# Mass balance for diffusion and analogy to heat conduction

1. Nov 22, 2014

### Maylis

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. Nov 22, 2014

### Staff: Mentor

ρ=ρ(0) at x = 0
ρ=ρ(L) at x = L

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