Question

5. A river with 20 ppm of a non-conservative (reactive) substance and an upstream flow of 25 m/s receives an agricultural discharge of 5 m/s carrying 2000 ppm of the same reactive substance. The chemical in the agricultural stream mix instantaneously with the main river flow. The reactive substance has a first-order decay rate of 0.25/day and the river has a cross sectional area of 20 mperpendicular to the direction of flow. A municipality located 40 km downstream of the agricultural stream discharge point withdraws water for municipal water supply purpose. Draw a schematic diagram showing the control volume. Find: (a) the steady-state concentration in the water withdrawn 40 km downstream? (b) the treatment requirement in the agricultural discharge to achieve 50 mg/L concentration in the withdrawal location 40km downstream?

Answer 1

Schematic diagram showing the control volume

CONTROL VOLUME ! & Alla - 20m² FLOW gr = 20m/ in a smals Ta HOKAM Agriculture municipal withobiawas discharge O point point.

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(a). River discharge, Qr = 25 m3/s

Pollutant concentration in river, Cr = 20 ppm

Agriculture discharge, Qa = 5 m3/s

Pollutant concentration in agriculture discharge, Ca = 2000 ppm

Using mixture law, the pollutant concentration in river-agriculture discharge mix, Co = (QrCr + QaCa) / (Qr + Qa) = (25*20 + 5*2000) / (25 + 5) = 350 ppm

Velocity of river, v = total discharge / area = (25+5)/20 = 1.5 m/s

Time taken by the pollutant to travel from the agriculture discharge point to the municipal withdrawal point, t = distance / speed = 40*1000 / 1.5 = 26666.67 s = 0.3086 days.

For the first-order kinetics reactions, the concentration at any time 't' is calculates using the following relation:

C(t) = Co e -kt

here, k is the decay constant, Co is the initial concentration.

Hence, the pollutant concentration at the withdrawal point will be:

C(t) = 350 e -0.25 * 0.3086 = 324 ppm

Hence, the steady-state concentration of the pollutant in the water withdrawn 40 km downstream will be 324 ppm.

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(b). If the desired concentration of the pollutant at the withdrawal point is 50 ppm [i.e C(t) = 50 ppm], the initial concentration at the agriculture discharge point shall be calculated as:

C(t) = 50 = Co * e -0.25 * 0.3086

Co = 54 ppm

Co = (QrCr + QaCa) / (Qr + Qa)

54 = (25*20 + 5*Ca) / (25 + 5)

Ca = 224 ppm

Hence, the concentration of pollutant in the agriculture discharge should be 224 ppm

Efficiency of treatment method required = [Cactual - Ca(required)] *100/ Cactual = (2000 - 224) * 100 / 2000 = 88.8%

Hence, a treatment unit having an efficiency of 89% with respect to the removal of this particular chemical pollutant should be designed for the agriculture discharge in order to acheive the desired concentration of chemical pollutant at the withdrawal point.

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