Question #231855

Calculate the size of a rectangular sedimentation tanks to treat a flow of 12,000 m3/d. Adopt an overflow rate of 20 m3/m2-d and detention time of 4 hours. Use two units in parallel. What will be the minimum weir length required if the weir overflow rate is not to exceed 300 m3/m-d? Also check the horizontal velocity.


1
Expert's answer
2021-09-03T02:00:07-0400

Discharge:12,000m3/dayOverflowRateOFR=20m/dayDepth=3mSurface area of rectangular tank Surface area=QOFR=1200020m2Area=600Let us assume,length to width Ratio be4:14x×1x=600x2=6004x2=150mX=3.162m4x=12.649mLet us adope the dimension of(13×4)m2Detention time,td=Dvs=320daytd=32024td=3.6hrswear overflow rate or weir Loading RateWFR:Flow RateLength=1200013m3s1d1=312.5m3/mdayDischarge : 12,000 m^3/day\\ Overflow Rate OFR = 20 m/day\\ Depth = 3 m\\ Surface \space area \space of \space rectangular \space tank \space Surface \space area =\frac{Q}{OFR} = \frac{12000}{ 20} m^2\\ Area = 600 \\ Let \space us \space assume, length \space to \space width \space Ratio \space be 4:1\\ 4x \times1x = 600\\ x^2 = \frac{600}{4}\\ x^2 = 150 m \\ X = 3. 162 m \\ 4x = 12.649 m \\ Let \space us \space adope\space the \space dimension \space of ( 13 \times 4 ) m^2 \\ Detention \space time, td = \frac{D}{vs} = \frac{3}{20} day \\ td = \frac{3}{20}*24 \\ td = 3.6 hrs \\ wear \space overflow \space rate \space or \space weir \space Loading \space Rate\\ WFR : \frac{Flow \space Rate}{ Length} \\ = \frac{12000}{ 13} m^3 s^{-1} d^{-1}\\ = 312. 5 m^3/ mday


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