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A city discharges 1400 liters per second of sewage into stream whose minimum rate of flow is 5000 liters per second.

A city discharges 1400 liters per second of sewage into stream whose minimum rate of flow is 5000 liters per second. The temperature of sewage as well as Water is $20^{\circ}C$. The 5 day B.O.D. at $20^{\circ}C$ for sewage is 180 mg/lit and that of river water is 1 mg/lit. The D.O. content of sewage is zero. And that of the stream is 90% of the saturated D.O. If the minimum of D.O. to be maintained in the stream is 4.2 mg/lit, find out the degree of sewage treatment, required assume the de-oxygenation coefficient is 0.1, and re- oxygenation coefficient is 0.3.

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From the table, the value of saturation

D.O. at $20^{\circ}C$ = 9.2 mg/l (assume)

D.O. content of the stream= 90% of saturation D.O.=$\left( \cfrac { 90 }{ 100 } \right) \times 9.2=8.28$mg/l

D.O. of mix at the start point (i.e at t=0)=$\left( \cfrac { 0 \times 1400+08.28\times 5000 }{ 5000+1400 } \right) =6.47 mg/l$

D$_{\circ}$ = initial D.O. deficit= D.O.$_{sat}$ – D.O.$_{mix}$= 9.2 – 6.47 = 2.73 mg/l

Minimum D.O. to be maintained = 4.2 mg/l

Critical D.O. deficit=D$_{c}$ = 9.2 – 4.2 = 5 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$_{\circ}$ = 2.73 mg/l $\hspace{20mm}$ D$_{c}$ = 5 mg/l

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

we get, ${ \left( \cfrac { L }{ 5\times 3 } \right) }^{ 3-1 }=3\left[ 1-(3-1)\times \cfrac { 2.73 }{ L } \right]$

$L$=22.63 mg/l ${ Y }_{ l }=L[1-{ 10 }^{ -{ k }_{ D }l }]=22.63(1-{ 10 }^{ -0.1\times 5 })=15.47mg/l$ Now using the equation, $BOD_{mix}$ $15.47=\cfrac { { C }_{ w }\times 1400+1\times 5000 }{ 1400+5000 }$ $C_{w}$ = 67.14 mg/l Degree of treatment required (%) = $\cfrac { Original\quad B.O.D.-Permissible\quad B.O.D. }{ Original\quad B.O.D. } \times 100=\cfrac { 200(Assumed)-67.14 }{ 200 } \times 100$ = 66.43%