Material balance for multiple units without chemical equation

Material Balance for

Multiple Units without

Chemical Reaction.

Index

• Material Balance

• System

• System Boundary

•Open System

• Closed System

• Accumulation

• Steady State System

• Steps to solve the snags

• Related Problems

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.

Law of the Conversation of Mass

“Matter is neither

created nor destroyed”.

General Expression

Input through system boundary – Output through system boundary + Generation within system boundary – Consumption within system boundary = Accumulation

Applications

Why Material Balance is so Important?

System

Any Arbitrary part OR

A Whole process is called a System.

System Boundary

Imaginary Limit OR Boundary of the system.

Open System

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

Closed System

A system in which no material crosses the system boundary.

Accumulation

The difference between Input and Output is called Accumulation.

Negative Accumulation

A reduction of material in the system is called negative accumulation.

Steady State System

A system in which all the conditions (temperature, pressure, amount of material) remains constant.

In Steady State System Accumulation = 0

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.

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.

Numerical

BASIS = F =100kg=hourOverall Material BalanceAs we know that INPUT =OUTPUTSO, F=P+WNow, Component KCL Balance0.2(100) =0.95(P) +0P= 21.05kg/hourNow, Component Water BalanceF =P+W0.8(100) = 0.05(21.05) +1(W)W =78.84kg/hourNow, Total Material Balance on Unit 1F+C =A

100 +C = A Equation (1)Now, KCL Balance0.2(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) =5C =62.5kg/hour Putting value of C in Equation (1)100+62.5=AA=162.5kg/hourNow, 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

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

Solution

Basis = 1400kg =1hrOverall Material Balance

W+G=A+B+DUnit 1 Material Balance

W+G=A+FAcetone Balance

1400(0.03) =F (0.19) In which F=42/0.19 So F=221.054Air Balance

1400(0.95) =0.995A A=1336.7Water Balance

1400(0.02) +W (1.00) =F (0.81) +A (0.005)28+W=0.81*221.05+1336.7*0.005

W=157.68Now we have to find D and B. For this we have to apply Material Balance on unit 2

+ unit 3Material balance unit 2+ unit 3

F=D+BAcetone Balance

221.05*0.19=0.99D+0.04B Equation 1Water Balance

0.81*221.05=0.01D+0.96B Equation 2Now comparing both equations

By this method we find the valueB= 188.08 D= 34.83

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

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 = 644.57kgComponent B Balance on 1 & 2:

0.3(1200) = (0.3) (255.43) + (0.15) (644.57) + B (300)In Stream W₂ WA + WB + WC 0.002 + 0.62+ C = 1WC = 37.8 %A = 0.2 % B = 62 % C = 37.8 %Total Material Balance on Unit 1 F = P₁ + W₁1200 + 255.43 + W₁W₁ = 944.57 kgW₁/F = 0.787

Basis 1000 lb/hr = FOverall Balance F = P + M + D1000 = P + M + D M = 1000 – P – D (1)Xylene – Balance0.2 (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 – 222.2M = 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/hrPutting the value of “P” in Equation (2)M = 777.8 – 11.80M = 766 lb/hr

BASIS= 14670=P3Overall BalanceF=V1 +V2+V3+P3NaCl Balance0.25(F) =0+0+0+14670(0.97)F=56919.6Total Material Balance on Unit 1F=V1+P1NaCl Balance0.25(56919.6) =0+0.33(P1)P1=43120.9Total Material Balance on Unit (2)P1=V2+P2NaCl Balance0.33(43120.9) =0+0.50(P2)P2=28459.8