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

[erobz]

That concerns me too. I suspect that the flowing water exerts a torque (it's viscous after all) that rotates the sphere about the point of support at the screw.

That's why I doubted the centre would lie on the line of the string, but I'd not expected the effect to be as strong as shown in the picture.

I was thinking with as thin as that flow is around the sphere pressure gradients should be very small in any direction, but baffled that there apparently is this torque!

 

[haruspex]

I was thinking with as thin as that flow is around the sphere pressure gradients should be very small in any direction, but baffled that there apparently is this torque!!!

Pressure gradient? I was thinking of the drag.

 

[erobz]

Pressure gradient? I was thinking of the drag.

Which manifests as a pressure loss along the flow. In the equation setup, drag is an internal force inside the control volume. What the simplified equation I use assumes "out of existence" is pressure distributions (other than uniform) at (and across) the inlet and outlet of the control volume.

 

[haruspex]

Which manifests as a pressure loss along the flow

But that is offset to some extent, perhaps mostly, by the loss of height.

 

[erobz]

But that is offset to some extent, perhaps mostly, by the loss of height.

My thoughts were the water is practically in free fall acceleration around the sphere so hydrostatic increase in pressure would be practically null from top to bottom?

It is accelerating in the direction normal to the sphere, but again it basically a film thickness. Significant changes in pressure radially would seem to be a ghost too.

 

[haruspex]

My thoughts were the water is practically in free fall acceleration around the sphere

But that is not consistent with the high drag demonstrated. The linear flow rate could reduce at first, as it suddenly becomes subject to the drag, then maybe increase a little as it descends around the steepest part, then slow again as it curves around towards the horizontal.
I note that the stream before contact shows no obvious narrowing, suggesting it is moving quite fast.

 

[erobz]

If we knew the weight of the sphere, screw, ect... we could work out the magnitude of this torque. Maybe that could provide some whisper of a path out of the weeds. If the water is not accelerating at $g$ in the vertical direction as it rounds the sphere, then perhaps I must double back on neglecting the weight of the flow inside the control volume. That could be the apparent torque. Its working in the right direction to produce it (at least).

 

[kuruman]

Another piece of information that we lack is exactly how @printereater measured the displacement x. The photo in post #176 shows a meter stick in the background. It was probably placed there to note the horizontal displacement of the end of the string in the foreground. That would work to determine the angle that the string makes relative to the vertical except, as already noted, the CM of the sphere is not along that line.

 

[printereater]

What is the weight of the screw in comparison to the sphere?
View attachment 341313

Maybe there is some issue there that escapes me between the model and reality, maybe it's a nothing burger...

BvU: Looked at the calculations - don't understand the column Z in data for beta. Why so high ? How much is weight W ?
$$Z={\frac{\ A_{o}}{\sin(\theta) }\left (\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}}\right )+\frac{W\rho A_{o}}{\dot{m}^{2}\sin(\theta)}}$$first term is of the order of 10-20 so I smell a rat in your calculation for the second term ...

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@erobz @BvU I just realised my ridiculous mistake. For the mass of the sphere, I didn’t want to disrupt my set-up incase I wanted to collect more data. So, what I did was to look for a sphere of similar size and material and used its mass for my calculations which was 42 grams. I found a spare sphere I bought together with the sphere used in the experiment and its mass turned out to be 7 grams. And just like that the percentage error is down to 420%. I am so sorry, this is so embarrassing

 

[printereater]

That concerns me too. I suspect that the flowing water exerts a torque (it's viscous after all) that rotates the sphere about the point of support at the screw.

@erobz @kuruman @haruspex What I meant was the screw was aligned with the line of the string when the sphere was in its initial position.

There’s definitely some torque involved once the sphere reaches the equilibrium position. In fact, solving for moments was my initial approach but I didn’t know if I could assume the inlet flow force to be acting directly vertically downwards.

 

[printereater]

Another piece of information that we lack is exactly how @printereater measured the displacement x. The photo in post #176 shows a meter stick in the background. It was probably placed there to note the horizontal displacement of the end of the string in the foreground. That would work to determine the angle that the string makes relative to the vertical except, as already noted, the CM of the sphere is not along that line.

