Upload
mxr-3
View
270
Download
5
Embed Size (px)
Citation preview
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 1/66
CHEMICAL REACTION
STOICHIOMETRY
CHAPTER 3- Material balance for
SIMPLE reactive system
BALANCES REACTIVE
PROCESSES
COMBUSTION REACTIONS
SEPARATION & RECYCLE
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 2/66
1) Stoichiometry
CHAPTER 3- Material balance for
SIMPLE reactive system
CHEMICAL REACTIONSTOICHIOMETRY
! Stoichiometry – theory of the proportions in which chemical speciescombine with one another.
*Note: Refer Felder pp.116.
! The Stoichiometric Equation of chemical reaction – statement of therelative number of molecules or moles of reactants and products thatparticipate in the reaction – must be balanced (no. of atoms of eachatomic species must be the same on both sides of the equation).Example:
2 SO2 + O2 → 2 SO3
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 3/66
CHAPTER 3- Material balance for
SIMPLE reactive system
CHEMICAL REACTIONSTOICHIOMETRY
! The Stoichiometric ratio two molecular species participating in areaction – ratio of their stoichiometric coefficients in the balancedequation.
1) Stoichiometry
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 4/66
CHAPTER 3- Material balance for
SIMPLE reactive system
CHEMICAL REACTIONSTOICHIOMETRY
! The limiting reactant – reactant that would run out if a reactionproceeded to completion, other reactants are termed excess reactant.
*Note: Refer Felder pp.118.
! A reactant is limiting if it is present in less than its stoichiometric proportionrelative to every other reactants.
! Fractional excess of reactant is the ratio of the excess to thestoichiometric requirement.
*Note: Refer Felder pp.120 for Example 4.6 – 1.
2) Limiting and Excess Reactants, Fractional Conversion and Extent
of Reaction
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 5/66
CHAPTER 3- Material balance for
SIMPLE reactive system
CHEMICAL REACTIONSTOICHIOMETRY
! Fractional conversion of a reactant is the ratio:
2) Limiting and Excess Reactants, Fractional Conversion and Extent of Reaction
( ) ( )
( )
of Anal excess x Fractio Aexcess of Percentage
n
nn Aexcess of Fractional
stoich A
stoich A feed A
100=
−
=
sion of Anal conver x Fraction of Aconversio Percentage
moles feed
ted moles reac n of Aconversio Fractional
100=
=
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 6/66
CHAPTER 3- Material balance for
SIMPLE reactive system
CHEMICAL REACTIONSTOICHIOMETRY
! In chemical processes, some reactants can usually combine in morethan one way, and the product once formed may react to yieldsomething less desirable.
*Note: Refer Felder pp.123 and Example 4.6 – 3 pp 124.
! For example, ethylene can be produced by the dehydrogenation ofethane:
C2H6 → C2H4 + H2
3) Multiple Reactions Yield and Selectivity
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 7/66
CHAPTER 3- Material balance for
SIMPLE reactive system
CHEMICAL REACTIONSTOICHIOMETRY
! Once hydrogen is produced, it can react with ethane to producemethane:
C2H6 + H2 → 2 CH4
• Moreover, ethylene can react with ethane to form propylene andmethane:
C2H4 + C2H6 → C3H6 + CH4
3) Multiple Reactions Yield and Selectivity
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 8/66
CHAPTER 3- Material balance for
SIMPLE reactive system
CHEMICAL REACTIONSTOICHIOMETRY
! Yield :
mpletelyreacted co
t had taniting reacd theactions anno side re
re weremed if thee been for would havmoles that
d duct formeesired promoles of d Yield
lim
=
! Selectivity :
med roduct for ndesired pmoles of u
d duct formeesired promoles of d ySelectivit =
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 9/66
MOLECULAR SPECIES BALANCE
CHAPTER 3- Material balance for
SIMPLE reactive system
ATOMIC SPECIES BALANCE
EXTENT OF REACTION
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 10/66
! Example
Methane is burned with air in a continuous steady state combustion reactorto yield a mixture of carbon monoxide, carbon dioxide and water. Thereaction taking place are:
CH4 + 3/2 O2 → CO + 2 H2O
CH4 + 2 O2 → CO2 + 2 H2O
The feed to the reactor contains 7.80 mole% CH4 , 19.4 % O2 , and 72.8% N2.The percentage conversion of methane is 90.0%, and the gas leaving thereactor contains 8 mol CO2/mol CO.
Calculate the molar composition of the product stream.
*Note: Refer felder pp.131
CHAPTER 3- Material balance for
SIMPLE reactive system
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 11/66
HO W ? ? ?
Analyze the information….
