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32 Part I Liquid Drilling Systems

Illustrative Example 2.3
A 10.5-ppg Power Law fluid with a consistency index of 20 cp equivalent and
flow behavior index of 0.8 is circulating at 250 gpm in an 8-¾-in-diameter
wellbore. Determine the flow regime inside a 4½-in OD, 16.60-lb/ft drill pipe
(3.826-in ID), and in the drill pipe/hole annulus.
Solution
Inside the drill pipe:
v = 250 = 6:98 ft/s
2
2:448 × ð3:826Þ
0:8
N Re = 89,100 × 10:5 × 6:98 ð2−0:8Þ × 0:0416 × 3:826 = 34,788
20 3 + 1/0:8

N Rec − L am = 3,470 − 1,370 × 0:8 = 2,374
N Rec − T ur = 4,270 − 1,370 × 0:8 = 3,174

Since N Re > 3,174, turbulent flow exists inside the drill pipe.
In the annulus:
v = 250 = 1:81 ft/s
2 2
2:448 × ð8:75 − 4:5 Þ
0:8

N Re = 109,000 × 10:5 × 1:81 ð2−0:8Þ × 0:0208 × ð8:75 − 4:5Þ = 6,523
20 2 + 1/0:8
Since N Re > 3,174, turbulent flow exists in the annular space.

For the region between transitional and turbulent flow, the critical
Reynolds number is
N Rec = 4,270 − 1370n (2.36)

Herschel-Bulkley Fluids
For Herschel-Bulkley fluids, the Reynolds number can be calculated
using the following equations in U.S. field units.
Inside the drill pipe:

2 n 3
ρν ð2−nÞ d
N Re = 2ð3n + 1Þ 6 2 7 (2.37)
7
6
n 4 n 3n + 1 n 5
d
τ y + K
2ν nC c