Episode 61 : MATERIAL BALANCE FOR REACTING SYSTEM
RATE OF CHEMICAL REACTION
participating in a chemical reaction
Stoichiometric equation of chemical reaction:
– Showing the relative number of molecules/moles of components participating in the chemical reaction
Reactants– components that react with each other in a chemical reaction
Products – components that are produced by a chemical reaction
Chemical reactor- equipment in which chemical reactions occur
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
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 61 : MATERIAL
BALANCE FOR REACTING
SYSTEM
2. COMPONENT MATERIAL BALANCE FOR
REACTING SYSTEMS
System
CONTROL
VOLUME
Fi1
wi1k
Fi2
ENVIRONMENT
Fo1
wo1k
Fo2
wo2k
Fo3
wo3k
wi2k
j1 j1
wo1k Fo1 wo2k Fo2 wo3k Fo3 wi1k Fi1 wi2k Fi2 Mk rk
Mkrk
Reactor
Compressor
Phase
separatior
Distillation
Mk rk
M L
Foj wokj Fij wikj
3. COMPONENT MATERIAL BALANCE FOR
REACTING SYSTEMS
System
CONTROL
VOLUME
Ni1
xi1k
Ni2
ENVIRONMENT
No1
xo1k
No2
xo2k
No3
xo3k
xi2k
rk
Reactor
Compressor
Phase
separator
Distillation
rk
M L
j1 j1
xo1k No1 xo2k No2 xo3k No3 xi1k Ni1 xi2k Ni2 rk
Noj xokj Nij xikj
4. RATE OF CHEMICAL REACTION
• Rate of chemical reaction of component k : rk
• Net rate of generation of component k
component k per unit time
in units of moles of
• Obtained from stoichiometric balance of chemical reactions
• Stoichiometry = relative proportions of chemical components
participating in a chemical reaction
• Stoichiometric equation of chemical reaction:
– Showing the relative number of molecules/moles of components
participating in the chemical reaction
• Reactants– components that react with each other in a
chemical reaction
• Products – components that are produced by a chemical
reaction
• Chemical reactor- equipment in which chemical reactions occur
6. EXAMPLE
• Example 3.1 Stoichiometric
equations
• SO3 synthesis
reaction 2SO2 +
O2 2SO3
• 2 moles of SO2 reacting with 1 mole of O2 to produce 2 moles
of SO3
• Ammonia synthesis
reaction N2 +
3H2 2NH3
• 1 mole of N2 reacting with 3 moles of H2 to produce 2 moles of
NHO3
Reactants Product
2SO2 + (1)O2 2SO3
7. STOICHIOMETRIC BALANCE
• Stoichiometric equation
• S = total number of components
• k = stoichiometric coefficient
•
•
Ck = molecular formula of component k
Sign Convention: k + for products & - for reactant
k Ck 0
k1
S
2SO2 + (1)O2 2SO3
2SO3 - 2SO2 - (1)O2 = 0
N2 + 3H2 2NH3
2NH3 - N2 - 3H2 = 0
8. STOICHIOMETRIC BALANCE
• Material balance
Total mass of reactants = mass of products in Stoich. Equation
Conservation of mass
S
• Mk = MW of components k
2SO2 + (1)O2 2SO3
2MST – 2MSD - (1)MO = 0
2(64) – 2(48) - (1)(32) = 0
N2 + 3H2 2NH3
2MA – MN - 3MH = 0
2(17) - (28) - 3(2) = 0
0k M k
k1
9. STOICHIOMETRIC BALANCE
• Elemental balance= total element in reactants is equal to the
total element in the product in the stoichiometric equation
S
•
k1
• mkl = number of element atom in a molecule of komponent k.
