balances on reactive systems - جامعة نزوى€¦ · · 2016-02-294/2/2012 1 balances on...
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Balances on Reactive SystemsMaterial balance no longer takes the form g
INPUT = OUTPUT
Must account for the disappearance of reactants and appearance of products through stoichiometry.
Stoichiometric EquationsThe stoichiometric equation of a chemical reaction is a statement qof the relative amounts of reactants and products that participate in the reaction.
2 SO2 + O2 → 2 SO3
A stoichiometric equation is valid only if the number of atoms of q yeach atomic species is balanced.
2 S → 2 S4 O + 2 O → 6 O
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Stoichiometric Equations
The stoichiometric equation of a chemical reaction is a qstatement of the relative amounts of reactants and products that participate in the reaction.
2 SO2 + O2 → 2 SO3
A stoichiometric ratio of two molecular species participating in a reaction is the ratio of their stoichiometric coefficients:stoichiometric coefficients:
2 mol SO3 generated / 1 mol O2 consumed2 mol SO3 generated / 2 mol SO2 consumed
Stoichiometric Equations
C4H8 + 6 O2 → 4 CO2 + 4 H2OIs this stoichiometric equation balanced?What is the stoichiometric coefficient of CO2?What is the stoichiometric ratio of H2O to O2?How many lb‐mol O2 react to form 400 lb‐mol CO2?
22
2 Olbmol600COlb l4O lbmol 6
COlbmol 400 =×
100 lbmol/min C4H8 is fed and 50% reacts. At what rate is water formed?
22
2 COlbmol4
minOH lbmol
200HC lbmol 1OH lbmol 4
50.0min
HClbmol 100 2
84
284 =××
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Limiting and Excess ReactantsTwo reactants are said to be in stoichiometric proportion if the p pratio (moles A present/moles B present) equals the stoichiometric ratio from the balanced reaction equation.
2 SO2 + O2 → 2 SO3
the feed ratio that would represent stoichiometric proportion is nSO2/nO2 = 2:1f f d h d hIf reactants are fed in stoichometric proportion, and the reaction proceeds to completion, all reactants are consumed.
Limiting and Excess ReactantsStoichiometric Proportion – Reactants are present in a ratio p pequivalent to the ratio of the stoichiometric coefficients.
A + 2B→ 2C
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Limiting and Excess ReactantsLimiting reactant – A reactant is limiting if it is present g g pin less than stoichiometric proportion relative to every other reactant.
A + 2B→ 2C
Excess reactant – All other reactants besides the limiting reactant.
Limiting and Excess Reactants
fractional excess (fXS) – ratio of the excess to the stoichiometricf ( fXS)proportion.
A + 2B→ 2C
( ) ( )( )
25.0445n
nnf
stoichA
stoichAfeedAXS
=−
=
−=
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Limiting and Excess Reactants
fractional conversion (f) – ratio of the amount of a fractional conversion (f) ratio of the amount of a reactant reacted, to the amount fed.
A + 2B→ 2C ( )( )fedA
reactedA
nn
f =
0.050fA == 0.0
80fB ==
Limiting and Excess Reactants
fractional conversion (f) – ratio of the amount of a reactant f ( f)reacted, to the amount fed.
A + 2B→ 2C
( )( )fedA
reactedA
nn
f =
2.051fA == 25.0
82fB ==
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Limiting and Excess Reactants
fractional conversion (f) – ratio of the amount of a reactant f ( f)reacted, to the amount fed.
A + 2B→ 2C
( )( )fedA
reactedA
nn
f =
4.052fA == 5.0
84fB ==
Limiting and Excess Reactants
fractional conversion (f) – ratio of the amount of a reactant f ( f)reacted, to the amount fed.
A + 2B→ 2C( )
( )fedA
reactedA
nn
f =
6.053fA == 75.0
86fB ==
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Limiting and Excess Reactantsfractional conversion (f) – ratio of the amount of a reactant f ( f)reacted, to the amount fed.
A + 2B→ 2C( )
( )fedA
reactedA
nn
f =
8.054fA == 0.1
88fB ==
Extent of Reaction
extent of reaction (ξ) – an extensive quantity describing the progress of a chemical reaction .p g
ν – stoichiometric coefficients: νA = ‐1, νB = ‐2, νC = 2
A + 2B→ 2C ni = ni0 + ν iξ
0=ξ
nA = nA0 − ξ nB = nB0 − 2ξ nC = nC0 + 2ξ
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Extent of Reaction
extent of reaction (ξ) – an extensive quantity describing the progress of a chemical reaction .
ν – stoichiometric coefficients: νA = ‐1, νB = ‐2, νC = 2
A + 2B→ 2C ni = ni0 + ν iξ ξ = 0
nB = 8 − 2ξ = 8 nC = 0 + 2ξ = 0 nA = 5− ξ = 5
Extent of Reaction
extent of reaction (ξ) – an extensive quantity describing the progress of a chemical reaction .
ν – stoichiometric coefficients: νA = ‐1, νB = ‐2, νC = 2
A + 2B→ 2C ni = ni0 + ν iξ ξ = 1
nB = 8 − 2ξ = 6 nC = 0 + 2ξ = 2 nA = 5− ξ = 4
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Extent of Reaction
extent of reaction (ξ) – an extensive quantity describing the progress of a chemical reaction .
ν – stoichiometric coefficients: νA = ‐1, νB = ‐2, νC = 2
A + 2B→ 2C ni = ni0 + ν iξ ξ = 2
nB = 8 − 2ξ = 4 nC = 0 + 2ξ = 4 nA = 5− ξ = 3
Extent of Reaction
extent of reaction (ξ) – an extensive quantity describing the progress of a chemical reaction .p g
ν – stoichiometric coefficients: νA = ‐1, νB = ‐2, νC = 2
A + 2B→ 2C ni = ni0 + ν iξ ξ = 4
nB = 8 − 2ξ = 0 nC = 0 + 2ξ = 8 nA = 5− ξ = 1
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Assume an equimolar reactant feed of 100 kmol:What is the limiting reactant?
