Page 94 - Practical Well Planning and Drilling Manual

Section 1 revised 11/00/bc 1/17/01 2:56 PM Page 70

[ ] Well Design
1.4.10

In Case 1, the net upward force is equal to the difference between
the upward and downward pressures acting on the cross-sectional area.
This net upward force can in this case be calculated by using the
(weight of steel x [1 - buoyancy factor]). The net upward force is the
same in Case 2, except here the force results from varying differential
pressures acting on the diameter of the casing instead of on the end.
In Case 3, the net upward force is equal to Cases 1 and 2, the dif-
ference is that the net force comes from the vector addition of the net
lateral force and the net axial force. Again the total buoyant weight can
be calculated by weight of steel x buoyancy factor. Since calculating the
net axial force is easy (cross-sectional area x difference in hydrostatic
head), we can deduce the net lateral force by vector subtraction.
Calculating it directly would be difficult, involving differential calculus.
In Case 4, we have taken the section of casing from Case 3 and
joined it on to casing extending to surface. This has no effect on the lat-
eral force. Therefore, by treating the section of casing independently (as
if it were freely suspended in mud), the net axial force subtracted using
vectors from the total buoyant weight gives us the net lateral force.
For example, let us assume that you want to consider a section of cas-
ing, 9 /8 in x 40 lbs/ft, 100 ft long, immersed in a mud of 0.5 psi/ft gradi-
5
ent (buoyancy factor 0.85; ppg 9.62). Inclination is 40˚. True vertical depth
of the lower end is 5000 ft. Calculate the net axial and lateral forces.

1. For the section of casing in question, use the buoyancy factor to
establish the buoyant weight as if it were freely suspended in mud.
This would be 40 lbs/ft x 100 ft x 0.85 = 3400 lbs.
Net upward force = weight in air - buoyant weight = 600 lbs.
2. Calculate the axial forces acting at the bottom and at the top of this
section if freely suspended in mud.
TVD at bottom (at centerline) is 5000 ft; cross-sectional area
is .7854(D 2 - d 2 ) = 11.454 in 2 ; hydrostatic pressure is 2500 psi
so the total force action upwards axially is 28,635 lbs. This fig-
ure can be used later to calculate net axial force throughout the
casing string.

TVD at top is 5000 ft - (100 Cos 40) = 4923.4 ft. Hydrostatic
pressure is 2462 psi and axial load acting downwards is
28,196 lbs.

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