A wastewater treatment plant has a rupture in one of their flow tubes, thereby l

Question

A wastewater treatment plant has a rupture in one of their flow tubes, thereby leading to the discharge of waste material into a local lake system. The rupture goes unnoticed for a long enough time that the concentration of the waste in the lake reaches a steady state. Assume that the lake can be modeled as a continuously stirred tank reactor, with inflow, outflow, and complete mixing, and that the lake flows so that the volume remains constant at 107 m3. Pollutant-free water enters the lake at a volumetric flowrate of 45 m3/s. The waste from the wastewater treatment plant (that flows into the lake) has a waste concentration of 120 mg/L and a volumetric flowrate of 6 m3/s. Thus, there are two inlets to the lake, one with unpolluted water and the other with polluted water. The waste naturally decays according to the rate relationship, k*Cwaste , where k = 0.15 day-1. Determine the steady state concentration of waste in the lake, in units of mg/L (milligrams per liter). Express your answer to one decimal place.

Explanation / Answer

Rate of change in waste concentration in lake = Rate of inflow of waste(both the inputs) - Rate of consumption or decay of waste - Rate of outflow of waste

The equation is given by   

VdC/dt = Qin1*Cin1 + Qin2*Cin2 - V(k*C) - Qout*C

V= volume of the lake = 107 m3

dC/dt = Change of concentration with respect to time

Qin1 = Flow rate of pollutant free water m3/sec

Cin1 = Concentration of pollutant in pollutant free water = 0 mg/L = 0 g/m3

Qin2 = Flow rate of waste water = 6 m3/sec

Cin2 = Concentration of waste in waste water = 120 mg/L = 120 g/m3

Qout = Exit flow rate of water = Qin1 + Qin2 = 45 + 6 = 51 m3/sec

C = concentration of waste in lake at any time t.

k = rate constant = 0.015 day-1 = 0.015/86400 sec-1 = 1.736 * 10-7 sec-1 .

Now at steady state, dC/dt = 0. and we know that Cin1 = 0. Then above equation becomes,

0 = Qin2*Cin2 - V(k*C) - Qout*C

C = Qin2*Cin2 / ( V*k + Qout) = (6*120)/(107 * 1.736 * 10-7 + 51 )

C = 14.117 g/m3 = 14.117 mg/L

In one decimal place,

Concentration of waste at steady state = 14.1 mg/L

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.