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

[printereater]

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.

Yep, that's fair. Thank you so much. Is the net outflow rate of momentum through control surface relevant for my experiment?

 

[erobz]

Yep, that's fair. Thank you so much. Is the net outflow rate of momentum through control surface relevant for my experiment?

Yes, momentum is entering and exiting the control volume, so those terms are relevant. Use the control volume I gave you, sketch in inlet/outlet area vectors.

 

[printereater]

Alright, thank you. I have important deadlines and exams coming up soon, I will continue working on this experiment once I become a bit more free. Btw, may I know which software you used to produce the free body diagram please

 

[erobz]

Alright, thank you. I have important deadlines and exams coming up soon, I will continue working on this experiment once I become a bit more free. Btw, may I know which software you used to produce the free body diagram please

MsPowerPoint

 

[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),
$$\frac{\dot{m}^{2}\sin(\theta)\cos(\beta )}{\rho A_{o}\sin(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}=\frac{\dot{m}^{2}}{\rho }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+W$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\tan(\beta )}=\frac{\dot{m}^{2}\rho A_{o}}{\rho }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+W\rho A_{o}$$
$$\frac{1}{\tan(\beta )}=\frac{\dot{m}^{2}\rho A_{o}}{\dot{m}^{2}\rho\sin(\theta) }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+\frac{W\rho A_{o}}{\dot{m}^{2}\sin(\theta)}$$
$$\tan(\beta )=\frac{1}{\frac{\ A_{o}}{\sin(\theta) }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+\frac{W\rho A_{o}}{\dot{m}^{2}\sin(\theta)}}$$
$$\beta =\arctan[\frac{1}{\frac{\ A_{o}}{\sin(\theta) }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+\frac{W\rho A_{o}}{\dot{m}^{2}\sin(\theta)}}]$$ I am not sure if there is a nicer way to express $\beta$

@erobz Hello! I am finally free to work on this experiment again. Can you guide me through working out the calculations uncertainties please

 

[printereater]

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.

I don't understand how \vec{v} and \vec{V} are different. Does \vec{V} refer to the velocity of the flow when it comes in contact with the sphere and is inertial frame of reference referring to the velocity of the flow right before it touches the sphere?

 

[erobz]

I don't understand how \vec{v} and \vec{V} are different. Does \vec{V} refer to the velocity of the flow when it comes in contact with the sphere and is inertial frame of reference referring to the velocity of the flow right before it touches the sphere?

$\vec{v}$ is the velocity of the flow w.r.t. an inertial frame. For example a frame fixed to the ground ( or in the lab), and $\vec{V}$ is the velocity of flow w.r.t. the control surface. They can be different from each other in situations involving moving control volumes. However, Your control volume is not in motion, so they coincide.

 

[printereater]

ohh alright got it. From my research I found out that $\frac{d}{dt} \int_{cv} \vec{v} \rho ~d V\llap{-}$ becomes $\dot{m}v_o-\dot{m}v_i$, I am not entirely sure why though.

I also found out that $\int_{cs} \vec{v} \rho \vec{V} \cdot d \vec{A}$ is taken to be 0 under the conditions that the flow of water is steady and there are no mass sources or sinks, as in the mass flow rate into the conrtol volume is equal to the mass flow rate exiting the control volume. Can you verify this please

 

[erobz]

ohh alright got it. From my research I found out that $\frac{d}{dt} \int_{cv} \vec{v} \rho ~d V\llap{-}$ becomes $\dot{m}v_o-\dot{m}v_i$, I am not entirely sure why though.

I also found out that $\int_{cs} \vec{v} \rho \vec{V} \cdot d \vec{A}$ is taken to be 0 under the conditions that the flow of water is steady and there are no mass sources or sinks, as in the mass flow rate into the conrtol volume is equal to the mass flow rate exiting the control volume. Can you verify this please

You have the explanations reversed.

 

[printereater]

You have the explanations reversed.

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?

 

[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 ...)

$\$

 

[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