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Jet nozzles effect in seawater

  1. Apr 14, 2009 #1
    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?
  2. jcsd
  3. Apr 14, 2009 #2


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    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.
  4. Apr 14, 2009 #3
    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?
  5. Apr 14, 2009 #4


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    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.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
    Last edited by a moderator: May 4, 2017
  6. Apr 14, 2009 #5


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    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]
    [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.
    Last edited: Apr 14, 2009
  7. Apr 15, 2009 #6
    Thanks a lot Minger and Astronuc! This were of great help!
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