# Estimating the force exerted by a water jet on a flat plate - Bernoulli vs Momentum Equation

• Pra_S
Pra_S
Hi,

Case 1:

Imagine a steady liquid water jet with unit cross sectional area - incompressible and inviscid. density D and uniform velocity V. Operating atmospheric pressure is 0.0 Pa.

The jet is in air ( no air resistance ) and hits a stationary flat plate perpendicularly.

Applying momentum equation gives - the force on the plate by the water jet as : D*V*V = DVV.

Hence the pressure on the plate = DVV / Area = DVV. (Area = 1)

Case 2.

For the same scenario the Stagnation pressure is = 0.5*D*V*V = 0.5 D V^2
The stagnation pressure is the pressure just before the plate, when the fluid completely stops on impact.

However, the two answers are different.

I have read the followings in the literature.

1. In practice, when the water first hit the plate it creates very high pressure (water hammer), which lasts for few micro seconds.

2. Then the pressure reduces and settles at the stagnation pressure.

My question is:

What quantity ( relevance in practice) is calculated using the momentum equation.

PS. - I have seen that , the initial water hammer on the plate can be calculated by Density*V*Speed of sound in water. Which is different than what is estimated though the momentum equation.

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It depends on where the energy goes in the collision and flow. The momentum equation by itself is not the whole story...we need to do energy accounting as well. The Momentum Equation works in concert with the First Law of Thermodynamics. Fundamentally, you "think" you are finding "strange things" because you are applying one without the other.

You also have to be careful about mass conservation. If a flow is hitting a flat plate it’s being deflected. Bringing the flow to a complete halt (as you do in one of your scenarios ) you are unintentionally accelerating the flow from ##v## to 0. Where is the mass going? That is not steady flow, you have to account for momentum accumulation in the control volume as well. In that scenario there must be non-conservative work done on the flow. All the energy it had on the way in must be lost to heat.

We can try to work some basic problems to see the process If you would like.

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jack action, berkeman and DeBangis21
Hi erobz. Many thanks for your insight.

Probabaly this paper will provide an ansewer.

Yuan, G. Y., Ni, B. Y., Wu, Q. G., Xue, Y. Z., & Han, D. F. (2022). Ice breaking by a high-speed water jet impact. Journal of Fluid Mechanics, 934, A1.

My aim was to find out the pressure on the wall to asses the baility of the wall material in withstanding the water jet for a ceratin application.
Apperenty, stage 1 and stage 2 can be controlled suitably by grdually increasing the jet pressure.
Stage 3 depnds on the final supply pressure.

I disagree with the statement that you must solve the energy equation. The nice thing about incompressible flows is that the energy equation is decoupled from the mass and momentum conservation equations, i.e., you can get forced without the energy equation.

The reason the Bernoulli equation approach outlined above does not work is that it only applies at a stagnation point. That's a single point rather than the full redirected flow against a surface. For the total force on the plate, you need to look at the total momentum change normal to the plate.

To remove ice from a wall using a water jet, you might separate the ice from the wall by a diagonal impact, not perpendicular. A diagonal impact launches an acoustic wave in the faster medium, being followed later by a wave in the slower medium. Those two waves shear the ice-wall boundary, so large sheets separate and fall away. That should minimise the energy needed to remove the ice from the wall, but it may leave the paint on the ice. You need to know the speed of sound in ice, the wall material and the paint. A diagonal jet should also do less damage to the wall material.

I disagree with the statement that you must solve the energy equation. The nice thing about incompressible flows is that the energy equation is decoupled from the mass and momentum conservation equations, i.e., you can get forced without the energy equation.
Let me pick your brain, to see if I can understand. In any real fluid there is going to be viscous loses and even some small gravitational effects. Thats going to determine the outlet area, velocity and pressure distributions in the outflow?

The typical (undergraduate) problem of an impinging jet on a flat plate, we have steady flow, and we assume uniform distributions of ##P## and ##V## and neglect ##z##. From Bernoulli this ends up in a restatement of ##V_{1} = V_{2}##. Did we need it... I guess not because mass conservation and unform distribution already demanded it. However, if we are actually to find a more detailed picture of ##P## and ##V## etc..., how is that done without energy considerations?

I feel like the decoupling is the exception, not the rule. I agree that I can find ##\vec{F}## without considering energy here:

But I also think this is only possible because of all the "strictly untrue" assumptions I made.

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For an incompressible flow, kinematics and thermodynamics are decoupled. Since density is constant (i.e., no longer coupled to temperature), you can solve the velocity and pressure fields directly from continuity and momentum conservation and ignore temperature altogether.

The other key is that @Pra_S specified inviscid flow, so any effects of viscous dissipation are neglected. Really, though, viscous dissipation is going to be a pretty small component of the overall energy balance in a problem like this anyway. Gravity is similar unless the ##\Delta h## is reasonably large (and what constitutes "reasonably" will depend on what fluid is being considered).

For an incompressible flow, kinematics and thermodynamics are decoupled. Since density is constant (i.e., no longer coupled to temperature), you can solve the velocity and pressure fields directly from continuity and momentum conservation and ignore temperature altogether.
Could just these equations be used to tell me the conditions/geometry in which a flow might transition from laminar to turbulent? Take my example, I could specify laminar flow inlet. Could be laminar flow could be turbulent on the way out with a viscous flow. How can the distributions be solved?

Is the flow involved a transient flow (sudden start-up of the jet against the wall) or is it steady flow? If it is steady flow, then the issue is resolved by noting that, by the Bernoulli equation, the kinetic energy is constant; the axial kinetic energy of the jet before impact becomes the radial kinetic energy of the sheet of fluid spreading radially on the plate. The fluid jet does not stagnate at the contact. The axial momentum equation is also satisfied by condition that the force exerted by the plate on the fluid is equal to ##\rho v^2 a##. How this axial force is distributed on the plate is a more complicated issue.

erobz said:
Could just these equations be used to tell me the conditions/geometry in which a flow might transition from laminar to turbulent? Take my example, I could specify laminar flow inlet. Could be laminar flow could be turbulent on the way out with a viscous flow. How can the distributions be solved?
In theory, yes.

In practice, there is no way to definitively predict the exact location/conditions leading to transition. It's a problem that is extremely sensitive to initial and boundary conditions.

For incompressible flows, you might take a look at the Orr-Sommerfeld equation.

For incompressible flows, you might take a look at the Orr-Sommerfeld equation.
I'd have to do a lot of studying, and likely wouldn't be able to put it together. I'll have to take your word for it, and that my intuition about friction and heat setting up the various gradients is flawed in some way.

erobz said:
I'd have to do a lot of studying, and likely wouldn't be able to put it together. I'll have to take your word for it, and that my intuition about friction and heat setting up the various gradients is flawed in some way.
Viscous dissipation can certainly impact the flow temperature, but for an in compressible flow, it doesn't impact the kinematics. For a compressible flow, it does.

erobz

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