! Feed: 0.078 mol CH4/mol, 0.194 mol O2/mol, 0.728 mol N2/mol
! Percentage conversion of CH4: 90.0% @ 0.90
! Product gas: 8 mol CO2/mol CO
CHAPTER 3- Material balance for SIMPLE reactive system
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 12/66
! Basis of calculation: 100 mol of feed
! Process flow chart:
Process
unit
100 mol
0.078 mol CH4/mol
0.194 mol O2/mol
0.728 mol N2/mol
n1 mol CH4
n2 mol O2
n3 mol N2
n4 mol CO
8n4 mol CO2
n5 mol H2O
CHAPTER 3- Material balance for
SIMPLE reactive system
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 13/66
! Write system equation andoutline a solution procedure
Additional information – fractionalconversion CH 4 = 0.90
( )
4
4
4
4
4
4
CH moles
CH unreactedof Moles
mol7.020.901000.078reactedCH Moles
1000.078reactedCH moles
fedCH molesreactedCH moles
7800
027807
900
1
1
.n
..n
. f
=
−=
=××=
=
×
==
Process
unit
100 mol
0.078 mol CH4/mol
0.194 mol O2/mol
0.728 mol N2/mol
n1 mol CH4
n2 mol O2
n3 mol N2
n4 mol CO
8n4 mol CO2
n5 mol H2O
Additional information – N 2 in = N 2 out = 0.728 x 100 = 72.8 mol N 2
n3 = 72.8 mol N 2
CHAPTER 3- Material balance for
SIMPLE reactive system
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 14/66
CH 4 Balance – From first reaction:- Process
unit
100 mol
0.078 mol CH4/mol
0.194 mol O2/mol
0.728 mol N2/mol
n1 mol CH4
n2 mol O2
n3 mol N2
n4 mol CO
8n4 mol CO2
n5 mol H2O
CH4 + 3/2 O2 → CO + 2 H2O
! 1 mol CH4 → 1 mol CO
! x mol CH4 → x mol CO = n4 → x = n4 → Eq.(1)
CH 4 Balance – From second reaction:-
CH4 + 2 O2 → CO2 + 2 H2O
! 1 mol CH4 → 1 mol CO2 ! (7.02 – x) mol CH4 → (7.02 – x) mol CO2 = 8n4
→ 8n4 = 7.02 – x→ Eq.(2)
! Insert Eq. 1 into 2 COmol
9
780
9
027
0270278
4
444
.
.
n
.nn.n
==
=→−=
CHAPTER 3- Material balance for
SIMPLE reactive system
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 15/66
O2 Balance – From first reaction:- Processunit
100 mol
0.078 mol CH4/mol
0.194 mol O2/mol
0.728 mol N2/mol
n1 mol CH4
n2 mol O2
n3 mol N2
n4 mol CO
8n4 mol CO2
n5 mol H2O
CH4 + 3/2 O2 → CO + 2 H2O
! 3/2 mol O2 → 1 mol CO
! 0.78 mol CO → 0.78 x 3/2 mol O2 = 1.17 mol O2 consumed
O2 Balance – From second reaction:-
CH4 + 2 O2 → CO2 + 2 H2O
! 2 mol O2→ 1 mol CO2 ! 6.24 mol CO2→ 6.24 x 2 mol O2 = 12.48 mol O2 consumed
mol
12.48-1.17-0.194x100consumed-
22
222
755 O.n
O Fed On
=
==
CHAPTER 3- Material balance for
SIMPLE reactive system
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 16/66
H 2O Balance – From first reaction:- Processunit
100 mol
0.078 mol CH4/mol
0.194 mol O2/mol
0.728 mol N2/mol
n1 mol CH4
n2 mol O2
n3 mol N2
n4 mol CO
8n4 mol CO2
n5 mol H2O
CH4 + 3/2 O2 → CO + 2 H2O
! 2 mol H2 O→ 1 mol CO
! 0.78 mol CO → 0.78 x 2 mol H2 O = 1.56 mol H2 O produced
H 2O Balance – From second reaction:-
CH4 + 2 O2 → CO2 + 2 H2O
! 2 mol H2 O → 1 mol CO2 ! 6.24 mol CO2→ 6.24 x 2 mol H2 O = 12.48 mol H2 O produced
O H .n
O H n
mol
12.481.56 produced
22
25
0414=
+==∑
CHAPTER 3- Material balance for
SIMPLE reactive system
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 17/66
Processunit
100 mol
0.078 mol CH4/mol
0.194 mol O2/mol
0.728 mol N2/mol
n1 mol CH4
n2 mol O2
n3 mol N2
n4 mol CO
8n4 mol CO2
n5 mol H2O
Component Mole Composition (%)
O2 5.75 5.72
CO 0.78 0.78
CO2 6.24 6.22
CH4 0.78 0.78
H2O 14.04 13.99
N2 72.8 72.51
Total 100.39 100%
A NS W E R
CHAPTER 3- Material balance for
SIMPLE reactive system
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 18/66
! Example
Methane is burned with air in a continuous steady state combustion reactorto yield a mixture of carbon monoxide, carbon dioxide and water. Thereaction taking place are:
CH4 + 3/2 O2 → CO + 2 H2O
CH4 + 2 O2 → CO2 + 2 H2O
The feed to the reactor contains 7.80 mole% CH4 , 19.4 % O2 , and 72.8% N2.The percentage conversion of methane is 90.0%, and the gas leaving thereactor contains 8 mol CO2/mol CO.
