Discharge of pollutants and toxic materials into rivers and lakes have put the a

Question

Discharge of pollutants and toxic materials into rivers and lakes have put the aquatic life at risk. Longtime exposure of a fish to these pollutants results in accumulation of pollutants in the body of fish. When the concentration of pollutants exceeds the standard limits, the fish cannot be used as food.

Pollutants may enter and leave fish by various mechanisms. The uptake of pollutants is carried out through gills and from food. The pollutants can be lost from fish via gills, egestion, metabolism. Growth dilution is assumed negligible. As a first approximation it can be assumed that the rates of uptake and discharge are directly proportional to pollutant concentration in water and in the fish body respectively. ( dC/dt=kC, first order)

Develop a model to estimate concentration of pollutant against time in a fish in a lake where the pollutant concentration is equal to Cp mg/L.

The concentration of pollutant in fish body is CF.

State your assumptions.

Verify the model.

If concentration of the pollutant in water Cp = 0.1 mg/L and kf = 0.01 d-1, k1 = 0.006 d-1 k2 = 0.0001 d-1 , ke= 0.00004 d-1 , km= 0.00002 d-1 Calculate the time required to reach to CF = Cp and CF = 1.478 mg/L

(K1=uptake from gills,K2=loss from gills, Kf=uptake from food, Ke=loss from egestion, Km=loss from motabolism)

Explanation / Answer

Assume uptake of pollutants as positive and release of pollutants negative. If Cp=concentration of water pollutants.==>rate of uptake of water pollutants... (i)from gills=dCg/dt=k1Cp (ii)from food=dCf/dt=kfCp ........................loss of pollutants (i)from gills=-dCg1/dt=-k2Cp (ii)from metabolism=dCm/dt=-kmCp (iii)from egestion=dCe/dt=-keCp................ Therefore concentration of pollutant against time in a fish in a lake where the pollutant concentration is equal to Cp mg/L.=dCf/dt= (k1+kf-k2-km-ke)Cp.......................ANS b)If concentration of the pollutant in water Cp = 0.1 mg/L and kf = 0.01 d-1, k1 = 0.006 d-1 k2 = 0.0001 d-1 , ke= 0.00004 d-1 , km= 0.00002 d-1 ................................................................................................. dCf/dt=1.584*10^-3 d-1 *mg/L ==>dt=dCf/1.584*10^-3=1.478/1.584*10^-3=933.08s..................ANS

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