Page 412 - Air and Gas Drilling Manual

9-16 Air and Gas Drilling Manual
Substituting the above values for, D hy, V m1,
Reynolds number for the volumetric flow rate derived from the laminar flow
terminal velocity equation. Equation 9-6 gives
0
(.293 ) ( .770 ) e1 into Equation 9-6 yields the
1
N R1
. 0 0002691
N R1 1 928
,
The Reynolds number calculated above is below 2,000 and, thus, the volumetric
flow rate of 190.8 gal/min produces laminar flow conditions in largest cross-section
of the annulus. Therefore, the use of Equation 9-3a and the actual flow conditions it
yielded are consistent.
Since the incompressible drilling fluid in this illustrative example is a drilling
mud with plastic fluid properties, turbulent flow conditions are assumed to exist at
Reynolds numbers greater than 2,000 (i.e., the transition flow condition Equation 9-
4a is not needed for this example). In order to complete the analysis to determine
the minimum volumetric flow rate for the incompressible fluid, turbulent flow
conditions must also be assessed.
Substituting the above values for, s , m, and D c into Equation 9-5a yields the
terminal velocity for turbulent flow conditions. Equation 9-5a gives
1
168 5 . 75 0 . 2
.
.
V t3 5 35 0 0165
75 0 .
.
V 0 767 ft/sec
t3
Using Equation 9-1, the total velocity of the fluid is
.
.
V m3 0 417 0 767
.
V 1 184 ft/sec
m3
The volumetric flow rate in the annulus section is
.
0 240) (
Q a3 (. 1 184)
3
.
Q a3 0 284 ft /sec
In field units, the above is
3
.
0 284 ( 60) 12
q
a3
231