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Mud Hydraulics Fundamentals 41

Illustrative Example 2.5
Determine the system pressure loss for the following well:

Total depth: 9,950 ft (3,036 m)
Casing: 9⅝ in, 43.5 lb/ft (8.755-in ID), set at 6,500 ft (1,982 m)
Open hole: 8½ in from 6,500 ft to 9,950 ft
Drill pipe: 9,500 ft of 4½ in, 16.6 lb/ft (3.826-in ID)
Drill collar: 450 ft of 6¾-in OD and 2¼-in ID
Surface equipment: Combination 3
Mud weight: 10.5 ppg
Plastic viscosity: 35 cp
Yield point: 6 lb/100 ft 2
Mud flow rate: 300 gpm
Solution
According to Table 2.2 shown earlier, the pressure loss through the surface
equipment is equivalent to that through 479 ft of 4½ in, 16.6-lb/ft drill pipe.
Inside the drill pipe:
v = 300 = 8:37 ft/s
2
2:448ð3:826Þ
μ = 35 + 6:66ð6Þð3:826Þ = 53 cp
a
8:37
N Re = 928 ð10:5Þð8:37Þð3:826Þ = 5,887 > 2,100, turbulent flow
53
0:75 1:75 0:25
Δp f = ð10:5Þ ð8:37Þ ð35Þ ð9,500 + 479Þ = 605 psi
1:25
1,800ð3:826Þ
Inside the drill collar:
v = 300 = 24:2 ft/s
2
2:448ð2:25Þ
μ = 35 + 6:66ð6Þð2:25Þ = 39 cp
a
24:2
N Re = 928 ð10:5Þð24:2Þð2:25Þ = 13,604 > 2,100, turbulent flow
39
0:75 1:75 0:25
Δp f = ð10:5Þ ð24:2Þ ð35Þ ð450Þ = 340 psi
1:25
1,800ð2:25Þ
(Continued )