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P. 158
126 Part II Gas Drilling Systems
Substituting Eqs. (6.9) and (6.10) into Eq. (6.23) yields
" 2 #
S g P
dP = 1 + 0:074d b S s R p + 76:3ðS x Q x + S l Q f Þ
S g Q go
53:3½T s + G cos ðI s ÞS
8 2 9
> Q go ½T s + G cos ðI s ÞS >
> f 9:77 >
> >
< AP =
× cos ðI s Þ± dS (6.24)
2gD H
> >
> >
> >
: ;
which can be simplified as
a cos ðI s ÞP
dP = + ab½T s + G cos ðI s ÞS dS (6.25)
T s + G cos ðI s ÞS P
Using the boundary condition at the top of the segment—that is, P = P s
at S = 0—the solution to Eq. (6.25) for nonhorizontal flow is
v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2a
u
u T s + G cos ðI s ÞS G
P = t P s + ab T s 2
2
T s
(6.26)
ða −GÞ cos ðI s Þ
r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
− ab ½T s + G cos ðI s ÞS 2
ða −GÞ cos ðI s Þ
It is obvious that this equation degenerates to Eq. (6.18) for vertical
flow when the inclination angle is zero. But it is not valid for horizontal
flow where the inclination angle is 90 degrees. For horizontal flow, Eq.
(6.25) becomes
dP = abT s dS (6.27)
P
Using the boundary condition at the exit point—that is, P = P s at S = 0—
the solution of Eq. (6.27) for horizontal flow is
2
2
P = P + 2abT s S (6.28)
s
If the angle-building section has a constant radius of curvature R,
there is no need to divide the curve section into a series of slant hole
segments with different inclination angles. Guo and colleagues (1994)
