625 Lesson 2A

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AT M A T M

Lesson 2A

Kicks & Gas Migration

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AT M A T MHarold Vance Department of

Petroleum Engineering


Density of real gasses Equivalent Mud Weight (EMW) Wellbore pressure before and after kick Gas migration rate - first order approx.

Gas migration rate - w/mud compressibility

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Read Ch 1 & 2 Watson Ch 8 Schubert Homework: 2.1-2.10

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Density of Real Gasses

M = molecular weight m = mass

n = no. of moles gg = S.G. of gas

ZRT

p

M M

M

ZRT pM

V M

ZRT pV

ZRT pV

n

V nM

V m

g g

air g

g

g

g

g

29

29

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What is the density of a 0.6 gravity gas at10,000 psig and 200 degF?

From Lesson 2, Fig. 1 p pr = p/p pc = 10,015/671 = 14.93 T

pr = (200+460)/358 = 1.84

Z = 1.413

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g = (29*0.6*10,015)/(1.413*80.28*660)

g = 2.33 ppg

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Equivalent Mud Weight, EMW

The pressure, p (psig) in a wellbore, at adepth of x (ft) can always be expressed interms of an equivalent mud density orweight.

EMW = p/(0.052*x) in ppg

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Depth Pressureft psia

0 500 010 505.98 10

100 559.8 100150 589.7 150200 619.6 200300 679.4 300

400 739.2 400500 799 5001000 1098 10002000 1696 20003000 2294 30004000 2892 40005000 3490 50006000 4088 60007000 4686 70008000 5284 80009000 5882 9000

10000 6480 1000011000 7078 11000

11600 7436.8 11600

0

2000

4000

6000

8000

10000

12000

0 2000 4000 6000 8000

After Kick

Before Kick

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Gas Migration

Gas generally has a much lower density than thedrilling 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 (ignoringtemp, fluid loss to formation, compressibility ofgas, mud, and formation)

This causes pressures everywhere in the wellboreto increase.

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Gas Migration

Problem 2.6: A 0.7 gravity gas bubbleenters the bottom of a 9000 ft vertical well

when the drill collars are being pulledthrough the rotary table. Flow is noted andthe well is shut in with an initial recorded

casing pressure of 50 psig. Influx height is350 ft, Mud weight =9.6 ppg.

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Gas Migration Assume surface temperature of 70 deg F.

Temp gradient = 1.1 deg F/100 ft. Surface pressure =14 psia

Determine the final casing pressure if thegas bubble is allowed to reach the surfacewithout expanding

Determine the pressure and equivalentdensity at total depth under these finalconditions

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Solution

First assumption - is that BHP is brought tothe surface

Pressure at the top of the bubble =14 + 50+.052*9.6*(9000-350) = 4378 psia T9000 = 70 + (1.1/100)*(9000 - 350) = 629 deg R

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Solution

p pc = 666 psia T pc = 389 deg R p pr = 4378/666 = 11.08 T pr = 629/389 = 1.62

Z = 0.925

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Solution

Assume, at first, that Z f = 1.0 (at thesurface)

Then, 4378 * V = p f * V________0.925 * 629 1.0 * (70 + 460)

so, p f = 3988 psia

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Solution

At surface: P pr = 3988 / 666 = 6.00

T pr = 530 / 389 = 1.36 Zf = 0.817

Pf = 3258 psia

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AT M A T MHarold Vance Department ofPetroleum Engineering

Solution

A few more iterative steps result in Zf = 0.705 and p f = 2812 psia

At the surface f = 0.7*2812/(2.77*0.705*530) = 1.9 ppg

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Solution

BHP =2812+0.052*1.9*350+.052*9.6*9650

BHP = 7160 psia

EMW = (7160 - 14)/(0.052 * 9000) EMW = 15.3 ppg

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Compression of Mud in Annulus

DV = compressibility * volume * D p = -6 * 10 -6 (1/psi) * 0.1(9000-350)*2626

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 Boyles Law, p2 * 48.63 = 2815 * 35

p2 = 2024 psia p8650 poA PoB PoCConsider: 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 1000

psig. One hour later the pressure is 2000 psig.

Ann Cap = 0.1 bbl/ft

MW = 14 ppg TD = 10,000

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Gas Migration Rate

How fast is the kick migrating?

What assumptions do we need to make?

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1 hr

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First Attempt

If the kick rises x ft. in 1 hr and the pressurein the kick = constant, then the pressure

increases everywhere, D p = 0.052*14*x

x = (2000-1000)/(0.052*14) x = 1374 ft Rise velocity = 1374 ft/hr

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Gas Migration Rate Field rule of thumb ~ 1000 ft/hr Laboratory studies ~ 2000 - 6000 ft/hr

Who is right?

Field results?

Is the previous calculation correct?

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Second Attempt

Consider mud compressibility Ann cap = 0.1 bbl/ft * 10000 ft= 1000 bbl of mud Volume change due to compressibility and

increase in pressure of 1000 psi, DV = 6*10 -6 (1/psi) * 1000 psi * 1000 bbl

= 6 bbl

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Second Attempt

i.e. gas could expand by 6 bbl, to 36 bbl Initial kick pressure=1000 + 0.052 * 14 * 10000= 8280 psig= 8295 psia

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Second Attempt

How far did it migrate in 1 hour? The pressure reduction in kick fluid=8260-6621=1659 psi The kick must therefore have risen an

additional x 2 ft, biven by:1659 = 0.052 * 14 * x 2 x2 = 2279 ft

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Second Attempt

2nd estimate = 1374 + 22793653 ft/hr What if the kick size is only 12 bbl? What about balooning of the wellbore? What about fluid loss to permeable

formations?

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Solution

hr f t v

v

hr t t f t

psi g

psi p pv

slip

slip

slip

/1154

5.00.10052.0

500800

12

12

Ignoringcompressibility andother effects

What factors affect gasslip velocity, ormigration rate?

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Gas slip velocity

The bubble size, and the size of the gas voidfraction, will influence bubble slip velocity.

The void fraction is defined as the ratio(or percentage) of the gas cross-sectionalarea to the total flow area.

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Gas slip velocity

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Gas slip velocity

Bubbles with a voidfraction > 25% assumea bullet nose shapeand migrate upwardsalong the high side ofthe wellboreconcurrent with liquid

backflow, on theopposite side of thewellbore

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625 Lesson 2A