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EPQ: Mathematical model of tea bag diffusion

  1. Aug 9, 2014 #1
    I've done some research, and I've looked at the derivation of the diffusion equation, but I don't want to reproduce even a more simple version of it. I'd like something that can I do almost exclusively myself. The point of this is not to plot some graphs, and extract some line of best fit or exponential equation, but to work toward a model (tea concentration as a function of time, probably) that can fits data, with improvements being made accordingly. I've tried the Rayleigh method of dimensional analysis as a start, but with few variables, I couldn't see it though. So, are there any mathematical techniques/methods that you guys suggest I could use, or anything else? I'm not looking at something exact, but maybe an approximation for a cylinder, or sphere, maybe. Thanks so much.
     
  2. jcsd
  3. Aug 9, 2014 #2
    I would solve for concentration left in the bag C(t) by solving the ODE
    ##\frac{dC}{dt} = -kC## where k is a constant.
    That is the simplest solution I can think of for the situation without the diffusion eq.
     
  4. Aug 9, 2014 #3
    I'm in high/secondary school, so I'm know expert in solving, but I know what it involves. Anyway, I had considered that, but would it not just give something of the form c(t)=c(0)*e^(-kt)? Or are you saying that k is a function of all of my other variables (temp, volume..)?
     
  5. Aug 9, 2014 #4
    K would be based off of the data, say you had 60% of the tea left After 40 seconds. That would allow k to be found. With all of those variables, I don't think an OdE would be sufficient.
     
  6. Aug 9, 2014 #5
    Ah, I see, but having second thoughts, I don't thing it will work because it models a drop in concentration, whereas diffusing tea would result in conc increasing, and thereafter plateauing, right?
     
  7. Aug 9, 2014 #6
    Well, the concentration of the tea left in the bag is decreasing as it dilutes with the water. It would have a minimum, when the water has the same concentration as the bag has left. If we were talking about the waters concentration, you would be correct. ( it would increase and plateau)
    Either way is a valid model.
     
  8. Aug 9, 2014 #7
    Yeah, I;m talking about the water's conc
     
  9. Aug 9, 2014 #8
    I've been sitting here all day, not really sure where to go (well, I have a few ideas), but are you sure there isn't anything else you know of? It doesn't have to be simple
     
  10. Aug 9, 2014 #9
    If I knew pde'a possibly. I'm trying to find a good diff eq to model it.
    Edited for spelling.
     
    Last edited: Aug 9, 2014
  11. Aug 9, 2014 #10
    Not sure what that means, but, ok
     
  12. Aug 9, 2014 #11
    Something such as ##\frac{dC}{dt} = -k(C-100)## where I chose 100 because I wanted it to max at 100%.
    Graph the solution of that. Does that look like you would expect? I can graph it for you if you can't.

    pDE are partial differential equations, like the diffusion equation with ##\partial## derivatives. It's higher level math.

    Edited for correct signs.
    My previous post should've said good model not gold lol.
     
    Last edited: Aug 9, 2014
  13. Aug 9, 2014 #12
    Thanks, will look at it. What is y? Just a constant?
     
  14. Aug 9, 2014 #13
    Well y is the concentration of tea in the water. You can use c' and c if that makes it clearer for you.

    Edit my bad. I see what you meant. I fixed my previous post.
     
  15. Aug 9, 2014 #14
    If I integrate, I get c(t)=kt(100-y). I already have a c term. So is y the same thing? Thanks
     
  16. Aug 9, 2014 #15
    So you know, the solution is ##C =p*e^{-kt}+100 ## where p is a constant, and e the natural base =2.7.... . Since at t=0 there is no tea in the water, we have p*e^0t+100=0 which is the same as p+100=0. So p=-100.
     
    Last edited: Aug 9, 2014
  17. Aug 9, 2014 #16

    These integrations are more complex than that. Have you heard of wolfram alpha? It should give a step by step solution so you can see. P.s. At high school leve, this math is totally unnecessary. It's 2nd/3rd year college lvl.
     
  18. Aug 9, 2014 #17
    For this project, I have to go beyond what I know
     
  19. Aug 9, 2014 #18
    Ok. Well the graph should look like this; is it what you expect?
    ImageUploadedByPhysics Forums1407606257.679452.jpg
     
  20. Aug 9, 2014 #19
    Wow. Thanks
     
  21. Aug 9, 2014 #20
    Your welcome. I hope that was what you're looking for! :)
     
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