Question

A tannery with a wastewater flow of 0.011 m^3/s and a BOD_s of 590 mg/L discharges into a creek. The creek has a flow of 1.7 m^3/s. Upstream of the tannery, the BOD_s of the creek water is 0.6 mg/L. The BOD rate constants (k) are 0.115 day^-1 for the tannery and 3.7 day^-1 for the creek. Calculate the initial ultimate BOD alter mixing.

Answer 1

Solution.

The ultimate BOD of tannery wastewater is

$L_0 = \frac{590}{1-e^{-0.115\cdot 5}} = 1349.2 \ \frac{mg}{L};$

The ultimate BOD of a creek is

$L_0 = \frac{0.6}{1-e^{-3.7\cdot 5}} = 0.6 \ \frac{mg}{L};$

The initial ultimate BOD can be calculated using an equation

$L_a = \frac{Q_wL_w+Q_rL_r}{Q_w+Q_r};$

$L_a = \frac{0.011\cdot 1349.2+1.7\cdot 0.6}{0.011+1.7} = 9.27 \ \frac{mg}{L}.$