THIS INCLUDE MATERIAL BALANCE INFO OPEN CLOSE SYSTEM AND

2. Material Balance for
Multiple Units without
Chemical Reaction.

3. Index
• Material Balance
• System
• System Boundary
• Open System
• Closed System
• Accumulation
• Steady State System
• Steps to solve the snags
• Related Problems

4. Material Balance
Material Balance is nothing more than
the application of the Law of the
Conversation of Mass. Material
Balance is also called Mass Balance.

5. Law of the Conversation of Mass
“Matter is neither created
nor destroyed”.

6. General Expression
Input through system boundary – Output through system boundary + Generation
within system boundary – Consumption within system boundary = Accumulation

7. Applications
Why Material Balance is so Important?

8. System
Any Arbitrary part
OR
AWhole process is
called a System.

9. System Boundary
Imaginary Limit OR Boundary
of the system.

10. Open System
A system in which material enters or
crosses the system boundary. Open
system is also called Flow System.

12. Closed System
A system in which no
material crosses the
system boundary.

13. Accumulation
The difference between Input and Output is called Accumulation.
Negative Accumulation
A reduction of material in the system is called negative accumulation.

14. Steady State System
A system in which all the conditions (temperature, pressure,
amount of material) remains constant.
In Steady State System
Accumulation = 0

15. Un-Steady State System
A system in which one or more of the conditions (pressure,
temperature, amount of material) of the system vary with time. It is
also known as a transient system
Continuous Process
A system in which the material enters or leaves the system
without break.

16. Strategy for solving material balance problems
1. Read and understand the problem statement.
2. Draw a sketch of process and specify the system boundary.
3. Place labels for unknown variables and values for non-variables on the sketch.
4. Obtain any missing needed data.
5. Choose the basis.
6. Determine the number of unknowns.
7. Determine the numbers of independent equations and carry out degree of freedom analysis.
8. Write down the equation to be solved.
9. Solve the equation and calculate the quantities asked for.

17. Numerical

18. BASIS = F =100kg=hour
Overall Material Balance
As we know that
INPUT =OUTPUT
SO,
F=P+W
Now,
Component KCL Balance
0.2(100) =0.95(P) +0
P= 21.05kg/hour
Now,
Component Water Balance
F =P+W
0.8(100) = 0.05(21.05) +1(W)
W =78.84kg/hour
Now,
Total Material Balance on Unit 1
F+C =A

19. 100 +C = A Equation (1)
Now,
KCL Balance
0.2(100) + 0.33(C) = 0.25(A)
20 +0.33(C) =0.25(100 + C)
20+ 0.33(C) = 25+0.25(C)
0.08(C) =5
C =62.5kg/hour
Putting value of C in Equation (1)
100+62.5=A
A=162.5kg/hour
Now,
Total Material Balance on Unit (2)
A =W+B
Now,
KCL Balance
0.25(A) = 0(W) + 0.5 (B)
0.25(162.5) = 0.5(B)
B = 81.25kg/hour

20. Figure
Acetone is used in the manufacturer
of many chemicals and also as a
solvent. In its latter role, many
restrictions are placed on the release
of acetone vapour to the
environment.You are asked to design
an acetone, recovery system having
the flow sheet illustrated in figure.All
the concentrations shown in figure of
both gases and liquids are specified in
weight percent in this special case to
make the calculations simpler.
Calculate A, F,W, B and D per hour.
Numerical

21. Solution
Basis = 1400kg =1hr
Overall Material Balance
W+G=A+B+D
Unit 1 Material Balance
W+G=A+F
Acetone Balance
1400(0.03) =F (0.19) In which F=42/0.19 So F=221.054
Air Balance
1400(0.95) =0.995A A=1336.7
Water Balance
1400(0.02) +W (1.00) =F (0.81) +A (0.005)
28+W=0.81*221.05+1336.7*0.005
W=157.68
Now we have to find D and B. For this we have to apply Material Balance on unit 2 + unit 3
Material balance unit 2+ unit 3
F=D+B
Acetone Balance
221.05*0.19=0.99D+0.04B Equation 1
Water Balance
0.81*221.05=0.01D+0.96B Equation 2
Now comparing both equations
By this method we find the value
B= 188.08 D= 34.83

22. Calculate all the streams while W2 is 25%of feed and also calculate unknown composition
in W2 also find the ratio W1 to F
Numerical

23. Feed = F =1200 kg
Overall Material Balance:
Input = Output
F = P1 + P2 +W2
1200 = P1 + P2 + 0.25 (1200)
900 = P1 + P2
Component A Balance on 1 & 2 unit:
0.15(F) = 0.45(P1) + 0.1(P2) + 0.002(W2)
0.15 (1200) = 0.45 (P1) + 0.1(900-P1)
P1 = 255.43kg & P2 = 644.57kg
Component B Balance on 1 & 2:

24. 0.3(1200) = (0.3) (255.43) + (0.15)
(644.57) + B (300)
In Stream W₂
WA +WB +WC
0.002 + 0.62+ C = 1
WC = 37.8 %
A = 0.2 % B = 62 % C = 37.8 %
Total Material Balance on Unit 1
F = P₁ +W₁
1200 + 255.43 +W₁
W₁ = 944.57 kg
W₁/F = 0.787

26. Basis
1000 lb/hr = F
Overall Balance
F = P + M + D
1000 = P + M + D
M = 1000 – P – D (1)
Xylene – Balance
0.2 (F) = 0 (P) + 0(M) +0.90(D)
0.2(1000) =0.9(D)
D = 222.2 lb/hr
Putting the value of “D” in Equation (1)
M = 1000 – P – 222.2
M = 777.8 – P (2)
Benzene Balance
0.4(F) = 0.99(P) + 0.05(M) + 0(D)
Using the Equation (2)
0.4(1000) = 0.99(P) + 0.05(777.8-P)
400 = 0.99(P) + 388.9 – 0.005 (P)
11.1 = 0.94(P)
P = 11.80 lb/hr
Putting the value of “P” in Equation (2)
M = 777.8 – 11.80
M = 766 lb/hr

28. BASIS= 14670=P3
Overall Balance
F=V1 +V2+V3+P3
NaCl Balance
0.25(F) =0+0+0+14670(0.97)
F=56919.6
Total Material Balance on Unit 1
F=V1+P1
NaCl Balance
0.25(56919.6) =0+0.33(P1)
P1=43120.9
Total Material Balance on Unit (2)
P1=V2+P2
NaCl Balance
0.33(43120.9) =0+0.50(P2)
P2=28459.8