\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{\mathrm{sec} } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{\mathrm{arccot} } \) \( \newcommand{\arcsec}{\mathrm{arcsec} } \) \( \newcommand{\arccsc}{\mathrm{arccsc} } \) \( \newcommand{\sech}{\mathrm{sech} } \) \( \newcommand{\csch}{\mathrm{csch} } \) \( \newcommand{\arcsinh}{\mathrm{arcsinh} } \) \( \newcommand{\arccosh}{\mathrm{arccosh} } \) \( \newcommand{\arctanh}{\mathrm{arctanh} } \) \( \newcommand{\arccoth}{\mathrm{arccoth} } \) \( \newcommand{\arcsech}{\mathrm{arcsech} } \) \( \newcommand{\arccsch}{\mathrm{arccsch} } \)

17Calculus Differential Equations - Mixing and Chemical Concentration in Fluids

1st Order

2nd/Higher Order

Laplace Transforms

Applications

Additional Topics

Tools

Calculus Tools

Additional Tools

Articles

On this page we discuss one of the most common types of differential equations applications of chemical concentration in fluids, often called mixing or mixture problems. The idea is that we are asked to find the concentration of something (such as salt or a chemical) diluted in water at any given time. Usually we are adding a known concentration to a tank of known volume. Sometimes the tank is being drained at the same time. There are many variations to this problem but they are usually easily set up and solved using basic differential equations techniques.

Basic Equation

There is only one basic equation you need and setting it up is the tricky part. We highly recommend that you carry your units with you as you set it up. Sometimes, mistakes can be caught when the units don't work out. The equation you need is

\(\displaystyle{ \frac{dA}{dt} = ( rate~in ) - ( rate~out ) }\)

Notes
1. Since this is a rate problem, the variable of integration is time t.
2. \(A\) is the amount or quantity of chemical that is dissolved in the solution, usually with units of weight like kg.
3. The rates (rate in and rate out) are the rates of inflow and outflow of the chemical. The units are usually weight per unit time, like kg/min.

As you work these problems, you will notice that there are two rates involved, the rate of flow of chemical and the rate of flow of the fluid. It is important that you understand the difference when you are working these problems. We will separate the flow rates like this.
chemical-in = rate of chemical coming into the tank with units weight/time, like kg/min.
fluid-in = rate of the fluid entering the tank with units volume/time, like litres/min = L/min.
[similarly for flow rates out of the tank]
Notice in the table above, the rate in and rate out are chemical-in and chemical-out rates. Rewriting that equation we have \(\displaystyle{ \frac{dA}{dt} = ( \textit{ chemical-in-rate } ) - ( \textit{ chemical-out-rate } ) }\).

The tricky part is knowing how to set up the rates of chemical-in-rate and chemical-out-rate. The chemical-in-rate is usually the easiest, so let's look at that first.

Chemical-In Rate

Usually we are given the concentration of the fluid coming in and the rate at which it is flowing in. For example, one of the practice problems gives the rate in as 10L/min of pure water (with no chemical or salt). There is no chemical in the solution (since it is pure water), so the amount of chemical is 0kg/L. The rate of inflow of the chemical is modeled as
\( \textit{ chemical-in } = (0 kg/L) (10 L/min) = 0 kg/min \).
Notice that we wrote our units in this equation, so that we were easily able to see that the units L canceled, leaving the units kg/min, which is what we want.

Chemical-Out Rate

Okay, so now let's look at outflow. This is usually where the variable we call \(A(t)\), or just \(A\), comes in. The variable \(A\) is the amount (in weight, like kg) of the chemical. It is dependent on t, so it is changing. The concentration in the tank at any time t is just \(A\) divided by the volume of fluid in the tank. The volume is calculated using this equation.
volume of fluid in the tank at any time = initial amount + ( rate coming in - rate going out ) times t
Notes
1. We assume the rates coming in and going out are constant. If they are not, the above equation would need to be adjusted accordingly.
2. If the fluid is coming in and going out at the same rate, the volume of fluid in the tank is constant and equal to the initial amount.

So, for example, if we have a tank that starts out with 1000L and fluid is coming in at a rate of 10L/min and going out at a rate of 12L/min, we have
volume of fluid in the tank = 1000L + (10L/min - 12L/min)t = 1000-2t L
This equation describes the amount of fluid in the tank at any time t and this is what is divided into A to get the rate of chemical out of the tank.

