Episode 62 : MATERIAL BALANCE FOR REACTING SYSTEM
Many chemical reactions are irreversible and occur in one direction only, namely forward
Reversible reactions occur in both directions i.e. forward and backward
SAJJAD KHUDHUR ABBAS
Ceo , Founder & Head of SHacademy
Chemical Engineering , Al-Muthanna University, Iraq
Oil & Gas Safety and Health Professional – OSHACADEMY
Trainer of Trainers (TOT) - Canadian Center of Human
Development
ROBOETHICS-CCS345 ETHICS AND ARTIFICIAL INTELLIGENCE.ppt
Episode 62 : MATERIAL BALANCE FOR REACTING SYSTEM
1. SAJJAD KHUDHUR ABBAS
Ceo , Founder & Head of SHacademy
Chemical Engineering , Al-Muthanna University, Iraq
Oil & Gas Safety and Health Professional – OSHACADEMY
Trainer of Trainers (TOT) - Canadian Center of Human
Development
Episode 62 : MATERIAL
BALANCE FOR REACTING
SYSTEM
2. EQUILIBRIUM REACTIONS
• Many chemical reactions are irreversible and occur in one
direction only, namely forward
Reversible reactions occur in both directions i.e. forward and
backward
Example: hydrolysis reaction of ethylene (E) to ethanol (A)
• C2H4 + H2O C2H5OH
• There are actually 2 opposing reactions
•
•
•
C2H4 + H2O C2H5OH
C2H4 + H2O C2H5OH
r1 = k1xExW
r2 = k2xA
When the reaction rate of the forward direction is equal to
the backwards reaction r1 = r2 , chemical equilibrium is
achieved
• Equilibrium constant
Ke
k1xE xW
r2 k2 xA
r1
1 K
E W
k1 xA
e
k x x
2
3. MATERIAL BALANCE FOR
EQUILIBRIUM REACTIONS
• In general for equilibrium reaction:
•
xk = equilibrium mole fraction of component k
• Material balance of equilibrium reactor
Equilibrium constant
S
Ke x
• Mole fraction of component k
out of reactor
k Ck 0
k1
S
S1 S
S1 S
k
k1
x x x ...x
1 2
1 2
Nik Nok
xik
Ni
= Nik + k r
xok
No
Equilibrium
reactor
Nok
Nik k r
o
S
k
ok
ok
N
N
x
4. MATERIAL BALANCE FOR
EQUILIBRIUM REACTIONS
• Substitute mole fraction into the equilibrium equation
• and
•
Non linear equation of r
• If the number of moles of reactants is equal to the number
of moles of products in the stoichiometry, then
s
N r ... N rS1
N r
N mo
SiS(S 1)i(S 1)i2i1
m
o
s
k1
k
k1
e
N r
N
x K
S
k
k
21
21
S
m k
k1
K N r
k1
kike
k
5. EXAMPLE
• Example 3.11 Let the following water shift reaction
• CO(g) + H2O(g) CO2(g) + H2(g)
occurs until equilibrium is achieved at T = 1105 K. At that
temperature, Ke for the reaction is 1.0. If the flow rates in of CO
(M) is 1.0 mole h-1 and water (W) is 2.0 mole h-1 and both CO2
(C) and H2 (H) are not present in the incoming stream, calculate
the equilibrium composition in the reactor and the equilibrium
conversion of the limiting reactant.
NiM = 1.0 mole h-1
NiW = 2.0 mole h-1
xoM
xoW
xoC
xoH
No
Water shift
reactor
Equilibrium
conversion Xe
6. EXAMPLE
• Four unknown :
– mole fraction of CO xoM out of the reactor
– mole fraction of CO2 xoC out of the reactor
– mole fraction of H2 xoH out of the reactor
– mole fraction of water, xoW out of the reactor
• 4 components and 4 independent mass balance equations
• Degree of freedom= 4 – 4 = 0
•
•
Calculate m
Then
S
m k 1111 0
k1
K N r
k1
N 1rN 1r
1rNiW 1rNiM 1 r2 r
r
1 0
2
.
