Unformatted text preview: on has properties of
water. Initial estimate of column diameter is 30 inches. Try 2inch Intalox saddles packing.
Find: (a)
(b)
(c)
(d)
(e)
(f)
(g) Minimum caustic solutiontoair molar flowrate ratio.
Maximum possible concentration of CO2 in caustic solution.
Nt at L/V = 1.4 times minimum.
Caustic solution rate.
Pressure drop per ft of column packed height.
NOG
Packed height for KGa = 2.5 lbmol/hft3atm. Analysis: Gas flow rate = 5,000(60)/379 = 792 lbmol/h.
Therefore, air in entering gas = V' = 0.97(792) = 768 lbmol/h
The CO2 in the entering gas = 0.03(792) = 24 lbmol/h
The CO2 in the exiting gas = 0.03(24) = 0.72 lbmol/h
The CO2 absorbed in the leaving liquid = 24  0.72 = 23.28 lbmol/h
The mole ratios are Yin = 24/768 = 0.03125, Yout = 0.72/768 = 0.0009375, Xin = 0.0 (a) From Eq. (611), L'min = V'K(fraction absorbed) = 768(1.75)(0.97) = 1,304 lbmol/h
(b) The maximum possible CO2 concentration in the caustic solution occurs at infinite stages
with the minimum liquid rate. The leaving liquid is in equilibrium with the entering gas.
Therefore,
Xout = Yin/K = 0.03125/1.75 = 0.0179 mol CO2 / mol caustic solution.
(c) Let L/V = L'/V' . Then, L = 1.4 (1,304) = 1,826 lbmol/h and L/G = 1,826/768 = 2.38.
For this liquid flow rate, a material balance for CO2 is,
768(0.03125  0.0009375) = 23.28 = 1,826(Xout  0.0). Solving, Xout = 0.01275
In the YX plot on the next page, the equilibrium line, Y = 1.75X, and the straight operating line,
passing through the column YX end points {Yin = 0.03125, Xout = 0.01275} and {Yout =
0.0009375, Xin = 0.0} are shown, with the equilibrium stages stepped off, giving Nt = 7.3.
(d) From Part (c), caustic rate = 1,826 lbmol/h.
(e) Assume the use of 2inch of ceramic Intalox saddles. From Table 6.8, FP = 40 ft2/ft3.
Use Fig. 6.36(a). Assume for the dilute caustic solution, f{ρL} = 1.0 and f{µL} = 1.0.
Take MV = 29, ML = 18, ρL = 62.4 lb/ft3, ρV = 29/379 = 0.0765 lb/ft3
Using conditions at the bottom of the column, Exercise 6.31 (conditions)
Analysis: (d) and (e) (continued) LM L ρV
X=
VM V ρ L 1/ 2 (1,849)(18) 0.0765
=
(792)(29)
62.4 1/ 2 = 0.051 From Fig. 6.36(a), at flooding, Y = 0.18
Now compute Y for the suggested column diameter of 30 inches = 2.5 ft.
From the continuity equation, uo = m/SρV = [(792)(29)/3600]/[3.14(2.5)2/4](0.0765) = 17 ft/s
From Table 6.8, the packing factor, FP = 40 ft2/ft3.
2
uo FP ρV
17 2 (40) 0.0765
Y=
=
= 0.440 , which is much greater than 0.18
g ρH 2O
32.2
62.4 Therefore, a diameter of 30 inches places the operation badly into the flooding region and a
pressure drop calculation is meaningless. A larger diameter is necessary. Exercise 6.31 (continued)
(f) From Eq. (695), with A = L/KV = (1,826)/1.75(768) = 1.36,
NOG = Nt = A ln(1/A)/(1 A) = 7.3(1.36) ln (1/1.36)/(1  1.36) = 8.5
(g) From Table 6.7, HOG = V/KGaPS, with V = 792 lbmol/h,
KGa = 2.5 lbmol/hft3atm, P = 1 atm
For column cross sectional area, S, assume 50% of flooding. At flooding, with Y = 0.18,
by ratio with the Y of 0.440 for a 2.5 ft...
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 Spring '11
 Levicky
 The Land

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