Query 04 redo

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course Mth 279

11:52 pm 6/26

query 04#$&*

course Mth 279

6/20 10:15 pm

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:

c_i = .03

V(0) = 1000

Q(0) = .05

Q(8) = .035

r_o = r_i = ???

dQ/dt = r_i*c_i - r_o * Q/V(t)

y' = dQ/dt

y = Q

y' + r_o/V (t) * y = r_i*c_i

p(t) = r_o/V(t)

at t = 0, V = 1000, c_i = .03, Q = .05

This already seems like too much work.

So I think I can do this:

dQ/dt = .035/8 = 0.0044

@&
dQ/dt will indeed average .0044, but dQ/dt is not constant so this value doesn't help much.
*@

.0044 = r_i * .03 - r_o * .05/1000

r_o = r_i = r

.0044 = .03r - .00005r

r = .1469 gallons/hour

You've written the correct symbolic equation

dQ/dt = r_i*c_i - r_o * Q/V(t)

The tank is full so the rate of outflow is equal to the rate of inflow, and the volume is constant at 1000 gallons.

Plug in the numbers. You get

dQ/dt = 5 * .03 - 5 * Q / 1000, or more simply

dQ/dt = .15 - .005 Q.

In this form I believe you'll be able to solve the equation without much trouble.

There is of course a little more to the problem, but setting up and solving the equation is a very good start.

@&
You've written the correct symbolic equation

dQ/dt = r_ic_i - r_o Q/V(t)

The tank is full so the rate of outflow is equal to the rate of inflow, and the volume is constant at 1000 gallons.

Plug in the numbers and let r stand for the rate of inflow (which is equal to the rate of outflow). You get

dQ/dt = r * .03 - r * Q / 1000, or more simply

dQ/dt = r ( .03 - Q/ / 1000 ).

This equation is separable.

Q is the amout of salt in the tank, which is initially 5% of 1000 gallons, or 50 gallons. So

Q(0) = 50.

After 8 hours the amount of salt should be 3.5% of 1000 gallons, or 35 gallons, so

Q(8) = 35.

*@

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Where did you get 5 from? Aren't we solving for r, the flow rate???

dQ/dt + .005 Q = .15

p(t) = .005

u = e^(.005t)

e^(.005t) Q' + .005e^(.005t) Q = .15e^(.005t)

integrate

e^(.005t) Q = .15/.005 e^(.005t) + C

Q = 30 + Ce^(.005t)

@&
This is a good solution to that equation, except that I believe the solution would be 30 + C e^(-.005 t), since C / e(^.005 t) = C e^(-.005 t).

You would have to use the initial condition on Q to evaluate C.
*@

From here, how do I solve for flow rate? Doesn't Q represent the amount of salt/solution???

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I have never been good at these tank problems.

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Self-critique rating:

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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:

V(0) = 500 gallons

Q(0) = .05

Q(8) = .035

c_i = ?

c_o = ?

r_i = ?

r_o = ?

lots of unknowns.

but, dQ/dt is a rate. We have the start points and end points. We can average them like in the assignment with the lifting weight guys doing push ups.

but we have gallons, hours and amount of salt. we need gallons per hour. we don't need salt?

This is very tricky. We don't know when it starts to overflow, so we can't just take the average of the two points, because if we did we would be assuming the inflow stops when the tank is full. But that's not what's happening.

Unlike the last one, r_i does not equal r_o because until the tank is full, there is no outflow.

I'm not sure how to do this problem.

dQ/dt = r_i*c_i - r_o * Q/V(t)

y' + r_o Q/V(t) = r_i*c_i

But I'm not sure where to go from here

This actually breaks into two phases, each with its own equation.

In the first phase there is no outflow, because the tank is still filling. This phase ends when the tank becomes full.

In the second phase there is an outflow and the volume remains constant.

