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Mud Hydraulics Fundamentals 47

Like the calculation in Illustrative Example 2.4, the values of the Reynolds number
N Re , the critical Reynolds number N Rec ,y,z,and C c are calculated following:
N Re = 14,563, N Rec = 1,537, N Re > N Rec , turbulent flow
C c = 0:7481, y = 0:0767, z = 0:2638
−0:2638
f c = 0:0767ð0:7481 × 14,563Þ = 0:0066
2
Δp f = 0:0066 × 0:6684 × 10:5 × 7:48 ð9,500 + 479Þ = 491 psi
5
1421:22 × ð3:826/12Þ
Inside the drill collar:
v = 300 = 24:2 ft/s
2
2:448ð2:25Þ
N Re = 65,119, N Rec = 1,537, N Re > N Rec , turbulent flow
C c = 0:8666, y = 0:0767, z = 0:2638
−0:2638
= 0:0043
f c = 0:0767ð0:8666 × 65,119Þ
2
Δp f = 0:0043 × 0:6684 × 10:5 × 7:48 ð450Þ = 204 psi
5
1421:22 × ð2:25/12Þ
In the cased-hole annulus:
v = 300 = 2:17 ft/s
2 2
2:448ð8:755 − 4:5 Þ
N Re = 1,818, N Rec = 2,737, N Re < N Rec , laminar flow

C a = 0:5487
(

Δp f = 4 × 0:04177 × 6
0:04177
14400ð0:73 − 0:375Þ
0:8 )

+ 16ð2 × 0:8 + 1Þ 0:6684 ð6,500Þ = 42 psi
2 2
0:8 × 0:5487ð0:73 − 0:375Þ
πð0:73 − 0:375 Þ
In the open-hole/drill pipe annulus:
v = 300 = 2:36ft/s
2 2
2:448ð8:5 −4:5 Þ
N Re = 2,080,N Rec = 2,737,N Re <N Rec , laminarflow
C = 0:5589

a
(

Δp f = 4×0:04177 × 6
0:04177
14400ð0:708−0:375Þ
0:8 )
+ 16ð2×0:8+1Þ 0:6684 ð3,000Þ = 21psi
2 2
0:8×0:5589ð0:708−0:375Þ
πð0:708 −0:375 Þ
(Continued )