s
iC iH
kike
k
7. EXAMPLE
• Solve forr
• Then r = 0.6667 mol j-1
Total flow rate out No
No Nik k r NiM 1r NiW 1r 1r 1r
•
Equilibrium composition
• CO
• H2O
• CO2
• H2
• Limiting reactant
• CO is limiting reactant
• Equilibrium CO
conversion
2 3r r2
r2
N N 1 2 3 mole h1
k1
iM iW
S
NiM
NiW
NiC
M r No
W r No
C r No
1 10.6667 3 0.111
2 10.6667 3 0.444
0 10.6667 3 0.222
0 10.6667 3 0.222
xoM
xoW
xoC
xoH
NiH H r No
1 N
M 1
1iMN 2
W 1
2iW
r 10.667 0.667
N 1
X
iM
M
M
8. MATERIAL BALANCE WITH MULTIPLE
REACTIONS
• Industrial chemical processes often involve more than 1 reaction
• The reaction producing the desired product competes with side
reaction producing undesired products
• Example production of ethylene by dehydrogenation of ethane
• C2H6 C2H4 + H2 r1
• Hydrogen reacts with ethane to produce methane
• C2H6 + H2 2CH4 r2
• Ethylene reacts with ethane to produce propylene and methane
C2H4 + C2H6 C3H6 + 2CH4 r3
• Only the first reaction produces the desired product
• Reactions 2 & 3 produce undesired products
9. MATERIAL BALANCE WITH MULTIPLE
REACTIONS
•
•
•
• C2H6 (E) is used in all reactions. Rate of reaction of C2H6 :
rE r1 r2 r3
C2H4 (L) is produced by reaction 1 and used in reaction 3. Rate of
reaction of C2H4 :
rL r1 r3
• H2 (H) is produced by reaction 1 and used in reaction 2. Rate of
reaction of H2 :
rH r1 r2
• CH4 (M) is produced by reactions 2 and 3. Rate of reaction of CH4
rM 2r2 2r3
• Not all reactions are used to produce ethylene!!
C2H6
C2H6
C2H4
C2H4 + H2
2CH4
r1
r2
r3
+ H2
+ C2H6 C3H6 + 2CH4
10. MATERIAL BALANCE WITH MULTIPLE
REACTIONS
• Three additional unknowns r1 r2 r3
• Ethane conversion
where NiEt = flow rate in of ethane
• We need 2 more equations to get degree of freedom zero:
Information of production of any 2 products per mole of reacted
ethane
• Ethylene selectivity:
Hydrogen selectivity:
• Selectivity of CH4 :
• Selectivity of C3H6:
r r1 r2 r3
1 3
Ethylene selectivity SL
r r r
E
L
rE r1 r2 r3
r1 r2
Hydrogen Selectivity S
rH
H
rM
2r2 r3
E 1 2 3
Methane selectivity S
r r r r
M
rE r1 r2 r3
3
Propylene selectivity S
rrP
P
XE r1 r2 r3 NiE
11. STOICHIOMETRIC BALANCE
OF MULTIPLE REACTIONS
reactant
• Total rate of reaction of component k in R
R
reactions
r1
• Component mole balance of multiple reactions
• Yield
• If there is one inlet and one outlet streams:
R
Nik Nok kr rr
r1
• Conversion of limiting reactant
R
r1
• Selectivity of reaction product h with respect to the limiting
kr rr
rk
Nikj Nokj kr rr
j1 j r1
L M R
X N N N pr r ipp ip op ip r N
R
hr rr X p Nip
r1 Nop
Nip
Nih
Noh
Sh
R
prrr rqmaks
r1
rqmaks
p
Ypq
r
12. STOICHIOMETRIC BALANCE
OF MULTIPLE REACTIONS
• Solution strategy for material balance of single reaction by
calculating the rate of reaction r, adds an unknown r and
degrees of freedom = 1
• Conversion Xp imposes relationship constraint between r and
molar flow rates in and out of the limiting reactants that reduce
the degree of freedom = 0
• The same strategy can be extended to the multi-reaction system
• If there are S components involved in R reactions, component
balances produce S independent equation with additional R
unknown: rr with r = 1, 2, 3, ..., R..
• For systems of R reactions, at least R - 1 additional equations
Or product yield
are required to solve it
Sh
• Product selectivity
X N p ip
R
hr rr
r1 Nop
Nip
Nih
Noh
rqmaks
R
prrr
r1
rqmaks
p
Ypq
r
13. EXAMPLE
• Example 3.12 Ethylene (L) can be produced by dehydrogenation
of ethane. Two important reactions involved are
C2H6 C2H4 + H2
C2H6 + H2 2CH4
•
•
• An ethane feed contains 85% ethane (E) and the rest are inert
components (I). Ethane conversion, XE is 0.501 and selectivity of
ethylene, SL is 0.471 mole of ethylene produced per mol of
ethane feed. Calculate the molar composition of the gas
products NoE
Conversion of C2H6 = 0.501
• Seven unknowns: molar flow rate out C2H6, NoE molar flow rate
out C2H4, NoL molar flow rate out H2, NoH molar flow rate out
CH4, NoM molar flow rate out I, NoI & 2 reaction rates, r1 and r2
Ni = 100 mole h-1
xi E = 0.85
xiI = 0.15
NoL
NoH
NoM
NoI
Ethane
dehydrogenation
reactor
14. EXAMPLE
• 5 components and 5 independent mass balance equations
• 2 reactions, ethane conversion & selectivity of ethylene
• Degree of freedom= 7 – 7 = 0 , the system has a unique solution
•
•
•
•
•
Basis of calculation 100 mole h-1 ethane flow
•
•
•
Then
Ethane conversion X
Ethylene selectivity
r N r r N x i iE Er r iE E1 1 E2 2
r1
E
2
r1 r2 42.585
2
r N r r N x Lr r iE L1 1 L2 2 i iE
r1
L S
0.471 1r 0r 0.851001 2
r1 40.035
r2 2.55
0.501 1r 1r 0.851001 2
15. EXAMPLE
• Component balance
• C2H6
• C2H4
• H2
N 85 42.585 42.415 mol h-1
oE
NoL L1r1 L2r2
NiL
N 40.035 mol h-1
oL
H1r1 H 2r2
NiH NoH
N 37.485 mol h-1
oH
NiE NoE E1r1 E2r2
0.85100 NoE 140.035 12.55
0 NoL 140.035 02.55
0 NoH 140.035 12.55
16. EXAMPLE
• Component balance
• CH4
•
• Check with total mass balance
Balanced!
