Mixing Problem (variable volume)

  • Thread starter Thread starter Tyst
  • Start date Start date
  • Tags Tags
    Mixing Volume
Tyst
Messages
26
Reaction score
0

Homework Statement


A tank contains 150L of water in which initially 30g or salt is dissolved. Brine runs in at a rate of 4 L/m and each litre contains 1.5g of dissolved salt. The mix in the tank is kept uniform. Brine runs out at 3L/m. What is that amount of salt in the tank at any time t?
Note that the volume of brime in the tank is not constant in time


Homework Equations



y' +p(x)y = r(x)

General solution to first order, linear ODE

y(t) = exp(-\intp(p(t)) dt) [\int r(t)\exp(\intp(p(t))dt) dt +C]



The Attempt at a Solution



Input/Output Gives

y(t)=(6-3y)/(150+t) <br />
-
y&#039;+3y = 6/(150+t)

Substituting these values into the general equation gives me

y(t) = e^(-3t)[\nt(6/(150+t)*e^(3t) dt) + C]


And from here ... well... some sort of a clue as to where to go next would be very nice.
Basically i am having trouble with that integral, is it possible to take \int(ln((6/(150+t)*\exp^(3t) dt) somehow? I'm not sure if/how to make this step.
I can take the log of the other parts of the equation...

ln(y(t)) = ln(exp^3t)+ [this is the bit I'm not sure of] + ln(C)
 
Last edited:
Physics news on Phys.org
Yes, thanks for showing me that - perhaps it's just the tiredness setting in, or perhaps I'm just completely dense! But i am still not able to work this out.
 
Well, I'm an idiot :P solved it now! Thanks!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top