Question $2$: Design an activated sludge process for a biological wastewater treatment

that can meet an effluent standard of $10\mathrm{.}0m\frac{g}{L}BO{D}_5.$

Assume the following data are provided:

Flow $=Q=0\mathrm{.}2\frac{m^3}{s}$

$BO{D}_5=S_0=90\mathrm{.}0m\frac{g}{L}$

$K{s}_s=100m\frac{g}{L}$BOD

$m=2\mathrm{.}5d^{-1}$

$K_d=0\mathrm{.}060d^{-1}$

$Y=0\mathrm{.}80mgVS\frac{S}{m}gBO{D}_5$ removed

$MLVSS=x=3000m\frac{g}{L}$

Calculate F$/$M ratio?

Calculate the mass of activated sludge production each day from the plant?

$S=\frac{K_s(1+k_d{}_c)}{{}_c({}_m-k_d)-1}$

$x=\frac{{}_{c}Y(S_o-S)}{(1+k_d{}_c)}$

$=\frac{V}{Q}$

$\frac{F}{M}=\frac{Q{S}_o}{Vx}$

$Y_{\text{o}\text{b}\text{s}}=\frac{Y}{1+k_d{}_c}$

$\{$:$P_x=Y_{\text{o}\text{b}\text{s}}Q(S_0-S){10}^{-3}\frac{kg}{g})$