OVERPRESSURES.pdf

O O V E R P R E S S U R E S OV VE ER RP PR RE ES SS SU UR RE ES S

OVERPRESSURES : AN INITIATION

1. PRESSURE CONCEPTS 1.1. DEFINITIONS

A pressure is a force divided by the surface where this force applies. where 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 5 1 atm = 1,013 *10 Pascal 2

A practical unit on the rig is the kgf/cm 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….. 1 155

Bars = Psi =

14.50 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 Vertical height height of fluid (m) g = gravity acceleration (9.81 m/s2) Using well site units, the formula becomes : Ph = H*d 10 J C 11 1/ 2 J CD D 1 /2 20 00 02

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NB : the term 10 is approached, for precision, you you should use 10.2 with 2 pressure in bars and 9.6 for pressure in kg/cm In API, the formula is: Ph = 0.052 * H * d

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

H

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 J C 11 1/ 2 J CD D 1 /2 20 00 02

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NB : the term 10 is approached, for precision, you you should use 10.2 with 2 pressure in bars and 9.6 for pressure in kg/cm In API, the formula is: Ph = 0.052 * H * d

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

H

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 J C 11 1/ 2 J CD D 1 /2 20 00 02

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Exercise 2:

Calculate Ph in the following examples Height of fluid (m): Height of fluid (ft):

1000 1000

Fluid SG (kg/l): Fluid SG (ppg):

1.5 10

Ph bar: Ph psi:

150.00

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) : Mud Weight (kg/l):

9.1 1.08

Slug volume (l): Extra volume in pit (l):

2000

Slug SG (kg/l):

1.6

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) : Mud Weight (kg/l):

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9.1 1.10

Slug volume (l): Height of air gap in pipes (m)

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3000 179.82

Slug SG (kg/l):

1.7


1.1.2. OVERBURDEN: S S = H * ρb 10

With

S = Overburden stress (kg/cm2) ρ b = Formation average bulk density (no unit) H = Vertical thickness thickness of overlying sediments (m)

In API, the formula is: S=H*

ρb * 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.

ρb

= (φ * ρf ) + (1-φ) * ρm

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

Sand

Shale Depth

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Relationship between bulk density and depth On shore

2.31

Off shore

ρ b

2.31

ρ b

Sea botttom

Depth

Depth

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

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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:

C A B

Water SG (kg/l) Hydrocarbon SG (kg/l) Point A & C depth (m) Point B depth (Hydrocarbon/water contact)(m) 2 Calculate Pf in A (kg/cm ) J C 11 1/ 2 J CD D 1 /2 20 00 02

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1.05 0.25 2000 2590 210.00


2

Calculate Pf in B (kg/cm ) Calculate Ph of the column of water AB 2 (kg/cm ) Calculate Ph of the hydrocarbon column 2 (kg/cm ) 2 Calculate the Pf in C (kg/cm ) Pressure anomaly at top of hydrocarbon 2 (kg/cm )

271.95 61.95 14.75 257.20 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) Distance from B to formation (m) Pf (kg/cm2) Equilibrium MW for rig A Equilibrium MW for rig B

2000 1200 200 1.00 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) Mud weight in hole (kg/l) 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) During a Leak Off Test, the pressure reached (kg/cm2) : You start circulating, the pressure losses in the annulus is (kg/cm2):

2000 1.20

100

1.14

10

1.25

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 Mud losses Swabbing Fluid influx in hole (Kick)

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Equivalent MW > MW Leak Off Test Circulating (ECD) Surge


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

∆P Pressure 2 kg/cm

G = ∆P in kg/cm2/10m ∆H

∆H

Depth m

Exercise 8:

Mud weight in hole (kg/l) Calculate the hydrostatic pressure gradient 2 (kg/cm /10m):

1.20 1.20

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

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Pressure

(kg/cm2)

Depth (m)

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GPf MW FRAC S


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) 2) P= Formation pressure at depth Z (bar or kg/cm Z = Subsea depth (m)

P*10 d

H

Z

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Sea level

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1.2.4. PLOT OF PRESSURE VS DEPTH

May be used with an overlay of pressure gradient

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

Depth (m)

1.03

1.20

1.30

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.

