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ChE 201 Section 3 Material Balance Material Balance: is accounting of material

ChE 201 Section 3 Material Balance Material Balance: is accounting of material is normally carried around a system What is a system? : It is a portion or whole of a process (or a plant) to be analyzed.

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ChE 201 Section 3 Material Balance Material Balance: is accounting of material

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  1. ChE 201 Section 3 Material Balance • Material Balance: • is accounting of material • is normally carried around a system • What is a system?: It is a portion or whole of a process (or a plant) to be analyzed. • What is a process? : It is one action or a series of actions or operations or treatments that result in an end. Dr Iskanderani Fall 2005

  2. BANK ACCOUNT MONEY BALANCE • On 10/8/1426 • Ali has SR 5,000 in his account in the bank • On 25/8/1426 • The bank deposited to his account SR 990 (his monthly salary) • On 28/8/1426 • He paid by telephone from his account • for his mobile tel bill (SR 345.34) • He also paid for his home electric bill (SR 230.89) • He received a check from his friend for a loan he gave to him(SR 500) • WHAT IS THE BALANCE OF HIS ACCOUNT as of 28/8/1426? Dr Iskanderani Fall 2005

  3. BANK ACCOUNT MONEY BALANCE ALI’s ACCOUNT SR 5000 + SR 990 - SR 345.34 - SR 230.89 + SR 500 BALANCE = 5000 + 990 + 500 – 345.34 – 230.89 = SR 5913.77 Dr Iskanderani Fall 2005

  4. BANK ACCOUNT MONEY BALANCE + SR 990 ALI’s ACCOUNT SR 5000 - SR 345.34 + SR 500 - SR 230.89 Dr Iskanderani Fall 2005

  5. SR 990 SR 345.34 SR 500 SR 230.89 BANK ACCOUNT MONEY BALANCE ALI’s ACCOUNT SR 5000 BALANCE = 5000 + 990 + 500 – 345.34 – 230.89 = SR 5913.77 ACCUMULATION =5000+990+500–345.34–230.89 = SR 5913.77 Dr Iskanderani Fall 2005

  6. Examples of operations: Dr Iskanderani Fall 2005

  7. For Systems: We must define the boundary of the system Systems are 2 types : closed and open Closed system : material is not crossingthe boundary Open system: material is crossing the boundary Dr Iskanderani Fall 2005

  8. Examples Distillation Column Closed ? Or open ? Water Tank Dr Iskanderani Fall 2005

  9. MB around a system : • we apply the Law of conservation of mass • Material Input - material output = Accummulation • Example of accummulation • - ve accummulaion +ve accummulation • At steady state, variables do not change with time ; and the eq becomes: • Material Input = material Output • In this course (and in most processes), systems are at steady state Let’s explain steady state Dr Iskanderani Fall 2005

  10. 100 kg/min 7000 kg Water Tank 100 kg/min WHAT WILL HAPPEN TO THE LEVEL OF WATER IN THE TANK BY TIME? IT WILL NOT CHANGE WITH TIME WE CALL IT STEADY STATE Dr Iskanderani Fall 2005

  11. 60 kg/min 7000 kg Water Tank 200 kg/min WHAT WILL HAPPEN TO THE LEVEL OF WATER IN THE TANK after 5 minutes? Dr Iskanderani Fall 2005

  12. 80 kg/min 6600 kg Water Tank After 5 minutes 200 kg/min The System CHANGED WITH TIME WE CALL IT UNSTEADY STATE Dr Iskanderani Fall 2005

  13.  If we have no reaction,the MB equation can be put as: Mass In = Mass out at steady state And also, Moles in = Moles out WHY? Dr Iskanderani Fall 2005

  14.  If we have no reaction,the MB equation can be put as: Mass In = Mass out at steady state And also, Moles in = Moles out WHY? Dr Iskanderani Fall 2005

  15.  If we have no reaction,the MB equation can be put as: What goes in must come out at steady state no reaction Mass in = mass out Moles in = Moles out Accumulation = 0 at steady state Dr Iskanderani Fall 2005

  16. Batch System 9000 kg 100% H2O 1000 kg 100% NaOH Initial state System boundry 10,000 kg 90% H2O 10% NaOH Final state Dr Iskanderani Fall 2005

  17. Batch System 9000 kg 100% H2O 1000 kg 100% NaOH 10,000 kg 90% H2O 10% NaOH System boundry Batch system represented as an open system Dr Iskanderani Fall 2005

  18. HCl ? H2SO4 ? H2O ? Dr Iskanderani Fall 2005

  19. HCl ? H2SO4 ? H2O ? Dr Iskanderani Fall 2005

  20. HCl ? H2SO4 ? H2O ? Dr Iskanderani Fall 2005

  21. HCl ? H2SO4 ? H2O ? Dr Iskanderani Fall 2005

  22. NOTE : If we divide by MWt , then the equations become also valid for moles. • Remember : no reaction here Dr Iskanderani Fall 2005

  23. 2 Dr Iskanderani Fall 2005

  24. SOLVE • A = 968.4kg and B = 31.6 kg • How many equations have we used? • How many equations are available ? • Are they all independent? • How many components do we have in the problem? Number of independent equations = no. of components in the system Dr Iskanderani Fall 2005

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