Unformatted text preview: (1.59)(0.0143)(0.148)(0.957)
KW =
= 3, 287
(1 − ε ) µV
(1 − 0.919 ) [(0.018)(0.000672)] From Eq. (6113), Ψ o = C p 64
1.8
64
1.8
+ 0.08 = 0.698
+
= 0.671
3, 287 3, 287 0.08
N ReV N ReV From Eq. (2), with the introduction of gc because of the use of American
Engineering units, ∆Po
a u2ρ 1
33.9 1.592 (0.148)
1
= Ψo 3 V V
= 0.671
= 0.164 lbf/ft3 or 0.00114 psi/ft
3
lT
ε 2 g c KW
0.919
2(32.2)
0.957
From Eq. (697), hL = 12 N FrL
N Re L 1/ 3 ah
a 2/3 Superficial liquid velocity = uL = m/SρL = (5,000/60)/(5.41)(62.4) = 0.247 ft/s
From Perry's Handbook, liquid viscosity = 0.95 cP
2
u L a (0.247) 2 (33.9)
From Eq. (699), N FrL =
=
= 0.064
g
32.2
uρ
(0.247)(62.4)
From Eq. (698), N Re L = L L =
= 712
aµ L (33.9)(0.95)(0.000672)
a
0.
0
From Eq. (6101), h = 0.85Ch N Re25 N Fr.1 = 0.85(0.593)(712) 0.25 (0.247) 0.1 = 2.27
L
L
a (3) Exercise 6.28 (continued)
Analysis: (d) (continued)
0.064
From Eq. (3), hL = 12
712 1/ 3 2.27 2 / 3 = 0.177 m3/m3 or ft3/ft3 From Eq. (1),
∆P
ε
=
∆Po
ε − hL 3/ 2 exp 1/ 2
13300
N
3/ 2 ( FrL )
a ∆P
0.919
= ( 0.00114 )
lT
0.919 − 0.103 3/ 2 exp 13300
1/ 2
0.064 )
= 0.0241 psi/ft or 0.669 in. H2O/ft
3/ 2 (
111.1 This is only 10% higher than the suggested value of 0.612 in. H2O/ft Exercise 6.29
Subject: Masstransfer coefficients for conditions of Exercise 6.26.
Given: Suggested masstransfer coefficients in Exercise 6.26.
Find: Masstransfer coefficients kLa and kGa from BilletSchultes correlation.
Analysis: Coefficient kLa is obtained from a rearrangement of the leftmost part of Eq. (6131),
uL
kL a =
(1)
H L ( aPh / a )
with HL from Eq. (6132) and (aPh /a) from (6136).
From Table 6.8, CL = 1.239 , ε = 0.919, and a = 111.1 m2/m3. From the results of Exercise 6.28,
come the following parameters and properties:
hL = 0.177 ft2/ft3, uL = 0.247 ft/s, ρL = 62.4 lb/ft3, µL = 0.95 cP
Take the surface tension, σ = 0.005 lbf/ft or 0.005(32.2) = 0.161 lb/s2
To compute (aPh /a), we need for the liquid phase, the Reynolds, Weber, and Froude numbers
from equations (6138), (6139), and (6140), respectively, based on the packing hydraulic
diameter from (6137).
d h = packing hydraulic diameter = 4
From Exercise 6.28,
Reynolds number = N ReL ,h = ε
0.919
=4
= 0.0331 m = 0.108 ft
a
111.1 uL d h ρ L (0.247)(0.108)(62.4)
=
= 2, 607
[(0.95)(0.000672)]
µL Weber number = N WeL ,h ( 0.247 ) (62.4)(0.108) = 2.55
u 2ρ d
= L L h=
σ
0.161 Froude number = N FrL , h ( 0.247 ) = 0.0175
u2
= L=
gd h 32.2(0.108) 2 2 From (6136), ( aPh
−1/ 2
= 1.5 ( ad h )
N Re L ,h
a ) (N
−0.2 = 1.5 [ (111.1)(0.0331) ] −1/ 2 ) (N )
0.75 We L , h −0.45 FrL , h ( 2607 ) ( 2.55 ) ( 0.0175 )
−0.2 0.75 −0.45 = 2.02 Exercise 6.29 (continued)
Analysis: (continued)
From Exercise 6.23, correcting for temperature, DL = 9.6 x 106(298/294) = 9.73 x 106 cm2/s or
1.05 x 108 ft2/s
From (6132), using American Engineering Units, 11
HL =
CL 12 1/ 6 1
1
=
1.239 12 From (1), k L a = 4hL ε
DL au L
1/ 6 1/ 2 uL a
a a ph 4(0.177)(0.919)
(1.05 × 10−8 )(33.9)(0...
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 Spring '11
 Levicky
 The Land

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