Query 04

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course Mth 279

With some guidance on a couple I should be able to finish the others. I didn't want to fill them up with clutter if I didn't know what I was doing. submitted 3/15 8

query 04

2.5.

1. A 3% saline solution flows at a constant rate into a 1000-gallon tank initially full of a 5% saline solution. The solutions remain well-mixed and the flow of mixed solution out of the tank remains equal to the flow into the tank.

What constant rate of flow is necessary to dilute the solution in the tank to 3.5% in 8 hours?

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Your solution:

the rate in flow will equal the rate outflow, the the salinity will be different.

This problem was a bit difficult for me to solve in one sitting, but this is my logic.

i'm letting Q(t) = the number of gallons of salt in my 1,000 gal water tank

so Q(0) is 50 galons of salt because 5% of 1,000 is 50

the rate at which salt leaves is r(t)c(t) = r(t)*Q(t)/V(t) where r is rate and c is concentration

so Q(t) - r(t)(.03) - ( Q(t) / 1,000 gal )

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.03 r - r * Q / 1000 is the rate of change of the amount of salt, so the equation is

dQ/dt = .03 r - r * Q / 1000

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take derivative wrt t to get that

dQ(t)/dt = .03 - .05Q(t)

solve for dt and take the integral

ln(.03 - .05Q(t) ) = t + c

going by the book here i want to say that .03-.05Q = Ae^(t/1,000)

what here is A, therefore i can solve for Q(t) and have one variable left to find, which is the rate at which the water leaves and enters.

I used 50 for A because of 50 gallons of salt, which i'm unsure of, but solving for Q(480 min) = 30.34 gal/min. which seems very reasonable.

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Question: 2. Solve the preceding question if the tank contains 500 gallons of 5% solution, and the goal is to achieve 1000 gallons of 3.5% solution at the end of 8 hours. Assume that no solution is removed from the tank until it is full,

and that once the tank is full, the resulting overflow is well-mixed.

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Your solution:

This question will be more refined as the preceding question is answered correctly, then its a matter of solving for variables.

but using the 30 gal/ min time, we start with 25 lbs of salt because its only 500 gal of water, at 30 gal/min, it will take 16.67 minutes to till up the rest of the way to 1,000 gal.

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To achieve the goal we need to adjust the rate. So we can't assume the same rate as before.

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Q(t) = 25e^(480/500gal) = 65.29 or 6.5%.

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You are adding a more dilute solution to the original; the concentration will not increase.

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now we need to get this 6.5% down to 3.5% which is a 3% loss in salt,

now concentration is flowing out at 30 gal/min at a decreasingly concentrated rate. which is of coarse an integral. and is also, where I think i'm screwing this up.

I understand the concept, but am having a bit of trouble getting the details down.

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Question: 3. Under the conditions of the preceding question, at what rate must 3% solution be pumped into the tank, and at what rate must the mixed solution be pumped from the tank,

in order to achieve 1000 gallons of 3.5% solution at the end of 8 hours, with no overflow?

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Your solution:

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Given Solution:

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Question: 4. Under the conditions of the first problem in this section, suppose that the overflow from the first tank flows into a second tank, where it is mixed with 3% saline solution.

At what constant rate must the 3% solution flow into that tank to achieve a 4% solution at the end of 8 hours?

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Your solution:

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Question: 5. In the situation of Problem #1, suppose that solution from the first tank is pumped at a constant rate into the second, with overflow being removed, and that the process continues indefinitely.

Will the concentration in the second tank approach a limiting value as time goes on? If so what is the limitng value? Justify your answer.

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Now suppose that the flow from the first tank changes hour by hour, alternately remaining at a set constant rate for one hour, and dropping to half this rate for the next hour before returning to the

original rate to begin the two-hour cycle all over again. Will the concentration in the second tank approach a limiting value as time goes on? If so what is the limiting value? Justify your answer.

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Answer the same questions, assuming that the rate of flow into (and out of) the tank is 10 gallons / hour * ( 3 - cos(t) ), where t is clock time in hours.

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Your solution:

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Question: 6. When heated to a temperature of 190 Fahrenheit a tub of soup, placed in a room at constant temperature 80 Fahrenheit, is observed to cool at an initial rate of 0.5 Fahrenheit / minute.

If at the instant the tub is taken from the oven the room temperature begins to fall at a constant rate of 0.25 Fahrenheit / minute, what temperature function T(t) governs its temperature?

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Your solution:

So this means the room temperature will decrease as a function of time as well as the temperature of the soup.

T(t) = (190 - 0.25(t))

R(t) = 80 - .5t

by setting these equal and solving for t, we get that t = 440 minutes, so after 440 min, the temp should be the same, which is the limiting factor

solving for t = 440 minutes to cool to room temperature

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Question: "

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You were on the right track on the early questions. Check my notes.

Then check out the document at

http://vhmthphy.vhcc.edu/tests/shell_01/differential_equations/q_a_04_solngksoddfu9e9lz.htm

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