4 Casing Design

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1

MASTER

Well Drill ing

Casing Design



2

Casing Design

Why Run Casing?

Types of Casing Strings

Classification of Casing

Wellheads Burst, Collapse and Tension

Example

Effect of Axial Tension on Collapse Strength

Example



3

Read Applied Drilling Engineering, Ch.7



4

Casing Design

Why run casing?

1. To prevent the hole from caving in2. Onshore - to prevent contamination of

fresh water sands

3. To prevent water migration toproducing formation

What is casing? Casing

Cement



5

Casing Design - Why run casing, cont’d

4. To confine production to the wellbore

5. To control pressures during drill ing

6. To provide an acceptable environment for

subsurface equipment in producing wells

7. To enhance the probability of drilling to total

depth (TD)

e.g., you need 14 ppg to control a lower zone,

but an upper zone will fracture at 12 lb/gal.

What do you do?



6

Functions of CasingIndividually

Drive pipeProvides a means of

nippling up diverters

Provides a mudreturn path

Prevents erosion of ground below rig

Conductor pipeSame as Drive pipe

Supports the weight

of subsequent casingstrings

Isolates very weakformations



7

Functions of Casing

IndividuallySurface casing

Provides a means of

nippling up BOPProvides a casing

seat strong enoughto safely close in a

well after a kick.Provides protection

of fresh water sands

Provides wellborestabilization

Intermediate orprotective casing

Usually set in thefirst abnormallypressured zone

Provides isolation of

potentiallytroublesome zones

Provides integrity to

withstand the highmud weightsnecessary to reach TD or next csg seat



8

Functions of Casing IndividuallyProduction casing

Provides zonalisolation (prevents

migration of water toproducing zones,isolates differentproduction zones)

Confines productionto wellbore

Provides the

environment to installsubsurfacecompletionequipment

Liners

Drilling liners

Same as

Intermediate orprotective casing

Production liners

Same as production

casing

Tieback liners

Tie back drilling orproduction liner to the

surface. Convertsliner to full string of casing



9

Types of Strings of Casing

1. Drive pipe or structural pile{Gulf Coast and offshore only}

150’-300’ below mudline.

2. Conductor string. 100’ - 1,600’

(BML)

3. Surface pipe. 2,000’ - 4,000’

(BML)

Diameter Example

16”-60” 30”

16”-48” 20”

8 5/8”-20” 13 3/8”



10

Types of Strings of Casing

4. Intermediate String

5. Production String (Csg.)

6. Liner(s)

7. Tubing String(s)

7 5/8”-13 3/8” 9 5/8”

Diameter Example

4 1/2”-9 5/8” 7”



11

Example Hole and String Sizes (in)

Structural casing

Conductor string

Surface pipe

IntermediateString

Production Liner

Hole Size

30”20”

13 3/8

9 5/8

7

Pipe Size

36”26”

17 1/2

12 1/4

8 3/4



12


Structural casing

Conductor string

Surface pipe

IntermediateString

Production Liner

Hole Size

30”20”

13 3/8

9 5/8

7

Pipe Size

36”26”

17 1/2

12 1/4

8 3/4



13


Structural casing

Conductor string

Surface pipeIntermediateString

Production Liner

250’

1,000’

4,000’

Mudline



14

Classification of CSG.

1. Outside diameter of pipe (e.g. 9 5/8”)

2. Wall thickness (e.g. 1/2”)

3. Grade of material (e.g. N-80)

4. Type to threads and couplings (e.g. API LCSG)

5. Length of each joint (RANGE) (e.g. Range 3)

6. Nominal weight (Avg. wt/ft incl. Wt. Coupling)

(e.g. 47 lb/ft)



15

σ

ε



16

Length of Casing Joints

RANGE 1 16-25 ft

RANGE 2 25-34 ft

RANGE 3 > 34 ft.



17

Casing Threads and Couplings

API round threads - short {CSG }API round thread - long {LCSG }

Buttress {BCSG }Extreme line {XCSG }

Other …

See Halliburton Book...



