Geoscience Reference
In-Depth Information
N
OG
= ln(
y
1
/
y
2
)
where
y
1
= Concentration of inlet gas.
y
2
= Concentration of outlet gas.
Use the peak value for inlet gas concentration:
N
OG
= ln(
y
1
/
y
2
) = ln(500/25) = 3.0
Determine the total number of transfer units provided by a tower with six spray sections. Remember
that each lower spray has only 60% of the efficiency of the section above it (due to a back mixing of
liquids and gases from adjacent sections). Spray section
N
OG
values are derived accordingly:
Top spray
N
OG
= 0.7 (given)
2nd spray
N
OG
= 0.7 × 0.6 = 0.42
3rd spray
N
OG
= 0.42 × 0.6 = 0.252
4th spray
N
OG
= 0.252 × 0.6 = 0.1512
5th spray
N
OG
= 0.1512 × 0.6 = 0.0907
Inlet
N
OG
= 0.5 (given)
To t a l
N
OG
= 0.7 + 0.42 + 0.252 + 0.1512 + 0.0907 + 0.5 = 2.114
This value is below the required value of 3.0. Now calculate the outlet concentration of gas:
N
OG
= ln(
y
1
/
y
2
)
y
1
/
y
2
= exp(
N
OG
) = exp(2.114) = 8.28
y
2
= 500/8.28 = 60.4 ppm
Does the spray tower meet the HCl regulation? Because
y
2
is greater than the required emission
limit of 25 ppm, the spray unit is not satisfactory.
18.4.3 p
aCKed
t
oWer
C
alCulations
■
EXAMPLE 18.9
Problem:
Pollution Unlimited, Inc., has submitted plans for a packed ammonia scrubber on an air
stream containing NH
3
. The operating and design data are given by Pollution Unlimited, Inc. We
remember approving plans for a nearly identical scrubber for Pollution Unlimited, Inc., in 1978.
After consulting our old files we find all the conditions were identical except for the gas flow rate.
What is our recommendation (USEPA, 1984b, p. 102)?
Given:
Tower diameter = 3.57 ft
Packed height of column = 8 ft
Gas and liquid temperature = 75°F
Operating pressure = 1.0 atm
Ammonia-free liquid flow rate (inlet) = 1000 lb/ft
2
-hr
Gas flow rate = 1575 acfm
Gas flow rate in the 1978 plan = 1121 acfm
Inlet NH
3
gas composition = 2.0 mol%
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