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34 Part I Liquid Drilling Systems

Illustrative Example 2.4
A 10.5-ppg Herschel-Bulkley fluid with a consistency index of 20 cp equivalent, a
2
flow behavior index of 0.8, and a yield stress of 5 lb/100 ft is circulating at
250 gpm in an 8¾-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
2:448ð3:826Þ

C c = 1 − 1 5 = 0:7513
2 × 0:8 + 1 " ð3 × 0:8 + 1Þ0:557 # 0:8
5 + 0:04177
3
0:8 × πð0:319/2Þ
0:8
2 ð2 − 0:8Þ 0:319 3

10:5 × 7:48 × 6:98
N Re = 2ð3 × 0:8 + 1Þ 6 2 5 = 11,975
7
0:8 4 0:319 0:8 3 × 0:8 + 1 0:8
5 + 0:04177
2 × 6:98 0:8 × 0:7513
The critical Reynolds number N Rec inside the drill pipe is calculated as follows:
logð0:8Þ + 3:93
y = = 0:0767
50
z = 1:75 − logð0:8Þ = 0:2638
7
1

N Rec = 4ð3 × 0:8 + 1Þ 1 − 0:2638 = 1,537
0:8 × 0:0767

Since N Re > 1,537, turbulent flow exists inside the drill pipe.
In the annulus:
v = 250 = 1:81 ft/s
2 2
2:448ð8:75 − 4:5 Þ

C = 1 − 1 5

a
0:8 + 1 ( ) 0:8
5 + 0:04177 2 × 0:557ð2 × 0:8 + 1Þ
2 2
0:8 × π½ð0:729/2Þ − ð0:375/2ފ½ð0:729/2Þ − ð0:375/2Þ Š
C = 0:552

a
2 3
ð2 − 0:8Þ 0:729 − 0:375 0:8

10:5 × 7:48 × 1:81
6 2 7
N Re = 4ð2 × 0:8 + 1 Þ 6 7= 1,506
7
0:8 6 0:729 − 0:375 0:8 0:8 5
4
5 + 0:04177 2ð2 × 0:8 + 1Þ
2 × 1:81 0:8 × 0:552