In the tracker app, you can calibrate the scale of length in the video by putting an object of known length in the background and telling it its length. Once I calibrated the length, I used the screw as the reference point. I recorded the x-coordinate of the screw at its initial position ($x_0$) and also recorded the x-coordinate of the screw at its equilibrium position ($x_1$). I measured $D$ using $D=x_1-x_0$.

For $\theta$, I manually extended the inlet/outlet flow paths and measured the angle between at the point of intersection of the 2 lines

 

[erobz]

What do you get if you take the inlet area equal to the outlet area, and increase $\theta$ to be the angle the water first begins to leave the sphere?

 

[kuruman]

And just like that the percentage error is down to 420%.

The discrepancy (not error) is better than that if you do it right. The plot below shows the % Discrepancy between theory and experiment defined as $$ Discrepancy=\frac{{Experiment-Theory}}{{Experiment}}\times 100$$where
$Experiment = \dfrac{D}{\sqrt{L^2-D^2 }}~$ is the measured value of $\tan\beta$
and $Theory$ is the theoretical expression for the tangent derived from your mathematical expression from the FBD. This is a live homework problem so what I reveal has to be limited. You have to do the work.

However, I would strongly recommend that in your expressions you use directly what you measured not what you derived from what you measured. An example is the way I wrote the tangent as the ratio of measured quantities. The tangent is sufficient to check the agreement between theory and experiment. Calculating the arctangent is superfluous. Another example is the way you found the areas. You divided the diameter by two to get the radius and then you squared and multiplied by $\pi$. No. $Area=\pi\frac{d^2}{4}.$ The reason for doing it this way will become apparent if you care to do error propagation analysis in this or other experimental endeavors.

You might also want to look into the matter of the torque acting on the sphere at equilibrium and how it might affect $\beta.$

Anyway here is my plot.

(Edited to fix typo in equation for $\tan\beta.$)

 

[BvU]

And I'm grateful for the opportunity to learn about google docs. A moving target ...

No longer grateful. It's got hold of my email, there is no way to get out, it feels like a pitch dark escape room with gas hissing somewhere



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[printereater]

What do you get if you take the inlet area equal to the outlet area, and increase $\theta$ to be the angle the water first begins to leave the sphere?

When I took the inlet area to be equal to the outlet area, the percentage error turned out to be 620%. I didn't get to change $\theta$ yet. I most probably won't until next Tuesday as I have 2 important deadlines coming up soon:(

 

[printereater]

If we knew the weight of the sphere, screw, ect... we could work out the magnitude of this torque. Maybe that could provide some whisper of a path out of the weeds. If the water is not accelerating at $g$ in the vertical direction as it rounds the sphere, then perhaps I must double back on neglecting the weight of the flow inside the control volume. That could be the apparent torque. Its working in the right direction to produce it (at least).

The weight of the sphere is $0.0738N$
The weight of the screw is $0.00949N$

 

[Steve4Physics]

The weight of the sphere is $0.0740402075N$
The weight of the screw is $0.0095124505N$

You must know ‘g’ (acceleration due to gravity) to a very high precision at your location! And you must be using a 9 digit display electronic balance!

 

[printereater]

The discrepancy (not error) is better than that if you do it right. The plot below shows the % Discrepancy between theory and experiment defined as $$ Discrepancy=\frac{{Experiment-Theory}}{{Experiment}}\times 100$$where
$Experiment = \dfrac{D}{\sqrt{L^2-x_{ave}^2 }}~$ is the measured value of $\tan\beta$
and $Theory$ is the theoretical expression for the tangent derived from your mathematical expression from the FBD.

Sorry for the confusion. The discrepancy is what we call the %error here. My plan was to experimentally find the length of the string and compare it with the actual value i.e. $Percentage error=\frac{\left | {L_{Exp}-L_{Actual}} \right |}{{L_{Actual}}}\times 100$.

I am not so sure what you are doing when you did $\sqrt{L^2-x_{ave}^2$. Is $x_{ave}$ supposed to be $D$?

Calculating the arctangent is superfluous

Although that is true, it allows me to show my application of the Reynolds Transport Theorem.

