Question

1. (30 points) Determine the 1-day BOD (BOD) and ultimate CBOD (BOD, or Lo) for a wastewater whose 5-day BOD at 20 °C is 200 mg/L, when -0.23 day.

Answer 1

For BOD calculation, let the ultimate BOD be L_o. Then the formula for t-day BOD (L_t) will be given as,

$L_t=L_0(1-\exp(-kt))$

For 5-day BOD, we have t=5. Thus,

$L_5=L_0(1-\exp(-k\times 5))$

Putting in the values, k= 0.23/day and L₅ = 200 mg/L, we have,

$200 mg/L=L_0(1-\exp(-0.23\times 5))\\ \Rightarrow 200 mg/L=0.683L_0\\ \Rightarrow L_0=292.7 mg/L$

Now that we have calculated ultimate CBOD, we will use it to calculate 1-day BOD. For 1-day BOD, t=1. Thus, we will have,

$L_1=L_0(1-\exp(-k\times 1))$

Putting in the values, k= 0.23/day and L₅ = 292.7 mg/L, we have,

$L_1=(292.7mg/L)\times(1-\exp(-0.23\times 1))\\ \Rightarrow L_1=60.1mg/L$

Thus, 1-day BOD is 60.1 mg/L and the ultimate BOD is 292.7 mg/L