OHCOHC 42242 22 →+
g
ethylene
A reactant is limiting if it is present in less than stoichiometricproportion relative to every other reactant.
12
nn
stoichO
HC
2
42 =⎟⎟⎠
⎞⎜⎜⎝
⎛
11
nn
feedO
HC
stoich
2
42
2
=⎟⎟⎠
⎞⎜⎜⎝
⎛
⎠⎝
Assume an equimolar reactant feed of 100 kmol:
OHCOHC 42242 22 →+
qWhat is the percentage excess of each reactant?
( ) ( )( )
%10000150100
nnn
fstoichO
stoichOfeedOO,XS
2
22
2
−
−=
%10000.150
===
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Assume an equimolar reactant feed of 100 kmol:If the reaction proceeds to completion:
2C2H4 + O2 → 2C2H4O
If the reaction proceeds to completion: (a) how much of the excess
reactant will be left?
(b) H h C H O
( )50
21000
nn424242 HC
oHCHC
=ξξ−+=
ξν+= ( )50n
501100n
nn
2
2
222
O
O
OoOO
=
−+=
ξν+=
nn OHCo
OHCOHC 424242ξν+=
(b) How much C2H4O will be formed?
(c) What is the extent of reaction?
( )100n
5020n
OHC
OHC
OHCOHCOHC
42
42
424242
=
+=
Assume an equimolar reactant feed of 100 kmol:If the reaction proceeds to a point where the fractional
2C2H4 + O2 → 2C2H4O
p pconversion of the limiting reactant is 50%, how much of each reactant and product is present at the end? What is ξ?
( )( ) 5.0nn
ffedHC
reactedHC
42
42 == ( )25
210050100
nn424242 HC
oHCHC
=ξξ−+=−
ξν+=
nn o ξν+( )2520n
nn
OHC
OHCo
OHCOHC
42
424242
+=
ξν+=
5.0nn
f HCoHC 4242 =
−=
( )75n
251100n
nn
2
2
222
O
O
OOO
=
−+=
ξν+=50n OHC 42
42
=5.0
nf o
HC 42
50n
5.0100n100
42
42
HC
HC
=
=−
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Assume an equimolar reactant feed of 100 kmol:
2C2H4 + O2 → 2C2H4O
Assume an equimolar reactant feed of 100 kmol:If the reaction proceeds to a point where 60 mol of O2 is left, what is the fractional conversion of C2H4? What is ξ?
( )20n
402100n
nn
42
42
424242
HC
HC
HCoHCHC
=
−+=
ξν+=
( )40
110060
nn222 O
oOO
ξξ−+=
ξν+=
( )( ) 8.0
10020100
nn
ffedHC
reactedHC
42
42 =−
==
40=ξ
Reaction StoichiometryAcrylonitrile produced by reaction of ammonia, propylene, and O2 at 30% conversion of limiting reactant:
C3H6 + NH3 + 3
2O3 → C3H3N + 3H2O
limitingdetermine limiting reactant
nNH3nC3H6( )0
= 0.120 ×100 0.100 ×100( )= 1.20
nNH3nC3H6( )stoich
= 1 1( )= 1
nO2nC3H6( )0
= 0.780 × 0.21×100 0.100 ×100( )= 1.64
nO2nC3H6( )stoich
= 1.5 1( )= 1.5
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Acrylonitrile produced by reaction of ammonia, propylene, and O2 at 30% conversion of limiting reactant:
f 0 20determine fractional excesses
C3H6 + NH3 + 3
2O3 → C3H3N + 3H2O
fXS = 0.20
fXS = 0.093
determine fractional excesses limiting
nNH3( )stoich= 10.0 mol C3H6
1 mol NH3
1 mol C3H6
⎛ ⎝ ⎜ ⎞
⎠ ⎟ = 10.0 mol NH3
fXS( )NH3=
NH3( )0− NH3( )stoich
NH3( )stoich
= 12.0−10.010.0
= 0.20
nO2( )stoich= 10.0 mol C3H6
1.5 mol O2
1 mol C3H6
⎛ ⎝ ⎜ ⎞
⎠ ⎟ = 15.0 mol O2
fXS( )O2=
O2( )0− O2( )stoich
O2( )stoich
= 16.4−15.015.0
= 0.093
Acrylonitrile produced by reaction of ammonia, propylene, and O2 at 30% conversion of limiting reactant:
fXS = 0.20limiting
C3H6 + NH3 + 3
2O3 → C3H3N + 3H2O
fXS = 0.093
fXS 0.20
use fractional conversion to determine amount of propylene that leaves the reactor
limiting
nC3H6
= 1− f( ) nC3H6( )0= 1− 0.30( )10.0 mol C3H6( )= 7.0 mol C3H6
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Acrylonitrile produced by reaction of ammonia, propylene, and O2 at 30% conversion of limiting reactant:
C3H6 + NH3 + 32
O3 → C3H3N + 3H2O fXS = 0.093
fXS = 0.20limiting
ni = ni0 + ν iξ
2
nC3H6= 7.0 mol C3H6
ξ = 3 mol
determine extent ofreaction by applying molebalance to propylene
nC3H6= nC3H6( )0
+ −1( )ξ7.0 mol = 10.0 mol − ξ
ξ = 3 mol
Acrylonitrile produced by reaction of ammonia, propylene, and O2 at 30% conversion of limiting reactant:
fXS = 0.093fXS = 0.20
OHNHCONHHC 233223
363 3+→++
( )( ) ( )( ) 93110012.0 =−+=NHn
nC3H6= 7.0 mol C3H6
ξ = 3 mol ni = ni0 + ν iξapply mole balance to all remaining species
( )( ) ( )( )( )( )( ) ( )( )
( ) ( )( )( )( )( ) ( )( )( ) ( )( ) 90.330
6.613010078.079.0
30.310
9.11310078.021.0
93110012.0
2
2
33
2
3
23
=+=
=+=
=++=
=−+=
+
OH
N
NHC
O
NH
n
n
n
n
n
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Fuel for motor vehicles other than gasoline are being eyedbecause they generate lower levels of pollutants thandoes gasoline Compressed propane has been suggesteddoes gasoline. Compressed propane has been suggestedas a source of economic power for vehicles. Suppose thatin a test 22 kg of C3H8 is burned with 400 kg of air toproduce 44 kg of CO2 and 12 kg of CO.a. What was the percent excess air?b. The components in the product stream.
