Question 6.5.2: Material Balances on a Crystallizer An aqueous potassium nit......
Material Balances on a Crystallizer
An aqueous potassium nitrate solution containing 60.0 wt% KNO_{3} at 80°C is fed to a cooling crystallizer in which the temperature is reduced to 40°C. Determine the temperature at which the solution reaches saturation and the percentage of the potassium nitrate in the feed that forms crystals.
The feed concentration must be converted to a solute/solvent ratio to use Figure 6.5-1. Since 100.0 g of the solution contains 60.0 g KNO_{3} and 40.0 g H_{2}O, the desired ratio is
\frac{60.0 g KNO_{3}}{40.0 h H_{2}O} = 1.50 \frac{g KNO_{3}}{g H_{2}O} = \frac{150 g KNO_{3}}{100 g H_{2}O}
From Figure 6.5-1, the saturation temperature of this solution is \boxed{74°C} .
Here is a flowchart of the process for an assumed basis of 100 kg feed.
There are three unknowns on the chart (m_{1}, x, m_{2}). We will assume that the solution leaving the crystallizer is saturated at 40°C. Accordingly, the value of x may be determined from the known solubility of KNO_{3} at that temperature, and the remaining two variables may be determined from material balances. From Figure 6.5-1, the solubility at 40°C is roughly 63 kg KNO_{3}/100 kg H_{2}O. The calculations follow.
x= \frac{63 kg KNO_{3}}{(63 + 100)kg solution} = 0.386 kg KNO_{3}/kg
H_{2}O Balance: \begin{array}{c|c}100 kg &0.400 kg H_{2}O \\ \hline & kg\end{array} =\begin{array}{c|c}m_{1}(kg)&(1 – 0.386 ) kg H_{2}O \\ \hline & kg \end{array} \Longrightarrow m_{1} = 65.1 kg
Mass Balance: 100 kg = m_{1} + m_{2} \overset{m_{1} = 65.1 kg}{\left. \Large{\Longrightarrow } \right.} m_{2} = 34.9 kg KNO_{3}(s)
The percentage of the potassium nitrate in the feed that crystallizes is therefore
\frac{34.9 kg KNO_{3} crystallized}{60.0 kg KNO_{3} fed} \times 100\% = \boxed{58.2 \%}
