Introduction to Chemical Engineering Processes/Problem Solving with Multiple Components

General Strategies for Multiple-Component Operations

The most important thing to remember about doing mass balances with multiple components is that for each component, you can write one independent mass balance. What do I mean by independent? Well, remember we can write the general, overall mass balance for any steady-state system:

$\mathrm{\Sigma }{\dot{m}}_{in}-\mathrm{\Sigma }{\dot{m}}_{out}=0$

And we can write a similar mass balance for any component of a stream:

$\mathrm{\Sigma }{\dot{m}}_{a,in}-\mathrm{\Sigma }{\dot{m}}_{a,out}+m_{a,gen}=0$

This looks like we have three equations here, but in reality only two of them are independent because:

1. The sum of the masses of the components equals the total mass
2. The total mass generation due to reaction is always zero (by the law of mass conservation)

Therefore, if we add up all of the mass balances for the components we obtain the overall mass balance. Therefore, we can choose any set of n equations we want, where n is the number of components, but if we choose the overall mass balance as one of them we cannot use the mass balance on one of the components.

The choice of which balances to use depends on two particular criteria:

1. Which component(s) you have the most information on; if you don't have enough information you won't be able to solve the equations you write.
2. Which component(s) you can make the most reasonable assumptions about. For example, if you have a process involving oxygen and water at low temperatures and pressures, you may say that there is no oxygen dissolved in a liquid flow stream, so it all leaves by another path. This will simplify the algebra a good deal if you write the mass balance on that component.

Multiple Components in a Single Operation: Separation of Ethanol and Water

Following the step-by-step method makes things easier.

Step 1: Draw a Flowchart

The first step as always is to draw the flowchart, as described previously. If you do that for this system, you may end up with something like this, where x signifies mass fraction, [A] signifies molarity of A, and numbers signify stream numbers.

Step 2: Convert Units

Now, we need to turn to converting the concentrations into appropriate units. Since the total flowrates are given in terms of mass, a unit that expresses the concentration in terms of mass of the components would be most useful. The vapor stream compositions are given as mass percents, which works well with the units of flow. However, the liquid phase concentration given in terms of a molarity is not useful for finding a mass flow rate of ethanol (or of water). Hence we must convert the concentration to something more useful.

In order to convert the molarity into a mass fraction, then, we need the molecular weight of ethanol and the density of a 4M ethanol solution. The former is easy if you know the chemical formula of ethanol: $C{H}_{3}C{H}_{2}OH$. Calculating the molecular weight (as you did in chem class) you should come up with about $46\frac{g}{mol}$.

Calculating the density involves plugging in mass fractions in and of itself, so you'll end up with an implicit equation. Recall that one method of estimating a solution density is to assume that the solution is ideal (which it probably is not in this case, but if no data are available or we just want an estimate, assumptions like these are all we have, as long as we realize the values will not be exact):

$\frac{1}{{\rho }_{SLN}}=\mathrm{\Sigma }(\frac{x_k}{{\rho }_k})$

In this case, then,

$\frac{1}{{\rho }_{SLN}}=\frac{x_{EtOH}}{{\rho }_{EtOH}}+\frac{x_{H2O}}{{\rho }_{H2O}}$

We can look up the densities of pure water and pure ethanol, they are as follows (from Wikipedia's articles w:Ethanol and w:Water):

${\rho }_{EtOH}=0.789\frac{g}{c{m}^3}=789\frac{g}{L}$

${\rho }_{H2O}=1.00\frac{g}{c{m}^3}=1000\frac{g}{L}$

Therefore, since the mass fractions add to one, our equation for density becomes:

$\frac{1}{{\rho }_{sln}}=\frac{x_{EtOH}}{789\frac{g}{L}}+\frac{1-x_{EtOH}}{1000\frac{g}{L}}$

From the NOTE above, we can now finally convert the molarity into a mass fraction as:

$x_{EtOH}=[EtOH]*\frac{(MW{)}_{EtOH}}{{\rho }_{SLN}}=4\frac{mol}{L}*46\frac{g}{mol}*(\frac{x_{EtOH}}{789\frac{g}{L}}+\frac{1-x_{EtOH}}{1000\frac{g}{L}})$

Solving this equation yields:

$x_{EtOH}=0.194$ (unitless)

Step 3: Relate your Variables

Since we are seeking properties related to mass flow rates, we will need to relate our variables with mass balances.

Remember that we can do a mass balance on any of the N independent species and one on the overall mass, but since the sum of the individual masses equals the overall only $N-1$ of these equations will be independent. It is often easiest mathematically to choose the overall mass balance and $N-1$ individual species balances, since you don't need to deal with concentrations for the overall measurements.

Since our concentrations are now in appropriate units, we can do any two mass balances we want. Lets choose the overall first:

${\dot{m}}_1-{\dot{m}}_2-{\dot{m}}_3=0$

Plugging in known values:

${\dot{m}}_2=100\frac{kg}{s}-20\frac{kg}{s}$

${\dot{m}}_2=80\frac{kg}{s}$

Now that we know ${\dot{m}}_2$ we can do a mass balance on either ethanol or water to find the composition of the input stream. Lets choose ethanol (A):

${\dot{m}}_{A1}={\dot{m}}_{A2}+{\dot{m}}_{A3}$

Written in terms of mass fractions this becomes:

$x_{A1}*{\dot{m}}_1=x_{A2}*{\dot{m}}_2+x_{A3}*{\dot{m}}_3$

Plugging in what we know:

$x_{A1}*100\frac{kg}{s}=0.8*80\frac{kg}{s}+0.194*20\frac{kg}{s}$

$x_{A1}=0.68$

Hence, the feed is 68% Ethanol and 32% Water.

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