Page 458 - Air and Gas Drilling Manual

9-62 Air and Gas Drilling Manual
1
f
i3
.
2 log 0 319 114 2
.
.
0 00015
.
f i3 0 016
Equation 6-82 for the third increment inside the drill string can be solved for the
pressure at the top of the increment, P i2. This involves selecting this lower limit by
a trial and error procedure. The magnitude of the lower 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 lower
limit pressure is
,
.
p i2 2 814 0 psia
P i2 p 144
i2
2
,
P i2 405 216 lb/ft abs
Substituting the values of ˙ w , ˙ w , Q g, Q m, P g, P i2, P i3, T g, T av3, D 7, f i3, and g
g m
into the left side of Equation 6-82 gives
,
! 565 157
#
# dP
,
# 2 2 375
,
# 40 123 1 .
.
# 33 20 0 016 P 0 426
.
.
∀ 405 216 40 123 1 . 1
,
,
.
.
0 426 20 54 0 319 2
.
P 4

Substituting the values of H 1, H 2, and H 3 into the right side of Equation 6-82 gives
,
6 650 350 2 375
,
,
1 dh 2 375
6 650 350
,
As can be seen in the above, the right and left hand sides of Equation 6-82 yield the
same answer. This shows that the lower limit pressure is correct.