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Episode 61 : MATERIAL BALANCE FOR REACTING SYSTEM

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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

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Episode 61 : MATERIAL BALANCE FOR REACTING SYSTEM

  1. 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. 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. 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. 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. 5. CHEMICAL REACTOR Ammonia Reactor Acid Nitric Reactor
  6. 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. 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. 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. 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. 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. 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. 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. 13. BIOREACTOR/FERMENTOR Bioreactor/fermentor
  14. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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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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

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