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I am trying to calculate at what pressure and speed water from a high pressure water jet will hit an object 100 millimeters away from the nozzle opening, but i can't figure it out.

The nozzle will be used at approximately 100 meter below sea level in salt water. Its a circular nozzle with a diameter = 2 millimeters. The water from the water jet is at 400 bar and with a flow of about 60 liters per minute.

Does anyone know what formula to use?

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The formula to use for the nozzle velocity is:

V = (gpm * .321) / A where: gpm = gallons per minute V = Velocity in ft/sec A = Area of nozzle in sq. inches.

A 2mm nozzle = 0.0787 inches. Area A = .785(0.0787^2) = 0.00486 in^2

Flow of 60 L/M = 15.7 gpm.

Nozzle velocity V = (gpm * .321) / A

            V = (15.7 * .321) / 0.00486 = 1036 ft/sec.

Impact force F = (Pn - Po) * A where: Po = pressure at ocean depth of 100 meters or 330 feet or 147 psi units. Pn = nozzle pressure of 400 bar in Psi = 5880.

                                   F = pounds force

Ocean water pressure Df at 330 feet depth is: Po = (Df * SG) / 2.31

               Where: SG = Ocean water specific gravity = 1.03
                       H = one psi for each 2.31 feet of head.

    Thus: Po = (330 * 1.03/ 2.31 = 147 psi.

    F = (Pn - Po) * A:  F = (5880 - 147) * 0.00486 = 5733 pounds.

Keep in mind that F is an impingement on only .00486 in^2 area.

Because type of nozzle not known, I assumed nozzle eff = 100 percent. In reality the nozzle eff. Will be some where between 60 and 90 percent.

By experience, in a stand-off distance of only 100 mm in subsea environment, the water jet stream will remain tight. Also, water speed velocity deterioration is assumed to be negligible.

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Even with a constant flow of 60 litres per min you need to provide at least the nozzle entry area(one other than the 2mm one) or the nozzle inlet velocity. Otherwise you will be left with one unkown in the Bernoulli's equation. If it is a constant area nozzle then the inlet and the outlet velocity will be the same because of the continuity equation(assuming incompressible flow). But that would imply that the change in kinetic energy term will be 0 while the pressure changes if you assume different pressures for inlet as well as outlet(the potential energies will cancel out if heights are the same). This leads to a contradiction as bernoullis equation would dictate the pressure change be equal to 0 to maintain equality. As far as outlet static pressure is concerned it should be close to the sea pressure at that depth.

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  • $\begingroup$ This leads to a contradiction as bernoullis equation would dictate the pressure change be equal to 0... Why a contradiction? If pressure from the source (i.e., the pump driving water to the nozzle) exceeds the surrounding water pressure, then water will flow out of the nozzle. If it does not, then either nothing moves (equal pressures) or there is a back-flow (unless there is no vacant space within the apparatus). $\endgroup$ Apr 20, 2016 at 16:56
  • $\begingroup$ I was talking just about the nozzle. Sure if the pump pressure is greater than surrounding pressure the fluid will flow. But due to constant area of nozzle there won't be any pressure or velocity change across it. $\endgroup$ Apr 21, 2016 at 3:17

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