments increases with depth, the rate of penetration normally declines with depth.ROP does not depend only on lithology and on compaction, another important factor is the Differential Pressure (P).

Differential Pressure is the difference between the Formation Pressure (Pf) and the pressure exerted by the column of mud in the well .

Let us remember how a rock bit works on bottom:

Impact Formation of a crater of crushed rock

The differential pressure will affect the cleaning of the crushed rock, if P is too high (Ph of mud bigger than Pf), the pressure will tend to “glue” the cuttings on bottom and ROP will reduce do to bad cleaning of the hole.But if P is low or even negative, the Pf will expel the cuttings from the bottom and ROP will increase.

So, in theory, we should have a ROP curve similar to the compaction trend:

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WOB

Bit tooth

WOB

Crushed rock


This could be the perfect tool, but ROP is mainly affected by the drilling parameters and this method could work only if RPM, WOB and bit diameter were constant.The solution is in the next chapter…

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ROP (mn/m)

Compaction trend

Top of overpressure

Depth


3.4.1.2. ‘D’ EXPONENT

The drilling exponent is a way to normalise ROP by eliminating the effect of drilling parameters variation: it gives an idea of the “drillability” of the formation. It has been optimised for shale (with less than 5% of sand content)

The first formula was:

1.26 – log10 ROP d = RPM

1.58 - log10 WOB Bit diam

with ROP in m/hour WOB in tons

Bit diameter in inches

In order to include the effect of differential pressure, a correction has been made:

Dc = d * Normal hydrostatic gradient ECD

With: Dc : corrected d exponentd : d exponentNormal hydrostatic gradient (from 1 to 1.08)ECD: Equivalent Circulating Density

Another correction for the bit wear has been proposed, but quite unsatisfactory.

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“d” exponent interpretation

First problem: What is the best position for the normal trend (Dcn)?

A: the trend is on the right part of the Dc curve.B: the trend is in the middle of the curveC: the trend is on the left part of the curve.

We know that Dc should work with pure shale, as the sand is easier to drill, the Dc in the sand will be to the left, so it is better to put the trend on the right points of the Dc curve, as they correspond to the best shale points.(A)

To understand the importance of the trend position, you must know that a calculation of the Formation pressure can be made, using the distance between Dc curve and its normal trend Dcn.(See chapter 4).Roughly, the Pf curve will be a “mirror image ” of the distance between Dc and Dcn.

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dcndc

Pf gradient

Normal gradient (ie 1.03)Depth

A B C

dcn

dc

Shale point Sand point


The use of this graph is to know what mud weight will counterbalance the Pf.In that case, we can see that some points of the Pf curve go over the MW curve, but these points correspond to the sand points, and are not to be taken in consideration. Only the red dot line (passing through te shale points Pf) is important and is used as reference for MW selection.

If you look at the following interpretation, the trend is on the left part of the Dc curve, consequently the Pf curve is shifted to the left and has no point over the MW curve, which seems to be less scary, but in that case, the actual Pf will be over the MW before we can detect it

²

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dcndc

Pf gradient & MW

Depth

dcndc

Pf gradient & MW

Depth

Kick


Second problem: Are we sure that a deflection of the Dc corresponds to an overpressure ?If we have a shale gradually passing to a sand, the Dc will have a deflection towards left, this can be mistaken for the starting of a transition zone.

If we compare the lithographic column and the Dc, we note that the trends in the shale and in the sand are parallel, there is only a shift. This shows the importance of the cuttings analysis, which is fundamental for the Dc interpretation.

Exercise 9: (Answer is at the end)

What other different parameters will bring a shift in the Dc trend?

3.4.1.3. SIGMA LOG

Sigma log was developed by Agip and Geoservices to replace the D exponent in the carbonates.“Sigma” stands for “rock strength parameter”.The drilling parameters used in the formula are the same that for Dc and the interpretation is similar, with a shift of the normal trend every time the lithology changes…


Decide if the following parameters increase or decrease when an overpressure by undercompaction occurs. And why !

3.4.1.4. TORQUE

or ?

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Shale

Sandy Shale passing to Shaly Sand

Sand

Dc

Dcn in shale

Dcn in sand


3.4.1.5. OVERPULL & DRAG

or ?

3.4.1.6. HOLE FILL

or ?

3.4.1.7. PIT LEVEL – DIFFERENTIAL FLOW – STAND PIPE PRESSURE

or ?

3.4.1.8. L.W.D

A useful parameter given by the “Logging While Drilling” tool is the formation resistivity and the gamma ray. Those 2 logs allows to check the behaviour of the shale resistivity.


In case of undercompaction, shale resistivity or ? Why?

