austinL
  • austinL
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200L of dye solution with a concentration of 1 g/L. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 L/min, the well-stirred solution flowing out at the same rate. Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
austinL
  • austinL
@Bella♥ Take a crack at it if you like :)
austinL
  • austinL
Ok, I am gonna try and talk myself through it on here.
austinL
  • austinL
\(\large{\dfrac{dQ}{dt}=rate~in-rate~out}\)

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anonymous
  • anonymous
Lets see hmm....
austinL
  • austinL
The concentration of the liquid entering the tank is \(\dfrac{0g}{liter}\) at a rate of \(\dfrac{2liters}{min}\)..... so it goes in at \(\dfrac{0g}{min}\).
austinL
  • austinL
I am properly confused by this problem.
austinL
  • austinL
@terenzreignz If you have any comments, feel very very free to share :P
anonymous
  • anonymous
Mee too! LAME!!! okay okay lets see if I can work thi out on paper hold upp!
terenzreignz
  • terenzreignz
I was hoping you wouldn't say that... hang on, let's see if I can somehow forget that it's a Friday xD
anonymous
  • anonymous
200L * 1g/L = 200g of dye at the beginning. 2L of the solution are flowing out and 2L of fresh water is flowing in, so the tank is losing (1g/L)(-2L/min) = -2g/min of dye. 1% of 200g = 2g, so to reach that dilution the tank has lost 198g of dye. This will take (1min/2g)*(198g) = 99 min = 1 hour 39 minutes. I think is how u do this! Hope I am right ahah I am not sure!
terenzreignz
  • terenzreignz
this fails to take into account the 'well-stirred' -ness of the solution... It only works if the drains drain a solution with 1g/L concentration every time, which isn't true...
austinL
  • austinL
Unfortunately no, that isn't how you are supposed to do it I believe. The answer in the back of my book is \(\approx460.5~min\)
anonymous
  • anonymous
Lame! Sorry than :/
austinL
  • austinL
Thank you for trying :)
anonymous
  • anonymous
¯t = −100 ln(1/100) m = 100 ln 100 m. If you use a calculator, you see that is is about 460.51701859880916 m, or a bit less than 8 hours.
anonymous
  • anonymous
The volume of liquid in the tank is always 200 L. Let Q(t) be the amount of die in the tank, measured in grams. The concentration of die in the tank at any given time is given by Q(t)/200. We have Q(0) = 200 L × 1 g/L = 200 g. The rate of change of Q(t) is given by dQ(t) dt = rate of die going into tank − rate of dye leaving tank = 0 g/L × 2 L/m − Q(t) 200 g/L × 2 L/min = − 1 100 Q(t) g/m Solving for Q(t) gives Q(t) = 200e−t/100 g. There will be 2 g of die in the tank when it is 1% of the original value. So we want to find the time ¯t when 2 g = 200 e−¯t/100 g, or The answer up above :)
anonymous
  • anonymous
I think you will find this answer is right :P
austinL
  • austinL
Don't mean to look the proverbial "gift horse" in the mouth or anything, but you should really learn to use \(\LaTeX\). It makes your responses much, MUCH easier to decipher. As is, I am having some problems following your work in the above response.
austinL
  • austinL
@terenzreignz I am gonna link in help that I received on another site. If you could help explain a later part in the answer of ChrisLT521 it would be appreciated. http://mathhelpboards.com/differential-equations-17/modeling-first-order-equation-6296.html#post28617
terenzreignz
  • terenzreignz
last part... starts from where?
austinL
  • austinL
\(\text{"Once you solve this simple ODE, you're then left with finding t such that x(t)=.01x(0)"}\) This is confusing me and I feel it shouldn't be.
austinL
  • austinL
Once you solve this simple ODE, you're then left with finding t such that x(t)=.01x(0
terenzreignz
  • terenzreignz
HAVE you solved the ODE?
austinL
  • austinL
\(\dfrac{dx}{dt}=-\dfrac{x}{100}\)? I am sorry, that is what I meant, how would I go about that? I feel really, really dumb right now.
terenzreignz
  • terenzreignz
Probably because it's Friday...rearrange... \[\Large \frac{dx}x = -\frac{dt}{100}\] Integrate both sides.
austinL
  • austinL
\(\int\dfrac{1}{x}dx = \int -\dfrac{1}{100}dt\) Right?
terenzreignz
  • terenzreignz
Yup.
austinL
  • austinL
\(\ln|x|=-\dfrac{t}{100}+C\)?
terenzreignz
  • terenzreignz
I prefer it like this: \[\Large \ln(x) = -\frac{t}{100}+ \ln(C)\] ln(C) is just as arbitrary a constant, and it makes this easier: Raise e to both sides of the equation, you get \[\Huge x = Ce^{-\frac{t}{100}}\]
terenzreignz
  • terenzreignz
Now, you need to find C.
terenzreignz
  • terenzreignz
ideas? :)
austinL
  • austinL
Not really unfortunately. I am not sure where the "Initial Value" comes in.... God I feel like a dunce right now.
terenzreignz
  • terenzreignz
Well, you do know what x should be when t = 0. Before mixing in any fresh water, the concentration was 1g/L and there were 200L
austinL
  • austinL
200g yes?
terenzreignz
  • terenzreignz
Yes.... so x(0) = 200. Work it :D
austinL
  • austinL
derp...
terenzreignz
  • terenzreignz
So C = 200. \[\Huge x = 200 e^{-\frac{t}{100}}\]
terenzreignz
  • terenzreignz
Now, for the concentration to be 1% of its original concentration (200), x has to be equal to 1% of 200, which is 2. \[\Huge \color{blue}2 = 200e^{-\frac{t}{100}}\]
austinL
  • austinL
I got it!!!! YEEEEEESSS!
terenzreignz
  • terenzreignz
Awesome ^_^
austinL
  • austinL
I hope every problem isn't this hard on this assignment.

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