Consider a lake, in which the residence time of water is 10 years. Into this lake...

 

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Consider a lake, in which the residence time of water is 10 years. Into this lake a highly soluble, non-co-distilling pollutant enters at a constant rate, a, and out of the lake pollutant exits, in the outflow water stream. Thus dM/dt = a - bM(t), where M(t) is the stock of pollutant in the lake. The rate of water evaporation from the lake is equal to the rate of stream outflow. Suppose that at a particular time the pollutant is not in steady state, but rather is observed to be increasing at a rate of 2%/year. By what percent will you have to reduce the rate of pollutant loading, a, in order to instantly bring M(t) to steady state? Consider a lake, in which the residence time of water is 10 years. Into this lake a highly soluble, non-co-distilling pollutant enters at a constant rate, a, and out of the lake pollutant exits, in the outflow water stream. Thus dM/dt = a - bM(t), where M(t) is the stock of pollutant in the lake. The rate of water evaporation from the lake is equal to the rate of stream outflow. Suppose that at a particular time the pollutant is not in steady state, but rather is observed to be increasing at a rate of 2%/year. By what percent will you have to reduce the rate of pollutant loading, a, in order to instantly bring M(t) to steady state?

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dMdt a bMt Residence time of water 10 years Rate of pollut increase 2 per year kno... View the full answer