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

## Systems with Recycle: a Cleaning Process

### Problem Statement

Example:

Consider a process in which freshly-mined ore is to be cleaned so that later processing units do not get contaminated with dirt. 3000 kg/hr of dirty ore is dumped into a large washer, in which water is allowed to soak the ore on its way to a drain on the bottom of the unit. The amount of dirt remaining on the ore after this process is negligible, but water remains absorbed on the ore surface such that the net mass flow rate of the cleaned ore is 3100 kg/hr.

The dirty water is cleaned in a settler, which is able to remove 90% of the dirt in the stream without removing a significant amount of water. The cleaned stream then is combined with a fresh water stream before re-entering the washer.

The wet, clean ore enters a dryer, in which all of the water is removed. Dry ore is removed from the dryer at 2900 kg/hr.

The design schematic for this process was as follows:

a) Calculate the necessary mass flow rate of fresh water to achieve this removal at steady state.

b) Suppose that the solubility of dirt in water is ${\displaystyle 0.4{\frac {g{\mbox{ dirt}}}{cm^{3}{\mbox{ H}}_{2}O}}}$. Assuming that the water leaving the washer is saturated with dirt, calculate the mass fraction of dirt in the stream that enters the washer (after it has been mixed with the fresh-water stream).

### 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 (${\displaystyle x_{O1},{\dot {m}}_{2},x_{D2},{\dot {m}}_{3},x_{D3},x_{O4}}$), three independent mass balances (ore, dirt, and water), and one solubility. The washer has 2 DOF.
• Around the dryer: 2 independent unknowns (${\displaystyle x_{O4},{\dot {m}}_{5}}$) and two independent equations = 0 DOF.
 Note: Since the dryer has no degrees of freedom already, we can say that the system variables behave as if the stream going into the dryer was not going anywhere, and therefore this stream should not be included in the "in-between variables" calculation.
• Around the Settler:5 independent unknowns (${\displaystyle {\dot {m}}_{3},x_{D3},{\dot {m}}_{7},{\dot {m}}_{8},x_{D8}}$), 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 (${\displaystyle {\dot {m}}_{2},x_{D2},{\dot {m}}_{8},x_{D8},{\dot {m}}_{9}}$) 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 (${\displaystyle x_{O1},{\dot {m}}_{2},x_{D2},{\dot {m}}_{3},x_{D3}}$), 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 (${\displaystyle x_{O1},{\dot {m}}_{7},{\dot {m}}_{9}}$) since ${\displaystyle {\dot {m}}_{5}}$ 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 ${\displaystyle x_{O1},{\dot {m}}_{7},}$ and ${\displaystyle {\dot {m}}_{9}}$.

• Around the settler again: since we know m7 the settler now has 0 DOF and we can solve for ${\displaystyle {\dot {m}}_{3},x_{D3},{\dot {m}}_{8},}$ and ${\displaystyle x_{D8}}$.
• Around the washer again: Now we know m8 and xD8. How many balances can we write?
 Note: If we try to write a balance on the ore, we will find that the ore is already balanced because of the other balances we've done. If you try to write an ore balance, you'll see you already know the values of all the unknowns in the equations. Hence we can't count that balance as an equation we can use (I'll show you this when we work out the actual calculation).

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 ${\displaystyle 0.4{\frac {g{\mbox{ dirt}}}{cm^{3}{\mbox{ H}}_{2}O}}}$. However, since we know the density of water (or can look it up), we can convert this to ${\displaystyle {\frac {kg{\mbox{ dirt}}}{kg{\mbox{ H}}_{2}O}}}$ as follows:

${\displaystyle 0.4{\frac {g{\mbox{ dirt}}}{cm^{3}{\mbox{ H}}_{2}O}}*1{\frac {cm^{3}{\mbox{ H}}_{2}O}{g{\mbox{ H}}_{2}O}}=0.4{\frac {g{\mbox{ dirt}}}{g{\mbox{ H}}_{2}O}}=0.4{\frac {kg{\mbox{ dirt}}}{kg{\mbox{ H}}_{2}O}}}$

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: ${\displaystyle {\dot {m}}_{4}={\dot {m}}_{5}+{\dot {m}}_{6}}$
• Ore Balance: ${\displaystyle {\dot {m}}_{4}*x_{O4}={\dot {m}}_{5}*x_{O5}+{\dot {m}}_{6}*x_{O6}}$

Substituting the known values:

• Overall: ${\displaystyle 3100={\dot {m}}_{5}+2900}$
• Ore: ${\displaystyle x_{O4}*3100=1*2900}$

Solving gives:

 ${\displaystyle {\dot {m}}_{5}=200{\frac {kg}{hr}}}$ ${\displaystyle x_{O4}=0.935{\frac {kg}{hr}}}$

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

• Water Balance: ${\displaystyle {\dot {m}}_{9}={\dot {m}}_{5}}$
• Ore Balance: ${\displaystyle x_{O1}*{\dot {m}}_{1}={\dot {m}}_{6}}$
• Dirt Balance: ${\displaystyle (1-x_{O1})*{\dot {m}}_{1}={\dot {m}}_{7}}$
 ${\displaystyle {\dot {m}}_{9}=200{\frac {kg}{hr}}}$, ${\displaystyle x_{O1}=0.967}$, ${\displaystyle {\dot {m}}_{7}=100{\frac {kg}{hr}}}$

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: ${\displaystyle {\dot {m}}_{3}={\dot {m}}_{7}+{\dot {m}}_{8}}$
• Dirt Balance: ${\displaystyle {\dot {m}}_{3}*x_{D3}={\dot {m}}_{7}*x_{D7}+{\dot {m}}_{8}*x_{D8}}$
• Efficiency of Removal: ${\displaystyle {\dot {m}}_{7}=0.9*{\dot {m}}_{3}*x_{D3}}$

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: ${\displaystyle {\dot {m}}_{3}*x_{D3}=0.4*{\dot {m}}_{3}*(1-x_{D3})}$

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

• ${\displaystyle {\dot {m}}_{3}=100+{\dot {m}}_{8}}$
• ${\displaystyle {\dot {m}}_{3}*x_{D3}=100+{\dot {m}}_{8}*x_{D8}}$
• ${\displaystyle {\dot {m}}_{3}*x_{D3}=111.11}$
• ${\displaystyle {\dot {m}}_{3}*x_{D3}=0.4*{\dot {m}}_{3}*(1-x_{D3})}$

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

 ${\displaystyle {\dot {m}}_{3}=388.89{\frac {kg}{hr}},{\dot {m}}_{8}=288.89{\frac {kg}{hr}},}$ ${\displaystyle x_{D3}=0.286,x_{D8}=0.0385}$

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

• Overall: ${\displaystyle {\dot {m}}_{8}+{\dot {m}}_{9}={\dot {m}}_{2}}$
• Dirt: ${\displaystyle {\dot {m}}_{8}*x_{D8}={\dot {m}}_{2}*x_{D2}}$

From which the final unknowns are obtained:

 ${\displaystyle {\dot {m}}_{2}=488.89{\frac {kg}{hr}}}$ ${\displaystyle x_{D2}=0.0229}$

Since the problem was asking for ${\displaystyle {\dot {m}}_{2}}$, we are now finished.