Diff Eq tank problem not working out

[Indychus]

I know you guys must get tired of the tank problem, but I just can't seem to get this figured out. This is my first post here, so hopefully this is in the right subforum and is presented properly.

"A brine solution of salt flows at a constant rate of 4L/min into a large tank that initially held 100L of pure water. The solution inside the tank is perfectly mixed at all times, and flows out of the tank at a rate of 3L/min. If the concentration of salt in the brine entering the tank is .2kg/L, determine the mass of salt in the tank after t minutes. At what time will the concentration of salt in the tank reach .1kg/L. Assume the tank is infinitely large."

To solve this, I set up my D.E. as follows. I omitted units to cut down on clutter:

dx/dt = input-output (x being concentration per time t)
dx/dt = (4 * .2) - (3 * (x(t)/100 + t))

I then put it into standard form: dx/dt + (3/(t+100))x=.8
Since this is linear, I solved using an integrating factor |(t+100)^3|

This gave me an equation x = .2(t+100)+(C/(t+100)^3)
I found C to be -2 x 10^7 using initial conditions x(0)=0

Plugging .1 in for x, I solved for t and obtained .1252 min or 7.506 sec. This seems very fast to raise the concentration so drastically. I have reworked it several times, but can not find my mistake. I will admit my integration skills are a bit rusty. Also, I neglected the absolute value sign in the integrating factor since negative time is not possible, maybe that is my error? Test tomorrow, appreciate any help!

 

[vela]

Your math is fine, but the way you have it set up, x(t) isn't the concentration — it's the total amount of salt in the tank. The concentration at time t would be x(t) divided by the volume of solution in the tank at time t.

 

[Indychus]

Wow, haha, such a simple mistake has eluded me for the better part of an hour. Thanks!

 

[lanedance]

this might be easier to do in terms of volumes

the rate of change of water in the tank will be,
V'(t) = v_in - v_out = (4-3)

V(t) = t + 100


the change in total salt water will be
s'(t) = s_in - s_out = v_in*c_in - v_out*c_out
= 4*0.2 - 3 *(s(t)/V(t))
s'(t) = 0.8-3s(t)/(100+t)

or in terms of the concentration
c(t) = s(t)/V(t)

then differntiating using the quotient rule
c'(t) = s'(t)/V(t) - s(t)/(V(t))^2

 

[Indychus]

Thanks for the quick responses guys, I've got a handle on it now. I'll probably be back in an hour or so with a heating/cooling problem though, so bear with me :)