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This is what I've gathered for the two phases:

Filling phase

V(0) = 500

Q(0) = .05

r_o = 0

overflow phase

V(8) = 1000

V_0(t) = 1000 (not sure how to state that the beginning volume of this phase does not occur at t = 0, but it is equal to V(8) )

Q(8) = .035

r_i = r_o

We are still solving for the rate of flow into the tank, r.

in the first phase, r_o = 0, so we get the equation

dQ/dt = r_i c_i - 0

since they are both constant we can get

Q = r_i*c_i* t + C

plugging in initial value

.05 = r_i*c_i *0 +C

Q = r_i*c_i * t + .05

@&
The initial value of Q, which is the amount of salt, is 5% of 500, or 25. So C will be 25.
*@

in the second phase, we can also get an equation

dQ/dt = r*c_i - r Q/V

@&
c_i is the concentration of the inflowing solution, which I believe is .03.
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dQ/dt + r/V Q = r*c_i

r and V are constants in the second phase

e^(r/V t) dQ/dt + r/V e^(r/V t) Q = r* c_i * e^(r/V t)

integrate

e^(r/V t) Q = V/r * r *c_i * e^(r/V t) + C

Q = Vc_i + Ce^(-r/V t)

when t = 8

.035 = (1000) c_i + Ce^(-r/1000 (8))

two equations, 3 unknowns. I don't know how to solve the second equation in a way that will let me find C

@&
You know c_i. See my preceding note.
*@

???

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

Also just realized, what do you mean by 5% solution. does that mean in 500 gallons there is 25 units of salt???

@&
Good.
*@

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Self-critique rating:

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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:

c_i = .03

c_o = 0

V(8) = 1000

Q(8) = .035*1000 = 35

r_i = ?

r_o = ?

dQ/dt = r_i*c_i - r_o * Q/V(t)

from previous problem we get Q(0) = 25, V(0) = 500

I can find dQ

dQ would be the incremental increase in solute within the tank.

What you have here is average rate at which the amount of solute in the tank increases.

However the amount of solute will not increase at a constant rate, so this result is of limited value

The volume of liquid in the tank, on the other hand, does increase at a constant rate you can easily calculate.

(35-25)/8 hours = 1.25 units per hour

1.25 = r_i*.03 - r_o Q/V

at 8 hours

1.25 = .03r_i - 35/1000 r_o

V is not constant. The amount of solution in the tank increases at the

constant rate you can calculate, in order to find the function V(t).

The rest of your solution is based on incorrect assumptions, which I believe you'll be able to correct in a revision.

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With no overflow, it means the tank is filling from 500 to 1000

V(0) = 500

V(8) = 1000

c_i = .03

Q(0) = .05

Q(8) = .035

Looking for two different rates, r_i and r_o are not equal

dQ/dt = r_i*c_i - r_o Q/V(t)

V(t) = r_i - r_o

@&
V(t) starts at 500 gallons and increases at a constant rate to 1000 gallons in 8 hours.

The constant rate at which V(t) increases, which you can easily determine, is equal to r_i - r_o.

So you can express either r_i or r_o in terms of the other, eliminating a variable.
*@

@&
You know c_i, so this will give you an equation with only one unknown constant.
*@

I'm not sure where to go from here. I can't solve the first equation because I don't have enough information. I can't solve for volume because I don't know what either flow rate is. I could put r_i - r_o into the first equation but I still have too many unknowns. Plus that would get rid of V(t) which I have values for

???

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at 0 hours

1.25 = .03r_i - 25/500r_0

subtracting the two equations

0 = 0 + .015 r_o

r_o = 0

1.25 = .03r_i

r_i = 41.6 gallons per hour.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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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:

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.

r_o = r_i = r = ?

Q(0) = 50

V(0) = 1000

c_i = .03

Q(8) = .04V

dQ/dt = r_i*c_i - r_o * Q/V(t)

at t = 0

dQ/dt = r*.03 - r* 50/1000

dQ/dt = -.02 r

integrate with respect to t

Q(t) = -.02r t

I'm getting confused between the two tanks I think.

The first one starts empty? or does it?

Wait. Am I finding when the first tank has 4% solution or when the second tank has 4%? I don't think that's the same. ???

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I do not understand this section. The book doesn't have any examples of a two tank system. Perhaps the DVDs will be helpful? I will be redoing this assignment when I receive the DVDs. Maybe I did better at the first few problems.

