Fluid force from a fluid flow vs. from a fluid jet

[Ola Sweden]

In fluid flow (such as wind), the force exerted on a perpendicular surface is derived from kinetic energy and is ρAv^2/2. But in problems involving a jet of fluid that strikes a plate, the force is derived from momentum and is explained to be ρAv^2, lacking the 1/2 factor. Can someone please explain this?

 

[PeroK]

Can you be more specific about the scenarios you are analysing?

My first thought is that if an object reflects a stream of particles, then the force is twice that for the case where the particles are absorbed. I.e. there is twice the momentum transfer in reflection compared to absorption.

 

[Ola Sweden]

Can you be more specific about the scenarios you are analysing?

My first thought is that if an object reflects a stream of particles, then the force is twice that for the case where the particles are absorbed. I.e. there is twice the momentum transfer in reflection compared to absorption.

These 2 links explain the different solutions. Dynamic pressure vs momentum change. As I see it, it is the same problem. A flow of a fluid causes a force on a surface perpendicular to the flow, as it stops the flow. But the solutions differ by this factor 1/2.

https://www.engineeringtoolbox.com/wind-load-d_1775.html

https://link.springer.com/chapter/10.1007/978-1-349-00870-4_16

 

[PeroK]

The first link is far less explicit about the calculations, and doesn't explain why only half of the momentum transfer applies.

The second explicity assumes the momentum of the water jet is lost. That looks correct to me.

 

[Ola Sweden]

I guess it is considered pretty basic physics and they don´t go into it that deep. This wikipedia link offers some insight:

https://en.wikipedia.org/wiki/Drag_equation

I was wondering if anyone has had the same question as I do and found out why the two approaches yields different results, it is very disturbing to me :(

 

[PeroK]

I guess it is considered pretty basic physics and they don´t go into it that deep. This wikipedia link offers some insight:

https://en.wikipedia.org/wiki/Drag_equation

I was wondering if anyone has had the same question as I do and found out why the two approaches yields different results, it is very disturbing to me :(

If a large rectangular surface has a drag coefficient of $2$, then the equations agree.

https://en.wikipedia.org/wiki/Drag_coefficient

 

[Ola Sweden]

If a large rectangular surface has a drag coefficient of $2$, then the equations agree.

https://en.wikipedia.org/wiki/Drag_coefficient

Yes, I have also had the thought that the answer to this is related to that. But- sometimes the drag coefficient is even >2 which means the force would be even larger than the momentum equation would yield... confusing.

 

[PeroK]

Yes, I have also had the thought that the answer to this is related to that. But- sometimes the drag coefficient is even >2 which means the force would be even larger than the momentum equation would yield... confusing.

Greater than $2$ might be an curved surface (curved the wrong way aerodynamically), where the wind cannot escape laterally and is forced back. The maximum coefficient would be $4$, perhaps, corresponding to total reflection of the incoming fluid.

 

[jack action]

I would say it depends on the problem. Do we assume that the mass or the velocity is constant?

If a mass of fluid is moving from point A to point B and its velocity changes then the force is $\frac{d(mv)}{dt} = m\frac{dv}{dt}$ because $m$ is constant where $m = \rho V$. The kinetic energy is thus $\frac{1}{2}\rho V v^2$, which agrees with the principle behind the Bernoulli equation, i.e. conservation of energy. And the dynamic pressure $p_k$ is defined as the kinetic energy per volume, thus $p_k = \frac{1}{2}\rho v^2$.

With a jet of water, the problem is seen differently. The jet comes out of some pipe with a known cross-sectional area $A$ and outlet pressure $p$ and velocity $v$. Here the velocity doesn't change, therefore the force is $\frac{d(mv)}{dt} = v\frac{dm}{dt}$ because $v$ is constant and it is a small mass of water that is constantly added, which is equal to $dm = \rho dV = \rho A dx$, thus $F = v \frac{\rho A dx}{dt} = \rho A v^2$. And then $p = \frac{F}{A} = \rho v^2$.

Maybe others have a better explanation.