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Component Amount(kg)
Molecular weight
Input(kg-mol)
Output(kg-mol)
C H 22 44 0 50 ?
C3H8 + 5O2 3CO2 + 4H2O
C3H8 22 44 0.50 ?Air 400 29 13.80 ?O2 0.21*13.80
=2.90?
N2 0.79*13.80=10.9
?
CO2 44 44 - 1
30
CO 12 28 - 0.43
H2O. 18 - ?
Total 29.87 ?
Percent excess air = (O2 in – O2 used)/(O2 theory)= (2.90 – 5*0.5)/(5*0.5)=16%
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Component Input(kg-mol)
Consumed [1]
Consumed [2]
Generated [1]
Generated [2]
Total Output(kg-mol)
C3H8 + 5O2 3CO2 + 4H2O [1]
C3H8 + 7/2O2 3CO + 4H2O [2]
(kg-mol) [ ](kg-mol)
[ ](kg-mol)
[ ](kg-mol)
[ ](kg-mol)
(kg-mol)
C3H8 0.50 0.33 0.43/3=0.14 - - 0.03
O2 0.21*13.80=2.90
5*0.33= 1.65 3.5*0.14= 0.49
- - 0.76
N2 0.79*13.80=10.9
- - - - 10.9
CO2 - - - 1 - 1
31
CO - - - - 0.43 0.43
H2O. - 4*0.33= 1.32
4*0.14= 0.56
1.88
Total 29.87 15
Example
A gas stream containing 80 mol% C2H6 and 20 mol% O2 is burned inA gas stream containing 80 mol% C2H6 and 20 mol% O2 is burned in an engine with 200% excess air. If 80% of the ethane goes to CO2, 10% goes to CO and the remaining unburned, determine the amount of the excess air per 100 moles of the feed gas.
C2H6 + 7/2O2 2CO2 + 3H2O [1]C2H6 + 5/2O2 2CO + 3H2O [2]
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Component Input Gas Stream Output
C2H6 + 7/2O2 2CO2 + 3H2OBasis=100 mole feed gas
[1]
(kg-mol)
C2H6 80
O2 20 + 0.21*Gair= 799.7
Air Gair= 4.76*780 = 3,712.8
N2 0.79*Gair= 2,933.1
CO
Percent excess air (based on air):
(O2 in – O2 used)/(O2 theory)=2O2 theory (Eq 1) = 3.5*80=280O2 used (air) =O2 theory (Eq 1)-20=260 moles(O2 in (air)-260)/(260)=2O2in (air)=260+(2*260)=780molesN2 in = 3.76*780=2,932.8
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CO2
CO
H2O.
Total
N 3.76 780 ,93 .8Air stream=2,932.8 + 780=4.76*780=3,712.8
Component Input(mol)
Consumed [1]
Consumed [2]
Generated [1]
Generated [2]
Total Output(mol)
C2H6 + 7/2O2 2CO2 + 3H2O [1]C2H6 + 5/2O2 2CO + 3H2O [2]
(mol) [ ](mol)
[ ](mol)
[ ](mol)
[ ](mol)
(mol)
C2H6 80 64 8 - - 8
O2 799.7 224 20 - - 204
N2 2,933.1 - - - - 2933.1
CO2 - - - 128 - 128
CO - - - 16 16
34
H2O. - - - 192 24 216
Total 3812.8 3,505.1
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Chemical EquilibriumGivenGiven
a set of reactive species, andreaction conditions
Determine1. the final (equilibrium) composition of the reaction
mixture2. how long the system takes to reach a specified state
short of equilibrium
Chemical EquilibriumIrreversible reactionIrreversible reaction
reaction proceeds only in a single direction A → Bconcentration of the limiting reactant eventually approaches zero (time duration can vary widely)
Equilibrium composition of an irreversible reaction is that which corresponds to complete conversion.
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Chemical EquilibriumReversible reactionReversible reaction
reaction proceeds in both directions A ↔ Bnet rate (forward – backward) eventually approaches zero (again, time can vary widely)
Equilibrium composition of a reversible reaction is that which corresponds to the equilibrium conversion.
Equilibrium Composition
An equilibrium reaction proceeds ( ) HCO 22
yy
yyTK =An equilibrium reaction proceeds
to an extent at temperature T based on the equilibrium constant, K(T).