2SO2 + (1)O2 2SO3
Balance on S: 2mSTS – 2mSDS - (1)mOS = 0
2(1) – 2(1) - (1)(0) = 0
Balance for o2: 2mSTO – 2mSDO - (1)mOO = 0
2(3) – 2(2) - (1)(2) = 0
0k mkl
10. STOICHIOMETRIC BALANCE
N2 + 3H2 2NH3
Balance of N: 2mAN – mNN – 3mHN = 0
2(1) – 1(2) - 3(0) = 0
Balance of H: 2mAH – mNH - 3mHH = 0
2(3) – 1(0) - 3(2) = 0
• Element balance can be used to determine the stoichiometric
coefficients provided that both the reactants and the
products are known
• If L elements are involved in the stoichiometric equations,
then there are L independent element balance equation a
• If S components and L elements are involved the
stoichiometric equations , degree of freedom = S – L
11. EXAMPLE
Example 3.2 Balance the stoichiometric equations of a reaction
between As2S5 and HNO3.
-1As2S5 - 2HNO3 3H3AsO4 + 4H2SO4 + 5H2O + 6NO2
The stoichiometric equation is rewritten as:
1As2S5 + 2HNO3 + 3H3AsO4 + 4H2SO4 + 5H2O + 6NO2 = 0
There are 6 species & 5 elements. Degree of freedom= 6 – 5 = 1
Balance of each element
O 32 + 43 + 44 + 5 + 26 = 0
As 21 +
3
= 0
S 51 +
4
= 0
H 2 +
33
+24 + 25 = 0
N 2 + 6 = 0
12. STOICHIOMETRY OF BIOCHEMICAL REACTIONS
• Biotechnological products are produced in fermentation
proses involving cell growth and bioproduiction
• Bioreactor/fermentor
• Biochemical transformation processes involved thousands
of biochemical reactions in the cell.
• Its stoichiometry is represented by a simple
pseudochemical reaction equation
• Stoichiometric balance of pseudochemical reaction:
– Elemental balance = ordinary chemical reactions
– Electron balance = different from ordinary chemical
reactions = involves energy transition
– Yield coefficient of biomass
– Yield coefficient of product
14. STOICHIOMETRY OF BIOCHEMICAL REACTIONS
• Biochemical reaction involves
– Substrate = glucose (CHmOn), oxygen & ammonia
– Products: cell mass (CHON), biochemical product (CHxOyNz),
water and & carbon dioxide
-1CHmOn - 2O2 -3NH3 4CHON +5CHxOyNz + 6H2O + 7CO2
• Value of coefficients m and n depends on substrate
•
•
•
•
Example: glucose m = 2 and n = 1.
Value of coefficients , and depends on microbe
Example: yeast, = 1.66, = 0.13 and = 0.40.
Divide the stoichiometric equations with 1
-CHmOn - ’1O2 -’2NH3 ’3CHON +’4CHxOyNz + ’5H2O + ’6CO2
• ’j = j-1/1 and j = 1, 2, 3, …6.
• Number of elements = 4; Number of components/stoichiometric
coefficients = 6
Degree of freedom of biochemical Stoichiometry = 6 – 4 = 2
15. STOICHIOMETRY OF BIOCHEMICAL REACTIONS
• Biochemical transformation involves electron transfer
determined by an electron balance
• Additional independent balance equations!
• Degree of reduction of component k, k is used in the electron
balance
• Degree of reduction k = number of equivalents of available
electrons per atom C
• Available electrons = electrons transferred to oxygen after
organic compound is oxidized to carbon dioxide, water and
ammonia in biochemical reactions
• Degree of reduction of organic compounds = sum of all the
product of element valency and element atomic number
divided by the number of C atoms in the compound
•
Vl = element valency in component l, mkC = number of carbon
atoms in component
L
vl mkl mkC
k1
k
16. STOICHIOMETRY OF BIOCHEMICAL REACTIONS
• Degree of reduction of several common organic materials: :
Methane CH4 = [1(4) + 4(1)]/1 = 8
= [6(4) + 12(1) + 6(-2)]/6 = 24/6 = 4Glucose C6H12O6
Ethanol C2H5OH = [2(4) + 6(1) + 1(-2)]/2 = 12/2 = 6
Glucose CHmOn
Cell mass CHON
s = [1(4) + m(1) + n(-2)]/1 = 4 + m - 2n
b = [1(4) + (1) + (-2) + (-3)]/1 = 4 + -
- 2 - 3
p = [1(4) + x(1) + y(-2) + z(-3)]/1 = 4 + x
- 2y – 3z
Product CHxOyNz
• The degree of reduction of water, ammonia & carbon dioxide = 0
• The degree of reduction of oxygen = -4
• S
•
Electron balance equation:
Additional independent equations!
