**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

$V=137.875m^3$