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- Homework Statement
- A tank contains 700 liters of water,to begin the tank is filled with 55 kilograms of salt.35 liters per minute of pure water is flowing in and 35 liters per minute of the water is flowing out.Describe the system using a differential equation (assume the mixing is homogenous).After how many minutes will the salt content be reduced to 15%
- Relevant Equations
- Differential equations
Hello!
First I tried modelling it like most mixing problems.
$$ \frac{dA}{dt} = rate coming in - rate coming out $$ where dA is the volume and dt is the time
rate coming in/out can be describe as; contrencation * flow rate.
Now if we plug that all on
$$ \frac{dA}{dt} = 35 * 0 - \frac{A}{700} * 35 $$
Now, we can make simplifications,separate the variables,than integrate and this should come out.
$$ ln(A) = \frac{-t}{20} + C $$
Now since a initial condition was given.55 kilograms are in the tank at time 0. we can do this.
A(0) = 55
We should get that C = ln(55);
Plug that back in and use e on both sides gets us too
$$ A = 55 e^{\frac{-t}{20}} $$
Now I think this is how can calculate the volume of the tank at any given point t.
Now to get at what point,or after what time salt content will be at 15% I tried the following.
15% of 55 is 8.25 kg,now we know that we want the value of A to be 8.25 but just not when.So we can write it like this.
A(t) = 8.25 kg
If we plug that in;
$$ ln(8,25) = 55 e^{\frac{-t}{20}} + C $$
than we can use ln on both sides,multiply by 20 to get rid of the fraction and we should get
t = 37,94 m;
Now I am not sure that this second part is correct.My logic is,I have a formula that will give me the A at any time point,simply reararange to get the time,but I am not sure if it can work that way.Thanks!
First I tried modelling it like most mixing problems.
$$ \frac{dA}{dt} = rate coming in - rate coming out $$ where dA is the volume and dt is the time
rate coming in/out can be describe as; contrencation * flow rate.
Now if we plug that all on
$$ \frac{dA}{dt} = 35 * 0 - \frac{A}{700} * 35 $$
Now, we can make simplifications,separate the variables,than integrate and this should come out.
$$ ln(A) = \frac{-t}{20} + C $$
Now since a initial condition was given.55 kilograms are in the tank at time 0. we can do this.
A(0) = 55
We should get that C = ln(55);
Plug that back in and use e on both sides gets us too
$$ A = 55 e^{\frac{-t}{20}} $$
Now I think this is how can calculate the volume of the tank at any given point t.
Now to get at what point,or after what time salt content will be at 15% I tried the following.
15% of 55 is 8.25 kg,now we know that we want the value of A to be 8.25 but just not when.So we can write it like this.
A(t) = 8.25 kg
If we plug that in;
$$ ln(8,25) = 55 e^{\frac{-t}{20}} + C $$
than we can use ln on both sides,multiply by 20 to get rid of the fraction and we should get
t = 37,94 m;
Now I am not sure that this second part is correct.My logic is,I have a formula that will give me the A at any time point,simply reararange to get the time,but I am not sure if it can work that way.Thanks!
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