Another example is the way you found the areas. You divided the diameter by two to get the radius and then you squared and multiplied by π

Oh right, it's definitely better to do it directly. I will change it

 

[printereater]

You must know ‘g’ (acceleration due to gravity) to a very high precision at your location! And you must be using a 9 digit display electronic balance!

I used a 2-digit display electronic balance but I used an online Weight calculator which used a very precise value of g. I will round it off to correct d.p

 

[printereater]

No longer grateful. It's got hold of my email, there is no way to get out, it feels like a pitch dark escape room with gas hissing somewhere



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Oh no:( Did your email get accidentally leaked?

 

[BvU]

When I took the inlet area to be equal to the outlet area, the percentage error turned out to be 620%. I didn't get to change $\theta$ yet. I most probably won't until next Tuesday as I have 2 important deadlines coming up soon:(

Carefully review your priorities. This experiment can wait.

We (PF) on the other hand, are intrigued

With the masses you (almost) give , I get values of $Z$ from 84 down to 60.

Your $D = L\sin\beta$ suggests $D=0$ if $\beta = 0$. But the intercept is signifcantly negative; what to think of that ?

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[kuruman]

Sorry for the confusion. The discrepancy is what we call the %error here. My plan was to experimentally find the length of the string and compare it with the actual value i.e. $Percentage error=\frac{\left | {L_{Exp}-L_{Actual}} \right |}{{L_{Actual}}}\times 100$.

I am not so sure what you are doing when you did $\sqrt{L^2-x_{ave}^2$. Is $x_{ave}$ supposed to be $D$?

Although that is true, it allows me to show my application of the Reynolds Transport Theorem.


Oh right, it's definitely better to do it directly. I will change it

What I am doing is
(a) find the experimental value of $\tan\beta$ given by $$\tan\beta=\frac{D}{\sqrt{L^2-D^2}}$$ which is what Nature tells you is the case. (I edited the expression to fix a typo in post #193.)
(b) find the theoretical value for $\tan\beta$ as predicted by the Reynolds Transport Theorem which is what you think is the case.
(c) Compare the two numbers. If they are the same to within experimental error, then the application of the Reynolds Transport Theorem to the phenomenon you observed is valid. If they are not the same to within experimental error, then either you did something wrong or the Reynolds Transport Theorem does not apply to what you did.

Note that there are two separate ideas here. One is the difference between what Nature tells you is the case (experimental value) and what you think is the case (theoretical value). That's the discrepancy. The second idea is how well (accuracy) you can measure things. That will give you a $\pm$ number of how close your experimental value can get to what Nature tells you is the case. That's the experimental error. Experimental values are rarely (if at all) identical to theoretical values. However, if the discrepancy is within the experimental error, i.e. if you cannot measure things any better than the difference between the two, then you can conclude that the theory matches the experiment and that you have successfully described mathematically what Nature tells you is the case. That is why you need to distinguish discrepancy from error regardless of what you call error "there."

Also, there is a more accurate way to determine the length of the string than using the Reynolds Transport Theorem. Just take a meter stick or use the picture and your software. The experiment is about predicting the angle by which the sphere tips, not about determining the length of the string.

 

[printereater]

Oh, I understand what you mean now. So when you plotted the % discrepancy graph against measurement number, the discrepancy was always able to remain below 100%? That's interesting. How do I determine the overall % discrepancy of the experiment though? Do I take the average of the % discrepancy of each trial.

We call the percentage discrepancy % error and the $\pm$ number % uncertainty

Also, there is a more accurate way to determine the length of the string than using the Reynolds Transport Theorem. Just take a meter stick or use the picture and your software. The experiment is about predicting the angle by which the sphere tips, not about determining the length of the string.

Alright

 

[printereater]

Carefully review your priorities. This experiment can wait.

We (PF) on the other hand, are intrigued

With the masses you (almost) give , I get values of $Z$ from 84 down to 60.

Your $D = L\sin\beta$ suggests $D=0$ if $\beta = 0$. But the intercept is signifcantly negative; what to think of that ?

