Page 335 - Air and Gas Drilling Manual

8-18 Air and Gas Drilling Manual
volumetric flow rate. Compressor units are selected to give a total volumetric flow
rate that exceeds the determined minimum volumetric flow rate.
Figure E-13 gives the minimum volumetric flow rates of air for a 7 7/8 inch
borehole with a 4 1/2 inch drill pipe. The flow rate is determined for the maximum
depth of the well which is 10,000 ft. The annulus around the drill collars at the
bottom of the drill string will have higher air velocities than in the annulus around
the drill pipe. This means the lowest velocity in the annulus is at the bottom of the
drill pipe. Therefore, by making the assumption that the borehole has no drill
collars, a conservative estimate is made of the approximate minimum volumetric
flow rate for the borehole. Therefore, Figure E-13 can be used to determine the
approximate minimum volumetric flow rate of air for this example drilling operation.
Using Figure E-13, the approximate minimum volumetric flow rate for drilling
at a depth of 10,000 ft is approximately 1,588.8 scfm (also found in Illustrative
Example 5.1). This is the minimum flow rate for sea level (API standard
conditions) air. But the drilling location is at 4,000 ft above sea level and the
drilling operation actual atmospheric temperature is 60˚ F. Thus, the above
minimum volumetric flow rate must be adjusted for the atmospheric conditions that
exist at the drilling location (i.e., to obtain the actual cubic feet per minute, acfm).
In Illustrative Example 8.1 the specific weight of air at API standard conditions was
3
found to be 0.0763 lb/ft . Table 4-1 gives an average atmospheric pressure of
12.685 psia for a surface location of 4,000 ft above sea level (mid latitudes North
America) (also see Appendix D). The actual atmospheric pressure for the air at the
drilling location (that will be utilized by the compressor), P at, is
.
p = 12 685 psia
at
P at = p 144
at
2
,
P at = 1 827 lb/ft abs
The actual atmospheric temperature of the air at the drilling location, T at, (that will
be used by the compressor) is
o
t at = 60 F
.
T at = t at + 459 67
o
.
T at = 519 67 R
Thus, P g and T g become
P = P at = 1 827 lb/ft abs
2
,
g
T = T at = 519 67 R
o
.
g