Jet nozzles effect in seawater

It is not clear what one is asking. Is the high-pressure nozzle operating underwater (as opposed to shooting into the air)?

If one uses the appropriate equation, then one simply uses the density and viscosity of sea-water.

A water jet underwater will dissipate quickly in the surrounding water, but it depends on the velocity of the jet. If the nozzle is used to propel a craft then the crafts forward speed would need to be considered.

 

Sounds like a jet cleaning system to clean underwater surfaces.

I see if I can find an example. I would imagine one wants some equation that is a function of distance and angle to surface, both of which will effect the momentum of the fluid at the surface.

For seawater, I'd recommend a 6 Moly SS, like ALX-6N or 254 SMO or the more recent Avesta 654 SMO.

http://www.alleghenyludlum.com/Ludlum/documents/AL_6XN_SourceBook.pdf

http://www.avestapolarit.com/upload/documents/technical/datasheets/AVPHighAlloyed.pdf [Broken]

See - http://www.avestapolarit.com/upload/documents/technical/acom/acom92_2.pdf [Broken]

Stepping outside the standards for higher corrosion resistance

 

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.