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Old Jun 22nd 2016, 12:05 PM   #1
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Effect of orifice diameter on force of water jet

Just realised I posted this in the high school section. (While the question is quite basic, I thought due to the practical application, it might be better here)

I am working on a job in wastewater treatment which involves rotary arms which distributes water onto filter media (Trickling Filter).

The rotation of the arms is driven by the force of the water jets exiting orifice holes along the length of each arm.

Would a decrease in the diameter of these holes reduce the force of the water jet, and hence reduce speed, or the other way round?

Its been a while since I studied this stuff, but I think that a reduction in orifice diameter decreases the force presented by the water jet despite the increase in velocity.

(I remember sitting in a university lab with a water jet hitting a plate, just cannot remember my findings!)

Hope you can help
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Old Jun 22nd 2016, 04:48 PM   #2
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Effect of orifice diameter on force of water jet

I'll give you my quick guess

If you think of a tank of water that is constantly being refilled and imagine place on the side of the tank. If the water level is maintained at a constant level then the pressure on the side of the tank at a given height will be constant and the force F on the wall of the tank will be given by

F=PxA
since P is constant then the force F is proportional to A. If there is suddenly a hole at this point then water will experience this force and be pushed out with that force. For a circular hole then I would expect the force on the water particles to vary in proportion to the square of the radius.

ma=P.pi .r^2
mdv=(P.pi. r^2)dt
so v=(P.pi r^2)t (v=0 at t = 0)

so(assuming we only consider the jet of water for 1s because gravity will slow the droplets down rapidly) then the change of velocity of the water particles being emitted per second is
v/t=(P.pi r^2)

Hence I think v is proportional to r^2 and r is proportional to sqrt(v).
Decreasing r will decrease v.

All this assumes P is constant i.e. the head of water in the tank is maintained constant).
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Old Jun 22nd 2016, 06:17 PM   #3
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Summation of moments-of-momentum...

There's a lot going on here. What is wanted? - rotate faster, pressure-clean with the jet? Some objective-specific assumptions are needed. First, physics thinking will help.

Rotation of the sprinkler arms happens as a consequence of the combined "moment of momentum" of the water jets that issue from the orifices.

It is reasonable to assume the sprinkler arm inside diameter and its rotation rate as such that water properties "prior to the orifice" are the same for all orifices in an arm. I mean the pressure of water in the arm, hub to last, outermost orifice --- Same Pressure. This narrows things a little.

Moment of momentum (of the jets) sum to be a torque acting on the arm. There is friction ~ the faster the rotation the more. Constant rotation rate happens.

The momentum of each jet is mV (mass water times velocity). Proportional to the mass leaving, times its velocity (non-linear). The moment of momentum for each jet is its momentum times its moment-arm.

This is what you're talking about... No small situation...

Good Luck, JP

Good Luck JP
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Old Jun 23rd 2016, 06:10 AM   #4
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Let me try an explanation. What you are describing is similar to a rotating lawn sprinkler. The force that causes the arms to rotate is due to the change in momentum of the water as it flows out along the length of the arm and then turns 90 degrees to flow out the orifice on the side of the arm. Assuming that the orifices are perpendicular to the arms, that change in momentum is equal to mass flow rate (Kg/s) times exit velocity. The mass flow rate is dependent on the pressure of the water and the diameter of the orifices - if you decrease the diameter of the orifices the mass flow rate also decreases, assuming that the pressure of the water in the arms remains the same. And in general the exit velocity remains essentially the same (again assuming no change in pressure). Hence if the water pressure inside the arms remains constant when you reduce the size of the orifice then the force (actually the torque) produced by the water streaming out of the orifices is reduced.

However (there is always an however) in practice it may be that decreasing the orifice causes an increase in pressure in the arms, which would mean the exit velocity of the water increases. This is what happens when you put your thumb over the end of a garden hose - the pressure within the hose increases due to the constriction caused by your thumb. So the final answer is - it depends on whether the internal pressure of the water inside the arms remains constant as you decrease the diameter of the orifices.
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Old Jun 23rd 2016, 01:42 PM   #5
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Thanks, I thought about it more today. And came to the conclusion that its actially quite a difficult question to answer.


The smaller holes will reduce the driving force, but because it is fed from a chamber by gravity, there will surely be some increase in pressure.

Thanks for the help, I dont often get a chance to think about the fundamental physics behind problems anymore, so all your responses are very interesting
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