Aeration dominant position

Aeration dominant position                


        Evaporation leaded by the heat science

If the air velocity is very important, we can simplify the problem, it may still be necessary to consider that the convection transport is very effective. Then, in that case, it is the conduction inside the fluid which could master the mechanism. In other words, the convection takes the vapour faster than the vapour is produced.

At the interface, we can made a thermal flux balance. As we know, the interface is in a balance state :

$$ 0=q_s+q_L$$

$$0=-\lambda_{liq} S_{flux} \frac{\partial T}{\partial y}\mid_{y=h}+\dot{m_{vap}}h_{fg}$$

with that we can write

$\dot{m}=\rho_l v_l S_{flux}=\rho_s v_s S_{flux} $

and $v_s$ is a boundary condition for the air at the interface liquid-vapour. This is brought by the mass conservation.

Now we can write $v_l=-\frac{d h}{d t}$

with $h$ the effective length of the liquid film.

We have $q_s=q_l$

afterwards : $$\frac{\lambda_l \Delta T}{h}=\rho_l \frac{d h}{d t} h_{fg}$$

We can deduce : $$h^2(t)={h_0}^2-\frac{2\lambda_l \Delta T}{\rho_l h_{fg}} t$$

with a characteristic time $$ \tau=\frac{h_0 \rho_l h{fg}}{2 \lambda_l \Delta T}$$


           Numerical Application :

for $\Delta=T_{wall}-T_{sat}=1 K, h_0=100\mu m, h_{fg}=2400 kJ/kg, \lambda_l=0,67 W/Km, \rho_l=1000 kg/m^3$.

We obtain:

$$\tau=17,9 s$$