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Unformatted text preview: from the value above for the top of the column,
DV = 0.052(382/354)1.75 = 0.059 cm2/s
Use the following additional properties and parameters, together with µV = 0.0092 cP and ρV =
0.179 lb/ft3 , where from Eq. (6134), N ReV =
From (6135),) NScV = uV ρV
(5.9)(0.181)
=
= 6,150
aµV (28.1)(0.0092)(0.000672) µV
(0.0092)(0.000672)
=
= 0.54
ρV DV
0.059
(0.181)
(2.54) 2 (12) 2 From Eq. (6133),
1
1/ 2 4ε
HG =
( ε − hL )
CV
a4 1/ 2 (N ) (N )
−3/ 4 ReV −1/ 3 ScV 1
1/ 2 4(0.977)
=
( 0.977 − 0.0412 )
0.408
92.34 1/ 2 uV a
DV aPh ( 6,150 ) −3/ 4 ( 0.54 ) −1/ 3 (1.80)(92.3)
= 0.14 m
(0.059 ×10 −4 )(191) At the bottom of the column, vapor rate = V =437 lbmol/h and liquid rate = L = 612 lbmol/h.
Therefore, V/L = (437)/(612) = 0.714
mV
From Eq. (7.51), HOG = HG +
HL
L
Therefore, values of HOG depend on the slope of the equilibrium line, which varies widely from
the top of the column to the bottom. At the top, assuming HG and HL are constant,
mV
H L = 0.23 + 1.63(0.154) m = 0.23 + 0.25 m
L
At the bottom, assuming HG and HL are constant,
H OG = H G + H OG = H G + mV
H L = 0.14 + 0.714(0.126) m = 0.14 + 0.09 m
L (d) The packed height is given by Eq. (6127), lT = HOGNOG = HETP Nt
Apply these equations in a stagebystage fashion, using the equilibrium stages in Fig. 7.15. For
each stage, convert the value of HOG to HETP with Eq. (7.53) as in Table 7.7 for Example 7.9,
HETP = HOG lnλ/(λ1) Analysis: (d) (continued) Exercise 7.53 (continued) Using a spreadsheet, the following results are obtained: Rectifying Section:
Stage
m
1
0.47
2
0.53
3
0.61
4
0.67
5
0.72
6
0.80
Total height = λ = mV/L HOG , m
0.77
0.35
0.86
0.36
0.99
0.38
1.09
0.40
1.17
0.41
1.30
0.43
2.31 m HETP, m
0.39
0.39
0.38
0.38
0.38
0.38 Stage
m
7
0.90
8
0.98
9
1.15
10
1.40
11
1.70
12
1.90
0.2 of 2.20
13
Total height = λ = mV/L
0.64
0.70
0.82
1.00
1.21
1.36
1.57 HETP, m
0.27
0.27
0.27
0.27
0.27
0.27
0.2*0.27
= 0.05 Stripping Section: HOG , m
0.22
0.23
0.24
0.27
0.29
0.31
0.34 1.67 m Note that although HOG varies significantly with stage in both sections, HETP does not.
The total packed height = 2.31 + 1.67 = 3.98 m or 13.1 ft.
(e) Compute pressure drops for the packed sections from the correlation of Billet and Schultes.
The pressure drop per unit height of packed bed is given by Eq. (6115),
1.5 1/ 2
∆Po ε − hL
∆P
13300
=
exp
( N FrL )
lT
lT
ε
a 3/ 2
This equation will be used in each section. ∆Po
a u 2ρ 1
= Ψo 3 o G
lT
ε 2 gc KW
The required packing parameters are as follows from Table 6.8 and above:
FP = 16 ft2/ft3, a = 92.3 m2/m3, ε = 0.977 m3/m3, Ch = 0.876, and Cp = 0.421
From Eq. (6110), (1) (2) Exercise 7.53 (continued) Analysis: (e) (continued)
1− ε
1 − 0.977
From Eq. (6112), DP = 6
=6
= 0.00150 m or 0.0049 ft
a
92.3
Rectifying section of the column:
Column diameter = DT = 3.8 ft
1
2 1 DP
2
1
0.0049
From Eq. (6111),
= 1+
= 1+
= 1037
.
KW
3 1 − ε DT
3 1 − 0.977
3.8
Therefore, KW = 1/1.037 = 0.964
Superficial gas velocity at 70% of flooding fro...
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 Spring '11
 Levicky
 The Land

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