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MTH 279
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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I'm working on the problem set for section 2-5 and Qu 2 is bothering me. The next box is the complete question. I took out my response to Qu 1 but I have worked it without a problem so it's just for reference.
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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 in necessary to dilute the solution in the tank to 3.5% in 8 hours?
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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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I know that r_o = 0 until V= 1000 so dQ/dt= r_i*c_i then Q=rct+Q_0. I solve this second one for r and get 41.667 gal/hr. However, at this rate it takes 12 hours to fill the tank and the Q gets larger. In order to fill the tank in 8 hours exactly the rate in needs to be 62.5 gal/hr but that makes it too concentrated. I'm having trouble reconciling these values. I think about having a fast flow rate that will fill quickly but then run out quickly to lower the Q back down since it will be over the desired amount. I don't have anymore time to think about it tonight so any help would be good. I also have a suspicion I'm not correctly figuring the salt content/volume exactly right.
@& See if you agree with the discussion I've inserted below.*@
@& In the first place we want to be sure it is possible to get 1000 gallons of 3.5% solution from the process.
If we just added 500 gallons of 3% solution to the existing 500 gallons of 5% solution, we would have 1000 gallons of 4% solution. So if we continue to add 3% solution, the concentration will continue to decrease, approaching a limiting value of 3%.
So we will reach the 5% level.
If r_in is the inflow rate, then the time until the tank is full will be 500 gal / r_in. At this point we have 1000 gallons of 4% solution, and the time we have left is 8 hrs - 500 gal / r_in.
We can set up an equation for the concentration as a function of time, starting from this instant.
If q is the quantity of salt in the tank at time t, then
dq / dt = .03 r_in - q / 1000 * r_out
and since r_out = r_in we have
dq/dt = (.03 - q(t) / 1000) * r_in.
Our solution for q(t) will contain r_in as one of its parameter, and will also contain an integration constant.
At t = 0 we have q = .04 * 1000 = 40, and at t = 8 - 500 / r_in, q must be .035 * 1000 = 35.
If we set q(8 - 500 / r_in) equal to 35, and q(0) = 40, we can find our integration constant as well as r_in.*@