3. kicks and gas migration.ppt

8/14/2019 3. Kicks and Gas Migration.ppt

1/42

PETE 625Well Control

Lesson 3

Kicks and Gas Migration


2/42


3/42

Assignments

Homework #2:

Ch 1, Problems 1.11-1.21

Read: All of Chapter 1


4/42

4

Density of Real Gases

M = molecular

weight

m = mass

n = no. of moles

gg= S.G. of gas

ZRT

pVn

V

nM

V

mg

29

M

M

M

ZRT

pM

V

M

ZRT

pV

air

g

g

ZRT

pgg

g29

(Real Gas Law)


5/42

5


What is the density of a 0.6 gravity gas at

10,000 psig and 200oF?

From Lesson 2, Fig. 1

ppr= p/ppc= 10,015/671 = 14.93

Tpr= (200+460)/358 = 1.84

Z = 1.413


6/42

6

1.413

14.93

1.84


7/42

7


g= 2.33 ppg

TRZ

p29 gg

g

p = 10,000 psig

T = 200oF

gg= 0.6

{660)28.80(413.1

015,10)6.0(29g


8/42

8

Equivalent Mud Weight,

EMW

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

a depth of x(ft) can always beexpressed in terms of an equivalent

mud density or weight.

EMW = p / (0.052 * x) in ppg


9/42

9

EMW

EMW is the density of themud that, in a column ofheight, x(ft) will generate the

pressure, p (psig) at thebottom, if the pressure at top= 0 psig

or, at TD:

p = 0.052 * EMW * TVD

0po=0

TVDp

x


10/42

10

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

Dep

th,

ft

Depth

pEMW

*052.0


11/42

11

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

Depth,

ft After Kick

Before Kick

SICP = 500 psig


12/42

12

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.


13/42

13p1 = p2 = p3 ??


14/42

14

Gas Migration

Example 1.7: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 recorded casing pressure of 50

psig. Influx height is 350 ft.

Mud weight = 9.6 ppg.


15/42

15

Gas Migration

Assume surface temperature of 70 oF.Temp gradient = 1.1

oF/100 ft. Surface

pressure = 14 psia

Determine the final casing pressure ifthe gas bubble is allowed to reach the

surface without expanding

Determine the pressure and equivalent

density at total depth under these final

conditions


16/42

16

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


17/42

17

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

pseudocritical - pseudoreduced


18/42

18

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

gg29


19/42

19

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

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

46070*0.1

p

629*925.0

378,4 o

BOTTOM SURFACE


20/42


21/42

21

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

g29


22/42

22

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


23/42

23

1. 4,413 psia2. 4,378

3. 3,988 (T)

4. 3,258 (Z)

5. 2,812 (Z)

6. 2,024 (mud)


24/42

24

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


25/42

25

Compression of Mud in Annulus

From Boyles Law, pV = const

p2* 48.63 = 2,812 * 35

p2= 2,024 psiap8650 poA poB poC

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

2nd iteration ? . 3rd

or, Is there a better way?


26/42

26

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.

Ann Cap = 0.1 bbl/ft

MW = 14 ppg

TVD = 10,000 ft


27/42

27

Gas Migration Rate

How fast is the kick migrating?

What assumptions do we need to make

to analyze this question?


28/42

28

1 hr


29/42

29

First Attempt

If the kick rises x ft. in 1 hr and the

pressure in the kick = constant, then

the pressure increases everywhere,

Dp = 0.052 * 14 * x

x = (2,000 - 1,000) / (0.052 * 14)

x = 1,374 ft

Rise velocity = 1,374 ft/hr


30/42

30

Gas Migration Rate

Field rule of thumb ~ 1,000 ft/hr

Laboratory studies ~ 2,0006,000 ft/hr

Who is right?

Field results?

Is the previous calculation correct?


31/42

31

Second Attempt

Consider mud compressibility

Ann. 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


32/42

32

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


33/42

33

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 morethan 1,374 ft!


34/42

34

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 x2ft, given by:

1,659 = 0.052 * 14 * x2

x2= 2,279 ft


35/42

35

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?...


36/42

36


37/42

37

Example 1.9

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


38/42

38

Simple Solution

hrttftpsi

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?


39/42

39

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 flowarea.


40/42

40

Gas slip velocity


41/42

41

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


42/42

Gas slip velocity

Large bubbles rise faster than smallbubbles

Other factors:

Density differencesHole geometry

Mud viscosity

Circulation rateHole inclination

One lab study showed max. rate at 45o.

3. kicks and gas migration.ppt

Documents