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, Qw = Discharge of waste water, QR = Discharge of stream water
Sw = BOD of waste water, SR = BOD of stream water
Ultimate BOD is calculated as;
\({\left( {BO{D_5}} \right)_{mix}} = {l_o}(1 - {e^{ - Kt}})\)
where, lo = Ultimate BOD and K = Oxygen consumption rate of the mix
Calculation:
Given, Qw = 0.2 m3/sec, QR = 4 m3/sec, Sw = 100 mg/L, SR = 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 (lo) = 9.76 mg/L