NiM NoM M 1r1 M 2r2
N 5.10mol h-1
M
2550 15MI 2550 15MI
1000.85301000.15MI
42.41530 40.03528 37.4852 5.11615MI
Ni xiI MI NoE ME NoLML NoH M H NoM MM Ni xiI M I
Ni xiE M E
0 N 040.035 22.55oM
17. EXAMPLE
• Example 3.13 Methane (M) is burned in a continuous burner to
produce a mixture of CO (X), CO2 (C) & water (W).
•
•
• The feed contains 7.8% mole methane, 19.4% mole oxygen and
CH4 + (3/2)O2 CO + 2H2O
CO2 + 2H2OCH4 + 2O2
•
the remainder nitrogen. Methane conversion is 90%. Ratio CO2
product /CO product = 8
NoM
Conversion CH4 = 0.90
Eight unknown :molar flow rate out CH4, NoM molar flow rate out
CO, NoX molar flow rate out CO2, NoC molar flow rate out O2, NoO
molar flow rate out air, NoW molar flow rate out nitrogen, NoN and
2 rates of reaction, r1 and r2
Ni = 100 mole h-1
xiM = 0.078
xiO = 0.194
xiN = 0.728
NoC /NoX
NoN = NiN
NoO
= 8
NoW
Natural gas burner
18. EXAMPLE
• 6 components and 6 independent mass balance equations
• 2 reactions & ethane conversion & product ratio CO2/CO
• Degree of freedom= 8 – 8 = 0
•
•
•
•
Basis 100 mole h-1 ethane stream.
Methane conversion
Or
• CO2 balance
•
r r N x M1 1 M 2 2 i iMXM Mr rr NiM
r1
2
0.9 1r 1r 0.0781001 2
r1 r2 7.02
NoC C1r1 C2r2
NiC
r2
NoC
0 NoC 0r1 1r2
19. EXAMPLE
• CO balance
•
But Then
• In matrix form
Hence
• CO2 balance
• CO balance
• CH4 balance
NoX X1r1 X 2r2
NiX
r1
NoX
8NoX
NoC r2 8r1
r 7.02 9 0.78 mole h1
1
r 80.78 6.24 mole h
1
2
0 NoX 1r1 0r2
1 1r1 7.02
01r2 8
-1
6.24mole hNoC
N 0.78 mole h-1
NoM M1r1 M 2r2
oX
NiM
7.8 NoM 10.79 16.24
N 0.78 mole h-1
oM
20. EXAMPLE
• O2
•
• H2O
•
•
N2
• Check with total balance:
• Balanced!
19.4 NoO 1.5r1 2r2
N 5.75 mole h-1
NoW W1r1 W 2r2
oO
0 NoW 2r1 2r2
N 14.04 mol h-1
oW
N N 72.8 mol h-1
oN iN
765.984 765.984
NoW MW
NoOMO NoN M N
NoC MC NoLMLL NoM MM
Ni xiM MM Ni xiOMO
Ni xiN M N
1000.078161000.19432 6.244 0.7828 0.781814.0418
1000.72828
5.75321000.72828
NiO NoO O1r1 O2r2
NiW
21. EXAMPLE
•
•
•
• 60% of a stream of pure 1-butena is converted in a reactor into
• Example 3.14 1-butene (B) is converted into 2 other isomers:
cis-2-butene (C) & trans-butene (T) on alumina catalyst:
1-butene
cis-2-butene
cis-2- butene
trans-2-butene
1-butene
trans-2- butenae
a product containing 25% mole cis-2-butene. Calculate the
concentration of other components
•
Conversion 60 %
Six unknowns : total molar flow rate out, No mole fraction out 1-
butene, xoB mole fraction out trans-2-butene, NoT and 3 rates of
reaction r1 r2 and r3
1-butene
Ni = 100 mole h-1
xiB = 1.0
No
xoB
xoC = 0.25
xoT
Isomerizer
22. EXAMPLE
• 3 components and 3 independent mass balance equations
• 3 reactions, 1 conversion & concentration of cis-2-butene
• Degree of freedom = 6 – 5 = 1
• Specify basis so that Degree of freedom becomes zero
•
•
•
Basis 100 mole h-1 1-butene stream.
