Query 04

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course Mth 279

6/23 12

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:

Q'(t) = 0.03*r - Q(t)*r/1000

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

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

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

integrate both sides

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

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

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

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

30 + C(1) = 50

C = 20

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

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

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

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

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

ln(0.25) = -r * 480/1000

r = - ln(0.25) / 0.48

r =approx. 2.89 gal/min

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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:

At the time when the tank is full, it will contain 500 gallons of 5% solution and 500 gallons 3% solution. This is equivalent to a

1000 gal 4% solution.

Q'(t) = 0.03*r - Q(t)*r/1000

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

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

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

integrate both sides

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

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

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

Q(0) = 30 + Ce^0 = 0.04(1000) = 40

30 + C(1) = 40

C = 10

Q(t) = 30 + 10e^(-r * t/1000)

Q(480) = 30 + 10e^(-r * 480/1000) = 0.035(1000) = 35

10e^(-r * 480/1000) = 5

e^(-r * 480/1000) = 5/10

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

ln(0.5) = -r * 480/1000

r = - ln(0.5) / 0.48

r =approx. 1.44 gal/min

After solving I realized this equtation does not take into account the time to fill the tank completely. I believe the time would

be t=500/r but I am unsure how to implement this.

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At this instant the solution will have concentration 4%.

So you model the second phase with rate r, and the condition that after an additional 8 hours - 500 / r has passed, the concentration will be 3.5%.

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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:

r1 and r2 are flow in and out respectively. The amount of water in the tank at any time t is 500*(r1-r2). If r1-r2 is negative the

level is dropping and if positive it is filling up. The tank will be filled in 8 hrs or 480 min.

480 = 500*(r1-r2)

(r1-r2) = 0.96

r2 = r1 - 0.96

My best guess would be dQ/dt = (r1-(r1-0.96))Q but I don't think that is right

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If r_in is the rate at which the 3% solution flows in and r_out the rate at which mixed solution flows out, we have

q ' = .03 r_in - (q / (500 + 62.5 * t ) * r_out.

The equation can be rearranged to the form

q ' - (r_out / (500 + 62.5 * t ) ) * q = .03 r_in

The integrating factor for this equation is

e^(integral((r_out / (500 + 62.5 * t ) ) dt)

= e^(r_out / 62.5 * ln | 500 + 62.5 * t | )

= (500 + 62.5 * t) ^ (r_out / 62.5).

The equation becomes

( (500 + 62.5 * t) ^ (r_out / 62.5) q ) ' = .03 r_in * (500 + 62.5 * t) ^ (r_out / 62.5).

Integrating the right-hand side we get

.03 r_in * 62.5 / (r_out / 62.5 + 1)(500 + 62.5 * t) ^ (r_out / 62.5 + 1) + c, where c is constant

[ If this expression gives you a headache, consider that .03 r_in is a constant and the rest can be written as (a + b t) ^ c for appropriate values of a, b and c. Letting u = a + b t we end up integrating a multiple of u^c, a simple power function with antiderivative 1 / (c + 1) u^(c + 1). ]

So

q = ( .03 r_in * 62.5 / (r_out / 62.5 + 1)(500 + 62.5 * t) ^ (r_out / 62.5 + 1) + c ) / ( (500 + 62.5 * t) ^ (r_out / 62.5))

= .03 r_in * 62.5 / (r_out / 62.5 + 1) (500 + 62.5 * t) + c / (500 + 62.5 * t) ^ (r_out / 62.5).

q(0) = 25

(this is 5% of 500 gal)

q(8) = 35

(3.5% of 1000 gal)

q_in + q_out = 62.5

provide us with three simultaneous equations from which we can find q_in, q_out and c.

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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:

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

integrate both sides

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

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

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

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

30 + C(1) = 50

C = 20

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

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

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

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

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

ln(0.5) = -r * 480/1000

r = - ln(0.5) / 0.48

r = approx. 1.44 gal/min

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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 tank would eventually reach the concentration of the first tank, but never drop below it.

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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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If overflow is still removed then it will eventually reach the same concentration and limiting value as the first tank, it will

simply take longer.

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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:

If the rate into and out of the tank are always equal, and the flow out is well mixed, then the concentration will still drop until

it reaches the concentration of the first tank.

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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:

at T_diff = 110 temp change is -.05 F/min

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

T_room = 80 - 0.025*t

dT/dt = (0.05 / 110) * (T - 80 -.025*t)

dT/dt - (0.05 / 110)*T = -(0.05/110*.025)*t - (80*0.05/110)

u(t) = e^(-0.05/110*t)

multiply through by u(t) and integrate

T/e^(.05/110*t) = (0.025*t + 135)/e^(.05/110*t) + C

divide by u(t)

T(t)= 0.025*t + Ce^(.05/110*t) + 135

initial condition T(0) = 190

190 = 0.025*(0) + Ce^(.05/110*(0)) + 135

190 = 0 + Ce^(0) + 135

C = 55

T(t) = 0.025*t + 55*e^(0.05/110*t) +135

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Question: "

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Question: "

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&#This looks good. See my notes. Let me know if you have any questions. &#