query 4

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course Mth 279

2/19/14

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:

Our 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

integrating both sides yeilds

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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Self-critique (if necessary):

OK

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Self-critique rating:

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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:

time to fill the take is t_1 = 500 / r

saline in the tank will be Q(t_1) = Q(500 / r) = .05 * 500 + .035 * r * t_1 = .05 * 500 + .03 * r * 500 / r = .05 * 500 + .03 * 500 = 40

concentration is .04

.05 * 500 + .03 * 500

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The concentration will be as you say, but you haven't determined t_1 or r, and the process still has 8 hours - t_1 to go.

You have to set up the equation for the remaining time, solve it, and use the solution to find t_1 and the rate r.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):Not sure if I did it right

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Self-critique rating:3

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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:

r_1 the rate at which solution flows in and r_2 the rate of outflow

500 gal * (r_1 - r_2) after 8 hours tank reaches 1000 gal capacity

don’t know where to go from here

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From this you can find the difference r_1 - r_2 of the two rates. You are left with the need to find one of the rates.

dQ/dt is the rate at which solute enters the system, minus the rate at which it exits the system.

In terms of r_1 and r_2, which is equal to r_1 minus the difference between the two rates (found previously), what is the rate at which solute enters the system?

In terms of r_1, r_2 and Q, what is the rate at which solute is removed from the system?

What therefore is the differential equation for this situation?

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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 rating:

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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 yeilds

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

Changing this last part a bit:

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.

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Not a bad attempt but:

The amount of solution in the second tank is r_1 * t + r_2 * t, since water flows into it from the first tank at rate r_2 and 3% solution flows in at rate r_1.

You appear to have treated the second tank as if the volume of solution was always 1000 gallons.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:

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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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As the time advances, the concentration would increase until they both reach the concentration of the amount being pumped. If not then one tank would reach it before the other.

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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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The first tank would reach the concentration of what is being pumped into it first, the second would take twice as long

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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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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

not sure how to set up

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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:

dT / dt = (.05 / 110) / min * T_diff

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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This is a good solution, as far as it goes, for the situation if room temperature remains constant, but it does not address what happens if the room temperature changes at the given rate.

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For the situation of constant room temperature, you are given the initial rate of temperature change.

It is k that determines how fast the temperature changes. So if you know how fast the temperature changes at any instant, you can find k.

You have the temperature function, so you can find its derivative. Its derivative at t = 0 is equal to -0.5 F / min. This allows you to solve for k.

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I might be able to figure out k, but I'm unsure of this method below

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

0.50 = k(80 - 190) -> T(0) = 190, T '(0) = 0.5

k = 0.5/-110

k = 0.25/-110

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Note that you still need to solve the problem for the case of falling room temperature.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):Think I was headed in the right direction

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Self-critique rating:2

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Question: "

Self-critique (if necessary):

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Question: "

Self-critique (if necessary):

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Self-critique rating:

#*&!

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You're on the right track. The more challenging problems gave you some trouble, but you made credible attempts.

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You might want to modify some of your work and submit it. If you do, use a copy of this document, leave everything (including my notes), and use **** before and after to mark your insertions.

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