Mass Balance- in Non-Reactive System 1 (Single unit)

LEARNING OBJECTIVES By the end of this topic, you should be able to: Performed material balance for distillation column

EXAMPLE 1 A liquid. mixture of benzene (B) and toluene (T) containing 55% B by mass is fed continuously to a distillation column with a feed rate of 100 kg/h. A product stream leaving the top of the column (overhead product) contains 85% B and a bottom product stream contains 10.6% B by mass. Determine the mass flow rate of the overhead product stream and the mass flow rate of the bottom product stream.

EXAMPLE 1 STEPS 2, 3 & 4 mV kg/h 0.850 kg B/kg 0.150 kg T/kg 100 kg/h  kg/h 100 kg/h 0.550 kg B/kg 0.850 kg B/kg 0.106 kg B/kg 0.450 kg T/kg 0.150 kg T/kg 0.894 kg T/kg DISTILLATION COLUMN mL  kg/h

EXAMPLE 1 STEPS 5-DoF DoF must be zero to be solvable DoF = unknown - independent eqn Unknown = 2 Independent equation = 2 DoF must be zero to be solvable mV  mL & Material Balance for B and T DoF = unknown - independent eqn = 2 - 2 = 0 (problem solvable)

Take the basis of calculation = 100 kg/h of feed EXAMPLE 1 STEPS 6 & 7 Take the basis of calculation = 100 kg/h of feed Since this operation is at steady state and non-reactive system, hence  Input = output Total balance: 100 kg/h = mV  mL + A DC 0.450 kg T/kg 0.150 kg T/kg 0.894 kg T/kg 100 kg/h 0.550 kg B/kg 0.850 kg B/kg 0.106 kg B/kg mV  kg/h mL Benzene balance: 100 (0.550) = 0.850 mV  0.106 mL B +

EXAMPLE 1 Solve equation A and B simultaneously The results are  mV  mL = 59.7 kg/h = 40.3 kg/h

EXAMPLE 2 1000 kg/hr of mixture containing equal parts by mass of methanol and water is distilled. Product streams leave the top and the bottom of the distillation column. The flow rate of the bottom stream is measured and found to be 673 kg/hr and the overhead stream is analyzed and found to contain 96.0 wt% methanol. Draw and label a flowchart of the process and do the degree of freedom analysis. Calculate the mass and mole fractions of methanol and the molar flow rates of methanol and water in the bottom product stream. Suppose the bottom product stream is analyzed and the mole fraction of methanol is found to be significantly higher than the value calculated in part (b), list as many possible reasons for the discrepancy as you can think of.

EXAMPLE 2 STEPS 2, 3 & 4 mV kg/h 4 wt % W 96 wt % M 1000 kg/h 50 wt% W  kg/h 1000 kg/h 50 wt% W 4 wt % W X wt% W 50 wt% M 96 wt % M Y wt% M DISTILLATION COLUMN  mL=673 kg/h

EXAMPLE 2 STEPS 5-DoF DoF must be zero to be solvable DoF = unknown - independent eqn Unknown = 3 Independent equation = 3 DoF must be zero to be solvable mV, X, Y  Material Balance for, Overall and specific balance for W and M DoF = unknown - independent eqn = 3 -3 = 0 (problem solvable)

EXAMPLE 2 mV kg/h 4 wt % W 96 wt % M F=1000 kg/h 50 wt% W mL=673 kg/h  kg/h F=1000 kg/h 50 wt% W 4 wt % W X wt% W 50 wt% M 96 wt % M Y wt% M DISTILLATION COLUMN  mL=673 kg/h Basic of calculation, F=1000 kg/h Overall Balance Specific Balance for Water

EXAMPLE 2 Basic of calculation, F=1000 kg/h Stream Water Methanol Total (kg/h) wt% INPUT, F 500 0.5 1000 TOP PRODUCT, MV 13.08 0.04 313.92 0.96 327 BOTTOM PRODUCT, ML 484.56 0.72 188.44 0.28 673

EXAMPLE 2 Molecular weight H2O=18 kg/kmol, CH3OH=32 kg/kmol Stream Water Methanol Total (kg/h) (kmol/h) (mol%) INPUT, F 500 27.78 0.64 15.62 0.36 1000 43.40 TOP PRODUCT, MV 13.08 0.73 0.07 313.92 9.81 0.93 327 10.54 BOTTOM PRODUCT, ML 484.56 26.92 0.82 188.44 5.89 0.18 673 32.81

EXAMPLE 2 Suppose the bottom product stream is analyzed and the mole fraction of methanol is found to be significantly higher than the value calculated in part (b), list as many possible reasons for the discrepancy as you can think of.

CLASS EXERCISE 1 300 gallons of a mixture containing 75 wt% ethanol and 25 wt% water (mixture SG=0.877) and a quantity of a 40.0 wt% ethanol-60 wt% water mixture (SG=0.952) are blended to produce a mixture containing 60.0 wt% ethanol. Draw and label a flowchart on the mixing process and do the degree of freedom analysis. Calculate the input stream containing 40.0 wt % ethanol.

TAKE HOME ACTIVITY 300 ga1 75 wt% Eth F1 25 wt% H2O P1 60 wt% Eth F2 Basis of calculation F1=100 kg F1+F2=P1 100+F2=P1 [1] Eth Balance 75+0.4F2=0.6P1 [2] Solve Eq [1] and [2] simultaneously F2=75 kg, and P1=175 kg

TAKE HOME ACTIVITY 300 gal 1 m3 264.17 gal =1.13 m3 or 991 kg Given, SG for 75 wt% Eth=0.877, density = 877 kg/m3 300 gal 1 m3 264.17 gal =1.13 m3 or 991 kg Hence, for 100 kg F1, required 75 kg F2, then for 991 kg of F1, will required (991/100)x75=743.25 kg