Formula derivation connecting vertical water flowrate & horizontal distance moved by a suspended sphere

[printereater]

I think not, that force is important in the transient state but in the steady (equilibrium) state is negligible.

EDIT :Ok to correct my self that force is not exactly negligible but I believe it is cancelled by the pressure of water on the ball .

Hi, I just saw your edit. How do we know that that force is cancelled by the pressure of water

 

[printereater]

I suspect it's not a trivial problem to solve theoretically. But experimentally, you should be able to find some functional relationship between the outflow angle and the mass flowrate. you have to measure ( control) incoming flowrate, and measure the angle of deflection.

You are going to be working with a control volume that is something like this:

View attachment 339090

And you apply Reynolds Transport Theorem - "The Momentum Equation" in fluid mechanics.

Which says for steady flow with uniform velocity distribution (one inlet-one outlet):

$$ \sum \vec{F} = \dot m \vec{v_o} - \dot m \vec{v_i} $$

Can you teach how to simplify RTT to this formula please

 

[erobz]

Hi, I just saw your edit. How do we know that that force is cancelled by the pressure of water

Examine the boundary of the control volume... I've already explained this to you. Atmospheric pressure acts over the entire boundary with the possible exception of the small areas inside the fluid jet as it enters and exits the control volume. We have also assumed the fluid jet experiences negligible change in pressure between inlet\outlet. There is simply no room for an unbalanced pressure force term.

 

[Delta2]

Hi, I just saw your edit. How do we know that that force is cancelled by the pressure of water

well if it wasn't cancelled by the pressure from water it would keep pushing the ball and the ball would exit water from the other side but this doesn't happen. @erobz said that the water is at atmospheric pressure but I think this is kind of wrong, the water pressure at the interface with the ball must be a bit bigger than the atmospheric pressure so that it cancels the force from the atmospheric pressure on the other side of the ball, and also provides a force that cancels the net force from the tension and weight.

 

[erobz]

@erobz said that the water is at atmospheric pressure but I think this is kind of wrong, the water pressure at the interface with the ball must be a bit bigger than the atmospheric pressure so that it cancels the force from the atmospheric pressure on the other side of the ball

You are talking about microscopic precision though. The pressure and velocity distributions are also nonuniform all throughout this flow and the flow is accelerating as it rounds the sphere. Its viscous flow over an unspecified spherical boundary in a vertical gravitational field. Its troubles me that this very small pressure difference over very small areas stands as "the theoretical issue" here...

I don't think going into all this helps the OP understand how a control volume is analyzed.

 

[Delta2]

Hmmm, ok not sure about the pressure of water on the interface it would be more correct to say that the sum of the forces is zero since we are at equilibrium. The (vector) sum of the forces is the force from the atmospheric pressure, the tension, the weight, and the force from water. Not sure how exactly they are cancelled , my guess was as what i say in post #109

 

[Delta2]

Its troubles me that this very small pressure difference over very small areas stands as "the theoretical issue"...

Sorry what exactly do you want to say here?

 

[erobz]

Sorry what exactly do you want to say here?

Why all the other complexity involving the inlet/outlet velocity, pressure distributions ignored, but undoubtedly very small difference in pressure in the flow between inlet an outlet is an issue.

 

[Delta2]

Why all the other complexity involving the inlet/outlet velocity, pressure distributions ignored, but undoubtedly very small difference in pressure in the flow between inlet an outlet is an issue.

Hmmm, I meant that the pressure distribution of the water on the contact surface with the ball must be in general a bit bigger than the atmospheric pressure.

 

[Delta2]

@erobz to state it explicitly, what about the force from the atmospheric pressure on the clean side of the ball, why don't we take this force into the equilibrium force balance?

 

[erobz]

This is again( approximately) the pressure distribution acting externally over the entire control volume boundary. The only microscopically questionable pressure distribution areas are inside the flow integrated across the inlet and outlet.

 

[Delta2]

Hmm ok you view it as one system, the ball and the water beam.

 

[Delta2]

Viewing it as one system puts the pressure distribution of water in the interface with the ball "under the mat" since it an internal force of the system.

