# 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