# Differential equation mixing problem

• Lord Anoobis
In summary, there was a computation error and a few typos in the given calculations. After correcting these, the correct solution was found to be ##y = 120 - 40e^{-t/100}## with ##t = 1## giving a value of ##y = 80.398...##, which is the correct answer for the percentage of alcohol after an hour.
Lord Anoobis

## Homework Statement

A vat with 2000L of beer contains 4% alcohol (by volume). Beer with 6% alcohol is pumped into the vat at a rate of 20L/min and the mixture is pumped out at the same rate. What is the percentage of alcohol after an hour?

## The Attempt at a Solution

At ##t = 0## the amount of alcohol is 80L, let the amount at time ##t## be ##y##.
##\frac{dy}{dt} = rate in - rate out##
##rate in = (20L/min)(0.06) = 1.2L/min##
##rate out = \frac{y}{2000}(20L/min) = \frac{y}{100}L/min##

##\frac{dy}{dt} = \frac{1200 - y}{t}##

##\int\frac{dy}{1200 - y} = \int\frac{dt}{100}##

##-\ln|1200 - y| = \frac{t}{100} + k##

##1200 - y = Ae^{-t/100}##, yielding
##y(t) = 1120e^{-t/100}##

Substituting ##t = 1## gives ##y \approx 91.14## and converting to a percentage gives about ##4.6%## while the actual answer is ##4.9##. The general solution works fine for ##t = 0## which implies that the value of ##A## is correct. Where lies the error here?

Lord Anoobis said:

## Homework Statement

A vat with 2000L of beer contains 4% alcohol (by volume). Beer with 6% alcohol is pumped into the vat at a rate of 20L/min and the mixture is pumped out at the same rate. What is the percentage of alcohol after an hour?

## The Attempt at a Solution

At ##t = 0## the amount of alcohol is 80L, let the amount at time ##t## be ##y##.
##\frac{dy}{dt} = rate in - rate out##
##rate in = (20L/min)(0.06) = 1.2L/min##
##rate out = \frac{y}{2000}(20L/min) = \frac{y}{100}L/min##

##\frac{dy}{dt} = \frac{1200 - y}{t}##

##\int\frac{dy}{1200 - y} = \int\frac{dt}{100}##

##-\ln|1200 - y| = \frac{t}{100} + k##

##1200 - y = Ae^{-t/100}##, yielding
##y(t) = 1120e^{-t/100}##

Substituting ##t = 1## gives ##y \approx 91.14## and converting to a percentage gives about ##4.6%## while the actual answer is ##4.9##. The general solution works fine for ##t = 0## which implies that the value of ##A## is correct. Where lies the error here?
You made a computation error, as ##\frac{dy}{dt} = \frac{1200 - y}{t}## is wrong.

There a further error(s) down the line, so after fixing the formula for ##\frac{dy}{dt}##, carefully redo your calculations.

Samy_A said:
You made a computation error, as ##\frac{dy}{dt} = \frac{1200 - y}{t}## is wrong.

There a further error(s) down the line, so after fixing the formula for ##\frac{dy}{dt}##, carefully redo your calculations.
I made a typo there, it's supposed to be

##\frac{dy}{dt} = \frac{1200 - y}{100}##

Lord Anoobis said:
I made a typo there, it's supposed to be

##\frac{dy}{dt} = \frac{1200 - y}{100}##
Still wrong.

Samy_A said:
Still wrong.
Multiplication error...

##\frac{dy}{dt} = \frac{120 - y}{100}##

Lord Anoobis said:
Multiplication error...

##\frac{dy}{dt} = \frac{120 - y}{100}##

Still a balls-up. I just don't see it.

Lord Anoobis said:
Multiplication error...

##\frac{dy}{dt} = \frac{120 - y}{100}##
Yes, that's correct.

Now just continue from this.
This will lead to ##120 - y = Ae^{-t/100}## (exactly as in your first computation, with 120 instead of 1200).

Now use y(0)=80 to get A.

Samy_A said:
Yes, that's correct.

Now just continue from this.
This will lead to ##120 - y = Ae^{-t/100}## (exactly as in your first computation, with 120 instead of 1200).

Now use y(0)=80 to get A.
More than one typo, actually. Anyway, now I get

##y = 120 - 40e^{-t/100}## with ##t = 1## giving ##y = 80.398...## which is not correct. That's what I meant by the persistent balls-up.

Lord Anoobis said:
More than one typo, actually. Anyway, now I get

##y = 120 - 40e^{-t/100}## with ##t = 1## giving ##y = 80.398...## which is not correct. That's what I meant by the persistent balls-up.
How many minutes are there in 1 hour?

Samy_A said:
How many minutes in an hour?
Problem solved and the need for a break confirmed. I can't believe I missed that. Much appreciated.

## What is a differential equation mixing problem?

A differential equation mixing problem is a type of mathematical problem that involves modeling the flow of a substance or fluid within a container. It uses differential equations to describe the rate of change of the substance's concentration over time.

## What are the main components of a differential equation mixing problem?

The main components of a differential equation mixing problem include the container or system in which the mixing is taking place, the substance being mixed, the rate at which the substance is entering or leaving the system, and the initial conditions of the system.

## What is the purpose of solving a differential equation mixing problem?

The purpose of solving a differential equation mixing problem is to understand and predict the behavior of the substance being mixed within the system. This can be useful in various fields such as chemistry, physics, and engineering.

## What are some common techniques for solving a differential equation mixing problem?

Some common techniques for solving a differential equation mixing problem include separation of variables, the method of undetermined coefficients, and using Laplace transforms. These methods can help find a general solution or a particular solution for the problem.

## What are some real-world applications of differential equation mixing problems?

Differential equation mixing problems have various real-world applications, such as modeling the spread of pollutants in a river, predicting the concentration of a drug in the body over time, and understanding the flow of air in a ventilation system. They are also commonly used in chemical engineering to optimize mixing processes in industrial settings.

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