CHEN 2800- Prof. Ring
Fundamentals of Process Engineering
Name:_______Solution ________________________
Open Book and Open Notes Exam
Score
Problem 1
______/20
Problem 2
______/30
Problem 3
______/50
Total
______________/100
1
CHEN 2800- Prof. Ring
Fundamentals of Process Engineering
Name:_______Solution ________________________
Problem 1: (20 points) Calculate the specific volume of propane at a pressure of 6000
kPa and a temperature of 230ºC.
2
CHEN 2800- Prof. Ring
Fundamentals of Process Engineering
Name:_______Solution ________________________
Problem 2: (30 points) A mixture of 50 mole % benzene and 50 mole % toluene is
contained in a cylinder at 39.36 in Hg absolute pressure. Calculate the temperature range
in which a two phase system can exist.
3
CHEN 2800- Prof. Ring
Fundamentals of Process Engineering
Name:_______Solution ________________________
Problem 3: (50 points) A pelletized mixture of Fe2O3 (hematite) and carbon at 298ºK is
reacted with oxygen at 298ºK to produce CO2(g) and Fe(l) by the following reaction
sequence:
1)
C(s) + ½ O2(g)  CO
2)
½ Fe2O3 (s) + 3/2 CO  Fe(s) + 3/2 CO2
Oxygen is fed to the reactor with 0% excess. You are to answer A and B assuming that
the reactor is adiabatic.
A) Determine the amount of carbon required for this process per mole of Fe(l)
produced, if all the Fe2O3 is reacted and the products leave the separator at
1900ºK. Note, that excess carbon may be reacted in reaction 1 to provide any
needed heat.
B) Determine the heat, Q, required for the separator per mole of Fe(l) produced, if all
the Fe2O3 is reacted and the products leave the separator at 1900ºK.
O2(g)
C(s)
Fe2O3(s)
298ºK
Reactor
Separator
CO(g)
CO2(g)
1900ºK
Fe(l)
1900ºK
Data for Problem:
Heat of Formation
Hf º (kJ/mole)
O2(g)
0
N2(g)
0
CO(g)
-110.599
CO2(g)
-393.798
Fe2O3(s)
-824.8
Fe(l)
Fe(s)
0
C(s)
0
Heat Capacity
(J/(mole ºK))
34.8
32.8
34.0
55.3
42.5
40.2
24.5
8.5
4
Melting Point Heat of Fusion
(ºK)
(kJ/gmole)
1838ºK
83.3 kJ/gmole
1809ºK
>2300ºK
18.8 kJ/gmole
-
CHEN 2800- Prof. Ring
Fundamentals of Process Engineering
Name:_______Solution ________________________
5
CHEN 2800- Prof. Ring
Fundamentals of Process Engineering
Name:_______Solution ________________________
6
CHEN 2800- Prof. Ring
Fundamentals of Process Engineering
Name:_______Solution ________________________
Solution
1)
Basis 1 kg propane
Tc=370K, Pc=4255kPa, Mw= 44.09 kg/kmol, R=8.314(kPa m3/kmol K)
Tr=(230+273.15)/370=1.36
Pr=6000/4255=1.408
From General compressibility chart z ~ 0.8
V^/Mw=zRT/(PMw)
V^/Mw= 0.8*8.314(kPa m3/kmol K)*(503K)/(6000*kPa*44.09 kg/kmole)=0.0127 m3/kg
2)
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CHEN 2800- Prof. Ring
Fundamentals of Process Engineering
Name:_______Solution ________________________
3)
Solution
Basis: 1 mole Fe(l) Product at 1900K
set T.ref = 298K therefore ΔHrxn must be supplied by extra C in reaction 1
1)
3/2C(s) + 3/4 O2(g)  3/2 CO
2)
½ Fe2O3 (s) + 3/2 CO  Fe(s) + 3/2 CO2
ΔHºrxn = 3/2*(-110.51 kJ/gmole)
ΔHºrxn = -12.87 kJ/gmole
3)
Fe(s)  Fe(l)
3/2C(s) + 3/4 O2(g) +½ Fe2O3 (s)  Fe(l) + 3/2 CO 2
3
C p_rxn  Cp_Fe_liq 
H0 rxn 
2
 Cp_CO2 
3
2
 Cp_C 
3
4
 Cp_O2 
ΔHºrxn = + 18.8 kJ/gmole
overall reaction is exothermic
1
2
 Cp_Fe2O3  63.35
1
mol K
J
1
kJ
 Hf_CO2    Hf_Fe2O3  178.297
2
mole
2

3

H rxn ( T)  H0 rxn  

T
C p_rxn d t  H f_Fe
298 K
H rxn( 1900 K)  58.01
kJ
mole
At 1900K the overall reaction is exothermics,
so no extra C is required. Actually cooling is required
either in the reactor or at the separator.
Ans A: 3/2 moles C required!
Ans B: must cool Q = 58.01 kJ.
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