ChE 201 Material Balance Lecture 1 Material Balance: is accounting of material is normally carried around a system

Transcription

1 ChE 201 Material Balance Lecture 1 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. 1 Fall 2011 ChE 201

2 BANK ACCOUNT MONEY BALANCE On 10/8/1432 Ali has SR 5,000 in his account in the bank On 25/8/1432 The bank deposited to his account SR 990 (his monthly allowance) On 28/8/1432 He paid by telephone from his account for his mobile tel bill (SR ) He also paid for his home electric bill (SR ) 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 2 Fall 2011 ChE /8/1432?

3 BANK ACCOUNT MONEY BALANCE ALI s ACCOUNT SR SR SR SR SR BALANCE = = SR Fall 2011 ChE 201 3

4 BANK ACCOUNT MONEY BALANCE + SR SR SR 500 ALI s ACCOUNT SR SR Fall 2011 ChE 201 4

5 BANK ACCOUNT MONEY BALANCE schematically SR 990 SR 500 ALI s ACCOUNT SR 5000 SR SR BALANCE = = SR ACCUMULATION = = SR Fall 2011 ChE 201 5

6 Remember:Examples of operations: Example Type Fluid transport (in a pipe ) Physical change Heat transport Physical change Distillation column Physical change Chemical reaction Chemical change Drying Physical change Filling a tank of water Physical change Mixing Physical change 6 Fall 2011 ChE 201 6

7 Remember: For Systems, We must define the boundary of the system Systems are 2 types : closed and open Closed system : material is not crossing the boundary Open system: material is crossing the boundary 7 Fall 2011 ChE 201

8 Examples D Reaction Water tank F Closed system? Or open system? Distillation Column Water Tank 8 Fall 2011 ChE 201

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 9 Fall 2011 ChE 201

10 100 kg/min Case 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 10 Fall 2011 ChE

11 60 kg/min Case kg Water Tank 200 kg/min WHAT WILL HAPPEN TO THE LEVEL OF WATER IN THE TANK after 5 minutes? 11 Fall 2011 ChE 201

12 80 kg/min Case kg Water Tank 200 kg/min The System CHANGED WITH TIME WE CALL IT UNSTEADY STATE 12 Fall 2011 ChE 201

13 at steady state If we have no reaction, the MB equation can be put as: Mass In = Mass out (no accumulation) And also, Moles in = Moles out WHY? 13 Fall 2011 ChE 201

14 at steady state 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 14 Fall 2011 ChE

15 Batch System 9000 kg 100% H 2 O 1000 kg 100% NaOH Initial state 10,000 kg 90% H 2 O 10% NaOH final state 15 Fall 2011 ChE 201

16 Batch System 9000 kg 100% H 2 O 1000 kg 100% NaOH System boundry System boundry Batch system represented as an open system 10,000 kg 90% H 2 O 10% NaOH 16 Fall 2011 ChE 201

17 Example kg H 2 O 50 kg HCl 50 kg H 2 O P? HCl? H 2 SO 4? H 2 O? 40 kg H 2 SO kg H 2 O Let us carry MB: Type of MB Mass IN = Mass OUT Total balance = P HCl balance 50 = HCl in P = 50 kg H 2 SO 4 balance 40 = H 2 SO 4 in P = 40 kg H 2 O balance = H 2 O in P =510 kg THEREFORE : Total mass in Mass of Component 1 in Mass of Component 2 in Mass of Component 3 in = Total mass out = Mass of Component 1 out = Mass of Component 2 out = Mass of Component 3 out 17 Fall 2011 ChE 201

18 300 kg H 2 O 50 kg HCl 50 kg H 2 O P? HCl? H2SO4? H2O? 40 kg H 2 SO kg H 2 O Let us carry MB: Type of MB Mass IN = Mass OUT Total balance = P HCl balance 50 = HCl in P = 50 kg H 2 SO 4 balance 40 = H 2 SO 4 in P = 40 kg H 2 O balance = H 2 O in P =510 kg THEREFORE : Total mass in Mass of Component 1 in Mass of Component 2 in Mass of Component 3 in = Total mass out = Mass of Component 1 out = Mass of Component 2 out = Mass of Component 3 out 18 Fall 2011 ChE 201

19 300 kg H 2 O 50 kg HCl 50 kg H 2 O P? HCl? H2SO4? H2O? 40 kg H 2 SO kg H 2 O Let us carry MB: Type of MB Mass IN = Mass OUT Total balance = P HCl balance 50 = HCl in P = 50 kg H 2 SO 4 balance 40 = H 2 SO 4 in P = 40 kg H 2 O balance = H 2 O in P =510 kg THEREFORE : Total mass in Mass of Component 1 in Mass of Component 2 in Mass of Component 3 in = Total mass out = Mass of Component 1 out = Mass of Component 2 out = Mass of Component 3 out 19 Fall 2011 ChE 201

20 300 kg H 2 O 50 kg HCl 50 kg H 2 O P? HCl? H2SO4? H2O? 40 kg H 2 SO kg H 2 O Let us carry MB: Type of MB Mass IN = Mass OUT Total balance = P HCl balance 50 = HCl in P = 50 kg H 2 SO 4 balance 40 = H 2 SO 4 in P = 40 kg H 2 O balance = H 2 O in P =510 kg THEREFORE : Total mass in Mass of Component 1 in Mass of Component 2 in Mass of Component 3 in = Total mass out = Mass of Component 1 out = Mass of Component 2 out = Mass of Component 3 out 20 Fall 2011 ChE 201 %HCl=

21 NOTE : If we divide by the respective molecular weights, then the equations become also valid for moles. Remember : no reaction here 21 Fall 2011 ChE 201

22 Example 2: MB with no reaction 5 % nitrocellulose 95% water A 8 % nitrocellulose 92% water C B Pure nitrocellulose Type of MB Mass in = Mass out Total balance A + B = 1000 kg nitrocellulose balance 0.05 A + B = 0.08 x 1000 H 2 O balance 0.95 A + 0 = 0.92 x Fall 2011 ChE 201

23 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 23 Fall 2011 ChE 201