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

[printereater]

TL;DR Summary: I am Highschool student writing a 4000 word research paper on Bernoulli's principle and the coanda effect. I need help with derivation of a formula that connects flow rate of water and distance moved by the sphere in my experiment.

I am a high school student writing a 4000 word research paper on Bernoulli's principle and coanda effect. In the experiment set-up shown here

a sphere made of Styrofoam is suspended in the air next to the tap. When the tap is turned on, the sphere gets 'attracted' towards the jet stream and the water starts to flow along the curvature of the sphere. For my experiment, I am trying to derive a formula that connects the flow rate of water and the distance moved by the sphere i.e., How does the varying flow rate of water affect the distance moved by the sphere.

I am not entirely sure how I can create a Freebody diagram which can help me create an equation. From my research so far, I understand that the sphere moves towards the jet-stream of water due to the difference in pressure (relatively higher air pressure on the outer side of the sphere and lower pressure where the sphere comes in contact with the water). Does this mean force of air plays a huge role in the equation?

So far, I found the document attached below which has an experiment in page 55 which is kind of similar to mine. It has a derivation to obtain the force acting on a cylindrical object but I am not entirely sure of how to use it. For anyone who is interested in this document, I translated it to English using google translate here.

Can someone please help me with the derivation of the formula that connects these 2 variables please (distance moved by the sphere and the flow rate of water). I am hoping to obtain a linear equation at the end. It is alright if the derivation requires math that is much more complicated than high school math as we are expected to go way beyond our syllabus.

My research paper consists of many parts like background information, analysis of data, evaluation of data etc. However, it is really difficult for me to proceed if I do not even have a slightly working equation. As such, please provide your input. Anything helps! Thank you so much in advance.

 

[haruspex]

The force the water exerts on the ball is equal and opposite to the force the ball exerts on the water. The latter deflects the flow by an observable angle and at a known rate.
The question is how the flow rate affects the angle.
Can you think of a way to analyse that?

 

[erobz]

I think the basic idea is there is a force on the sphere due to the change in momentum of the flow. A mass flowrate enters the control volume vertically, water adheres to the sphere and leaves the control volume at some angle.

 

[printereater]

The force the water exerts on the ball is equal and opposite to the force the ball exerts on the water. The latter deflects the flow by an observable angle and at a known rate.
The question is how the flow rate affects the angle.
Can you think of a way to analyse that?

Yea, I am able to analyse that data roughly using the recorded videos of each trial. I have no idea how I can link flow rate and the deflected angle though

 

[erobz]

I have no idea how I can link flow rate and the deflected angle though

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:

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):

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

 

[Delta2]

The problem has two phases or should I say two states:

The transient state when the ball is attracted towards the flow of water and enters the stream of water till it stops due to equilibrium of forces. To fully analyze what is happening during this phase is I believe in the scope of a post graduate research paper.

The equilibrium state where the ball doesn't move anymore but of course the water keeps moving along the surface of the sphere. If we assume that the flow of water is laminar and of constant velocity we can indeed make an equilibrium equation relating the velocity of water, force on sphere due to atmospheric pressure and water pressure and the distance the ball has travelled from the vertical position.

I think the qualitative explanation is that the faster the flow of water, the less pressure the water will exert on the ball (bernoulli principle), hence the more distance the ball will travel or should i say the bigger the angle from the vertical.

 

[erobz]

I think the qualitative explanation is that the faster the flow of water, the less pressure the water will exert on the ball (bernoulli principle), hence the more distance the ball will travel or should i say the bigger the angle from the vertical.

For an incompressible fluid jet I believe the flow is at atmospheric pressure. I think Bernoulli's would just state $v_o = v_i$. In reality I think we are going to get some change in velocity (via flow area change), via shear stress that causes a velocity gradient as it rounds the bend.

In my opinion even the steady state solution is a complex fluids problem. I certainly don't know how to go about it.

 

[Delta2]

For an incompressible fluid jet I believe the flow is at atmospheric pressure. I think Bernoulli's would just state $v_o = v_i$. In reality I think we are going to get some change in velocity (via flow area change), via shear stress that causes a velocity gradient as it rounds the bend.

In my opinion even the steady state solution is a complex fluids problem. I certainly don't know how to go about it.

Hmm, yes even the equilibrium state might be a complex problem if we don't make a simplifying assumptions.

The way I see this problem is like we have the wing of a plane, the major difference is that in one side of wing we have flow of water and in the other side flow of air or static air. There is certainly pressure difference between the two sides of the sphere.

 

[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):

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

Hi, I will look into Reynolds Transport Theorem, thank you. I have been using screenshots from the recorded videos and measuring the angles using an online protractor like this. I have a feeling that the angles are super inaccurate. Is this a bad idea?

Also, do you think the document I attached has any relevance to my experiment?