3.4.2. LAGGED METHODS

3.4.2.1. GAS

A good indicator of an increase of Pf is the gas “sucked” from the formation during a trip or a pipe connection (by swabbing).

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Time

A: Positive differential pressureB: Positive, decreasing differential pressure (transition zone)C: Negative differential pressure (Background gas is also increasing)

A B C


The problem with this method is that it depends on the velocity of the hook when the string is pulled up; two different drillers will give two different pipe connection gases.A much better system is to check the “Pump off Gas”: the driller stops the pumps without moving the string, so there is no swabbing. But you lose the pressure losses in the annulus; the equivalent mud weight in hole drops from ECD to MW. In that case, a gas show means that the differential pressure is close to be negative.

Check also the gas ratios! If you have more heavy gases (ie C2/C3 is decreasing), you enter a transition zone.


Decide if the following parameters increase or decrease when an overpressure by undercompaction occurs. And why !

3.4.2.2. MUD WEIGHT

or ?

3.4.2.3. MUD TEMPERATURE

or ?

3.4.2.4. MUD RESISTIVITY

or ?

3.4.3. CUTTING ANALYSIS

Direct analysis of the cuttings can bring information on an eventual overpressure:

Cuttings are bigger and tend to be concave, lot of caving.

Shale density: plotting shale density (ie using a microsol) versus depth can show an undercompaction:

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Depth

Shale density

Top of undercompaction


3.5. METHODS AFTER DRILLING


Here is a list of electrical logs done at the end of the drill phases, which will give indications on overpressure ? In what direction is the deflection of the curve in an undercompacted zone?

1. Sonic2. Resistivity3. Gamma ray4. Spontaneous potential5. Conductivity6. Caliper7. Density log8. Neutron log

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4. QUANTITATIVE PRESSURE EVALUATION

4.1. FORMATION PRESSURE EVALUATION

Most methods to calculate Pf will compare the undercompacted shale with a normal compaction state. A normal compaction trend is necessary.

4.1.1. EQUIVALENT DEPTH METHOD

To every undercompacted point A correspond a normally compacted point B: those two points have the same compaction but at different depths!Depth HB is called “the equivalent depth”.

We can apply Terzaghi equation for Overburden pressure, (S = + Pf) .We know that we have the same compaction in A and B, so the stress must be the same in both points ! AB

We can writeB = SB – PfB andA = SA – PfA

As AB we have : SB – PfB = SA – PfA

Or PfA = PfB + (SA - SB)

Exercise 14: (Answer is at the end)Replace the four factors by their Pf equivalent densities or Overburden gradient in the last equation to obtain the final formula:

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Depth

A

B

Dc or Shale density or porosity log, etc

HB

HA

B


with: S = H * GS and Pf = H * Deq

10 10

DeqA = GSA – HB * (GSB – DeqB)HA

With DeqA : Equilibrium density in A (= Formation pressure gradient)DeqB : Equilibrium density in B (= normal hydrostatic gradient)ZB : Equivalent depthZA : Depth of undercompacted shaleGSA: Overburden gradient at AGSB: Overburden gradient at B ( very often, GSA and GSB can be considered equals)

This method can be used to calculate Pf from the following parameters:Dc, Shale density, Shale resistivity, Sonic, Density and porosity logs.

Give best results for Pf grad> 1.4

Exercise 15:

Calculate Pf gradient with the equivalent depth method:

Mud weight in hole (kg/l) 1.30HA 2900HB 1900Normal gradient 1.08Dc in A 1.3Overburden gradient 2.31Pf Gradient

4.1.2. RATIO METHOD

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Depth

Dc or Shale density or porosity log, etc

A

XA XB

B

Depth

A

B

Dc

HB

HA

B


In that case, we must found the value of the parameter (ie the Dc) at the same depth on the normal trend:

DeqA = DeqB * XB

XA

With DeqA : Equilibrium density in A (= Formation pressure gradient)DeqB : Equilibrium density in B (= normal hydrostatic gradient ie 1.03)XB : Theorical value on normal trendXA : Actual value

This method can be used to calculate Pf from the following parameters:Dc, Shale density, Shale resistivity, Sonic, Density and porosity logs.

