part 3 kicks and gas migration

8/19/2019 Part 3 Kicks and Gas Migration

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Contents

• Density of real gases

• Equivalent Mud Weight (EMW)

• Wellbore pressure before and after kick

• Gas migration rate - first order approx.

• Gas migration rate – with temperature,mud compressibility and Z-factor

considerations

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Density of Real Gases

M = molecular

weight

m = mass

n = no. of moles

gg = S.G. of gas

3

ZRT

pVn

V

nM

V

mg

29

M

M

M

ZRT

pM

V

M

ZRT

pV

air

g

g

ZRT

pgg

g

29

(Real Gas Law)


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• What is the density of a 0.6 gravity gas at

10,000 psig and 200oF?

• From Lesson 2, following figure

• ppr = p/ppc = 10,015/671 = 14.93

• Tpr = (200+460)/358 = 1.84

• Z = 1.413

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5

1.413

14.93

1.84


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g = 2.33 ppg

TRZ

p29 gg

g

6

p = 10,000 psig

T = 200oF

gg = 0.6

{660)28.80(413.1

015,10)6.0(29g


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Equivalent Mud Weight, EMW

• The pressure, p (psig) in a wellbore, at a

depth of x (ft) can always be expressed interms of an equivalent mud density or

weight.

• EMW = p / (0.052 * x) in ppg

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EMW

• EMW is the density of the mudthat, in a column of height, x (ft)will generate the pressure, p

(psig) at the bottom, if thepressure at top = 0 psig

or, at TD:

• p = 0.052 * EMW * TVD

8

0po=0

TVDp

x


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9

0

2,000

4,000

6,000

8,000

10,000

12,000

0.0 10.0 20.0 30.0 40.0 50.0

EMW, ppg

D e p

t h ,

f t

Depth

pEMW

*052.0


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10

0

2,000

4,000

6,000

8,000

10,000

12,000

0 1,000 2,000 3,000 4,000 5,000 6,000 7,000 8,000

Annulus Pressure, psig

D e p

t h ,

f t After Kick

Before Kick

SICP = 500 psig


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Gas Migration

• Gas generally has a much lower density than the

drilling mud in the well, causing the gas to rise

when the well is shut in.

• Since the gas, cannot expand in a closed wellbore,

it will maintain its pressure as it rises (ignoring

temp, fluid loss to formation, compressibility of

gas, mud, and formation)• This causes pressures everywhere in the wellbore

to increase.

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12

P1

P2

P3

P1 = P2 = P3 ???

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Gas Migration

Example 1: A 0.7 gravity gas bubble enters the bottom of a

9,000 ft vertical well when the drill collars are being pulled

through the rotary table.

Flow is noted and the well is shut in with an initial recordedcasing pressure of 50 psig. Influx height is 350 ft. Mud weight

= 9.6 ppg. Assume surface temperature of 70oF. Temp

gradient = 1.1oF/100 ft. Surface pressure = 14 psia

• Determine the final casing pressure if the gas bubble isallowed to reach the surface without expanding

• Determine the pressure and equivalent density at total

depth under these final conditions

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Gas Properties at Bottom

• First assumption: BHP is brought to the surface

• Pressure at the top of the bubble

• P8,650 = 14 + 50 + 0.052 * 9.6 * (9,000-350)= 4,378 psia

• T9,000 = 70 + (1.1/100) * 9,000 + 460

= 629oR

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Gas Properties at Bottom

• ppc = 666 psia

• Tpc = 389 deg R

• ppr = 4,378/666 = 6.57

• Tpr = 629/389 = 1.62

• Z = 0.925

15 pseudocritical - pseudoreduced


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Bottomhole Pressure

• g = 29*0.7*4,378 / (0.925 * 80.28 * 629)

= 1.90 ppg

• DpKICK = 0.052 * 1.9 * 350 = 35 psi

• BHP = 4,378 + 35

BHP = 4,413 psia (~surface press.?

ZRTpg

gg

29

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Pressure at Surface

• Assume, at first, that Zf = 1.0 (at the surface)

• Then,

46070*nR*0.1Vp

629*nR*925.0V378,4

ZnRTpV o

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so, po = 3,988 psia (with Temp. corr.)

46070*0.1

p

629*925.0

378,4 o

BOTTOM SURFACE


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Solution with Z-factor Corr.

