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Mass balance for diffusion and analogy to heat conduction

  1. Nov 22, 2014 #1

    Maylis

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    Gold Member

    Hello,
    I just began learning mass transfer, and I am trying to use analogies from heat transfer to help me solve problems. For example, if you have one dimensional heat transfer through a plane wall, I would start with a general energy balance.

    $$\frac {dE}{dt} = \dot Q_{x} - \dot Q_{x + \Delta x} + \dot e_{gen}A \Delta x$$
    $$ \rho A \Delta x \hat c_{p} \frac {dT}{dt} = \dot Q_{x} - \dot Q_{x + \Delta x} + \dot e_{gen}A \Delta x$$
    dividing through and taking the limit as x approaches 0,
    $$\rho \hat c_{p} \frac {dT}{dt} = - \frac {1}{A} \frac {d \dot Q}{dx} + \dot e_{gen}$$

    Then using Fourier's law ##\dot Q = -kA \frac {dT}{dx}## I substitute and get
    $$ \rho \hat c_{p} \frac {dT}{dt} = \frac {d}{dx} (k \frac {dT}{dx}) + \dot e_{gen}$$

    Now I want to extend this to mass transfer, and I find my textbook to be sorely lacking in even properly showing a derivation from the beginning, so I try myself
    $$ \frac {dm}{dt} = \dot m_{x} - \dot m_{x + \Delta x} + \dot m_{gen} A \Delta x$$
    $$ A \Delta x \frac {d \rho}{dt} = \dot m_{x} - \dot m_{x + \Delta x} + \dot m_{gen} A \Delta x$$
    Diving through and the limit as delta x approaches 0
    $$ \frac {d \rho}{dt} = - \frac {1}{A} \frac {d}{dx} (-D_{AB}A \frac {d \rho}{dx}) + \dot m_{gen} $$
    Now I can remove the generation term to get
    $$ \frac {d \rho}{dt} = \frac {d}{dx} (D_{AB} \frac {d \rho}{dx})$$
    Assume steady state, so then I integrate once
    $$ D_{AB} \frac {d \rho}{dx} = C_{1} $$
    $$ D_{AB} \rho = C_{1}x + C_{2} $$
    Now what boundary conditions are appropriate to use?
     
    Last edited: Nov 22, 2014
  2. jcsd
  3. Nov 22, 2014 #2
    ρ=ρ(0) at x = 0
    ρ=ρ(L) at x = L
     
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