Question 2.4: Selection of system boundaries and organisation of the solut...

Only those flows necessary to illustrate the choice of system boundaries and method of
calculation are given in the Solution.

Basis: 1 hour
Take the first system boundary round the filter and dryer.

With 1 per cent loss, polymer entering sub-system

=\frac{10,000}{0.99}=\underline{\underline{10,101} } kg

Take the next boundary round the reactor system; the feeds to the reactor can then be calculated.
At 90 per cent conversion, monomer feed

=\frac{10,101}{0.9}=\underline{\underline{11,223} } kg

Unreacted monomer = 11,223 – 10,101 = \underline{\underline{ 1122} } kg

Short-stop, at 0.5 kg/1000 kg unreacted monomer

= 1122 \times 0.5 \times 10^{-3} =\underline{\underline{ 0.6 } } kg

Catalyst, at 1 kg/1000 kg monomer

= 11,223 \times1 \times 10^{-3} =\underline{\underline{ 11 } } kg

Let water feed to reactor be F1, then for 20 per cent monomer

0.2=\frac{11,223}{F_{1}+ 11,223}
F_{1}=\frac{11,223\left(1-0.2\right) }{0.2} =\underline{\underline{44,892} } kg
Now consider filter-dryer sub-system again.
Water in polymer to dryer, at 5 per cent (neglecting polymer loss)

=10,101\times 0.05 =\underline{\underline{505} } kg

Balance over reactor-filter-dryer sub-system gives flows to recovery column.

water, 44,892 + 10,101 – 505 = \underline{\underline{54,448} } kg

monomer, unreacted monomer, = \underline{\underline{1122} } kg

Now consider recovery system

With 98 per cent recovery, recycle to reactor

= 0.98\times 1122 =\underline{\underline{1100} } kg

Composition of effluent 23 kg monomer, 54,488 kg water.
Consider reactor monomer feed

Balance round tee gives fresh monomer required

= 11,223-1100=\underline{\underline{10,123} } kg