the reaction solution. It can be assumed that the reaction rate is first$-$order in both the

concentration of $A(c_A)$ and $B(c_B),$ and so second$-$order overall:

$-r_{A}k{c}_A{c}_B$

The batch process consists of the following FOUR steps:

Initially, the tank is empty.

Fill tank with solution $($ethanol$/$A$/$B$).$ Increase temperature of the feed from $20C$ to

$50C(1$ hour$)$

Reaction takes place isothermally at $50C(4$ hours, to give conversion of $50\%)$

Filtration of reaction solution to give product $P(2$ hours$)$

Drying with hot nitrogen gas followed by the removal of powder product $(3$ hours$)$

Information on stages

It can be assumed that the heating time for Stage $1(t_1)$ is proportional to the change in

temperature $($which is $30C$ in the case above$).$ This would assume a constant$-$power

heater. The heat capacity of the solution can be taken to be $500k\frac{J}{K}.$

For a batch reactor, assuming an overall second$-$order dependency, the reaction time

$(t_2)$ is related to the fractional conversion $(x)$ by:

$t_2\frac{1}{k(c_A{)}_0}(\frac{x}{1-x})$

where $k$ is the rate constant $($which increases with temperature$)$ and $(c_A{)}_0$ is the initial

concentration of component A$.$

At the completion of the reaction, the final concentration and amount of product are $(c_p{)}_{\text{e}\text{n}\text{d}}$

and $(n_p{)}_{\text{e}\text{n}\text{d}}$ respectively.

For batch filtration, the filtration time $(t_3)$ is defined as the time taken for the entire

solvent to pass through the filter case, which will increase with volume of filtrate. As the filter

cake increases in depth, so does the resistance to flow which will give a reduction in

volumetric flowrate. From Darcy's law and constant $P$ filtration, it can be shown that:

$t_3$prop$(n_P{)}_{\text{e}\text{n}\text{d}}{V}_{\text{e}\text{n}\text{d}}$

The final filtrate $(V_{\text{e}\text{n}\text{d}})$ is assumed to be $500L$ for the above case. As an example, if the

volume is doubled, keeping the concentration the same, the filtration time will be quadrupled.

Because an excess amount of hot gas is used $-$ the product should be bone dry for

quality control purposes $-$ the drying time $(t_4)$ can be considered to be independent of the

final cake thickness, and the large value is for product quality reasons.

For ethanol, the maximum solubility values are $0\mathrm{.}6mo\frac{l}{L}$ and $0\mathrm{.}25mo\frac{l}{L}$ for components

A and $B$ respectively at $50C.$ These are the maximum amounts that will dissolve in the

solvent. If more A and $B$ is added, this will remain in the solid phase.

Based on the information given previously, we would like to quantify the process and

this is obtained by using the cycle time $($CT$)$ to generate the equivalent continuous process:

$($i$)$ Calculate the cycle time $($CT$)$ and production rate for product $P($moles$)$ per $24$ hours.

$($ii$)$ Show how the process can be represented on a Gantt chart.

$($iii$)$ Will the fact that product $P$ is sparingly soluble affect the calculation of production rate?

Hint: your first step should be to quantify this type of solubility and estimate the amount of

product $P$ that will be left in solution.