Diffusion dominant position

Diffusion dominant position                


 

In this part, the context is the same as in the precedent part  but the air charged in vapour is fixed. Then the molecular diffusion is the only mechanism which transports the vapour into the air.

For this, we will simplify the developed equations in the previous section depending on this specific case.

 

         1- Mass balance equation 

$$ \frac{d}{dt}(hDl\rho)=-\rho_S v_S S_{flux}$$

In this equation the liquid volume variation is equal to the vapour flux out going from the surface.

 

          2- Interface mass balance equation 

At the surface there is :

  • a production of vapour $\rho_S v_S S_{flux}$,
  • a transport by convection and diffusion (Fick's law)

The convection velocity is $v_c=(1-\omega)v_a+\omega v_v$ with $v_a$ air velocity and $v_v$ vapour velocity. The convection velocity in the case of fixed air is $v_c=\omega v_v$

At the surface $v_{cS}=\omega_S v_S$.

 

The output vapour is considered on the interface as : $\rho_S v_{cS} S=\rho_S \omega_S v_S S$.

The balance equation corresponds to "production=diffusion+evaporation",  then :

$$\rho_S v_S S=-D_S \rho_S (\frac{\partial \omega}{\partial y})_{y=h} S+\rho_S\omega_S v_S S$$

We then develop this expression and we obtain:

$$v_S=\frac{D_S (\frac{\partial \omega}{\partial y})_{y=h}}{\omega_S-1}$$

Afterwards the Spalding parameter $b=\frac{\omega}{\omega_S-1}$ then $v_S=D_S(\frac{db}{dy})_{y=h}  (8)$

After calculations, it's simplified to:

$$j_S=\rho_S \frac {D_S}{L-h}ln(1+(b_L-b_S))$$

and becomes:

$$j_S=\rho_S \frac {D_S}{L-h}ln(1+\frac{(\omega_S-\omega_L)}{1-\omega_S})$$

 

         3- Conclusion

We can observe on this evaporation model that the flux depends on the boundary conditions and thermodynamic balance.

The model is available for a mono-constituent. For a particular case we can deduce the thermodynamic balance parameters $T_{Sat}=T_{wall}$ and $\rho_{Sat}$ with tables.

The comparison inside the logarithm from the last formula ( the sign of the difference between $b_L$ and $b_S$)  gives us the sign of the flux:

  • If the mass concentration is more important far from the interface than the mass concentration at the interface, then flux is negative : there is condensation.
  • Else if $b_S>b_L$ there is evaporation. Else $b_L=b_S$ the system is in total balance state.

Moreover, we can deduce a characteristic time of the problem:

$\tau=\frac{L}{v_S}$

To have an order of magnitude, here there's a numerical application in the water case : 

$T_w=300K, \rho_S=2.38*10^{-2}kg/m^3, L=50 cm, D_S=5*10^{-5} m^2/s$

we obtain $\tau=60$ hours.