Page 514 - Air and Gas Drilling Manual

Therefore, Equation 6-92 becomes
2
1 Chapter 10: Stable Foam drilling 10-39
f a2
0 668 0 552
.
.
.
2 log 114
.
0 00015
f 0 021
.
a2
Equation 6-88 for the second increment in the annulus can be solved for the
pressure at bottom of the increment, P a2. This involves selecting this upper limit of
the intergral by a trial and error procedure. The magnitude of the upper limit
pressure on the left side of the equation is selected to allow the left side integral to
equal the right side integral. The integration can be carried out on the computer
using one of the commercial analytic software programs. The trial and error
magnitude of the upper limit pressure that satisfies Equation 6-88 for this annulus
section is
.
p a2 541 9 psia
P a2 p a2 144
2
,
P a2 78 039 lb/ft abs
Substituting the values of ˙ w , Q g, Q w, P g, P a1, P a2, T g, T av2, D 1, D 3, f a2, and g
t
into the left side of Equation 6-88 gives
,
∃ 78 039
&
& dP
& 2 ! 350
& 55 635 3 . 0 234
,
.
& 17 37 0 021 P
.
.
% 70 502 1 ∀
,
55 635 3 .
,
.
0 234 747 0 668 2 0 552 2
.
.
.
P 4
#
Substituting the value of H 1 and H 2 into the right side of Equation 6-88 gives
,
6 650 350
1 dh 350
6 650
,
As can be seen in the above, the right and left hand sides of Equation 6-88 yield the
same answer. This shows that the upper limit pressure is correct.