I am having difficulties with a mixing problem using DE: A tank initially has 200 liters of water with 30 grams of salt dissolved in it. A solution with a gram per liter of salt is pumped with a rate of 4L/min; the solution is pumped out at the same rate. Find the number of grams of salt in the tank at a given instant t.

I am having difficulties to setup the problem, should it be:
dQ/dt = rate in * initial concentration - rate out * concentration at a given time t?

Question

I am having difficulties with a mixing problem using DE: A tank initially has 200 liters of water with 30 grams of salt dissolved in it. A solution with a gram per liter of salt is pumped with a rate of 4L/min; the solution is pumped out at the same rate. Find the number of grams of salt in the tank at a given instant t.

I am having difficulties to setup the problem, should it be:
dQ/dt = rate in * initial concentration - rate out * concentration at a given time t?

anonymous · Answer

My initial try was: $$\frac{dQ}{dt} = 4\frac{30}{200} - 4 \frac{Q(t)}{200 + 4t} $$But I don't think this is the way.

anonymous · Answer

Albeit the units are correct.

experimentx · Answer

?? is solution being pumped out as well as pumped in too??

anonymous · Answer

Yup, the solution is being pumped in and then the water is being pumped out. I am sorry for the confusion.

anonymous · Answer

I think that the equation should be: $$\frac{dQ}{dt} = \frac{3}{5} - \frac{4Q}{200} $$Since the rate in = rate out, the volume is constant.

experimentx · Answer

somehow i get a feeling that we will get a graph that degenerates in to a single line at t=$\infty$

anonymous · Answer

My problems are with the constants. My answer was correct in some sense, that e^(-t/50) is the time dependant variation. The constants, tho, I can't get them to solve the IVP.

anonymous · Answer

May I know what Q stands for @bmp

anonymous · Answer

Q(t) is the amount of salt for a given time t. :-)

anonymous · Answer

The answer is Q(t) = 200 - 170e^(-t/50)