Calculate the molar composition of the product stream.
*Note: Refer felder pp.131
CHAPTER 3- Material balance for
SIMPLE reactive system
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 19/66
HO W ? ? ?
Analyze the information….
! Feed: 0.078 mol CH4/mol, 0.194 mol O2/mol, 0.728 mol N2/mol
! Percentage conversion of CH4: 90.0% @ 0.90
! Product gas: 8 mol CO2/mol CO
CHAPTER 3- Material balance for
SIMPLE reactive system
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 20/66
! Basis of calculation: 100 mol of feed
! Process flow chart:
Process
unit
100 mol
0.078 mol CH4/mol
0.194 mol O2/mol
0.728 mol N2/mol
n1 mol CH4
n2 mol O2
n3 mol N2
n4 mol CO
8n4 mol CO2
n5 mol H2O
CHAPTER 3- Material balance for
SIMPLE reactive system
C f
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 21/66
! Write system equation andoutline a solution procedure
Additional information – fractionalconversion CH 4 = 0.90
( )
4
4
4
4
4
4
CH moles
CH unreactedof Moles
mol7.020.901000.078reactedCH Moles
1000.078reactedCH moles
fedCH molesreactedCH moles
7800
027807
900
1
1
.n
..n
. f
=
−=
=××=
=
×
==
Processunit
100 mol
0.078 mol CH4/mol
0.194 mol O2/mol
0.728 mol N2/mol
n1 mol CH4
n2 mol O2
n3 mol N2
n4 mol CO
8n4 mol CO2
n5 mol H2O
Additional information – N 2 in = N 2 out = 0.728 x 100 = 72.8 mol N 2
n3 = 72.8 mol N 2
CHAPTER 3- Material balance for
SIMPLE reactive system
CHAPTER 3 M i l b l f
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 22/66
! Atom C: 1 unk. (n4)
! Atom H: 1 unk. (n5)
! Atom O: 3 unk. (n2, n4, n5)
Analyze the atomic balance for:
mol0.78
9
mol7.02 0.780-7.89 80.7807.8
CO mol1
Cmol1COmol8
COmol1
Cmol1COmol
CH mol1
Cmol1CH mol0.780
CH mol1
Cmol1CH mol.8
2
2
4
4
4
4
==→=→++=
×+×+×=×
4444
447
nnnn
nn
Solve the atomic balance on C…
CHAPTER 3- Material balance for
SIMPLE reactive system
Processunit
100 mol
0.078 mol CH4/mol
0.194 mol O2/mol
0.728 mol N2/mol
0.780 mol CH4
n2 mol O2
72.80 mol N2
n4 mol CO
8n4 mol CO2
n5 mol H2O
mol14.042mol28.08 3.12-31.22 23.1231.2
OH mol1
H mol2OH mol
CH mol1
H mol4CH mol0.780
CH mol1
H mol4CH mol.8
2
2
4
4
4
4
==→=→+=
×+×=×
555
57
nnn
n
Solve the atomic balance on H…
CHAPTER 3 M t i l b l f
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 23/66
mol5.752
mol11.5 11.52 14.042.480.78238.8
H mol1Omol1H mol14.04
CO mol1Omol2COmol6.24
COmol1
Omol1COmol0.78
Omol1
Omol2Omol
Omol1
Omol2Omol9.4
2
2
2
2
2
2
2
2
==→=→+++=
×+×+
×+×=×
222
2
1
1
nnn
OO
n
Solve the atomic balance on O…
CHAPTER 3- Material balance for
SIMPLE reactive system
Process
unit
100 mol
0.078 mol CH4/mol
0.194 mol O2/mol
0.728 mol N2/mol
0.780 mol CH4
n2 mol O2
72.80 mol N2
0.78 mol CO
6.24 mol CO2
14.04 mol H2O
CHAPTER 3 M t i l b l f
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 24/66
Component Mole Composition (%)
O2 5.75 5.72
CO 0.78 0.78
CO2 6.24 6.22
CH4 0.78 0.78
H2O 14.04 13.99
N2 72.8 72.51
Total 100.39 100%
A NS W E R
CHAPTER 3- Material balance for
SIMPLE reactive system
Process
unit
100 mol
0.078 mol CH4/mol
0.194 mol O2/mol
0.728 mol N2/mol
0.780 mol CH4
5.75 mol O2
72.80 mol N2
0.78 mol CO
6.24 mol CO2
14.04 mol H2O
CHAPTER 3 M t i l b l f
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 25/66
! Example
Methane is burned with air in a continuous steady state combustion reactorto yield a mixture of carbon monoxide, carbon dioxide and water. Thereaction taking place are:
CH4 + 3/2 O2 → CO + 2 H2O
CH4 + 2 O2 → CO2 + 2 H2O
The feed to the reactor contains 7.80 mole% CH4 , 19.4 % O2 , and 72.8% N2.The percentage conversion of methane is 90.0%, and the gas leaving thereactor contains 8 mol CO2/mol CO.
Calculate the molar composition of the product stream.
