Differential equation mixing problem

[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?

Homework Equations

The Attempt at a Solution

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

Which leads to:
$\frac{dy}{dt} = \frac{1200 - y}{t}$

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

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

$1200 - y = Ae^{-t/100}$, 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?

 

[Samy_A]

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?

Homework Equations

The Attempt at a Solution

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

Which leads to:
$\frac{dy}{dt} = \frac{1200 - y}{t}$

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

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

$1200 - y = Ae^{-t/100}$, 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.

 

[Lord Anoobis]

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

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

 

[Samy_A]

I made a typo there, it's supposed to be

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

Still wrong.

 

[Lord Anoobis]

Still wrong.

Multiplication error...

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

 

[Lord Anoobis]

Multiplication error...

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

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

 

[Samy_A]

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.

 

[Lord Anoobis]

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.

 

[Samy_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.

How many minutes are there in 1 hour?

 

[Lord Anoobis]

How many minutes in an hour?

Problem solved and the need for a break confirmed. I can't believe I missed that. Much appreciated.