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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 Presentation
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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.
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?
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
ALI’s ACCOUNT
SR 5000
+ SR 990 - SR 345.34
- SR 230.89+ SR 500
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
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
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
Dr Iskanderani Fall 2005 8
Examples F Reaction Water tank
D
Closed ? Or open ?
Distillation Column
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 OutputIn this course (and in most processes), systems are at steady state
Let’s explain steady state
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
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
Dr Iskanderani Fall 2005 12
6600 kg
Water
Tank
80 kg/min
200 kg/min
The System CHANGED WITH TIME
WE CALL IT UNSTEADY STATE
After 5
minutes
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
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
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
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
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
Dr Iskanderani Fall 2005 18
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 ?
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 ?
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 ?
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 ?
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
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
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
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