$(1$ point$)$ The slurry from this scrubber is sent to a coagulation system to separate out the solid and liquid waste for disposal. The slurry composition has the same flow rate of water with a solids concentration defined by the previous step. The coagulation and flocculation process removes $80\%$ of the suspended solids $($We define control efficiency the same for air and water$),$ and uses $1m\frac{g}{L}$ alum. What is the dry mass of sludge produced? Include PM removed from the gaseous exhaust and subsequently removed through coagulation as well as the solids generated in the coagulation process.

Coagulation: $A{l}_2(S{O}_4{)}_3*14H_{2}O+6HC{O}_3^{-}$longrightarrow$2Al(OH{)}_3(s)+6C{O}_2+14H_{2}O+3S{O}_4^{2-}$

$Q_{\text{s}\text{l}\text{u}\text{r}\text{r}\text{y}}$ is conserved

influent to coagulation is slurry effluent above

$$ is same as above

but concentrations in water not air

Solids influent to coagulation $($see slurry$)$

s$/$d

Solids in effluent water $($think $)$

$\frac{g}{d}$

Solids removed $($mass conserved$)$

g$/$d

Solids created $(Al(OH{)}_3)$

$k\frac{g}{d}$

solids removed $($sum$)$

sg$/$d

$(1$ point$)$ You also need to dispose of these solids. Assuming the solids have a density $500k\frac{g}{m^3}($and mass rate from above$)$ and that you will dispose of it in a landfill with $E=1\mathrm{.}2,$ how much landfill space would you need to budget for during a $15$ year operation?

Hint: Since this isn't related to population, treat $P*C$ as equal to the rate of solids calculated above.

$V_{LF}=\frac{P*E*C}{D_c}=\frac{\text{d}\text{a}\text{i}\text{l}\text{y}\text{m}\text{a}\text{s}\text{s}*E}{D_c}.$

Daily Volume

Total volume:

$\frac{m^3}{?}$ day

$\frac{m^3}{15}yrs$