Question

Q2-A binary solution containing 57.5 % of MVC by weight is to be separated by flash distillation at a pressure of 1 atm. Two equilibrium stages (chamber) are connected in series and the bottom product of the first chamber is sent to the second one. If one-third of the feed streams are evaporated at each chamber, For two chambers find the followings Flash Chamber-1 Flash Chamber-II Temperature of the chamber (C) Composition of the bottom product (X) Composition of the top product lýp) Percentage of Recovery (P.R.) Overall recovery Table Q2. Vapor-Liquid Equilibrium Data at 1 atm Mass fraction of MVC in liquid Boiling temperature T(C) x Mass fraction of MVC in vapor y 0.00 100 0.00 88.2 0.05 0.623 80.0 0.10 0.784 71.9 0.20 0.802 67.6 0.30 0.910 65.2 0.40 0.925 63.4 0.50 0.933 62.0 0.60 0.940 60.8 0.70 0.946 59.7 0.80 0.953

Answer 1

We are given a binary solution having 57.5% MVC which is separated using flash distillation at 1 atm pressure in two chambers.

The bottom of the first chamber is fed to the second chamber.

One-third of feed is evaporated in each chamber.

We are given the vapour-liquid equillibrium data at 1 atm. From this we can form a relation between y (vapour mole fraction) and x (liquid mole fraction).

Let $F_{1}$ be the feed to first chamber $z_{F_{1}}$ the mole fraction of MVC in the feed to first chamber. It is given as, $z_{F_{1}}=0.575$ .

Let the top/distillate from the first chamber be represented by
Di
and the bottom/residue be represented by $W_{1}$ . $y_{D_{1}}$ and $x_{W_{1}}$ represents the respective mole fraction of MVC.

The overall mole balance on the chamber can be written as,

F1 =W1+ D1 ---(1)

The MVC balance on the chamber gives,

$F_{1}z_{F_{1}}=W_{1}x_{W_{1}}+D_{1}y_{D_{1}}---(2)$

The fraction of feed can be written as, $f=\frac{D}{F}$

The equations can be rearranged to arrive the equation,

$(W_{1}+D_{1})z_{F_{1}}=W_{1}x_{W_{1}}+D_{1}y_{D_{1}}$

$\frac{y_{D}-z_{F}}{x_{W}-z_{F}}=\frac{-W}{D}=\frac{f-1}{f}---(3)$

$\frac{D_{1}}{W_{1}}=\frac{f}{1-f}---(4)$

So here, we are given $f=\frac{1}{3}$

So we get,

$\frac{y_{D}-0.575}{x_{W}-0.575}=\frac{-W}{D}=\frac{1/3-1}{1/3}$

$\Rightarrow \frac{y_{D}-0.575}{x_{W}-0.575}=-2$

$\Rightarrow y_{D}-0.575=-2x_{W}+1.15$

$\Rightarrow y_{D}=-2x_{W}+1.725$

So this line is the operating line which can be drawn by taking a Feed point (0.575, 0.575) on equilibrium and drawing a line with slope -2 as shown, This line intersecting with equillibrium diagram gives the equillibrium concentration and thus from diagram we can find the temperature of the chamber.

T X у 100 0 0 88.2 0.05 0.623 80 0.1 0.784 71.9 0.2 0.802 67.6 0.3 0.91 65.2 0.4 0.925 63.4 0.5 0.933 62 0.6 0.94 60.8 0.7 0.946 59.7 0.8 0.953

We can see from the graph that the liquid composition in the first chamber is 39.5% MVC and the vapour composition as 92% MVC.

1 0.9 operating line 0.8 0.7 y 0.6 0.5 0.4 0.3 0.2. 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 X 1.7 0.8 0.9 1 1.1

110 100 T 90 80 70 60 50 0 0.1 0.2 0.3 0.4 0.5 X, Y 6 0.7 0.8 0.9 1
So we can see that the temperature of the first flash chamber is 65⁰C from the graph.

Now the recovery is defined as,

$R=\frac{MCV\; in \; vapour}{MCV\; in \; feed}=\frac{D_{1}y_{D_{1}}}{F_{1}z_{F_{1}}}=f\frac{y_{D_{1}}}{z_{F_{1}}}=(1/3)\frac{0.92}{0.575}=0.5333$

So the recovery in the first chamber is 53.33%

Now in the second Chamber.

The bottom from the first chamber is fed as feed to the second chamber, so $F_{2}=W_{1}$

$z_{F_{2}}=x_{W_{1}}=0.395$

Only the feed point of diagram changes. The
f
remains the same that is $f=\frac{D_{2}}{F_{2}}=\frac{1}{3}$ .

The new feed point is the liquid composition of the bottom of the first chamber. That is (0.395, 0.395).

1 0.9 0.8 0.7 у 0.6 operating line 0.5 0.4 0.3 0.2 0.1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 X 0.7 0.8 0.9 1 1.1

We can see from the graph that the liquid composition in the second chamber is 19% MVC and the vapour composition as 79.7% MVC.

110 100 T 90 80 70 60 50 O 0.1 0.2 0.3 0.4 0.5 X, Y .6 0.7 0.8 0.9 1
So we can see that the temperature of the second flash chamber is 72.5⁰C from the graph.

Now the recovery is defined as,

$R=\frac{MCV\; in \; vapour}{MCV\; in \; feed}=\frac{D_{2}y_{D_{2}}}{F_{2}z_{F_{2}}}=f\frac{y_{D_{2}}}{z_{F_{2}}}=(1/3)\frac{0.797}{0.395}=0.6726$

So the recovery in the first chamber is 67.26%

Now overall recovery in the both chambers can be calculated as,

$R=\frac{MCV\; in \; vapour\; from\: 1\; and\; 2}{MCV\; in \; feed}=\frac{D_{1}y_{D_{1}}+D_{2}y_{2}}{F_{1}z_{F_{1}}}=\frac{D_{1}y_{D_{1}}}{F_{1}z_{F1}}+\frac{D_{2}y_{D_{2}}}{F_{1}z_{F1}}$

From (1) and (4)

$\Rightarrow \frac{D_{2}}{F_{1}}=\frac{F_{2}*f}{W_{1}+D_{1}}=\frac{F_{2}*f}{W_{1}(1+\frac{D_{1}}{W_{1}})}=\frac{F_{2}*f}{F_{2}(1+\frac{f}{1-f})}=\frac{f}{1/(1-f)}=f(1-f)=\frac{2}{9}$

$R=0.5333+\frac{2}{9}*\frac{0.797}{0.575}$