# System of diff eqs modeling tanks of brine

• nate9519
In summary, the problem involves two large tanks partially filled with brine and initially containing 5000 ounces and 10000 ounces of salt, respectively. The tanks are connected and 4 gallons per minute is pumped from tank A into tank B, while 3 gallons per minute is pumped from tank B into tank A. This process continues until one tank overflows or one tank becomes empty. The equations used to model this system are based on the rates of change of salt content in each tank and the flow of salt between the tanks. However, there seems to be some confusion regarding the initial volume and the units used in the equations.
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. Homework 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 I am stuck. I have no idea how to integrate the right side. is this the result of a previous error ?

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?

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

Astronuc said:
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?

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

nate9519 said:
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
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

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

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.

HallsofIvy said:
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.
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

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.

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

Chet

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

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.

Chestermiller said:
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

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

## 1. What is a system of differential equations?

A system of differential equations is a set of equations that describe the behavior of a dynamic system over time. It involves equations with derivatives, which represent the rates of change of various variables in the system.

## 2. How are differential equations used to model tanks of brine?

Differential equations can be used to model tanks of brine by describing the rate of change of the brine levels in the tanks over time. This can be done by setting up equations that take into account the inflow and outflow rates of the brine, as well as any other factors that may affect the levels.

## 3. What are the variables typically involved in a system of differential equations for tanks of brine?

The variables involved in a system of differential equations for tanks of brine may include the volume or level of brine in the tank, the inflow and outflow rates of brine, the concentration or salinity of the brine, and any other factors that may affect the levels such as temperature or pressure.

## 4. How are initial conditions and boundary conditions used in modeling tanks of brine with differential equations?

Initial conditions refer to the values of the variables at the starting point of the system, while boundary conditions refer to the constraints on the system at the boundaries. These conditions are used in the differential equations to solve for the behavior of the system over time and determine the levels of brine in the tanks at different points.

## 5. Are there any limitations to using a system of differential equations to model tanks of brine?

Like any mathematical model, a system of differential equations may have limitations in accurately representing the behavior of the real-world system. This can be due to simplifications made in the equations or assumptions about the system that may not hold true in all situations. It is important to validate the model and consider other factors when using it to make predictions or decisions.

• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
3K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
4K
• Calculus
Replies
2
Views
2K
• Calculus and Beyond Homework Help
Replies
16
Views
3K