(Solved): B. Determine the following for a sedimentation tank: i. Surface area ii. Dimensions iii. Volume iv. ...

B. Determine the following for a sedimentation tank: i. Surface area ii. Dimensions iii. Volume iv. Weir length. Specifications - The sedimentation basin to be designed hereunder will be a rectangular sedimentation basin with the following specifications: - Rectangular basin - Depth: 7-16 $$\mathrm{ft}$$ - Width: 10-50 ft - Length: $$4\times$$ width - The treatment plant flow is $$3\mathrm{MGD}$$. - Influent baffle to reduce flow momentum - Slope of bottom toward sludge hopper $$>1\%$$ - Continuous sludge removal with a scraper velocity $$<15\mathrm{ft}/\min$$ - Generally the detention time: 4-8 hours - Consider a tank with automatic sludge removal, so the detention time should be 4 hours - Flow through velocity: $$40.5\mathrm{ft}/\min$$ - Overflow rate: 500-1,000 gal/day-ft2 - Weir loading: 15,000-20,000 gal/day-ft Note: 1 gallon $$=0.133681$$ cubic feet Hint: Always consider more than one tank in your design Formulas: - $$\mathrm{A}={\mathrm{Q}}_c/$$ O.R. where O.R. is the overflow rate - $$Q=$$ volume / time - Volume of a rectangular: $$\mathrm{V}={\mathrm{L}}^{?}\mathrm{W}?\mathrm{d}$$ - cross-sectional area of the tank: $$\mathrm{Ax}={\mathrm{W}}^{?}\mathrm{d}$$ - flow through velocity of the tank: $$\mathrm{V}=\mathrm{Qc}/\mathrm{Ax}$$ - Weir length: Lw = Qc / W.L Where, W.L is weir loading