*Note: Refer felder pp.131
CHAPTER 3- Material balance for
SIMPLE reactive system
CHAPTER 3 M t i l b l f
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 26/66
HO W ? ? ?
Analyze the information….
! Feed: 0.078 mol CH4/mol, 0.194 mol O2/mol, 0.728 mol N2/mol
! Percentage conversion of CH4: 90.0% @ 0.90
! Product gas: 8 mol CO2/mol CO
CHAPTER 3- Material balance for
SIMPLE reactive system
CHAPTER 3 M t i l b l f
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 27/66
! Basis of calculation: 100 mol of feed
! Process flow chart:
Process
unit
100 mol
0.078 mol CH4/mol
0.194 mol O2/mol
0.728 mol N2/mol
n1
mol CH4
n2 mol O2
n3 mol N2
n4 mol CO
8n4 mol CO2
n5 mol H2O
CHAPTER 3- Material balance for
SIMPLE reactive system
CHAPTER 3 M t i l b l f
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 28/66
!Write system equation andoutline a solution procedure
Additional information – fractionalconversion CH 4 = 0.90
( )
4
4
4
4
4
4
CH moles
CH unreactedof Moles
mol7.020.901000.078reactedCH Moles
1000.078reactedCH moles
fedCH molesreactedCH moles
7800
027807
900
1
1
.n
..n
. f
=
−=
=××=
=
×
==
Processunit
100 mol
0.078 mol CH4/mol
0.194 mol O2/mol
0.728 mol N2/mol
n1 mol CH4
n2 mol O2
n3 mol N2
n4 mol CO
8n4 mol CO2
n5 mol H2O
Additional information – N 2 in = N 2 out = 0.728 x 100 = 72.8 mol N 2
n3 = 72.8 mol N 2
CHAPTER 3- Material balance for
SIMPLE reactive system
CHAPTER 3 M t i l b l f
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 29/66
From first reaction:-
Process
unit
100 mol
0.078 mol CH4/mol
0.194 mol O2/mol
0.728 mol N2/mol
0.78 mol CH4
n2
mol O2
72.8 mol N2
n4 mol CO
8n4 mol CO2
n5 mol H2O
CH4
+ 3/2 O2
→ CO + 2 H2
O
From second reaction:-
CH4 + 2 O2 → CO2 + 2 H2O
CHAPTER 3- Material balance for
SIMPLE reactive system
∑+= jiji0i ξvnn:equationFrom
1ξ
2ξ
( ) ( )[ ]
( )
( ) ( )
( ) ( ) 3Eq. 0
2Eq. 2--19.419.4
1Eq. mol7.02 0.780--7.80
mol0.780 n,calculatioFrom
--7.807.80
→=→=×++==
→=⎥⎦
⎤⎢⎣
⎡×−+⎟
⎠
⎞⎜⎝
⎛×−+==
→=+→=
=
=×−+×−+==
14114
21212
2121
1
21211
1
2
32
2
3
11
2
4
ξ ξ ξ
ξ ξ ξ ξ
ξ ξ ξ ξ
ξ ξ ξ ξ
nnn
nn
n
nn
CO
O
CH
CHAPTER 3 Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 30/66
CHAPTER 3- Material balance for
SIMPLE reactive system
( ) ( )
( )
( ) ( )
( ) ( )
( ) ( )( ) ( )[ ] ( )
( ) ( )
( ) ( ) mol14.046.2420.7822ξ2ξnn
mol5.756.242-0.782
3 -19.42ξ-ξ
2
3 -19.4nn
mol6.240.7888nξn
mol0.78ξnnmol0.78n mol7.029n mol7.028nn mol7.02ξξ
(1)Equationinto(4)and(3)EqSubstitute
5Eq. 2ξ2ξn 2ξ2ξξ2ξ20nn
4Eq. ξ8n ξξ108nn
3Eq. ξn ξξ10nn
2Eq. 2ξ-ξ2
3 -19.4ξ2ξ
2
3 19.4nn
1Eq. mol7.02ξξ 0.780ξ-ξ-7.80mol0.780n n,calculatioFrom
ξ-ξ-7.80ξ1ξ17.80nn
215OH
212O
42CO
14CO
444421
21521215OH
24224CO
14114CO
21212O
2121
1
21211CH
2
2
2
2
2
2
4
=+=+==
====
=×===
===
=→==+→=+
→+=→+=×++×++==
→=→=×++==
→=→=×++==
→=⎥⎦
⎤⎢⎣
⎡×−+⎟
⎠
⎞⎜⎝
⎛×−+==
→=+→=
=
=×−+×−+==
CHAPTER 3 Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 31/66
Component Mole Composition (%)
O2 5.75 5.72
CO 0.78 0.78
CO2 6.24 6.22
CH4 0.78 0.78
H2O 14.04 13.99
N2 72.8 72.51
Total 100.39 100%
A NS W E R
CHAPTER 3- Material balance for
SIMPLE reactive system
Process
unit
100 mol
0.078 mol CH4/mol
0.194 mol O2/mol
0.728 mol N2/mol
0.780 mol CH4
5.75 mol O2
72.80 mol N2
0.78 mol CO
6.24 mol CO2
14.04 mol H2O
CHAPTER 3 Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 32/66
1) Overall Conversion and Single Pass Conversion
CHAPTER 3- Material balance for
SIMPLE reactive system
! Overall Conversion :
SEPARATION & RECYCLE
processthetoinput reactant
processthe fromoutput reactant processthetoinput reactant ConversionOverall
−
=
! Single – Pass Conversion :
reactor thetoinput reactant
processthe fromoutput reactant reactor thetoinput reactant Conversion Pass-Single
−
=
CHAPTER 3 Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 33/66
CHAPTER 3- Material balance for
SIMPLE reactive system
! Example (Felder pp.135)
Dehydrogenation of propane
Propane is dehydrogenated to form propylene in a catalytic reactor:
C3 H8 → C3 H6 + H2 The process is to be designed for a 95% overall conversion of propane. The
reaction products are separated into two streams:
1st : contains H2 , C3H6 and 0.555% of the propane that leaves the reactor,
is taken off as product
2nd : contains balance of the unreacted propane and 5% of the propylene in the first stream.