Give best results for Pf grad< 1.4

Exercise 16:

Calculate Pf gradient with the ratio method:

Depth 2900Normal gradient 1.08Dc in A 1.3Dcn in B 1.7Pf Gradient

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4.1.3. EATON’S METHOD

This method is used frequently and is adapted to the different parameters:

Shale resistivity:

DeqA = GS – (GS – DeqN) * RshO1.2

RshN

With DeqN : Equilibrium density in B (= normal hydrostatic gradient) RshN : Theorical shale resistivity on normal trend (B)

RshO : Observed value of shale resistivity (A) GS: Overburden gradient observed at observed depth

D exponent:

DeqA = GS – (GS – DeqN) * DcO1.2

DcN

With DeqN : Equilibrium density in B (= normal hydrostatic gradient) DcN : Theorical Dc on normal trend (B)

DcO : Observed value of Dc (A) GS: Overburden gradient observed at observed depth

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Depth

Dc or Shale resistivity

A

XA XB

B


For conductivity and Sonic, the ratio is opposite (the trend of these parameters is towards left):

Sonic:

DeqA = GS – (GS – DeqN) * tN 3.0

tO

With DeqN : Equilibrium density in B (= normal hydrostatic gradient) tN : Theorical transit time on normal trend (B)tO : Observed value of transit time (A)

GS: Overburden gradient observed at observed depth

Conductivity:

DeqA = GS – (GS – DeqN) * CN 1.2

CO

With DeqN : Equilibrium density in B (= normal hydrostatic gradient) CN : Theorical conductivity on normal trend (B)

CO: Observed value of tranconductivity (A) GS: Overburden gradient observed at observed depth

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Depth

Sonic or conductivity

A

XAXB

B


Exercise 17:

Calculate Pf gradient with Eaton’s method:

Depth 2900Overburden gradient 2.25Normal gradient 1.05Dc in A 1.3Dcn in B 1.7Pf Gradient from DcObserved shale resistivity in A (Ohm.m) 0.68Shale resistivity on normal trend in B (Ohm.m) 3.50Pf Gradient from Shale resistivityObserved t sonic in A (sec/ft) 100t sonic on normal trend in B (sec/ft) 80Pf Gradient from t sonic

4.1.4. DIRECT OBSERVATION OF DIFFERENTIAL PRESSURE

The following parameters can give a good indication of the differential pressure: Gas (cf pump off gas), Mud losses (indicates that differential pressure is too high), Kick (allows direct evaluation of Pf).

4.1.5. FORMATION TESTS

Give a direct evaluation of the Pf

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4.2. OVERBURDEN GRADIENT EVALUATION

The formula for overburden pressure is (cf 1.1.2)

S = H * b

10If we have an electric log for the formation density, we can use it to calculate the overburden:Divide the log in intervals of depth with similar density;

Then fill the following form to calculate the overburden at the end of each depth interval:

Exercise 18:

Interval bottom

(m)

Thickness (m)

Bulk density (kg/l)

Overburden pressure in the interval

(bar)

Total overburden

pressure (bar)

Overburden gradient

(bar/10m)

150 150 1.06 S=150*1.06/10 = 15.9

15.9 GS= 15.9*10/150

= 1.06400 250 1.70 S=250*1.70

/10 = 42.515.9 + 42.5 = 58.4

GS= 58.4*10/400

= 1.46

700 300 1.80

1070 370 1.89 182.3

1210 140 2.05 211.0

1400 190 2.02 249.4

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0 Formation density 3

100

200

300

400

500

600

Depth

Average density in the interval


Plotting the overburden gradient versus depth gives a curve:

The equation of the curve is:

S = a (Ln(Depth))2 + bLn(depth) + c

The coefficients a, b and c are regional characteristics

If no density log is available, a ”hard formation” or a “soft formation” set of coefficients a-b-c are used.

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Depth

S gradient (bar/10m)


4.3. FRACTURATION GRADIENT EVALUATION

4.3.1. APPLICATIONS

Determination of casing points

Determination of maximum mud weight

Computation of MAASP (Maximum Allowable Annulus Surface Pressure during a kick)

4.3.2. LEAK OFF TEST : L.O.T

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Pf MW FRAC (bar)

Depth

Previous casing shoe

Set casing shoe here when MW reaches FRAC gradient

Pump

in a well with closed BOP’s

until the pressure in the well reaches fracturation pressure of the formation


The recording of the pump pressure gives the following curve:

The Fracturation Pressure is:

PFRAC = PLOT + Mud Hydrostatic Pressure in the well

Exercise 19:

Calculate the FRAC (Fracturation gradient ) using the following data:

Casing shoe vertical depth ( = LOT depth) (m) 1500MW in hole (kg/l) 1.5LOT pressure (bar) 25Fracturation pressure (Bar)FRAC (bar/10m)

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LOT Pressure

Pumping | Stop pumping Time

PressureThe pressure increases until fracturation pressure is reached


From Terzaghi equation, we know that: S = Pf +

In the case of a fracturation, PFRAC = S3

As it is impossible to know the value of 3, we assume that 3 = 1 multiplied by a coefficient K (= Ratio of vertical to horizontal stress)As we know S1 (Overburden), we just need to calculate K to know S3 (and automatically PFRAC).

