Unformatted text preview: n.
2/3 qL
From Eq. (651), hl = φ e hw + C
Lw φ e 40,000
= 0.72 2.5 + 0.362
385 0.72 2/3 = 9.0 in. of nC8 From Eq. (655), with maximum bubble size of 1/4 inch = 0.00635 m,
hσ = 6σ / gρ L DBmax = 6(20 / 1000) / (9.8)(703)(0.00635) = 0.00274 m = 0.11 in. nC8
From Eq. (2), ht = hd + hl + hσ = 0.045 + 9.0 + 0.11 = 9.2 in. nC8 = 0.23 psi/tray. Excessive!
(c) Apply criterion of Eq. (664). hd + hσ = 0.045 + 0.11 = 0.155 in. < hl = 9.0 in.
Therefore, weeping will occur.
(d) From Fig. 6.28, since FLV = 1.64, entrainment will be very low.
(e) Because entrainment is very low, EMV will not decrease.
(f) From Eqs. (670) and (672), hdf =(ht + hl + hda) / 2 (3) Area of downcomer opening = Ada = Lwha Take ha = 2 in., Ada = (32.1)(2/12) = 5.4 ft2
qL
From Eq. (671), hda = 0.03
100 Ada 2 40, 000
= 0.03
100(5.4) 2 =164 in. Very excessive Exercise 6.21
Subject: Column performance for 40% of flooding.
Given: Data in Examples 6.5, 6.6, and 6.7.
Find: (a) Column diameter in Example 6.5 for f = 0.4.
(b) Vapor pressure drop in Example 6.6 for f = 0.4.
(c) Murphree vaporpoint efficiency in Example 6.7 for f = 0.4
Analysis: (a) Example 6.5:
In this example, a value of f = 0.80 was used, giving DT = 2.65 ft
From Eq. (644), by ratioing values of f,
0.80
DT = 2.65
0.40 1/ 2 = 3.75 ft = 1.15 m (b) Example 6.6:
In this example, a tower diameter of 1 m gives a vapor pressure drop = 0.093 psi/tray,
with a vapor hole velocity of 47.9 ft/s, a weir length of 0.73 m, an active area vapor velocity of
5.99 ft/s, and Ks = 0.265 ft/s.
Vapor hole velocity varies inversely with the square of the column diameter. Therefore,
uo = 14.6(1/1.15)2 = 11.0 m/s = 36 ft/s
From Eq. (650), hd is directly proportional to hole velocity squared. Therefore,
hd = 1.56(36/47.9)2 = 0.88 in. of liquid
Weir length is proportional to column diameter. So, Lw = 0.73(1.15/1) = 0.84 m = 0.33 in.
Vapor velocity based on active area varies inversely with the square of column diameter.
Therefore, Ua = 5.99(1/1.15)2 = 4.53 ft/s.
From Eq. (653), Ks is proportional to Ua. Thus, Ks = 0.265(4.53/5.99) = 0.20 ft/s
From Eq. (652), φe = exp(4.257Ks0.91) = exp[4.257(0.20)0.91] = 0.37
The value of C in Eq. (651) remains at 0.362.
From Eq. (651),
qL
hl = φ e hw + C
Lw φ e 2/3 12.9
= 0.37 2 + 0.362
33 0.37 2/3 = 0.88 in. hσ = 0.36 in. (no change from Example 6.6)
From Eq. (649), ht = hd + hl + hσ = 0,88 + 0.88 + 0.36 = 2.12 in.
Tray pressure drop = htρL = 2.12(0.0356) = 0.076 psi/tray
(c) Example 6.7:
In this example, a tower diameter of 1 m gives a EOV = 0.77. Must redo all calculations.
DT = 1.15 m, A = 1.038 m2 = 10,380 cm2, Aa = 0.8(1.038) = 0.83 m2 = 8,300 cm2
Lw = 33 in. = 0.84 m, φe = 0.37, hl = 0.88 in. = 2.24 cm
Ua = 4.53 ft/s = 137 cm/s, Uf = 10.2 ft/s, f = Ua/Uf = 4.53/10.2 = 0.44 Analysis: (c) (continued) Exercise 6.21 (continued) F = UaρV0..5 = 1.37(1.92)0.5 = 1.90 (kg/m)0.5/s ,
From Eq. (664), t L =
From Eq. (665), t G = qL = 812 cm3/s hl Aa 2.24(10,380)
=
= 28.6 s
qL
812
1 − φ e hl (1 −...
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 Spring '11
 Levicky
 The Land

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