# Governing equations of discharge

Dimensioning of an hydraulic system

for water bombing from an aircraft

Governing equations of discharge

Geometry of the tank

The tank considered in the study is the following : The size of the tank and of the doors are defined by:

• $A_{in}=L_TW_T$
• $A_D=L_DW_D$

With :

• $H_T=2 m, L_T=2 m$ and $W_T=2 m$ the height, length and width of the tank
• $L_D=0.5 m$ and $W_D=0.5 m$ the length and width of the door

Variables and physical properties

We consider a fluid whose properties are :

• $\mu_L \sim 10^{-3} - 1.6 Pa.s$
• $\rho_L \sim 1000 - 1090 kg/m^3$
• $g$ the constant of gravity
• $P_{atm}$ the atmospheric pressure which is the value of pressure for inlet and outlet of the tank
• $h$ the height of water in the tank varying in time
• $u_{in}, u_{out}$ the inlet and outlet vertical fluid velocities

$u_{in}=-\frac{dh}{dt}$ velocity of the free surface

• $Q, V$ the flow rate of water dropped and its volume

Definition of the problem

The flow rate is relevant to determine the characteristics of the hydraulic system. To that end, it is essential to provide an accurate knowledge of the volume of liquid released per time.

This study leads to calculate the fluid position thanks to the vertical velocity efflux $u_{out}$.

In the tank considered, the fall of the liquid is free : the outflow from the aircraft tank to the end of the discharge will change from acceleration-dominated to steady-state. The tank doors can be suddenly opened to the maximum exit area or progressively open in order to control the flow-rate (this situation is in practice used to provide an uniform flow-rate : see Opening law of doors and Law for constant flow-rate)

Acceleration-dominated situation

From the Navier-Stokes equation for the conservation of momentum :

$\frac{\partial \rho U}{\partial t} + \nabla. \rho UU = -\nabla P + \rho g + \mu \nabla. T$

The acceleration-dominated velocity, which includes a pressure-related term is the following :

$u_{A}=u_{out}+\Delta t \left(g+\frac{\Delta P}{\rho h} \right)$

$u_{out}$ is the efflux velocity in the previous time-step (at $t= 0 s \Rightarrow u_{out}=0 m/s)$

After a given time, we can neglect the acceleration, so the flow becomes steady-state (and no longer acceleration-dominated). By assuming that the vertical flow in the tank is inviscid, incompressible and irrotational along a streamline, we can describe it with by Bernoulli equation :

$\frac{1}{2}\rho U^2 + \rho g z + P = constant$

We can apply this principle between two points : the top of the surface of the tank and the bottom. So that :

$\frac {1}{2} \rho u_{in}^2 + \rho g z_{in} + P_{in} = \frac{1}{2} \rho u_{out}^2 + \rho g z_{out} + P_{out}$

Hence

$u_{out}=\sqrt{\frac{2g\rho h+2\Delta P}{\rho}+u_{in}^2}$

with $\Delta P$ the difference of pressure between the inlet and the outlet of the tank and $h$ the height of water in the tank (time-varying).

We compare the two regimes by ploting the outlet velocity : The two regims tend to quasi steady states close to each other.
The difference regards the first second of simulation: indeed, the fluid accelerates to tend to maximum speed and then decelerates.
This phase is shown by the curve of the outlet velocity in acceleration
state ($u_{out}=u_{out A}$), but is not represented on the steady state curve ($u_{out}=u_{out B}$).

However, the acceleration phase is quite short compared to the steady state phase and it was easier for us to calculate the outlet velocity thanks to the Bernoulli equation, therefore we choose to neglect the acceleration phase and to define :

$u_{out}=u_{out}B$

Relevant fluid mechanical principles

Given the outlet velocity, the conservation of mass leads to a value of flow-rate :

$Q=u_{out}A_{out}=u_{in} A_{in}$

and the volume of fluid released :

$V=A_{in}(H_T-h)$

Results for the base case

Here are the results for the case of a discharge of the tank with a sudden opening of the doors and without the acceleration phase:

Outlet velocity

Inlet velocity

Water flow

Height of water

Volume dropped

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