In this example problem, we're going to show how
to use a ternary phase diagram.
And we're gonna do that by looking at an extraction process,
so the idea's we have an extraction unit.
There's a feed coming in.
We'll list the properties,
and it has a solvent flow rate coming in.
And leaving is raffinate, which means what's left
of the feed after we extracted the solute
and also leaving is extract.
So, this is the solvent plus the solute that we've tried
to transfer from the feed.
And I've drawn the R and E flow streams as purple
because they're in equilibrium.
This is important that the two streams leaving the system
we are assumed are in equilibrium.
So, now on the diagram, we are gonna write
down the values that we know,
and then we'll start using the phase diagram.
So, you can see on the phase diagram,
we've already located the feed here, 35% of solute,
so 30 here, so you can see 35%.
And it has no solvent, so it's right
on the side of the triangle.
Since this is a pure solvent feed,
then the solvent is here.
So, the next step is to locate the overall composition.
If we just mix the solvent and the feed together,
let's determine that composition.
Well, so what I'm going to do is write down the contents
of the mixture, and then calculate the mass fractions.
So, I'm going to write down the number of kilograms
of each of these three components.
If we're to mix everything together,
and then calculate the mass fractions for that mixture.
So, you can see here I took the flow rate of the feed stream
multiplied by the mass fraction, and it's 35% solute
to determine the number of kilograms of solute,
likewise, the number of kilograms of carrier.
Add these three streams together,
and it says the total is 349.4 kilograms per hour
that's going into the system.
So, these are all kilograms per hour, also.
So, I calculate the mass fractions,
and then I've located this mixing point,
which turns out has to be on a line
connecting the feed and the solvent.
And you can see that 80, 85% solvent is here,
and then 9.3% of carrier, so slightly less than 10
along that line.
So, it has to be on a line between these two points.
So, once I have the overall composition,
I can say what two compositions are in equilibrium.
If we have a point located at the mixing point.
So, what we want to do is draw a tie line,
so let me show the tie line with the phase diagram.
So, in the phase diagram, I've indicated the tie line
that's connecting two equilibrium phases:
the extract, which is solvent-rich,
and the raffinate, which is carrier-rich,
and then I can read the values off the diagram.
So, we look at, let's look at the raffinate
and the mass fraction of solute,
so it looks like, if I look here to here's 10%,
so it's less than 10%.
I'm gonna estimate that 8% or fraction of .08.
And then the carrier mass fraction,
so it's, looks like almost right
on the line 85%
or a mass fraction of 0.85.
And that means the mass fraction of the solvent is 0.07,
since the fractions have to add up to 1.
Why I can do the same thing for the extract.
Trying to read off the value here.
As you can see, it's mostly solvent,
so these are reasonable estimates we're reading
off the graph.
And now we can do a mass balance
to determine what the flow rates
of the raffinate and the extract are.
And the mass balances are, of course,
that the total amount leaving must
be the total amount entering.
And so, R+E=349.4 that we calculated earlier
for the total amount.
And then I can do a balance on the carrier,
and the carrier and the raffinate 0.85
times the raffinate flow rate
plus the carrier in the extract 0.02
times the extract flow-rate
equals the total amount of carrier that we had.
And we calculated that earlier .65 times the feed
or 32.5 in kilograms per hour.
And then the solvent,
so again I'm doing 0.07 this value times the raffinate
plus 0.94 times the extract has to equal the amount
of solvent that entered the system, kilograms per hour.
So, now we need to do is
solve these equations simultaneously,
so what I'm going to do is make some substitutions.
So, we can solve these mass balances independently for R,
so we're gonna get two solutions,
and they're gonna differ slightly just
because we made estimates.
So, let me rewrite these equations.
We made estimates when we read the ternary phase diagram.
So, the first equation gives its value
of 30.7 kilograms per hour, the raffinate flow rate,
so we get a slightly different number,
and we're just gonna take the average of these two values.
So the average value then for R is 32.7 kilograms per hour,
and then E is 349.4 minus 32.7,
and so E is equal to 316.7 kilograms per hour.
And considering the significant figures that we used
in the calculations, particularly reading the values,
then we would say R is 33 kilograms per hour,
and E is 317 kilograms per hour.
These are the final flow rates
leaving the single stage extraction unit.