Expert: Alon Mandes - 4/27/2009

Question
A polluted river with a nutrient concentration of 100 g/m^3 is flowing at a rate of 100m^3/day into an estuary of volume 1000m^3.At the same time,water from the estuary is flowing into the ocean at 100m^3/day.The initial nutrient concentration in the estuary is 25g/m^3.
1.Let N(t)be the amount of the nutrient in the estuary at time t.Write down and solve an appropriate differential equation for N(t) along with the appropriate initial condition.
2.After a long time,what is te concentration of nutrient in the estuary?
3.It is known t hat if the nutrient concentration in the estuary reaches 80g/m^3 an algal bloom will occur.How many days odes it take for the nutrient concentration to reach this treashold.

I thank you in advance.

Answer
Let's 1st of all build the diffEqua of the Amount of nutrient is
the estuary N(t) :
To do so, we build an equation for the rate of change of N(t) since
we are given information per day , say N'(t).
Thus,
Per day, or each day we have amount of nutrient In to the estuary &
amount of nutrient Out from the estuary :
In : 100gr/m^3 * 100 m^3/day = 10^4 per day
Out : [N(t)/1000]* 100m^3/day = 0.1N(t) per day
Thus,
N'(t)=10^4-0.1N(t)
This is the differential equation, with initial condition N(0)=25.
the solution is :
N(t)=10^5+[25-10^5]e^(-0.1t) .
After a long time (t->∞) : N(∞)=10^5 gr.
3. We solve the equation : N(t)/1000=80 .
10^5+[25-10^5]e^(-0.1t)=80000
[25-10^5]e^(-0.1t)=-20000
e^(-0.1t)=-800/-99975
Ln [ e^(-0.1t) ] = Ln [0.008]

-0.1t=-4.82
t=48 days .

Alon.