18

Rounded Threads

* 8 per inch

~ Square Threads

* Longer* Stronger

Integral Joint

* Smaller ID, OD

* Costs more

* Strong



19



20

23 lb/ft

26 lb/ft

N-80



21

API Design Factors (typical)

Collapse 1.125

Tension 1.8

Burst 1.1

Required

10,000 psi

100,000 lbf

10,000 psi

Design

11,250 psi

180,000 lbf

11,000 psi



22

Normal Pore Pressure Abnormal Pore Pressure

0.433 - 0.465 psi/ft gp

> normal

Abnormal



23Design from bottom



24

X-mas TreeWing Valve

Choke Box

Master

Valves

Wellhead

• Hang Csg. Strings

• Provide Seals

• Control Productionfrom Well

Press. Gauge



25

Wellhead



26

Wellhead



27

Casing Design

Burst: Assume full reservoir pressure all along the wellbore.

Collapse: Hydrostatic pressure increases with depth

Tension: Tensile stress due to weight of string is highest at top

STRESS

Tension

Burst

Collapse

Collapse

Tension

Depth

Burst



28

Casing Design - Tension



29

Tensile force balance on pipe body

A*Fsyieldten

σ=



30


Example 7.1:

Compute the body-yield strength for 20-

in., K-55 casing with anominal wall thicknessof 0.635 in. and a

nominal weight perfoot of 133 lbf/ft.

A*Fsyieldten

σ=



31


Solution: This pipe has a minimum

yield strength of 55,000 psiand an ID of:

.in730.18)635.0(200.20d =−=

K55

A*F syieldten σ=



32


Thus, the cross-sectional area of steel is

and a minimum pipe-body yield

is predicted by Eq. 7.1 atan axial force of:

.in.sq 63.38)73.1820(4

A 22s =−=

π

lbf 000,125,2)63.38(000,55Ften ==

A*F syieldten σ=



33

Pipe Body Yield Strength

where

p

22

y Y)dD(4P −=

inpipe,of diameter insided

inpipe,of diameter outsideD

psistrength,yieldminimumspecifiedY

lbf strength,yieldbodypipeP

p

y

=

=

=

=



34

Casing Design - Burst

(from internal pressure)

4 Internal Yield Pressure for pipe

4 Internal Yield Pressure for couplings

4 Internal pressure leak resistance

pp

Internal

Pressure



35

Internal Yield Pressure for Pipe (Burst)

where

⎥⎦

⎤

⎢⎣

⎡

= D

t Y2

875.0Pp

inpipe,of O.D.D

inthickness,wallnominalt

psistrength,yieldminimumY

psipressure,yieldinternalP

p

=

=

=

=

FP = DLP

F T = 2tLYP

DLP = 2tLYP

⎥⎦

⎤⎢⎣

⎡=

D

t2YP

p

FP

FT



36

Example

For 7”, 26 #/ft P-110 pipe

9,955

7*26.276)-(7*110,000*2*0.875

=

=

psi960,9P = (to the nearest 10 psi)

…agrees with Halliburton Tables.

⎥⎦

⎤⎢⎣

⎡=

D

tY2875.0P

p



37

Casing Design - Burst

Example

Design a 7” Csg. String to 10,000 ft.

Pore pressure gradient = 0.5 psi/ft

Design factor, Ni=1.1

Design for burst only.



38

Burst Example

1. Calculate probable reservoir pressure.

psi000,5ft000,10*ft

psi5.0pres ==

2. Calculate required pipe internal yieldpressure rating

psi500,51.1*000,5N*pp iresi ===

Ni = API Design Factor for BURST = 1.1



39

Example

3. Select the appropriate csg. grade and wt.

from the Halliburton Cementing tables:

Burst Pressure required = 5,500 psi

7”, J -55, 26 lb/ft has BURST Rating of 4,980 psi7”, N-80, 23 lb/ft has BURST Rating of 6,340 psi

