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Gas Compressors 143

E p = polytropic efficiency (0.7 ∼ 0.9), fraction
p o = gas pressure at outlet condition, psia or N.m 2
It is important to know that the efficiency drops significantly with
compression ratio. In field practice, the pressure ratio seldom exceeds 4.
When the compression ratio is greater than 4, the compression is broken
into multiple stages, with the compression ratio in each stage being less
than 4. In gas drilling operations where the required gas pressure is on
the order of a few hundred psia, three-stage compression is often ade-
quate for the primary compressors. More stages of compression are used
in the secondary compressors (boosters).

Illustrative Example 6.5
For the air drilling conditions specified in Illustrative Example 6.4, determine
the required compressor horsepower. Assume that the efficiency of the com-
pressor is 0.71.

Solution
The dry air requirement is calculated as

Q a = 0:0283ð75 + 460Þ ð1615Þ = 1,789 ft /min
3
13:665
The water vapor saturation pressure can be estimated as

1750:286
6:39416−
p w = 10 217:23 + 0:555ð75Þ = 0:4291 psia
The humid air requirement is calculated to be

Q h = 13:665 ð1,789Þ = 1,833 ft /min
3
13:665 − ð0:95Þð0:8Þð0:4291Þ
The overall compression ratio is
p o 136
= = 9:92 > 4
p i 13:665
If two stages of compression are used:
1/2
= 9:92 = 3:15 < 4OK
p o p ffiffiffiffiffiffiffiffiffi
p i
Equation (6.50) gives a required compressor horsepower of
1:25−1
" #
136
HP c = ð2Þð1:25Þð13:665Þð1,833Þ ð2Þð1:25Þ − 1 = 397 hp
13:665
229:17ð1:25 − 1Þð0:71Þ