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Unformatted text preview: ripping of water.
Find: (a) Minimum water rate.
(b) Column diameter and height for water rate of 2 times minimum and 50% of flooding
gas velocity if 38mm ceramic Berl saddles are used.
(c) Same as Part (b) for 50mm Pall rings.
(d) Packing recommendation.
Analysis: First compute the ammonia material balance.
Flow rate of NH3free air = 2,000/29 = 69 lbmol/h
Flow rate of NH3 in the entering gas = 12(69)/(76012) = 1.107 lbmol/h
Flow rate of NH3 in the exiting liquid = 0.996(1.107) = 1.1026 lbmol/h
Flow rate of NH3 in the leaving gas = 1.107  1.1026 = 0.0044 lbmol/h
(a) Operation in dilute region, where we can apply Henry's law. From Fig. 6.49, using results
from Exercise 6.33, when x=0.0108, y=0.007. Therefore, the equilibrium equation is y = 0.65x.
Therefore, K = 0.65. Entering gas rate = V = 69 + 1.107 = 70.1 lbmol/h
From Eq. (611), Lmin = VK(fraction absorbed) = 45.4 lbmol/h
(b) L = 2Lmin = 2(45.4) = 90.8 lbmol/h
Compute column diameter for 50% of flooding, using Fig. 6.36.
Take properties of gas and liquid as those of air and water, respectively.
From the ideal gas law, ρG = PM/RT = (1)(29)/(0.7302)(528) = 0.0752 lb/ft3
LM L ρV
In Fig. 6.36(a), X =
VM V ρ L 1/ 2 (45.4)(18) 0.0752
=
(70.1)(29) 62.4 1/ 2 = 0.014 Take f{ρL} and f{µL} =1.0. From Fig. 6.36(a), at flooding, Y = 2
uo FP ρV
= 0.22
g ρH 2O (1) For 38mm Berl saddles, the packing factor, FP, is not included in Table 6.8. However, it can be
found in Perry's Chemical Engineers' Handbook, which gives FP = 65 ft2/ft3.
Yg ρH2 O
(0.22)(32.2) 62.4
2
Solving Eq. (1), uo =
=
= 90.4
FP ρV
65
0.0752
Therefore, uo = 9.5 ft/s. The fraction of flooding = f = 0.5.
From Eq. (6103), the column diameter is,
4VM V
DT =
fuo πρV 1/ 2 4(70.1/ 3, 600)(29)
=
(0.5)(9.5)(3.14)(0.0752) 1/ 2 = 1.42 ft and S=πDT2/4 =3.14(1.42)2/4=1.58 ft2 Exercise 6.34 (continued) Analysis: (b) (continued)
For column height, use Eq. (689), because masstransfer is controlled by the gas phase.
For NOG , use Eq. (688) with xin = 0. Absorption factor = A = L/KV = (90.8)/(0.65)(70.1) = 2
From Eq. (693), using yin = 1.107/70.1 = 0.0158, yout = 0.0044/69 = 0.0000638
N OG = ln ( A − 1) / A ( yin ) / ( yout ) + (1 / A)
( A − 1) / A = ln (2 − 1) / 2 0.0158 / 0.0000638 + (1 / 2)
(2 − 1) / 2 = 9.65 For HOG, literature data can be used, as found on page 1638 of Perry's Chemical Engineers'
Handbook, 6th edition or on page 663 of "EquilibriumStage Separation Operations in Chemical
Engineering" by Henley and Seader, as shown below.
To use this plot, L" = 90.9(18)/1.58 = 1,040 lb/hft2 and V”=G" = 70.1(29)/1.58 = 1,290 lb/hft2
Then, HOG = 1.9 ft and column height = lT = NOGHOG = (9.65)(1.9) = 18.3 ft.
AirAmmoniaWater system with 1.5inch Berl Saddles, AE units Exercise 6.34 (continued)
Analysis: (c) (continued)
(c) Assume the 50mm Pall rings are metal. From Table 6.8, FP = 27 ft2/ft3.
2
Solving Eq. (1), uo = Yg ρH 2O
(0.22)(32.2) 62.4
=
= 235
FP ρV
27
0.0752 Therefore, uo = 15.3 ft/s. The fraction of flooding = f = 0.5.
From Eq. (6103), the column d...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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