 

[erobz]

Hmm ok you view it as one system, the ball and the water beam.

Thats how the control volume approach works. We have to capture the flow momentum. We can make the boundary arbitrarily tight the whole way around the ball and flow so we can ignore the weight of the air captured inside as an external force. Why Ignored the weight of the water in the cv is because its more or less in free fall.

 

[Delta2]

Well we put under the mat the question of the OP which is about the same as the question at post 115.

My explanation is that

• On the ball: The total force from water on the ball balances the sum of tension and weight and the atmospheric pressure force (on the clean side of the ball)
• On the water beam: The sum of the atmospheric pressure force on water beam and the force from the ball on water beam causes the curvature of water beam.

What do you think?

 

[erobz]

Well we put under the mat the question of the OP which is about the same as the question at post 115.

My explanation is that

• On the ball: The total force from water on the ball balances the sum of tension and weight and the atmospheric pressure force (on the clean side of the ball)
• On the water beam: The sum of the atmospheric pressure force on water beam and the force from the ball on water beam causes the curvature of water beam.

What do you think?

I think viscosity is the key player. I don't believe the flow pressure significantly deviates from atmospheric. Just my opinion, but examine the actual experiment control volume ( not my crude diagram):

The thickness of the jet is reduced to a film around the ball. The outside boundary of the flow is atmospheric. The inside (where the ball is touching it) is just a fraction of a millimeter away. I simply don't believe there is a massive pressure gradient hidden there. What I believe is there is friction or surface tension grabbing the flow and distributing it, and changing its direction.

If you are looking at just the flow enclosed as the cv, I have trouble drawing a thin enough boundary around it. It seems unlikely the pressure significantly changes across its boundary.

In my opinion, no friction (viscosity) and\or surface tension, no more effect.

 

[erobz]

Something else the OP could try to improve on is the flow appears to come in laminar velocity distribution and leaves turbulent. That changes the application of the RTT (for momentum) a bit to the integral version.

No doubt it is not an easy experiment to carry out with much precision.

 

[printereater]

Something else the OP could try to improve on is the flow appears to come in laminar velocity distribution and leaves turbulent. That changes the application of the RTT (for momentum) a bit to the integral version.

No doubt it is not an easy experiment to carry out with much precision.

Is there anything I can do to ensure that the water leaves as laminar flow when it exits?

 

[printereater]

I think viscosity is the key player. I don't believe the flow pressure significantly deviates from atmospheric. Just my opinion, but examine the actual experiment control volume ( not my crude diagram):

What I believe is there is friction or surface tension grabbing the flow and distributing it, and changing its direction.

In my opinion, no friction (viscosity) and\or surface tension, no more effect.

If this is the case. Is it wrong for me to explain the effect the way they explained in this article?

"A consequence of the Bernoulli theorem is that a fast-flowing stream drags and accelerates some air around it creating a velocity and a pressure gradient in the air: the faster the air, the lower the pressure around the stream. The low pressure tends to be compensated by some air coming from the nearby space. However, if a surface is located in close proximity, no air can arrive and the low pressure tends to bring together the two opposite sides, i.e. the stream and the surface that are pushed together by the ambient pressure exerted on the outer sides of the stream and the surface. Once the stream has adhered to the surface, the external pressure will continue keeping this situation"

 

[erobz]

Is there anything I can do to ensure that the water leaves as laminar flow when it exits?

I don't know, could be a function of ball smoothness.

 

[erobz]

If this is the case. Is it wrong for me to explain the effect the way they explained in this article?

"A consequence of the Bernoulli theorem is that a fast-flowing stream drags and accelerates some air around it creating a velocity and a pressure gradient in the air: the faster the air, the lower the pressure around the stream. The low pressure tends to be compensated by some air coming from the nearby space. However, if a surface is located in close proximity, no air can arrive and the low pressure tends to bring together the two opposite sides, i.e. the stream and the surface that are pushed together by the ambient pressure exerted on the outer sides of the stream and the surface. Once the stream has adhered to the surface, the external pressure will continue keeping this situation"

They are explaining how the transient stage initiates. What has been done here is the steady state solution (i.e. once the ball is immersed in the fluid stream and comes to rest).

