Mixing Tank, Differential Equations Problem

[bdh2991]

Homework Statement

a larg tank is filled with 500 gals of water with 400lbs of salt. pure water is pumped into the tank at a rate of 3gal/min. the well mixed solution is pumped out at a rate of 7 gal/min.

I need help finding the Differential

Homework Equations

The Attempt at a Solution

I tried coming up with the D.E. and this is what i got dA/dt = 3 - 7A/(500-t)

except after i did solved the d.e. i kept getting more salt than i started with which makes no sense...

Please Help lol

 

[fauboca]

Homework Statement

a larg tank is filled with 500 gals of water with 400lbs of salt. pure water is pumped into the tank at a rate of 3gal/min. the well mixed solution is pumped out at a rate of 7 gal/min.

I need help finding the Differential

Homework Equations

The Attempt at a Solution

I tried coming up with the D.E. and this is what i got dA/dt = 3 - 7A/(500-t)

except after i did solved the d.e. i kept getting more salt than i started with which makes no sense...

Please Help lol

Your denominator is wrong. How is rate out defined?

 

[bdh2991]

Your denominator is wrong. How is rate out defined?

not sure lol...i went off of a similar problem that was gaining instead of losing...

i know that it is supposed to be the flow rate out multiplied by the concentration out but i don't understand how to get the concentration of salt exiting the tank...

 

[fauboca]

not sure lol...i went off of a similar problem that was gaining instead of losing...

i know that it is supposed to be the flow rate out multiplied by the concentration out but i don't understand how to get the concentration of salt exiting the tank...

R_{out} = -t(7gal/min - 3gal/min) = -4t

Where you have 500-t. I didn't check over the rest of you equation but that was what was immediate obvious.

 

[bdh2991]

ok so instead i should have A/(-4t) rather than A/(500-t)?

 

[fauboca]

ok so instead i should have A/(-4t) rather than A/(500-t)?

No, why would you disregard the amount of solution?

 

[bdh2991]

i guess I'm misunderstanding what you are saying...it would help more if you just showed me what the equation should look like

 

[fauboca]

i guess I'm misunderstanding what you are saying...it would help more if you just showed me what the equation should look like

I could have phrased it better.

When you are dealing with mixtures, you denominators is

\frac{A}{sol+(R_{in}-R_{out})t}

When inflow and out flow are the same, we just have A/sol in gallons.

Your R_{in}=3 and out=7 so (3-7)t = -4t.

So what is the denominator?

 

[bdh2991]

A/500-(-4t)

so A/500+4t?

 

[bdh2991]

it can't be that because i just tried it and it didn't come out right

 

[fauboca]

I could have phrased it better.

When you are dealing with mixtures, you denominators is

\frac{A}{sol+(R_{in}-R_{out})t}

When inflow and out flow are the same, we just have A/sol in gallons.

Your R_{in}=3 and out=7 so (3-7)t = -4t.

So what is the denominator?

500+(-4t) = 500-4t

 

[bdh2991]

ok i made some dumb mistakes but i finally got what i needed thanks!