where yi is the mole fraction of species i
OHCO 2yy
ξν+= i0ii nn
total
ii n
ny =
Water‐gas shift reaction:Assume 1 mole CO and 2 mole H2OK(1105 K) = 1.00
( ) ( ) ( ) ( )g2g2g2g HCOOHCO +⇔+
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Equilibrium Composition
( ) HCO 22yy
TK =( ) ( ) ( )11nn 0COCO ξ−=ξ−+= ( )OHCO 2
yyTK( ) ( ) ( )
( ) ( ) ( )( ) ( )
( ) ( )3nnnnn
1nn
1nn
21nn
11nn
222
22
22
22
HCOOHCOtotal
0HH
0COCO
0OHOH
0COCO
=+++=
ξ=ξ++=
ξ=ξ++=
ξ−=ξ−+=
ξξ+
ξν+= i0ii nntotal
ii n
ny =
Water‐gas shift reaction:Assume 1 mole CO and 2 mole H2OK(1105 K) = 1.00
222
( ) ( ) ( ) ( )g2g2g2g HCOOHCO +⇔+
Equilibrium Composition
( ) HCO 22
yy
yyTK =
1n ξ−=( )( )
( )( )==
ξξξξ
1213330= OHCO 2
yy
ξν iii nn += 0
total
ii n
ny =
3n
n
n
2n
1n
total
H
CO
OH
CO
2
2
2
=
ξ=
ξ=
ξ−=ξ−= ( )( )ξ−ξ− 21
( )( )
mol667.0
22
2122
2
=ξξ+ξ−ξ−=ξ
ξ−ξ−=ξ
667.0
667.0
333.1
333.0
====
Water‐gas shift reaction:Assume 1 mole CO and 2 mole H2OK(1105 K) = 1.00
( ) ( ) ( ) ( )g2g2g2g HCOOHCO +⇔+
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Equilibrium Composition( )( )
( )( ) 121
=−− ξξ
ξξ ( )OHCO
HCO 22
yy
yyTK =
3n
222.03/667.0y
222.03/667.0y
444.03/333.1y
111.03/333.0y
total
H
CO
OH
CO
2
2
2
=
==
==
====
( )( )21 ξξ
( )( )
mol667.0
22
2122
2
=ξξ+ξ−ξ−=ξ
ξ−ξ−=ξ
OHCO 2yy
ξν+= i0ii nntotal
ii n
ny =
Water‐gas shift reaction:Assume 1 mole CO and 2 mole H2OK(1105 K) = 1.00
( ) ( ) ( ) ( )g2g2g2g HCOOHCO +⇔+
limiting reactant is CO
ξν+nnat equilibrium111.03/333.0yCO ==
ξν+= i0ii nn ξ = 0.667mol
( )( )mol 333.0
667.011nCO
=−+=
at equilibrium,
fractional conversion at equilibrium
667.0f 00.1333.000.1 == −
3n
222.03/667.0y
222.03/667.0y
444.03/333.1y
total
H
CO
OH
2
2
2
=
==
==
==
Water‐gas shift reaction:Assume 1 mole CO and 2 mole H2OK(1105 K) = 1.00
( ) ( ) ( ) ( )g2g2g2g HCOOHCO +⇔+
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Multiple Reactions
C2H4 + 12
O2 → C2H4OC H 3O 2CO 2H O C2H4 + 3O2 → 2CO2 + 2H2O
( ) ∑ ξν+=j
jij0ii nnfor j reactions of i species,mole balance becomes
( ) ( ) ( ) 21HCHC 11nn ξ−+ξ−+= ( ) ( ) ( )( ) ( ) ( )
( ) ( ) 10OHCOHC
2121
0OO
210HCHC
1nn
3nn
11nn
4242
22
4242
ξ++=
ξ−+ξ−+=
ξ+ξ+( ) ( )( ) ( ) 20OHOH
20COCO
2nn
2nn
22
22
ξ++=
ξ++=
Multiple Reactions
C2H4 + 12
O2 → C2H4OC2H4 + 3O2 → 2CO2 + 2H2O 2 4 2 2 2
reactionssidenowithconversion100%atformed molesformed product desired moles
yield =
( ) ∑ ξν+=j
jij0ii nnfor j reactions of i species,mole balance becomes
formed product undesired molesformed product desired moles
yselectivit =
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Multiple Reactions
100 moles A fed to a batch reactorproduct composition: 10 mol A, 160 B, 10 C
What is1. fA?2. YB?3. SB/C?
CA
B2A
→→
10100−4. ξ1, ξ2 9.0
10010100
fA ==
Multiple Reactions
100 moles A fed to a batch reactorproduct composition: 10 mol A, 160 B, 10 C
What is1. fA?2. YB?3. SB/C?
160
CA
B2A
→→
4. ξ1, ξ2( )( ) 889.0
10100160
Y12B =
−=
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100 moles A fed to a batch reactorproduct composition: 10 mol A, 160 B, 10 C
What is1. fA?2. YB?3. SB/C? 16
10160
/ ==CBS
CA
B2A
→→
4. ξ1, ξ2 10
100 moles A fed to a batch reactorproduct composition: 10 mol A 160 B 10 Cproduct composition: 10 mol A, 160 B, 10 C
What is1. fA?2. YB?
S ? nn 22A11AAA ξν+ξν+=nn ξν+
CA
B2A
→→
3. SB/C?4. ξ1, ξ2 1090
10010
nn
12
21
22A11AAoA
=ξ−=ξξ−ξ−=
ξν+ξν+
80
20160
nn
1
1
11BBoB
=ξξ+=
ξν+=
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Balances on Reactive Processes
Continuous, steady‐state dehydrogenation of ethane, y y gTotal mass balance still has INPUT = OUTPUT formMolecular balances contain consumption/generationAtomic balances (H and C) also have simple form
24262 HHCHC +→
Balances on Reactive Processes
Continuous, steady‐state dehydrogenation of ethaneFirst consider molecular balances:
Molecular H2 balance: generation = output generation H2 = 40 kmol H2/min
C2H6 balance: input = output + consumption 100 kmol C2H6/min = n1 + (C2H6 consumed)
C2H4 balance: generation = output(C H t d) (C2H4 generated) = n2
24262 HHCHC +→
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Balances on Reactive Processes
Continuous, steady‐state dehydrogenation of ethaneAt i C b l i t t t HHCHC +→Atomic C balance: input = output
Atomic H balance: input = output
( ) ( ) ( )426262 HC mol 1
C mol 22HC mol 1
C mol 21HC mol 1
C mol 262 nnHC mol 100 && +=
( ) ( ) ( ) ( )Hmol4Hmol6Hmol2Hmol 6 40mol100 nnHC && ++=
24262 HHCHC +→
( ) ( ) ( ) ( )4262262 mol 12 mol 11 mol 1 mol 162 40 mol 100 HCHCHHC nnHC ++=
Independent Equations
To understand the number of independent species balances in areacting system requires an understanding of independent algebraicequations.Algebraic equations are independent if you cannot obtain any of themby adding/subtracting multiples of the others.
