Assume that lake Erie has volume of 500 km3 and that its rat

Assume that lake Erie has volume of 500 km^3 and that its rate of inflow (from Lake Huron) and outflow (to Lake Ontario) are both 330 km^3 per year. Suppose that at time t = 0 years, the pollutant concentration of Lake Erie is six times that of Lake Huron (which is assumed to have a constant pollutant concentration. Assuming that the outflow is perfectly mixed lake water, how long will it take to reduce the pollutant concentration in Lake Erie to twice that o\'Lake Huron?

Solution

Let rate ofinflow from Huron to Erie =r_in

Rate of outfolw from Erie to Ontario=r_out

Given that the concentration of Erie to Huron C_0= 6*inflow concentration

C_0=6*c_in

C_in=(1/6)*c_0

Let x_0be the pollutant in kg

X be the pollotant out kg

V is volume constant in lake Erie =500km^3

Given inflow=outflow rate

r_in=r_out

C_out=x/v

C_in=(1/6)*c_0=(1/6)*(x_0/v)

Rate of change of pollutants with respect to time

dx/dt=r_in*c_in-r_out*c_out

dx/dt={(r*x_0)/6*v}-{r*x/v}

dx/dt=(r/v)*{(x_0/6)-x}

Now separate thevariable

[dx/{(x_0/6)-x}]=(r/v)*dt

Multiply with ( -) on bothsides then

{dx/[(-x_0/6)+x]}=-(r/v)*DT

Integrating on both sides

log[x-(x_0/6)]=-(r/v)*t+c

x-(x_0/6)=k*e^{-(r/v)*t}

x=(x_0/6)+k*e^{(-r/v)*t}

Therefore x(t)=(x_0/6)+k*{-(r/v)*t

Let t=0,then

x_0=(x_0/6)+k

K=x_0-(x_0/6)

K=(5/6)*x_0

Therefore x(t)=x_0/6+(5/6)*x_0*e^[-(r/v)*t]

For calculation of timr to take to reduce the pollutant concentration in lake Erie to twice of lake Huron.

2X(t*)=x_0let us consider for t*

x(t*)=x_0/2

X(t*)=x_0/2

x(t*)=

x_0/2=x_0{(1/6)+(5/6)*e^(-(r/v)*(t*))

1/2=1/6+(5/6)*e^((-r/v)*(t*))

(1/2)-(1/6)=(5/6)*e^((-r/v)*(t*))

(2/6)*(6/5)=e^((-r/v)*(t*))

2/5=e^((-r/v)*(t*))

-(v/r)*(t*)=log(2/5)

t*=-(v/r)*log(2/5)

t*=-(500/330)*log(2/5)

t*=-(50/33)*(-0.397)

t*=0.601

.

t*=-(50/33)\"log(2/5)