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A test bottle containing just seeded dilution water has its DO level drop by 0.8 mg/L in a 5-day test. A 300 mL BOD bottle filled with 30 mL of wastewater and the rest with seeded dilution water experiences a drop of 7.3 mg/L in the same period (5-day). Calculate the BOD5 of the wastewater.

Respuesta :

Answer:

BOD_5 = =65.8 mg/l

Explanation:

dilution water DO level = 0.8 m/l

BOD level drop to 7.3 mg/l

we know that BOD at 5th day can be clculated by using following relation[tex]BOD_5 = \frac{(D_1 -D_2) -(B_1-B_2)(1-P)}{P}[/tex]

[tex]D_1 -D_2[/tex] - DO drop in BOD bottle

[tex]B_1-B_2[/tex] - dilution water drop

P= 30/300 = 0.1

[tex]BOD_5 = \frac{(7.3) -(0.8)(1-0.1)}{0.1}[/tex]

[tex]BOD_5 = =65.8 mg/l[/tex]

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