Revised Query 4

#$&*

course MTH 279

I was able to correct and find a solution for all but one of the problems, I think. 2/21 530 pm

Query 4#$&*

course MTH 279

2/4 5 pmI will be completely honest with you; I had no idea how to even tackle this section. Modeling has always been a struggle for me. I wish I could tell you what I didn't understand in each question, but I’m at a complete loss in almost all of these scenarios of how to even start out. In general, I don’t know how to apply the given information to the equation model. I tried reading the book, but it’s hard for me to get anything useful out of it. I also watched videos from the DVD's. If I had more time, I'd probably look through the class notes a bit. If you have any tips as to how I could organize the information in problems like this or maybe a method for setting up the problems, I would greatly appreciate it. I may be better able to understand it.

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:

93.75 gal/hr

Q’ = r * .03 - r * (50/1000)

-15/8 = -.02r

r = 93.75

The concentration in the tank does not remain at 5%; that is, the quantity of salt in the tank does not remain at 50 gallons.

Q ' is the rate of change of the amount of salt in the tank. The amount of salt in the tank must therefore be Q.

If you replace that static 50 with the dynamic Q, you'll get an equation you can easily solve.

50 is the initial number of gallons of salt in the tank, which gives you the initial condition Q(0) = 50.

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r = -125ln(.25)

@&
Which is around 170, indicating that a flow rate of about 170 gal / hr is necessary.

In the 8 hours this means that somewhat over 1000 gallons flows into the tank as somewhat over 1000 gallons of mixed solution flows out.

1000 gallons of 3% mixed with 1000 gallons of 5% would yield 1000 gallons of 4% solution. However the concentration of the outflow solution being higher than the concentration of the inflow, salt concentration will fall more rapidly than if the two concentrations were just mixed.

So the solution makes sense.
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Starting with : Q' = .03r -r(Q/1000)

Solving as a 1st order linear nonhomogeneous: Q = 30+Ce^[(-r/1000)t]

Imposing initial condition Q(0) = 50, I found C = 20, so

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

Solved r by using Q(8) = 35

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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:

No difference.

There is a difference. The tank only contains 500 gallons of solution at the initial instant, so no salt leaves the tank until it is full.

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After I sent you the question form about this problem, I realized I had written it down sloppy and that's why I thought it said 500r instead of 500/r. Anyways,

r= -125ln(.5) + 125/2

I solved Q' = .03r-r(Q/1000) as 1st order nonhomogenous and imposed initial condition Q(0) = 40 to arrive at the equation:

Q = 30+10e^[(-r/100)t]

And knowing that the time at which the tank is full is (500/r) from the fact that Q = .03rt and you've added 500*.03 = 15 lbs of salt (15/(.03r) = t)

Lets you solve the equation for r when you plug in the time remaining, which would be 8 - (500/r)

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@&
Good.

Here's a solution for your reference:

When the tank first reaches 1000 gallons it will contain equal amounts from the 3% and the 5% solutions, so the concentration will be 4%. If r is the rate of flow into the container, then the time required to achieve this state is the time for the 500 gallons of solution to flow in, which is 500 / r.

At that time the amount of solution flowing out is equal to the amount flowing in. The concentration of the incoming solution is 3%, and if q(t) is the amount of salt in the container the concentration of the exiting solution is q / (1000 gallons). Our equation will be

q ' = .03 r - (q / 1000) * r,

which we rearrange to the standard form

q ' + (r / 1000) * q = .03 r.

The solution of this equation is found by the usual method to be

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

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.

The q function is now

q = 30 + 17 e^(-r/1000 * t),

and we apply the given condition that q(8) / 1000 = 3.5%, which becomes q(8) = 35
:
35 = 30 + 17 e^(-r/1000 * 8).

Solving for r we get

-r/1000 * 8 = ln(5/17)

so that

r = 125 ln(5/17) = 150.

This means that the tank will reach 1000 gallons after 500 / 150 hours = 3.7 hours

with over 4 hours remaining, during which over 600 gallons of additional 3% solution will flow into the full tank.

To check the solution, we use the function

q(t) = 30 + 17 e^(-150 / 1000 * t) = 30 + 17 e^(-.15 t), t > 500/150 hours

which uses some approximations, and which applies only to t values greater than the 500 / 150 hours required to first finish filling the tank.

q(500/150) should therefore be 40, since at the time the tank becomes full it contains 40 gallons of salt. Evaluating we get

q(500/150) = 30 + 17 e^(-.15 * 500/150) = 40.3.

q(8) should be 35, and we get q(8) = 30 + 17 e^(-.15 * 8) = 35.1.

Within the accuracy of our approximations, these results agree with the conditions of the problem.

*@

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I’m sure there is a difference, but I don’t know how having the 500 gallons initially and 1000 gallons at the end would affect the solution to the problem, or how you would factor that into the equation.

Figure out how much salt will be in the tank when it reaches 1000 gallons, and how many of the 8 hours you then have left. From that point on the equation is very similar to the preceding, but with slightly different conditions.

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Self-critique rating: 3

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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:

41.6 gal/hr

Q’ = r_i * c_i and no outflow makes r_o * c_o be zero

10/8 = r * .03

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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 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?

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Your solution:

No attempt

The first equation gives you the Q(t) function for the first tank. Knowing Q(t) and the outflow rate you can find the rate at which salt enters the second tank, as a function of t.

The second tank is large, so it simply accumulates the inflow from the first tank, and from the 3% solution flowing in.

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This is the one I couldn't quite get. I got that

Q' = r_1*c_1 + r_2*c_2

I know that r_1 is constant; it is the outflow rate from the first tank. I know that c_1 would be Q/V, both of which are changing wrt time. I further know that Q is the Q from the first problem and V would be (I think) (t^2/2)(r_1 + r_2). I am solving for r_2 and I know that c_2 is constant at .03.

So I arrived at the equation

Q' = r_1[(30 + 20e^(-r_1/1000t) )/( (t^2/2)*(r_1 + r_2) ) ]

Which is a huge mess. Which makes me think it's terribly wrong, because A: I got rid of my Q and I really don't think attempting to integrate that thing would be the appropriate thing to do. B: I wouldn't know how to tell what the final volume of solution and therefore amount of salt in the tank would be...since I don't know both the rates.

@&
You do know the rate at which solution flows into that second tank from the first, which is the rate (around 150 gal/hr) found in the earlier problem.

Here's much of the solution:

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 )

*@

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

In this problem, I’m assuming the c_i is changing constantly as it is added to the second tub, but I don’t know how to factor that into the equation.

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Self-critique rating: 3

Revised Query 4

@& You've done a good job of wrestling with these tough problems. I've inserted a couple of full or partial solutions for your reference. *@

@&
You've done a good job of wrestling with these tough problems.

I've inserted a couple of full or partial solutions for your reference.
*@