CHE 201 Tutorial 6
Q1
A sodium chloride solution at a rate of 20000 kg/h (9% NaCl) is mixed with a recycle
stream which is 20% NaCl. The mixed stream is then fed to evaporator where it is
concentrated to 25% NaCl. This concentrated solution is then fed to a crystallizer where
sodium chloride crystals product is separated which is 90 % NaCl and 10% H2O. The rest
constitutes the recycle stream. Calculate
a. The crystal production rate in kg/h.
b. Water removed in the evaporator per hour.
c. Recycle stream rate in kg/h.
d. Composition of mixed stream feed entering
Q2
A liquid mixture containing 30 mol% benzene (B), 25% toluene (T) and the balance
xylene (X) is fed to a distillation column. The bottoms product contains 98 mole% X and
no B, and 96% of the X in the feed is recovered in this stream. The overhead product is
fed to a second column. The overhead product from the second column contains 97% of
the B in the feed to this column. The composition of this stream is 94 mole % B and the
balance T.
a. Calculate the molar flow rates and the mole fractions in each product stream from
both columns.
b. Calculate the percentage of B in the process feed (i.e. feed to first column) that
emerges in the overhead product from the second stream.
nt4
y4B = 0.94
y4T = 0.06
nt2
n1t = 100 mol/h y2B
y2T
y1B = 0.3 y2X
y1T = 0.25 n t5
y1X = 0.45
nt3 y5 B
y5 T
y3X = 0.98 y5 X
y3T = 0.02
CHE 201 Tutorial # 6
A liquid mixture containing 30 mol% benzene (B), 25% toluene (T) and the balance
xylene (X) is fed to a distillation column. The bottoms product contains 98 mole% X and
no B, and 96% of the X in the feed is recovered in this stream. The overhead product is
fed to a second column. The overhead product from the second column contains 97% of
the B in the feed to this column. The composition of this stream is 94 mole % B and the
balance T.
c. Calculate the molar flow rates and the mole fractions in each product stream from
both columns.
d. Calculate the percentage of B in the process feed (i.e. feed to first column) that
emerges in the overhead product from the second stream.
Solution:
nt4
y4B = 0.94
y4T = 0.06
nt2 II
n1t = 100 mol/h y2B
I y2T
B
y1 = 0.3 y2X
y1T = 0.25 n t5
y1X = 0.45
nt3 y5 B
y5 T
y3X = 0.98 y5 X
y3T = 0.02
Relations
1- 96% of X in stream 1 → X in stream 3
i.e. 0.96 0.45 n1 0.98 nt3
t
2- 97% of b in stream 2 → B in stream 4
i.e. 0.97 y B nt2 0.94 nt4
2
Degree of freedom table
Unit I II Overall
Unknown 8 8 10
Flow 0 Basis 0 0
Comp. 3 1 3
M.B. 3 3 3
Relations 1 1 1
D.O.F 1 Zero 3 3
Part (a)
M.B. Column I
Total: n1 100 mol / h n t2 n t3
t
(1)
B: 0.3(100) y n 0
B
2
t
2 (2)
X: 0.45(100) y n 0.98 n
X
2
t
2
t
3 (3)
Relation 1 0.96(0.45)(100) 0.98 n t3
n t3 44.08 mol / h
From(1) n t2 55.92 mol / h
0.3(100)
From(2) y B
2 0.536
55.92
0.45(100) 0.98(44.08)
From(3) y 2
X
0.032
55.92
y T 1 0.032 0.536 0.432
2
M.B. Column II
Total: n t2 55.92 mol / h n t4 n t5 (1)
B: y n 0.536(55.92) 0.94 n y n
B
2
t
2
t
4
B
5
t
5 (2)
T: 0.432(55.92) 0.06 n y n t
4
T
5
t
5 (3)
Relation 2 0.97(0.536)(55.92) 0.94 n t
4
n t4 30.93 mol / h
From(1) n t5 55.92 30.93 24.99mol / h
0.536(55.92) 0.94(30.93)
From(2) y B
5 0.036
24.99
0.432(55.92) 0.06(30.93)
From(3) y T
5 0.892
24.99
y 5 1 0.892 0.036 0.072
X
Part (b)
The percentage of B in the process feed (i.e. feed to first column) that emerges in the
overhead product from the second stream:
B in stream 4 0.94 (30 .93)
100 100 97 %
B in stream 1 0.3(100 )