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

[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 ?

$\$

 

[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?