# Effect of viscosity

Dimensioning of an hydraulic system

for water bombing from an aircraft

Effect of viscosity

Pressure losses

Given the geometry  of the tank, it is essential to take into account the pressure losses. These can be linear or singular. • Regular pressure losses are caused by water movement along the walls of the tank (or the pipe).

$\Delta Pr= \lambda \rho \frac{u_{out}^2}{2} \frac{h}{L_T}$

where $\lambda$ (pressure loss coefficient) depends on the nature of the flow, laminar or turbulent. We calculate the Reynolds number to know in which regime we are: $Re=\frac{u_{out} L_T}{\nu}$.

If $Re<2000$ the flow is laminar: $\lambda=\frac{64}{Re}$.

If $2000<Re<10^5$ the flow is turbulent: $\lambda=\frac{0.316}{Re^{0.25}}$  according to Blasius.

change of the fluid viscosity has effects on the pressure losses because a higher viscosity leads to more friction on the walls of the tank.

We can see this effect on the formulas: if $\nu$ increases, hence $Re$ diminishes and $\lambda$ increases: the friction is more important.

• Singular pressure losses are caused by all the other local phenomena, such as section or direction change in the pipe (in our study we are dealing with sudden narrowing at the exit of the tank that creates turbulence, so loss of energy).

$\Delta Ps= \xi \frac{u_{out}^2}{2} \rho$

where $\xi=0.5$

To include the losses of pressure, we adapt the Bernoulli formula:

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

with $\Delta P_{pd}$ the total pressure drop $(\Delta Ps$ and $\Delta Pr)$ and in our study : $P_{in}=P_{out}=P_{atm}$

$\Delta P_{pd}=\lambda \rho \frac{u_{out}^2}{2} \frac{h}{L_T}+\xi \frac{u_{out}^2}{2} \rho$

$\Delta P_{pd}= \rho \frac{u_{out}^2}{2} ( \lambda \frac{h}{L_T}+\xi )$

$u_{out}= \sqrt {\frac{u_{in}^2+2gh}{1+\lambda \frac{h}{L_T}+\xi }}$

Analysis of results

In the base case, we use only one door which opens instantly. We don't consider singular and regular losses of pressure so, as a consequence, we neglect the effects of viscosity.

We simulate two more cases:

-in a second case, we add the losses of pressure for water viscosity $\mu=1.e^{-3}$ Pa.s

-in the third case, we increase the value of the viscosity, to simulate the case of a drop with fire retardants. $\mu=1.6$ Pa.s

First, we compare the base case with the case with water viscosity: We can see that the pressure drops slow the fluid: the orange curve has an inferior maximum compared to the grey one, and, as the fluid moves more slowly, the drop time is longer.

Then, we compare case two, viscous water, with case three, water with retardant, more viscous: The difference between these two cases is slight, we can zoom a time t=0 to observe it: When the viscosity of the fluid varies, the impact on outlet velocity is negligible, therefore a drop of water with fire retardant is similar to a drop of pure water.

We call back the obtained curves for the base case:

Outlet velocity

Inlet velocity

Water flow

Height of water

Volume dropped

If we add the pressure losses in the tank, we obtain:

Outlet velocity

Inlet velocity

Water flow

Height of water

Volume dropped

If we consider a drop with fire retardant:

Outlet velocity

Inlet velocity

Water flow

Height of water

Volume dropped

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