Maple Solutions to the Chemical Engineering Problem Set

Material Balances for Col2
Material Balances for Col3
F x = F x + F x
1 S, 1 2 S, 2 3 S, 3
F x =
1 T, 1
F x =
1 B, 1
F x =
2 X, 2
F x =
2 S, 2
F x =
2 T, 2
F x =
2 B, 2
F x =
3 X, 3
F x =
3 S, 3
F x =
3 T, 3
F x =
3 B, 3
F x + F x
2 T, 2 3 T, 3
F x + F x
2 B, 2 3 B, 3
F x + F x
4 X, 4 5 X, 5
F x + F x
4 S, 4 5 S, 5
F x + F x
4 T, 4 5 T, 5
F x + F x
4 B, 4 5 B, 5
F x + F x
6 X, 6 7 X, 7
F x + F x
6 S, 6 7 S, 7
F x + F x
6 T, 6 7 T, 7
F x + F x
6 B, 6 7 B, 7
In the above Maple construction we have created the component material balances for each component
(the inner loop) and for each process unit (the outer loop).
The total molar flows are given the symbol F and x refers to the mole fraction of some component.
The first index of the component mole fraction identifies the component in question, the second
associates that quantity with a particular process stream. This double loop will work with all simple
material balances regardless of complexity provided we have identified the components, the units and
the inputand output streams associated with each unit.
The mole fraction summation equations can be created as follows:
> SumEqn:='SumEqn': i:='i':
for j in Streams do
SumEqn[j]:=add(x[i,j],i=components)=1;
print(SumEqn[j]);
od:
x + x + x + x = 1
X, 1 S, 1 T, 1 B, 1
x + x + x + x = 1
X, 2 S, 2 T, 2 B, 2
x + x + x + x = 1
X, 3 S, 3 T, 3 B, 3
x + x + x + x = 1
X, 4 S, 4 T, 4 B, 4
x + x + x + x = 1
X, 5 S, 5 T, 5 B, 5
x + x + x + x = 1
X, 6 S, 6 T, 6 B, 6
x + x + x + x = 1
X, 7 S, 7 T, 7 B, 7
The component material balances and the mole fraction summation equations comprise the complete
Page 10