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Unformatted text preview: ights are approximate because they are based on values of HOG that are
assumed constant over the packed sections, which is not the case because m varies.
(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),
3/ 2 1/ 2
∆Po ε − hL
∆P
13300
=
exp
( N FrL )
lT
lT
ε
a 3/ 2
This equation will be used for each packing in each section. ∆Po
a u2ρ 1
= Ψo 3 V V
lT
ε 2 g c KW
The required packing parameters are as follows from Table 6.8 and above:
NOR PAC Montz
23
FP , ft /ft
14
33
2
3
a, m /m
86.8
300
0.947
0.930
ε, m3/m3
Ch
0.651
0.482
0.350
0.295
Cp
From Eq. (6110), (1) (2) 1− ε
1 − 0.947
=6
= 0.00366 m or 0.012 ft for NOR PAC
a
86.8
1− ε
1 − 0.930
DP = 6
=6
= 0.0014 m or 0.00459 ft for Montz
a
300 From Eq. (6112), DP = 6 Rectifying section of the column:
For NOR PAC , column diameter = DT = 4.3 ft 1
2 1 DP
2
1
0.012
= 1+
= 1+
= 1035
.
KW
3 1 − ε DT
3 1 − 0.947 4.3
Therefore, KW = 1/1.035 = 0.966 From Eq. (6111), Analysis: (e) (continued) Exercise 7.52 (continued) For Montz, column diameter = DT = 5.3 ft
1
2 1 DP
2
1
0.00459
From Eq. (6111),
= 1+
= 1+
= 1008
.
KW
3 1 − ε DT
3 1 − 0.930
5.3
Therefore, KW = 1/1.008 = 0.992
Superficial gas velocity at 70% of flooding from above = 13 ft/s for NOR PAC.
Superficial gas velocity at 70% of flooding from above = 8.4 ft/s for Montz
Gas viscosity = 0.011 cP and gas density = 0.0688 lb/ft3.
From Eq. (6114), for NOR PAC,
u Dρ
(13)(0.012)(0.0688)
N ReV = V P V KW =
(0.966) = 26,500
(1 − ε ) µV
(1 − 0.947 ) (0.011)(0.000672)
For Montz, N ReV = (8.4)(0.00459)(0.0688)
(0.992) = 5, 090
(1 − 0.930 ) (0.011)(0.000672) From Eq. (6113), for NOR PAC,
Ψo = Cp 64
1.8
64
1.8
+ 0.08 = 0.350
+
= 0.280
N ReV N ReV
26, 500 26,5000.08 For Montz,
Ψo = 0.295 64
18
.
+
= 0.272
5,090 5,0900.08 From Eq. (2), for NOR PAC,
2
∆Po
a uV ρV 1
86.8(0.3048) 132 (0.0688)
1
= Ψo 3
= 0.280
= 1.63 lbf/ft3 or 0.0113 psi/ft
3
lT
ε 2 g c KW
0.947
2(32.2)
0.966
For Montz,
∆Po
300(0.3048) 8.4 2 (0.0688)
1
= 0.272
= 2.35 lbf/ft3 or 0.0163 psi/ft
3
lT
0.930
2(32.2)
0.992
From Eq. (1), for NOR PAC, with hL = 0.0162 m3/m3 and liquid Froude number = 6.63x105 ∆P
0.947 − 0.0162
= 0.0113
lT
0.947 1.5 exp 1/ 2
13300
6.63 ×10 −5 )
= 0.0126 psi/ft
3/ 2 (
86.8 Exercise 7.52 (continued)
Analysis: (e) Rectifying section (continued)
From Eq. (1), for Montz, with hL = 0.0241 m3/m3 and liquid Froude number =9.91x105 ∆P
0.930 − 0.0241
= 0.0163
lT
0.930 1.5 exp 1/ 2
13300
9.91× 10−5 )
= 0.0161 psi/ft
3/ 2 (
300 Stripping section of the column:
For NOR PAC , column diameter = DT = 3.4 ft
1
2 1 DP
2
1
0.012
From Eq. (6111),
= 1+
= 1+
= 1044
.
KW
3 1 − ε DT
3 1 − 0.947 3.4
Therefore, KW = 1/1.044 = 0.957
For Montz, column diameter = DT = 4.2 ft
1
2 1 DP
2
1
0.00459
From Eq. (6102),
= 1+
= 1+
= 1010
.
KW
3 1 − ε DT
3 1 − 0.930
4.2
Therefore, KW = 1/1.010 = 0.990
Superficial gas velocity at 70% of flooding from above = 20 ft/s for NOR PAC.
Superficial ga...
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 Spring '11
 Levicky
 The Land

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