Mixture

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March 30th 2011, 10:20 PM
Naples

Mixture

A tank holds 300 gallons of a brine solution, the rate at which brine is pumped into the tank is 3 gal/min with a concentration of salt of 2 lb/gal and the well stirred solution is pumped out at a rate of 2 gal/in. It stands to reason that since brine is accumulating in the tank at the rate of 1 gal/in, any finite tank must eventually overflow. Suppose that the tank has an open top and has a total capacity of 400 gallons and A(0)=50.

(a) When will the tank overflow?
(b) What will be the number of pounds in the tank at the instant it overflows?
(c) Assume that although the tank is overflowing, brine solution continues to be pumped in at a rate of 3 gal/min and the solution continues to be pumped out at a rate of 2 gal/min. Devise a method for determining the number of pounds of salt in the tank at t = 150 minutes.
(d) Determine the number of pounds of salt in the tank as t approaches infinity. Does your answer agree with your intuition?
(e) Use a graphing utility to plot the graph of A(t) on the interval [0,500)

First I set up a DE using the information provided
$DA/dt = 6 - (2A/(300 + t))$
$DA/dt + (2A/(300 + t)) = 6$
Used an integrating factor of $(300 + t)^2$
Multiplied by IF and integrated and cleaned it up to get:
$A= 600 + 2t + C/(300+t)^2$ used initial value A(0)=0 to solve for C
$A= 600 + 2t - (4.95x10^7)/(300+t)^2$

(a) tank overflows when 300+t = 400 so when t=100 minutes

(b) A(100)=490 lbs

(c) This is where I get iffy, I'm pretty sure you can't just use A(150) because you need to take into account that the tank is overflowing, I just don't know how to.

(d) I did the limit as t approaches infinity of A(t) so:
$600 + 2(inf) - 0 = infinity$
obviously since the container has a finite capacity, it's impossible for the amount of salt to be infinity, the problem is that the equation doesn't account for the container's limited capacity

(e) I plugged my equation A(t) into my calculator and basically just got a straight line going diagonally across the screen

Thanks in advance
March 31st 2011, 01:31 AM
Ackbeet

In my copy of Zill, 4th edition, this is problem 3.1.26, exactly. If that's also true for you, I think maybe your answer to (b) is incorrect. Zill did an example in the text (it's example 5 in my copy), where he gives you the solution to the problem before it starts overflowing. The solution he gives is

$A(t)=600+2t-(4.95\times 10^{7})(300+t)^{-2}.$

Double-check what you get when you plug in $t=100$ into this equation.

As for (c), you always have this:

$\dot{A}(t)=\text{rate of salt entering}-\text{rate of salt leaving}.$

How much salt is entering? How much is leaving? I don't think it's wise to continue from here, until you have the correct DE and have solved it correctly.
March 31st 2011, 04:24 AM
Naples

Quote:

Originally Posted by Ackbeet

In my copy of Zill, 4th edition, this is problem 3.1.26, exactly. If that's also true for you, I think maybe your answer to (b) is incorrect. Zill did an example in the text (it's example 5 in my copy), where he gives you the solution to the problem before it starts overflowing. The solution he gives is

$A(t)=600+2t-(4.95\times 10^{7})(300+t)^{-2}.$

Double-check what you get when you plug in $t=100$ into this equation.

As for (c), you always have this:

$\dot{A}(t)=\text{rate of salt entering}-\text{rate of salt leaving}.$

How much salt is entering? How much is leaving? I don't think it's wise to continue from here, until you have the correct DE and have solved it correctly.

The equation you gave me for A(t) is the same one that I derived in my original post.
For (c) can I just solve for A(150) then using that equation for A(t) or does the overflowing of the container change something?
March 31st 2011, 04:39 AM
Ackbeet

Looks like I made a mistake in my working. Your figure for (b) is correct. You can't, however, just plug in 150 into that formula, because the DE changes at t = 100. The overflowing of the container definitely changes the DE. So, I go back to my previous question: how much salt is leaving, and how much is entering?
March 31st 2011, 07:57 AM
Naples

Rate in is still 6 lb/gal
Rate out would be 2a/400 then cause after it overflows the amount in the container remands at 400
So dA/dt + 2A/400 = 6
March 31st 2011, 08:00 AM
Ackbeet

The rate out would be 3A/400, because the tank is leaking. Otherwise, your new DE is correct. What's the initial condition?
March 31st 2011, 08:03 AM
Naples

Initial condition would just be the amount when tank first started to overflow so A(0)=491
March 31st 2011, 08:07 AM
Ackbeet

You could do it that way, I suppose. You're "resetting your clock" by doing that, which is probably not what the original problem had in mind. I would probably do y(100) = 491. That way, the 150 in the expression y(150) will turn out right. Doing y(100) = 491 will also make it easier to graph.