'k k 0
k1
17. STOICHIOMETRY OF BIOCHEMICAL REACTIONS
• Respiratory quotient RQ molar basis
• RQ 7 2 '6 '1
• Yield of cell biomass mass basis
•
• Mb = formula MW of biomass & Ms = formula MW of substrate
• Yield of product mass basis
•
•
•
Mp = formula MW of product
Value of RQ is obtained from experiment
MS 4MB 1MS '3 MB
YX / S
MS 5MP 1MS '4 MP
YP / S
18. STOICHIOMETRY OF BIOCHEMICAL REACTIONS
• Element balance equations (4 equations) plus
– Electron balance equation
– Respiratory quotient equation
– Yield of biomass
– Yield of product
• 8 equations & 6 unknown variables
•
•
Degrees of freedom = 6-8 = -2
Two equations are not independent and can be used to check the
balance stoichiometry
• Balance of elements
C
H -m + 3’2 + ’3 + x’4 + 2’5 = 0
N
O
-1+’3 + ’4 + ’6 = 0
’2 + ’3 + z’4 = 0
-n + 2’1 + ’3 + y’4 + ’5 + 2’6 = 0
19. STOICHIOMETRY OF BIOCHEMICAL REACTIONS
• Electron balance
•
•
•
•
•
-s - 4’1 + b’3 + p’4 = 0
s = degrees of reduction of substrate
b = degrees of reduction of biomass
p = degrees of reduction of product
H and O element balances involve water and there is so much water
• Both balances are difficult to use
• Only the C, N and electron balances are used
C -1+’3 + ’4 + ’6 = 0
N
-s
’2 + ’3 + z’4 = 0
- 4’1 + b’3 + p’4 = 0
20. EXAMPLE
Example 3.3 Aerobic growth of S. cerevisiae (yeast) on ethanol
-CH3O0.5 - ’1O2 -’2NH3 ’3CH1.704O0.408N0.149 + ’5H2O + ’6CO2
• Determine the values of ’1, ’2, ’3, and ’6 if RQ = 0.66, Yield of
biomass on substrate & Yield of biomass on oxygen
• Degree of reduction of substrate & biomass
Ethanol
Biomass
CH3O0.5
CH1.704O0.408N0.149
S = [1(4) + 3(1) + (0.5)(-2)]/1 = 6
B = [1(4) + 1.704(1) + 0.408(-2) +
0.149(-3)]/1 = 4.441
• Element balance of C & N, electron balance and RQ.
