Question

A wastewater treatment plant disposes of its effluent in a surface stream. Characteristics of the stream and effluent are shown below.

Parameter	Waste water	Stream water	Waste water mix stream water
Flow (m3/sec)	0.2	4
Dissolved oxygen, mg/L	1	7
BOD5 at 20°C, mg/L	100	2
Oxygen consumption rate (K1 at 20°C) (1/day)	0.2	0.2	0.23
Oxygen reaeration rate (K2 at 20°C) (1/day)	-	0.3	0.3

For 20°C stream water temperature, the equilibrium concentration of oxygen = 9.17 mg/L. Assuming no temperature correction is required, answer the following: Calculate ultimate BOD of wastewater and stream water mix water?

1. 8.5 mg/L
2. 9 mg/L
3. 9.5 mg/L
4. 9.76 mg/L

Answer 1

Correct Answer - Option 4 : 9.76 mg/L

Concept:

BOD of the mix, \({\left( {BO{D_5}} \right)_{mix}} = \frac{{{Q_w}.{S_w} + {Q_R}.{S_R}}}{{{Q_w} + {Q_R}}}\)

Where, Q_w = Discharge of waste water, Q_R = Discharge of stream water

S_w = BOD of waste water, S_R = BOD of stream water

Ultimate BOD is calculated as;

\({\left( {BO{D_5}} \right)_{mix}} = {l_o}(1 - {e^{ - Kt}})\)

where, l_o = Ultimate BOD and K = Oxygen consumption rate of the mix

Calculation:

Given, Q_w = 0.2 m³/sec, Q_R = 4 m3/sec, S_w = 100 mg/L, S_R = 2 mg/L

\({\left( {BO{D_5}} \right)_{mix}} = \frac{{{Q_w}.{S_w} + {Q_R}.{S_R}}}{{{Q_w} + {Q_R}}}\)

⇒ \({\left( {BO{D_5}} \right)_{mix}} = \frac{{0.2 \times 100 + 4 \times 2}}{{0.2 + 4}} = 6.67\ mg/L\)

∵ We know, \({\left( {BO{D_5}} \right)_{mix}} = {l_o}(1 - {e^{ - Kt}})\)

For waste water mix stream water K = 0.23 day^-1 (From given table)

⇒ \(6.67 = {l_o}(1 - {e^{ - 0.23 \times 5}})\)

Ultimate BOd (l_o) = 9.76 mg/L