OVERPRESSURES.pdf

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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 Pascal It 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..

1 Bars = 14.50 psi

155 Psi = 10.69 bars

1.1.1. HYDROSTATIC PRESSURE : Ph

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)

Using well site units, the formula becomes : Ph = H*d 10

OVERPRESSURES : AN INITIATION


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NB : the term 10 is approached, for precision, you should use 10.2 with pressure in bars and 9.6 for pressure in kg/cm2

In API, the formula is: Ph = 0.052 * H * d 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 !

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

H

{


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Exercise 2:

Calculate Ph in the following examples

Height of fluid (m):

1000 Fluid SG (kg/l):

1.5 Ph bar:

150.00

Height of fluid (ft):

1000 Fluid SG (ppg):

10 Ph psi:

520.00

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):

962.96

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)

179.82


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1.1.2. OVERBURDEN: S

S = H * rb

10 With S = Overburden stress (kg/cm2) rb = Formation average bulk density (no unit)

H = Vertical thickness of overlying sediments (m) In API, the formula is: S = H * rb * 0.433

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

rb = (f * rf) + (1-f) * rm

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

Relationship between porosity and depth

Porosity

Depth

Sand

Shale


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rb

Depth

2.31 rb

Depth

2.31

Sea botttom

Relationship between bulk density and depth

On shore Off shore

1.1.3. FORMATION PRESSURE: Pf

Also called Pore pressure :Pp

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

1.1.3.1. NORMAL Pf: 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 Pf ANOMALY: Pf


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10 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

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

(kg/cm2) 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) 1.05 Hydrocarbon SG (kg/l) 0.25 Point A & C depth (m) 2000 Point B depth (Hydrocarbon/water contact)(m) 2590 Calculate Pf in A (kg/cm2) 210.00

A C

B


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Calculate Pf in B (kg/cm2) 271.95 Calculate Ph of the column of water AB (kg/cm2)

61.95

Calculate Ph of the hydrocarbon column (kg/cm2)

14.75

Calculate the Pf in C (kg/cm2) 257.20 Pressure anomaly at top of hydrocarbon (kg/cm2)

47.20

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. MW = P * 10 H

Exercise 6: Two rigs are drilling a well in the same overpressured formation, calculate the equilibrium MW for each rig:


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Distance from A to formation (m) 2000 Distance from B to formation (m) 1200 Pf (kg/cm2) 200 Equilibrium MW for rig A 1.00 Equilibrium MW for rig B 1.67

In a well, the mud weight can be different than the equivalent MW. The following exercise will show you some examples:

Exercise 7:

Well total depth (m) 2000 Mud weight in hole (kg/l) 1.20 Calculate 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)

100

1.14

During a Leak Off Test, the pressure reached (kg/cm2) :

10

1.25

You start circulating, the pressure losses in the annulus is (kg/cm2):

1

1.21

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 > MW Mud losses Leak Off Test Swabbing 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) 1.20 Calculate the hydrostatic pressure gradient (kg/cm2/10m):

1.20

As the figures are similar, MW and Pressure gradient may be plotted on the pressure gradient, Fracturation gradient and Overburden gradient.

Pressure kg/cm2

DP

DH

Depth m

G = DP in kg/cm2/10m DH


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Pressure (kg/cm2)

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 or kg/cm2)

Z = Subsea depth (m)

Sea level

H

P*10 d

Z


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1.2.4. PLOT OF PRESSURE VS DEPTH

May be used with an overlay of pressure gradient

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.

1.03 1.20 1.30

Pf & Ph (due to MW) in kgf/cm2

Depth (m)

S3

S1 (overburden)

S2


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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 + s (Terzaghi equation) With S = Total stress (kgf/cm2)

Pp = Pore pressure ( or formation pressure) (kgf/cm2) s = Effective stress (on the grains of the rock) (kgf/cm2)

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 shales. Water elimination from shales 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)

Terzaghi experiment

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

S S S

A C B

Pf

s


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

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

Pf = Ph S = s

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 of different concentrations, until the concentration of each solutions 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 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) D CaSO4 (Anhydrite) + 2H2O 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 .

Clay layer

Pure water

Salt water

Osmotic flow Pf decreases

Pf increases


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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, thermochemical 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.

2.8. MISCELLANEOUS

2.8.1. CARBONATE COMPACTION Normally, carbonates do not have problems of undercompaction. 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.

Phytoplankton with its shield of coccoliths

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


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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.

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 is the normal compaction trend.

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.

Impervious barrier

Hydrostatic pressure

Overpressure

Depth

Pressure


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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.


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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 (DP). 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 DP 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.

WOB

Bit tooth

WOB

Crushed rock


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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

In order to include the effect of differential pressure, a correction has been made: Dc = d * Normal hydrostatic gradient MW With: Dc : corrected d exponent d : d exponent Normal hydrostatic gradient (from 1 to 1.08) MW: mud weight in hole 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 curve C: 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.

dcn dc

Pf gradient

Normal gradient (ie 1.03) Depth

A B C

dcn

dc

Shale point Sand point


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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

dcn dc

Pf gradient & MW

Depth

dcn dc

Pf gradient & MW

Depth

Kick


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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

Exercise 10: (Answer is at the end) Decide if the following parameters increase or decrease when an overpressure by undercompaction occurs. And why !