• C
• N
• Electron
• RQ
-1+’3 + ’6 = 0
’2 + 0.149’3 = 0
-6 - 4’1 + 4.41’3 = 0
'6 0.66'1
21. EXAMPLE
• 4 unknowns & 4 equations & degree of freedom= 4 – 4 = 0
• Substitute last equation with first equation
’3 - 0.66’1 = 1
and
• Then
4.41’3 - 4’1 = 6
and
• Yield of biomass on oxygen
Formula biomass MW MB
• Formula ethanol MW MS
• Yield of biomass on substrate
1 4 6 0.66'3
1 4 4.41 0.66 0.0367 '1 1 0.0367 0.66 1.4595
'2 0.1490.0367 0.0055 '6 0.661.4595 0.96327
12 1.7041 0.14914 0.40816 22.318
12 31 0.516 23
M 0.036722.318 23 0.0356 g g-1
X / S 3 BY ' M S
Y '3 MB '1 MO 0.036722.318 1.459532 0.0175 g g1
X / O2
6
1
11 0.663
4.41 4
22. MATERIAL BALANCE WITH SINGLE
REACTION
• COMPONENT MATERIAL BALANCE FOR REACTING SYSTEMS
Molar Mass
• Rate of chemical reaction of component k rk
• Ammonia synthesis reaction :
•
•
•
•
N2 + 3H2 2NH3
If rate of reaction of nitrogen = -rN
Rate of reaction of hydrogen = -rH
(negative: nitrogen is used)
= (H /N)(-rN )= (-3/-1)(-rN)
Rate of reaction of ammonia = rA = (A /N)(-rN )= (2/-1)(-rN)
Then
Rate of reaction r is fixed for a given reaction stoichiometric equation
• Rate of reaction of component
• COMPONENT MATERIAL BALANCE FOR REACTING SYSTEMS
Molar Mass
rk Mk rk
M L M L
j1 j1 j1
Noj xokj Nij xikj Foj wokj Fij wikj
j1
r
rA
rH
rN
rk
A H N k
rk k r
Mkk rFoj wokj Fij wikj
M L
j1 j1
k rNoj xokj Nij xikj
M L
j1 j1
23. MATERIAL BALANCE WITH SINGLE REACTION
• Example 3.4 Lets say for the SO3 synthesis reaction, 15 mole h1 O2 (A)
40 mole h-1 SO2 (B) and 0 mole h-1 SO3 (C) is fed into a reactor. If the
flow rate out of O2 is 8 mole h-1 calculate the flow rates of other
components
• Basis 55 mole j-1 feed O2 + 2SO2 2 SO3
• 3 components & 3 independent material balance equations
• Degree of freedom= 3 – 3 = 0
• Choose component mole balance that has most information to get r :O2
• SO2 mole balance
• SO3 mole balance
• Substituting r in
SO2 & SO2 balances:
NiA = 15 mole h-1
NiB = 40 mole h-1
NiC = 0 mole h-1
NoA =8 mole h-1
NoB
NoC
Reactor SO3
NoA Ar
NoB Br
NoC C r
NiA
NiB
NiC
N 40 2r 40 27 26 mole h1
oB
NoC 2r 27 14 mole h
1
1
r 7 mole h15 8 1r
40 NoB 2r
0 NoC 2r
24. EXAMPLE
• Example 3.5 Growth of S. cerevisiae on glucose is described by the
following equation
•
• In a batch bioreactor of volume 105 L, the yeast concentration required
C6H12O6 + 3O2 + 0.48NH3 0.48C6H10NO3 + 4.32H2O + 3.12CO2
is 50 g dry mass L-1.
• Calculate the yield of biomass/substrate YX/S Yield of biomass /oxygen
YX/G and respiratory quotient RQ. Calculate the required concentration
and total amount of glucose and (NH3)2SO4 in the nutrient media.
• How much oxygen is required and carbon produced by the bioreaction ?
• If the growth rate at exponent phase is r = 0.7 gdm L-1 h-1, determine
the rate of oxygen utilization.
• MW glucose = 180, MW oxygen = 32, MW ammonia = 17, MW
(NH4)2SO4 = 116, MW biomass = 144, MW carbon dioxide = 44 and MW
water = 18.
25. EXAMPLE
CoO = 4 mg L-1
CoC = 2 mg L-1
Batch Bioreactor
Glucose feed
CiG
CiO = 0
CiC = 0
CiB = 0
Gas exhaust
Feed (NH4)2SO4
CiA
Biomass product
CoB= 50 gdm L-1
CoG = 5.0 g L-1
CoA = 1.0 g L-1
O2 Feed
26. EXAMPLE
• 6 components and six independent mass balance equations
• Water balance is not used because the presence of a lot of water
• Degrees of freedom = 6 – 5 = 1
• Yield of biomass/glucose YX/S
• Yield of biomass/oxygen YX/O2.
