Question

A city discharges 100 (cum. Sec) m3/sec of sewage into a river, which is fully saturated with oxygen and flowing at the rate of 150 m3/sec during its lean days with a velocity of 0.1 m/sec. The 5th day BOD of sewage at the given temperature is 280 mg/L. Find when and where the critical DO deficit will occur in the downstream portion of the river and what is its amount. Assume f = 4.0 (Self-purification constant and KD = 0.1. *

Answer 1

Total solids produced = 1000 kg (dry mass)

Volatile solids = 70% total solids = 700kg

Non-volatile solids = 30% T.S. = 300kg

Volatile solids removed indigestion

Volatile solids left in digested sludge = 350kg

Non-volatile solids in digested sludge

Mass of water in wet digested sludge =90%

10% mass of solids

10kg of solids make

Therefore, 10kg of solids contain

Or 650 kg of solids contain $\frac{90}{10}*650=5850 kg$

Density of volatile solids= $1000*1.05*(p_w*s_w)=1050kg/m^3$

Similarly density of non-volatile solids =$1000*2.5=2500kg/m^3$

Volume of volatile solids in wet sludge $\frac{350}{1050}=0.333 m^3$

Vol. of non-volatile solids in wet sludg $\frac{300}{2500}=0.12 m^3$

Vol. of water in wet sludge $\frac{5850}{1000}=5.85 m^3$

Hence, total volume of wet sludge $0.333+0.12+5.85 =6.303 m^3$