Page 107 - Practical Well Planning and Drilling Manual

Section 1 revised 11/00/bc 1/17/01 2:56 PM Page 83

1.4.14
Casing Design [ ]

1. Identify the top of the supported casing. This may be at the top of
cement (if no centralizers above TOC) or at the top centralizer, if
centralizer spacing above TOC is relatively small—say one per two
joints in gauge hole.
2. Calculate F and F once the cement has set at the top supported
b s
depth, with the actual external and internal pressures. Unless very
heavy mud is inside then F should exceed F .
s
b
3. Calculate or estimate the maximum internal pressure in service
(due to heavier mud only; or completion fluid hydrostatic plus sur-
face pressure if applicable) and use this to recalculate F .
s
4. Calculate the increase in F due to thermal expansion (refer
b
“Revised F due to thermal expansion” following in this section).
b
For tubings, also calculate the increase in F due to pump out force
b
at the packer if a nonanchored seal assembly is used.
5. If the revised F exceeds the revised F then the casing will buckle.
s
b
In this case calculate how much extra tension applied at the sur-
face would cause F to reduce below F ; this force can then be
s
b
applied as an overpull after the cement has set by setting the cas-
ing in slip type hangers.
Revised F due to thermal expansion. The coefficient of thermal
b
expansion (a) for steel gives the thermal strain in a uniform body sub-
jected to uniform heating (refer to Section 1.4.7, “Mechanical
Properties of Steel”).
-6
Strain e due to uniform thermal expansion of steel = 6.9 x 10 /°F
-5
(1.24 x 10 /°C)
Young’s Modulus of Elasticity E = 30 x 10 6

With these two pieces of information we can calculate the effect on
the neutral point if we assume a uniform increase in temperature
throughout a length of pipe.
Assuming a length of 13 /8 in 72# N80 casing with the top of
3
cement at 10,000 ft, subjected to an overall temperature increase of
20˚C. The thermal strain (stretch ÷ original length) = 20 x 0.0000124
= 0.00025, and if this is multiplied by the original length of 10,000 ft
then the increase in length = 20 x 0.0000124 x 10,000 = 2.48 ft.
Knowing Young’s Modulus, we can calculate the equivalent stress
6
by multiplying strain (0.00025) by E (30 x 10 ), which equals 7440 psi.

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