Study Tip

As you work these problems, it may help to set up a table containing everything you know and everything you need to find. In the first practice problem, we set one up as an example. As you study, try different formats for the table until you come up with one that makes sense for you.

Well-Mixed Tank

When we discussed the outflow of the tank above, we had to use the idea that the tank was well mixed and that the chemical was dispersed evenly throughout the tank. If we didn't have this assumption, we would have needed a term in the equation that was a function of time describing how the chemical was dispersed in the tank. This is an extremely complicated concept and one that you will probably not come across in differential equations.

Before working some practice problems, let's watch a quick video explaining these types of problems in a bit more detail. This is pretty theoretical but it will give you another perspective that may be helpful.

Commutant - well-mixed tank

video by Commutant

Practice

Instructions - - Solve the following problems, giving your answers in exact form. Unless otherwise stated, assume the tanks are well-stirred.

Basic Problems

A tank contains 1000L of fluid and 15kg dissolved salt. Fresh water enters at 10L/min and the tank is draining at 10L/min. How much salt is in the tank at t minutes and after 20 minutes?

Problem Statement

A tank contains 1000L of fluid and 15kg dissolved salt. Fresh water enters at 10L/min and the tank is draining at 10L/min. How much salt is in the tank at t minutes and after 20 minutes?

Solution

One way to get your head around this problem is to build a table of the information you have and what you need. Here is one way to build a table. If this doesn't help you, try another format.

inflow

outflow

fluid flow rate

10 L/min

10 L/min

chemical concentration

0 kg/L

A/vol kg/L

chemical flow rate

(10 L/min)(0 kg/L) = 0 kg/min

(10 L/min)(A/vol kg/L) = 10A/vol kg/min

From the table, we can now see that we need to determine the volume of the fluid.
volume = initial amount + (rate in - rate out)t = 1000L + (10 L/min - 10 L/min)t = 1000L
Adding the two expressions in the last row, we have \(\displaystyle{ \frac{dA}{dt} = (0 ~kg/min) - (A/100~ kg/min) }\).

In this video, she shows how carrying the units along is helpful and she explains the solution clearly. However, she does something strange with the constant of integration by replacing \(e^c\) with \(\pm c\). This is confusing since there are actually two mistakes here. First, \(e^c\) is always positive, so the \(\pm\) sign is incorrect. Secondly, since \(e^c\) can never equal \(c\), it just doesn't make sense. What she is actually doing is representing the unknown constant \(e^c\) with another unknown constant but she doesn't change her notation. We would probably write \(k=e^c\) to make it clear that they are different.

625 video

video by Krista King Math

close solution
A tank with 200 gallons of brine solution contains 40 lbs of salt. A concentration of 2 lb/gal is pumped in at a rate of 4 gal/min. The concentration leaving the tank is pumped out at a rate of 4 gal/min. How much salt is in the tank after 1 hour? How much salt is in the tank after a very long time?

Problem Statement

A tank with 200 gallons of brine solution contains 40 lbs of salt. A concentration of 2 lb/gal is pumped in at a rate of 4 gal/min. The concentration leaving the tank is pumped out at a rate of 4 gal/min. How much salt is in the tank after 1 hour? How much salt is in the tank after a very long time?

Solution

627 video

video by MIP4U

close solution
A tank initially holding 80gals of fluid with 40lbs of salt has two sources of fluid being pumped in. One source is pumping pure water in at 8 gal/min and the second source is pumping in a salt solution of 0.5lbs/gal at 2 gal/min. The tank is emptying at 10gal/min. Find the amount of salt in the tank at any time t.

Problem Statement

A tank initially holding 80gals of fluid with 40lbs of salt has two sources of fluid being pumped in. One source is pumping pure water in at 8 gal/min and the second source is pumping in a salt solution of 0.5lbs/gal at 2 gal/min. The tank is emptying at 10gal/min. Find the amount of salt in the tank at any time t.

Solution

1361 video

video by Kent Mearig

close solution
A 100 gallon tank with 50 gals of pure water has a solution of 0.4lbs/gal being pumped in at a rate of 3 gals/min. The well-mixed solution leaves the tank at a rate of 2 gals/min. Find an expression for the salt concentration at any time t.

