# Homework Help: System of diff eqs modeling tanks of brine

1. Feb 26, 2015

### nate9519

1. The problem

Two very large tanks are each partially filled with 2000 gallons of brine. Initially 5000 ounces of salt is dissolved in tank A and 1000 ounces of salt is dissolved in tank B. The system is closed in that the well stirred liquid is only pumped between the two tanks. Four gallons per minute is pumped from tank A into tank B, while three gallons per minute is pumped from tank B into tank A. This continues until one tank overflows or one tank is empty.

2. Relevant Equations

This is the system I have constructed which is consistent with the way my book models this type of problem. If it looks wrong please let me know.

dx/dt = (3y / (2000 - t)) - (4x / (2000 + t))

dy/dt = (4x / (2000 + t)) - (3y / (2000 - t))

3. Attempt at solution

my problem is when I try to solve the system.

(dx/dt) + (4x / (2000 + t)) = (3y / (2000 - t))

4 / (2000 + t) is my integrating factor which turns to (t + 2000)^4 because of the ln. after multiplying through I get

d/dt [ x (t + 2000)^4] = (3y / (2000 - t)) (t + 2000)^4

this is where im stuck. I have no idea how to integrate the right side. is this the result of a previous error ?

2. Feb 28, 2015

### Astronuc

Staff Emeritus
How does one get (2000 - t) or (2000 + t)? If 2000 refers to the initial volume, then one should be not be adding or subtracting time (t) from volume.

What is the volume of the tanks, i.e., when does tank B start overflowing?

3. Mar 1, 2015

### Staff: Mentor

The mass balance equations look wrong. I get the following:

$$\frac{d[(2000-t)x]}{dt}=3y-4x$$
$$\frac{d[(2000+t)y]}{dt}=4x-3y$$

$$(2000-t)\frac{dx}{dt}=3(y-x)$$
$$(2000+t)\frac{dy}{dt}=4(x-y)$$

Chet

4. Mar 2, 2015

### nate9519

I think subtracting and adding time comes from the fact that the volume fluctuates. this equations are identical to the example in my book so I don't understand how they could be wrong

5. Mar 2, 2015

### Staff: Mentor

Please present the derivation of the equations in your book. I've had a lot of industry experience with this kind of thing, and I'm totally confident with the equations that I wrote. In my first two equations, the left hand sides are the rates of change of salt content in each tank; i.e., the time derivative of the product of tank volume with salt concentration. The right hand sides are the flow of salt into and out of each tank; i.e., flow rate in times salt concentration in minus flow rate out times salt concentration out (the later also being equal to the concentration of salt in the tank).

Chet

6. Mar 2, 2015

### Staff: Mentor

I just realized that, what I'm calling x, you are calling (2000-t)x, and what I'm calling y, you are calling (2000+t)y, so your equations are correct after all. If you add your two differential equations together, you get
$$\frac{d(x+y)}{dt}=0$$
This says that the total amount of salt in the two tanks is always constant.

So, it follows that (x + y) = 6000 for all time.

So, in your first equation, substitute y = 6000-x, and you have a first order ODE soley in terms of x. See how this works out for you.

Chet

7. Mar 2, 2015

### HallsofIvy

It should, at least, occur to you that your "2000" is the 2000 gallons of brine while "t" is time measured in minutes. "2000 gallons minus t minutes" and "2000 gallons plus t minutes" are meaningless.

8. Mar 2, 2015

### Staff: Mentor

In Nate9519's defense:

2000 (gal) + 1 (gpm) x t (min) = (2000 + t) gal (of solution in tank after t minutes)

2000 (gal) - 1 (gpm) x t (min) = (2000 - t) gal (of solution in tank after t minutes)

So, it's not as meaningless as you think.

Chet

9. Mar 3, 2015

### HallsofIvy

A valid point, but where would we get "1 gallon per minute" from? The brine is being transported from tank A to tank B at 4 gallons per minute so the amount of brine in tank A after t minutes would be 2000- 4t, not 2000- t. If not meaningless then it is wrong.

10. Mar 3, 2015

### Staff: Mentor

What does this mean to you: "while three gallons per minute is pumped from tank B into tank A"

Chet

11. Mar 3, 2015

### nate9519

*Important note. 1000 is suppose to be 10000. my instructor had missed a zero and I just learned of this this morning.

sorry chet but there is no derivation. its just an answer key.
the way my book teaches it is that the system is the rates of change of the salt for each tank which is the input rate minus the output rate. the input rate for tank A is 4 gal/min times x amount of salt per gallon( but this gallon amount is fluctuating so it must be a variable which is where the t comes from). as chet hinted at the net change for tank A and B is 1 and -1 gal/min respectively.

12. Mar 3, 2015

### nate9519

this makes perfect sense. I will try this and post the result