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Episode 62 : MATERIAL BALANCE FOR REACTING SYSTEM

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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

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Episode 62 : 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 62 : MATERIAL BALANCE FOR REACTING SYSTEM
  2. 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. 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 k1 S S1 S S1 S k k1 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. 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  rS1 N  r N mo SiS(S 1)i(S 1)i2i1 m o s k1   k k1 e N  r N x K  S k k   21 21 S m  k k1 K  N  r k1 kike k
  5. 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. 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  1111  0 k1 K  N  r k1 N  1rN  1r  1rNiW  1rNiM 1 r2  r r  1 0 2 . s iC iH kike k
  7. 7. EXAMPLE • Solve forr • Then r = 0.6667 mol j-1 Total flow rate out No No  Nik k r NiM  1r  NiW  1r  1r  1r • 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 h1 k1 iM iW S  NiM  NiW  NiC M r No W r No C r No  1 10.6667 3  0.111  2  10.6667 3  0.444  0  10.6667 3  0.222  0  10.6667 3  0.222 xoM xoW xoC xoH  NiH H r No 1 N M  1 1iMN 2 W  1  2iW  r  10.667  0.667 N 1 X  iM M M
  8. 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. 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. 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  2r2  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. 11. STOICHIOMETRIC BALANCE OF MULTIPLE REACTIONS reactant • Total rate of reaction of component k in R R reactions r1 • Component mole balance of multiple reactions • Yield • If there is one inlet and one outlet streams: R Nik  Nok  kr rr r1 • Conversion of limiting reactant R r1 • Selectivity of reaction product h with respect to the limiting  kr rr rk Nikj  Nokj  kr rr j1 j r1 L M R X  N  N N    pr r ipp ip op ip  r N    R  hr rr X p Nip r1 Nop Nip  Nih Noh Sh  R  prrr rqmaks r1 rqmaks p Ypq r 
  12. 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 r1 Nop Nip  Nih Noh   rqmaks R  prrr r1 rqmaks p Ypq r 
  13. 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. 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 r1 E 2    r1  r2  42.585 2  r N   r  r N x Lr r iE L1 1 L2 2 i iE r1 L S  0.471  1r  0r 0.851001 2 r1  40.035 r2  2.55 0.501  1r  1r 0.851001 2
  15. 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 0.85100 NoE  140.035 12.55 0  NoL  140.035 02.55 0  NoH  140.035 12.55
  16. 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 1000.85301000.15MI  42.41530 40.03528 37.4852 5.11615MI  Ni xiI MI  NoE ME  NoLML  NoH M H  NoM MM  Ni xiI M I Ni xiE M E 0  N  040.035 22.55oM
  17. 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. 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 r1 2 0.9  1r  1r 0.0781001 2 r1  r2  7.02  NoC C1r1 C2r2 NiC  r2 NoC 0  NoC  0r1  1r2
  19. 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 h1 1 r  80.78 6.24 mole h 1 2 0  NoX  1r1  0r2      1 1r1  7.02   01r2 8 -1  6.24mole hNoC N  0.78 mole h-1  NoM M1r1 M 2r2 oX NiM 7.8  NoM  10.79 16.24 N  0.78 mole h-1 oM
  20. 20. EXAMPLE • O2 • • H2O • • N2 • Check with total balance: • Balanced! 19.4  NoO  1.5r1   2r2 N  5.75 mole h-1  NoW W1r1 W 2r2 oO 0  NoW  2r1  2r2 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 1000.078161000.19432 6.244 0.7828 0.781814.0418 1000.72828    5.75321000.72828     NiO  NoO O1r1 O2r2 NiW
  21. 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. 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 r1  Br rBX    3 r1  r3  65  NoC C1r1 C2r2  0  r1  r2 NoC 0.65  1r  0r  1r 100 1 2 3 0  NoC  1r1  1r2  0
  23. 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  1r2  1r3 NoT  100  r  r  r  r  r  r  100 mole h 1 1 3 1 2 2 3N  NoB  NoC  NoTo 100  NoB  1r1  0r2  1r3 o  NoC oC N x 100 r1  r2  25 0.25  r1  r2
  24. 24. EXAMPLE • Then • Composition of product • Check 10056 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  0.25100 25 mole h1 oC ooC  N  N  100  25  40  35 mole h1NoB  No oToC  35 100  0 35xoB xoT  40 100  0.40 100 100  NoB M B  NoC MC  NoT MT NiB M B
  25. 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. 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 r1 2 2r1  r2  2.5 0.25   2r  1r 0.11001 2
  27. 27. EXAMPLE • Ethylene balance at this condition • Then • Yield of ethylene oxide • Hence • Ethylene balance  NoE E1r'1E2r'2 NiE 0.1100  7.5  2r'10 1 r'1  1.25 mole h 1  2r'1  2.5 mole hrEmaks 0.8  2r  0r  2.51 2 r1 1.0 mole h rEmaks R r1 YL E  Lr rr 1 r  0.5 mole h1 2 NiE  NoE E1r1 E2r2 N  7.5 mole h1 oE 0.1100  NoE   21.0 10.5
  28. 28. EXAMPLE • Ethylene oxide • Oxygen • Water • CO2 0  NoL  21 00.5 N  2.0 mole h1 oL 0.210.9100  NoO  11  30.5 N  16.4 mole h1  NoW W1r1 W 2r2 oO 0  NoW  01.0 20.5 N  1.0 mole h1 oW  NoL L1r1 L2r2 NiL  NoO O1r1 O2r2 NiO NiW 0  NoC  01 20.5 C1r1 C 2r2 NiC  NoC N  1.0 mole h1 oC
  29. 29. EXAMPLE • N2 • Total flow rate product stream • Composition of product streams  0.790.9100 71.1 mole h1NiN  NoN N  7.5  2.0 16.4 1.0 1.0  71.1  99 mole h1 o  7.5 99  0.0758 16.4 99  0.1657 1.0 99  0.0101  1.0 99  0.0101  71.1 99  0.7182 xoL xoE xoO xoW xoC xoN  2.0 99  0.0202
  30. 30. Thanks for Watching Please follow me / SAJJAD KHUDHUR ABBAS
  • DeependraKumarYadav3

    Mar. 26, 2019
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    Sep. 16, 2017
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    Mar. 14, 2017

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

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