Page 278 - Materials Chemistry, Second Edition

Groundwater Remediation 261

For reciprocating compressors, the efficiencies (E) are generally in the
range of 70% to 90% for isentropic and 50% to 70% for isothermal compres-
sion. The actual horsepower requirement can be found as:

hp
= (6.36)
hp actual
E
Example 6.19: Determine the Power Requirement for Air Sparging
Three air-sparging wells were installed into the contaminant plume of the aquifer
3
described in Example 6.17. The injection air flow rate into each well is 5 ft /min.
A compressor is to serve all three wells. Head loss of the piping system and the
injection head was found to be 1 psi. Using the calculated air injection pressure
from Example 6.18, determine the required horsepower of the compressor.

Solution:
(a) The required injection pressure = the final delivery of the com-
pressor, P 2
= minimum injection pressure + head loss = 4.37 + 1.0
= 5.37 psig
= (5.37 + 14.7) psi = 20.1 psi = (20.1)(144) = 2,890 lb/ft 2

(b) Assuming isothermal expansion, use Equation (6.34) to determine
the theoretical power requirement as:

×
= 3.03 10 −5 PQ ln P 2
hp theoretical 1 1
P 1
2,890
×
−5
= 3.03 10 [(14.7)(144)][(3)(5)]ln = 0.3 hp
(14.7)(144)
Assuming an isothermal efficiency of 60%, the actual horsepower
required is determined by using Equation (6.36):
= hp theoretical = 0.3 = 0.5 hp
hp actual
E 50%

(c) Assuming isothermal expansion, use Equation (6.35) to determine
the theoretical power requirement as:

3.03 × 10 −5 k   P  ( k−1)/ k 
2
= PQ  −  1
hp theoretical 1 1  
k − 1    P   
1
−5
(3.03 × 10 )(1.4)   2,890  (1.4 −1)/1.4 
= [(14.7)(144)][(5.0)(3)]    −  1
1.4 − 1    (14.7)(144)   
= 0.31 hp