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
j1 j1
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
j1 j1
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

5. CHEMICAL REACTOR
Ammonia Reactor
Acid Nitric Reactor

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
k1
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
 0k M k
k1

9. STOICHIOMETRIC BALANCE
• Elemental balance= total element in reactants is equal to the
total element in the product in the stoichiometric equation
S
•
k1
• 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
 0k 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 32 + 43 + 44 + 5 + 26 = 0
As 21 +
3
= 0
S 51 +
4
= 0
H 2 +
33
+24 + 25 = 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

13. BIOREACTOR/FERMENTOR
Bioreactor/fermentor

14. STOICHIOMETRY OF BIOCHEMICAL REACTIONS
• Biochemical reaction involves
– Substrate = glucose (CHmOn), oxygen & ammonia
– Products: cell mass (CHON), biochemical product (CHxOyNz),
water and & carbon dioxide
-1CHmOn - 2O2 -3NH3  4CHON +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  ’3CHON +’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
k1
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 CHON
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
k1

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  0.1490.0367 0.0055 '6  0.661.4595 0.96327
 12 1.7041 0.14914 0.40816 22.318
 12  31 0.516 23
M  0.036722.318 23  0.0356 g g-1
X / S 3 BY  ' M S
Y  '3 MB '1 MO   0.036722.318 1.459532 0.0175 g g1
X / O2
  6
  1   
11  0.663
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
j1 j1 j1
Noj xokj  Nij xikj Foj wokj  Fij wikj
j1
 r
rA

rH

rN

rk
A H N k
rk  k r
 Mkk rFoj wokj  Fij wikj
M L
j1 j1
 k rNoj xokj  Nij xikj
M L
j1 j1

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  27 26 mole h1
oB
NoC  2r  27 14 mole h
1
1
r  7 mole h15  8  1r
40  NoB   2r
0  NoC  2r

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
0.48144
1180
1
 0.384 g g 
 M
YX / S
G G
BMB
 M

0.48144

332
1
X / O2 0.72 g gY 
O O
B B
 M
RQ 
C

3.12
1.04 mole mole1
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 L150
r 
CiBV

CoBV
 rV
M M
B
B B
5010  0.48r105

5
144
0 
N  NoG G rViG
C 105
 5105
 50105

180 1440.48180
iG
CiGV

CoGV
 rV
M M
G
G G
50180
 5 
1440.48 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
50116
2144
1
 1 20.139  21.139 g LCiA  1
N  N  A
rViA oA
2

EXAMPLE
• One mole NH3
• (NH4)2SO4
C 105
 0.4850105

21440.48116116
iA
CiAV

CoAV

A
rV
M A M A 2
OrV
MO
C VNiO  NoO  oO
0.00410  35010 

1440.48 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
0.00210  3.125010 
1440.48
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
400.9
 
  4
1
9 mole h

r 
NiA X A
A
 NoA ArNiA
1
 4 mol hNoA
 NoO OrNiO
N  15 mol h1
oO
N rNiN  NoN
N  36 mol h1
oN
 NoW W rNiW
1
 54 mol hNoW
40  NoA   49
60  NoO   59
0  NoNO  49
0  NoW  69
• 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
500.8
  5
1
 8 mole h

r 
O
NiO XO
 NoA ArNiA
 NoO OrNiO
 NoN N r
 NoW W r
NiN
N  50   48 18 mole h1
oA
N  50   58 10 mole h
1
oO
N  0  48 32 mole h1
oN
 0  68 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)101
(1)10 1
  0.2
NiA A NiP
EA 
A iP P N 
P 16.38  (1.5)101
(1.5)10 1
 0.092
O2 NiP P

NiO O NiP
EO 
0.310
 
 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  13 9 mole h
1
oA
N  16.38  1.53 11.88 mole h
1
oO
 0  13 3 mole h
1NoC
N  0  33 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
41 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 1.0G
M  1
1
 1.0 kmole hr 
N  N  8.216 kmole h1
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   21 0.184 kmole h
1
oO
 0  11 1.0 kmole h
1NoC
N  0  21 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 0.48r105

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  0.072338105

188 188
104
CiA
 10 1 0.480.072338
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   30.072338105

 NoC CrVNiC
0  N  3.120.072338105
oC
22.4 L
 50323.47 L min-1
h 60 min mole
1hQ 134795
mole
oU

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