Calculate the composition of the product, the ratio (moles recycled)/(molefresh feed) and the single pass conversion.
CHAPTER 3 Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 34/66
CHAPTER 3- Material balance for
SIMPLE reactive system
HO W ? ? ?
Basis of calculation.. reactor theto fed H C mol 100 3 8
! 95% overall conversion of propane
Analyze the information..
! Sketch and insert the known and unknown values….
CHAPTER 3 Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 35/66
CHAPTER 3- Material balance for
SIMPLE reactive system
Reactor Separator 100 mol C3H8
n1 mol C3H8
n2 mol C3H6
n3 mol C3H8
n4 mol C3H6
n5 mol H2
5.5x10-3 n3 mol C3H8
0.95 n4 mol C3H6
n5 mol H2
0.99445 n3 mol C3H8
0.05 n4 mol C3H6
CHAPTER 3 Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 36/66
CHAPTER 3- Material balance for
SIMPLE reactive system
! Overall conversion of propane = 0.95
83
3
3
90.900
100
1055.5100950
H C mol n
n .
3=
×−
=
−
! Mol balance at fresh feed – recycle on propane:
( )
mol n
n
nn
9.995
90.90099445.0100
99445.0100
1
1
13
=
=+
=+
CHAPTER 3 Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 37/66
CHAPTER 3- Material balance for
SIMPLE reactive system
! Mol balance at fresh feed – recycle on propylene:
2405.0 nn =
! Atomic Balance at Reactor:
C: n4
H : n5 , n4
! Atomic Balance on C:
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )64
44
4341
100
339.900305.039.995
33305.03
H C mol n
nn
nnnn
3=
+=+
+=+
62
24
510005.0
05.0
H C mol n
nn
3
=×=
=
CHAPTER 3 Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 38/66
CHAPTER 3- Material balance for
SIMPLE reactive system
!Atomic Balance on H:
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
25
5
54321
95
2610089.9006589.995
26868
H mol n
n
nnnnn
=
++=+
++=+
! Product gas outlet:
( )
( )mol n H
mol n H C
H C mol n H C 3
95
9510095.095.0
59.9001055.51055.5
52
462
8
3
3
3
83
==
===
=×=×=
−−
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 39/66
1) Combustion Chemistry
CHAPTER 3- Material balance for
SIMPLE reactive system
! Combustion – the rapid reaction of a fuel with oxygen.
*Note: Refer Felder pp.142.
! When a fuel is burned, carbon in the fuel reacts to form either CO2 orCO, hydrogen forms H2O, and sulfur form SO2.
! Partial or incomplete combustion – a combustion reaction in whichCO is formed from a hydrocarbon.
COMBUSTION REACTIONS
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 40/66
CHAPTER 3- Material balance for
SIMPLE reactive system
! Example:
C + O2 → CO2 Complete combustion of carbon
C3 H8 + 5O2 → 3CO2 + 4H2 O Complete combustion of propane
C3 H8 + 3.5O2 → 3CO + 4H2 O Incomplete combustion of propane
COMBUSTION REACTIONS
! Composition on a wet basis – mole fraction of a gas that containswater.
! Composition on a dry basis (Orsat Analysis)– mole fraction of thesame gas that without the water.
! Stack gas or flue gas – product gas that leaves a combustion furnace.
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 41/66
2) Theoretical and Excess Air
CHAPTER 3- Material balance for
SIMPLE reactive system
! Theoretical Oxygen – the moles (batch) or molar flow rate(continuous) of O2 needed for complete combustion of all the fuel fedto the reactor, assuming that all carbon in the fuel is oxidized to CO2 and all hydrogen is oxidized to H2O.
! Theoretical Air – the quantity of air that contains theoretical oxygen.
! Percent Excess Air:
*Note: Refer Felder pp.145.
