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Unformatted text preview: YY*) Area 0.0103
0.0103
0.0207
0.0206
0.0206
0.0413
0.0412
0.0413
0.0336 74.9
42.5
27.5
17.8
13.2
9.88
7.43
6.18
5.50 0.772
0.438
0.569
0.367
0.272
0.408
0.306
0.256
0.185 1
versus Y. The area under the curve is the value of the
(Y − Y *)
integral for NOG . In the above table, the trapezoidal method is used to break the curve into 9
segments of ∆Y each, varying in width from 0.0103 to 0.0413. For each segment (width), the
average value of 1/(YY*) is listed and the product = Area = ∆Y times 1/(YY*) is given in the
last column. The sum of these areas is NOG = 3.573. The following is a plot of Exercise 6.36
Subject: Absorption of ammonia from air into water with a packed column.
Given: Operating conditions in Example 6.15. Entering gas is 100 lbmol/h containing 40
mol% NH3. Absorbent is 488 lbmol/h of water. Absorb 95% of the NH3. Equilibrium data in
Fig. 6.44. Column operates at 1 atm and 25oC.
Assumptions: No stripping of water. No absorption of air. Liquid has properties of water.
Find: Column diameter and height if packing is 1.5inch metal Pall rings, using conditions at
bottom of tower.
Analysis: From Example 6.15, the conditions at the bottom of the tower are:
Entering gas: 60 lbmol/h of air and 40 lbmol/h of NH3
Exiting liquid: 488 lbmol/h of water and 38 lbmol/h of NH3
Operating temperature is 298 K (536oF) and pressure is 1 atm
To compute column diameter use Fig. 6.36 to obtain flooding gas velocity, assume operation at
50% of flooding to obtain operating gas velocity, and use continuity equation to obtain column
cross sectional area and column diameter.
Average gas molecular weight = [(60)(29) + (40)(17)]/100 = 24.2
Gas density = ρV = PM/RT = (1)(24.2)/(0.7302)(536) = 0.0618 lb/ft3
From the abscissa of Fig. 6.36(a),
LM L ρV
X=
VM V ρ L
From Fig. 6.36(a), Y = 0.105 = 1/ 2 = (526)(17.9) 0.0618
(100)(24.2) 62.4 1/ 2 = 0.123 2
uo FP ρV
f {ρ L } f {µ L }
g ρH2 O (1) For 1.5inch (35 mm) metal Pall rings, from Table 6.8, FP = 40 ft2/ft3
At 25oC, from Figs. 6.36(b) and (c), f{ρL} = 1.0 and f{µL} = 0.98
From a rearrangement of Eq. (1),
2
uo = Yg ρH2 O
FP ρV 1
(0.105)(32.2) 62.4
1
=
= 87 ft2/s2
f {ρ L } f {µ L }
40
0.0618 (1)(0.98) Then, uo = 9.3 ft/s
Take the operating superficial velocity = uV = fuo = 0.5 uo = 0.5(9.3) = 4.65 ft/s
From Eq. (699),
4GM G
DT =
fuo πρG 1/ 2 4(100 / 3, 600)(24.2)
=
(0.5)(9.3)(3.14)(0.0618) 1/ 2 = 1.73 ft Analysis: (continued) Exercise 6.36 (continued) (2)
From Eq. (6127), column height = lT = HOGNOG
From Example 6.15, NOG = 3.46 and from above, column cross section = S = πD2/4 = 2.35 ft2.
Masstransfer data for 2inch metal Pall rings for the absorption of NH3 from air into water are
presented in Figs. 6.42 and 6.43. In the solutions to Exercise 6.34, these data are fitted to the
(3)
equation,
kGa = 43.4 F0.7 uL0.45 , s1
0.5
0.5
1/2 1
where the gas capacity factor = F = uV(ρV) = f uo (ρV)
in kg s m1/2 and from the
continuity equation, uL = m/SρL in m/s or m3/m2s
For our case, F = (0.5)(9.3)(0.0618)0....
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 Spring '11
 Levicky
 The Land

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