7”, N-80, 26 lb/ft has BURST Rating of 7,249 psi

Use N-80 Csg., 23 lb/ft



40



41

23 lb/ft

26 lb/ft

N-80



42

Casing Design - Collapse



43

Collapse Pressure

The following factors are important:

4 The collapse pressure resistance of a pipedepends on the axial stress

4 There are different types of collapsefailure



44

Collapse Pressure

There are four different types of collapse

pressure, each with its own equation forcalculating the collapse resistance:

4 Yield strength collapse

4 Plastic collapse

4 Transition collapse4 Elastic collapse



45

If Axial Tension is Zero:

Yield Strength Plastic Transition Elastic

→ )t/D(

J -55 14.81 25.01 37.31

N-80 13.38 22.47 31.02

P-110 12.44 20.41 26.22




46

Casing Design

Collapse pressure - with axial stress

1.

⎪

⎭

⎪⎬⎫

⎪

⎩

⎪⎨⎧

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −=

P

A

2/12

P

APPA

Y

S5.0

Y

S75.01 Y Y

YPA = yield strength of axial stressequivalent grade, psi

YP = minimum yield strength of pipe, psi

SA = Axial stress, psi (tension is positive)



47


2. Calculate D/t to determine proper equation

to use for calculating the collapse pressure

Yield Strength

Collapse :

Plastic Collapse:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟ ⎠ ⎞⎜

⎝ ⎛

−⎟ ⎠

⎞⎜⎝

⎛

=2p YP

tD

1t

D

Y2P

CB

t

DA YP pp −

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

⎟ ⎠

⎞⎜⎝

⎛ =

C C ’



48

Transition

Collapse:

Elastic

Collapse:

⎥⎥⎥

⎥

⎦

⎤

⎢⎢⎢

⎢

⎣

⎡

−⎟ ⎠

⎞⎜⎝

⎛ = G

t

D

F YP p T

2

6

E

1t

D

t

D

10X95.46

P⎥⎦

⎤⎢⎣

⎡−⎟

⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ =

Casing Design - Collapse, cont’d

E l 2



49

Example 2

Determine the collapse strength of 5 1/2”O.D., 14.00 #/ft J -55 casing under zero

axial load.

1. Calculatethe D/t ratio: ( )

booknHalliburtoFrom

54.22

012.5500.52

1500.5

tD

↑

=−

=

C C



50

If Axial Tension is Zero:

Yield Strength Plastic Transition Elastic

→ )t/D(

J -55 14.81 25.01 37.31

N-80 13.38 22.47 31.02

P-110 12.44 20.41 26.22


Example 2



51

Example 2

2. Check the mode of collapse

Table (above) shows that,

for J -55 pipe,with 14.81 < D/t < 25.01

the mode of failure is plastic collapse.

54.22=t

D

C i D i C ll



52


Calculate D/t to determine proper equation touse for calculating the collapse pressure

Plastic Collapse: CB

t

DA YP pp −

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

⎟ ⎠

⎞⎜⎝

⎛ =

Example 2



53

Example 2

The plastic collapse is calculated from:

206,10541.054.22

991.2000,55

CBt/D

A YP pp

−⎥⎦

⎤⎢⎣

⎡−=

−⎟ ⎠ ⎞⎜

⎝ ⎛ −=

psi117,3P p =Halliburton Tablesrounds off to 3,120 psi

Example 3



54

Example 3

Determine the collapse strength for a 5 1/2” O.D.,14.00 #/ft, J -55 casing under axial load of 100,000lbs

The axial tension will reduce the collapse pressureas follows:

