# Homework Help: Mixing (DE) problem

1. Jun 25, 2005

### EvLer

I am not really sure how to get right values for variables/constants:

A tank contains 100 gallons of salt water which contains 10 lbs of salt. A salt solution of 2 lbs salt per gallon enters the tank at a rate of 3 gallons perminute while a flow of fresh water runs into the tank at rate 5 gallons per minute.
The well-stirred solution runs out of the tank at rate of 7 gall/min. Let A(t) be the amount of salt in lbs at time t in mins. The initial value problem for A(t) is ...
(to be found as a DE with IC).

V(0) = 100
r2(flow rate) = 7 gall/min

A(0) = 10 gal.
The underlined section is where I need some help, at least then I can start on the problem and see how I end up.

Last edited: Jun 25, 2005
2. Jun 25, 2005

### saltydog

EvLer, I'll take a chance on this one because I'm not sure. See what you think and maybe others can help too:

Isn't the time rate of change of the amount of salt s(t) in the tank:

$$s^{'}(t)=\quad\text{salt in rate}\quad-\quad\text{salt out rate}$$

I tell you what, that fresh water just makes it more complicated. Isn't it the same as if they were just one pipe putting in 8 gal/min with a total of 6 pounds salt/minute?

So the rate of salt in is just 6.

Now for the rate out: Well, it's the number of gallons out times the concentration of salt in the tank at time t. The concentration is the amount of salt which is s(t) divided by the total volume. Note that 8 goes in but only 7 goes out. So isn't the volume a function of t? It's 100+t in fact. Can you put it all together? Note the initial conditions: s(0)=10. Use your best judgement; I'm not sure.

3. Jun 26, 2005

### EvLer

We have 2 equations that are used in these problems:

dV/dt = rate_in - rate_out;
so: V = (rin - rout)t + C;
C is not hard to find given V(0) = 100.

Then second equation is: (A is amount of stuff in the liquid, r - rate, c - concentration)

dA/dt = rincin - routcout;
where cout = A(t)/V(t);

So, basically I need to figure out rates and cin.
I wasn't sure if I could just add rate of fresh water and salty water but i think it makes sense, like you said. But don't you think that increasing the rate of flow with salty water being diluted in this case, it's concentration is going to decrease rather than increase per unit of volume at a point in time? Although I think it's supposed to be constant, concentration that is, given rates are constant.

Or maybe I am turned around

4. Jun 26, 2005

### saltydog

Well Evler, haven't worked mixing problems in a while but interesting nevertheless. This is what I come up with:

Using the analysis I gave above, I get for the "amount" of salt in the tank as a function of time s(t):

$$s^{'}=6-7\left(\frac{s}{100+t}\right);\quad s(0)=10$$

little of this, little of that, we obtain:

$$s(t)=\frac{3}{4}(100+t)+\frac{c}{(100+t)^7};\quad c=-65(100)^7$$

The first plot is the amount of salt in the tank as a function of time. Now, the concentration as a function of time is:

$$\frac{s(t)}{100+t}$$

The second plot is the concentration of salt as a function of time.

Edit: Oh yea, if I'm right, don't just copy my work ok. Make sure you understand the concepts, why I set up the equations like I did, how to integrate it, and does the resulting equations make "intutitive" sense.

#### Attached Files:

File size:
4.4 KB
Views:
72
• ###### conc of salt.JPG
File size:
4.3 KB
Views:
65
Last edited: Jun 26, 2005
5. Jun 26, 2005

### HallsofIvy

Yes, that's correct. As saltydog told you, salt is coming in at 6 pounds per minute and going out at 7A(t)/(100+t)

The differential equation for the amount of salt A(t) is

dA/dt= 6- 7A/(100+t).

No, the concentration will not be a constant- whether it is increasing or decreasing depends on the original concentration and I haven't calculated the solution to see what happens. In this problem, the initial concentration of salt is 1/10 pound of salt per gallon of water. The concentration of salt coming in is 2 pounds in 8 gallons of water (yes, you can treat this as a single pipe): 1/4 pound per gallon. The concentration of salt in the tank will increase toward a limit of 1/4 pound per gallon. If the concentration of salt in the tank initially had been greater than 1/4 pound per gallon, it would decrease toward that limit.

6. Jun 26, 2005

### saltydog

You know what EvLer, I think I did your homework for you and I very much don't want to go against forum rules. Really, you know what, I say you do a different one now:

Initially, a 200 gal tank contains 40 lbs salt. A pipe going into the tank is flowing at 5 gal/min with a conc of salt at 2 lbs/gal. Another pipe is flowing out at the same rate. Find the amount of salt s(t) in the tank at time t. Yep, full report by morning.

7. Jun 26, 2005

### EvLer

although this isn't really homework, it's a practice exam.
Thanks much for help!

:grumpy:
Deal!
Here's what I have:

dV/dt = rin - rout;

V(t) = C, V(0) = 200 => V(t) = 200;

dA/dt = rincin - routA(t)/V(t);

A(t) = e(-5/200)t[400*e(5/200)t + C];

given A(0) = 40 => C = -360

A(t) = e(-5/200)t[400*e(5/200)t - 360];

Are you taking points off for abscense of plots?

Last edited: Jun 26, 2005
8. Jun 26, 2005

### saltydog

Very good EvLer. Actually I do reqest a plot, that is if you have Mathematica handy. You know LaTex code makes everything look nicer and easier to read. Like this:

$$A(t)=400-360e^{-0.025t}$$

9. Jun 26, 2005

### EvLer

Erm, I think I just got me a "B" :sad:

We have MATLAB on school machines, but I need to get there first. Perhaps if you give me some time...

Yeah, I can do simple things with LaTex, can you direct me to a short tutorial on how to use it for various kinds of expressions, like fractions, matrices, etc. If you hold your cursor w/t clicking on the thread that uses LaTex you can see the code used, but tutorial would be nice.
Thanks again!

Last edited: Jun 26, 2005
10. Jun 26, 2005

### saltydog

EvLer, don't go out of your way for the plot. It's just a suggestion. There is a thread in the General Physics forum called "Introducting LaTex". It has lots of examples and some web references.