• Respiratory quotient
•
• Choose Basis =500 kg dry biomass = 50 gdm L-1 in a 105 L bioreactor
• Choose component balance with the most information to get r
• Biomass balance
0.48144
1180
1
0.384 g g
M
YX / S
G G
BMB
M
0.48144
332
1
X / O2 0.72 g gY
O O
B B
M
RQ
C
3.12
1.04 mole mole1
O 3
NoB BrVNiB
27. EXAMPLE
• Biomass balance
• Hence the rate of reaction
• Glucose balance
• Total amount of glucose required = 13,520.8 kg
144 0.48
0.772 mole L150
r
CiBV
CoBV
rV
M M
B
B B
5010 0.48r105
5
144
0
N NoG G rViG
C 105
5105
50105
180 1440.48180
iG
CiGV
CoGV
rV
M M
G
G G
50180
5
1440.48 5 130.208 135.208 g L
1CiG
28. requires 1/2 mole of (NH4)2SO4.
•
• Total (NH4)2SO4 required = 2,113.9 kg
• O2 balance
• Total O2 utilization Nio - Noo = 217.026 kmole oxygen
50116
2144
1
1 20.139 21.139 g LCiA 1
N N A
rViA oA
2
EXAMPLE
• One mole NH3
• (NH4)2SO4
C 105
0.4850105
21440.48116116
iA
CiAV
CoAV
A
rV
M A M A 2
OrV
MO
C VNiO NoO oO
0.00410 35010
1440.48 217.026x10
3
55
32
NiO NoO
1
1
05
29. EXAMPLE
• CO2 balance
• NoCO2 = 225.689 kmole carbon dioxide = 9930.316 kg carbon dioxide
• The dissolved gas concentrations are very small and will be neglected in
fermenter balances
C rV
MC
CoCV
NoC NiC
0.00210 3.125010
1440.48
3
55
225.689x10
44
NiC NoC
30. CONVERSION & LIMITING REACTANT
• Common measure of course of reaction is the fractional conversion /
conversion of the limiting reactant
• Conversion links the outlet flow rate with the inlet flow rate of the
same component = additional independent equation!
• Reaction rate r
•
•
The limiting reactant finishes first if the reaction is left to react by itself
If the reaction is left to react, the rate of reaction r increases to reach
the value of rlimiting when Nok = 0
• Reactant with the lowest value for Nik/(-k) finishes first
• Limiting reactant=reactant that has the lowest Nik/(-k)
• Other reactants= excess reactant
Excess fraction of component k
ik
Nik Nok
k
N
X Nik X k Nik Nok
k
r
Nik Xk
k
Nik
Limitingr
k ip p
k Nip p
Nik
E k
N
Nik Xk k r
31. EXAMPLE
• Example 3.6 The reaction between ammonia (A) and oxygen (O) on Pt
catalyst produces nitric acid and water (W). The stoichiometric
equation is given by
4NH3 + 5O2 4NO + 6H2O•
• Under certain conditions, conversion of NH3 into NO (N) can achieve
90% at ammonia flow rate NH3 40 mole h-1 and O2 60 mole h-1 .
Calculate the other flow rate
• Basis is 100 mole h-1 feed.
• 4 components & 4 independent material balance equations .
• Degree of freedom= 4 - 4 = 0
NiA = 40 mole j-1
NiO = 60 mole h-1
NiN = 0 mole h-1
NiW = 0 mole h1
NoA
NoO
NoN
NiW
Acid nitric
Reactor
Conversion 90%
32. EXAMPLE
• Stoichiometric coefficient
• Use conversion of ammonia
to get Rate of reaction r
• Component mole balance
• NH3
• O2
• NO
• H2O
400.9
4
1
9 mole h
r
NiA X A
A
NoA ArNiA
1
4 mol hNoA
NoO OrNiO
N 15 mol h1
oO
N rNiN NoN
N 36 mol h1
oN
NoW W rNiW
1
54 mol hNoW
40 NoA 49
60 NoO 59
0 NoNO 49
0 NoW 69
• NH3 A = -4 O2 O = -5
• NO N = 4 H2O W = 6
33. EXAMPLE
Example 3.7 If the reaction in Example 3.6 achieves 80% conversion with
equimolar ammonia and oxygen feed that is fed at 100 mole h-1. Calculate
the flow rate out of all components
• Stoichiometric equation is given by
•
•
4NH3 + 5O2 4NO + 6H2O
Choose basis 100 mole h-1 feed
• Determination of the limiting reactant
• Limiting reactant= Oxygen because it has the smallest Nik/(-k)
• Conversion information is for conversionm of oxygen!
