Happy Valley Pond is currently populated by yellow perch. The pond is fed by two

Question

Happy Valley Pond is currently populated by yellow perch. The pond is fed by two springs: spring A contributes 50 gallons of water per hour during the dry season and 80 gallons of water during the rainy season. Spring B contributes 60 gallons of water per hour during the dry season and 75 gallons of water during the rainy season. During the dry season an average of 110 gallons of water per hour evaporates from the pond, and an average of 90 gallons per water evaporates during the rainy season. There is a small spillover dam at one end of the pond and any overflow will go over the dam into Bubbling Brook. When the pond is full (i.e. the water level is the same height as the top of the spillover dam) it contains 475,000 gallons of water.

            Spring B has become contaminated with salt and is now 10% salt. The perch will start to die if the concentration of salt in the pond rises to 1%. Assume that the salt will not evaporate but will mix thoroughly with the water in the pond. There was no salt in the pond before the contamination of spring B. Your group has been called upon by the Happy Valley Bureau of Fisheries to try and save the perch.

I. Assume that it is the dry season and that at time t = 0 the pond contains 400,000 gallons and spring B became contaminated.   Let S(t) be the amount of salt in the pond.

a. Write a differential equation for S.

b. Solve and graph the equation.

c. Do the fish die? If so, when do they start to die?

II. Keeping all other assumptions the same, answer a-c above, assuming the pond was full at time t = 0.

III. Answer a-c above, assuming the pond was full at time t = 0 and the contamination of spring B occurred at the beginning of the rainy season.

d. Draw a graph of the salt in the pond for the next three months.

e. How much salt will be in the pond in the long run?

IV. Answer questions a-e, if the contamination of spring B occurred during the dry season but just two weeks before the rainy season started.

V. It is very difficult to find where the contamination of spring B originates so the Bureau of Fisheries proposes to flush 100 gallons per hour of pure water through the pond. Analyze this plan and suggest any modifications or improvements which could save the perch

Please answer 3e, part 4 and part 5

Explanation / Answer

Solution :

After contamination by salt (lets call the amount of salt S), spring B still contributes 60 gallons/hr to the pond.
But 10% of that is now salt.
10% of 60 is 6.
The equivalent of 6 gallons/hr of salt is entering the pond. If we write this in the language of mathematics, then that will be our answer for question (a), we have:

a) dS/dt=6 (units in gallons/hr)

b) To solve, we rewrite the equation above as dS= 6.dt then integrate:
dS= 6.dt

From which:
S= 6t+c (c is arbitrary constant of integration)
At t=0, the amount of salt in the pond is S=0
So, finding for c, we have c= S-6t= 0-6(0)=0

Therefore the buildup of salt in the pond over time is:
S= 6t (Units of salt equivalent to gallons, units of time in hours)

c) After contamination of spring B, the amount of water entering the pond from both spring A and spring B is 50+54 gallons/hr.
The amount of water being lost from the pond is 110 gallons/hr through evaporation.
So, summing the amounts of water being added and lost gives us 50+54-110= -6 gallons/hr
The pond loses 6 gallons/hr of water.
But the amount of salt entering the pond is equivalent to 6 gallons/hr, replacing the water being lost.

What we have is the amount of salt and water in the pond stays at 400,000 gallons. But the water content is effectively being turned into salt at a rate of 6 gallons/hr.

What we need to find is when 1% of the 400,000 gallon mix is salt.
1% of 400,000 gallons is 4000 gallons.
We have an equation for the amount of salt, we set the amount of salt at 4000, then solve for t:
S=6t 4000=6t
t= 4000/6= 2000/3= 666.7 hrs (to 1 decimal place)
The pond will reach a salt content of 1% on the 22nd day, then the perch will start to die

ll) For this question, the same maths applies. The amounts of water entering and leaving the pond is the same, the amount of salt entering the pond is the same.
In this case however, 475,000 gallons of water is effectively being turned into salt at a rate of 6 gallons/hr.
1% of 475,000 is 4750. Now we have from our equation:
4750=6t
From which t=4750/6=791.7 hrs (to 1 decimal place)
The perch will start to die just over a month after contamination begins.

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.