Question

Some wastewater has a BOD5 of 150 mg/L at 20° C. The reaction rate k at that temperature has been determined to be 0.20/day. Find BOD5 (in mg/L) at 15° C.

Answer 1

Answer:

The value of $\mathbf{BOD_5 \ at \ 15^0 \ C \ is \simeq 130.14 \ mg/L}$

Explanation:

From the given information:

To find the BOD5 (in mg/L) at 15° C; we need to know the ultimate carbonaceous BOD and the reaction rate coefficient at  15° C.

So, to start with the calculation of the ultimate carbonaceous BOD by using the formula:

$BOD_t = L_o (1-e^{-k*t})$

where;

$BOD _t$ = Biochemical oxygen demand at t days

$L_o$ = the ultimate carbonaceous

∴

$150= L_o (1-e^{-0.20*5})$

$L_o =\dfrac{150}{ (1-e^{-0.20*5})}$

$L_o =\dfrac{150}{ (1-0.3679)}$

$L_o =237.30 \ mg/L$

Thus, the ultimate carbonaceous BOD = 237.30 mg/L

For the reaction rate coefficient; we use the formula:

$k_{\tau} = k_{20} \times \theta^{\tau - 20}$

where;

$k_{\tau}$ = rate of the reaction constant at various temperature (T) = 15

$k_{20}$ = rate of the reaction constant at standard laboratory = 0.20

$\theta$ = constant = 1.047

∴

$k_{15} =0.20 \times \theta^{15 - 20}$

$k_{15} =0.20 \times 1.047^{-5}$

$k_{15} =0.20 \times0.7948$

$k_{15} = 0.1590 / day$

Thus, at 15° C, the reaction constant (k) = 0.1590 / day

Finally, the BOD5 (in mg/L) at 15° C can be calculated by using the formula:

$BOD_t = L_o (1 - e^{-k*t})$

$BOD_t = 237.30 \times (1 - e^{-0.1590*5})$

$BOD_t = 237.30 \times (1 -0.45158)$

$BOD_t = 237.30 \times (0.54842)$

$\mathbf{BOD_5 \simeq 130.14 \ mg/L}$