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Unformatted text preview: h Benzoic acid
Nitrogen
Water
Total 0.634
5.770
0.150
6.554 0.00519
0.20600
0.00832
0.21951 Mol fraction 0.0236
0.9385
0.0379
1.0000 Partial
Pressure if no
condensation,
torr
17.9
713.3
28.8
760.0 Exercise 17.37 (continued)
Now determine the exit temperature and the amount of BA desublimed.
Assume that water does not condense.
Let x = fraction of BA desublimed and T be the exiting temperature
Assume a thermodynamic path of all water evaporated at 25oC and BA condensed at the final
temperature (however, it doesn’t matter because BA and N2 specific heats are assumed constant
regardless of the phase state).
An enthalpy balance can be written as:
Cooling of feed gas to T and desublimation of BA = vaporization of added water and
heating of gas of added water and added N2 to T. Using cgs units, with a heat of vaporization for
water = 583 cal/g at 25oC,
2,760(0.32)(177.5 – T) + 634x(134) =
150(583) + 150(0.44)(T – 25) + 1,354(0.32)(T – 105) + 2,290(0.32)(T – 25) (1) This balance must be consistent with a partial pressure of nondesublimed BA in the gas
equal to the vapor pressure at T.
For a set of assumed values of T, compute x and compare the partial pressure of BA to its
vapor pressure, using a spreadsheet. We need an interpolating equation for the vapor pressure.
A fit of the ClausiusClapeyron equation from 96 to 105oC with Polymath, gives,
s
log10 ( PBA in torr ) = 9.679 − T( o 3572
C ) + 273 The spreadsheet results are:
T, C x Vapor
Pressure,
torr kmol/h BA
in vapor Partial
Pressure,
torr 96
97
98
99
100
101
102
103
105 0.804
0.829
0.854
0.878
0.903
0.928
0.953
0.978
1.028 1.00
1.06
1.12
1.19
1.27
1.34
1.42
1.51
1.70 0.001018
0.000889
0.000760
0.000631
0.000502
0.000372
0.000243
0.000114
0.000145 3.59
3.14
2.69
2.23
1.77
1.32
0.86
0.40
0.51 The vapor pressure = the partial pressure at an exit temperature of 101oC with 93% of the BA
desublimed.
Check to determine if any water is condensed at this temperature. The vapor pressure of
water at 101oC exceeds 760 torr. Since the partial pressure of water in the exiting gas is much
less than this, the assumption of no condensation of water is valid. Exercise 17.38
Subject: Desublimation on the outside of a heat exchanger tube.
Given: Heat exchanger tube with a coolant flowing through the inside and desublimation
occurring on the outside surface of the tube.
Assumptions: Rate of desublimation controlled by heat conduction through the deposited
crystal layer. Sensible heat effect is negligible compared to the latent heat of desublimation.
Temperatures of the gas and the coolant assumed constant over tube length.
Find: Derive (1775), which gives the time for a crystal layer to grow from the outside tube
radius, ro, to a radius, rs.
Analysis: Equation (1775) gives, t= ρc ∆H s
rs2
r
1
ln s − ( rs2 − ro2 )
ro
4
kc (Tg − Tc ) 2 (1) Rate of heat released by desublimation = rate of heat conduction through the crystal layer.
Therefore, ∆H s dm
dr kc ALM (Tg − Tc )
= ∆H s 2πrLρc
=
dt
dt
r − ro (2) where, ∆H s = heat o...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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