Page 66 -

42 Part I Liquid Drilling Systems

Illustrative Example 2.5 (Continued )
In the cased-hole annulus:

v = 300 = 2:17 ft/s
2 2
2:448ð8:755 − 4:5 Þ
μ = 35 + 5ð6Þð8:755 − 4:5Þ = 94 cp
a
2:17
N Re = 757 ð10:5Þð2:17Þð8:755 − 4:5Þ = 781 < 2,100, laminar flow
94
" #
Δp f = ð35Þð2:17Þ + 6 ð6500Þ = 73 psi
2
200ð8:755 − 4:5Þ
1,000ð8:755 − 4:5Þ
In the open-hole/drill pipe annulus:

v = 300 = 2:36 ft/s
2 2
2:448ð8:5 − 4:5 Þ
μ = 35 + 5ð6Þð8:5 − 4:5Þ = 86 cp
a
2:36
N Re = 757 ð10:5Þð2:36Þð8:5 − 4:5Þ = 872 < 2,100, laminar flow
86
" #
Δp f = ð35Þð2:36Þ + 6 ð3,000Þ = 38 psi
2
200ð8:5 − 4:5Þ
1,000ð8:5 − 4:5Þ
In the open-hole/drill collar annulus:

v = 300 = 4:59 ft/s
2 2
2:448ð8:5 − 6:75 Þ
μ = 35 + 5ð6Þð8:5 − 6:75Þ = 46 cp
a
4:59
N Re = 757 ð10:5Þð4:59Þð8:5 − 6:75Þ = 1,388 < 2,100, laminar flow
46
" #
Δp f = ð35Þð4:59Þ + 6 ð450Þ = 31 psi
2
200ð8:5 − 6:75Þ
1,000ð8:5 − 6:75Þ
The total system pressure loss is

Δp d = 605 + 340 + 73 + 38 + 31 = 1,087 psi = 7,394 kPa