 

[erobz]

Hi, I will look into Reynolds Transport Theorem, thank you. I have been using screenshots from the recorded videos and measuring the angles using an online protractor like this. I have a feeling that the angles are super inaccurate. Is this a bad idea?

I don't know...you will have to see what happens when fitting the data.

Also, do you think the document I attached has any relevance to my experiment?

Yes, that is Reynolds Transport Theorem. They are not assuming a jet of constant area (velocity) in that solution, but they do appear to be assuming $v$ is uniformly distributed over the inlet and outlet.

 

[printereater]

The problem has two phases or should I say two states:

The transient state when the ball is attracted towards the flow of water and enters the stream of water till it stops due to equilibrium of forces. To fully analyze what is happening during this phase is I believe in the scope of a post graduate research paper.

The equilibrium state where the ball doesn't move anymore but of course the water keeps moving along the surface of the sphere. If we assume that the flow of water is laminar and of constant velocity we can indeed make an equilibrium equation relating the velocity of water, force on sphere due to atmospheric pressure and water pressure and the distance the ball has travelled from the vertical position.

I think the qualitative explanation is that the faster the flow of water, the less pressure the water will exert on the ball (bernoulli principle), hence the more distance the ball will travel or should i say the bigger the angle from the vertical.

I get what you mean. I honestly regret choosing this experiment now but it's too late:( Can you tell me more about the equilibrium equation please. I did mention that the flow is assumed to be laminar and the velocity remains constant in my paper

 

[printereater]

I don't know...you will have to see what happens when fitting the data.

Yes, that is Reynolds Transport Theorem. They are not assuming a jet of constant area (velocity) in that solution.

Got it, thank you:)

 

[erobz]

I get what you mean. I honestly regret choosing this experiment now but it's too late:( Can you tell me more about the equilibrium equation please. I did mention that the flow is assumed to be laminar and the velocity remains constant in my paper

To get the solution in the paper assume uniformly distributed velocity, incompressible flow.

You will use continuity $\dot m = \rho A_iv_i = \rho A_o v_o$

I updated the diagram to specify a few more parameters and a coordinate system that makes the math less sign error prone:

You write the equation of motion I gave for the control volume in each coordinate direction ( two independent equations). You will be able to solve for $\beta$.

 

[Delta2]

I get what you mean. I honestly regret choosing this experiment now but it's too late:( Can you tell me more about the equilibrium equation please. I did mention that the flow is assumed to be laminar and the velocity remains constant in my paper

You shouldn't regret in my opinion, doing a problem that is not a typical conventional routine should be something good.. I think @erobz updated post#5 gives the equilibrium state equation where $\sum F$ is the total force on the ball, and $\dot m$ is the mass flow rate of water and $v_0,v_i$ are the velocity (beware the velocity not the speed) of water before and after deflection.

 

[Delta2]

@erobz I see a slight problem, I can count 3 equations (together with continuity) and 4 unknowns , the two angles, the magnitude of tension and the magnitude of $v_o$. Unless we take $|\vec{v_o}|=|\vec{v_i}|$ as assumption.

 

[erobz]

@erobz I see a slight problem, I can count 3 equations (together with continuity) and 4 unknowns , the two angles, the magnitude of tension and the magnitude of $v_o$. Unless we take $|\vec{v_o}|=|\vec{v_i}|$ as assumption.

My line of thinking is they are measuring $\theta$, controlling $\dot m$ and measuring/estimating $A_i,A_o$. Weight is measured.

As far as I can tell I'm left with two equations two unknows $T,\beta$?

 

[printereater]

You write the equation of motion I gave for the control volume in each coordinate direction ( two independent equations). You will be able to solve for $\beta$.

Hi, which equation of motion are you referring to? From What I understand about Reynold's transport theorem so far, is it correct to conclude that the column of water until the water comes in contact with the sphere is system and the water that gets deflected is control volume?

 

[erobz]

Hi, which equation of motion are you referring to? From What I understand about Reynold's transport theorem so far, is it correct to conclude that the column of water until the water comes in contact with the sphere is system and the water that gets deflected is control volume?

Reynolds Transport simplifies ( or can simplify under certain assumptions) to the equation at the bottom of post #5.

 

[printereater]

You shouldn't regret in my opinion, doing a problem that is not a typical conventional routine should be something good.. I think @erobz updated post#5 gives the equilibrium state equation where $\sum F$ is the total force on the ball, and $\dot m$ is the mass flow rate of water and $v_0,v_i$ are the velocity (beware the velocity not the speed) of water before and after deflection.

Alright, hopefully everything goes well. I am really grateful for the both of you for guiding me:)

 

[erobz]

Alright, hopefully everything goes well. I am really grateful for the both of you for guiding me:)

Feel free to try to write down the equations of motion in each component direction so we can verify you have it correct.