S1 (overburden)

S2

S3 J C 11 1/ 2 J CD D 1 /2 20 00 02

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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 +

σ

(Terzaghi equation)

With S = Total stress (kgf/cm2) Pp = Pore pressure ( or formation pressure) (kgf/cm2) σ = 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

Pf

σ J C 11 1/ 2 A J CD D 1 /2 20 00 02

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C


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

Pf = Ph S=σ

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.

Pf decreases

Osmotic flow

Pure water Clay layer

Pf increases

Salt water

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.

in overpressure generation, specially if interlayed M ajor role with shale.

2.5.2. SULFATE DIAGENESIS

Gypsum is the precipitated form of CaSO 4, 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 . J C 11 1/ 2 J CD D 1 /2 20 00 02

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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 .

in overpressure generation in confined series of I mportant r ole shaly sands or carbonate. 2.7. TECTONICS 2.7.1. R ELIEF & STRUCTURING

As already seen in chapter 1.1.3, relief can be the cause of pressure anomalies (ie: artesian well). 2.7.2. R EORGANISATION OF STRESS FIELD

Sediments are subjected to overburden and to horizontal tec tonic 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

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


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. Pressure

Hydrostatic pressure

Overpressure Impervious barrier Depth

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3.3. PREDICTION

By prediction, we intend methods before drilling… 3.3.1. R EGIONAL 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. R EAL TIME METHODS 3.4.1.1. R ATE 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:

WOB WOB Bit tooth

Impact

Crushed rock

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.

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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. Shale point

Sand point

dc dcn A

B

C

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. Pf gradient

Depth

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dc

dcn

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Normal gradient (ie 1.03)


Pf gradient & MW

Depth

dc

dcn

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

Pf gradient & MW

Kick

Depth

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dc

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dcn


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.

Dc

Shale Sandy Shale passing to Shaly Sand

Dcn in shale

Sand Dcn in sand

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

æ or è J C 11 1/ 2 J CD D 1 /2 20 00 02

? -24-


3.4.1.5. OVERPULL & DRAG

æ or è

?

3.4.1.6. HOLE FILL

æ or è

?

3.4.1.7. PIT LEVEL – DIFFERENTIAL FLOW – STAND PIPE PRESSURE

æ

or è ?

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:


In case of undercompaction, shale resistivity æ or è

?

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

B

A

C

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

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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:


Decide if the following parameters increase or decrease when an overpressure by undercompaction occurs. And why ! 3.4.2.2. MUD WEIGHT

æ

or è ?

3.4.2.3. MUD TEMPERATURE

æ

or è ?

3.4.2.4. MUD RESISTIVITY

æ or è

?

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: Shale density

Top of undercompaction Depth J C 11 1/ 2 J CD D 1 /2 20 00 02

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3.5. METHODS AFTER DRILLING Exercise 13:


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. 2. 3. 4. 5. 6. 7. 8.

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Sonic Resistivity Gamma ray Spontaneous potential Conductivity Caliper Density log 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

Dc or Shale density or porosity log, etc

B

HB

HA

A

Depth

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 = σ + Pf) . We know that we have the same compaction in A and B, so the stress σ must be the same in both points ! σA = σB We can write σB = SB – Pf B

and σA = S A – Pf A

As σA = σB we have : SB – Pf B = SA – Pf A Or Pf A = Pf B + (SA - S B)

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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: Dc

B

HB

HA

A

Depth

Mud weight in hole (kg/l) HA HB Normal gradient Dc in A Overburden gradient Pf Gradient J C 11 1/ 2 J CD D 1 /2 20 00 02

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1.30 2900 1900 1.08 1.3 2.31 1.50


4.1.2. R ATIO METHOD

Dc or Shale density or porosity log, etc XA

XB

A

B

Depth

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 Normal gradient Dc in A Dcn in B Pf Gradient