NiA = 50 mole j-1
NiO = 50 mole j-1
NiN = 0 mole j-1
NiW = 0 mole j-1
NoA
NoO
NoN
NiW
Acid nitric
reactor
Conversion 80%
12.5
4
50
A
NiA
5
10
50
O
NiO
34. EXAMPLE
• Use conversion of oxygen to get the rate of reaction:
• Component balance
• NH3
• O2
• NO
• H2O NiW
500.8
5
1
8 mole h
r
O
NiO XO
NoA ArNiA
NoO OrNiO
NoN N r
NoW W r
NiN
N 50 48 18 mole h1
oA
N 50 58 10 mole h
1
oO
N 0 48 32 mole h1
oN
0 68 48 mole h
1NoW
35. EXAMPLE
Example 3.8 Acrylonitrile (C) is produced by the following reaction:
C3H6 + NH3 + (3/2)O2 C3H3N + 3H2O
The feed contains 10% mole propylene (P), 12% mole ammonia (A) and
78% mole air. Conversion of limiting reactant is 30%. By choosing 100
mole h-1 feed as the basis, determine the limiting reactant, fractional
excess of other reactants and flow rate out of all components.
6 unknown & 6 independent material balance equations
Degree of freedom= 6 – 6 = 0
Determine the limiting reactant
Propylene is the limiting reactant
Ni = 100 mole h-1
xiA = 0.12
xiP = 0.10
NiO = (0.21)(0.78)
NiN = (0.79)(0.78)
NoA
NoP
NoO
NiN
NoC
NoW
Acrylonitrile
Reactor
Conversion 30%
12
1
12
W P
NiW 10
1
10
NiP
16.38
1.5
10.92
O
NiO
36. EXAMPLE
• Fractional excess of other reactants
• NH3
• O2
• From conversion,
calculate rate of reaction
• NH3
• O2
• C3H3N
• H2O
• N2
P 12 (1)101
(1)10 1
0.2
NiA A NiP
EA
A iP P N
P 16.38 (1.5)101
(1.5)10 1
0.092
O2 NiP P
NiO O NiP
EO
0.310
1
1
3 mole h
r
XP NiP
P
NoA ArNiA
NoO OrNiO
C rNiC NoC
1
NiN 61.62 mole hNoN
NoW W rNiW
N 12 13 9 mole h
1
oA
N 16.38 1.53 11.88 mole h
1
oO
0 13 3 mole h
1NoC
N 0 33 9 mole h
1
oW
37. EXAMPLE
Example 3.9 Natural gas containing methane only is burnt in an
incinerator CH4 + 2O2 CO2 + 2H2O
Calculate all the outgoing molar flow rate of components, total molar
flow rate and fractional excess ofair
4 unknown & 4 independent material balance equations
Degree of freedom= 4 – 4 = 0
• Convert the flow rate units into mole units.
•
•
• FU NU
NUO =
NUN =
FG NG
Air MW= 0.21(32) + 0.79(28) = 28.84
0.21(10.4) = 2.184 kmole h-1
0.79(10.4) = 8.216 kmole h-1
FG = 16 kg h-1
wGM = 1.0
No
NoC
NoN
NoO
NoW
Natural
gas burner
FU = 300 kg h-1
xUO = 0.21
xUN = 0.79
41 kmole h-1
CH
4
16 kg 1 kmole CH4
h 16 kg CH
10.40kmole h air
h 28.84 kg air
300 kg 1 kmole air -1
38. EXAMPLE
• Basis is 1.0 kmole h-1 natural gas
• Stoichiometric coefficients
• CH4 CH4 = -1 O2 O2 = -2 CO2
•
CO2 = 1 H2O H2O = 2
Assume complete combustion = all methane reacted.
