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Unformatted text preview: 0.130 = 1.508 kmol/h = 184 kg/h.
The ratio of mass of entering gas mixture to entering BA mass = Mr = 28.02 ( 0.992 ) + 122.12 ( 0.008 ) 122.12 ( 0.008 )
Sensible heat + heat of fusion per gram of BA desublimed = = 29.45 M r CPg (Tin − Tout ) g + ∆H s = 29.45(0.32)(130 − 80) + 134 = 605 cal/g
1.316(605)
From (1775), time = t =
1.4(40)
Number of tubes = 184 (14.2 )
1.94 = 1,350 ( 2.5 )
2 2 ln 2.5
1
− ( 2.52 − 1.252 ) = 14.2 h
1.25
4 Exercise 17.37
Subject: Bulkphase desublimation of benzoic acid from N2 by injection of water followed by
two steps of nitrogen quench.
Given: 3 m3/h of gas at 1 atm and a temperature of 10oC above the dew point and containing 6.4
mol% benzoic acid (BA) and 93.6 mol% N2. 150 cm3/h of water at 25oC is added. Then the gas
is further cooled in two steps with N2 at 1 atm, first with 1.5 m3/h at 105oC, and secondly with
2.0 m3/h at 25oC. The following are properties of BA:
Melting point = 122.4oC
Specific heat of solid and gas = 0.32 cal/goC
Heat of sublimation = 134 cal/g. Therefore, desublimation is exothermic at 134 cal/g
Vapor pressure of BA:
96
105
119.5
132.1
146.7
162.6
172.8
Temp., oC
Vapor
1
1.7
5
10
20
40
60
Pressure,
solid
solid
solid
liquid
liquid
liquid
liquid
torr
Assumptions: Equilibrium. Desublimation occurs in the bulk gas. Adiabatic process.
Find: The final gas temperature and the fractional yield of solid benzoic acid crystals.
Analysis: A plot of the vapor pressure data, using a spreadsheet is shown on the following page.
Note the change in slope at the melting point.
MW of BA = 122.12
MW of H2O = 18.02
MW of N2 = 28.01
First find the entering gas temperature by computing the dew point temperature.
Partial pressure of BA in entering gas = 0.064(760) = 49 torr. From the vapor pressure curve,
this is in the liquid region with a temperature of 167.5oC. Therefore, the entering gas
temperature is 167.5 + 10 = 177.5oC
Next, calculate the mass flow rate of the entering gas, using SI units.
MW of entering gas = M = 0.064(122.12) + 0.936(28.01) = 34.03
Temperature = 177.5 + 273.2 = 450.7 K
Pressure = 1 atm = 101.3 kPa
Therefore, from the ideal gas law, ρ= PM (101.3) ( 34.03)
=
= 0.920 kg/m3
RT ( 8.314 )( 450.7 ) Entering gas flow rate = 3(0.920) = 2.76 kg/h or 2.76/34.03 = 0.0811 kmol/h
This gives 0.064(0.0811) = 0.00519 kmol/h or 0.634 kg/h of BA
and 2.76 – 0.634 = 2.126 kg/h of N2 Exercise 17.37 (continued) Vapor Pressure, torr 100 10 1
80 90 100 110 120 130 140 150 160 170 180 Temperature, C Water at a flow rate of 150 cm3/h or 0.150 kg/h is now added.
Now add 1.5 m3/h of N2 at 105oC and 1 atm. From the ideal gas law,
PM (101.3) ( 28.02 )
ρ=
=
= 0.903 kg/m 3
RT ( 8.314 )( 378.2 )
Therefore the mass flow rate of N2 added = 1.5(0.903) = 1.354 kg/h
Next, add 2.0 m3/h of N2 at 25oC and 1 atm. From the ideal gas law,
PM (101.3) ( 28.02 )
ρ=
=
= 1.145 kg/m3
RT ( 8.314 )( 298.2 )
Therefore the mass flow rate of N2 added = 2.0(1.145) = 2.290 kg/h
Thus, the final exiting combined gas and solid is as follows:
Component
kg/h
kmol/...
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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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