A city discharges 100 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 ?
Answers
Answer:
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
10
90
∗650=5850kg
Density of volatile solids= 1000*1.05*(p_w*s_w)=1050kg/m^31000∗1.05∗(p
w
∗s
w
)=1050kg/m
3
Similarly density of non-volatile solids =1000*2.5=2500kg/m^31000∗2.5=2500kg/m
3
Volume of volatile solids in wet sludge \frac{350}{1050}=0.333 m^3
1050
350
=0.333m
3
Vol. of non-volatile solids in wet sludg \frac{300}{2500}=0.12 m^3
2500
300
=0.12m
3
Vol. of water in wet sludge \frac{5850}{1000}=5.85 m^3
1000
5850
=5.85m
3
Hence, total volume of wet sludge 0.333+0.12+5.85 =6.303 m^30.333+0.12+5.85=6.303m
3