Question

The oxygen sag curve caused by a food processing plant reaches a minimum DO equal to 3.0 mg/L. Upstream from the plant, the river DO is saturated at 10.0 mg/L, and it has no BOD of its own. Just downstream from the discharge point, the DO is still essentially saturated (i.e., consider the initial oxygen deficit to be zero so the downstream deficit is proportional to initial BOD). By what percentage should the BOD of the plant waste be reduced to assure a healthy stream with at least 5.0 mg/L DO everywhere?

Answer 1

28.57%

Step-by-step explanation:

Given that:

`DO_(sat)` = 10.0 mg/L

`DO_(min)` = 3.0 mg/L

We can calculate for the Original maximum deficit can be calculated which can be illustrated as follows:

`DO_(max) = DO_(sat)-DO_(min)`

where;

`DO_(max)` represents the maximum dissolved oxygen

`DO_(sat)` represents the saturated dissolved oxygen

`DO_(min)` represents the minimum dissolved oxygen

`DO_(max)` = 10.0 mg/L - 3.0 mg/L

`DO_(max)` = 7.0 mg/L

We can also calculate the Desired maximum deficit by using the above expression:

`DO_(max) = DO_(sat)-DO_(min)`

In which ;

`DO_(sat)` = 10.0 mg/L

`DO_(min)` = 5.0 mg/L

Now that we have all we need to replace the value; we have:

`DO_(max)` = 10.0 mg/L - 5.0 mg/L

`DO_(max)` = 5.0 mg/L

Finally, the percentage needed by the BOD for removing the waste can be determined using the following expression;

percentage =
`((desired DO_(max))/(Original DO_(max))) *100`%

where;

Desired
`DO_(max)` = 5.0 mg/L

Original
`DO_(max)` = 7.0 mg/L

Then we have;

percentage =
`(5.0)/(7.0)*100`%

=71.43%

Required BOD = 100% - 71.43%

Required BOD = 28.57%

Hence, The percentage necessary for the BOD of the plant waste to be reduced to assure a healthy stream with at least 5.0 mg/L DO everywhere = 28.57%