17Calculus Differential Equations - Mixing and Chemical Concentration in Fluids

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.

Well-Mixed 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.

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

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

Basic

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

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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

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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

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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 consecutive 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

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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

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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

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A vat with 500 gallons of beer contains 4% alcohol (by volume). Beer with 7% alcohol is pumped into the vat at a rate of 5 gal/min and the mixture is pumped out at the same rate.
(a) What is the amount of alcohol after an hour?
(b) what is the percentage of alcohol after an hour?

Problem Statement

A vat with 500 gallons of beer contains 4% alcohol (by volume). Beer with 7% alcohol is pumped into the vat at a rate of 5 gal/min and the mixture is pumped out at the same rate.
(a) What is the amount of alcohol after an hour?
(b) what is the percentage of alcohol after an hour?

Solution

3861 video

video by blackpenredpen

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A tank containing 60 gallons of a solution composed of 85% water and 15% alcohol. A second solution containing half water and half alcohol is added to the tank at a rate of 4 gallons per minute. At the same time, the tank is being drained at the same rate. Assuming the tank is being stirred constantly, how much alcohol will be in the tank after 10 minutes?

Problem Statement

A tank containing 60 gallons of a solution composed of 85% water and 15% alcohol. A second solution containing half water and half alcohol is added to the tank at a rate of 4 gallons per minute. At the same time, the tank is being drained at the same rate. Assuming the tank is being stirred constantly, how much alcohol will be in the tank after 10 minutes?

Solution

3883 video

video by blackpenredpen

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Intermediate

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

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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

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You CAN Ace Differential Equations

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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 Basic Equation Chemical-In Rate Chemical-Out Rate Well-Mixed Tank Practice

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