BALANCE MASICO UNILLANOS-1 ENERGY AND MASS BALANCE IN CHEMICAL PROCESSSES

ENERGY AND MASS BALANCE IN CHEMICAL PROCESSSES

1. Chemical Process Calculation 3 rd Semester Lecture#5 Introduction to material balances

2. Material balance involves calculations the quantities of all materials that enter and leave any system or process which are based on the principle of the "law of conversation of mass". This law states that matter is neither created nor destroyed in the process and the total mass remains unchanged. Introduction to material balance Total mass entering a process/Unit – Total mass leaving a process/Unit = Mass accumulation in a process/Unit Output Input Output system Output 1 2

3. Material Balance without Chemical Reactions Introduction The knowledge of material balance with or without chemical reaction is very essential for process design. In similar way, energy balance with or without chemical reaction is also essential for process design related to heat transfer unit operations. Thus, we can say that material and energy balances are the basic tools of process calculations. The design of equipment starts only after the completion of material and energy balance of the process. The material balance calculations give the information regarding the amount of material coming in, material going out, material accumulated or depleted during the unit operations. Based on these calculations, the feasibility of the process or performance of the equipment can be judged. The law of conservation of mass is the basis for material balance calculations. It states that the total mass of various components involved in a unit process remains constant. Thus, for any unit process Input – Output = Accumulation For steady state unit process where accumulation is constant, we have Input – Output = 0

4.   In the case that there is NO Chemical Reaction   Total mass entering a process/unit – Total mass leaving a process/unit = Mass accumulation in a process/unit Classification of Material Balance without Chemical Reaction The material balance problems without chemical reaction can be classified as follows: 1. Material balance at steady state operation. 2. Material balance at unsteady state operation.

5. Steady state process: The steady state process can be defined as that process in which all the operating conditions ( temperature, pressures, compositions and flow rate) remains constant with time. In such process there is NO accumulation in the system, and the equation of material balance can be written as: Input = Output Input - Output= Accumulation For steady state unit process where accumulation is constant, we have: Input- Output= 0 5

6. Various Important Operations Carried Out in Industry The various important operations carried out in industry are as follows: 1. Distillation 2. Absorption 3. Extraction 4. Crystallization 5. Drying 6. Mixing 7. Evaporation.

7. Various important operations carried out in industry Distillation It is a method of separation of mixture based on differences in volatilities of components in a boiling liquid mixture. It is a unit operation or physical separation process. The product obtained from the top is called distillate or overhead product. The product removed from the bottom is called residue or bottom product. 7

8. Distillation Distillate (D) Bottom product (W) Material balance on distillation column 1. Overall material Balance: Where F: feed rate kg/hr D: distillate kg/hr W: residue or bottom product kg/hr 2. Component Balance: Material balance of component ‘ A ’ in distillation column can be written as: where xA = Mole fraction of A in feed, Distillate and bottom product. F xA = D xA + W xA A in feed = A in distillate + A in bottoms Input = Output F = D + W 8 f Feed(F)

9. Steps for calculation Draw the block diagram Put a letter or a symbol Putting equation Choosing a basis Checking solution input=output   Over all material balance In = Out Feed = Top product + Bottom product+ Product * If we take component A material balance on A A× F = A × T

10. EXAMPLE 10,000 kg/h of solution containing 20% methanol is continuously fed to a distillation column. Distillate is found to contain 98% methanol and waste solution from the column carries 1%methanol. All percentage are by weight. Calculate: (i) The mass flow rates of distillate and bottom product (ii) The percentage loss of methyl alcohol. Solution: 10,000 kg/h of 20% methanol feed solution Let x = Mass flow rates of distillate and y = Mass flow rates of (bottom) waste solution.

11. Overall material balance: 10,000 = x + y ………(i) Material balance of methanol: 10,000 × 0.2 = x × 0.98 + y × 0.01 2000 = 0.98 x + 0.01 y ………(ii) Now x + y = 10,000………(i) Also , 0.98 x + 0.01 y = 2000………(ii) Multiplying Eq. (i) by 0.98, we get 0.98 x + 0.98 y = 9800………(iii) Subtracting Eq. (ii) from Eq. (iii), we get Therefore, y = 8041.2 kg/h and x = 1958.8 kg/h

12. Mass flow rate of distillate = 1958.8 kg/h Mass flow rate of bottoms = 8041.2 kg/h Methanol in waste solution = 8041.2 × 0.01 = 80.412 kg/h Methanol in feed solution = 10,000 × 0.2 = 2000 kg/h Therefore, percentage loss of methanol = = = 4.02

13. EXAMPLE A distillation column separates 20% C6H6, 50 % Toluene, 30 % Xylene into 95 % C6H6, 4% Toluene and 1% Xylene and waste product containing 2 % C6H6. Calculate the quantities of distillate and residue if 1000 kgmol /h of feed is fed. Solution: F= 1000 kgmol /h Over all material balance: F = D + W 1000 = D + W ………(i) Material balance of Benzene: 1000 × 0.2 = D × 0.95 + W × 0.02 200 = 0.95 D + 0.02 W ………(ii) Now, D + W = 1000………(i) Also, 0.95 D + 0.02 W = 200………(ii) Multiplying Eq. (i) by 0.02, we get 0.02 D + 0.02 W = 20………(iii)

14. Subtracting Eq. (iii) from Eq. (ii), we get Therefore, D = 193.55 kgmol /h W = 806.45 kgmol /h Mass flow rate of distillate = 193.55 kgmol /h Mass flow rate of residue = 806.45 kgmol /h Material balance for Toluene: 1000 × 0.5 = 193.55 × 0.04 + 806.45 × T T = 61%

15. Material balance for Xylene: 1000 × 0.3 = 193.55 × 0.01 + 806.45 × X X = 37% Quantities in distillate = 193.55 × (78 × 0.95 + 92 × 0.04 + 106 × 0.01) = 15259.48 kg Quantities in residue = 806.45 (78 × 0.02 + 92 × 0.61 + 106 × 0.37) = 78145 kg