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course MTH 279
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 in necessary to dilute the solution in the tank to 3.5% in 8 hours?
dQ/dt = 0.03r- Q(t)*r/1000
= r (0.03 - Q(t)/1000)
= (r/1000) * (30 - Q(t))
= (-r/1000) * (Q(t) - 30)
dQ/(Q(t) - 30) = -r/1000 dt
ln(Q(t) - 30) = -r*t/1000 + c
Q(t) - 30 = C * e^(-r*t/1000)
Q(t) = 30 + Ce^(-r*t/1000)
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You should be aware that this can also be solved as a linear nonhomogeneous equation:
dq/dt = .03 r - (r/1000) * q so
dq/dt + (r/1000) * q = .03 r .
Using e^(r/1000 * t) as integrating factor we obtain
(e^(r/1000 * t) * q) ' = .03 r e^(r / 1000 * t),
leading to solution
q = ( 1000 * .03 e^(r/1000 * t) + C ) / e^(r/1000 * t) = 30 + C e^(-r/1000 * t).
This result is identical to your solution so far.
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Q(0) = 30 + Ce^0 = 0.05(1000) = 50
30 + C(1) = 50
C = 20
Q(t) = 30 + 20e^(-r * t/1000)
Q(8*60) = 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 = 2.888 gal/min approx.
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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.
.05 * 500 + .03 * 500 = 40
at 4% solution the tank starts to overflow
dQ/dt = 0.03r- Q(t)*r/1000
= r (0.03 - Q(t)/1000)
= (r/1000) * (30 - Q(t))
= (-r/1000) * (Q(t) - 30)
dQ/(Q(t) - 30) = -r/1000 dt
ln(Q(t) - 30) = -r*t/1000 + c
Q(t) - 30 = C * e^(-r*t/1000)
Q(t) = 30 + Ce^(-r*t/1000)
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q = 30+ Ce^(-r/1000 * t), in agreement with your solution to this point.
However it doesn't appear that you consider the time elapsed while the tank fills.
See if the following makes sense:
The time required to get the 4% solution and full tank is t = 500 / r.
The initial condition for this phase of the flow can be written as q = 40 when t = 500 / r:
40 = 30 + C e^(-r / 1000 * 500 / r) = 30 + C e^-(1/2)
so that
C = 10 e^(1/2) = 17, very approximately.
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Q(0) = 30 + Ce^0 = 0.04(1000) = 40
30 + C(1) = 40
C = 10
Q(t) = 30 + 10e^(-r * t/1000)
Q(8*60) = 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 = 1.73 gal/min approx.
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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?
dQ/dt = 0.03R1- Q(t)*R2/(500+ .1042t)
R1 - (2.88-1.73) = dQ/dt
Integrate?
##I am lost on the rest of this problem##
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Here's a start:
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).
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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?
dQ/dt = 0.03r(t) - Q(t)r(t)/1000
Could be worked like:
= r (0.03 - Q(t)/1000)
= (r/1000) * (30 - Q(t))
= (-r/1000) * (Q(t) - 30)
dQ/(Q(t) - 30) = -r/1000 dt
Integrate
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(8*60) = 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 = 1.444 gal/min
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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.
Yes, the limiting value being 3% saline solution because it keeps taking in the 3% saline solution.
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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.
Yes, there is still 3% saline being put into the mixture with the overflow being discarded, so despite the speed it is still nearing a limiting value of 3%
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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.
It will still get closer to a value of 3% because cos(t) cannot be greater than 1 , thus 3% saline is still being placed into the tank with lost overflow.
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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?
(T+(-80-.25t)) * (.5/110) = dT/dt
(T+(-80-.25t)) * (.5/110) dt = dT
Integrate
(.5/110)(Tt) - 80(.5/110)t - .25(.5/220)t^2 = T + C
multiply by 220
Tt - 80t- .125t^2 = 220 T + 220C
T(0) = 190 K = 220C
-80t - .125t^2 -190 = K
Tt + 190 = 220T
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If the room temperature remains at 80 F the solution is
dT/dt = -k ( T - 80 )
dT/(T-80) = -k dt
ln|T - 80| = -k t + C
T - 80 = A e^(-kt)
T = 80 + A e^(-kt)
T(0) = 190 so A = 110
T ' = -k * 110 e^(-kt)
T ' (0) = -.5 = -110 k e^0 so k = 5 / 110.
T = 80 + 110 * e^(-5/110 * t).
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If room temperature falls at .25 F / min, then the right-hand side becomes
-k ( T - (80 - .25 t) )
and the equation is as you indicate. Writing the equation as
dT/dt = -k ( T - (80 - .25 t) )
it becomes clear that the equation cannot be separated, so we are left to solve it as a linear nonhomogeneous equation.
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You're doing OK, but be sure you understand how to solve these equations as linear nonhomogeneous equations. This is necessary on the last problem.
Check my notes and let me know if you have questions.
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