2][12y6x3
[1] 4y2x
=+=+
[3] 4y2x =+2][ 12y6x3 =+
[5] 6zy4
[4] 2zx2
=+=−
3×[1] = [2] 2×[3] – [4] = [5]
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Independent Equations
To understand the number of independent species balancesp pin a reacting system requires an understanding ofindependent algebraic equations.Algebraic equations are independent if you cannot obtainany of them by adding/subtracting multiples of the others.
2][ 12y6x3
[1] 4y2x
=+=+ ( )
( )1212
12y6y612
12y6y243
==+−=+−
Independent Species
If two molecular or atomic species are in the same ratio topeach other where ever they appear in a process and thisratio is incorporated in the flowchart labeling, balances onthose species will not be independent equations.
31 nn && =
31
31
n76.3n76.3 && =
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Independent Chemical Reactions
When using molecular species balances or extents ofg preaction to analyze a reactive system, the degree of freedomanalysis must account for the number of independentchemical reactions among the species entering and leavingthe system.
Independent Chemical Reactions
Chemical reactions are independent if the stoichiometricpequation of any one of them cannot be obtained by addingand subtracting multiples of the stoichiometric equationsof the others.
[2]CB
[1] 2BA
→→
[3] 2CA
[2] CB
→→
2×[2] + [1] = [3]
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Solving Reactive Systems
There are 3 possible methods for solving balances around a reactivesystem:
1. Molecular species balances require more complex calculations thanthe other methods and should be used only for simple (singlereaction) systems.
2. Atomic species balances generally lead to the most straightforwardsolution procedure, especially when more than one reaction isinvolved
3. Extents of reaction are convenient for chemical equilibriumproblems.
Molecular Species Balances
To use molecular species balances to analyze a reactive system, the balances must contain generation and/or consumption terms.
The degree‐of‐freedom analysis is as follows:# unknown labeled variables
+ # independent chemical reactions‐ # independent molecular species balances
# h i l i k i bl‐ # other equations relating unknown variables# of degrees of freedom
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Molecular Species Balances
2 unknown labeled variables+ 1 independent chemical reactions
at steady state24262 HHCHC +→
‐ 3 independent molecular species balances‐ 0 other equations relating unknown variables0 degrees of freedom
Molecular Species Balances
at steady state24262 HHCHC +→
H2 Balance: generation = output
2 402
HkmolgenH =
C2H6 Balance: input = output + consumption
( )( )1 1
min
1min
2
62262 40 1000 HkmolHCkmolHkmolHCkmol n += & ( )( )
min
162
2
60 HCkmoln =&
C2H4 Balance: generation = output
( )( )min
HC kmol2
H kmol 1HC kmol 1
minH kmol
2
42
2
422
40n
40n
=
=
&
&
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Atomic Species Balance
24262 HHCHC +→
All atomic species balances take the form INPUT = OUTPUT
Degree‐of‐freedom analysis, ndf =# unknown labeled variables
‐ # independent atomic species balances p p‐ # molecular balances on independent nonreactive species‐ # other equations relating unknown variables
Atomic Species Balance
All atomic species balances take the form INPUT = OUTPUT
Degree‐of‐freedom analysis, ndf = 0 =2 unknown labeled variables
‐ 2 independent atomic species balances ‐ 0 molecular balances on independent nonreactive species
‐ 0 other equations relating unknown variables
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Atomic Species Balance
HHCHC +→
C Balance: input = output
( )( ) ( ) ( )21
1 2
2 1 2
1 1 2
min
mol 100
100426262
42
nnk
nn HCkmolCkmol
HCkmolCkmol
HCkmolCkmolHCkmol
&&
&&
+=
+=
24262 HHCHC +→
H Balance: input = output
( )( ) ( )( )( ) ( )
21
HC kmol 1H kmol 4
2HC kmol 1H kmol 6
1
H kmol 1H kmol 2
minH kmol
HC kmol 1H kmol 6
minHC kmol
n4n6+molk 80molk 600
nn
40 100
4262
2
2
62
42
&&
&&
+=
++
=
Atomic Species BalanceHHCHC +→
Solve simultaneously
n4n6+molk80molk600:H
nnmolk 100 :C 21
+=+=
&&
&&
24262 HHCHC +→
min/HC kmol 40n
min/HC kmol 60n
n4n6+molk 80molk 600 :H
422
621
21
==
+=
&
&
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Extent of Reaction
The 3rd method by which to determine molar flows in a reactive system is using expressions for each species flow rate in terms of extents of reaction (ξ).
Degree‐of‐freedom analysis for such an approach:ndf = # of unknown labeled variables
( ) ∑ ξν+=j
jij0ii nn
+ # independent reactions‐ # independent nonreactive species‐ # other relationships or specifications
Incomplete Combustion of CH4
Methane is burned with air in a continuous steady‐state reactor to yield a mixture of carbon monoxide, carbon dioxide, and water.
The feed to the reactor contains 7.80 mol% CH4, 19.4 mol% O2, 72.8 mol% N2. Methane undergoes 90.0% conversion, and the effluent gas contains 8 mol CO per mole CO.
CH4 + 32O2 → CO+2 H2O
CH4 +2 O2 → CO2 +2 H2O
contains 8 mol CO2 per mole CO.
