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A city discharges 1600 lit/sec of sewage into a stream whose minimum rate of flow is 6000 lit/sec.

A city discharges 1600 lit/sec of sewage into a stream whose minimum rate of flow is 6000 lit/sec. The temperature of sewage as well as water is 20$^{\circ}$C. The 5 day BOD at 20 0 for sewage is 200 mg/lit and that of river water is mg/lit. The DO content of sewage is zero and that of the stream is 90% of the saturation DO. If the minimum DO to be maintained in the stream is 4.5 mg/lit, find out degree of sewage treatment required. Assume the de-oxygenation coefficient as 0.1 and re-oxygenation coefficient as 0.3.

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From the table, the value of saturation

D.O. at $20^{\circ}C$ = 9.17 mg/l (fix value assume)

D.O. content of the stream= 90% of saturation D.O.=$\left( \cfrac { 90 }{ 100 } \right) \times 9.17=8.25$mg/l

D.O. of mix at the start point (i.e at l=0)=$\left( \cfrac { 8.25\times 6000+0\times 1600 }{ 6000+1500 } \right) =6.6 mg/l$

D$_{\circ}$ = initial D.O. deficit= D.O.$_{sat}$ – D.O.$_{mix}$= 9.17 – 6.6 = 2.57 mg/l

(Assume instantaneous mixing)

Minimum D.O. to be maintained in the stream = 4.5 mg/l

Max. permissible saturation deficit

D$_{c}$ = 9.17 – 4.5 = 4.67 mg/l

From equation given below,

${ \left( \cfrac { L }{ { D }_{ c }f } \right) }^{ f-1 }=f\left[ 1-(f-1)\cfrac { { D }_{ \circ } }{ L } \right]$

D$_{\circ}$ = 2.57 mg/l $\hspace{20mm}$ D$_{c}$ = 4.67 mg/l

${ f=\cfrac { { K }_{ R } }{ { K }_{ D } } }=\cfrac { 0.3 }{ 0.1 } =3$

we get, ${ \left( \cfrac { L }{ 4.67\times 3 } \right) }^{ 3-1 }=3\left[ 1-(3-1)\times \cfrac { 2.57 }{ L } \right]$

$L$=21.1 mg/l ${ Y }_{ l }=L[1-{ 10 }^{ -{ k }_{ D }l }]=21.1(1-{ 10 }^{ -0.1\times 5 })=14.43mg/l$ Now, $BOD_{mix}$ $14.43=\cfrac { { C }_{ 5 }\times 1600+1\times 1600 }{ 1600+6000 }$ $C_{5}$ = 64.8 mg/l Degree of treatment required (%) = $\cfrac { Original\quad B.O.D.-Permissible\quad B.O.D. }{ Original\quad B.O.D. } \times 100=\cfrac { 200-64.8 }{ 200 } \times 100$ = 67.6%