 

[erobz]

They are explaining how the transient stage initiates. What has been done here is the steady state solution (i.e. once the ball is immersed in the fluid stream and comes to rest).

I would like to discuss this with anyone that is still listening? They try to invoke Bernoulli's, and say that the pressure is lower in the moving air by dint of it simply being in motion. They are acting as if energy is conserved when Bernoulli's is called into action, but it clearly isn't. Work is being done on the air by the fluid jet, there is an external force acting on the air. The fluid jet is grabbing air (via friction), pulling it down at the interface with the jet as it falls. It's not as if the air had a certain energy and spontaneously changed its velocity, demanding a change in pressure because nothing else is able to change in the equation (no remaining energy reservoirs to trade energy with). The energy of in the air necessary to transition to higher velocity comes from friction with the fluid jet.

The way I see it, we do not need a pressure gradient to describe this force that manifests. Air being dragged by the fluid jet, sticks to the ball and is deflected, the same as the water jet. The force would manifest as a reaction to its rate of momentum change, just as the water does, albeit much smaller (the object must be quite close to the fluid jet for this to initiate).

What am I missing?

 

[printereater]

They are explaining how the transient stage initiates. What has been done here is the steady state solution (i.e. once the ball is immersed in the fluid stream and comes to rest).

oh so, their explanation is for how it gets started and yours is for how it stays at that equilibrium?

 

[printereater]

I would like to discuss this with anyone that is still listening? They try to invoke Bernoulli's, and say that the pressure is lower in the moving air by dint of it simply being in motion. They are acting as if energy is conserved when Bernoulli's is called into action, but it clearly isn't. Work is being done on the air by the fluid jet, there is an external force acting on the air. The fluid jet is grabbing air (via friction), pulling it down at the interface with the jet as it falls. It's not as if the air had a certain energy and spontaneously changed its velocity, demanding a change in pressure because nothing else is able to change in the equation (no remaining energy reservoirs to trade energy with). The energy of in the air necessary to transition to higher velocity comes from friction with the fluid jet.

The way I see it, we do not need a pressure gradient to describe this force that manifests. Air being dragged by the fluid jet, sticks to the ball and is deflected, the same as the water jet. The force would manifest as a reaction to its rate of momentum change, just as the water does, albeit much smaller (the object must be quite close to the fluid jet for this to initiate).

What am I missing?

I am quite confused too. This is one of the few documents I could find related to my experiment and it keeps talking about it as if it's the pressure gradient that keeps the sphere at equilibrium.

"the external pressure will continue keeping this situation"

That was why I was wondering if the air pressure needs to be considered in the free body diagram. But after reading your explanation, I think yours makes more sense

 

[printereater]

If this is the case. Is it wrong for me to explain the effect the way they explained in this article?

"A consequence of the Bernoulli theorem is that a fast-flowing stream drags and accelerates some air around it creating a velocity and a pressure gradient in the air: the faster the air, the lower the pressure around the stream. The low pressure tends to be compensated by some air coming from the nearby space. However, if a surface is located in close proximity, no air can arrive and the low pressure tends to bring together the two opposite sides, i.e. the stream and the surface that are pushed together by the ambient pressure exerted on the outer sides of the stream and the surface. Once the stream has adhered to the surface, the external pressure will continue keeping this situation"

@Delta2 What is your opinion on this?

 

[Delta2]

@Delta2 What is your opinion on this?

Yes this seems to be a qualitative explanation of the transient state, when the ball is attracted to the stream of water. According to this once we reach equilibrium the "external" atmospheric pressure (in one side of the ball, and one side of the water stream) helps to maintain the equilibrium.

 

[Delta2]

I am not sure I understand very well @erobz post #127

 

[Delta2]

Another way to see this is that though we are at equilibrium we say $$\sum F=\dot m \vec{v_o}-\dot m\vec{v_i}\neq 0$$ though since we are at equilibrium we should set $$\sum F =0$$.