Rate of methane reaction
• Component mole balances
• N2
• O2
• CO2
• H2O
•
•
•
Total flow rate out No = 11.4 kmole h-1
Air fractional excess:
or 9.2%
N 1.0G
M 1
1
1.0 kmole hr
N N 8.216 kmole h1
oNUN
NoO Or
NoC C r
NoW W r
NUO
NGC
NGW
M 2.184 (2)1.0 1
(2)1.0 1
0.092
NUO O NG
EO
O G M N
N 2.184 21 0.184 kmole h
1
oO
0 11 1.0 kmole h
1NoC
N 0 21 2.0 kmole h
1
oW
39. EXAMPLE
Example 3.10 Yeast in example 3.2 reacts with glucose, oxygen
and ammonium sulfate according to the same stoichiometry in a
chemostat bioreactor with volume V = 105 L in the figure below.
The rate of ventilation is 50,000 L min-1. Dilution rate D = Qi/V =
0.1 j-1.
Feed
Qi = 10000 L h-1
CiA
CiG
CiB = 0
Product
Qo = Qi = 10000 L h-1
CoG = 5 g L-1
CoB = 50 gdm L-1
CoA= 1 g L-1
Air feed
QiU = 50,000 L min-1
NiU = 133928.6 mole h-1
N =28125.0 mole h-1
iO
N =105803.6 mole h-1
iN
N =0iC
Exhaust air
NoU
QoU
NoO
NoN
NoC
40. EXAMPLE
• Chemostat = continuous bioreactor
• At steady sate, substrate, other nutrient and oxygen are fed and
products are withdrawn at the same volumetric flow rate
• Volumetric flow rate Qi = VD
•
•
•
= (105)(0.1) = 10000 L h-1
5 equations for 5 unknowns & Degree of freedom= 5 – 5 = 0
Basis 10000 L h-1 volumetric flow rate
Chose component balance with most information to get r:
Biomass
• Then
144
r = (50)/[(10)(144)(0.48)] = 0.072338 mole L-1 h-1
NoB BrVNiB
0 104
50 0.48r105
BrV
M
Q
CiB
Q
CoB
MB B
ii
41. EXAMPLE
• Glucose
• Then CiG = 5 + (188)(10)(0.072338) = 140.99 g L-1
• One mole NH3 requires 1/2 mole of (NH4)2SO4. Then (NH4)2SO4
balance
• Then CiA = 1 + (116)(0.48)(10)(0.072338)/2 = 21.14 g L-1
• Molar flow rate of air feed
• Then
NoG GrVNiG
N N
A
rV
iA oA
2
GrV
M
Q
CiG
Q
CoG
MG G
ii
104
Cig
104
5 0.072338105
188 188
104
CiA
10 1 0.480.072338
2
10
116 116
5
4
A
rV
M A 2
Q
CiA
Q
CoA
M A
i i
N 133928.571 mole h-1
min 1 j 22.4 L
60min molQ 50000
L
iUiU
42. EXAMPLE
• Hence
•
• Oxygen
NiO = (0.21)( 133928.571) = 28125.0 mole h-1
NiN = (0.79)( 133928.571) = 105803.6 mole h-1
or
oO
NoO =28125.0–21701.4 = 6423.6 mole oxygen h-1
• Nitrogen: NoN = NiN = 105803.6 mole h-1
• Carbon dioxide
• Rate of CO2 production NoCO2 = 22569 mole carbon dioxide h-1
• Rate of gas out
NoU=NoO+NoN+NoC=6423.6+105803.6+22569.0=134795 mole h-1
NoO OrVNiO
28125.0 N 30.072338105
NoC CrVNiC
0 N 3.120.072338105
oC
22.4 L
50323.47 L min-1
h 60 min mole
1hQ 134795
mole
oU