#$&* course MTH 279 query 042.5.
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: 2. Solve the preceding question if the tank contains 500 gallons of 5% solution, and the goal is to achieve 1000 gallons of 3.5% solution at the end of 8 hours. Assume that no solution is removed from the tank until it is full, and that once the tank is full, the resulting overflow is well-mixed. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Well, we know that there will be 0.035*1000 = 35 lb of salt in the solution at 8 hrs. We also know that there is 0.05*500 = 25 lb of salt in the solution at the start. These will be our initial conditions. The amount of water at any time in the tank begins at 500 gal and increases over time by the inflow rate, up to the maximum of 1000gal. Q’ = 0.03r - Qr/(500+rt) Q’ + Qr/(500+rt) = 0.03r µ = e^(ln(rt+500)) = rt+500 Q’(rt+500) + Qr(rt+500)/(rt+500) = 0.03r(rt+500) ʃd/dt((rt+500)Q)dt = ʃ0.03r(rt+500)dt (rt+500)Q = 0.015rt(rt+1000) + C Q = (0.015r^2t^2+15rt)/(rt+500) + C/(rt+500) Q(0) = 25 >> C = 12500 Q(t) = (0.015r^2t^2+15rt)/(rt+500) + 12500/(rt+500) Q(8) = 35 >> r ≈ 193.57 Q(t) = (0.015(193.57)^2t^2+15(193.57)t)/((193.57)t+500) + 12500/((193.57)t+500) The flow rate necessary for this situation would be about 193.57 gal/hr. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: r = rate out s = rate in Net flow required: 1000 = 500 + 8r >> r = net flow rate of 62.5 gal/hr in order to reach exactly 1000 gallons in 8 hours with no overflow. s = 62.5+r Q’ = 0.03s - Qr/(500+62.5t). Q’ + r/(500+62.5t)Q = 0.03s. µ = e^(ʃr/(500+62.5t)dt) = e^(0.016ln(t+8)r) = (t+8)^(0.016r). Q’*(t+8)^(0.016r) + (t+8)^(0.016r)*(r/(500+62.5t))*Q = 0.03(t+8)^(0.016r)*s ʃd/dt((t+8)^(0.016r)*Q) dt = ʃ0.03(t+8)^(0.016r)*s dt (t+8)^(0.016r)*Q = [1.875*s*(t+8)^(0.016r+1)]/(r+62.5) + C Q = [1.875*s*(t+8)]/(r+62.5) + C(t+8)^(-0.016r) Q(0) = 25 >> C = 10*(1.03383073625^r) Q(8) = 35 = [1.875*(62.5+r)*(t+8)]/(r+62.5) + (10*(1.03383073625^r))(t+8)^(-0.016r) >> r = 62.5 gal/hr If r = 62.5 gal/hr, then s = 125 gal/hr. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: 4. Under the conditions of the first problem in this section, suppose that the overflow from the first tank flows into a second identical tank, which is initially empty. At what constant rate must a 3% solution flow into that tank to achieve a tank full of 3.5% solution at the end of 8 hours, with no overflow? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first problem in this section says that the flow into tank A is equal to the flow out of tank A, so how would there be overflow into tank B? I’ll assume that the outflow from tank A goes straight into tank B. The net flow into tank B, in order to reach 1000gal in 8 hours must be 125 gal/hr. If the flow into and out of Tank A is 250ln(2) [which we found before], then the flow into tank B must be 250ln(2) gal/hr and the flow out of tank B to maintain the 1000gal must be (250ln(2) - 125)gal/hr. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I really don’t understand this problem very well. As I noted, if the flow into and out of tank A are equal, how can we have overflow into tank B? Is the 3.5% solution at the end of the 8 hours supposed to be in both tanks, or only in the first, or only in the second? I just took a stab in the dark and did a net flow analysis, which made sense to me, given the constraints of the first problem; however, I’m not positive that I went in the direction you had intended with this question. ------------------------------------------------ Self-critique rating: 3 ********************************************* 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 limiting value? Justify your answer. **** #$&* Now suppose that the flow from the first tank changes hour by hour, alternately remaining at a set constant rate for one hour, and dropping to half this rate for the next hour before returning to the original rate to begin the two-hour cycle all over again. Will the concentration in the second tank approach a limiting value as time goes on? If so what is the limiting value? Justify your answer. **** #$&* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The concentration in both tanks will tend towards 3% as time goes on, which is the limiting value. In the second case, the limiting value will still be 3%. As time goes on, the salt concentration will come closer and closer to the input concentration, as will the concentration in tank A. The limiting concentration shouldn’t be dependent on the flow rate. As time goes on, the concentration should be limited by the concentration that is being input into the system. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: 6. When heated to a temperature of 190 Fahrenheit a tub of soup, placed in a room at constant temperature 80 Fahrenheit, is observed to cool at an initial rate of 0.5 Fahrenheit / minute. If at the instant the tub is taken from the oven the room temperature begins to fall at a constant rate of 0.25 Fahrenheit / minute, what temperature function T(t) governs its temperature? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T_ambient(t) = 80 - 0.25t T’ = -k(T-T_ambient) T’ = -k(T - (80 - 0.25t)) T’ = -kT + 80k - 0.25kt T’ + kT = 80k - 0.25kt µ = e^(ʃkdt) = e^(kt) e^(kt)T’ + e^(kt)kT = e^(kt)80k - 0.25kte^(kt) ʃd/dt(e^(kt)T)dt = ʃ(e^(kt)80k - 0.25kte^(kt))dt e^(kt)T = -0.25te^(kt) + 0.25e^(kt)/k + 80e^(kt) + C T = -0.25k + 0.25/k + 80 + Ce^(-kt) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ________________________________________ #$&*