Question

A facultative pond system for sewage treatment consists of two ponds in series. The hydraulic retention time (HRT) of each pond is 6 days. The total BOD\(_5\) reduction through the entire pond system is 90 %. If the ponds are considered to be completely mixed, then the rate constant describing the BOD\(_5\) removal is _______ day\(^{-1}\) (rounded off to two decimal points). Assume that the entire rate constant is the same for both the ponds.

Hint:

The rate constant for BOD\(_5\) removal can be calculated from the hydraulic retention time and the percentage reduction in BOD.

Answer 1

Correct Answer: 0.34

For a completely mixed system, the rate constant \( k \) can be calculated using the following relationship: \[ k = \frac{1}{HRT} \ln \left( \frac{C_0}{C_t} \right), \] where:
- \( HRT \) is the hydraulic retention time,
- \( C_0 \) is the initial BOD concentration,
- \( C_t \) is the concentration at time \( t \).
The total BOD\(_5\) reduction is 90 %, so the remaining BOD\(_5\) after treatment is 10 % of the initial concentration, i.e. \( C_t/C_0 = 0.1 \). Thus, the equation becomes: \[ k = \frac{1}{6} \ln \left( \frac{1}{0.1} \right) = \frac{1}{6} \ln(10) \approx 0.0347 \, \text{day}^{-1}. \] Thus, the rate constant for BOD\(_5\) removal is \( 0.35 \, \text{day}^{-1} \).