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Balances on Atoms
Example, balance on H & O, not H2O
Class 11 – Friday, September 17th
Example, balance on C & H, not C2H6
Example, balance on H, not H2
Balances using atomic species
Analysis For Atomic Balances
• Draw & Completely Label a Flowchart
• Count the unknown labeled variables
• Count the number of independent atomic
species balances
• Count the number of independent non-reactive
molecular species balances
• Count the number of other equations relating
unknown variables
This method is useful for reactive systems since
atoms cannot be generated or consumed in
chemical processes!
What’s the general balance for atomic species
balances?
a)
b)
c)
d)
Input = output
Input + generation = output + consumption
Input = output + accumulation
Input + generation = output
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Degree of freedom analysis
Number of degrees of freedom =
# unknowns
- # independent atomic species balances
- # independent non-reactive molecular species
- # other equations
What’s missing from atomic species
degree of freedom analysis compared
to extent of reaction and molecular
species balances?
a) A generation unknown
b) A consumption unknown
c) A reaction unknown
hopefully zero!!
Dehydrogenation of Ethane
Dehydrogenation of Ethane
100 kmol C2H6/min
C2H6 Î C2H4 + H2
40 kmol H2/min
100 kmol C2H6/min
n1 kmol C2H6/min
n2 kmol C2H4/min
ndf =
C2H6 Î C2H4 + H2
40 kmol H2/min
n1 kmol C2H6/min
n2 kmol C2H4/min
ndf =
2 unknown labeled variables
- 2 independent atomic species (C, H)
- 0 independent non-reactive molecular species balances
- 0 other equations relating unknown variables
= 0 (can be solved)
2 unknown labeled variables
- 2 independent atomic species (C, H)
- 0 independent non-reactive molecular species balances
- 0 other equations relating unknown variables
= 0 (can be solved)
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A mixture of 75 mole % propane & 25 mole %
hydrogen is burned with 25 % air. Combustion is
incomplete. Fractional conversions of 90 % for the
propane & 85 % for the hydrogen are achieved. What
is the molar composition of the product stream?
Sometimes moles are conserved as
well as mass in reactions (under
special circumstances). Are they in this
reaction?
Dehydrogenation of Ethane
100 kmol C2H6/min
a) Yes
b) No
c) Not enough
information
C2H6 Î C2H4 + H2
40 kmol H2/min
n1 kmol C2H6/min
n2 kmol C2H4/min
ndf =
2 unknown labeled variables
- 2 independent atomic species (C, H)
- 0 independent non-reactive molecular species balances
- 0 other equations relating unknown variables
a)
b)
c)
d)
0
1
2
3
= 0 (can be solved)
A mixture of 75 mole % propane & 25 mole % hydrogen is burned with 25 % excess
air. Fractional conversions of 90 % for the propane & 85 % of the hydrogen are
achieved; of the propane that reacts, 95 % reacts to form CO2 & the rest reacts to
form CO. What is the molar composition of the product stream?
100 mol/hr
.75 P .25H
How many degrees of freedom are there?
C3H8 + 5O2 → 3CO2 + 4H2O
2H2 + O2 → 2H2O
no
0.79 N2
0.21 O2
nC3H8
nH2
nO2
nN2
nH2O
nCO2
nCO
A mixture of 75 mole % propane & 25 mole % hydrogen is burned with 25 %
air. Fractional conversions of 90 % for the propane & 85 % of the hydrogen
are achieved; of the propane that reacts, 95 % reacts to form CO2 & the rest
reacts to form CO. What is the molar composition of the product stream?
100 mol/hr
.75 P .25H
no
0.79 N2
0.21 O2
C3H8 + 5O2 → 3CO2 + 4H2O
2H2 + O2 → 2H2O
nC3H8
nH2
nO2
nN2
nH2O
nCO2
nCO
Why do we need the reaction for CO?
I. To figure out the excess air. II. To do the C and O balances.
III. To use the fractional conversions IV. We don’t need the reaction for CO.
a) I & II
b) I & III
c) II & III
d) I, II & III
e) IV
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