 

[printereater]

oh I see

 

[printereater]

Feel free to try to write down the equations of motion in each component direction so we can verify you have it correct.

By equations of motion in each direction, are you referring to the suvat equations

 

[erobz]

By equations of motion in each direction, you mean the individual components of the \dot{m}v_{i} and \dot{m}v_{0}?

Yeah, What I mean is the equation applies in each coordinate direction independently. The coordinates are labeled $\hat i,\hat j$ in the diagram in post #13. Don't forget about the external forces, they are vectors too.

Oh, and put your latex code in delimiters. For code in line with text:

## \dot{m}\vec{v_i} ##

For stand alone centered equation:

$$ \dot{m}\vec{v_i} $$

 

[printereater]

Oh alright, got it. Is it okay if I write it out and send a photo? Sorry, it is taking me forever to type out the code. Also, I am wondering if I need to include the force of Air on the outer side of the sphere because this is the force that is pushing the sphere towards the water right?( due to the difference in pressure)

 

[Delta2]

Also, I am wondering if I need to include the force of Air on the outer side of the sphere because this is the force that is pushing the sphere towards the water right?

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 .

 

[erobz]

Oh alright, got it. Is it okay if I write it out and send a photo? Sorry, it is taking me forever to type out the code. Also, I am wondering if I need to include the force of Air on the outer side of the sphere because this is the force that is pushing the sphere towards the water right?( due to the difference in pressure)

No, the control volume completely surrounds the system. It is atmospheric pressure the whole way around its boundary, with the possible exception (under a microscopic precision) of the tiny areas where the flow enters and exits the control volume. I think the deviation from atmospheric pressure inside this flow stream is miniscule at worst, and incalculable at best.

 

[printereater]

No, the control volume completely surrounds the system. It is atmospheric pressure the whole way around its boundary, with the possible exception (under a microscopic precision) of the tiny areas where the flow enters and exits the control volume. I think the deviation from atmospheric pressure inside the flow stream is miniscule at worst, and incalculable at best.

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

Alright, here's the net forces in each direction

 

[printereater]

Alright, here's the net forces in each direction

oops I forgot to add the other components, hold on

 

[erobz]

Alright, here's the net forces in each direction

They don't like links to external sources like that. You can upload pics directly on the site by clicking the insert image.

In all honesty, hand written equations are strongly discouraged, and sometimes completely ignored. You were so close on the latex. Just take bit to figure it out, I'm assuming it's not due tomorrow.

 

[printereater]

oh Alright then I don't mind typing on latex, I just feel that I am wasting your time because I am taking forever to type

 

[erobz]

oh Alright then I don't mind typing on latex, I just feel that I am wasting your time because I am taking forever to type

Don't worry about my time, I choose when or when not to waste it!

 

[printereater]

In x-coordinate direction,
$$-Tsin(\beta )-\dot{m}v_{0}sin\left ( \Theta \right )$$

In y-coordinate direction,
$$-W-\dot{m}v_{i}-\dot{m}v_{0}cos\left ( \Theta \right )$$

 

[erobz]

In x-coordinate direction,
$$-Tsin(\beta )-\dot{m}v_{0}sin\left ( \Theta \right )$$

In y-coordinate direction,
$$-W-\dot{m}v_{i}-mv_{0}cos\left ( \Theta \right )$$

Equal signs are missing. Use code \theta for $\theta$, and for trig use \sin , \cos for $\sin, \cos$.

There is a missing force component in the y direction, and it doesn't seem like you are using the convention in the diagram...which is fine... but just state what your "positive convention" in each direction is clearly.

 

[printereater]

Equal signs are missing. Use code \theta for $\theta$, and for trig use \sin , \cos for $\sin, \cos$.

There is a missing force component in the y direction, and it doesn't seem like you are using the convention in the diagram...which is fine... but just state what your "positive convention" in each direction is clearly.

In x-coordinate direction,
$$=T\sin(\beta )+\dot{m}v_{0}\sin\left ( \theta \right )$$

In y-coordinate direction,
$$=W+\dot{m}v_{i}+\dot{m}v_{0}\cos\left ( \theta \right )+T\sin(\beta )$$

 

[erobz]

In x-coordinate direction,
$$=T\sin(\beta )+\dot{m}v_{0}\sin\left ( \theta \right )$$

In y-coordinate direction,
$$=W+\dot{m}v_{i}+\dot{m}v_{0}\cos\left ( \theta \right )+T\sin(\beta )$$

I'm not really sure what is happening with these equations.

It should be like $something = something~else$ not $blank = something$

And state your assumption for positive coordinate in each direction...they aren't making sense. I strongly suggest using the coordinate I have established in post #13 where down is positive y , and left is positive x.