Query_04

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course MTH-279

06/21 around 5:48AM. I had a bit if difficulty on this assignment. Some light-shedding would be much appreciated. Thanks in advance.

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 equations becomes dQ/dt = 0.03r(t) - Q(t)r(t)/1000

= r(t) (0.03 - Q(t)/1000)

= (r(t)/1000) * (30 - Q(t))

= (-r(t)/1000) * (Q(t) - 30)

dQ/(Q(t) - 30) = -r(t)/1000 dt

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integrating both sides yields

ln(Q(t) - 30) = -r(t)*t/1000 + c

Q(t) - 30 = C * e^(-r(t)*t/1000)

Q(t) = 30 + Ce^(-r(t)*t/1000)

Q(0) = 30 + Ce^0 = 0.05(1000) = 50

30 + C(1) = 50

C = 20

Q(t) = 30 + 20e^(-r(t) * t/1000)

Q(8*60) = 30 + 20e^(-r(t) * 480/1000) = 0.035(1000) = 35

20e^(-r(t) * 480/1000) = 5

e^(-r(t) * 480/1000) = 5/20

e^(-r(t) * 480/1000) = 0.25

ln(0.25) = -r(t) * 480/1000

r(t) = - ln(0.25) / 0.48

r(t) = 2.888 gal/min approx.

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Given Solution:

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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:

????Not sure how to get this one started????

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Suggestions:

FIgure out the concentration when the tank first reaches 1000 gallons.

Then proceed as in the previous question.

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Given Solution:

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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:

Not sure since I can’t complete the prior solution.

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If you choose to submit the previous question, in light of my hints, using the Question Form, you can also include this one and any new thoughts you might have on its setup.

Until you make some progress on that one, there's no benefit to me giving you hints on this one. Once you do make some progress on the preceding question, this one might become more accessible.

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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 large 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 3.5% solution at the end of 8 hours?

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Your solution:

dQ/dt = 0.03r(t) - Q(t)r(t)/1000

= r(t) (0.03 - Q(t)/1000)

= (r(t)/1000) * (30 - Q(t))

= (-r(t)/1000) * (Q(t) - 30)

dQ/(Q(t) - 30) = -r(t)/1000 dt

integrating both sides

ln(Q(t) - 30) = -r(t)*t/1000 + c

Q(t) - 30 = C * e^(-r(t)*t/1000)

Q(t) = 30 + Ce^(-r(t)*t/1000)

Q(0) = 30 + Ce^0 = 0.05(1000) = 50

30 + C(1) = 50

C = 20

Q(t) = 30 + 20e^(-r(t) * t/1000)

Q(8*60) = 30 + 20e^(-r(t) * 480/1000) = 0.035(1000) = 40

20e^(-r(t) * 480/1000) = 10

e^(-r(t) * 480/1000) = 10/20

e^(-r(t) * 480/1000) = 0.5

ln(0.5) = -r(t) * 480/1000

r(t) = - ln(0.5) / 0.48

r(t) = 1.444 gal/min approx.

confidence rating #$&*:

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The concentration of the overflow from the first tank changes with time.

The solution to that problem is

q(t) = 30 + 20 e^(-r/1000 * t), with r = 180 gal / hr (approx.)

(this is consistent with your more accurate 2.88 gal / minute).

This will affect your equation and, of course, your solution.

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Given 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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The concentration in the second would approach the solution of that being pumped in, and I assume the 3% solution would be the limiting value?

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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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Regardless of the rate, the first solution of 3% should still be the limiting factor, as it is the baseline/bottomline solution. The rate at which this occurs is different from the prior situation.

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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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Given Solution:

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Self-critique (if necessary):

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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:

T '(t) = k(T_room - T(t))

T '(t) = k(80 - T(t))

integrate

T(t) = 80 + Ce^(-kt)

T(o) = 190

190 = 80 + C

C = 110

T(t) = 80 + 110e^(-kt)

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Your equation would apply if the room temperature stayed at 80 F.

Sticking with that assumption, we can solve the equation and evaluate C and k.

You have correctly evaluated C.

What bit of given information (other than the rate at which room temperature is falling, which we are for now ignoring) have you not yet used, and how would it be used to evaluate k?

Then:

How would the equation change given the rate at which the room temperature changes?

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique (if necessary):

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Self-critique rating:

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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:

T '(t) = k(T_room - T(t))

T '(t) = k(80 - T(t))

integrate

T(t) = 80 + Ce^(-kt)

T(o) = 190

190 = 80 + C

C = 110

T(t) = 80 + 110e^(-kt)

@&

Your equation would apply if the room temperature stayed at 80 F.

Sticking with that assumption, we can solve the equation and evaluate C and k.

You have correctly evaluated C.

What bit of given information (other than the rate at which room temperature is falling, which we are for now ignoring) have you not yet used, and how would it be used to evaluate k?

Then:

How would the equation change given the rate at which the room temperature changes?

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique (if necessary):

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Self-critique rating:

#*&!

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You've got the basic procedures down pretty well.

The more complex problems, though, seem to have pretty much eluded you.

If you choose to do more work on those problems, in light of my notes and hints, feel free to submit them using the Question Form.

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