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

[erobz]

oh yeaa oops. $\frac{d}{dt} \int_{cv} \vec{v} \rho ~d V\llap{-}$ is taken to be 0 and $\int_{cs} \vec{v} \rho \vec{V} \cdot d \vec{A}$ becomes $\dot{m}v_o-\dot{m}v_i$ and thus $\sum F=\dot{m}v_o-\dot{m}v_i$. What other assumptions do I need to make?

So you understand how those integrals reduce computationally? Care for a quiz?

 

[printereater]

So you understand how those integrals reduce computationally? Care for a quiz?

Not really:( I am just trying to figure out how to show that RTT simplifies to $\sum F=\dot{m}v_o-\dot{m}v_i$ as that is enough for my paper. I think I need to learn a lot more stuff before I can actually understand it. Can you try to explain it to me like I am 10 please

 

[erobz]

Not really:( I am just trying to figure out how to show that RTT simplifies to $\sum F=\dot{m}v_o-\dot{m}v_i$ as that is enough for my paper. I think I need to learn a lot more stuff before I can actually understand it. Can you try to explain it to me like I am 10 please

Not everything can be boiled down like that. I don't understand why they expect you to know this mathematics... writing a paper over your current level of understanding is just a waste of time. You're not learning anything.

 

[printereater]

yea I agree:( I think it is more of a test on researching skills and the ability to write a report

 

[printereater]

@erobz I processed all the data and plotted the graph only to realise that my percentage error is around 2100%.

Using the FBD,
$$\sin(\beta )=\frac{D}{L}$$
$$D=L\sin(\beta)$$

Where $D$ is the horizontal displacement of the sphere and $L$ is the length of the string. I plotted $D$ against $\sin(\beta)$ so that the gradient of this graph gives me the length of the string. The actual length is $0.25m$ while the experimental length is around $5.8m$

I checked all my equations numerous times, there are no errors and the derived equation theoretically makes sense right. I am not even sure what's going wrong. I am so demoralised as I spent so much time and effort on this, I don't even know what to do now:(

 

[erobz]

@erobz I processed all the data and plotted the graph only to realise that my percentage error is around 2100%.

Using the FBD,
$$\sin(\beta )=\frac{D}{L}$$
$$D=L\sin(\beta)$$

Where $D$ is the horizontal displacement of the sphere and $L$ is the length of the string. I plotted $D$ against $\sin(\beta)$ so that the gradient of this graph gives me the length of the string. The actual length is $0.25m$ while the experimental length is around $5.8m$

I checked all my equations numerous times, there are no errors and the derived equation theoretically makes sense right. I am not even sure what's going wrong. I am so demoralised as I spent so much time and effort on this, I don't even know what to do now:(

Please present all the inputs to the equation accompanying a diagram. I don’t want to guess what you are measuring, I want to see it.

Hopefully some of the others can help us find sources of error between the proposed model and what you have experimentally. Does your experiment match our expectations of it, if not what are the subtle differences.

 

[printereater]

Please present all the inputs to the equation accompanying a diagram. I don’t want to guess what you are measuring, I want to see it.

Hopefully some of the others can help us find sources of error between the proposed model and what you have experimentally. Does your experiment match our expectations of it, if not what are the subtle differences.

Sorry! To make everything easier, I made a document summarising everything that has happened so far with images. Please take a look at it when you happen to be free.

 

[erobz]

Sorry! To make everything easier, I made a document summarising everything that has happened so far with images. Please take a look at it when you happen to be free.

How did you measure the flow rate? And how/where did you measure the areas?

 

[printereater]

How did you measure the flow rate? And how/where did you measure the areas?