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2900 1.08 1.3 1.7 1.41

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4.1.3. EATON’S METHOD

Dc or Shale resistivity XA

A

XB

B

Depth

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

Shale resistivity: DeqA = GS – (GS – DeqN) *  RshO   RshN 

1.2

With DeqN : Equilibrium density in B (= normal hydrostatic gradient) Rsh N : 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) *  DcO   DcN  1.2

With DeqN : Equilibrium density in B (= normal hydrostatic gradient) Dc N : Theorical Dc on normal trend (B) DcO : Observed value of Dc (A) GS: Overburden gradient observed at observed depth

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

Sonic or conductivity XB

B

XA

A

Depth

Sonic: DeqA = GS – (GS – DeqN) *

 ∆tN 3.0  ∆tO 

With DeqN : Equilibrium density in B (= normal hydrostatic gradient) ∆t N : 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   CO 

1.2

With DeqN : Equilibrium density in B (= normal hydrostatic gradient) C N : Theorical conductivity on normal trend (B) CO: Observed value of tranconductivity (A) GS: Overburden gradient observed at observed depth

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Exercise 17:

Calculate Pf gradient with Eaton’s method: Depth Overburden gradient Normal gradient Dc in A Dcn in B Pf Gradient from Dc Observed shale resistivity in A (Ohm.m) Shale resistivity on normal trend in B (Ohm.m) Pf Gradient from Shale resistivity Observed ∆t sonic in A (µsec/ft) ∆t sonic on normal trend in B (µsec/ft) Pf Gradient from ∆t sonic

2900 2.25 1.05 1.3 1.7 1.38

0.68 3.50 2.08

100 80 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 * ρb 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;

Depth

0

Formation density

100

3

Average density in the interval

200 300 400 500

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)

150

150

1.06

400

250

1.70

Overburden pressure in the interval (kg/cm2) S=150*1.06 /10 = 15.9

Total overburden pressure (kg/cm2) 15.9

S=250*1.70 /10 = 42.5

15.9 + 42.5 = 58.4

Overburden gradient (kg/cm2/10m) GS= 15.9*10/150 = 1.06 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

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Plotting the overburden gradient versus depth gives a curve:

S gradient 2 (kg/cm /10m)

Depth

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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4.3. FRACTURATION GRADIENT EVALUATION 4.3.1. APPLICATIONS §

Determination of casing points

Pf MW FRAC (kg/cm )

Previous casing shoe

Set casing shoe here when MW reaches FRAC gradient

Depth

§

§

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

Pump

in a well with closed BOP’s

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:

Pressure

The pressure increases until fracturation pressure is reached

LOT Pressure

Pumping

|

Stop pumping

Time

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) MW in hole (kg/l) LOT pressure (kg/cm2) 2 Fracturation pressure (Kg/cm ) 2 FRAC (kg/cm /10m)

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1500 1.5 25 250.00 1.67


S1 (overburden)

The fracture plan is perpendicular to the weakest stress S3 S2

S3

From Terzaghi equation, we know that: S = Pf + σ 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 S 3 (and automatically PFRAC). S3 = PFRAC = P f + K* σ1

=

Pf + K* (S1 - Pf )

And K = PFRAC - P f S1 - P f 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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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 (∆P): Ph = H * MW + ∆P 10 Then calculate the Equivalent MW = Ph * 10 H Exercise 8:

For 1 m, the pressure increment is ∆P = ∆H * 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 æ : 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 æ : for the same reason that causes the torque to go up. Hole filling æ : Caving may fill the hole during tripping. Pit level æ : in case of kick Differential flow æ : in case of kick Pump pressure è : 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 è : The undercompacted shale contains more salted water , as salted water has a good electrical conductivity and so the resistivity decreases !

Exercise 12:

Mud Weight è : An influx with salted water will make the mud density decrease. Mud temperature æ : 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 è : An influx with salted water has a good electrical conductivity and so the resistivity decreases ! Exercise 13:

Sonic: á Resistivity: á Gamma ray: sometimes á, but interpretation doubtful Spontaneous potential Conductivityâ Caliper may show a shrinkage of the well diameter Density log: á Neutron log: á J C 11 1/ 2 J CD D 1 /2 20 00 02

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OVERPRESSURES.pdf

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