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You will see the assignment in the file and it will show you how to work on it . If you have any questions, ask meA→ B+C
n A, feed = 100 moles/sec
n I , feed = 0 → 100 moles/sec
ሶy
 ሶ ==200
100
A, ,
out
B+C
Reactor
Separator
A+ I
The process depicted above is used to react Component A into Components B and C. In
the process the feed stream which can contain some inert component, I, is mixed with
a recycle stream before entering the reactor. In the reactor, the reaction extent is given
in the process diagram which is a function of the outlet mole fraction of component A.
After the reactor, a separator is used to separate components B and C from unreacted
A and I. Some of A and I are recycled back to the beginning of the process and the rest
is purged.
A→ B+C

ሶy
 ሶ ==200
100
A, ,
out
n A, feed = 100 moles/sec
B+C
n I , feed = 0 → 100 moles/sec

Reactor

Separator

A+ I

ሶ ,3 = ሶ ,1 + ሶ ,2
ሶ ,3 = ሶ ,1 + ሶ ,2
ሶ ,3 = ሶ ,1 + ሶ ,2
ሶ ,3 = ሶ ,1 + ሶ ,2

ሶ ,4 = ሶ ,3 − ሶ
ሶ ,4 = ሶ ,3 + ሶ
ሶ ,4 = ሶ ,3 + ሶ
ሶ ,4 = ሶ ,3
ሶ 4 = ሶ ,4 + ሶ ,4 + ሶ ,4 + ሶ ,4
ሶ ,4
,4 =
ሶ 4
A→ B+C

ሶy
 ሶ ==200
100
A, ,
out
n A, feed = 100 moles/sec
B+C
n I , feed = 0 → 100 moles/sec

Reactor

Separator

ሶ ,5 = ሶ ,4 ; = 0
ሶ ,5 = ሶ ,4 ; = 1
ሶ ,5 = ሶ ,4 ; = 1
ሶ ,5 = ሶ ,4 ; = 0

ሶ ,2 = ሶ ,6
ሶ ,2 = ሶ ,6
ሶ ,2 = ሶ ,6
ሶ ,2 = ሶ ,6
ሶ ,6 = ሶ ,4 − ሶ ,5
ሶ ,6 = ሶ ,4 − ሶ ,5
ሶ ,6 = ሶ ,4 − ሶ ,5
ሶ ,6 = ሶ ,4 − ሶ ,5
ሶ ,7 = ሶ ,6 − ሶ ,2
ሶ ,7 = ሶ ,6 − ሶ ,2
ሶ ,7 = ሶ ,6 − ሶ ,2
ሶ ,7 = ሶ ,6 − ሶ ,2

A+ I

The cost of component A is \$0.25/mole. There is no cost for the inert compound.
The value of component B is \$1.00/mole and that of component C is \$0.80/mole.
There is a cost of recycling the unreacted A and I, which is \$0.20/mole of A and I
combined. Vary the amount of inert in the feed stream and show how much A
and I should be recycled to maximize the profit of the plant.
Moles Inert in Feed, moles/s
Optimum Flow Rate for Maximum Profit, moles/s
Successive Substitution for
Solving Recycle Problems
Fresh Water
m 2
m 4
m 1
1% Pulp
1000 kg/min
2% Pulp (solid)
0.1 kg/L Sulfate
(liquid basis)
Mixer
m 3
12% Pulp
Filter
Assume liquid has s.g. of 1.0
m 1, pulp = 0.02m 1 = 20 kg/min
m 1, sulfate = 0.1(0.98m 1 ) = 98 kg/min
m 1, water = 0.9(0.98m 1 ) = 882 kg/min
x1, sulfate =
98
= 0.10 liquid basis
980
m 5
Filtrate
m 1 + m 2 = m 3
m 1, pulp = m 3, pulp = 20 kg/min
m 3, pulp
m 3
= 0.01  m 3 = 2000 kg/min  m 2 = 1000 kg/min
m 3, water = m 1, water + m 2 = 1882 kg/min
m 3, sulfate = m 1, sulfate = 98 kg/min
m 3 = m 4 + m 5
m 3, pulp = m 4, pulp = 20 kg/min
m 4, pulp
m 4
= 0.12  m 4 = 166.7 kg/min  m 5 = 1833.3 kg/min
m 3, sulfate
m 3, water
=
m 5, sulfate
m 5, water
=
98
= 0.0521
1882
m 5 = m 5, sulfate + m 5, water = 0.0521m 5, water + m 5, water
m 5, water = 1742.6 kg/min  m 5, sulfate = 90.7 kg/min  m 4, sulfate = 7.3 kg/min
x5, sulfate
90.7
=
= 0.0495
1833.3
m 4, sulfate
m 4, pulp
=
7.3
= 0.365
20
Fresh Water
m 2 = 800 kg/min
m 4
m 1
1% Pulp
1000 kg/min
2% Pulp
0.1 kg/L Sulfate
(liquid basis)
Mixer
m 6
m 7
m 3
12% Pulp
Filter
m 5
Filtrate
Fresh Water
 2 = 800 kg/min
m
4
m
1
m
1% Pulp
1000 kg/min
2% Pulp
0.1 kg/L Sulfate
Mixer
6
m
7
m
3
m
12% Pulp
Filter
5
m
Filtrate
• Guess the sulfate mass fraction in Stream 5
• Perform material balances and calculate new
mass fraction of sulfate in Stream 5
• Rinse and repeat as necessary
1000 kg/min
2% Pulp
0.1 kg/L Sulfate
Fresh Water
1% Pulp
Mixer
Filter
12% Pulp
1% Pulp
Mixer
Filter
12% Pulp
Filtrate
Determine the sulfate to pulp ratio in exit
stream (leaving the filter)

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