The answer is 164 mmHg for the pressure. Demonstrate how to obtain this number by using the Antoine equation and Raoult's law.
The answer is 164 mmHg for the pressure. Demonstrate how to obtain this number by using the Antoine equation and Raoult's law.
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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The answer is 164 mmHg for the pressure. Demonstrate how to obtain this number by using the Antoine equation and Raoult's law.
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Example 8.4-4
Solution
Equipment Encyclopedia
evaporator
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individual enthalpy changes for the three steps.
Partial Vaporization of a Mixture
An equimolar liquid mixture of benzene (B) and toluene (T) at 10°C is fed continuously to a vessel in which
the mixture is heated to 50°C. The liquid product is 40.0 mole% B, and the vapor product is 68.4 mole% B.
How much heat must be transferred to the mixture per g-mole of feed?
Basis: 1 mol Feed
1 mol, 10°C
0.500 mol B/mol
0.500 mol T/mol
We start with a degree-of-freedom analysis:
Q Search
CD
EVAPORATOR
H
ny(mol), 50°C
0.684 mol B/mol
0.316 mol T/mol
Q(J)
n₁ (mol), 50°C
0.400 mol B/mol
0.600 mol T/mol
3 unknown variables (ny, nL, Q)
-2 material balances
-1 energy balance
= 0 degrees of freedom
We could count each specific enthalpy to be determined as an unknown variable, but then we would also
count the equations for each of them in terms of heat capacities and latent heats, leaving the number of
degrees of freedom unchanged.
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H₁
Ĥ₂
Substance
B(1)
T(1)
B(v)
T(v)
HA
CREATIVITY EXERCISE
10°C
The values of nout were determined from the known mole fractions of benzene and toluene in the outlet
streams and the calculated values of ny and n₁. We do not know the feed-stream pressure and so we assume
that AH for the change from 1 atm to Pfeed is negligible, and since the process is not running at an unusually
low temperature or high pressure, we neglect the effects of pressure on enthalpy in the calculations of Ĥ₁
through H4. (The pressure can be estimated from the Antoine equation and Raoult's law to be 164 mm Hg.)
The heat capacity and latent hea dats needed to calculate the outlet enthalpies are obtained from Tables B.1
and 2)
5
50°C
451
The formulas (including the APEx formulas) and values of the unknown specific enthalpies are given
below. Convince yourself that the formulas represent AH for the transitions from the reference states to the
process states,
10°C
of 695
nin
mol
(Cp)CH₂1dT
CP)C.H.qd [=Enthalp("benzene"/10,50,"C","T")] =(5.34 kJ/mol
H₂= pd [Enthalpytoluene" 10,50,"C","T"]=6.341/kJ/mol
80.1°C
(Cp)c.m. qd + (AH.)c.n. (80.1°C) + (Cp)C,H,(v)dT
50%
80.1°C
[=Enthalpy("benzene",10,80.1,"C","1") Hv("benzene")
110.62°C
0.500
0.500
10°C
The energy balance is
Hin
Hemol)
0
0
Enthalpy("benzene",80.1,50,"C","g")] =37.53 kJ/mol
50°C
10.62°C
(C₂)C₂H₂ndT
dT+(AĤy) C₂H₂ (110.62°C) + (Cp)C₂H₂(v)dT
[=Enthalpy("toluene",10,11082,"C","1") + Hv("toluene")
+Enthalpy("toluene",110.62,50,"C","g")]=42.94 kJ/mol
Q = AH = n₂Ĥ₁ -
nout
Hout
(mol) (kJ/mol)
out
H
0.259
0.389
0.241
0.111
Ĥ₁
Ĥ
H₂
A₁
in
- Σn;#₁⇒ [Q=17,7 kJ |
A gas emerges from a stack at 1200°C. Rather than being released directly to the atmosphere, it can
be passed through one or several heat exchangers, and the heat it loses can be put to use in a variety
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H₁
Ĥ₂
Substance
B(1)
T(1)
B(v)
T(v)
HA
CREATIVITY EXERCISE
10°C
The values of nout were determined from the known mole fractions of benzene and toluene in the outlet
streams and the calculated values of ny and n₁. We do not know the feed-stream pressure and so we assume
that AH for the change from 1 atm to Pfeed is negligible, and since the process is not running at an unusually
low temperature or high pressure, we neglect the effects of pressure on enthalpy in the calculations of Ĥ₁
through H4. (The pressure can be estimated from the Antoine equation and Raoult's law to be 164 mm Hg.)
The heat capacity and latent hea dats needed to calculate the outlet enthalpies are obtained from Tables B.1
and 2)
5
50°C
451
The formulas (including the APEx formulas) and values of the unknown specific enthalpies are given
below. Convince yourself that the formulas represent AH for the transitions from the reference states to the
process states,
10°C
of 695
nin
mol
(Cp)CH₂1dT
CP)C.H.qd [=Enthalp("benzene"/10,50,"C","T")] =(5.34 kJ/mol
H₂= pd [Enthalpytoluene" 10,50,"C","T"]=6.341/kJ/mol
80.1°C
(Cp)c.m. qd + (AH.)c.n. (80.1°C) + (Cp)C,H,(v)dT
50%
80.1°C
[=Enthalpy("benzene",10,80.1,"C","1") Hv("benzene")
110.62°C
0.500
0.500
10°C
The energy balance is
Hin
Hemol)
0
0
Enthalpy("benzene",80.1,50,"C","g")] =37.53 kJ/mol
50°C
10.62°C
(C₂)C₂H₂ndT
dT+(AĤy) C₂H₂ (110.62°C) + (Cp)C₂H₂(v)dT
[=Enthalpy("toluene",10,11082,"C","1") + Hv("toluene")
+Enthalpy("toluene",110.62,50,"C","g")]=42.94 kJ/mol
Q = AH = n₂Ĥ₁ -
nout
Hout
(mol) (kJ/mol)
out
H
0.259
0.389
0.241
0.111
Ĥ₁
Ĥ
H₂
A₁
in
- Σn;#₁⇒ [Q=17,7 kJ |
A gas emerges from a stack at 1200°C. Rather than being released directly to the atmosphere, it can
be passed through one or several heat exchangers, and the heat it loses can be put to use in a variety
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