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CH4 + 32O2 → CO+2 H2O
fCH4 = 0.9
The feed to the reactor contains 7.80 mol% CH4, 19.4 mol% O2, 72.8 mol% N2. Methane undergoes 90.0% conversion, and the effluent gas contains 8 mol CO2 per mole CO.
ndf = 5 unknowns + 2 independent reactions
CH4 +2 O2 → CO2 +2 H2O
p‐ 5 expressions for ξ (CH4, O2, CO, CO2, H2O)‐ 1 nonreactive species balance (N2)‐ 1 specified methane conversion
=0
CH4 + 32O2 → CO+2 H2O
CH4 +2 O2 → CO2 +2 H2O
fCH4 = 0.9
N2 balance: nonreactive species, INPUT = OUTPUT
CH4 conversion specification:
( ) 2molN mol
N N mol 8.72mol 100728.0n 2
2==
( )( )( )CHmol ll( )( )( ) 4molCHmol
CH CHmol780.0mol1000780.0900.01n 4
4=−=
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OHCOOCHOHCOOCH
2224
2223
4
2 2 2
+→+
+→+
Extent of reaction mole balances:
( ) ( ) ( )( ) ( )( ) ( )( ) ( ) ( ) 210OHOH
20COCO
10COCO
210CHCH
22nn
1nn
1nn
11nn
22
22
44
ξ++ξ++=
ξ++=
ξ++=
ξ−+ξ−+= 0.78 =7.80− ξ1 − ξ2nCO = ξ1nCO2
=8nCO = ξ2nH2O =2ξ1 +2ξ2( )
( ) ( ) ( ) 2123
0OO
0
2nn22
22
ξ−+ξ−+= nO2
=19.4− 32
ξ1 −2ξ2
Product Separation and Recycle
Two definitions of reactant conversion are used in the analysis of chemical reactors with product separation and recycle of unconsumed reactants.
⎟⎟⎞
⎜⎜⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛=
reactorfromoutput‐reactortoinputreactant
pass glesin
process to inputprocess from output ‐ process to input
reactantconversion
overall
⎟⎟⎠
⎜⎜⎝
=reactor to input
reactantconversion
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%75%100A/min mol 100
A/min mol 25 ‐ A/min mol 100conversion
pass glesin
%100%100A/min mol 75
0 ‐ A/min mol 75conversion
overall
=×⎟⎠⎞
⎜⎝⎛=
=×⎟⎠⎞
⎜⎝⎛=
Catalytic Propane Dehydrogenation
C3H8 → C3H6 +H2
95% overallconversion
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Catalytic Propane Dehydrogenation95% overallconversion
C3H8 → C3H6 +H2
Overall Process
n = 3 unknowns (n n n )ndf = 3 unknowns (n6, n7, n8) – 2 independent atomic balances (C and H) – 1 relation (overall conversion)= 0
consider n6, n7, n8 known for further DOF analyses
Catalytic Propane Dehydrogenation95% overallconversionndf = 2
C3H8 → C3H6 +H2
Mixingpoint
n = 4 unknowns (n n n n )ndf = 4 unknowns (n9, n10, n1, n2) – 2 balances (C3H8 and C3H6) = 2
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Catalytic Propane Dehydrogenation95% overallconversionndf = 2
ndf = 3
C3H8 → C3H6 +H2
reactor
n = 5 unknowns (n n n n n )ndf = 5 unknowns (n3, n4, n5, n1, n2) – 2 balances (C and H) = 3
Catalytic Propane Dehydrogenation95% overallconversionndf = 2
ndf = 3 ndf = 0
C3H8 → C3H6 +H2
separator
n = 5 unknowns (n n n n n )ndf = 5 unknowns (n3, n4, n5, n9, n10) – 3 balances (C3H8, C3H6, and H2)– 2 relations (reactant and product recovery fractions)= 0
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Catalytic Propane Dehydrogenation95% overallconversion
C3H8 → C3H6 +H2
overall
( )( ) HCl5l1009501
conversionrelationship
( )( ) 836 HCmol5mol10095.01n =−=
Catalytic Propane Dehydrogenation95% overallconversion
836 HCmol 5n =
C3H8 → C3H6 +H2
overallC atomic balance
( )( ) ( )( ) ( )637
HC mol 1C mol3
7HC mol 1C mol3
83HC mol 1C mol3
HC mol 95n
nHC mol 5mol 100638383
=
+=
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Catalytic Propane Dehydrogenation95% overallconversion
836 HCmol 5n =
C3H8 → C3H6 +H2
overall
H atomic balance
637 HCmol 95n =
( )( ) ( )( )( )( ) ( )
263
8383
H mol 1H mol 2
8HC mol 1H mol 6
63
HC mol 1H mol8
83HC mol 1H mol 8
nHC mol 95
HC mol 5mol 100
++
=
28 H mol 95n =
Catalytic Propane Dehydrogenation95% overallconversion
836 HCmol 5n =
C3H8 → C3H6 +H2
separator
given relations
637 HCmol 95n =
28 H mol 95n =
( )( ) 6310710
83336
HC mol 75.4nn0500.0n
HCmol900nn00555.0n
=→==→=
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Catalytic Propane Dehydrogenation95% overallconversion
833 HCmol900n = 836 HCmol 5n =
C3H8 → C3H6 +H2
separator
propane balance n10 = 4.75 mol C3H6
637 HCmol 95n =
28 H mol 95n =
n3 =n6 +n9 →n9 = 895 mol C3H8
Catalytic Propane Dehydrogenation95% overallconversion
833 HCmol900n = 836 HCmol 5n =
C3H8 → C3H6 +H2
mixerpropane balance
n10 = 4.75 mol C3H6
n9 = 895 mol C3H8
637 HCmol 95n =
28 H mol 95n =
100+n9 =n1 →n1 = 995 mol C3H8
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Catalytic Propane Dehydrogenation95% overallconversion
n1 = 995 mol C3H8
833 HCmol900n = 836 HCmol 5n =
C3H8 → C3H6 +H2
mixer
propylene balance n10 = 4.75 mol C3H6
n9 = 895 mol C3H8
637 HCmol 95n =
28 H mol 95n =
n10 =n2 →n2 = 4.75 mol C3H6
Catalytic Propane Dehydrogenation95% overallconversion
n1 = 995 mol C 3H8 833 HCmol900n = 836 HCmol 5n =
C3H8 → C3H6 +H2
reactor
C atomic balance n10 = 4.75 mol C3H6
n9 = 895 mol C3H8
n1 995 mol C 3H8
n2 = 4.75 mol C3H6
( )( ) ( )( )Cmol3Cmol3 HCl754HCl959 +
637 HCmol 95n =
28 H mol 95n =
( )( ) ( )( )( )( ) ( )( )
634
HC mol 1C mol 3
