Evaporation

 


Establishment of a model for the water bombing from an aircraft


So far, 100% of the liquid released reaches the ground. In reality, a portion of liquid is lost due to evaporation. To model this phenomenon, we relied on the Reveillon's equations, giving the evolution of the radius. We assumed that there is a large difference of temperature but no gradient at first step.

$ r\frac{dr}{dt}=4\frac{\rho_G  D}{\rho_L}  Sh_c  ln(B_M  +  1) $

$ Sh_c=\frac{0,55*{Re}^{\frac{1}{2}}*{Sc}^{\frac{1}{3}}}{{\left(1+\frac{1,232}{Re*{Sc}^{\frac{4}{3}}}\right)}^{\frac{1}{2}}} $

$ B_M={\left(B_T+1\right)}^{\frac{Nu_c  Sc}{Sh_c  Pr}}-1 $

$ B_T=\frac{\rho_L  U_p  Cp_L}{h_c} $

$ h_c=\frac{1,14 + 0,17}{d_p}{\left(T_f-T_s\right)}^{0,25} $

With :

  • $ r $ the droplet radius
  • $ B_M $ the mass Spalding number
  • $ B_T $ the thermal Spalding number
  • $ Nu_c $ the convective Nusselt number
  • $ Sc $ the Schmidt number
  • $ Sh_c $ the convective Sherwood number
  • $ Pr $ the Prandtl number
  • $ h_c $ the convective exchange coefficient
  • $ T_f=30 °C $ the fluid temperature
  • $ T_s=10 °C $ the droplet's surface temperature

At the moment, the drop does not undergo enough the evaporation because the radius diminishes from 4 mm at the beginning of the collapse to 3,9 mm when it reaches the ground.

Hence, the results are unsatisfactory and need to be dug more deeply.

 

 

 

 

 

See next : Wake capture

See also : Notations

 

 


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