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

[erobz]

Sorry, I just saw your message. I will try again today. The goal is it to factorise the right hand side and make $\beta$ the subject of the equation right

You are trying to solve for $\beta$ as a function of all the other variables you intend to measure. If that’s is what “making it the subject is “, then sure. You made an algebraic mistake when inverting both sides.

 

[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]

I am not sure if there is a nicer way to express $\beta$

Its good enough for now. Do you understand how to use it?

Latex tip. Use automatically sized parenthesizes

Instead of $$ ( 1 + \frac{1}{x} ) $$

use code

$$ \left( 1 + \frac{1}{x} \right) $$

to get

$$ \left( 1 + \frac{1}{x} \right) $$

 

[printereater]

Hi yes I understand it. How can I use it to link the distance moved by the sphere and the flow rate though

 

[erobz]

Hi yes I understand it. How can I use it to link the distance moved by the sphere and the flow rate though

$\beta$ is the steady state angle, you need all the other variables measured (or controlled) and the equation spits out the angle the string makes with vertical (getting how far it moved is just trig), you can determine how well that fits with the angle $\beta_{exp}$ you experimentally measure. There is going to be error in all the variables to account for etc...

 

[printereater]

$\beta$ is the steady state angle, you need all the other variables measured (or controlled) and the equation spits out the angle the string makes with vertical (getting how far it moved is just trig), you can determine how well that fits with the angle you measure. There is going to be error in all the variables to account for etc...

I need to plot a graph using the 2 variables though (Flow rate and distance moved by the sphere). If I use this formula, the distance moved by the sphere can be expressed this way, $\tan(\beta ) =\frac{D}{L} \therefore D={L}\tan(\beta )$, where D is the distance moved by the sphere and L is the length of the string. If I plot this function, I will be showing a relationship between $\beta$ and D. $\beta$ has 2 variables that are changing constantly $\theta$ and $\dot{m}$. Is there a way to overcome this and show the relationship between $\dot{m}$ and D without the $\theta$

 

[printereater]

There is going to be error in all the variables to account for etc...

Yes! Can you guide me through the calculations of uncertainties too please

 

[printereater]

I need to plot a graph using the 2 variables though (Flow rate and distance moved by the sphere). If I use this formula, the distance moved by the sphere can be expressed this way, $\tan(\beta ) =\frac{D}{L} \therefore D={L}\tan(\beta )$, where D is the distance moved by the sphere and L is the length of the string. If I plot this function, I will be showing a relationship between $\beta$ and D. $\beta$ has 2 variables that are changing constantly $\theta$ and $\dot{m}$. Is there a way to overcome this and show the relationship between $\dot{m}$ and D without the $\theta$

I just realised, there are 4 variables that are changing each time based on the flow rate of water $\ A_{o},\ A_{i}, \theta, \dot{m}$. So I have no choice but to plot D against $\beta$ right

 

[erobz]

I need to plot a graph using the 2 variables though (Flow rate and distance moved by the sphere). If I use this formula, the distance moved by the sphere can be expressed this way, $\tan(\beta ) =\frac{D}{L} \therefore D={L}\tan(\beta )$, where D is the distance moved by the sphere and L is the length of the string. If I plot this function, I will be showing a relationship between $\beta$ and D. $\beta$ has 2 variables that are changing constantly $\theta$ and $\dot{m}$. Is there a way to overcome this and show the relationship between $\dot{m}$ and D without the $\theta$

This is the steady state solution. $\dot m$ must be controlled. The idea is that if you control the flowrate to be constant, all other experimentally measured variables $\theta, A_i,A_o$ are constant too (for each flowrate). So, the equation predicts the angle of the string $\beta$. Then, you change the flow rate(but keep it steady), and re-measure the variables, and again the equation predicts the angle. You compare how well the equation matches what you experimentally measure for $\beta$ at each flowrate.

The goal is to see how well our theory about reality, matches reality. Will it be a decent model given our assumptions, and our inability to precisely measure the variables that contrive it? That is the purpose of an experiment.

 

[printereater]

Right, I get what you mean but for the graph I plot, am I going to plot $\beta$ against $\dot{m}$ or do I just plot $D=L\tan (\beta )$ to show the relationship between flowrate of water and sphere. With this formula I can show how $\beta$ changes with D to indirectly show the relationship between flowrate and D because what's affecting $\beta$ is $\dot{m}$ but I am wondering if there is a direct way to show the relationship flowrate and D.

 

[erobz]

Right, I get what you mean but for the graph I plot, am I going to plot $\beta$ against $\dot{m}$

Just do this. you are introducing more sources of error by adding new variables that need measured.

but I am wondering if there is a direct way to show the relationship flowrate and D.

Do you mean without measuring $\theta, A_i, A_o$?

 

[printereater]

Just do this. you are introducing more sources of error by adding new variables that need measured.

But then i would not be answering my research question though:(

Do you mean without measuring $\theta, A_i, A_o$?

I mean ideally I would love to have an equation in which I have D on the left hand side and $\dot{m}$ on the right hand side to directly express a relationship between these 2 variables.

 

[erobz]

I mean ideally I would love to have an equation in which I have D on the left hand side and $\dot{m}$ on the right hand side to directly express a relationship between these 2 variables.

A closed form solution of $\beta( \dot m )$...for all I know that could be a Nobel Prize in Physics.

 

[printereater]

oh Let's not go there then

 

[erobz]

oh Let's not go there then

By all means, give it shot if you are unsatisfied.

 

[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?