What is the formula for calculating water jet reaction force?

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SUMMARY

The discussion centers on calculating the reaction force of a water jet for maneuvering small boats using specific formulas derived from fluid dynamics principles. The primary formula presented is R = 0.157 * P * d^2, where R is the reaction force in Newtons, P is pressure in bar, and d is the nozzle diameter in mm. The user also derived the nozzle diameter using L = 2/3 * d^2 * sqrt P, leading to d = sqrt(L / (2/3 * sqrt P)). The calculations confirm that both methods yield consistent results for a pump operating at 1.52 bar and a flow rate of 450 l/min, resulting in a reaction force of 130 N.

PREREQUISITES
  • Understanding of fluid dynamics principles, particularly Bernoulli's equation.
  • Familiarity with the concepts of pressure, flow rate, and nozzle design.
  • Basic knowledge of Newton's laws of motion as they apply to fluid flow.
  • Ability to perform mathematical calculations involving square roots and unit conversions.
NEXT STEPS
  • Study Bernoulli's principle in detail to understand its application in fluid dynamics.
  • Learn about the effects of pipe work friction on fluid flow and pressure loss.
  • Explore advanced fluid dynamics simulations using software like ANSYS Fluent.
  • Investigate different nozzle designs and their impact on jet performance and reaction force.
USEFUL FOR

This discussion is beneficial for mechanical engineers, fluid dynamics researchers, and anyone involved in designing water propulsion systems for small vessels.

Acebaraka
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Hi all,

my first post on here and just wanted to check something I'm working on for a project.

The project is to use a pump to create a jet of water to manouvre small boat.

I have found a formula to work out the jet reaction force in a fire fighting textbook which gives:

R = 0.157 * P * d^2 where R = Reaction force in Newtons
P = Pressure in bar
d = nozzle diameter in mm

To avoid buying and testing different pumps and nozzles empirically to find the best reaction force I wanted to link a given pumps flow and pressure to find nozzle diameter and then using this and the pressure and the above formula to get the reaction force.

I have been using

L = 2/3 * d^2 * sqrt P where L = flow l/min
d = nozzle diameter in mm
P = pressure in bar

re-arranged to give

d = sqrt (L/ (2/3 * sqrt P))

this gives d in mm then putting this and the same pressure back into the above reaction formula to get reaction in Newtons.

as an example:

a pump giving P = 1.52 bar
Q = 450 l/min

d = sqrt (450/ (2/3 * sqrt 1.52))
= 23.4 mm

then

R = 0.157 * P * d^2
= 0.157 *1.52 *23.4^2
=130 N

Questions:

Are the formulas I'm using valid? If so could someone show me how to get to them from first principles or just tell me they're ok :) If not then why?Other losses: I have thought about losses due to pipe work friction but if the nozzle is underwater what sort of losses could I expect due to reduced flow because of higher pressure at/just after outlet

Thanks in advance for any help.
 
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Welcome to PF acebaraka!

You could try using: F = dp/dt = d(mv)/dt = v(dm/dt) + m(dv/dt). If the speed of the water jet is constant (ie. dv/dt=0), the force is given by f = v(dm/dt) where v is the speed of the water exiting the nozzle and dm/dt is the mass flow rate out of the nozzle. The mass flow rate is the volume flow rate x mass/unit volume of water (1 kg/l).

You can determine the speed of the water using Bernoulli's principle: ##\Delta \frac{1}{2}\rho v^2 = -\Delta P## (the change in kinetic energy per unit volume is equal and opposite to the change in pressure).

AM
 
Andrew Mason said:
Welcome to PF acebaraka!

You could try using: F = dp/dt = d(mv)/dt = v(dm/dt) + m(dv/dt). If the speed of the water jet is constant (ie. dv/dt=0), the force is given by f = v(dm/dt) where v is the speed of the water exiting the nozzle and dm/dt is the mass flow rate out of the nozzle. The mass flow rate is the volume flow rate x mass/unit volume of water (1 kg/l).

You can determine the speed of the water using Bernoulli's principle: ##\Delta \frac{1}{2}\rho v^2 = -\Delta P## (the change in kinetic energy per unit volume is equal and opposite to the change in pressure).

AM
Thanks for your reply, do I have this right ?

delta P would be:

system pressure at or just before the nozzle - (for the sake of ease at the moment) atmospheric pressure

which would be 1.52bar but in SI

so SqRt (P / (0.5*density)) would give v

so
v = SqRt (152000 / (0.5*1000))
=17.43 m/s

Then

F= v * (dm/dt) for 450l/min (dm/dt) = 450/60 = 7.5 kg/s
= 17.43 * 7.5
= 130N

which rather wonderfully comes out to the same as what I had with the other equation, to good to be true? :)
 
Acebaraka said:
Thanks for your reply, do I have this right ?

so
v = SqRt (152000 / (0.5*1000))
=17.43 m/s

Then

F= v * (dm/dt) for 450l/min (dm/dt) = 450/60 = 7.5 kg/s
= 17.43 * 7.5
= 130N

which rather wonderfully comes out to the same as what I had with the other equation, to good to be true? :)
Not at all. I expect that the formulas you had were derived the same way.

AM
 

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