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I Dissolution of a metal sphere

  1. Mar 7, 2016 #1
    Hello All

    I could get some help I would greatly appreciate it.

    I am trying to figure how to calculate the dissolution time of sphere undergoing constant corrosion at a rate
    of corrosion.

    Through a little google-fu, I found an article which gives me the solution ( http://arxiv.org/pdf/1208.5925.pdf )
    but I am having trouble understanding the proof.

    In this article there is a dissolving sphere

    From this article we know that

    the mass of the sphere is equal to dm/dt = -c*s(m) (1

    and the general solution is

    m(t)=mo - A*(mo^2/3)*t + (1/3)*(A^2)*(mo^(1/3))*(t^2) -1/27*(A^3)*(t^3)

    m(t) = (a-k*t)^3
    a = initial mass = (mo)^1/3
    k = (A/3).

    The article then gives an example where

    mo = initial mass = 10 grams
    p = density = 0.8 mg/mm^2
    c = corrosion rate = -0.003 mg/(s*mm^2)

    I am having difficulty understanding how to relate the rate of corrosion c to A.

    I know that c = (dm/dt)/s(m)

    The article shows a series of graphs for Mass, Radius, SA, and Volume vs time.

    I copied these graphs into excel and add used excel to find a trendline.

    m(t) = -2E-07*t^3 + 0.0002*t^2 - 0.079t+ 9.9879

    By plugging values into the trend line above I was able to find a solution to

    A = 0.0173.


    However, do not understand how to relate A to c(-0.003).

    Some help would be greatly appreciated.

    Attached Files:

  2. jcsd
  3. Mar 7, 2016 #2

    Simon Bridge

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    corrosion rate coefficient is given as a number c milligrams of mass dissolved per second from each square millimeter of surface area.
    Therefore, an object with surface area A square millimeters will have an instantaneous rate of mass loss of ##\dot m =cA## milligrams per second.
    You know how to find the surface area of a solid sphere from it's mass and density right?
    Last edited: Mar 7, 2016
  4. Mar 8, 2016 #3
    Hello Simon,

    Thank you very much for your response.

    Maybe I am not fully understanding your answer.

    The A, I am trying to find is not the surface area of the sphere but rather a constant found in the solution to

    dm/dt = -c*s(m) = -c*4πi*(3m/(4πi*ρ)^(2/3) (equation 1)

    where the solution to the equation listed above is

    m(t)=mo - A*(mo^2/3)*t + (1/3)*(A^2)*(mo^(1/3))*(t^2) -1/27*(A^3)*(t^3) (equation 2)

    which simplifies to

    m(t) = (mo^(1/3) - (A/3)t)^(1/3) (equation 3)

    Where A is constant. Maybe I am wrong here, but I do not believe that A equation 2 and 3 is equal to the surfaces area of the sphere.

    Am I wrong here?

    If I am only given the density, rate of corrosion, and diameter of sphere, how would I go about finding equation 3.

    Thank you very much for your help.
  5. Mar 8, 2016 #4

    Simon Bridge

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    Science Advisor
    Homework Helper

    Then you need to identify what physical property the article is using the letter A to stand for. So far you have A=3k .... but that begs the question: what is k? (Though it is possible that A is just the constant of integration...)
    Either that - or discard the article and work out your own solution.
    The second is probably faster.... I've already done it: took me a couple of minutes but then, I've had practise.
    Last edited: Mar 9, 2016
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