Mixing problem Helllllllllllllllp?

[Muteb]

a river flows into a lake of volume V at a rate of R bg/d with a concentration of C mg/L of fluoride. Initially the concentration of Fluoride in the lake is V0 mg/L. A dam on one side of the lake allows water to flow out via a river at a rate R1 and into a water treatment plant at rate R2. The flow through the dam and the rate of evaporation(E) keeps the volume of the lake constant.


1- Find an expression for the concentration of fluoride in the lake at any time?



2- The world health organization recommends drinking water should have a maximum concentration of fluoride of 1 mg/L. If the flow into the lake is 0.25 bg/d with a concentration of 3.6 mg/L of fluoride and if the concentration in the lake is initially 0.8 mg/L of fluoride and if the evaporation rate is 0.077 inches/ day and R2=0.2R1 and the area of the lake is 32 sq.miles and volume is 467 bg.

Find the time that the water plant should start taking corrective action on the amount of fluoride in the local communities drinking water.



I did the first part, but I could not do the second part.

The first:
Solution:
We use this formula: dv/dt=Input Rate – Output rate.

We have inputs: Outputs:
1-Volume of a rate of R bg/d. 1- Leaving rate= (R1+R2+Eev).
2-Concentration of C mg/L of fluoride. 2- Initial Concentration in the lake= V0 mg/L.


Finding the factor: U= e^t(R1+R2+Eevp)
Multiply the factor by the equation; , now after integrated the equation, we get the general solution: V(t)=(R*C)/(R1+R2+Eev) +(R^(-R1+R2+Eev)t)C0.


Co= V(t)(e^(R1+R2+Eev)t)-(e^9R1+R2+Eev)*(r*C)/(R1+R2+Eev)

So, V(t)= )=(R*C)/(R1+R2+Eev)[(1-(e^(R1+R2+Eev)t)].


Please help me with that I have to turn it tomorrow.

 

[jvicens]

2- Solution:
First, we need to find the concentration of fluoride in the lake at any given time using the formula we found in part 1:
C(t)=(R*C)/(R1+R2+Eev)[(1-(e^(R1+R2+Eev)t)].

Now, we can plug in the given values to find the concentration at any time t:
C(t)=(0.25*3.6)/(0.2R1+0.2R1+0.077)[(1-(e^(0.2R1+0.2R1+0.077)t)].

Next, we need to find the time at which the concentration of fluoride in the lake reaches the maximum recommended level of 1 mg/L. We can do this by setting the concentration equal to 1 mg/L and solving for t:

1=(0.25*3.6)/(0.2R1+0.2R1+0.077)[(1-(e^(0.2R1+0.2R1+0.077)t)].

Simplifying, we get: t= ln(1-0.2R1-0.2R1-0.077)/(-0.2R1-0.2R1-0.077).

Finally, we can plug in the given values to find the specific time at which the water treatment plant should start taking corrective action:
t= ln(1-0.2R1-0.2R1-0.077)/(-0.2R1-0.2R1-0.077)= ln(0.646)/(-0.577)= 0.931 days.

Therefore, the water treatment plant should start taking corrective action after approximately 0.931 days to ensure that the concentration of fluoride in the local communities' drinking water does not exceed the maximum recommended level.