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Episode 61 : MATERIAL BALANCE FOR REACTING SYSTEM
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Transcript of Episode 61 : MATERIAL BALANCE FOR REACTING SYSTEM
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SAJJAD KHUDHUR ABBAS
Ceo , Founder & Head of SHacademyChemical Engineering , Al-Muthanna University, IraqOil & Gas Safety and Health Professional – OSHACADEMYTrainer of Trainers (TOT) - Canadian Center of Human Development
Episode 61 : MATERIALBALANCE FOR REACTING
SYSTEM
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COMPONENT MATERIAL BALANCE FOR REACTING SYSTEMS
System
CONTROLVOLUME
Fi1
wi1k
Fi2
ENVIRONMENT
Fo1
wo1k
Fo2
wo2k
Fo3
wo3kwi2k
j1 j1
wo1k Fo1 wo2k Fo2 wo3k Fo3 wi1k Fi1 wi2k Fi2 M k rk
Mkrk
Reactor
Compressor
Phase
separatior
Distillation
M krk
M L
Fojwokj Fijwikj
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COMPONENT MATERIAL BALANCE FOR REACTING SYSTEMS
System
CONTROLVOLUME
Ni1
xi1k
Ni2
ENVIRONMENT
No1
xo1k
No2
xo2k
No3
xo3kxi2k
rk
Reactor
Compressor
Phase
separator
Distillation
rk
M L
j1 j1
xo1k No1 xo2k No2 xo3k No3 xi1k Ni1 xi2k Ni2 rk
Noj xokj N ij xikj
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RATE OF CHEMICAL REACTION• Rate of chemical reaction of component k : rk
• Net rate of generation of component kcomponent k per unit time
in units of moles of
• Obtained from stoichiometric balance of chemical reactions
• Stoichiometry = relative proportions of chemical componentsparticipating 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
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CHEMICAL REACTOR
Ammonia ReactorAcid Nitric Reactor
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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 molesof 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
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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 0k1
S
2SO2 + (1)O2 2SO3
2SO3 - 2SO2 - (1)O2 = 0
N2 + 3H2 2NH3
2NH3 - N2 - 3H2 = 0
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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
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STOICHIOMETRIC BALANCE• Elemental balance= total element in reactants is equal to the
total element in the product in the stoichiometric equationS
•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
0 k mkl
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STOICHIOMETRIC BALANCEN2 + 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
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EXAMPLEExample 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
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STOICHIOMETRY OF BIOCHEMICAL REACTIONS
• Biotechnological products are produced in fermentationproses involving cell growth and bioproduiction
• Bioreactor/fermentor
• Biochemical transformation processes involved thousandsof 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
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BIOREACTOR/FERMENTOR
Bioreactor/fermentor
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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 = 6Degree of freedom of biochemical Stoichiometry = 6 – 4 = 2
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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 availableelectrons per atom C
• Available electrons = electrons transferred to oxygen afterorganic compound is oxidized to carbon dioxide, water andammonia 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
vlmkl mkC
k1
k
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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 0k1
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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
M S 4M B 1M S '3 M BYX / S
M S 5M P 1M S '4 M PYP / S
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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
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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
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EXAMPLEExample 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
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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 M B
• Formula ethanol MW M S
• 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 -1X / S 3 BY ' M S
Y '3 M B '1 M O 0.036722.318 1.459532 0.0175 g g1X / O2
6
1
11 0.663
4.41 4
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MATERIAL BALANCE WITH SINGLEREACTION
• 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 SYSTEMSMolar Mass
rk M krk
M L M L
j1 j1 j1Noj xokj N ij xikj Fojwokj Fijwikj
j1
rrA
rH rN
rk
A H N k
rk kr
M kk rFojwokj Fijwikj
M L
j1 j1
k rNoj xokj N ij xikj
M L
j1 j1
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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 inSO2 & 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 Cr
NiA
N iB
N iC
N 40 2r 40 27 26 mole h1oB
NoC 2r 27 14 mole h1
1r 7 mole h15 8 1r
40 NoB 2r
0 NoC 2r
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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 /oxygenYX/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.
