Assignment 6

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course Mth 279

7/8

query 042.5.

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:

1. input rate: 0.03r

output rate of salt qr/v = qr/1000

Rate of change of q is dq/dt = r_in - r_out = 0.03r - qr/1000

2. dq/dt = (r/1000)(30 - q)

1/(30-q) dq = r/1000 dt

integrate

ln(30-q) = rt/1000 + C

30 - q = Ce^(rt/1000)

q(0) = 50

30 - 50 = Ce^0

C = -20

q(t) = -20e^(rt/1000) + 30

3. when t = 8

output rate: qr/1000 ; q=r*1000 = .035*1000 = 35

q(t) = -20e^(rt/1000) + 30

35 = -20e^(8r/1000) + 30

5/20 = -e^(8r/1000)

ln(1/4) = - 8r/1000

r = ln(1/4) * -1000/8 = 173 gal/hour

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Given Solution:

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Self-critique (if necessary):

I am not confident on this section. I don’t understand exactly what I am doing and I don’t understand the problem. I had to look up how to solve this problem.

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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:

flow in = .03r

flow out = qr/500

dq/dt = .03r - (qr/500)

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When the tank holds 500 gallons, there is no outflow. Outflow doesn't begin until the tank is full.

The tank immediately starts filling, so 500 gallons applies only to the first instant.

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dq/dt = (r/500)(15-q)

integrating

ln(15-q) = (rt/500) + C

q(t) = 500*.05 = 25

q(0)=25

15 - 25 = Ce^0

C = -10

q(t) = 10e^(rt/500) + C

q(8) = 1000 * .035 = 35

35 = 10 e^(8r/500) + 15

2 = e^(8r/500)

ln(2) = 8r/500

r = (500/8)ln(2) = 43 gal/hr

???not sure if this is correct???

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Your procedure isn't bad, but it isn't totally applicable to this problem.

Check my note above.

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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:

I have no idea what to really do on this one. I am unsure if my answer to the previous question is correct.

q(8) = 1000

r_1 = flow input

r_2 = flow output

500(r_1 - r_2) = q(t)

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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 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:

In the same boat as the last question. Again this section with the word problems really seems to be tripping me up.

Would it be possible to see how to solve this question???

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You first need to work the earlier problems. Then I can give you some hints and leading questions on this one.

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Given Solution:

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Self-critique (if necessary):

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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:

I am unsure how to answer these questions.

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Given 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:

T’(t) = k[80-T(t)]

integrate

T(t) = 80 + Ce^-kt

190 = 80 + Ce^(-0k)

190 = 80 + C

C = 110

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

T’(t) = k(80 - T(t))

.5 = k(80-190)

k = .5/-110

T(t) = 80 + 110e^(.5t/-110)

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Good solution.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

Not confident on this question and solution

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Self-critique (if necessary):

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Good work on the first and last problems.

The mixture problems are sequential, and each is more difficult than the preceding.

Check my notes.

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Naturally I'll be glad to look at additional work and/or answer additional questions.

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