Given:-
population equivalent =7000
loading rate=0.09 kg/capita/day
volatile solids in raw sludge =70%
moisture content of raw sludge-=96%
digestion period=25 days
$P_2$= volatile solids resolution during digestion = 50%
$P_2$=moisture content of digested sludge=92%
storage period required for digested sludge=90 days
2)To find volume of sludge per day:-
weight of sludge per day=7000*0.09=630kg
For all practical purposes, the specific gravity of the sludge can be taken equal to that of matter i.e. 1.0
$Let V_1$=volume of sludge collected in primary settling tank(PST)
Moisture content of raw sludge=96%
which means that 4 kg of dry solids will make 199 kg of wet sludge
$V_1$=volume of sludge collected in primary settling tank(PST)
$=\frac{70}{100}*\frac{630}{1}*\frac{100}{4}$
=11025kg
∴wet sludge produced per day=11025kg
As the specific gravity of the sludge is assumed to be unity i.e. 1.0
Volume of raw sludge produced per day $(V_1)=\frac{11025}{1000}m^3$
$V_1=11.025m^3$
2)Tofind volume of digested sludge:-
Let volume of digested sludge
$P_1$=moisture content of raw sludge=96%
$P_2$=moisture content of digested sludge=92%
Volume of digested sludge$=\frac{100-P_1}{100-P_2}*V_1$
Volume of digested sludge$=\frac{100-96}{100-92}*110.25$
$=frac{4}{8}*110.25=5.51m^3/kg$
volatile solid reduction during digestion=50%
∴Volume of digested sludge$=5.51*\frac{50}{100}=2.755m^3$
∴$V_2=Volume of digested sludge=2.755m^3$
3)To find volume of digester tank(V):-
V=volume of digester tank for the digestion period of 25 days
$=[V_1-\frac{2}{3}(V_1-V_2)$]*digestion period
$=[11.025-\frac{2}{3}11.025-2.755)$]*25
$=[11.025-\frac{2}{3}8.27)$]*25
=[11.025-5.51]*25
=5.515*25
$V=137.875m^3$