Jet nozzles effect in seawater

Is it possible to calculate a high-pressure nozzles effect/carrying length in seawater if the medium used in the high-pressure nozzle is seawater?

 

Thats right Astronuc, the high pressure nozzle operates under water and is not moving.
The diameter of the nozzle is Ø1,2mm and the media flushed through it is seawater (the same as the media it is submerged in). The flow is 5-10L/min.
What equation would you recommend to figure:
- The length of the spray?
- The effect of the spray(on a surface) if it has a distance to a surface of i.e. 10mm?

 

Oh man, you are lucky I'm in the middle of a numerical run right now.

From Viscous Fluid Flow(3rd ed.), Frank White, Chapter 4, Section 10.6.

If a round jet emerges from a circular hole with sufficient momentum, it remains narrow and grows slowly, the radial changes [tex] \partial /\partial r[/tex] being much larger than axial changes [tex]\partial / \partial x[/tex]
Continuity:
[tex] \frac{\partial u}{\partial x} + \frac{1}{r}\frac{\partial}{\partial r}(rv) = 0 [/tex]
x momentum
[tex]u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} = \frac{v}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)[/tex]

Schlichting reasoned that the jet thickness grew linearly, so the similarity variable is [tex]r/x[/tex]. he defined the stream function
[tex]\psi(r,x) = \nu x F(\eta)\,\, \eta = \frac{r}{x}[/tex]
From which the axisymmetrical velocity components are:
[tex]u = \frac{1}{r}\frac{\partial \psi}{\partial r} = \frac{\nu F'}{r} [/tex]
[tex]v = -\frac{1}{r}\frac{\partial \psi}{\partial x} = \frac{\nu}{r}(\eta F' - F)[/tex]
Substitution into the x-momentum equation gives the following third-order non-linear differential equation.
[tex] \frac{d}{d\eta}\left( F'' - \frac{F'}{\eta}\right) = \frac{1}{\eta^2} (FF'' - \eta F'^2 - \eta FF'') [/tex]
The boundary conditions are F(0) = F'(0) = F'(infinity) = 0. The exact solution is:
[tex]F = \frac{(C\eta)^2}{1 + (C\eta /2)^2}[/tex]
Where C is a constant determined from the momentum of the jet
[tex] J = \rho \int^\infty_0 u^2 2\pi r\,dr = \frac{16\pi}{3}\rho C^2 \nu^2 [/tex]
[tex] C = \left( \frac{3J}{16\pi \rho \nu^2}\right)^{1/2}[/tex]
The axial jet velocity is then:
[tex] u = \frac{3J}{8\pi \mu x}\left( 1 + \frac{C^2 \eta^2}{4}\right)^{-2}[/tex]
The term in parenthesis is the shape of the jet profile. The jet centerline velocity drops off as [tex]x^{-1}[/tex]. The mass flow rate across any axial section of the jet is:
[tex] \dot{m} = \rho \int^{\infty}_0 u2\pi r\,dr = 8\pi \mu x [/tex]

IF the jet is laminar, and we assume a simple plane laminar jet, we find that the velocity distribution is:
[tex] u(x,y) = u_{max} \sech^2 a\eta [/tex] Or:

[tex] u(x,y) = u_{max}\sech^2 \left[ 0.2752 \left(\frac{J\rho}{\mu^2 x^2}\right)^{1/3} y \right] [/tex]

Where J is the momentum flux. At this point, if we define width of the jet as twice the distance y where [tex]u = 0.001u_{max}[/tex] then:
[tex]\mbox{Width} = 21.8 \left(\frac{x^2 \mu^2}{J \rho}\right)^{1/3}[/tex]

That may be slightly more helpful on the width of the jet question. On the effect of the surface....no idea.

 

Thanks a lot Minger and Astronuc! This were of great help!