PETE 625Well Control

Lesson 3

Kicks and Gas Migration

2

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

Assignments

Homework #2:

Ch 1, Problems 1.11-1.21

Read: All of Chapter 1

4

Density of Real Gases

M = molecular weight

m = mass

n = no. of moles

g = S.G. of gas

ZRT

pVn

V

nM

V

mg

29

M

M

MZRT

pM

V

M

ZRT

pV

airg

g

ZRT

pgg

29

(Real Gas Law)

5

Density of Real Gases

What is the density of a 0.6 gravity gas at 10,000 psig and 200 oF?

From Lesson 2, Fig. 1

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

Tpr = (200+460)/358 = 1.84

Z = 1.413

6

1.413

14.93

1.84

7

Density of Real Gases

g = 2.33 ppg

TRZ

p29 gg

p = 10,000 psig

T = 200 oF

g = 0.6{

660)28.80(413.1

015,10)6.0(29g

8

Equivalent Mud Weight, EMW

The pressure, p (psig) in a wellbore, at a depth of x (ft) can always be expressed in terms of an equivalent mud density or weight.

EMW = p / (0.052 * x) in ppg

9

EMW

EMW is the density of the mud that, in a column of height, x (ft) will generate the pressure, p (psig) at the bottom, if the pressure at top = 0 psig

or, at TD:

p = 0.052 * EMW * TVD

0po=0

TVD p

x

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

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

Dep

th,

ft After Kick

Before Kick

SICP = 500 psig

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.

13p1 = p2 = p3 ??

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

Gas Migration

Assume surface temperature of 70 oF. Temp gradient = 1.1 oF/100 ft. Surface pressure = 14 psia

Determine the final casing pressure if the gas bubble is allowed to reach the surface without expanding

Determine the pressure and equivalent density at total depth under these final conditions

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

= 629 oR

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

Bottomhole Pressure

g = 29*0.7*4,378 / (0.925 * 80.28 * 629)= 1.90 ppg

pKICK = 0.052 * 1.9 * 350 = 35 psi

BHP = 4,378 + 35

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

ZRT

pgg

29

19

Pressure at Surface

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

Then,

46070*nR*0.1

Vp

629*nR*925.0

V378,4

ZnRT

pV o

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

46070*0.1

p

629*925.0

378,4 o

BOTTOM SURFACE

20

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

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

29

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

1. 4,413 psia2. 4,3783. 3,988 (T)4. 3,258 (Z)5. 2,812 (Z)6. 2,024 (mud)

24

Compression of Mud in Annulus vA = 0.1 bbl/ft)

V = compressibility * volume * p

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

V = -13.63 bbls

Initial kick volume = 0.1 * 350 = 35 bbls

New kick volume = 35 + 13.63

= 48.63 bbl

25

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?

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

Gas Migration Rate

How fast is the kick migrating?

What assumptions do we need to make to analyze this question?

28

1 hr

29

First Attempt

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

p = 0.052 * 14 * x

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

x = 1,374 ft

Rise velocity = 1,374 ft/hr

30

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?

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,

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

= 6 bbl

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

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 than 1,374 ft!

34

Second Attempt

How far did it migrate in 1 hour?The pressure reduction in kick fluid

= 8,260 - 6,621=1,659 psiThe kick must therefore have risen an

additional x2 ft, given by:

1,659 = 0.052 * 14 * x2

x2 = 2,279 ft

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

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

Simple Solution

hrtt

ftpsi

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

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.

40

Gas slip velocity

41

Gas slip velocityBubbles with a void

fraction > 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

Gas slip velocity Large bubbles rise faster than small

bubbles Other factors:

Density differences Hole geometry Mud viscosity Circulation rate Hole inclination

One lab study showed max. rate at 45o.