3.4.1.4. TORQUE

or ?

Shale

Sandy Shale passing to Shaly Sand

Sand

Dc

Dcn in shale

Dcn in sand


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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.

Exercise 11: (Answer is at the end) 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).

Time

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

A B C


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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.

Exercise 12: (Answer is at the end) 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:

Depth

Shale density

Top of undercompaction


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3.5. METHODS AFTER DRILLING

Exercise 13: (Answer is at the end) 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. Sonic 2. Resistivity 3. Gamma ray 4. Spontaneous potential 5. Conductivity 6. Caliper 7. Density log 8. Neutron log


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4. QUANTITATIVE PRESSURE EVALUATION

4.1. FORMATION PRESSURE EVALUATION Most methods to calculate Pf will compare the undercompacted shales 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 = s + Pf) . We know that we have the same compaction in A and B, so the stress s must be the same in both points ! sA = sB We can write sB = SB PfB and sA = SA PfA As sA = sB we have : SB PfB = SA PfA Or PfA = PfB + (SA - SB)

Depth

A

B

Dc or Shale density or porosity log, etc

HB

HA


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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: 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 depth ZA : Depth of undercompacted shale GSA: Overburden gradient at A

GSB: 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 > 1.4

Exercise 15: Calculate Pf gradient with the equivalent depth method:

Mud weight in hole (kg/l) 1.30 HA 2900 HB 1900 Normal gradient 1.08 Dc in A 1.3 Overburden gradient 2.31 Pf Gradient 1.50

Depth

A

B

Dc

HB

HA


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4.1.2. RATIO METHOD

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 trend XA : 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 < 1.4

Exercise 16:

Calculate Pf gradient with the ratio method:

Depth 2900 Normal gradient 1.08 Dc in A 1.3 Dcn in B 1.7 Pf Gradient 1.41

Depth

Dc or Shale density or porosity log, etc

A

XA XB

B


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4.1.3. EATONS 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

Depth

Dc or Shale resistivity

A

XA XB

B


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

DeqA = GS (GS DeqN) * DtN 3.0 DtO

With DeqN : Equilibrium density in B (= normal hydrostatic gradient) DtN : Theorical transit time on normal trend (B) DtO : 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

Depth

Sonic or conductivity

A

XA XB

B


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Exercise 17:

Calculate Pf gradient with Eatons method:

Depth 2900 Overburden gradient 2.25 Normal gradient 1.05 Dc in A 1.3 Dcn in B 1.7 Pf Gradient from Dc 1.38 Observed shale resistivity in A (Ohm.m) 0.68 Shale resistivity on normal trend in B (Ohm.m) 3.50 Pf Gradient from Shale resistivity 2.08 Observed Dt sonic in A (msec/ft) 100 Dt sonic on normal trend in B (msec/ft) 80 Pf Gradient from Dt sonic 1.64

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 * rb 10

If 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 (kg/cm2)

Total overburden

pressure (kg/cm2)

Overburden gradient

(kg/cm2/10m)

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

15.9 GS= 15.9*10/150

= 1.06 400 250 1.70 S=250*1.70

/10 = 42.5 15.9 + 42.5 = 58.4

GS= 58.4*10/400

= 1.46

700 300 1.80 54.0 112.4 1.61

1070 370 1.89 69.9 182.3 1.70

1210 140 2.05 28.7 211.0 1.74

1400 190 2.02 38.4 249.4 1.78

0 Formation density 3

100

200

300

400

500

600

Depth

Average density in the interval


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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.

Depth

S gradient (kg/cm2/10m)


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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

Pf MW FRAC (kg/cm2)

Depth

Previous casing shoe

Set casing shoe here when MW reaches FRAC gradient

Pump in a well with closed BOPs until the pressure in the well reaches fracturation pressure of the formation


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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) 1500 MW in hole (kg/l) 1.5 LOT pressure (kg/cm2) 25 Fracturation pressure (Kg/cm2) 250.00 FRAC (kg/cm2/10m) 1.67

LOT Pressure

Pumping | Stop pumping Time

Pressure The pressure increases until fracturation pressure is reached


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From Terzaghi equation, we know that: S = Pf + s In the case of a fracturation, PFRAC = S3

As it is impossible to know the value of S3, we assume that S3 = S1 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* s1 = 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.

S3

S1 (overburden)

S2

The fracture plan is perpendicular to the weakest stress S3


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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 SG Equivalent volume of mud = Weight of slug / MW Extra 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 slug Height annulus = Ph slug * 10 / MW Extra volume height = Height annulus Slug height Extra 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) * MW 10 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 Then calculate the Equivalent MW = Ph * 10 H

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

Ph = H * MW + DP 10 Then calculate the Equivalent MW = Ph * 10 H

Exercise 8:

For 1 m, the pressure increment is DP = DH * MW = 1 * MW = 0.1 * MW 10 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 doubtful Spontaneous potential Conductivity Caliper may show a shrinkage of the well diameter Density 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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