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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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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

1. 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

2. 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 of the tank =$\frac{capacity}{depth}$=$\frac{2560}{3.2}$=800$m^2$

Providing two units, surface area of one unit=400$m^2$

1. Tank dimensions: Let length of tank=4*width ∴ L=4B

Surface of the tank, A=LB=4BB=4$B^2$

4$B^2$=400

$B^2$=100

∴B=10m

L=4*B=40m

Provide 2 each Of size 40 m 10 3.5 (assuming free board=0.3m)