Question

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.

Answer 1

BOD_5 = =65.8 mg/l

Step-by-step 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
`BOD_5 = ((D_1 -D_2) -(B_1-B_2)(1-P))/(P)`

`D_1 -D_2` - DO drop in BOD bottle

`B_1-B_2` - dilution water drop

P= 30/300 = 0.1

`BOD_5 = ((7.3) -(0.8)(1-0.1))/(0.1)`

`BOD_5 = =65.8 mg/l`