Question

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.

Answer 1

@Bella♥
Take a crack at it if you like :)

Answer 2

Ok, I am gonna try and talk myself through it on here.

Answer 3

\(\large{\dfrac{dQ}{dt}=rate~in-rate~out}\)

Answer 4

Lets see hmm....

Answer 5

The concentration of the liquid entering the tank is \(\dfrac{0g}{liter}\) at a rate of \(\dfrac{2liters}{min}\)..... so it goes in at \(\dfrac{0g}{min}\).

Answer 6

I am properly confused by this problem.

Answer 7

@terenzreignz If you have any comments, feel very very free to share :P

Answer 8

Mee too! LAME!!! okay okay lets see if I can work thi out on paper hold upp!

Answer 9

I was hoping you wouldn't say that... hang on, let's see if I can somehow forget that it's a Friday xD

Answer 10

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!

Answer 11

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...

Answer 12

Unfortunately no, that isn't how you are supposed to do it I believe. The answer in the back of my book is \(\approx460.5~min\)

Answer 13

Lame! Sorry than :/

Answer 14

Thank you for trying :)

Answer 15

¯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.

Answer 16

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 :)

Answer 17

I think you will find this answer is right :P

Answer 18

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.

Answer 19

@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

Answer 20

last part... starts from where?

Answer 21

\(\text{"Once you solve this simple ODE, you're then left with finding t such that x(t)=.01x(0)"}\)
This is confusing me and I feel it shouldn't be.

Answer 22

Once you solve this simple ODE, you're then left with finding t such that x(t)=.01x(0

Answer 23

HAVE you solved the ODE?

Answer 24

\(\dfrac{dx}{dt}=-\dfrac{x}{100}\)?
I am sorry, that is what I meant, how would I go about that? I feel really, really dumb right now.

Answer 25

Probably because it's Friday...rearrange...
\[\Large \frac{dx}x = -\frac{dt}{100}\]
Integrate both sides.

Answer 26

\(\int\dfrac{1}{x}dx = \int -\dfrac{1}{100}dt\)
Right?

Answer 27

Yup.

Answer 28

\(\ln|x|=-\dfrac{t}{100}+C\)?

Answer 29

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
\[\Huge x = Ce^{-\frac{t}{100}}\]

Answer 30

Now, you need to find C.

Answer 31

ideas? :)

Answer 32

Not really unfortunately. I am not sure where the "Initial Value" comes in.... God I feel like a dunce right now.

Answer 33

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

Answer 34

200g yes?

Answer 35

Yes.... so
x(0) = 200.

Work it :D

Answer 36

derp...

Answer 37

So C = 200.

\[\Huge x = 200 e^{-\frac{t}{100}}\]

Answer 38

Now, for the concentration to be 1% of its original concentration (200), x has to be equal to 1% of 200, which is 2.
\[\Huge \color{blue}2 = 200e^{-\frac{t}{100}}\]

Answer 39

I got it!!!! YEEEEEESSS!

Answer 40

Awesome ^_^

Answer 41

I hope every problem isn't this hard on this assignment.