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x
1
* = Outlet concentration.
y
1
= Inlet gas mole fraction.
m
= Henry's law constant.
L
m
= Liquid molar flow rate (lb-mol/ft
2
-hr).
G
m
= Gas molar flow rate (lb-mol/ft
2
-hr).
x
1
* =
y
1
/
m
= (0.02)/(1.20) = 0.0167
Calculate
y
2
for 90% removal. Because it is required to remove 90% of NH
3
, there will be 10% of
NH
3
remaining in the outlet gas stream; therefore, by material balance,
y
2
= (0.1)(
y
1
)/[(1 -
y
1
) + (0.1)(
y
1
)]
where
y
1
= Inlet gas mole fraction.
y
2
= Outlet gas mole fraction.
y
2
= (0.1)(
y
1
)/[(1 -
y
1
) + (0.1)(
y
1
)] = (0.1)(0.02)/[(1 - 0.02) + (0.1)(0.02)] = 0.00204
Determine the minimum ratio of molar liquid flow rate to molar gas flow rate (
L
m
/
G
m
)
min
by a mate-
rial balance. Material balance around the packed column is calculated as
G
m
(
y
1
-
y
2
) =
L
m
(
x
1
* -
x
2
)
(
L
m
/
G
m
)
min
= (
y
1
-
y
2
)/(
x
1
* -
x
2
)
where
L
m
= Liquid molar flow rate (lb mol/ft
2
-hr).
G
m
= Gas molar flow rate (lb mol/ft
2
-hr).
y
1
= Inlet gas mole fraction.
y
2
= Outlet gas mole fraction.
x
1
* = Outlet concentration.
x
2
= Inlet liquid mole fraction.
(
L
m
/
G
m
)
min
= (
y
1
-
y
2
)/(
x
1
* -
x
2
) = (0.02 - 0.00204)/(0.0167 - 0) = 1.08
Calculate the actual ratio of molar liquid flow rate to molar gas flow rate (
L
m
/
G
m
). Remember that
the actual liquid flow rate is 25% more than the minimum based on the given operating conditions:
(
L
m
/
G
m
) = 1.25(
L
m
/
G
m
)
min
= 1.25 × 1.08 = 1.35
Calculate the value of (
y
1
-
mx
1
)/(
y
2
-
mx
2
), where
y
1
= Inlet gas mole fraction.
m
= Henry's law constant.
x
1
= Outlet liquid mole fraction.
y
2
= Outlet gas mole fraction.
x
2
= Inlet liquid mole fraction.
(
y
1
-
mx
1
)/(
y
2
-
mx
2
) = [(0.02) - (1.2)(0)]/[(0.00204) - (1.2)(0)] = 9.80
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