Page 144 - Air and Gas Drilling Manual

4-30 Air and Gas Drilling Manual
The actual shaft power to give the above theoretical shaft power is obtained from
Equation 4-39 (where ε is replaced by 1). For this example, Equation 4-39 gives
(with ε
m
=
˙
W = 0.90) 189 .1 v = 210 .1
as
. 090
The above determined 210.1 horsepower is the actual shaft power needed by the
compressor to produce the 240 psig maximum pressure output. This power level is
less than the prime mover’s rated input power of 300 horsepower, thus, this
compressor system is capable of operating at a sea level surface location.
b) Surface location at 6,000 ft above sea level.
This rotary compressor has a total fixed ratio. Therefore, the maximum pressure
output this compressor can achieve is reduced from what the compressor can output
at sea level conditions. This derated maximum pressure output is determined by
multiplying the total fixed ratio across all three stages (obtained in part a above, i.e.,
r st = 17.33) by the new atmospheric pressure condition. Table 4-1 gives the mid
latitudes atmospheric pressure at an elevation of 6,000 ft as
p = 11 769. psia
i
The derated maximum pressure output of the compressor is
.
p o = (17 33 ) (11 769. ) = 204 0. psia
The specific heat ratio for air is

k = 14
.
3
The rated volumetric flow rate into the compressor is 900 ft /min. For this
example the compressor is located at 6,000 ft above sea level, thus,
q = 900 acfm
i
With the above terms the theoretical shaft horsepower required to compress the
air moving through the machine is given by Equation 4-35a. Thus, the theoretical
shaft horsepower is )
. ( ) (11
() 31 4 .769 ) (900 )    204 0 . (04 . )  
˙
31
W = .  ()( 4 −1
s )  
. (04 229 .17   11 .769
   
˙
W = 151 .4
s
For this example