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Self-critique rating:

Don't worry too much about this problem until you've solved the first three. If you can get them, then it might be worth your time and effort to work on this one, and at that time I'll have a better idea of your thinking and will be better able to focus my advice.

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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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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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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:

The limiting value of Q(t) is c_i*V

.03*1000 = 30.0 units of salt

If the mixture is removing more salt than the tank is being filled with, the least amount the tank can contain is the amount that is flowing in.

In the second case, the concentration inflow is oscillating, similar to a sine function. Sine does not approach a limit because it is constantly changing between +A and -A. Therefore, there is no limiting value.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

Not at all sure about the first, but I think I got the second

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Self-critique rating:

Good.

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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:

Theta' = k (S-theta)

.5 = k(80-190)

k = -.004545

.25 = -.004545 (80 - theta(t))

theta = 135 ???

That's not a function though.

theta(t) = S + theta_0e^(-kt)

theta(t) = 80 + 190 e^(.004545t)

You've made a decent attempt.

I'll let T stand for temperature of the soup, T_room for the temperature of the room and t for clock time.

dT/dt = k ( T - T_room), as I believe you understand.

If T_room was constant, you would already have your equation.

However T_room changes, so T_room is itself a function of clock time t.

To get your differential equation, all you need to do is find that room temperature function and plug it in for T_room.

Then of course you'll need to try to solve the equation.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I'm sorry. I know I'm going to have to do this again

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Self-critique rating:

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Question:

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Self-critique (if necessary):

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Self-critique rating:

You're doing some good thinking and with the help of my notes I believe you'll be able to make good additional progress.

Self-critique (if necessary):

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Self-critique rating:

You're doing some good thinking and with the help of my notes I believe you'll be able to make good additional progress.

Self-critique (if necessary):

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Self-critique rating:

#*&!

@&
You've made good progress. Check my notes and see if you can come up with solutions.

In any case you have the basic idea, so don't let yourself get too bogged down on this.
*@

Query 04 redo

@& dQ/dt will indeed average .0044, but dQ/dt is not constant so this value doesn't help much. *@

@& This is a good solution to that equation, except that I believe the solution would be 30 + C e^(-.005 t), since C / e(^.005 t) = C e^(-.005 t). You would have to use the initial condition on Q to evaluate C. *@

@& The initial value of Q, which is the amount of salt, is 5% of 500, or 25. So C will be 25. *@

@& c_i is the concentration of the inflowing solution, which I believe is .03. *@

@& You know c_i. See my preceding note. *@

@& Good. *@

@& V(t) starts at 500 gallons and increases at a constant rate to 1000 gallons in 8 hours. The constant rate at which V(t) increases, which you can easily determine, is equal to r_i - r_o. So you can express either r_i or r_o in terms of the other, eliminating a variable. *@

@& You know c_i, so this will give you an equation with only one unknown constant. *@

@& You've made good progress. Check my notes and see if you can come up with solutions. In any case you have the basic idea, so don't let yourself get too bogged down on this. *@

@&
dQ/dt will indeed average .0044, but dQ/dt is not constant so this value doesn't help much.
*@

@&
This is a good solution to that equation, except that I believe the solution would be 30 + C e^(-.005 t), since C / e(^.005 t) = C e^(-.005 t).

You would have to use the initial condition on Q to evaluate C.
*@

@&
The initial value of Q, which is the amount of salt, is 5% of 500, or 25. So C will be 25.
*@

@&
c_i is the concentration of the inflowing solution, which I believe is .03.
*@

@&
You know c_i. See my preceding note.
*@

@&
Good.
*@

@&
V(t) starts at 500 gallons and increases at a constant rate to 1000 gallons in 8 hours.

The constant rate at which V(t) increases, which you can easily determine, is equal to r_i - r_o.

So you can express either r_i or r_o in terms of the other, eliminating a variable.
*@

@&
You know c_i, so this will give you an equation with only one unknown constant.
*@

@&
You've made good progress. Check my notes and see if you can come up with solutions.

In any case you have the basic idea, so don't let yourself get too bogged down on this.
*@