P

p

A2

p

APA Y

Y

S5.0

Y

S75.01 Y

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −=

( ) psi

Area

F S A A 820,24

012.55.54

000,100

22

=−

==π

Example 3 cont’d AA YS S

Y ⎥⎤

⎢⎡

⎟ ⎞

⎜⎛

⎟ ⎞

⎜⎛

507501

2



55

Example 3 cont d

The axial tension will reduce the collapsepressure rating to:

psi216,38

000,55000,55

820,245.0

000,55

820,2475.01 Y

2

PA

=

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡⎟ ⎠

⎞⎜⎝

⎛ −⎟

⎠

⎞⎜⎝

⎛ −=

Here the axial load decreased the J-55

rating to an equivalent “J-38.2” rating

P

p

A

p

APA Y

Y Y Y

⎥⎥⎥

⎦⎢⎢⎢

⎣

⎟⎟

⎠⎜⎜

⎝ −

⎟⎟

⎠⎜⎜

⎝ −= 5.075.01

Example 3 cont’d



56

Example 3 - cont d

551,243.70010x557.4

54.22

945.2216,38

CBt/D

A YP

2

PAp

=−⎥⎦

⎤⎢⎣

⎡−=

−⎟ ⎠

⎞⎜⎝

⎛ −=∴

−

psi550,2P p ≈

…compared to 3,117 psi with no axial stress!

Example 3 cont’d



57

Example 3 - cont d

We shall be using API Tables to correct for theeffect of axial tension on collapse strength of casing.

The Halliburton Cementing Tables list thecollapse resistance of 5 ½ -in, 14.00 lb/ft J -55casing at 3,120 psi.

The axial tension in this case would derate the

collapse strength to about 2,550 psi.



58

Combined Loading



59



60



61

Linear Interpolation



62

p

(iii) CmSP

(ii) CmSP

(i) CmSP

cmxy

22

11

+=

+=+=

+=




63

ea te po at o

⇒−=−− )SS(mPP )ii()iii( 1212

12

12

S S PPm

−−=

)()(P )()(1

12

12

11

S S S S

PPS S mPiii −

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

−

−=−=−−




64

p

( )12

12

11 PP

SSSSPP −⎟⎟

⎠ ⎞⎜⎜

⎝ ⎛

−−+=∴

With design factor:

( )⎥⎦

⎤⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛

−

−−= 21

12

11cc PP

SS

SSP

.F.D

1P



65

α = dogleg severity, deg/100 ft

= angle build rate, deg/100 ft

απ= 00018,RadiusBuild

Length of arc, L = R∆θRL



66

∆L = (R + r)∆θ - R∆θ

θ∆=θ∆=∆ 2drL n

( ) 180100122

d

L2

d

L

L nn πα=

θ∆=

∆=ε∆

nn

6

d218d

180400,2

10*30E α=α

π=ε∆=σ∆

nd218α=σ∆

∆θR + rR

g R

snAd218F α= (7.14a)

Figure 7.14 - Incremental stress caused by



67

bending of casing in a directional well

The area of steel, As, can be expressed

conveniently as the weight per foot of pipe divided by the density of steel.For common field units, Eq. 7.14a

becomes

)14.7.........(....................64 bwd F nab α =

ly.respectivelbf/ft,and in.,ft,0degrees/10lbf,

of unitshave and ,,, where wd F nab α

Example



68

)14.7.........(....................64 bwd F nab α =

/53 in7

100deg/5

,

ft lbf wd

ft

n

=

=

=α

Fab

= 64 * 5 * 7 * 35 = 74,400 lbf

Fab = 74,400 lbf

Casing Design Example



69

g g p

Example Problem

API Design Factors

“Worst Possible Conditions”

Effect of Axial Tension on Collapse Strength

Iteration and Interpolation

Design for Burst, Collapse and Tension

Casing Design Example



70

Design a 9 5/8-in., 8,000-ft combinationcasing string for a well where the mud wt.will be 12.5 ppg and the formation porepressure is expected to be 6,000 psi.

Only the grades and weights shown areavailable (N-80, all weights). Use API

design factors.

Design for “worst possible conditions.”

Casing Design - Solution



71

g g

Before solving this problem is it necessary tounderstand what we mean by “Design Factors”

and “worst possible conditions”.

API Design Factors

Design factors are essentially “safety factors”that allow us to design safe, reliable casingstrings. Each operator may have his own set

of design factors, based on his experience,and the condition of the pipe.

