Trajectory and envelope of the column

 


Establishment of a model for the water bombing from an aircraft


Assumptions :

  • the jet remains compact until the breakup location
  • at the breakup location, the disintegration of the jet happens instantly, representing the Kelvin-Helmholtz and Rayleigh-Taylor instabilities, the mass shedding...

 

   

Schematic of the jet element movement along the trajectory and the aerodynamic force (left) ; Scheme of the elliptical deformation of the liquid jet cross-section (right)

 

 

Elliptical deformation :

The jet cross-section deformation in the $ (x,y) $ plan follows the law :

$ k(t)=1+We  \frac{C_{F}}{C_{k}*C_{b}}\left[1-\exp\left(-\frac{t}{t_{D}}\right)\left(\cos\omega t+\frac{1}{\omega t_{D}}\sin \omega t\right)\right] $

$ We=\frac{\rho_G {W_x}^2 D_0}{2 \sigma} $

$ \frac{1}{t_D}=\frac{C_m}{2}  \frac{\mu_L}{\rho_L {R_0}^2} $

$ \omega=C_k  \frac{\sigma}{\rho_L {R_0}^3}-\frac{1}{t_D} $

$ C_F=\frac{1}{3}  ;  C_k=8  ;  C_m=5  ;  C_b=\frac{1}{2} $

 

First idea for the trajectory :

The equation for the jet curvature angle as a function of time is given by A. Mashayek :

$ \frac{d\theta}{dt}=\frac{\left(F_{aero}-F_{shear}\right)  \cos\theta}{\rho_L  \pi  ab  h  U_j} $

$ F_{aero}=C_D  a  h  \rho_G {\left(W_x  \cos\theta\right)}^2 $

$ F_{shear}=\pi  ab  \mu_L  U_j  \sin d\theta $

By combining the two equations, we get the evolution of the jet envelope as a function of time in 3D :

Jet trajectory in the $ (x,z) $ plan

Jet trajectory in the $ (x,y) $ plan

Jet trajectory in the 3D

As we can see it on those graphics, the trend of the trajectory $ \theta(t) $ is correct : the liquid jet is released from the plane, falls under its gravity and then, bends under the aerodynamic forces that apply on the jet.

Yet, we can notice the trajectory is too straight : the liquid column reaches the ground far too close from the discharge location $ (0,0,H) $. Indeed, on the figure "jet trajectory in the $ (x,z) $ plan", when the water reaches the ground $ (z=0  m) $, the liquid has only covered 1 meter on the horizontal axis.

We chose not to work more on that model since it would have been too complicated for the short time we got, but it would be interesting to work on it next years. Actually, it may be more accurate to study the influence of the aerodynamic forces upon the curvature of the jet much further.

 

Second idea for the trajectory :

Then, we focused on the correlation for the jet trajectory, given by Becker :

$ \frac{z}{D_0}=2,32  q^{0,09}  \left(\frac{x}{D_0}\right)^{0,32} $

This formula is correct for momentum ratios of 1-26, $We_{aero}$ of 360-2120, at normal temperatures and pressures from 1,5 to 15 bar. As for our study :

  • $ q=3,25 $
  • $ We_{aero}=98,7 $
  • with normal temperatures
  • $ P=P_{atm} $

Still, we decided to extrapolate Becker's study to our work since the trajectory seemed to be relevant, compared to the videos we watched about firefighting aircraft.

 

Jet envelope :

We used the formula given by Elshamy and Jeng to get the expression of the upper boundary of the envelope :

$ z=H-4,26 D_0 {|\frac{x}{D_0}-0,5|}^{0,349}  q^{0,408}  {We_{aero}}^{-0,30} $

Then, we deduced the lower boundary of the envelope as the symmetric curve of the upper boundary with respect to the jet trajectory :

Model for the jet trajectory, upper and lower boundaries

 

Jet envelope and trajectory :

By combining the elliptical deformation with the jet trajectory and the jet envelope equations, we got the representation of the liquid column after the breakup phenomenon :

Evolution of the liquid column in 3D

Evolution of the liquid column in the $ (x,z) $ plan

As we can see it on the graphics, the liquid column undergoes both a spanwise elliptical deformation in the $ (x,y) $ plan and an elliptical deformation in the $ (x,z) $ plan due to its own gravity. This complies with what we expected after having watched the bombing videos.

 

See next : Breakup location

See also : Notations

 

 


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