EXAMPLE $4\mathrm{.}10-$ Removal of $CuS{O}_4$ from a Pollution Control Residue.

A pollution$-$control residue containing copper oxide was leached with sulfuric acid to produce a slurry having a pulp density of $2\mathrm{.}20.$ The $wCuS{O}_4$ in the solution phase of the slurry is $0\mathrm{.}160.$ The slurry is washed counter$-$currently in three stages with water. The slurry moving between stages and discharge has the same volume fraction liquid as the entering slurry. The volumetric flow of liquid moving counter$-$currently to the slurry is the same as the volume of wash water. How much wash water should be used such that the discharge slurry contains $1\mathrm{.}0\%$ of the incoming amount of copper? The slurry solids have a density of $5,0.$

Data. The approximate specific gravity of $CuS{O}_4$ solutions is:

$CuS{O}_4$ solution $=1+w_{CuS{O}_4}$

Solution. Heretofore, we used a mass basis for material entering and traversing the system, but the sometimes the flow between mixer$-$settler devices is specified on a volume basis. This occurs when a pump is set for a certain volumetric flow. The assumption of conservation of volume is

Chapter $4$ Fundamentals of Material Balances in Non$-$Reacting Systems

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good when the solution is dilute, but in this example, the incoming solution is not dilute, having a density of $1\mathrm{.}16.$ Even though a constant$-$volume flow is specified between devices, volume is not necessarily conserved, so at some point, at least one stream volume must be left unspecified. In other words, it is not possible to specify arbitrarily the volume of every stream of a process.

For streams where the volume is specified, the mass balance equations require a volume$-$tomass conversion. Please review Example $1\mathrm{.}12$ for conversion arithmetic between density, volume, and mass of a slurry. Since no basis was specified, a volume basis of $1000L$ input slurry is chosen, which has a mass of $2200kg.$

The first step is to calculate the mass of each phase in the feed and discharge slurry. For the feed slurry:

Mass balance: $,$ mass of liquid $+$ mass of solid $=2200kg$

Volume balance: volume of liquid $+$ volume of solid $=1000L.$

Recognizing that the volume is equal to mass$/$density$,$ the above two equations can be solved for the mass and volume of each phase of the feed slurry.

$\backslash$table$[[$Feed slurry,mass, $kg,$volume, L$,$mass fraction$],[$solid$,1354,271,1\mathrm{.}0$ solid$],[$liquid$,846,729,0\mathrm{.}16$ CuSO$4$

Here $is$ the question and here $is$ the answer $.$ Solve using Mass balances

Table $4\mathrm{.}14$ Mass balance for $CuS{O}_4$ washing circuit $(kg)$

$\backslash$table$[[,\frac{M}{S}$ I,$\frac{M}{S}$ II$,\frac{M}{S}$ III$],[$mass fraction of $CuS{O}_4,0\mathrm{.}04215,0\mathrm{.}00990,0\mathrm{.}00192],[$mass of wash liquid out,$3179,3093,3069],[$mass of $CuS{O}_4$ in wash liquid,$314,30\mathrm{.}6,5\mathrm{.}9],[$mass of slurry liquid out,$760,736,730],[$mass of $CuS{O}_4$ in slurry liquid,$32\mathrm{.}0,7\mathrm{.}3,1\mathrm{.}4]]$