Unformatted text preview: nd p = partial pressure of water vapor in atm. Effective diffusivity of water
vapor in the adsorbent particle = 0.05 cm2/s. Exit humidity of water vapor to be 0.0009 lb
H2O/lb dry air.
Assumptions: Isothermal and isobaric operation.
Find: Time to breakthrough for the equilibrium model.
Time to breakthrough for the masstransfer model of Klinkenberg.
Set of breakthrough curves for the masstransfer model of Klinkenberg.
Approximate width of the MTZ.
Average loading of the bed at breakthrough.
Analysis: First determine the moisture content of the entering air. From a psychrometric chart,
the entering humidity for 80% R.H., 80oF, and 1 atm is 0.0177 lb H2O/lb dry air or a water vapor
content of 0.0177/1.0177 x 100% = 1.739 wt%. By Dalton's law, partial pressure of water vapor
in the entering air = (1)[0.0177/18.02]/[0.0177/18.02 + 1/28.97] = 0.0277 atm. The mole
fraction of water vapor in the entering air = 0.0277.
Assume a bed crosssectional area of 1 ft2. Then, entering gas rate = 100 ft3/min. Bed volume =
(1)(5) = 5 ft3. Bed contains 5(39) = 195 lb silica gel.
The MW of the entering gas = 18.02(0.0277) + 28.97(0.9723) = 28.67.
From the ideal gas law, density of entering gas = PM/RT = (1)(28.67)/[(0.7302)(540)] = 0.0727
lb/ft3. Therefore, the entering gas flow rate = 100(0.0727) = 7.27 lb/min with a water vapor flow
rate of 0.01739(7.27) = 0.1264 lb H2O/min. For the dilute conditions, the desired c/cF =
0.0009/0.0177 = 0.05. So use this as the breakthrough value.
Equilibrium model: Equilibrium loading based on entering feed = 15.9(0.0277) = 0.440 lb
H2O/lb silica gel.
Therefore, can adsorb 0.440(195) = 85.8 lb H2O. The ideal time for breakthrough = 85.8/0.1264
= 679 min = 11.3 h.
q S 0.440(195)
Alternatively, can apply Eq. (1592), with Lideal/LB = 1, so tideal = F =
= 679 min.
QF cF
0.1264
Masstransfer model using Klinkenberg equations:
2
Rp
Rp
1
From Eq. (15106),
=
+
(1)
kK 3k c 15De
Rp = 0.28/2 = 0.14 cm = 0.0014 m,
De = 0.05 cm2/s = 5 x 106 m2/s Exercise 15.26 (continued)
Analysis: (continued) Dp G
D
From Eq. (1565), k c = i 2 + 11
.
Dp
µ 0 .6 µ
ρDi 1/ 3 (2) From Perry's Handbook, at 32oF, the diffusivity of water vapor in air at 1 atm = 0.22 cm2/s.
From Eq. (336), the diffusivity is proportional to T to the 1.75 power. Therefore,
Di = 0.22[(80 + 460)/(32 + 460)]1.75 = 0.26 cm2/s = 0.26 x 104 m2/s.
From above, ρ = 0.0727 lb/ft3 = 1.16 kg/m3.
From a handbook, µ of air at 80oF = 1.75 x 105 kg/ms.
The mass velocity = G = 7.29/1.0 = 7.27 lb/minft2 = 0.592 kg/m2s
µ
1.75 × 10 −5
= 0.580
N Sc =
=
ρDi 116(0.26 × 10 −4 )
.
N Re = Dp G
µ = 0.0028(0.592)
= 94.7
175 × 10 −5
. 0.26 × 10 −4
0 .6
1/ 3
2 + 11 94.7
.
0.58
= 0149 m/s
.
0.0028
1
0.0014
(0.0014) 2
From Eq. (1),
=
+
= 0.0031 + 0.0261 = 0.0292 s
kK 3(0149) 15(0.05 × 10 −4 )
.
The equilibrium constant, K, the given value is 15 lb H2O/lb gelatm. We need to convert this to
units of (lb H2O/ft3 gel)/(lb H2O/ft3 gas).
From Table 15.2, assume smallpore silica gel with εb = 0....
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 Spring '11
 Levicky
 The Land

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