It seems to me that regarding the ball, the force from the atmospheric pressure minus the force from the water at the water interface equals the term $$\dot m \vec{v_o}-\dot m \vec{v_i}$$ so that if we transfer this term to the other side of the equation we can include it inside the forces and write $$\sum F'=0$$ which seems like more an equilibrium of forces condition.

 

[Delta2]

if you ask me the full math behind my conclusion "Force from atmospheric pressure on one side of the ball-(minus) Force from water pressure on the other side of the ball=$\dot m\vec{v_o}-\dot m\vec{v_i}$"" I am not able to provide them.

No this , certainly does not imply that "Force from atmospheric pressure=$\dot m \vec{v_o}$" and "Force from water pressure=$\dot m\vec{v_i}$".

 

[printereater]

In x-coordinate direction
$$T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}}$$
$$T=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}\rightarrow (6)$$

Substitute (5) into (6),
$$\frac{\dot{m}^{2}\sin(\theta)\cos(\beta )}{\rho A_{o}\sin(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}=\frac{\dot{m}^{2}}{\rho }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+W$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\tan(\beta )}=\frac{\dot{m}^{2}\rho A_{o}}{\rho }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+W\rho A_{o}$$
$$\frac{1}{\tan(\beta )}=\frac{\dot{m}^{2}\rho A_{o}}{\dot{m}^{2}\rho\sin(\theta) }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+\frac{W\rho A_{o}}{\dot{m}^{2}\sin(\theta)}$$
$$\tan(\beta )=\frac{1}{\frac{\ A_{o}}{\sin(\theta) }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+\frac{W\rho A_{o}}{\dot{m}^{2}\sin(\theta)}}$$
$$\beta =\arctan[\frac{1}{\frac{\ A_{o}}{\sin(\theta) }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+\frac{W\rho A_{o}}{\dot{m}^{2}\sin(\theta)}}]$$

@erobz @Delta2 Am I not allowed to use this formula now?

 

[erobz]

@erobz @Delta2 Am I not allowed to use this formula now?

The only debate that is happening is what is responsible for the force. The analysis is fine.

 

[erobz]

Can you provide the mass, diameter or the sphere, angle $\beta$, and angle $\theta$, in that experiment I'd like to do some calculation surrounding pressure gradients in the film (flow), via centripetal acceleration.

 

[printereater]

The diameter of the sphere is 10cm. I do not have the accurate measurements of $\theta$ and $\beta$ yet though, sorry:(

 

[printereater]

$$ \sum \vec{F} = \dot m \vec {v_o} - \dot m \vec {v_i}$$

Can you show me how to derive this from RTT please

 

[erobz]

Can you show me how to derive this from RTT please

I'm not going to outright show you, but I will try to facilitate you in developing it.

Here is the Reynolds Transport Theorem for the extensive property $m \vec{v}$. a.k.a. "The Momentum Equation"

$$ \sum \vec F = \frac{d}{dt} \int_{cv} \vec{v} \rho ~d V\llap{-} + \int_{cs} \vec{v} \rho \vec{V} \cdot d \vec{A} \tag{1}$$

On the LHS is the sum of the external forces acting on matter inside control volume, on the RHS (first integral) the time rate of change of momentum in control volume, and the next integral is the net outflow rate of momentum through control surface ( it's a summation over all flows entering and exiting the control surface).

$\vec v$ is the velocity the flow w.r.t. an inertial frame of reference
$\vec{V}$ is the velocity of the flow w.r.t. the control surface ($cs$). Just know that $\vec{v}$ and $\vec{V}$ are not necessarily the same under all circumstances. The distinction becomes relevant on moving control volumes.
$dV\llap{-} (= dx ~dy ~dz)$ implies to integrate $\vec{v}(x,y,z)$ over the control volume ( $cv$).
$d \vec{A}$ a differential area element directed outwards - perpendicular to plane of inlet/outlet. (Take note of the dot product in the second integral )
$\rho$ is the density of the flow. In general it could be $\rho(x,y,z)$, but for our purposes ( incompressible flow ), its a constant.

I think that covers definitions, so go ahead and familiarize yourself with the notation, and ask some questions to get the ball rolling.