I collected water below the sphere after the sphere moved to its equilibrium position using a beaker and measured the time taken($t$) for the beaker to become almost full. Then, I transferred the water to a measuring cylinder to find the volume($V$) of the collected water. I found $\dot{V}$ by dividing $V$ by $t$. For calculations, I converted $\dot{V}$ to $\dot{m}$ using $\dot{m}=\dot{v}\times \rho$

For the areas, I used the tracker application to measure the diameters of the inlet and outlet flows as shown in the tracker image and calculated areas using $A=\pi r^2$

 

[erobz]

I collected water below the sphere after the sphere moved to its equilibrium position using a beaker and measured the time taken($t$) for the beaker to become almost full. Then, I transferred the water to a measuring cylinder to find the volume($V$) of the collected water. I found $\dot{V}$ by dividing $V$ by $t$. For calculations, I converted $\dot{V}$ to $\dot{m}$ using $\dot{m}=\dot{v}\times \rho$

For the areas, I used the tracker application to measure the diameters of the inlet and outlet flows as shown in the tracker image and calculated areas using $A=\pi r^2$

I’m also somewhat concerned about the angle and area of the out flow. Where you are measuring it ( where it as condensed into a more readily measurable shaped stream) gravity may be having significant effects on the angle. Immediately after leaving the sphere it starts on its parabolic trajectory. The angle of the outflow with respect to vertical $\theta$ is changing as the water leaves the sphere. Decreasing.

 

[printereater]

I’m also somewhat concerned about the angle and area of the out flow. Where you are measuring it ( where it as condensed into a more readily measurable shaped stream) gravity may be having significant effects on the angle. Immediately after leaving the sphere it starts on its parabolic trajectory. The angle of the outflow with respect to vertical $\theta$ is changing as the water leaves the sphere. Decreasing.

Oh so should I measure them at a higher place like immediately after the water exits the sphere instead?

 

[erobz]

Oh so should I measure them at a higher place like immediately after the water exits the sphere instead?

You should, but what is the shape of the area there? It’s a difficult experiment to control and measure.

 

[printereater]

Oh yeaa that's a good point I have been taking each outlet area to be circular in shape. Do you think I should plot a function along its contour and integrate it to find the area? That is so much work though:(

 

[erobz]

Oh yeaa that's a good point I have been taking each outlet area to be circular in shape. Do you think I should plot a function along its contour and integrate it to find the area? That is so much work though:(

I don't know. I'm just throwing ideas at you. You have to decide if you are going to put in the effort or not. It's not my project.

 

[printereater]

I don't know. I'm just throwing ideas at you. You have to decide if you are going to put in the effort or not. It's not my project.

Got it. I think I will collect angles and areas a bit higher first and see how it goes from there. I will update you on it. Thank you for your help:)

Just wondering, the assumptions made for RTT equation would not affect the results to this extent right

 

[erobz]

Just wondering, the assumptions made for RTT equation would not affect the results to this extent right

Probably not, but they are simplifications. So, again I don’t know. I think what you will find is that instead of looking at the total error of a measurement by itself, the error in each measurement propagated through the computations will give you a relative wide range of values for the angle within the expected experimental error.

 

[erobz]

Perhaps one of the Scientist @BvU, @kuruman, @haruspex , @Steve4Physics, @Orodruin , etc…can explain to you how to deal with all the experimental error in all the measurements statistically in this computation or point you to good literature. It’s not something I am experienced in enough to know the best approach to use. It going to be tough, because It doesn’t seem like many people have been interested in the problem, and this is a very lengthy thread.

 

[printereater]

@BvU @kuruman @haruspex @Steve4Physics @Orodruin I would really appreciate it if you can provide your valuable input

 

[haruspex]

@BvU @kuruman @haruspex @Steve4Physics @Orodruin I would really appreciate it if you can provide your valuable input

I studied this problem a little at the start, but it is very complex.
What simplifications are reasonable? That the flow is constant speed over the surface? That the normal force is constant magnitude? That there's no tangential force? That the stream has constant width and thickness? That the centre of the sphere lies on the line of the string? ….

