che 201 section 3 material balance material balance: is accounting of material
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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. - PowerPoint PPT PresentationTRANSCRIPT
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Dr Iskanderani Fall 2005 1
ChE 201 Section 3 Material BalanceMaterial 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.
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Dr Iskanderani Fall 2005 2
BANK ACCOUNT MONEY BALANCEOn 10/8/1426Ali has SR 5,000 in his account in the bank
On 25/8/1426The 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?
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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
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Dr Iskanderani Fall 2005 4
BANK ACCOUNT MONEY BALANCE
ALI’s ACCOUNT
SR 5000
+ SR 990 - SR 345.34
- SR 230.89+ SR 500
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Dr Iskanderani Fall 2005 5
BANK ACCOUNT MONEY BALANCE
ALI’s ACCOUNT
SR 5000
SR 990
SR 500 SR 230.89
SR 345.34
BALANCE = 5000 + 990 + 500 – 345.34 – 230.89 = SR 5913.77
ACCUMULATION =5000+990+500–345.34–230.89 = SR 5913.77
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Dr Iskanderani Fall 2005 6
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
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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 crossing the boundary
Open system:material is crossing the boundary
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Dr Iskanderani Fall 2005 8
Examples F Reaction Water tank
D
Closed ? Or open ?
Distillation Column
Water
Tank
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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 OutputIn this course (and in most processes), systems are at steady state
Let’s explain steady state
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Dr Iskanderani Fall 2005 10
WHAT WILL HAPPEN TO THE LEVEL OF
WATER IN THE TANK BY TIME?
7000 kg
Water
Tank
100 kg/min
100 kg/min
IT WILL NOT CHANGE WITH TIME
WE CALL IT STEADY STATE
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Dr Iskanderani Fall 2005 11
WHAT WILL HAPPEN TO THE LEVEL OF
WATER IN THE TANK after 5 minutes?
7000 kg
Water
Tank
60 kg/min
200 kg/min
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6600 kg
Water
Tank
80 kg/min
200 kg/min
The System CHANGED WITH TIME
WE CALL IT UNSTEADY STATE
After 5
minutes
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Dr Iskanderani Fall 2005 13
If we have no reaction,the MB equation can be put as:
Mass In = Mass out
And also, Moles in = Moles out
WHY?
at steady state
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Dr Iskanderani Fall 2005 14
If we have no reaction,the MB equation can be put as:
Mass In = Mass out
And also, Moles in = Moles out
WHY?
at steady state
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Dr Iskanderani Fall 2005 15
If we have no reaction,the MB equation can be put as:
What goes in must come out
Mass in = mass out Moles in = Moles out
Accumulation = 0
at steady state
at steady stateno reaction
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Dr Iskanderani Fall 2005 16
Batch System
Initial state
9000 kg
100% H2O
1000 kg
100% NaOH
Final state
10,000 kg
90% H2O
10% NaOH
System boundry
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Dr Iskanderani Fall 2005 17
Batch System
9000 kg
100% H2O
1000 kg
100% NaOH
10,000 kg
90% H2O
10% NaOH
Batch system represented
as an open system
System boundry
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300 kg H2O
50 kg HCl50 kg H2O
P
40 kg H2SO4
160 kg H2O
Let us carry MB:Type of MB Mass IN = Mass OUT
Total balance 50 +50 +300 + 40 + 160 = PHCl balance 50 = HCl in P = 50 kgH2SO4 balance 40 = H2SO4 in P = 40 kgH2O balance 50 +300 + 160 = H2O in P =510 kg
THEREFORE :Total mass in = Total mass outMass of Component 1 in = Mass of Component 1 outMass of Component 2 in = Mass of Component 2 outMass of Component 3 in = Mass of Component 3 out
?HCl ?
H2SO4 ?
H2O ?
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Dr Iskanderani Fall 2005 19
300 kg H2O
50 kg HCl50 kg H2O
P
40 kg H2SO4
160 kg H2O
Let us carry MB:Type of MB Mass IN = Mass OUT
Total balance 50 +50 +300 + 40 + 160 = PHCl balance 50 = HCl in P = 50 kgH2SO4 balance 40 = H2SO4 in P = 40 kgH2O balance 50 +300 + 160 = H2O in P =510 kg
THEREFORE :Total mass in = Total mass outMass of Component 1 in = Mass of Component 1 outMass of Component 2 in = Mass of Component 2 outMass of Component 3 in = Mass of Component 3 out
?HCl ?
H2SO4 ?
H2O ?
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Dr Iskanderani Fall 2005 20
300 kg H2O
50 kg HCl50 kg H2O
P
40 kg H2SO4
160 kg H2O
Let us carry MB:Type of MB Mass IN = Mass OUT
Total balance 50 +50 +300 + 40 + 160 = PHCl balance 50 = HCl in P = 50 kgH2SO4 balance 40 = H2SO4 in P = 40 kgH2O balance 50 +300 + 160 = H2O in P =510 kg
THEREFORE :Total mass in = Total mass outMass of Component 1 in = Mass of Component 1 outMass of Component 2 in = Mass of Component 2 outMass of Component 3 in = Mass of Component 3 out
?HCl ?
H2SO4 ?
H2O ?
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Dr Iskanderani Fall 2005 21
300 kg H2O
50 kg HCl50 kg H2O
P
40 kg H2SO4
160 kg H2O
Let us carry MB:Type of MB Mass IN = Mass OUT
Total balance 50 +50 +300 + 40 + 160 = PHCl balance 50 = HCl in P = 50 kgH2SO4 balance 40 = H2SO4 in P = 40 kgH2O balance 50 +300 + 160 = H2O in P =510 kg
THEREFORE :Total mass in = Total mass outMass of Component 1 in = Mass of Component 1 outMass of Component 2 in = Mass of Component 2 outMass of Component 3 in = Mass of Component 3 out
?HCl ?
H2SO4 ?
H2O ?
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NOTE : If we divide by MWt , then the equations become also valid for moles.
Remember : no reaction here
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Example :
Pure nitrocellulose ?kg
5% nitrocellulose95% water
1000 kg
Type of MB Mass IN = Mass OUTTotal balance A + B = 1000 kgnitrocellulose balance 0.05 A + B = 0.08 x 1000H2O balance 0.95 A + 0 = 0.92 x 1000
8 % nitrocellulose92% waterA
B
2
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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