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Gas Compressors 137

Miska (1984) presents a formula to correct air humidity. If the water-
separating efficiency of compressors is considered, Miska’s formula can be
modified to

Q h = p a Q a (6.48)
p a – f w Φp w
where
3
3
Q h = volumetric flow rate of humid air, ft /min or m /min
f w = water separation efficiency, fraction
Φ = relative humidity, fraction
p w = water vapor saturation pressure, psia or kPa

The water vapor saturation pressure can be estimated based on temperature
from the following formula (Miska, 1984):
6:39416− 1750:286
p w = 10 217:23 + 0:555t (6.49)
where

t = temperature, °F

Illustrative Example 6.3
A well is to be drilled with air at an elevation of 4,000 ft. The site air relative
humidity is 0.8 at a temperature of 85°F. The compressor’s dewatering effi-
ciency is 95%. The minimum required air injection rate is estimated to be
2,485 scf/min to carry up the cuttings. Determine the minimum required com-
pressor capacity (in situ air flow rate) for the operation.
Solution
Based on Table 6.1, the site pressure is estimated to be 12.685 psia. The dry
air requirement is calculated as

Q a = ð0:0283Þð85 + 460Þ ð2,485Þ = 3,021 ft /min
3
ð12:685Þ
The water saturation pressure is
1750:286
6:39416
p w = 10 217:23 + ð0:555Þð85Þ = 0:5949 psia
The humid air requirement is calculated to be
3
Q h = 12:685 ð3,021Þ = 3,133 ft /min
12:685 − ð0:95Þð0:8Þð0:5949Þ