Query 04

#$&*

course MTH 279

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

capacity = 1000 gal

concentration in = c_i = 0.03 lb/gal

concentration out = c_o = 0.035 lb/gal

flow in = flow out = r lb/gal

Q(t) = amount of salt at time t (in hrs)

Q(0) = 0.05*1000 = 50 lb

Q(8) = 0.035*1000 = 35 lb

Q’ = c_i*r - c_o*r = c_i*r - r*Q(t)/1000

Q’ + (r/1000)Q = 0.03r

µ = e^(rt/1000)

Q’e^(rt/1000) + e^(rt/1000)(r/1000)Q = 0.03re^(rt/1000)

ʃd/dt(e^(rt/1000)Q)dt = ʃ0.03re^(rt/1000)dt

e^(rt/1000)Q = 30e^(rt/1000) + C

Q = 30 + Ce^(-rt/1000)

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

C = 20

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

Q(8) = 35 = 30 + 20e^(-8r/1000)

r = 250ln(2)

Q(t) = 30 + 20e^(-ln(2)t/4)

The flow rate required would be 250ln(2) gal/hr or about 173.29 gal/hr.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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 #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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 #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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 #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

@&

The statement of the problem was supposed to have been changed as follows, though the change apparently didn't make it to the posted version of the problem:

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?

The solution to this problem:

Recall that the solution to the first problem was

q(t) = 30 + 20 e^(-r/1000 * t), with r = 180 gal / hr (approx.)

So the concentration in the first tank is

q(t) / 1000 = (30 + 20 e^(-.18 * t )) / 1000.

Water flows out of the first tank at 180 gal / hr (the same as the rate of flow into that tank), so the rate at which salt flows from the first tank into the second is

180 * q(t) / 1000 = 5.4 + 3.6 e^(-.18 * t).

Let r_2 be the rate at which 3% solution flows into the second tank, and let q_2(t) be the amount of salt in that tank.

We assume that the second tank is large enough to contain all the solution that flows into it for 8 hours.

The rate at which salt enters the tank in the 3% solution is .03 * r_2, and the rate at which it enters in the first solution is our previous expression 5.4 + 3.6 e^(-.18 t). Thus

q_2 ' = .03 r_2 + 5.4 + 3.6 e^(-.18 t).

This amount doesn't depend on q_2 (note that it would if water was allowed to flow from the second tank).

The equation is easily separable

dq_2 = (.03 r_2 + 5.4 + 3.6 e^(-.18 t)) dt,

leading to solution

q_2(t) = (.03 r_2 + 5.4) t + 20 e^(-.18 t) + c.

The tank is presumably empty at t = 0 so c = -20

The rate at which solution increases in the tank is just 180 gal / hr + r_2, the sum of the two rates of flow into the tank, so at clock time t the volume of solution in the tank is (180 gal / hr + r_2) * t. The concentration is therefore

concentration = ( (.03 r_2 + 5.4) t + 20 e^(-.18 t) - 20 ) / ((180 + r_2) * t).

The concentration is .035 when t = 8 so

.035 = ( (.03 r_2 + 5.4) * 8 + 20 e^(-.18 * 8 ) - 20 ) / ((180 + r_2) * 8 )

This equation has no positive solution, meaning that if the equation is correct, it is not possible to achieve a 3.5% solution in 8 hours.

However a 3.5% solution will eventually be achieved. It should be obvious that when 3% solution is being added to both tanks, the concentrations in both will approach 3% as a limit. This is consistent with the concentration function

( (.03 r_2 + 5.4) t + 20 e^(-.18 t) - 20 ) / ((180 + r_2) * t),

which as t => infinity approaches the limit .03.

*@

------------------------------------------------

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 #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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 #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: OK"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

__________________

&#Good responses. See my notes and let me know if you have questions. &#