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EXAMPLE
CoO = 4 mg L-1
CoC = 2 mg L-1
Batch Bioreactor
Glucose feedCiG
CiO = 0
CiC = 0
CiB = 0
Gas exhaust
Feed (NH4)2SO4CiA
Biomass productCoB= 50 gdm L-1
CoG = 5.0 g L-1
CoA = 1.0 g L-1
O2 Feed
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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
0.481441180
1 0.384 g g
MYX / S
G G
BM B
M0.48144
332
1
X / O2 0.72 g gY O O
B B
M
RQ C
3.121.04 mole mole1
O 3
NoB BrVN iB
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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 MB
B B
5010 0.48r105
5
1440
N NoG GrViG
C 105 5105 50105
180 1440.48180iG
CiGV
CoGV rV
M MG
G G
50180 5
1440.48 5 130.208 135.208 g L
1CiG
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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
21441
1 20.139 21.139 g LCiA 1
N N A rViA oA2
EXAMPLE
• One mole NH3
• (NH4)2SO4
C 105 0.485010521440.48116116
iA
CiAV
CoAV A rV
M A M A 2
OrVM O
C VNiO NoO oO
0.00410 35010 1440.48
217.026x103
55
32NiO NoO
11
05
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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 infermenter balances
CrVM C
CoCV NoC NiC
0.00210 3.125010 1440.48
355
225.689x1044
NiC NoC
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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
N ik Nok
kN
X N ik X k N ik N ok
k
r N ik X k
k
Nik
Limitingr
k ip p
k N ip p
N ik
E k
N
N ik X k k r
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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 achieve90% 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 nitricReactor
Conversion 90%
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EXAMPLE
• Stoichiometric coefficient
• Use conversion of ammonia
to get Rate of reaction r
• Component mole balance
• NH3
• O2
• NO
• H2O
400.9
41
9 mole h
r NiA X A
A
NoA ArN iA1
4 mol hNoA
NoO OrN iON 15 mol h 1
oO
N rNiN NoNN 36 mol h1
oN
NoW W rN iW1
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
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EXAMPLE
Example 3.7 If the reaction in Example 3.6 achieves 80% conversion withequimolar ammonia and oxygen feed that is fed at 100 mole h-1. Calculatethe 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 nitricreactor
Conversion 80%
12.5 4
50
A
NiA
510
50
O
NiO
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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 ArN iA
NoO OrN iO
NoN N r
NoW W r
NiN
N 50 48 18 mole h 1oA
N 50 58 10 mole h1
oO
N 0 48 32 mole h1oN
0 68 48 mole h1NoW
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EXAMPLEExample 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 = 0Determine 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
AcrylonitrileReactor
Conversion 30%
12
112
W P
NiW 10
1 10
NiP16.38
1.5 10.92
O
NiO
![Page 36: Episode 61 : MATERIAL BALANCE FOR REACTING SYSTEM](https://reader030.fdocuments.net/reader030/viewer/2022020314/587203b21a28ab176b8b590f/html5/thumbnails/36.jpg)
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
0.2NiA ANiP
EA A iP P N
P 16.38 (1.5)10 1(1.5)10 1
0.092O2NiP P
NiO O NiP
EO
0.310
11
3 mole h
r X P NiP
P
NoA ArN iA
NoO OrN iO
CrN iC NoC
1 N iN 61.62 mole hNoN
NoW W rN iW
N 12 13 9 mole h1
oA
N 16.38 1.53 11.88 mole h1
oO
0 13 3 mole h1NoC
N 0 33 9 mole h1
oW
![Page 37: Episode 61 : MATERIAL BALANCE FOR REACTING SYSTEM](https://reader030.fdocuments.net/reader030/viewer/2022020314/587203b21a28ab176b8b590f/html5/thumbnails/37.jpg)
EXAMPLEExample 3.9 Natural gas containing methane only is burnt in anincinerator 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 equationsDegree 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
Naturalgas 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 airh 28.84 kg air
300 kg 1 kmole air -1
![Page 38: Episode 61 : MATERIAL BALANCE FOR REACTING SYSTEM](https://reader030.fdocuments.net/reader030/viewer/2022020314/587203b21a28ab176b8b590f/html5/thumbnails/38.jpg)
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 11
1.0 kmole hr
N N 8.216 kmole h1
oNUN
NoO Or
NoC Cr
NoW W r
NUO
NGC
NGW
M 2.184 (2)1.0 1(2)1.0 1
0.092NUO O NG
EO O G M N
N 2.184 21 0.184 kmole h1
oO
0 11 1.0 kmole h1NoC
N 0 21 2.0 kmole h1
oW
![Page 39: Episode 61 : MATERIAL BALANCE FOR REACTING SYSTEM](https://reader030.fdocuments.net/reader030/viewer/2022020314/587203b21a28ab176b8b590f/html5/thumbnails/39.jpg)
EXAMPLEExample 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
CiACiG
CiB = 0
ProductQo = Qi = 10000 L h-1
CoG = 5 g L-1
CoB = 50 gdm L-1
CoA= 1 g L-1
Air feedQiU = 50,000 L min-1
NiU = 133928.6 mole h-1
N = 28125.0 mole h-1iO
N = 105803.6 mole h-1iN
N =0iC
Exhaust airNoU
QoU
NoO
NoN
NoC
![Page 40: Episode 61 : MATERIAL BALANCE FOR REACTING SYSTEM](https://reader030.fdocuments.net/reader030/viewer/2022020314/587203b21a28ab176b8b590f/html5/thumbnails/40.jpg)
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
BrVM
QCiB Q
CoB
M B B
ii
![Page 41: Episode 61 : MATERIAL BALANCE FOR REACTING SYSTEM](https://reader030.fdocuments.net/reader030/viewer/2022020314/587203b21a28ab176b8b590f/html5/thumbnails/41.jpg)
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 GrVN iG
N N A rV
iA oA2
GrVM
QCiG Q
CoG
M G G
ii
104Cig 1045 0.072338105188 188
104CiA 10 1 0.480.0723382
10
116 116
54
A rV
M A 2Q
CiA Q CoA
M A
i i
N 133928.571 mole h -1
min 1 j 22.4 L
60min molQ 50000L
iUiU
![Page 42: Episode 61 : MATERIAL BALANCE FOR REACTING SYSTEM](https://reader030.fdocuments.net/reader030/viewer/2022020314/587203b21a28ab176b8b590f/html5/thumbnails/42.jpg)
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 3.120.072338105oC
22.4 L 50323.47 L min -1
h 60 min mole
1hQ 134795mole
oU
![Page 43: Episode 61 : MATERIAL BALANCE FOR REACTING SYSTEM](https://reader030.fdocuments.net/reader030/viewer/2022020314/587203b21a28ab176b8b590f/html5/thumbnails/43.jpg)
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