Introduction to Chemical Engineering Processes/Example: A simple system with recycle

First Step: Draw a Flowchart
A schematic is given in the problem statement but it is very incomplete, since it does not contain any of the design specifications (the efficiency of the settler, the solubility of soil in water, and the mass flow rates). Therefore, it is highly recommended that you draw your own picture even when one is provided for you. Make sure you label all of the streams, and the unknown concentrations.



Second Step: Degree of Freedom Analysis

 * Around the washer: 6 independent unknowns ($$x_{O1}, \dot{m}_2, x_{D2}, \dot{m}_3, x_{D3}, x_{O4}$$), three independent mass balances (ore, dirt, and water), and one solubility. The washer has 2 DOF.
 * Around the dryer: 2 independent unknowns ($$x_{O4}, \dot{m}_5$$) and two independent equations = 0 DOF.


 * Around the Settler:5 independent unknowns ($$\dot{m}_3, x_{D3}, \dot{m}_7, \dot{m}_8, x_{D8}$$), two mass balances (dirt and water), the solubility of saturated dirt, and one additional information (90% removal of dirt), leaving us with 1 DOF.
 * At the mixing point: We need to include this in order to calculate the total degrees of freedom for the process, since otherwise we're not counting m9 anywhere. 5 unknowns ($$\dot{m}_2, x_{D2}, \dot{m}_8, x_{D8}, \dot{m}_9$$) and 2 mass balances leaves us with 3 DOF.

Therefore, Overall = 3+2+1 - 6 intermediate variables (not including xO4 since that's going to the dryer) = 0

The problem is well-defined.

Devising a Plan
Recall that the idea is to look for a unit operation or some combination of them with 0 Degrees of Freedom, calculate those variables, and then recalculate the degrees of freedom until everything is accounted for.

From our initial analysis, the dryer had 0 DOF so we can calculate the two unknowns xO4 and m5. Now we can consider xO4 and m5 known and redo the degree of freedom analysis on the unit operations.


 * Around the washer: We only have 5 unknowns now ($$x_{O1}, \dot{m}_2, x_{D2}, \dot{m}_3, x_{D3} $$), but still only three equations and the solubility. 1 DOF.
 * Around the settler: Nothing has changed here since xO4 and m5 aren't connected to this operation.
 * Overall System: We have three unknowns ($$x_{O1}, \dot{m}_7, \dot{m}_9$$) since $$\dot{m}_5$$ is already determined, and we have three mass balances (ore, dirt, and water). Hence we have 0 DOF for the overall system.

Now we can say we know $$ x_{O1}, \dot{m}_7, $$ and $$ \dot{m}_9 $$.


 * Around the settler again: since we know m7 the settler now has 0 DOF and we can solve for $$ \dot{m}_3, x_{D3}, \dot{m}_8, $$ and $$ x_{D8} $$.


 * Around the washer again: Now we know m8 and xD8. How many balances can we write?

The washer therefore has 2 unknowns (m2, xD2) and 2 equations (the dirt and water balances) = 0 DOF

This final step can also be done by balances on the recombination point (as shown below). Once we have m2 and xD2 the system is completely determined.

Converting Units
The only given information in inconsistent units is the solubility, which is given as $$ 0.4 \frac{g \mbox{ dirt}}{cm^3 \mbox{ H}_2O } $$. However, since we know the density of water (or can look it up), we can convert this to $$ \frac{kg \mbox{ dirt}}{kg \mbox{ H}_2O } $$ as follows:

$$ 0.4 \frac{g \mbox{ dirt}}{cm^3 \mbox{ H}_2O} * 1 \frac{ cm^3 \mbox{ H}_2O}{g \mbox{ H}_2O} = 0.4 \frac{ g \mbox{ dirt}}{g \mbox{ H}_2O} = 0.4 \frac{ kg \mbox{ dirt}}{kg \mbox{ H}_2O} $$

Now that this information is in the same units as the mass flow rates we can proceed to the next step.

Carrying Out the Plan
First, do any two mass balances on the dryer. I choose total and ore balances. Remember that the third balance is not independent of the first two!


 * Overall Balance: $$ \dot{m}_4 = \dot{m}_5 + \dot{m}_6 $$
 * Ore Balance: $$ \dot{m}_4*x_{O4} = \dot{m}_5*x_{O5} + \dot{m}_6*x_{O6} $$

Substituting the known values:


 * Overall: $$ 3100 = \dot{m}_5 + 2900 $$
 * Ore: $$ x_{O4}*3100 = 1*2900 $$

Solving gives:

Now that we have finished the dryer we do the next step in our plan, which was the overall system balance:


 * Water Balance: $$ \dot{m}_9 = \dot{m}_5 $$
 * Ore Balance: $$ x_{O1}*\dot{m}_1 = \dot{m}_6 $$
 * Dirt Balance: $$ (1-x_{O1})*\dot{m}_1 = \dot{m}_7 $$

Next we move to the settler as planned, this one's a bit trickier since the solutions aren't immediately obvious but a system must be solved.


 * Overall Balance: $$ \dot{m}_3 = \dot{m}_7 + \dot{m}_8 $$
 * Dirt Balance: $$ \dot{m}_3*x_{D3} = \dot{m}_7*x_{D7} + \dot{m}_8*x_{D8} $$
 * Efficiency of Removal: $$ \dot{m}_7 = 0.9*\dot{m}_3*x_{D3} $$

Using the solubility is slightly tricky. You use it by noticing that the mass of dirt in stream 3 is proportional to the mass of water, and hence you can write that:


 * mass dirt in stream 3 = 0.4 * mass water in stream 3
 * Solubility: $$ \dot{m}_3*x_{D3} = 0.4*\dot{m}_3 * (1-x_{D3}) $$

Plugging in known values, the following system of equations is obtained:


 * $$ \dot{m}_3 = 100 + \dot{m}_8 $$
 * $$ \dot{m}_3*x_{D3} = 100 + \dot{m}_8*x_{D8} $$
 * $$ \dot{m}_3*x_{D3} = 111.11 $$
 * $$ \dot{m}_3*x_{D3} = 0.4*\dot{m}_3*(1-x_{D3}) $$

Solving these equations for the 4 unknowns, the solutions are:

Finally, we can go to the mixing point, and say:


 * Overall: $$ \dot{m}_8 + \dot{m}_9 = \dot{m}_2 $$
 * Dirt: $$ \dot{m}_8*x_{D8} = \dot{m}_2*x_{D2} $$

From which the final unknowns are obtained:

Since the problem was asking for $$ \dot{m}_2 $$, we are now finished.

Check your work
These values should be checked by making a new flowchart with the numerical values, and ensuring that the balances on the washer are satisfied. This is left as an exercise for the reader.