 

[printereater]

I studied this problem a little at the start, but it is very complex.
What simplifications are reasonable? That the flow is constant speed over the surface? That the normal force is constant magnitude? That there's no tangential force? That the stream has constant width and thickness? That the centre of the sphere lies on the line of the string? ….

yea:( It's tough to make any of those assumptions as the flow rate is not even constant to begin with. I think it is reasonable to assume that the center of the sphere lies on the string though, The string was attached to the sphere at the exact middle point on the top using a screw

 

[Steve4Physics]

@BvU @kuruman @haruspex @Steve4Physics @Orodruin I would really appreciate it if you can provide your valuable input

Hi @printereater. I can't face going through ~170 posts to catch up. So it’s not clear (to me) what, if any, problems have now been resolved in the posts and what problems remain. I may not be alone in this!

I’d suggest starting a new thread which provides a description of your experiment and your remaining questions.

For information, a Google search for “Coanda effect sphere equations water” brings up a number of hits, e.g. https://cdn.intechopen.com/pdfs/404...rical_investigations_on_the_coanda_effect.pdf

(Incidentally, ‘Coanda’ is a person’s name so (unlike units named after people) should be written using upper case ‘C’. )

 

[BvU]

@BvU @kuruman @haruspex @Steve4Physics @Orodruin I would really appreciate it if you can provide your valuable input

Flattered . One small hindrance: over 170 posts to read to find out how any further input can be helpful ...
(ah, already noticed ...)

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[printereater]

@BvU @Steve4Physics @haruspex @kuruman @Orodruin Sorry, I forgot to mention that I made a document summarising everything that has happened so far. I will make another thread. Meanwhile, Please look through this document

 

[kuruman]

@BvU @Steve4Physics @haruspex @kuruman @Orodruin Sorry, I forgot to mention that I made a document summarising everything that has happened so far. I will make another thread. Meanwhile, Please look through this document

I was going to suggest posting a summary but you preempted me. I looked over the document and I was frustrated by the lack of units in the data tables. Also, use powers of 10 instead of several zeroes after the decimal point. It makes the reading much easier. Please fix all that and repost. Thanks. Your humongous discrepancy might be due to unit conversion errors.

 

[printereater]

Oops, I made the document in a hurry yesterday I will try to get everything fixed tmr. I initially used powers of 10 and rounded off everything to 3sf but I thought people who are reading it would prefer to have full data

 

[erobz]

The string was attached to the sphere at the exact middle point on the top using a screw

What is the weight of the screw in comparison to the sphere?

Also, it doesn't appear that the line of action of the string passes through the spheres center of mass.

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

 

[BvU]

Well, I suppose 8 pages is better than 171 posts .

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


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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[Steve4Physics]

Oops, I made the document in a hurry yesterday I will try to get everything fixed tmr. I initially used powers of 10 and rounded off everything to 3sf but I thought people who are reading it would prefer to have full data

Here are some specific hints for the 1st table which may help.

You could have the volume column headed: $Vol~(\times 10^{-6}~m^3)$. In fact the standard formatting convention is not to use brackets but to put the units after an oblique stroke. So $Vol~/\times 10^{-6}~m^3$ would be even better. But follow your local convention. The values below the heading would then be 120, 126, 173, etc.

State the value you are using for the density of water. There is no real need for the volume flow rate column unless the values are specifically needed for some purpose. You could just have a mass flow rate column headed $Mass~flow~rate / \times 10^{-3}~kg s^{-1}$ with values of 5.66, 6.26, 8.49 etc (rounding to 3 sig. figs. seems appropriate).

I have no idea why you have written ‘(vol flow rate)’ beneath ‘Mass flow rate’. And I have no idea what ‘Mass flow rate 2’ means.

 

[kuruman]

Also, it doesn't appear that the line of action of the string passes through the spheres center of mass.

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.

have no idea why you have written ‘(vol flow rate)’ beneath ‘Mass flow rate’. And I have no idea what ‘Mass flow rate 2’ means.

It means (Mass flow rate)².

To @printereater: Please post the mass of the sphere. It's not in the Summary you provided and without it, we cannot check your numbers.

 

[haruspex]

I suspect that the flowing water exerts a torque

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.

 

[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?