 

[boneh3ad]

For one, the force on a submerged plate is not what your links say it is, i.e.
$$|F|\neq\frac{1}{2}A\rho v^2.$$
Engineering Toolbox is sort of renowned for wrong answers, in my experience. In reality, the force is generally proportional to dynamic pressure,
$$|F|\propto\frac{1}{2}A\rho v^2,$$
by way of the drag coefficient, $C_D$:
$$|F|=\frac{1}{2}C_D A\rho v^2.$$
The physics that determine the value of $C_D$ are, to put it mildly, very complex.

The most intuitive answer for why the impinging jet and immersed plate problems are different is because the flow fields are different. The immersed plate has fluid all around it, which affects the way momentum is transferred into the plate. It also has a much greater contribution to the final answer from the wake behind it while the jet problem has nearly zero contribution from the wake.

 

[Ola Sweden]

I would say it depends on the problem. Do we assume that the mass or the velocity is constant?

If a mass of fluid is moving from point A to point B and its velocity changes then the force is $\frac{d(mv)}{dt} = m\frac{dv}{dt}$ because $m$ is constant where $m = \rho V$. The kinetic energy is thus $\frac{1}{2}\rho V v^2$, which agrees with the principle behind the Bernoulli equation, i.e. conservation of energy. And the dynamic pressure $p_k$ is defined as the kinetic energy per volume, thus $p_k = \frac{1}{2}\rho v^2$.

With a jet of water, the problem is seen differently. The jet comes out of some pipe with a known cross-sectional area $A$ and outlet pressure $p$ and velocity $v$. Here the velocity doesn't change, therefore the force is $\frac{d(mv)}{dt} = v\frac{dm}{dt}$ because $v$ is constant and it is a small mass of water that is constantly added, which is equal to $dm = \rho dV = \rho A dx$, thus $F = v \frac{\rho A dx}{dt} = \rho A v^2$. And then $p = \frac{F}{A} = \rho v^2$.

Maybe others have a better explanation.

How do you mean the velocity doesn´t change? The velocity goes from v to zero in the direction of the flow does it not?

 

[Ola Sweden]

For one, the force on a submerged plate is not what your links say it is, i.e.
$$|F|\neq\frac{1}{2}A\rho v^2.$$
Engineering Toolbox is sort of renowned for wrong answers, in my experience. In reality, the force is generally proportional to dynamic pressure,
$$|F|\propto\frac{1}{2}A\rho v^2,$$
by way of the drag coefficient, $C_D$:
$$|F|=\frac{1}{2}C_D A\rho v^2.$$
The physics that determine the value of $C_D$ are, to put it mildly, very complex.

The most intuitive answer for why the impinging jet and immersed plate problems are different is because the flow fields are different. The immersed plate has fluid all around it, which affects the way momentum is transferred into the plate. It also has a much greater contribution to the final answer from the wake behind it while the jet problem has nearly zero contribution from the wake.

Ok this seems interesting.. First, Engineering Toolbox is not the only source for that equation, it is generally accepted when it comes to wind force, like in the Eurocodes for example. I see what you mean with the complexity though, but that sort of comes in with the shape coefficients and not the generic equation. There seems to be something fundamental I am missing here.

I´ve always thought about it as simply a mass m with a velocity v and thus a kinetic energy E has to be braked to a halt in a time t which correlates to a distance d. And so the braking force must be easily obtained from knowing the kinetic energy. But I see now that it might be a bit more complicated when it comes to fluids...

 

[jack action]

How do you mean the velocity doesn´t change? The velocity goes from v to zero in the direction of the flow does it not?

If you measure the velocity of the water at the outlet of the pipe, it will be constant even though the source of the water may be a very large tank where the velocity is assumed to be zero. What changes is that new water is always added to the jet. Once the jet hits the plate, I just assume it is the same thing in reverse. It's a question of choosing the appropriate control volume.

Again, maybe others may explain this better.

Engineering Toolbox is not the only source for that equation

In the link there is a note:

Note - in practice wind force acting on a object creates more complex forces due to drag and other effects.

it is generally accepted when it comes to wind force, like in the Eurocodes for example.

