Well Drilling Engineering

More Casing Design

Dr. DO QUANG KHANH

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 Casing Design - cont’d

 Casing Threads
 Using the Halliburton Cementing Tables
 Yield Strength of Casing (in tension)
 Burst Strength
 Effect of Axial Tension on Collapse Strength
 Effect of Pipe Bending
 Effect of Hydrogen Sulfide
 Selection of Casing Settling Depths

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 Read:
Applied Drilling Engineering, Ch.7

HW #

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Rounded Threads
* 8 per inch

~ Square Threads
* Longer
* Stronger

Integral Joint
* Smaller ID, OD
* Costs more
* Strong
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<--- BURST ---> <--- TENSION --->
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 Tensile force balance on pipe body

Example 7.1:
Compute the body-
yield strength for 20-
in., K-55 casing with a
nominal wall thickness
of 0.635 in. and a
nominal weight per
foot of 133 lbf/ft.
Ften   y ield * A s
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 Tensile force balance on pipe body
K55
Solution:
This pipe has a minimum
yield strength of 55,000 psi
and an ID of:
Ften   y ield * A s

d  20 .00  2 ( 0 .635)  18 .730in .

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 Tensile force balance on pipe body

Thus, the cross-sectional area of steel is


As  ( 20  18.73 )  38.63sq.in .
2 2

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and a minimum pipe-body yield
is predicted by Eq. 7.1 at
an axial force of:
Ften   y ield * A s

Ften  55,000 (38.63 )  2,125,000 lbf

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 Pipe Body Yield Strength

 2 2
Py  (D  d )Yp
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where
Py  pipe body yield strength, lbf
Yp  specified minimum yield strength, psi
D  outside diameter of pipe, in
d  inside diameter of pipe, in
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 Pipe Body Yield Strength

Example
What is yield strength of body of 7”, 26 #/ft,
P-110 casing?
 2
Py  (D  d2 )Yp
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 2
Py  (7  6.276 2 )110,000  830,402 lbf
4

Py  830,000 lbf (to the nearest 1,000 lbf).

…agrees with Halliburton
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Internal Yield Pressure for Pipe (Burst)

FT
 2Yp t 
P  0.875   FP
 D 

where FP = DLP
FT = 2tLYP
P  internal yield pressure, psi
DLP = 2tLYP
Yp  minimum yield strength, psi
t  nominal wall thickness, in  2Yp t 
P 
D  O.D. of pipe, in  D 
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 Example

For 7”, 26 #/ft P-110 pipe

 2Yp t 
P  0.875  
 D 
(7 - 6.276)
 0.875 * 2 * 110,000 *
2*7
 9,955

P  9,960 psi (to the nearest 10 psi)

…agrees with Halliburton Tables.
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Ellipse of
Plasticity

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TENSION

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Length of arc, L = RDqrad L

DL = (R + r)Dq - RDq Dq
R R+r
dn
DL  r Dq  Dq
2
DL dn Dq dn  
D   
L 2 L 212  100 180
30 * 10 6

D  E D  dn  218dn
2,400 180
D  218  dn F  218  dn A s (7.14a)
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Figure 7.14 - Incremental stress caused by
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
Fab  64  d n w.............................( 7.14b )
where Fab , , dn , and w have units of
lbf, degrees/100 ft, in., and lbf/ft, respectively.
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 Example

  5 deg/ 100 ft
dn,  7 in
w  35 lbf / ft

Fab  64  dn w .......... .......... .........( 7 .14b )

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

Fab = 74,400 lbf

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 Rc = 22

( Rc = 22 corresponds to a yield strength of about 90,000 psi

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)
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Production casing design load for burst.
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Production casing design load for collapse.
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 Tensile Strength of Casing

What is the maximum

length of N-80 casing that
can hang in an air-filled
wellbore
without exceeding the
minimum yield strength
of the pipe?

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 Tensile Strength of Casing
F
What is the maximum length of N-80
casing that can hang in an air-filled
wellbore without exceeding the
minimum yield strength of the pipe?

FMAX = 80,000 As = As LMAX 490/144

LMAX = 80,000 * 144/490

LMAX = 23,510 ft W

{ With a 1.8 design factor, LMAX = 13,060 ft }

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