Question

In an evaporator 25 ton / h of a solution of 10% NaOH, 10% NaCl, and 80% water by weight. During evaporation, the water evaporates and the salt precipitates like crystals They are allowed to settle and are removed. The outgoing concentrated solution of the Evaporator contains 50% NaOH, 2% NaCl and 48% water. Based on This information is requested:

1. Draw the process flow diagram, indicating each of its streams and compositions (known and unknown).

2. Calculate the kilograms of precipitated salt and the kilograms of solution concentrated for every hour of work.

Answer 1

1- Flow Diagram (file attached)

2- Kilograms of precipitated salt: 2400kg/h

Kilograms of solution concentrated per hour: 5000kg/h

Step-by-step explanation:

We have an Evaporator with 25ton/h of a solution with: 10% NaOH, 10% NaCl, and 80% water. That means that we have an input which is a stream with the following composition fractions:

Stream 1:

NaOH= 0.1

NaCl= 0.1

Water= 0.8

Then we have 3 outputs, 3 streams that leave the evaporator, which are:

Stream 2:

Only contains water so its composition fraction of water is 1.

Stream 3:

Only contains NaCl so its composition fraction of NaCL is 1.

Stream 4:

NaOH= 0.5

NaCl= 0.02

Water= 0.48

We know that stream 1 is 25 ton/h and enter the evaporator but we do not know the flow rate of stream 4 that is the concentrated solution leaving the evaporator, so we will make a particular mass balance of the component NaOH that is present in both streams:

fraction of NaOH in stream 1 ×flow rate of stream 1= fraction of NaOH in stream 4× flow rate of stream 4

0.1×25 ton/h = 0.5× flow rate of stream 4

flow rate of stream 4= (0.1×25 ton/h)/0.5= 5ton/h= 5000kg/h

Now to know the kilograms of precipitated salt, which is the flow rate of stream 3 we make a particular mass balance of the component NaCl:

(fraction of NaCl in stream 1 ×flow rate of stream 1)- (fraction of NaCl in stream 3×flow rate of stream 3)=flow rate of stream 4× fraction of NaCl in stream 4

(0.1×25 tn/h)- (1×flow rate of stream 3)= 5tn/h × 0.02

flow rate of stream 3= 2.4 tn/h =2400 kg/h