Casing Design



72

Casing Design

In PETE 661, we’ll use the design factorsrecommended by the API unless otherwise

specified.

These are the API design Factors:

Tension and J oint Strength: N T = 1.8

Collapse (from external pressure): Nc= 1.125Burst (from internal pressure): Ni = 1.1

Casing Design



73

g g

What this means is that, for example, if we

need to design a string where the maximumtensile force is expected to be 100,000 lbf ,we select pipe that can handle 100,000 * 1.8

= 180,000 lbf in tension.

Note that the Halliburton Cementing Tables

list actual pipe strengths, without safetyfactors built in.

Casing Design



74

Unless otherwise specified in a particularproblem, we shall also assume the following:

Worst Possible Conditions

1. For Collapse design, assume that thecasing is empty on the inside (p = 0 psig)

2. For Burst design, assume no “backup”fluid on the outside of the casing (p = 0 psig)

Casing Design



75

Worst Possible Conditions, cont’d

3. For Tension design,assume no buoyancy effect

4. For Collapse design,

assume no buoyancy effect

The casing string must be designed to stand up to the

expected conditions in burst, collapse and tension.

Above conditions are quite conservative. They are also

simplified for easier understanding of the basic concepts.




76

Burst Requirements (based on the expected pore

pressure)

The whole casing string must be capable of withstanding this internal pressure without failing inburst.

psi600,6P

1.1* psi000,6

Factor Design* pressure poreP

B

B

=

=

=

D e p t h

Pressure




77

Collapse Requirements

For collapse design, we start at the bottomof the string and work our way up.

Our design criteria will be based onhydrostatic pressure resulting from the 12.5ppg mud that will be in the hole when the

casing string is run, prior to cementing.

Casing Designe

p t h



78

Collapse Requirements, cont’d

severelessare

tsrequiremencollapsetheholetheupFurther

.bottomtheatd'reqpsi850,5P125.1*000,8*5.12*052.0

factordesign*depth*weightmud*052.0P

c

c

←=

=

=

D e

Pressure

Casing Design



79

Req’d: Burst: 6,600 psi Collapse: 5,850 psi

Casing Design



80

Note that two of the weights of N-80 casingmeet the burst requirements, but only the

53.5 #/ft pipe can handle the collapserequirement at the bottom of the hole (5,850psi).

The 53.5 #/ft pipe could probably run all theway to the surface (would still have to checktension), but there may be a lower costalternative.

Casing Design



81

To what depth might webe able to run N-80, 47#/ft? The maximumannular pressure that this

pipe may be exposed to,is:

psi231,4125.1

760,4

factor design

pipeof pressureCollapse

Pc===

D e p t h

Pressure

Casing Design



82

First Iteration

At what depth do we see this pressure (4,231psig) in a column of 12.5 #/gal mud?

ft509,65.12*052.0

231,45.12*052.0

Ph

h*5.12*052.0P

c1

1c

===∴

=



Casing Design



84

Weight, W1 = 53.5 #/ft * 1,491 ft= 79,769 lbf

This weight results in an axialstress in the 47 #/ft pipe

psi877,5

in13.572

lbf 769,79

areaend

weightSof

21 ===

Casing Design



85

The API tables show that the above

stress will reduce the collapse resistancefrom4,760 to somewhere between

4,680 psi (with 5,000 psi stress)

and 4,600 psi (with 10,000 psi stress)

Casing Design



86

Interpolation between these values showsthat the collapse resistance at 5,877 psi

axial stress is:

psi148,4125.1

666,4P

psi666,4)600,4680,4(*)000,5000,10(

)000,5877,5(680,4P

cc1

1c

==

=−−

−−=

With the design factor,

( )21

12

11c1P PP

S S

S S P −⎟⎟

⎠

⎞⎜⎜⎝

⎛

−

−−=

Casing Design



87

This (4,148 psig) is the pressure at a

depth

Which differs considerably from the

initial depth of 6,509 ft, so a seconditeration is required.

ft382,65.12*052.0

148,4h2 ==



88



89

Casing Design



90

Second Iteration

Now consider running the 47 #/ftpipe to the new depth of 6,382 ft.