COMBUSTION REACTIONS
( ) ( )
( )100%
airof moles
airof molesairof moles AirExcess%
ltheoretica
ltheoretica fed×
−
=
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 42/66
! Example (Felder pp.143)
A stack gas contains 60.0 mole% N2 , 15.0% CO2 , 10.0% O2 , and the balanceH2O. Calculate the molar composition of the gas on a dry basis.
CHAPTER 3- Material balance for
SIMPLE reactive system
HO W ? ? ?
Basis of calculation.. gaswetmol100
Comp.Mol (wet
basis)
O2 10
CO2 15
N2 60
H2O 15
Total 100 11.8%%O
10085
10%O
2
2
=
×=
Mol (Dry
basis)
% Mol
(Dry basis)
10 11.8
15 17.7
60 70.5
85 1.00
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 43/66
! Example (Felder pp.144)
An Orsat analysis (a technique for stack analysis) yields the following dry basis composition: 65.0 mole% N2 , 14.0% CO2 , 11% CO and 10.0% O2.
A humidity measurement shows that the mole fraction of H2O in the stackgas is 0.0700. Calculate the stack gas component on a wet basis.
CHAPTER 3 Material balance for
SIMPLE reactive system
HO W ? ? ?
Basis of calculation.. basisdrymol100
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 44/66
CHAPTER 3 Material balance for
SIMPLE reactive system
Comp.Mol (Dry
basis)
O2 10
CO2 14
N2 65
CO 11
H2O 0
Total 100
Mol (Wetbasis)
% Mol(Dry basis)
10
14
65
11
x 0.0700
100 + x 1.00
mol7.527x 70.93x
70.070x-x x0.07x7
x100
x0.070 nn y
total
OH
OH 2
2
=→=
=→=+
+
=→=
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 45/66
CHAPTER 3 Material balance for
SIMPLE reactive system
Comp.Mol (Dry
basis)
O2 10
CO2 14
N2 65
CO 11
H2O 0
Total 100
0930.==
107.527
10
y 2O
Mol (Wetbasis)
Mol fraction(Wet basis)
10 0.093
14 0.130
65 0.605
11 0.102
7.527 0.070
107.527 1.00
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 46/66
! Example (Felder pp.145)
100 mol/h of butane (C4H10) and 5000 mol/h of air are fed into a combustionreactor. Calculate the percentage of excess air.
CHAPTER 3 Material balance for
SIMPLE reactive system
HO W ? ? ?
Find the theoretical air needed
OH 5CO4O2
13 H C
222104+→+
22104
2104
Omol650Omol
2
13100 H Cmol100
Omol2
13 H Cmol1
=×→
→
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 47/66
CHAPTER 3 Material balance for
SIMPLE reactive system
( )
mol3095
0.21
mol650n
mol650nn0.21
mol650nl,theoreticaO
airltheoretica
Oairltheoretica
O2
2
2
==
==
=
1.6%airexcessof Percentage
100%3095
35 airexcessof Percentage
100%n
nn airexcessof Percentage
airltheoretica
airltheoreticaairexcess
6
095000
=
×
−
=
×
−
=
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 48/66
! Example (Felder pp.147)
Ethane is burned with 50% excess air. The percentage conversion of theethane is 90%; of the ethane burned, 25% react to form CO and the balancereact to form CO2. Calculate the molar composition of the stack gas on a dry basis and the mole ratio of water to dry stack gas.
CHAPTER 3 Material balance for
SIMPLE reactive system
OH 3CO2O2
5 H C
OH 3CO2O2
7 H C
2262
22262
+→+
+→+
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 49/66
CHAPTER 3 Material balance for
SIMPLE reactive system
HO W ? ? ?
Analyze the information….