4HC mol 1C mol3
83
HC mol 1Cmol3
63HC mol 1Cmol3
83
HC mol 75.99n
nHC mol 009
HCmol.754HC mol 959
6383
6383
=
+=
+
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Catalytic Propane Dehydrogenation95% overallconversion
n1 = 995 mol C 3H8 833 HCmol900n = 836 HCmol 5n =
C3H8 → C3H6 +H2
reactor
H atomic balance n10 = 4.75 mol C3H6
n9 = 895 mol C3H8
n1 995 mol C 3H8
n2 = 4.75 mol C3H6634 HCmol75.99n = 637 HCmol 95n =
28 H mol 95n =
( )( ) ( )( )( )( ) ( )( ) ( )
25
H mol 1H mol 2
5HC mol 1H mol 6
63HC mol 1H mol 8
83
HC mol 1H mol 6
63HC mol 1H mol 8
83
H mol 95n
nHC mol 75.99HC mol 009
HC mol .754HC mol 959
26383
6383
=
++=
+
Catalytic Propane Dehydrogenation95% overallconversion
n1 = 995 mol C 3H8 833 HCmol900n = 836 HCmol 5n =
C3H8 → C3H6 +H2
single‐passconversion
n10 = 4.75 mol C3H6
n9 = 895 mol C3H8
n1 995 mol C 3H8
n2 = 4.75 mol C3H6634 HCmol75.99n =
25 Hmol95n =637 HCmol 95n =
28 H mol 95n =
( ) ( )( ) %55.9%100
HC mol 959HCmol009HCmol959
f83
8383passsingle =×
−=−
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Catalytic Propane Dehydrogenation95% overallconversion
836 HCmol 5n =833 HCmol900n =n1 = 995 mol C 3H8
fsingle−pass =9.55%
C3H8 → C3H6 +H2
Recycle ratio
637 HCmol 95n =
28 H mol 95n =
n10 = 4.75 mol C3H6
n9 = 895 mol C3H8
n1 995 mol C 3H8
n2 = 4.75 mol C3H6634 HCmol75.99n =
25 Hmol95n =
( )( ) ( )
( ) feed fresh molrecycle mol109 0.9
mol 001mol .754mol 958
feed mol 001nn
R =+
=+
=
PurgingNecessary with recycle to prevent accumulation of a species that is both present in the fresh feed and is recycled rather than separated with the
dproduct.
mixed fresh feedand recycle is aconvenient
basis selection fsingle−pass = 60%
CO2 +3H2 → CH3OH+H2O
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Methanol Synthesis
ndf = 7 unknowns (n0, x0C, np, x5C, x5H, n3, n4) + 1 rxnp 5 5 3
‐ 5 independent species balances = 3
fsingle−pass = 60%
CO2 +3H2 → CH3OH+H2O
Methanol Synthesisndf = 4 unknowns (n1, n2, n3, n4) + 1 rxn
– 4 independent species balances 4 p p– 1 single pass conversion = 0
fsingle−pass = 60%
CO2 +3H2 → CH3OH+H2O
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ndf = 3 unknowns (n5, x5C, X5H)– 3 independent species balances
= 0
fsingle−pass = 60%
CO2 +3H2 → CH3OH+H2O
ndf = 3 unknowns (n0, x0C, nr)– 3 independent species balances
= 0
fsingle−pass = 60%
CO2 +3H2 → CH3OH+H2O
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ndf = 1 unknowns (np)– 1 independent species balance
investigatemole balances andtheir solution in the textp p
= 0
fsingle−pass = 60%
the text
CO2 +3H2 → CH3OH+H2O
Combustion Reactions
Combustion ‐ rapid reaction of a fuel with oxygen.p ygValuable class of reactions due to the tremendous amount of heat liberated, subsequently used to produce steam used to drive turbines which generates most of the world’s electrical power.Common fuels used in power plants:
coalcoalfuel oil (high MW hydrocarbons)gaseous fuel (natural gas)liquified petroleum gas (propane and/or butane)
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Combustion Chemistry
When a fuel is burnedC forms CO2 (complete) or CO (partial combustion)H forms H2OS forms SO2
N forms NO2 (above 1800°C)Air is used as the source of oxygen. DRY air analysis:
78.03 mol% N2ll f20.99 mol% O2
0.94 mol% Ar0.03 mol% CO2
0.01 mol% H2, He, Ne, Kr, Xe
usually safe to assume:79 mol% N221 mol% O2
Combustion Chemistry
Stack (flue) gas – product gas that leaves a furnace.(f ) g p gComposition analysis:
wet basis – water is included in mole fractionsdry basis – does not include water in mole fractions
Stack gas contains (mol) on a wet basis:60.0% N2, 15.0% CO2, 10.0% O2, 15.0% H2ODry basis analysis:Dry basis analysis:
60/(60+15+10) = 0.706 mol N2/mol15/(60+15+10) = 0.176 mol CO2/mol10/(60+15+10) = 0.118 mol O2/mol
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Combustion Chemistry
Stack gas contains (mol) on a dry basis:65% N2, 14% CO2, 10% O2, 11% CO
assume 100 mole dry gas basis7.53 mole H2O65 mole N2
14 mole CO2
xH2O = 7.53107.5
= 0.0700
xN2= 65
107.5= 0.605
xCO2= 14
107 5= 0.130
10 mole O2
11 mole COtotal = 107.5 mole
CO2 107.5
xO2= 10
107.5= 0.0930
xCO = 11107.5
= 0.102
Theoretical and Excess AirThe less expensive reactant is commonly fed in excess of stoichiometricratio relative to the more valuable reactant thereby increasingratio relative to the more valuable reactant, thereby increasingconversion of the more expensive reactant at the expense of increaseduse of excess reactant.In a combustion reaction, the less expensive reactant is oxygen,obtained from the air. Conseqently, air is fed in excess to the fuel.Theoretical oxygen is the exact amount of O2 needed to completelycombust the fuel to CO2 and H2O.Theoretical air is that amount of air that contains the amount ofTheoretical air is that amount of air that contains the amount oftheoretical oxygen.
