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A waste water stream (flow = 2 m3/sec, ultimate BOD = 90 mg/litre) is joining a small river (flow = 12 m3/s, ultimate BOD = 5 mg/litre). Both water streams get mixed up instantaneously. Cross-sectional area of the river is 50 m2. Assuming the de-oxygenation rate constant, K = 0.25/day the BOD (in mg/litre) of the river water, 10 km downstream of the mixing point is
1. 1.68
2. 12.63
3. 15.46
4. 1.37

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Correct Answer - Option 3 : 15.46

Concept:

Biochemical oxygen demand (BOD):

BOD is defined as amount of oxygen demanded by the micro-organisms C5H7NO2 (Bacteria) present in it to decompose biodegradable organic matter in wastewater under aerobic conditions.

BOD is measure of strength of waste water

5 day BOD @ 20°c = [(DO) I - (DO) F] × D.F

(DO) I = Initial dissolved oxygen

(DO) F = Final dissolved oxygen

D.F = Dilution factor

\({\rm{D}}.{\rm{F}} = \frac{{300}}{{{{\rm{V}}_{\rm{s}}}}} = \frac{{100}}{{{\rm{\% dilution}}}}\)

\({\rm{D}}{{\rm{O}}_{{\rm{initial}}}} = \frac{{{{\rm{Q}}_{\rm{S}}}{{\left( {{\rm{DO}}} \right)}_{\rm{S}}} + {{\rm{Q}}_{\rm{R}}}{{\left( {{\rm{DO}}} \right)}_{\rm{R}}}}}{{{{\rm{Q}}_{\rm{S}}} + {{\rm{Q}}_{\rm{R}}}}}\)

QS and QR is flow of water in stream and river respectively

(DO)S and (D)R is dissolved oxygen in stream and river respectively

Expression for BOD

Lo = Initial organic matter in waste water at time t =0

Lt = Organic matter remained in wastewater at any time t

Lo – Lt = Organic matter removed in time t

\({{\rm{L}}_{\rm{t}}} = {{\rm{L}}_{\rm{o}}}{{\rm{e}}^{ - {\rm{Kt}}}}\)

Where

K = BOD rate constant

Calculation:

Given,

QS = 2 m3/sec, (BOD)s = 90 mg/litre

QR = 12 m3/sec, (BOD)R = 5 mg/litre

Area = 50 m2, K = 0.25/day

Distance of downstream = 10 km

\({\rm{BO}}{{\rm{D}}_{{\rm{mix}}}} = \frac{{{{\rm{Q}}_{\rm{S}}}{{\left( {{\rm{BOD}}} \right)}_{\rm{S}}} + {{\rm{Q}}_{\rm{R}}}{{\left( {{\rm{BOD}}} \right)}_{\rm{R}}}}}{{{{\rm{Q}}_{\rm{S}}} + {{\rm{Q}}_{\rm{R}}}}}\)

\({\rm{BO}}{{\rm{D}}_{{\rm{mix}}}} = \frac{{\left( {2 \times 90} \right) + \left( {12 \times 5} \right)}}{{2 + 12}} = 17.14{\rm{\;mg}}/{\rm{lit}}\)

Now we know that

\({\rm{Velocity}} = \frac{{{{\rm{Q}}_{{\rm{mix}}}}}}{{\rm{A}}} = \frac{{2 + 12}}{{50}} = 0.28{\rm{\;}}{{\rm{m}}^3}\)

\({\rm{Time\;taken\;to\;travel\;}}10{\rm{\;km\;distance\;}} = \frac{{{\rm{Distance}}}}{{{\rm{velocity}}}} = \frac{{10000}}{{0.28}} = 35714.28{\rm{\;sec}} = 0.41{\rm{\;days}}\)

\({{\rm{L}}_{\rm{t}}} = {{\rm{L}}_{\rm{o}}}{{\rm{e}}^{ - {\rm{Kt}}}}\)

\({{\rm{L}}_{\rm{t}}} = 17.14 \times {{\rm{e}}^{ - 0.25 \times 0.41}} = 15.46{\rm{\;mg}}/{\rm{litre}}\)

Hence the BOD of river water, 10 km downstream of the mixing point = 15.46 mg/litre

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