Conversion 1-butene
or
•
• Cis-2-butene balance NiC
r N r r r NiBB3 3B2 2iB B1 1
r1
Br rBX
3
r1 r3 65
NoC C1r1 C2r2 0
r1 r2
NoC
0.65 1r 0r 1r 100
1 2 3
0 NoC 1r1 1r2 0
23. EXAMPLE
• 1-butenae
• Then
• Trans-2-butene
• Then
• Total mole going out of reactor by combining all relationship
• Mole fraction out cis-2-butene 0.25
•
• Hence
NoB B1r1 B2r2 B3r3
NiB
100 r1 r3
NoB
NiT NoT T 2r2 T 3r3
r2 r3
0 NoT 1r2 1r3
NoT
100 r r r r r r 100 mole h
1
1 3 1 2 2 3N NoB NoC NoTo
100 NoB 1r1 0r2 1r3
o
NoC
oC
N
x
100
r1 r2 25
0.25
r1 r2
24. EXAMPLE
• Then
• Composition of product
• Check
10056 35(56) 25(56) 40(56)
•
Balanced!
N r r
N
r r r r 65 25 40 mole h
1
2 3 1 3 1 2oT
x N 0.25100 25 mole h1
oC ooC
N N 100 25 40 35 mole h1NoB No oToC
35 100 0 35xoB
xoT 40 100 0.40
100 100
NoB M B NoC MC NoT MT
NiB M B
25. EXAMPLE
• Example 3.15 Ethylene oxide (L) is produced through partial
oxidation of ethylene (E) in air
• At ethylene conversion of 25%, feed contaminating 10%
ethylene produce an ethylene oxide yield of 80% from ethylene.
Calculate the composition of the product
• Eight unknowns: total molar flow rate out, No mole fraction out
ethylene, xoE mole fraction out ethylene oxide, xoL mole fraction
out oxygen, xoO mole fraction out nitrogen, xoN mole fraction out
water, xoW & 2 rate of reaction r1 & r2
Ni = 100 mol h-1
xiE = 0.1
xiO
xiN
No
xoE
xoL
x
x
oO
oN
xoW
xoC
Catalyrtic
reactor
Conversion 25 %
Yield of EtO 80%
• 2C2H4 + O2 2C2H4O
• C2H4 + 3O2 2CO2 + 2H2O
26. EXAMPLE
• 5 components and 5 independent mass balance equations
• 3 reactions, ethylene conversion, sum of mole fractions and
ethylene oxide composition of output stream
• Degree of freedom = 5 – 5 = 0
•
•
Basis 100 mole h-1 feed stream
Ethylene conversion
• Or
• In order to use information on the product yield, maximum rate
of reaction of ethylene is needed with other rates of reaction set
to zero e.g. rate of CO2
N'iC N'oC C 2r'2
0 0 r'2
r'2 0
r r N x E1 1 E2 2 i iEX E Errr NiE
r1
2
2r1 r2 2.5
0.25 2r 1r 0.11001 2
27. EXAMPLE
• Ethylene balance at this condition
•
Then
• Yield of ethylene
oxide
• Hence
• Ethylene balance
NoE E1r'1E2r'2
NiE
0.1100 7.5 2r'10
1
r'1 1.25 mole h
1
2r'1 2.5 mole hrEmaks
0.8 2r 0r 2.51 2
r1 1.0 mole h
rEmaks
R
r1
YL E Lr rr
1
r 0.5 mole h1
2
NiE NoE E1r1 E2r2
N 7.5 mole h1
oE
0.1100 NoE 21.0 10.5
28. EXAMPLE
• Ethylene oxide
• Oxygen
• Water
• CO2
0 NoL 21 00.5
N 2.0 mole h1
oL
0.210.9100 NoO 11 30.5
N 16.4 mole h1
NoW W1r1 W 2r2
oO
0 NoW 01.0 20.5
N 1.0 mole h1
oW
NoL L1r1 L2r2
NiL
NoO O1r1 O2r2
NiO
NiW
0 NoC 01 20.5
C1r1 C 2r2
NiC NoC
N 1.0 mole h1
oC