How Do You Calculate the Salt Concentration in a Tank Over Time?

[KillerZ]

Homework Statement

A tank contains 200 liters of fluid which 30 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 4 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.

Homework Equations

initial amount of liquid = 200L
A(t=0) = 30g
rate in = 4L/min
rate out = 4L/min
concentration of the inflow liquid = 1 g/L

The Attempt at a Solution

\frac{dA}{dt} = (1g/L)(4L/min) - \frac{(4L/min)(A(t))g}{(200L) + (4L/min - 4L/min)t}

\frac{dA}{dt} = (4) - \frac{4A}{200}

\frac{dA}{dt} + \frac{4A}{200} = (4)

\frac{dA}{dt} + \frac{A}{50} = (4)

I(t) = e^{\int\frac{A}{50}dt}

I(t) = e^{\frac{At}{50}}

f(t) = \frac{1}{e^{\frac{At}{50}}}[\int (e^{\frac{At}{50}})(4) dt + c]

f(t) = \frac{1}{e^{\frac{At}{50}}}[\frac{200e^{\frac{At}{50}}}{A} + c]

f(t) = \frac{200e^{\frac{At}{50}}}{Ae^{\frac{At}{50}}} + \frac{c}{e^{\frac{At}{50}}}]

 

[HallsofIvy]

Homework Statement

A tank contains 200 liters of fluid which 30 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 4 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.

Homework Equations

initial amount of liquid = 200L
A(t=0) = 30g
rate in = 4L/min
rate out = 4L/min
concentration of the inflow liquid = 1 g/L

The Attempt at a Solution

\frac{dA}{dt} = (1g/L)(4L/min) - \frac{(4L/min)(A(t))g}{(200L) + (4L/min - 4L/min)t}

\frac{dA}{dt} = (4) - \frac{4A}{200}

\frac{dA}{dt} + \frac{4A}{200} = (4)

\frac{dA}{dt} + \frac{A}{50} = (4)

I(t) = e^{\int\frac{A}{50}dt}

No, no, no! "A" is your dependent variable, not a constant. You find the integrating factor by integrating the coefficient of A. Your integrating factor is
e^{\int\frac{dt}{50}}= e^{\frac{t}{50}}

I(t) = e^{\frac{At}{50}}

f(t) = \frac{1}{e^{\frac{At}{50}}}[\int (e^{\frac{At}{50}})(4) dt + c]f(t) = \frac{1}{e^{\frac{At}{50}}}[\frac{200e^{\frac{At}{50}}}{A} + c]

And now you've completely lost me! Where did "f(t)" come from? There was no "f" before this line. You are supposed to be finding A(t).

Since the integrating factor is
e^{\int\frac{dt}{50}}= e^{\frac{t}{50}}
you have
e^{\frac{t}{50}}\frac{dA}{dt}-e^{\frac{t}{50}}\frac{A}{50}= 4e^{\frac{t}{50}}
\frac{d}{dt}\left(e^{\frac{t}{50}}A\right)= 4e^{\frac{t}{50}}[/itex]<br /> Integrate that to find A. The left side is, of course, just<br /> e^{\frac{t}{50}}A(t)<br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> f(t) = \frac{200e^{\frac{At}{50}}}{Ae^{\frac{At}{50}}} + \frac{c}{e^{\frac{At}{50}}}] </div> </div> </blockquote><br /> Frankly, I think it would be easier not to use the &quot;integrating factor&quot; method at all. This is a separable equation.<br /> <br /> \frac{dA}{dt}= 4-\frac{A}{50}= \frac{200- A}{50}<br /> \frac{dA}{200- A}= \frac{dt}{50}<br /> Integrate both sides of that.

 

[KillerZ]

opps I made some typos and I don't know what I was thinking doing the integrating factor .

Here is my next attempt:

\frac{dA}{dt} = (1g/L)(4L/min) - \frac{(4L/min)(A(t))g}{(200L) + (4L/min - 4L/min)t}

\frac{dA}{dt} = (4) - \frac{4A}{200}

\frac{dA}{dt} + \frac{4A}{200} = (4)

\frac{dA}{dt} + \frac{A}{50} = (4)

I(t) = e^{\int\frac{1}{50}dt}

I(t) = e^{\frac{t}{50}}

A(t) = \frac{1}{e^{\frac{t}{50}}}[\int (e^{\frac{t}{50}})(4) dt + c]

A(t) = \frac{1}{e^{\frac{t}{50}}}[200e^{\frac{t}{50}} + c]

A(t) = \frac{200e^{\frac{t}{50}}}{e^{\frac{t}{50}}} + \frac{c}{e^{\frac{t}{50}}}

 

[HallsofIvy]

A(t) = \frac{200e^{\frac{t}{50}}}{e^{\frac{t}{50}}} + \frac{c}{e^{\frac{t}{50}}}

I hope you understand that this can be simplified!

And, of course, use the fact that A(0)= 30 to find c.

 

[KillerZ]

A(t) = \frac{200e^{\frac{t}{50}}}{e^{\frac{t}{50}}} + \frac{c}{e^{\frac{t}{50}}}

A(t) = 200 + \frac{c}{e^{\frac{t}{50}}}

A(0) = 30 = 200 + \frac{c}{e^{\frac{0}{50}}}

c = 30 - 200 = -170

A(t) = 200 - \frac{170}{e^{\frac{t}{50}}}