Question

A wastewater sample has an ultimate BOD of 300 mg/L. BOD reaction rate constant is 0.22 per day at 20 ℃. If the temperature coefficient is 1.05, the 5-day BOD at 25 ℃, in mg/L, is:

Hint:

When calculating BOD at different temperatures, always apply the temperature correction factor to account for the increased microbial activity at higher temperatures.

Answer 1

Correct Answer: 220

Step 1: Use the Temperature Correction Formula
The BOD reaction rate constant at temperature \( T \) is given by: \[ k_T = k_{20} \times \theta^{(T - 20)} \] where:
\( k_T \) is the rate constant at temperature \( T \)
\( k_{20} \) is the rate constant at 20°C
\( \theta \) is the temperature coefficient
\( T \) is the desired temperature in °C
For \( T = 25^\circ C \), substitute the values: \[ k_{25} = 0.22 \times 1.05^{(25 - 20)} = 0.22 \times 1.05^5 \] Now calculate \( k_{25} \): \[ k_{25} = 0.22 \times 1.276 = 0.28 \, {per day} \] Step 2: Use the BOD Formula to Find the 5-Day BOD The 5-day BOD (\( BOD_5 \)) is given by: \[ BOD_t = L_0 \left( 1 - e^{-k_T t} \right) \] where:
\( BOD_t \) is the BOD at time \( t \) (in days)
\( L_0 \) is the ultimate BOD
\( k_T \) is the rate constant at temperature \( T \)
\( t \) is the time (5 days in this case)
Substitute the values: \[ BOD_5 = 300 \left( 1 - e^{-0.28 \times 5} \right) \] Calculate the exponent: \[ 0.28 \times 5 = 1.4 \] Now calculate the exponential term: \[ e^{-1.4} \approx 0.2466 \] Thus, the 5-day BOD is: \[ BOD_5 = 300 \left( 1 - 0.2466 \right) = 300 \times 0.7534 = 226.02 \, {mg/L} \]