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Twenty six million gallons per day of wastewater with a DO of 1.00 mg/L is discharged into a river with a DO of 6.00 mg/L. If the flow rate of the river is 165 x 106 gal/d and saturation value of dissolved oxygen is 9.17 mg/L, what is the oxygen deficit he two flows? (0 Points)

Respuesta :

Answer:

oxygen deficit = 3.851 mg/L

Explanation:

given data

flow rate of the river=  165 × [tex]10^{6}[/tex] gal/d

saturation value of dissolved oxygen = 9.17 mg/L

to find out

oxygen deficit he two flows

solution

we will apply here formula for dissolved oxygen content after dilution is

Do mix = [tex]\frac{Qw*(Do)w +Qr*(Do)r}{Qw+Qr}[/tex]     ..........................1

here Qw is rate of flow of waste water  i.e 26 ×[tex]10^{6}[/tex] gal/d

(Do)w is Do of waste waster i.e 1 mg/L

Qr is aret of flow of river i.e 165 ×[tex]10^{6}[/tex] gal/d

(Do)r is do of river water i.e 6 mg/L

so put all value in equation 1 we get

Do mix = [tex]\frac{26*10^6*1 +165*10^6*6}{(26+165)10^6}[/tex]

solve we get

Do mix = 5.319 mg/L

so

oxygen deficit =  saturation oxygen - (Do) mix     ..............2

oxygen deficit = 9.17 - 5.319

oxygen deficit = 3.851 mg/L

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