Differential Word problem help..

[Puk3s]

A tank contains 100 kg of salt and 2000 L of water. A solution of a concentration 0.025 kg of salt per liter enters a tank at the rate 5 L/min. The solution is mixed and drains from the tank at the same rate.

Find the amount of salt in the tank after 4.5 hours.


Basically I'm stumped from the beginning because I don't know how to set up the equation.. Any help would be great! Thanks

 

[galactus]

This problem will involve an integrating factor.

\(\displaystyle \frac{dy}{dt}=\text{rate in}-\text{rate out}\)

\(\displaystyle \text{rate in}=\left(\frac{1}{40} \;\ \frac{kg}{L}\right)\left(5 \;\ \frac{L}{min}\right)=\frac{1}{8} \;\ \frac{kg}{min}\)

At time t, the mixture contains y(t) kg of salt in 2000 L of water.

The concentration at time t is \(\displaystyle \frac{y(t)}{2000} \;\ \frac{kg}{L}\) and

\(\displaystyle \text{rate out}=\left(\frac{y(t)}{2000} \;\ \frac{kg}{L}\right)\left(5 \;\ \frac{L}{min}\right)=\frac{y(t)}{400} \;\ \frac{kg}{min}\)

So, we have:

\(\displaystyle \frac{dy}{dt}=\frac{1}{8}-\frac{y}{400}\)

\(\displaystyle \frac{dy}{dt}+\frac{y}{400}=\frac{1}{8}\)

Subjtect to the initial condition \(\displaystyle y(0)=100\)

This is the DE to solve. Start by finding your integrating factor.

After you solve and find the equation that models the amount of salt at time t, sub in t=4.5

 

[Puk3s]

This problem will involve an integrating factor.

\(\displaystyle \frac{dy}{dt}=\text{rate in}-\text{rate out}\)

\(\displaystyle \text{rate in}=\left(\frac{1}{40} \;\ \frac{kg}{L}\right)\left(5 \;\ \frac{L}{min}\right)=\frac{1}{8} \;\ \frac{kg}{min}\)

At time t, the mixture contains y(t) kg of salt in 2000 L of water.

The concentration at time t is \(\displaystyle \frac{y(t)}{2000} \;\ \frac{kg}{L}\) and

\(\displaystyle \text{rate out}=\left(\frac{y(t)}{2000} \;\ \frac{kg}{L}\right)\left(5 \;\ \frac{L}{min}\right)=\frac{y(t)}{400} \;\ \frac{kg}{min}\)

So, we have:

\(\displaystyle \frac{dy}{dt}=\frac{1}{8}-\frac{y}{400}\)

\(\displaystyle \frac{dy}{dt}+\frac{y}{400}=\frac{1}{8}\)

Subjtect to the initial condition \(\displaystyle y(0)=100\)

This is the DE to solve. Start by finding your integrating factor.

After you solve and find the equation that models the amount of salt at time t, sub in t=4.5

Thanks for your help! You had it all right (except t=270 but nbd ) Nice job I think i could set this up next time if I had to so thanks

 

[BigGlenntheHeavy]

\(\displaystyle Puk3s; \ Instead \ of \ having \ to \ put \ your \ thinking \ cap \ on \ when \ evaluating \ these\)

\(\displaystyle type \ of \ equations, \ just \ use \ Dr. \ Hippo's \ amazing \ formula. \ to \ wit:\)

\(\displaystyle \frac{dy}{dt}+\frac{r_2y}{v_0+(r_1-r_2)t} \ = \ q_1r_1.\)

\(\displaystyle Now, \ just \ plug \ you \ numbers \ into \ the \ above \ formula \ and \ you \ get:\)

\(\displaystyle \frac{dy}{dt}+\frac{5y}{2000} \ = \ (.025)(5) \ = \ \ .125-\frac{y}{400} \ = \ \frac{50-y}{400} \ = \ \frac{dy}{dt},\)

\(\displaystyle same \ as \ galactus \ above \ with \ y(0) \ = \ 100.\)

\(\displaystyle Also \ note: \ When \ the \ amount \ of \ solution \ going \ into \ said \ vat, \ tank, \ etc. \ is \ the\)

\(\displaystyle same \ amount \ flowing \ out, \ there \ isn't \ any \ integrating \ factor \ as \ they \ cancel \ out.\)

 

[galactus]

Very nice G man.

To make sure I got it all correctly:

r1 is the rate in

r2 is the rate out

q1 is the concentration

v0 is the volume of the tank.

I can see it is a general derivation. Very nice.

Who is Dr. Hippo?.

 

[BigGlenntheHeavy]

\(\displaystyle What!!! \ \ you \ never \ heard \ of \ Herr \ Gottfried \ von \ Hippo, \ the \ famous \ German \ mathematician.\)

\(\displaystyle Next \ you'll \ be \ saying \ that \ you \ never \ heard \ of \ Professor Ichikara \ Moto, \ the \ Japanese\)

\(\displaystyle mathematician \ of \ great \ fame.\)

\(\displaystyle Post Script: \ Y ou \ got \ it \ all \ correct \ galactus.\)