psi378,6in572.13lbf 563,86S

lbf 563,865.53*)382,6000,8(W

22

2

==

=−=

Casing Design

I t l ti i



91

Interpolating again,

This is the pressure at a depth of

( ) psi pcc 140,4600,4680,4*5000

5000378,6

680,4125.1

1

2=

⎭⎬

⎫

⎩⎨

⎧

⎥⎦

⎤

⎢⎣

⎡−

−−=

ft369,65.12*052.0

140,4h 3 ==

( )⎥

⎥

⎦

⎤

⎢

⎢

⎣

⎡−

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

−

−−=

2112

1

1c1 D.F.

1P PP

S S

S S P

Casing Design

Thi i ithi 13 ft f th d l If



92

This is within 13 ft of the assumed value. If more accuracy is desired (generally not

needed), proceed with the:Third Iteration

psi429,6572.13

259,87S

lbf 259,875.53*)369,6000,8(W'369,6h

3

3

3

==

=−==

Pcc3 = ?

Casing Design



93

Third Iteration, cont’d

2

3

140,4

)600,4680,4(*

000,5

000,5429,6680,4

125.1

1

cc

cc

P psi

Pthus

==

⎭

⎬⎫

⎩

⎨⎧

−−

−=

Casing Design



94

Third Iteration, cont’d

This is the answer we are looking for, i.e.,we can run 47 #/ftN-80 pipe to a depth of 6,369 ft, and 53.5 #/ft pipe between 6,369

and 8,000 ft.Perhaps this string will run all the way to thesurface (check tension), or perhaps an evenmore economical string would include some43.5 #/ft pipe?

Casing Design



95

At some depth the 43.5 #/ft pipe would be

able to handle the collapse requirements,but we have already determined that it willnot meet burst requirements.

! NO∴

N-80

43 5 #/ft?



96

N-8053.5 #/ft

N-8047.0 #/ft

43.5 #/ft?Depth = 5,057?

5,066?5,210?

Depth = 6,3696,3696,382

6,509

8,000

Tension Check



97

The weight on the top joint of casing

would be

With a design factor of 1.8 for tension, a

pipe strength of

weightactual602,386

)/#5.53*631,1()/#0.47*369,6(

lbs

ft ft ft ft

=

+

required islbf 080,695602,386*8.1 =

Tension Check



98

The Halliburton cementing tables give a

yield strength of 1,086,000 lbf for the pipebody and a joint strength of 905,000 lbf forLT & C.

surfacetoOK isft/#0.47∴

Casing Design Review



99

We have 4 different weights of casingavailable to us in this case:

1. Two of the four weights are unacceptableto us everywhere in the string becausethey do not satisfy the burstrequirements.

2. Only the N-80, 53.5 #/ft pipe is capable of withstanding the collapse requirementsat the bottom of the string




100

3. Since the 53.5 #/ft pipe is the mostexpensive, we want to use as little of it

as possible, so we want to use asmuch 47.0 #/ft pipe as possible.

4. Don’t forget to check to make sure thetension requirements are met; both for

pipe body, and for threads andcouplings (T&C).




101

The collapse resistance of N-80, 47 #/ft willdetermine to what depth it can be run. Twofactors will reduce this depth:

Design Factor

Axial Stress (tension)

“Halliburton” collapse resistance: 4,760 psi Apply design factor: psi231,4

125.1

760,4=




102

To determine the effect of axial stressrequires an iterative process:

1. Determine the depth capability without

axial stress

2. Determine axial stress at this point

ft509,65.12*052.0

231,4depth ==


3 Determine corresponding collapse resistance



103

3. Determine corresponding collapse resistance

4. Determine depth where this pressure exists

5. Compare with previous depth estimate

6. Repeat steps 2-6 using the new depthestimate

7. When depths agree, accept answer(typically 2-4 iterations) (agreement towithin 30 ft will be satisfactory)

Date post:	14-Apr-2018
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