! Feed: 50% excess air
! Percentage conversion of C2H6: 90.0% @ 0.90 – 25% form CO, balanceform CO2
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 50/66
! Basis of calculation: 100 mol of C2H6 fed
! Process flow chart:
Processunit
100 mol C2H6 n1 mol C2H6
n2 mol O2
n3 mol N2
n4 mol CO
n5 mol CO2
n6 mol H2O
50% excess air
n0 mol air
0.79 mol N2/mol
0.21 mol O2/mol
CHAPTER 3 Material balance for
SIMPLE reactive system
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 51/66
! Write system equation and
outline a solution procedure
Additional information – fractionalconversion C2H 6 = 0.90
( )
621
162
62
62
H Cmol10n
mol100.1100nH Cunreactedof Moles
mol900.90100reactedH C Moles
mol100 fedH C Moles
0.90 f
=
=×=
=×=
=
=
C 3 ate a ba a ce o
SIMPLE reactive system
Process
unit
100 mol C2H6 n1 mol C2H6
n2
mol O2
n3 mol N2
n4 mol CO
n5 mol CO2
n6 mol H2O
50% excess air
n0 mol air
0.79 mol N2/mol
0.21 mol O2/mol
Find the amount of air fed into the process unit:
! Calculate the theoretical O2 needed
! Calculate the theoretical air needed
! From % excess air, calculate the amount of excess air fed to the process unit
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 52/66
SIMPLE reactive system
Process
unit
100 mol C2H6 n1 mol C2H6
n2
mol O2
n3 mol N2
n4 mol CO
n5 mol CO2
n6 mol H2O
50% excess air
n0 mol air
0.79 mol N2/mol
0.21 mol O2/mol
( )
mol16670.21
Omol350
n
Omol350n0.21
2airltheoretica
2airltheoretica
==
=
Calculate the theoretical O2 needed
22262
262
OlTheoreticaOmol350Omol2
7 100 H Cmol100
Omol2
7 H Cmol1
:combustioncompleteFor
==×→
→
Calculate the theoretical air needed
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 53/66
SIMPLE reactive system
Process
unit
100 mol C2H6 n1 mol C2H6
n2
mol O2
n3 mol N2
n4 mol CO
n5 mol CO2
n6 mol H2O
50% excess air
n0 mol air
0.79 mol N2/mol
0.21 mol O2/mol
Calculate the amount of excess air fed to the process unit
2500moln
0.501667
1667n 0.50
n
nn
100%n
nnair%excess
0
0
airltheoretica
airltheoretica0
airltheoretica
airltheoretica0
=
=
−
→=
−
×
−
=
N 2 in = N 2 out = 0.79 x 2500 = 1975 mol N 2
n3 = 1975 mol N 2
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 54/66
SIMPLE reactive systemProcess
unit
100 mol C2H6 10 mol C2H6
n2 mol O2
1975 mol N2
n4 mol CO
n5 mol CO2
n6 mol H2O
50% excess air
2500 mol air
0.79 mol N2/mol0.21 mol O2/mol
Given 75% ethane reacted form CO2
COmol135n
formedCOmol135267.5 H Cmol67.5
COmol2 H Cmol1
:1reactionFrom
mol67.5mol900.75CO formtoreactH C Amount
25
262
262
262
=
=×→
→
=×=
Given 25% ethane reacted form CO
COmol45n
formedCOmol45222.5 Cmol22.5
COmol2 Cmol1
:2reactionFrommol22.5mol900.25CO formtoreactC
4
2
2
2
=
=×→
→
=×
=
6
6
6
H
H
H Amount
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 55/66
SIMPLE reactive system
Process
unit
100 mol C2H6 10 mol C2H6
n2 mol O2
1975 mol N2
45 mol CO
135 mol CO2
n6 mol H2O
50% excess air
2500 mol air
0.79 mol N2/mol
0.21 mol O2/mol
! Atom C: 0 unk.
! Atom H: 1 unk. (n6)
! Atom O: 2 unk. (n2, n6)
Analyze the atomic balance for:
mol2702
mol540 n 60-6002n 2n60600
OH mol1
H mol2OH moln
H Cmol1
H mol6H Cmol10
H Cmol1
H mol6H Cmol100
666
2
26
62
62
62
62
==→=→+=
×+×=×
Solve the atomic balance on H…
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 56/66
SIMPLE reactive system
Process
unit
100 mol C2H6 10 mol C2H6
n2 mol O2
1975 mol N2
45 mol CO
135 mol CO2
n6 mol H2O
50% excess air
2500 mol air
0.79 mol N2/mol
0.21 mol O2/mol
mol232.52
mol465 n 4652n 452n1050
OH mol1
Omol1OH mol
molCO1
Omol2COmol1
COmol1
Omol1COmol45
Omol1
Omol2Omoln
Omol1
Omol2Omol25000.21
222
2
2
2
2
2
2
2
2
==→=→+++=
×+×
+×+×=××
270270
27035
2
Solve the atomic balance on O…
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 57/66
Component Mole Composition (y)
O2 232.5 0.097
CO 45 0.019
CO2 135 0.056
C2H6 10 0.004
N2 1975 0.824
Total 2397.5 100%
A NS W E R
SIMPLE reactive system
Process
unit
100 mol C2H6 10 mol C2H6
232.5 mol O2
1975 mol N2
45 mol CO
135 mol CO2
270 mol H2O
50% excess air
2500 mol air
0.79 mol N2/mol
0.21 mol O2/mol
Moles ratio of water todry stack gas:
gasstackdrymolOH mol0.113
gasstackdrymol2397.5
OH mol270
2
2
=
=
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 58/66
! Example (Felder pp.149)
A hydrocarbon gas is burned with air. The dry basis product gas compositionis 1.5 mol%CO, 6.0 mol% CO2 , 8.2 % O2 , and 84.3% N2. There is no atomicoxygen in the fuel. Calculate the ratio of hydrogen to carbon in the fuel gasand speculate on what the fuel might be. Then calculate the percent excess airfed to the reactor.
SIMPLE reactive system
HO W ? ? ?
Analyze the information….
