Mud Clinging Constant

A well is experiencing lost circulation problems at the bottom of the casing string. A liner will be run into the well. If the liner is lowered at a maximum rate of 93 ft/min, will the surge pressures exceed the fracture gradient? Use the Bingham model and assume laminar annular How. In addition, assume the peak pipe velocity is the same as the average value.

casing depth casing ID open hole depth open hole OD liner size liner length drillpipe

  • 10,000 ft
  • 8.5 in.
  • 13,000 ft
  • 8.5 in.
  • 7 in. (flush joint) = 3,600 ft
  • 4.5 in.

mud = 16.6 lb/gal

  • 38 cp (PV)
  • 15 lb/100 ft- (YP) fracture gradient = 17.0 lb/gal pipe velocity = 93 ft/min = 1.55 ft/sec
  • Assume the liner has a closed end from a float shoe.) Solution:
  1. The maximum surge pressures occur when the bottom of the liner reaches the casing seat (sec Fig. 18-15).
  2. The mud How rate leaving the well when the liner shoe reaches the casing seat is:

3. Compute the annular velocities around the drillpipe (Va„) and the liner tvj

76.8

2.448 (8.5: - 7-)

  • 1.349 ft/sec
  • 80.9 ft/min

4,5-in. drillpipe

Mud 16.6 lb/gal

6.400 ft

Fracture gradient

7-in. liner

10,000 ft

93 ft/min

13,000 ft

  1. 18-15 Illustration for Example 18.10
  2. Refer to Fig. I 8-14 and determine tiie clinging constant, k, for the pipe and liner:

k ^ 0.38 liner - ratio " 7/8.5 - 0.823 k = 0.45

5. The effective annular velocities around the pipe (V^) and the liner (V,c) are:

- 0.603 - 0.38 (-1.55) = 1.192 ft/sec Vlc - 1.349 ft/sec - (0.45)( - 1.55) = 2.046 ft/sec

6, The pressure surge caused by the drillpipe is as follows (use tiq. 18.27):

p 1,500 (dH- - dp2) 225 (dH - d„) (38)(6,400X1.192) ( I5)<6,400)

1,500 (8.52 - 4,5-) 225 ( 8.5 - 4.5) = 3.716 psi + 106 psi = 110 psi

7. The pressure surge caused by the liner is:

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