Design a rectangular sedimentation tank using the following data 1) population 1 lakh (2) Overflow rate Of 25m/day (3)Depth of the tank=3.2m

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written 3.0 years ago by

sagarkolekar ★ 11k

Solution: Data : Population : 1 $0^5$ , Overflow rate=25m/day

Tank depth=3.2 m.

Assume the rate of water supply to be 200 lit/hr/day

Daily Demand=rate of water supplypopulation=2001 $0^5$ =2*1 $0^7$ lit/day= $\frac{2*10^7}{24}$ =833.33 $m^3$ /hr
Overflow rate= $\frac{Q}{A}$ =25m/day= $\frac{25}{24}$ m/hr

here, $\frac{Q}{A}$ = $\frac{25*100}{24*60*60}$ = $V_s$ = $\frac{25}{24*36}$ =0.028cm/sec

3.Detention time= $\frac{Depth of tank}{overflow rate}$ = $\frac{3.2}{25/24}$ =3.072 hrs

4.Capacity of the tank = Hourly DischargeD.T=833.333.072=2560 $m^3$

5.Surface area …

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