• At surface:

–ppr = 3,988 / 666 = 6.00

–Tpr = 530 / 389 = 1.36

– Zf = 0.817

p0 = 3,258 psia

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Solution with Z-factor

• A few more iterative steps result in

–Z0 = 0.705 and p0 = 2,812 psia

• At the surface

• f = 29*0.7*2,812 / (0.705*80.28*530)

= 1.9 ppg

ZRT

pgg

g

29

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New BHP & EMW

• New BHP = 2,812 + 0.052 * 1.9 * 350+ 0.052 * 9.6 * 8,650

• New BHP = 7,165 psia

• EMW = (7,165 - 14)/(0.052 * 9,000)

• EMW = 15.3 ppg

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21

1. 4,413 psia2. 4,378

3. 3,988 (T)

4. 3,258 (Z)

5. 2,812 (Z)

6. 2,024 (mud)Mud9.6 ppg

Gas

0.1 bbl/ft

7,004 psia

7,179 psia

7,378 psia

4,413 psia

629 R

8,650’

9,000’

64 psia

530 R2,812 psia

ΔPmud=2,626 psi


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Compression of Mud in Annulus vA = 0.1

bbl/ft)

• DV = compressibility * volume * Dp

• = -6 * 10-6 (1/psi) * 0.1(9,000-350)*2,626

• DV = -13.63 bbls

• Initial kick volume = 0.1 * 350 = 35 bbls

• New kick volume = 35 + 13.63

= 48.63 bbl

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Compression of Mud in Annulus

• From Boyle’s Law, pV = const

– p2 * 48.63 = 2,812 * 35

– p2 = 2,024 psia

p8650 poA poB poC

Consider: V,p,Z const. p,Z change mud comp.

2nd iteration ? ……………. 3rd

or, Is there a better way?

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Gas Migration Rate

• A well is shut in after taking a 30 bbl kick.

The SIDPP appears to stabilize at 1,000 psig.

One hour later the pressure is 2,000 psig.

Hole Cap = 0.1 bbl/ft

MW = 14 ppg

TVD = 10,000 ft

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Gas Migration Rate

• How fast is the kick migrating?

• What assumptions do we need to make

to analyze this question?

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26

P1

P2

P3

12 3

1 Hr


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Gas Migration Rate

• Field rule of thumb ~ 1,000 ft/hr

• Laboratory studies ~ 2,000 – 6,000 ft/hr

• Who is right?

• Field results?

• Is the previous calculation correct?

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Second Attempt

• Consider mud compressibility

• Hole capacity = 0.1 bbl/ft * 10,000 ft

= 1,000 bbl of mud

• Volume change due to compressibility and

increase in pressure of 1,000 psi,

• DV = 6*10-6 (1/psi) * 1,000 psi * 1,000 bbl

= 6 bbl

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Second Attempt

• i.e. gas could expand by 6 bbl, to 36 bbl

• Initial kick pressure

=1,000 + 0.052 * 14 * 10,000 (approx.)

= 8,280 psig

= 8,295 psia

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Second Attempt

• A 20% expansion would reduce the

pressure in the kick to ~ 0.8*8,295

= 6,636 psia

= 6,621 psig

•So, the kick must have migrated more than1,374 ft!

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Second Attempt

• How far did it migrate in 1 hour?

• The pressure reduction in kick fluid

= 8,260 - 6,621=1,659 psi

• The kick must therefore have risen an

additional x2 ft, given by:

1,659 = 0.052 * 14 * x2 x2 = 2,279 ft

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Second Attempt

• 2nd estimate = 1,374 + 2,279

= 3,653 ft/hr

• What if the kick size is only 12 bbl?

• What about balooning of the wellbore?

• What about fluid loss to permeable

formations? T? Z?...

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Bore Hole Ballooning

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Example

• Kick occurs. After shut-in, initial csg.

Press = 500 psig. 30 minutes later,

p = 800 psig

• What is the slip velocity if the kick volume

remains constant?

• MW = 10.0 ppg

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Simple Solution

36

hr ttftpsi

g

psippv

12

12slip

Ignoring

temperature,

compressibility and

other effects.

5.00.10052.0

500800vslip

hr /ft154,1vslip

What factors affect

gas slip velocity, or

migration rate?


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Gas slip velocity

• The bubble size, and the size of the gas void fraction,

will influence bubble slip velocity.

• The “void fraction” is defined as the ratio (or

percentage) of the gas cross-sectional area to the

total flow area.

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Gas Slip Velocity

Bubbles with a voidfraction > 25% assume a

bullet nose shape and

migrate upwards along

the high side of the

wellbore concurrent

with liquid backflow, on

the opposite side of the

wellbore

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Gas slip velocity

•Large bubbles rise faster than smallbubbles

• Other factors:

– Density differences

– Hole geometry

– Mud viscosity

– Circulation rate

– Hole inclination

• One lab study showed max. rate at 45o.

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End

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part 3 kicks and gas migration

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