$\$

It is true that there is a non-zero y-intercept but it is quite low tbh (-0.0156). Which means that $D = L\sin\beta$ is in fact working right

 

[printereater]

@erobz @kuruman @BvU @Steve4Physics @haruspex Thank you so much for your help, I will try to come back by next Wednesday hopefully. It's really insane how I can assemble all the physics gods within a few clicks. I love the internet HAHAHAHA

 

[printereater]

@erobz @kuruman @BvU @Steve4Physics @haruspex @Delta2 Hello everyone! Sorry it's been a while, I was caught up with the other assignments and barely had any time to work on this report. Here's the summary of the experiment again.

The latest update is that the theoretical model of this experiment gave me a percentage error of 480% and I explained about all the potential errors involved in the experiment which has caused such a huge percentage error.

What I need help with currently is really understanding how the coanda effect works and how the sphere is able to stay in its new equilibrium after getting attached to the jetstream.

I found an article related to this experiment which explained that:

" 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. If the surface is convex, the curvature causes a continuing acceleration of the stream, and thus a low pressure between the stream and the surface."

According to them it's the external pressure which continues to keep the sphere in its new equilibrium position.

However, what I understand from @erobz 's explanation is that it's the viscosity and surface tension which enables the sphere to stay in its equilibrium position.

So is the article wrong when it says that "Once the stream has adhered to the surface, the external pressure will continue keeping this situation. If the surface is convex, the curvature causes a continuing acceleration of the stream, and thus a low pressure between the stream and the surface."?

 

[erobz]

@erobz @kuruman @BvU @Steve4Physics @haruspex @Delta2 Hello everyone! Sorry it's been a while, I was caught up with the other assignments and barely had any time to work on this report. Here's the summary of the experiment again.

The latest update is that the theoretical model of this experiment gave me a percentage error of 480% and I explained about all the potential errors involved in the experiment which has caused such a huge percentage error.

What I need help with currently is really understanding how the coanda effect works and how the sphere is able to stay in its new equilibrium after getting attached to the jetstream.

I found an article related to this experiment which explained that:

" 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. If the surface is convex, the curvature causes a continuing acceleration of the stream, and thus a low pressure between the stream and the surface."

According to them it's the external pressure which continues to keep the sphere in its new equilibrium position.

However, what I understand from @erobz 's explanation is that it's the viscosity and surface tension which enables the sphere to stay in its equilibrium position.

So is the article wrong when it says that "Once the stream has adhered to the surface, the external pressure will continue keeping this situation. If the surface is convex, the curvature causes a continuing acceleration of the stream, and thus a low pressure between the stream and the surface."?

They are basically saying that there are two flows happening, the water and the air being drug by the water. I don’t know, but I expect the any effect from air flow to be dominated by the momentum change in the water flow. Try disrupting the air flow they claim with a spoon or a sheet of paper etc…by bringing its edge close to the water. If there is a significant air flow it’s going to divert and change the effect. Let us know what happens.

I think the air flow is involved in the start of the effect (to get the ball to touch the water), but I don’t suspect it’s significant in keeping the effect once the water has adhered.

 

[erobz]

@printereater

Lets also try to get ahold of an order of magnitude pressure difference from the outside of the wetted surface, to the inside of the wetted surface working radially in 2D.

1) What is the average velocity of the inlet based on your incoming volumetric/mass flowrate measurement?

2) What is the thickness radially of the flow in the radial direction in the middle of the wetted surface? (Assume linear velocity gradient)

$$ \frac{V_1}{r_i} = \frac{V_2}{r_o}$$

3) Appy equation for pressure variation in rotating flow ignoring elevation between points 1 and 2. Assume $V_2 = \bar v$, and $P_2 = 0 \text{gauge}$?
$$ P_1 + \cancel{\rho g z_1} - \frac{1}{2} \rho V_1^2 =\cancel{P_2}^0 + \cancel{\rho g z_2} - \frac{1}{2} \rho V_2^2$$

4) Approximately what fraction of the sphere surface area is wetted by this flow ( 1/4, 1/3)?

5) Calculate the net force $P_1 A_{proj.}$?

How does it compare with the other forces in the horizontal direction? Someone stop me if I'm leading them astray?