Mass Balance-

in Non-Reactive System 1 (Single unit)

1

3

LEARNING OBJECTIVES

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

12

EXAMPLE 2

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.

13

STEPS 2, 3 & 4

DISTILLATION COLUMN

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

kg/h

EXAMPLE 2

14

STEPS 5-DoF

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)

EXAMPLE 2

15

Take the basis of calculation = 100 kg/h of feed Since this operation is at steady state and non-reactive system, hence Input = output

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

kg/h

Total balance:

100 kg/h = mV mL

+ A

Benzene balance:

100 (0.550)

h

B kg

= 0.850 mV

0.106 mL

B +

STEPS 6 & 7

EXAMPLE 2

16

Solve equation A and B simultaneously

The results are

mV

mL

= 59.7 kg/h

= 40.3 kg/h

EXAMPLE 2

17

EXAMPLE 3

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.

a. Draw and label a flowchart of the process and do the degree of freedom analysis.

b. Calculate the mass and mole fractions of methanol and the molar flow rates of

methanol and water in the Top product stream.

c. 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.

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STEPS 2, 3 & 4

DISTILLATION COLUMN

50 wt% M

96 wt % M

Y wt% M

1000 kg/h

50 wt% W

4 wt % W

X wt% W

mV

kg/h

mL=673 kg/h

EXAMPLE 3

19

EXAMPLE 3

STEPS 5-DoF

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)

20

DISTILLATION

COLUMN

50 wt% M

96 wt % M

Y wt% M

F=1000 kg/h

50 wt% W

4 wt % W

X wt% W

mV

kg/h

mL=673 kg/h

EXAMPLE 3

327

6731000

V

V

LV

m

m

mmF

Basic of calculation, F=1000 kg/h

Overall Balance

28.072.01

72.0

)673()327(04.0500

04.0500

Y

X

X

mXm LV

Specific Balance for Water

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EXAMPLE 3

Stream Water Methanol Total

(kg/h) wt% (kg/h) wt% (kg/h)

INPUT, F 500 0.5 500 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

Basic of calculation, F=1000 kg/h

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EXAMPLE 3

Stream Water Methanol Total Total

(kg/h) (kmol/h) (mol%) (kg/h) (kmol/h) (mol%) (kg/h) (kmol/h)

INPUT, F 500 27.78 0.64 500 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

Molecular weight H2O=18 kg/kmol, CH3OH=32 kg/kmol

500/18

13.08/18

484.56/18

500/32

313.92/32

188.44/32

27.78/

(27.78+15.62)

0.73/

(0.73+9.81)

15.62/

(27.78+15.62)

9.81/

(0.73+9.81)

26.92/

(26.92+5.89)

5.89/

(26.92+5.89)

23

EXAMPLE 3

a.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.

24

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-60.0 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.

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TAKE HOME ACTIVITY

F1

F2

P1

300 gal 75 wt% Eth 25 wt% H2O

40 wt% Eth 60 wt% H2O

60 wt% Eth 40 wt% H2O

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

26

TAKE HOME ACTIVITY

300 gal

1 m3

264.17 gal

Given, SG for 75 wt% Eth=0.877, density = 877 kg/m3

=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