! Dry basis: 0.015 mol CO/mol, 0.06 mol CO2/mol,0.082 mol O2/mol, 0.843 mol N2/mol
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 59/66
! Basis of calculation: 100 mol of dry basis product
! Process flow chart:
Process
unit
n1 mol C
n2 mol H1.5 mol CO
8.2 mol O2
84.3 mol N2
6.0 mol CO2
n4 mol H2O
n3 mol air 0.79 mol N2/mol
0.21 mol O2/mol
SIMPLE reactive system
N 2 in = N 2 out = 84.3 mol N 2
Calculate the air fed to the unit
( )
mol106.70
0.79
mol84.3n
mol84.3n0.79
3
3
==
=
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 60/66
SIMPLE reactive system
Processunit
n1 mol C
n2 mol H1.5 mol CO
8.2 mol O2
84.3 mol N2
6.0 mol CO2
n4 mol H2O
106.70 mol air
0.79 mol N2/mol
0.21 mol O2/mol
! Atom C: 1 unk. (n1)
! Atom H: 2 unk. (n2, n4)
! Atom O: 1 unk. (n4)
Analyze the atomic balance for:
mol7.5n 61.5n
COmol1
Cmol1COmol6
COmol1
Cmol1COmol1.5Cmoln
11
2
21
=→+=
×+×=
Solve the atomic balance on C…
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 61/66
SIMPLE reactive system
Processunit
n1 mol C
n2 mol H1.5 mol CO
8.2 mol O2
84.3 mol N2
6.0 mol CO2
n4 mol H2O
106.70 mol air
0.79 mol N2/mol
0.21 mol O2/mol
mol14.9n n1.51216.444.814
OH mol1
Omol1OH moln
COmol1
Omol1COmol1.5
COmol1
Omol2COmol6
Omol1
Omol2Omol8.2
Omol1
Omol2Omol106.700.21
44
2
24
2
2
2
2
2
2
=→+++=
×+×
+×+×=××
Solve the atomic balance on O…
Solve the atomic balance on H…
mol9.8n
H mol1
H mol2H mol1H moln
2
2
22
2
94
=
×=
OO.
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 62/66
SIMPLE reactive system
Calculate the H/C in the fuel
)(CH methane CH/molmol4CH/molmol3.977.5
29.8
n
n4
C
H →≈==
Percentage of excess air
2actionReOH O2
1 2H
1Reaction CO OC
22
22
→+
→+
mol14.95n
H 2mol
Omol0.5H mol29.8
C1mol
Omol1Cmol7.5n
nnn
ltheoreticaO
22
ltheoreticaO
2ltheoreticaO1ltheoreticaOltheoreticaO
2
2
222
=
×+×=
+=
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 63/66
SIMPLE reactive system
( )
airmol71.190.21
Omol14.95n
Omol14.95n0.21
2
airltheoretica
2airltheoretica
==
=
Calculate the theoretical air needed
Calculate the amount of excess air fed to the process unit
49.9%100%71.19
71.19106.7air%excess
100%n
nnair%excess
airltheoretica
airltheoretica0
=×
−
=
×
−
=
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 64/66
! Exercise
A researcher burned n-Pentane (C5H12) with excess air in a continuouscombustion chamber.
The analysis on the product gas and report shows that product gas contains0.304 mole% pentane, 5.9 mole% oxygen, 10.2 mole% carbon dioxide and the balance nitrogen on a dry basis. Based on 100 mol of dry product gas,
i) Draw and label a flowchart for this process
(2 marks)
ii) Calculate the mol of n-Pentane and air fed in the combustion chamber,the percentage of excess air and the fractional conversion of n-Pentanein this process.
(8 marks)
SIMPLE reactive system
O H COO H C 222125
658 +→+
CHAPTER 3- Material balance for
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 65/66
! Exercise
Ethylene (C2H4) has been commercially used for production of ethanol(C2H5OH) by hydration process:
However, some of the product is converted to diethyl ether ((C2H5)2O) in theside reaction:
The feed to the reactor contains ethylene (C2H4), steam (H2O) and inert gas(G). A sample of the reactor effluent gas is analyzed and found to contain 43.3mole% ethylene, 2.5 mole% ethanol, 0.14% ether, 9.3% inert gas andthe balance water. Based on 100 mol of effluent gas,
i) Draw and label a flowchart for this process (3 marks)
ii) Calculate the molar composition of the reactor feed, the percentage conversion of ethylene, the fractional yield of ethanol and theselectivity of ethanol production relative to diethyl ether production.
(9 marks)
SIMPLE reactive system
OH H C O H H C 52242
→+
( ) ( ) O H O H C OH H C 2252522 +→
CHAPTER 4- ENERGY balance for non
7/28/2019 Reactive Mass Balance
http://slidepdf.com/reader/full/reactive-mass-balance 66/66
! A common practice is to arbitrarily designate a reference state for a
substance at which U or H is declared to equal zero, and then tabulateU and/or H for the substance relative to the reference state. *Note: ReferFelder pp. 339 and 359
! In Chapter 7 (Felder), U and H are state properties of a species; theirvalues depend only on the state of the species – primarily on itstemperature and state of aggregation (solid, liquid or gas) and, to alesser extent, on its pressure (and for mixtures of some species, on itsmole fraction in the mixture).
! When a species passes from one state to another, both ΔU and ΔH forthe process are independent of the path taken from the first state to
reactive system
Reference State