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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.
18
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
21
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
22
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.
25
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