Based on this document, no forces are estimated without a coefficient included. (Although there are 3 special cases where $C_sC_d = 1$)

 

[boneh3ad]

Ok this seems interesting.. First, Engineering Toolbox is not the only source for that equation, it is generally accepted when it comes to wind force, like in the Eurocodes for example. I see what you mean with the complexity though, but that sort of comes in with the shape coefficients and not the generic equation. There seems to be something fundamental I am missing here.

I´ve always thought about it as simply a mass m with a velocity v and thus a kinetic energy E has to be braked to a halt in a time t which correlates to a distance d. And so the braking force must be easily obtained from knowing the kinetic energy. But I see now that it might be a bit more complicated when it comes to fluids...

The static (thermodynamic) pressure in a moving fluid is $p$. This $p$ is the pressure "felt" by an object immersed in a fluid. The dynamic pressure in that fluid, which is essentially a measure of kinetic energy per unit volume, is $1/2\rho v^2$. If you combine the two, you get what we call total pressure or stagnation pressure,
$$p_0 = p + \frac{1}{2}\rho v^2.$$
Unless there is a dissipative process occurring, total pressure is constant in the flow, so when a fluid stagnates (comes to rest) against an object immersed in it like your plate, $v\to0$ and you end up with
$$p = p_0.$$
From above, you know that $p_0>1/2\rho v^2$, so by extension we know that the force on that side of the plate, $F=p_0 A$, has to be higher than that provided by the dynamic pressure alone.

Now, the net force on the plate is
$$F_{\mathrm{net}} = F_{\mathrm{front}} - F_{\mathrm{back}} = \left(p + \frac{1}{2}\rho v^2\right)A - p_{\mathrm{back}} A.$$
So, the only way that the net force is equal to $1/2 A\rho v^2$ is if $p_{\mathrm{back}} = p$. This is not generally true, however, because a wake will form behind the plate as the fluid moves over it, and that wake will generally produce static pressures that are lower than $p$, we know that the net force will also generally be greater than that provided by static pressure alone.

Of course, the front of the plate is not identically at $p_0$ across the entire acreage, either. The flow is only truly zero velocity (without the action of a dissipative process) at the stagnation point, so this is an imperfect analysis. But it does serve to illustrate that simply using the dynamic pressure doesn't work.

 

[256bits]

Ok this seems interesting.. First, Engineering Toolbox is not the only source for that equation, it is generally accepted when it comes to wind force

You are totally correct.
But it is not at all that interesting. Drag is actually misleading the answer to your question.
Engineering toolbox is accepted for the force of the wind on a surface.

Drag force is that for an object subjected to a moving fluid, and can contain skin effects due to viscous effects, as well as the force from the form of the object due to the cross section.
None of that applies to your question about the force on a surface.

You asked about the force on a surface, and why force obtained from Bernuoilli, and stagnation pressure, or from the impulse-momentum equation give different answers by 1/2.
Should you design your windows of your house by which equation so that they do not blow in.

 

[256bits]

Ok this seems interesting.. First, Engineering Toolbox is not the only source for that equation, it is generally accepted when it comes to wind force

You are totally correct.
But it is not at all that interesting. Drag is actually misleading the answer to your question.
Engineering toolbox is accepted for the force of the wind on a surface.

Drag force is that for an object subjected to a moving fluid, and can contain skin effects due to viscous effects, as well as the force from the form of the object due to the cross section.
None of that applies to your question about the force on a surface.

You asked about the force on a surface, and why force obtained from Bernuoilli, and stagnation pressure, or from the impulse-momentum equation give different answers by 1/2.
Should you design your windows of your house by which equation so that they do not blow in.

 

[cjl]

Engineering toolbox is accepted for the force of the wind on a surface

No, as boneh3ad said, Engineering Toolbox is wrong here (and they are frequently oversimplified). The correct answer will always involve a coefficient Cd, and will vary significantly depending on the geometry. In some cases, Cd will be close to 1, in which case it will be equal to the engineering toolbox equation, but that is not generally true.

As for the jet, boneh3ad already explained the differences quite well there, so I'll just refer to his post above.