Water Jet Reaction Force

1. Oct 21, 2015

Acebaraka

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 text book 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.

2. Oct 21, 2015

Andrew Mason

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

3. Oct 21, 2015

Acebaraka

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? :)

4. Oct 21, 2015

Andrew Mason

Not at all. I expect that the formulas you had were derived the same way.

AM