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Gas Drilling Operations 155

Substituting Eqs. (7.3) and (7.4) into Eq. (7.2) yields
2
S g TQ g
−5
E k = 4:49 × 10 (7.5)
2
A p
The gas pressure p depends on the hole configuration and the gas
injection rate, as shown in Eqs. (6.18), (6.26), and (6.29). For vertical bore-
holes, setting the left-hand side of Eq. (7.5) to be equal to E km and combin-
ing it with Eq. (6.18) give the minimum gas injection rate required by water

Illustrative Example 7.1
For the following well conditions, predict the minimum required air injection
rate for water removal:
Well geometry
Total measured depth: 6,000 ft
Bit diameter: 4.75 in
Drill pipe OD: 2.375 in
Material properties
Specific gravity of rock: 2.75 (water = 1)
Specific gravity of gas: 1 (air = 1)
Gas specific heat ratio: 1.25 (water = 1)
Specific gravity of oil: 1 (water = 1)
Specific gravity of water: 1.07 in
Pipe roughness: 0.0018 in
Borehole roughness: 0.1
Environment
Site elevation: 0 ft
Ambient pressure: 14.7 psia
Ambient temperature: 60°F fraction
Relative humidity: 0°F/ft
Geothermal gradient: 0.01
Operating conditions
Surface choke/flow line pressure: 14.7 psia
Rate of penetration: 60 ft/hour
Rotary speed: 50 rpm
Oil influx rate: 0 bbl/hour
Water influx rate: 8 bbl/hour
Interfacial tension: 60 dynes/cm

Solution
This problem can be solved using the chart in Figure B.10. The answer is
1,200 scf/min.