S3 PFRAC = Pf + K* Pf + K* S1 - Pf

And K = PFRAC - Pf

S1 - Pf

As we can calculate PFRAC every time we have a LOT, we can calculate K for that depth!

Plotting K versus depth gives a curve which equation is :

Ln(K) = aLn(depth) + b

As per overburden, if no LOT is available, a ”hard formation” or a “soft formation” set of coefficients a-b are used.

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S3

S1 (overburden)

S2

The fracture plan is perpendicular to the weakest stress S3


EXERCISES CORRECTIONS

Exercise 1:Multiply the bars by 14.4988 to get the psi.Divide the psi by 14.4988 to get the bars.

Exercise 2:Use the formula Ph = H * d to get the Ph in bars

10

Use the formula Ph = d * H * 0.052 to get the Ph in psi

Exercise 3:Weight of slug = slug volume * slug SGEquivalent volume of mud = Weight of slug / MWExtra volume = Equivalent volume of mud – Slug volume

Formula: Extra volume (metric) = Slug volume * (Slug SG 1) MW

Pipe internal volume is useless , unless you want to calculate directly from the Ph !!:

Void in pipes = Extra volume

Ph annulus Ph slug

Ph annulus = Ph slugHeight annulus = Ph slug * 10 / MWExtra volume height = Height annulus – Slug heightExtra volume = Extra volume height * Pipe internal volume

Exercise 4:Air gap in pipes is calculated by the formula (metric)

Air gap height = Slug height *(slug SG – MW)MW

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Exercise 5:Note that the Ph difference between A & B is the same than the Ph of the water column AB.Consequently, pressure in C is pressure in B minus the Ph of the column of hydrocarbon

Exercise 6:Use the following formula for both rigs:

MWequilibrium = Pf * 10 H

Exercise 7:

A. During a trip, the driller forgot to fill the hole and the mud level is lower than normal : calculate the Ph with the reduced column of mud:

Ph = (H - air gap) * MW10

Then calculate the Equivalent MW = Ph * 10 H

B. During a Leak Off Test : calculate the Ph and add the LOT pressure:

Ph = H * MW + LOT Pressure 10


C. You start circulating, calculate the Ph and add the pressure losses (P):

Ph = H * MW + P 10


Exercise 8:

For 1 m, the pressure increment is P = H * MW = 1 * MW = 0.1 * MW10 10

For 10m, we get G = MW

Exercise 9:

Change of bit type: if you run an unsuitable bit (ie a “hard bit” to drill soft shale), you may have to shift towards right.

Change of bit diameter: after a casing you will have to make a shift

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Drastic modification of the drilling parameters: the parameters should be optimised, Dexponent will not “correct” bad drilling parameters.

Bad hydraulics: the mud weight should not be too high. Geological unconformity: two different states of compaction are in contact. Deviated well: the WOB recorded is not the actual WOB at bottom (slack off).

Exercise 10:

Torque : The swelling of clay cause a decrease in hole diameter, accumulation of large cuttings or caving on the bit and stabilizers, all these problems are linked to negative differential pressure(MW too low).

Overpull and drag : for the same reason that causes the torque to go up. Hole filling : Caving may fill the hole during tripping. Pit level : in case of kick Differential flow : in case of kick Pump pressure : in case of kick, the annulus is filled with mud and light

fluid (ie gas), so the pressure losses in the annulus will be less than with a complete column of mud.

Exercise 11:Shale resistivity : The undercompacted shale contains more salted water , as

salted water has a good electrical conductivity and so the resistivity decreases !

Exercise 12:Mud Weight : An influx with salted water will make the mud density decrease.

Mud temperature : The formation temperature gradient will increase in an undercompacted zone.

Measuring mud temperature does not give a precise idea of the formation temperature as all actions at surface (new mud, water adding, mixing,

trips) will modify the mud temperature. Remember also that the mud has a cooling effect on the bit!

Mud resistivity : An influx with salted water has a good electrical conductivity and so the resistivity decreases !

Exercise 13:

Sonic: Resistivity: Gamma ray: sometimes , but interpretation doubtfulSpontaneous potentialConductivityCaliper may show a shrinkage of the well diameterDensity log: Neutron log:

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Exercise 14:

HA* DeqA = HB * DeqB + (GSA * HA – GSB * HB)10 10 10 10

HA* DeqA = HB * DeqB + (GSA * HA – GSB * HB)

DeqA = HB * DeqB + GSA * HA – GSB * HB

HA

DeqA = GSA * HA + HB * DeqB - GSB * HB

HA HA HA

DeqA = GSA + HB * DeqB - GSB * HB

HA HA

DeqA = GSA – HB * (GSB – DeqB) HA

Exercise 19:

PFRAC = PLOT + MW * H 10

FRAC = PFRAC * 10 H

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Documents

Over Pressures