Height of drop pattern

 


Dimensioning of an hydraulic system

for water bombing from an aircraft


 

Height of drop pattern

 

Required flow rate

We analyze the drop pattern for a water dropping previously studied. The situation is the following :

First of all, these are the variables used afterwards :

  • $U_g$  relative velocity between the aircraft and the ground
  • $W$  wind velocity
  • $U_r=U_g-W$  relative velocity between the airtanker and the air
  • $d_S=S^{1/2}$  characteristic length scale of the exit of the tank, based on the exit area $S$ (in our study $S=L_DW_D$)
  • $U_L=Q/S$  mean liquid velocity at the tank exit
  • $T=V/Q$  time of release
  • $\Phi=V_g/V$   fraction of recovered liquid
  • $\lambda$  width of the drop pattern
  • $L$  length of the drop pattern
  • $\rho_L, \rho_a$  density of liquid / air
  • $q=\frac{\rho_LU_L^2}{\rho_aU_r^2}$  jet/air momentum flux ratio

We use the following relations :

$L=U_gT+f_1\lambda$

$\lambda=f_2q^{1/5}d_S$

$f_1\approx2$ and $f_2\approx27$ (for gravity systems) have been set for light air wind conditions $(W\le 1.5 m/s)$

 

In our case, we choose to study the drop of the section : Effect viscosity, and to use data from simulation section : Law for constant flow-rate.

$T$ $14 s$
$Q$ $0.5 m^3/s$
$\Phi$ $0.80$
$\rho_a$ $1.2 kg/m^3$
$U_g$ $50 m/s$
$U_L=Q/S$ $3 m/s$

We are studying the maximum coverage level on the pattern centerline $\eta_{max}$ :

$\eta_{max}\approx4\sqrt{\frac{\ln(2)}{\pi}}\frac{\Phi V}{(L-2\lambda)\lambda}$

Its dependance with the flow-rate is the following :

For a flow-rate of $Q=0.5 m^3/s$ we know $\eta_{max}$.

Besides along the drop pattern, the height of water is given by :

$\eta=\eta_{max}exp\left( \frac{-\lambda_n^2}{2\lambda_0^2} \right)$

$\lambda_0=\frac{\lambda}{4ln(2)}$ the standard deviation

$\lambda_{\eta}$ the width of the coverage level $\eta$

This figure represents the height of drop pattern depending on the width of water on ground :

The calculation of the volume of water on ground gives : $V=6.3 m^3$, which is very close to the amount expected (we release $V=8 m^3$ of water and only 80 per cent of it reaches the ground).

At last, total heat which can be released by evaporation of water on ground is approximately : 525 kW for a 30 s drop. This value is an approximation and a thermal analysis of the vegetation on fire needs to be implement to improve the study.

This aspect will be developed by an other group of BEI on this website : Establishment of a model for the water bombing from an aircraft

Source : "Air tanker drop pattern" - D. Legendre, E. Alméras, A. Chassagne


Top of page