Page 129 - Air and gas Drilling Field Guide 3rd Edition

120 CHAPTER 5 Compressors and Nitrogen Generators

The actual shaft power _ W as is actual power needed to compress the air to a pres-
sure of 150 psig. Equation (5-39) becomes
131200
_ W as ¼ ¼ 147864 watts:
ð0:95Þð0:934Þ
The above determined 147.9 kW is the actual shaft power needed by the com-
2
pressor to match the back pressure of 103.5 N/cm abs. Because this power level
is less than the prime mover’s rated input power of 261 kW, this compressor
system is capable of operating at a sea level surface location.
(b) Surface location at 6000 ft above sea level (USCS units)
The volumetric efficiency e v can be determined using Equation (5-39).
n s ¼ 2
p i ¼ 11:769 psia
p o ¼ 150 þ 11:769 ¼ 161:769 psia:
Equation (5-37) becomes
1

161:769 2
r s ¼ ¼ 3:708:
11:769
The specific heat ratio for air is
k ¼ 1:4:

The clearance volume ratio for this compressor is
c ¼ 0:02:
With these values, Equation (5-38) becomes
1
8 39
2
< =
e v ¼ 0:96 1 0:02 ð3:708Þ1:4 1 5
4
: ;
e v ¼ 0:930:
3
The rated volumetric flow rate into the compressor is 950 ft /min. For this
example, the compressor is located at 6000 ft above sea level, thus
q i ¼ 950 acfm:
With these terms the theoretical shaft horsepower required to compress the air
moving through the machine is given by Equation (5-35a). Thus, the theoretical
shaft horsepower is
2 3
ð0:4Þ

ð2Þð1:4Þ ð11:769Þð950Þ 6 161:769 ð2Þð1:4Þ 7
_ W s ¼ 6 1 7
ð0:4Þ 229:17 4 11:769 5
_ W s ¼ 155:1 hp: