CHEN 2800- Prof. Ring
Fundamentals of Process Engineering
Name:____Solution____________________
Final Exam
Open Book and Open Notes Exam
Score
Problem 1
______/15
Problem 2
______/30
Problem 3
______/30
Problem 4
______/25
Total
______________/100
1
CHEN 2800- Prof. Ring
Fundamentals of Process Engineering
Name:____Solution____________________
Problem 1: (15 Points)
In homework problem 17.19, (P. 535 of text), you were given moist air at 25C,
100 kPa, and a dew point of 16C, and were asked to calculate the temperature
to which the air must be cooled to condense 50% of the initial water at a constant
pressure of 100 kPa.
Redo that problem by allowing the pressure to increase until 50% of the initial
water has condensed keeping the temperature constant at 25C. Use the
following Antoine constants in your calculation:
B
log10 P   A C T
where T is temperature (K), P* is vapor pressure (bar), A=4.65430, B=1435.264,
C=-64.848, for the temperature range is 255.8-373 K. Report your answer in
kPa.
A  4.65430
B  1435.264 K
To  273.15 K
C  64.848 K
A
P'( T)  1 bar 10
B
P1  100 kPa
C T
y water1 
P'( 16 K  To )  1.801 kPa
half water is removed from Air so exit air is given by
y water2
P2 
y water1
P'( 25 K  To )
2
P2
2 P'( 25 K  To )
y water1
 353.081 kPa
2
P'( 16 K  To )
P1
 0.018
CHEN 2800- Prof. Ring
Fundamentals of Process Engineering
Name:____Solution____________________
Problem 2: (30 points) Sea water is desalinized by reverse osmosis using the scheme
pictured below with the data given in the figure. Please determine the answers to the
following questions.
a) Determine the flow rate of waste brine (W) in kg/h.
b) Determine the flow rate of desalinized water (D) in kg/h.
c) Determine the flow rate of recycle ( R ) in kg/h.
Recycle ( R )
4.0%NaCl
Reverse Osmosis
Unit
Feed
Sea Water
1000 kg/h
3.1% NaCl
(D)
Desalinized
Water Product
500 ppm NaCl
3
(W)
Waste Brine
5.25% NaCl
CHEN 2800- Prof. Ring
Fundamentals of Process Engineering
Name:____Solution____________________
4
CHEN 2800- Prof. Ring
Fundamentals of Process Engineering
Name:____Solution____________________
Problem 3: (30 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
ΔHrxn = -12.399 kJ/gmole
Oxygen is fed to the reactor with 0% excess. The reactor operates at 1900K a
temperature just above the melting point of iron (Fe). The product gasses are separated
from the liquid Fe product in the separator which also operates adiabatically above the Fe
melting point at 1900ºK. As the plant engineer you are required to determine what to do
with the heat wasted in the product gas stream. If the exit gas is cooled to 600ºK in a heat
exchanger (Qex) and this heat used to superheat the inlet oxygen, how much cooling (QR)
would be required for the reactor per mole of Fe(l) produced?
298ºK
O2(g)
Qex
C(s)
Fe2O3(s)
298ºK
Reactor
Separator
600ºK
Fe(l)
1900ºK
QR
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
CO2(g)
1900ºK
Heat Capacity
(J/(mole ºK))
34.8
32.8
34.0
55.3
42.5
40.2
24.5
8.5
5
Melting Point Heat of Fusion
(ºK)
(kJ/gmole)
1838ºK
83.3
1809ºK
>2300ºK
18.8
-
CHEN 2800- Prof. Ring
Fundamentals of Process Engineering
Name:____Solution____________________
Data for Problem
Cp_O2  34.8 
J
Cp_CO  34.0 
mole K
J
Cp_CO2  55.3 
mole K
Heat of Formation Data
J
mole K
Δ Hrxn_1  26.416 
CO
Cp_Fe2O3  42.5 
Cp_C  8.5
J
mole K
Cp_Fe_liq  40.5 
J
Cp_Fe_s  24.5 
mole K
mole K
mole
kJ
Hf_CO2  393.798 
mole
kJ
Hf_CO  110.599 
mole
Δ Hrxn_2 
3/2C(s) + 3/4 O2(g)  3/2 CO
2)
½ Fe2 O3 (s) + 3/2 CO  Fe(s) + 3/2 CO2
Fe2O3
197 
3
2
 94.057
3
2
3
2
1
kJ
Hreactants   Hf_Fe2O3  412.4
2
mole

Tf
298 K

Cp_Fe_s d t  Δ Hf_Fe_liq  



H

2  f_CO2 
3

600 K
298 K
kcal
mole
 393.798
 824.8
1
mol
1
mol
 kJ
 kJ
kcal
mole
1
kcal  3 
kcal 
1
    197 
    26.416  mole   12.399 mol  kJ
mole  2 
2 

ΔHºrxn = + 18.8 kJ/gmole
overall reaction is exothermic
1
kJ
 Hf_CO2  Hf_Fe_s    Hf_Fe2O3  178.297
mole
2


Hproducts  1   

kcal
mole
ΔHºrxn = -12.87 kJ/gmole
Overall Enthalpy Balance 0=Q+W-ΔH using (eq. 25.6)
ΔH=Hproducts(Tout)-ΔHreactants(Tin)+ΔHrxn(298K)
based upon overall reaction stoichiometry, outlet conditions and T.ref=298K

94.057
ΔHºrxn = 3/2*(-110.51 kJ/gmole)
3)
Fe(s)  Fe(l)
3/2C(s) + 3/4 O 2(g) +½ Fe2 O3 (s)  Fe(l) + 3/2 CO 2
2
 kJ
Reaction heat
1)
3
1
mol
kJ
Hf_Fe2O3  824.8
mole
Solution
Basis: 1 mole Fe(l) Product at 1900K
set T.ref = 298K
Perform an overall MB and EB.
Δ Hrxn_298 
CO2
kJ
Hf_Fe_s  0 
mole
Tf  1900 K
kJ
 110.599
J
To  298  K
mole K
Δ Hf_Fe_liq  18.8 
J
kcal
mole

1900 K
Tf

Cp_Fe_liq d t ...  507.597


kJ
mole
Cp_CO2 d t 


kJ
QR  Hproducts  Hreactants  Δ Hrxn_298  273.494
mole
Reaction is exothermic
so cooling is required
Answer: QR=273.494 kJ
6
  94.057 

  94.057 

kcal
  1   197  kcal   0  kJ  178.297 1  kJ



2 
mole 
mole
mol
mole 
kcal
  1   197  kcal   18.8  kJ  159.497 1  kJ
 2 

mole 
mole
mol

mole 
CHEN 2800- Prof. Ring
Fundamentals of Process Engineering
Name:____Solution____________________
Problem 4 (25 points)
Potassium hydroxide is commonly produced from KCl by electrolysis. The process is
sketched below. All concentrations are percent by weight.
a)
What is the fractional conversion of KCl to KOH?
b)
How many kilograms of chlorine gas is produced per kilogram of product?
c)
How many kilograms water must be evaporated in the evaporator per kilogram of
product?
Cl2
KCl
Dissolve
30% KCl
Solution
H2
Electrolysis
H2O
Evaporation
water
Product
50% KOH
7%KCl
43% H2O
Data for problem
Mole Weight
KCl
74.56
KOH
56.11
Cl2
70.91
H2
2.01
H2O
18.01
7
CHEN 2800- Prof. Ring
Fundamentals of Process Engineering
Name:____Solution____________________
Reaction 1
KCl + H2O --> KOH + 1/2 H2 + 1/2 Cl2
Basis 0.3 kg KCl fed and 0.7 kg water fed
0.3 kg
M wKCl
xKCl 
0.3 kg
MwKCl
F 
0.3 kg

MwKCl

0.7 kg
 0.094
xH2O  1  xKCl
MwH2O
0.7 kg
 42.891 mol
MwH2O
Use Basis 1 kg of product to determine the mole fractions for KOH and KCl in product only
0.5 kg
M wKOH
y KOH 
 0.5 kg  0.07 kg  0.43 kg 
 Mw
MwKCl MwH2O 
KOH


 0.264
0.07 kg
y KCl 
M wKCl
 0.5 kg  0.07 kg  0.43 kg 
 Mw
MwKCl MwH2O 
KOH


 0.028
y H2O  1  y KOH  y KCl
Overall mole balance on K
0.5 kg
0.07 kg




M wKOH
M w KCl

 F x
P

KCl
  0.5 kg
0.07 kg
0.43 kg 
0.5 kg
0.07 kg
0.43 kg  









 Mw

Mw
Mw
Mw
Mw
Mw
KOH
KCl
H2O 
KOH
KCl
H2O  


0.5 kg
0.07 kg





MwKOH
MwKCl

 F x
P
KCl
  0.5 kg
0.07 kg
0.43 kg  




 Mw

Mw
Mw
KOH
KCl
H2O  

P 
F xKCl
0.5 kg
0.07 kg





MwKOH
MwKCl


  0.5 kg
0.07 kg
0.43 kg  



  Mw
MwKCl MwH2O 
KOH


Product flow rate
 13.777 mol
Wgt of Product Stream


WP  P y KOH MwKOH  y KCl MwKCl  y H2O MwH2O  0.408 kg
R  y KOH P  3.64 mol
FractionalConversion 
M Cl2 
1
2

Moles reacted in reaction 1


y KOH P
xKCl F
 0.905
Answer

 y KOH P
M Cl2 MwCl2  0.129 kg
M Cl2 MwCl2
WP
Wgt Cl2 produced
Answer
 0.316
Overall Mole Balance on Water
In-Out-reacted=0
xH2O F  P y H2O  1 E  1 R
0
E  F xH2O  R  P y H2O  25.474 mol
E MwH2O  0.459 kg
E MwH2O
WP
 1.123
Moles water evaporated
weight of water evaporated
Answer
8