Problem Statement

A 100 gallon tank with 50 gals of pure water has a solution of 0.4lbs/gal being pumped in at a rate of 3 gals/min. The well-mixed solution leaves the tank at a rate of 2 gals/min. Find an expression for the salt concentration at any time t.

Solution

This problem is solved in two videos. He sets up the differential equation in the first video and then solves it in the second video.

1362 video

video by SchoolOfChuck

1362 video

video by SchoolOfChuck

close solution
A 1000L tank starts out with 200L of fluid containing 10g/L of dye. Pure water is poured in at 20 L/min and the tank is being drained at a rate of 15L/min. Write the equation for the amount of dye in the tank at any time t.

Problem Statement

A 1000L tank starts out with 200L of fluid containing 10g/L of dye. Pure water is poured in at 20 L/min and the tank is being drained at a rate of 15L/min. Write the equation for the amount of dye in the tank at any time t.

Solution

1363 video

video by David Lippman

close solution
A tank contains 90 kg of salt and 2000 L of water. Pure water enters the tank at the rate of 8 L/min. The solution is mixed and drains from the tank at the rate of 4 L/min.
(a) What is the amount of salt in the tank initially?
(b) Find the amount of salt in the tank after 1.5 hours.
(c) Find the concentration of salt in the solution in the tank as time approaches infinity. (Assume your tank is large enough to hold all the solution.)

Problem Statement

A tank contains 90 kg of salt and 2000 L of water. Pure water enters the tank at the rate of 8 L/min. The solution is mixed and drains from the tank at the rate of 4 L/min.
(a) What is the amount of salt in the tank initially?
(b) Find the amount of salt in the tank after 1.5 hours.
(c) Find the concentration of salt in the solution in the tank as time approaches infinity. (Assume your tank is large enough to hold all the solution.)

Solution

2289 video

video by MIP4U

close solution

Intermediate Problems

Initially a tank contains 1kg of salt dissolved in 100L of water. Salty water containing 1/4kg/L at a rate of 3L/min is added to the tank and the (stirred) solution is draining from the tank at 3L/min. Determine an equation for how much salt is in the tank at any time t.

Problem Statement

Initially a tank contains 1kg of salt dissolved in 100L of water. Salty water containing 1/4kg/L at a rate of 3L/min is added to the tank and the (stirred) solution is draining from the tank at 3L/min. Determine an equation for how much salt is in the tank at any time t.

Solution

626 video

video by Dr Chris Tisdell

close solution
A tank holds 300 gallons of brine solution with 40 lbs of salt. A concentration of 2 lbs/gal is pumped in a rate of 4 gal/min. The concentration leaving the tank is pumped out at a rate of 3gal/min. How much salt is in the tank after 12min?

Problem Statement

A tank holds 300 gallons of brine solution with 40 lbs of salt. A concentration of 2 lbs/gal is pumped in a rate of 4 gal/min. The concentration leaving the tank is pumped out at a rate of 3gal/min. How much salt is in the tank after 12min?

Solution

1360 video

video by MIP4U

close solution

You CAN Ace Differential Equations

Topics You Need To Understand For This Page

Related Topics and Links

To bookmark this page and practice problems, log in to your account or set up a free account.

Calculus Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations Topics Listed Alphabetically

Precalculus Topics Listed Alphabetically

Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

how to read math books

Get great tutoring at an affordable price with Chegg. Subscribe today and get your 1st 30 minutes Free!

The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free.

Do NOT follow this link or you will be banned from the site!

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.

Links and banners on this page are affiliate links. We carefully choose only the affiliates that we think will help you learn. Clicking on them and making purchases help you support 17Calculus at no extra charge to you. However, only you can decide what will actually help you learn. So think carefully about what you need and purchase only what you think will help you.

We use cookies on this site to enhance your learning experience.

17calculus

Copyright © 2010-2020 17Calculus, All Rights Reserved     [Privacy Policy]     [Support]     [About]

mathjax.org
Real Time Web Analytics
17Calculus
We use cookies to ensure that we give you the best experience on our website. By using this site, you agree to our Website Privacy Policy.