What is the formula for calculating water jet reaction force?

AI Thread Summary
The discussion focuses on calculating the reaction force of a water jet for maneuvering a small boat using a pump. A formula from a firefighting textbook is presented: R = 0.157 * P * d^2, where R is the reaction force in Newtons, P is pressure in bar, and d is nozzle diameter in mm. The user also derives a formula for nozzle diameter based on flow rate and pressure, ultimately confirming that both methods yield the same reaction force of 130 N for a specific pump configuration. Additional considerations include potential losses due to pipe friction and pressure effects when the nozzle is underwater. The formulas discussed are validated through principles of fluid dynamics, reinforcing their accuracy.
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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