Query_04

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course Mth 279

3/17I had some real trouble with this one

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:

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

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It's misleading to call the rate r(t), since the rate is constant.

Expressing r as r(t) makes it appear that r(t) needs to be integrated on the right-hand side.

Of course r(t) can be a constant function, but it's best to leave of the implied t dependence in this case.
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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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Good solution.
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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:

I have hopelessly worked on this problem to no success.

Every couple of pages, I seem to find something that doesn't make sense and have to start over again.

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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. *********************************************Self-critique rating:------------------------------------------------ Self-critique (if necessary):&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& It's best to give me at least some indication of your thinking, without which it's difficult for me to know how to get you on track.

If the rate is r, then the time to fill the take is t_1 = 500 / r and the amount of 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 and the concentration is .04, or 4%. The last step .05 * 500 + .03 * 500 could have been written out directly, since you know that when the tank is initially full it contains 500 gallons of each solution.

You could at this point just consider a full tank that starts with 4% concentration, with 3% solution flowing in and mixed solution flowing out.

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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:

Without knowing the answer from the previous problem, this one seems impossible to figure out, since I don't even know how to set the last problem up.

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Given Solution:

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You will have two rates, r_1 the rate at which solution flows in and r_2 the rate of outflow.

The amount in the tank after time t will be 500 gal * (r_1 - r_2), and after 8 hours the tank will have just reached its 1000 gal capacity. From this you can calculate r_1 - r_2, so you'll have an expression for r_2 in terms of the still-unknown r_1.

You can how write the differential equation for dQ/dt in terms of r_1 and r_2, then use the previous information to express r_2 in terms of r_1. This will leave you with an equation having only one unknown.

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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:

I know that the problem would look similar to

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

Could be worked like:

= 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.

but im unsure if this would work.

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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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As time goes on, the concentration in both would increase, until they both reached the concentration of the amount being pumped in,

only if the rate comming into the first tank was the same as the rate going into the second. If not then one tank would reach this before the other.

As far as setting up a problem though, I am unsure. These concentration problems confuse me and waste alot more of my time thinking about them than is necissary to solve them.

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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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in this scenario, the first tank would reach the concentration of what is being pumped into it first, while the second tank would take twice as long.

As far as setting up a problem, I don't know how.

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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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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:

I am a little unsure of this problem, but I believe that it uses the equation:

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

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

integration yeilds:

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

T(o) = 190

190 = 80 + C

C = 110

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

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

But not sure, T ' (0) could be equal to 0.25

making k = 0.25/-110

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Given Solution:

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Good. Compare with the following:

Let T_diff be the difference between room temperature and the temperature of the soup.

From the initial information we know that when T_diff = 110 F the temperature changes at .05 F / min.

Assuming rate dT/dt to be proportional to T_diff we get

dT / dt = (.05 / 110) / min * T_diff, or suppressing units

dT / dt = (.05 / 110) * (T - T_room).

If T_room is constant this equation is easily solved by separating variables, yielding a familiar exponential solution.

However in this case T_room is changing. We easily enough see that

T_room = 80 F - .025 F / min * t,

where t is clock time.

Again suppressing units we have

dT/dt = (.05 / 110) * ( T - 80 - .025 t).

This is a nonhomogeneous linear differential equation, which can be solved in the usual manner.

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Query_04

@& It's misleading to call the rate r(t), since the rate is constant. Expressing r as r(t) makes it appear that r(t) needs to be integrated on the right-hand side. Of course r(t) can be a constant function, but it's best to leave of the implied t dependence in this case. *@

@& Good solution. *@

@& Check my notes. I suggest using a question form to submit attempts to the second and third problems, according to my notes. *@

@&
It's misleading to call the rate r(t), since the rate is constant.

Expressing r as r(t) makes it appear that r(t) needs to be integrated on the right-hand side.

Of course r(t) can be a constant function, but it's best to leave of the implied t dependence in this case.
*@

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Good solution.
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Check my notes. I suggest using a question form to submit attempts to the second and third problems, according to my notes.
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