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Theoretical and Excess Air
Excess air is the amount by which the air fed to the Excess air is the amount by which the air fed to the reactor exceeds the theoretical air.
% excess air =
moles air fed ‐ moles air theoretical
moles air theoretical×100%
Theoretical and Excess Air
C4H10 + 13
2O2 → 4CO2 +5H2O
nC4H10 = 100 mol/hr; nair = 5000 mol/hr
( )hrO mol
650HC molO mol 5.6
hrHC mol 100
n 2
104
2104ltheoreticaO2
=⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛=
( ) airmol3094
airmol.764Omol506n 2
⎟⎟⎞
⎜⎜⎛
⎟⎞
⎜⎛( )
hr3094
O molhrn
2
2ltheoreticaair =⎟⎟
⎠⎜⎜⎝
⎟⎠⎞
⎜⎝⎛=
% excess air =
5000−30943094
×100% =61.6%
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Combustion Reactors
Procedure for writing/solving material balances for a g/ gcombustion reactor1. When you draw and label the flowchart, be sure the outlet
stream (the stack gas) includesa. unreacted fuel (unless the fuel is completely consumed)b. unreacted oxygenc. water and carbon dioxide (and CO if combustion is
i l )incomplete)d. nitrogen (if air is used as the oxygen source)
2. Calculate the O2 feed rate from the specifed percent excess oxygen or air
3. If multiple reactions, use atomic balances
Combustion of Ethane
OHCOOHCOHCOOHC
5
22227
62
3232
+→+
+→+fC2H6 = 0.9OHCOOHC 222
562 32 +→+ C2H6 9
25% of the ethane burned forms CO
degree‐of‐freedomanalysis
ndf = 7 unknowns‐ 3 atomic balances 1 nitrogen balance‐ 1 nitrogen balance
‐ 1 excess air specification‐ 1 ethane conversion specification‐ 1 CO/CO2 ratio specification
= 0
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fC2H6 = 0.9excess air specification
OHCOOHCOHCOOHC
2225
62
22227
62
3232
+→+
+→+
25% of the ethane burned forms CO
( ) 262
262ltheoreticaO O mol 350
HC molO mol 5.3HC mol 100
n2
=⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
( )( )air mol 2500n
Omol 3505.1n21.0
0
20
==
( )( ) 62621 HCmol0.10HCmol10090.01n =−=
Ethane conversion specification
C H 7 O 2CO 3H O
fC2H6 = 0.9CO/CO2 ratio specification n0 = 2500 mol air
n1 =10.0 mol C2H6
C2H6 + 72O2 → 2CO2 +3H2O
C2H6 + 52O2 → 2CO+3H2O
25% of the ethane burned forms CO
( )( )( ) OC mol 0.45react HC mol 1gen CO mol 2
HC mol 1009.025.0n62
624 =⎟⎟⎠
⎞⎜⎜⎝
⎛=
Nitrogen balance
( )( ) 23 Nmol1975iramol250079.0n ==
( )( ) ( ) ( ) ( )25
CO mol 1C mol 1
5CO mol 1C mol 1
4HC mol 1C mol 2
1HC mol 1C mol 2
62
CO mol 135n
nnnHC mol 10026262
=
++=
Atomic C balance
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C2H6 + 72O2 → 2CO2 +3H2O
fC2H6 = 0.9
n1 =10.0 mol C2H6
n3 =1975 mol N2
C2H6 + 5
2O2 → 2CO+3H2O
atomic H balance
( )( ) ( )( ) ( )OH mol 270n
nHC mol 10HC mol 100
26
OH mol 1H mol 2
6HC mol 1H mol 6
62HC mol 1H mol 6
62 26262
=
+=
n0 = 2500 mol air
n4 =45.0 mol CO
n5 =135 mol CO2
( )( ) ( ) ( )( )( )( ) ( )( )
22
OH mol 1O mol 1
2CO mol 1O mol 2
2
CO mol 1O mol 1
O mol 1O mol 2
2O mol 1O mol 2
2
O mol 232n
OH mol 702CO mol 135
CO mol 54nO mol 255
22
22
=
++
+=
atomic O balance
C2H6 + 72O2 → 2CO2 +3H2O
C2H6 + 52O2 → 2CO+3H2O
fC2H6 = 0.9
n1 =10.0 mol C2H6
n3 =1975 mol N2
C2H6 9
25% of the ethane burned forms COatomic O balance
( )( ) ( ) ( )( )( )( ) ( )( )OHmol1
O mol 12COmol1
O mol 22
CO mol 1O mol 1
O mol 1O mol 2
2O mol 1O mol 2
2
OHmol702COmol135
CO mol 54nO mol 25522
++
+=
n0 = 2500 mol air
n4 = 45.0 mol CO
n5 =135 mol CO2
n6 = 270 mol H2O
( )( ) ( )( )22
OHmol12COmol12
O mol 232n22
=
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C2H6 + 72O2 → 2CO2 +3H2O
C2H6 + 52O2 → 2CO+3H2O
fC2H6 = 0.9
n1 =10.0 mol C2H6
n3 =1975 mol N2
n2 = 232 mol O2
C2H6 9
25% of the ethane burned forms CO
stack gas composition (dry basis)
sum=10+232+1974+45+135= 2396
n0 = 2500 mol air
n4 = 45.0 mol CO
n5 =135 mol CO2
n6 = 270 mol H2O
y1 =10 2396= 0.00417 mol C2H6 mol
y2 = 232 2396= 0.0970 mol O2 mol
y3 =1974 2396=0.824 mol N2 mol
y4 = 45 2396=0.019 mol CO mol
y5 =135 2396=0.0563 mol CO2 mol
270 mol H2O
2396 mol dry stack gas
=0.113mol H2O
mol dry stack gas