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

[erobz]

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?

I think viscosity is the key player. I don't believe the flow pressure significantly deviates from atmospheric. Just my opinion, but examine the actual experiment control volume ( not my crude diagram):

The thickness of the jet is reduced to a film around the ball. The outside boundary of the flow is atmospheric. The inside (where the ball is touching it) is just a fraction of a millimeter away. I simply don't believe there is a massive pressure gradient hidden there. What I believe is there is friction or surface tension grabbing the flow and distributing it, and changing its direction.

If you are looking at just the flow enclosed as the cv, I have trouble drawing a thin enough boundary around it. It seems unlikely the pressure significantly changes across its boundary.

In my opinion, no friction (viscosity) and\or surface tension, no more effect.

 

[erobz]

Something else the OP could try to improve on is the flow appears to come in laminar velocity distribution and leaves turbulent. That changes the application of the RTT (for momentum) a bit to the integral version.

No doubt it is not an easy experiment to carry out with much precision.

 

[printereater]

Something else the OP could try to improve on is the flow appears to come in laminar velocity distribution and leaves turbulent. That changes the application of the RTT (for momentum) a bit to the integral version.

No doubt it is not an easy experiment to carry out with much precision.

Is there anything I can do to ensure that the water leaves as laminar flow when it exits?

 

[printereater]

I think viscosity is the key player. I don't believe the flow pressure significantly deviates from atmospheric. Just my opinion, but examine the actual experiment control volume ( not my crude diagram):

What I believe is there is friction or surface tension grabbing the flow and distributing it, and changing its direction.

In my opinion, no friction (viscosity) and\or surface tension, no more effect.

If this is the case. Is it wrong for me to explain the effect the way they explained in this article?

"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"

 

[erobz]

Is there anything I can do to ensure that the water leaves as laminar flow when it exits?

I don't know, could be a function of ball smoothness.

 

[erobz]

If this is the case. Is it wrong for me to explain the effect the way they explained in this article?

"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"

They are explaining how the transient stage initiates. What has been done here is the steady state solution (i.e. once the ball is immersed in the fluid stream and comes to rest).

 

[erobz]

They are explaining how the transient stage initiates. What has been done here is the steady state solution (i.e. once the ball is immersed in the fluid stream and comes to rest).

I would like to discuss this with anyone that is still listening? They try to invoke Bernoulli's, and say that the pressure is lower in the moving air by dint of it simply being in motion. They are acting as if energy is conserved when Bernoulli's is called into action, but it clearly isn't. Work is being done on the air by the fluid jet, there is an external force acting on the air. The fluid jet is grabbing air (via friction), pulling it down at the interface with the jet as it falls. It's not as if the air had a certain energy and spontaneously changed its velocity, demanding a change in pressure because nothing else is able to change in the equation (no remaining energy reservoirs to trade energy with). The energy of in the air necessary to transition to higher velocity comes from friction with the fluid jet.

The way I see it, we do not need a pressure gradient to describe this force that manifests. Air being dragged by the fluid jet, sticks to the ball and is deflected, the same as the water jet. The force would manifest as a reaction to its rate of momentum change, just as the water does, albeit much smaller (the object must be quite close to the fluid jet for this to initiate).

What am I missing?

 

[printereater]

They are explaining how the transient stage initiates. What has been done here is the steady state solution (i.e. once the ball is immersed in the fluid stream and comes to rest).

oh so, their explanation is for how it gets started and yours is for how it stays at that equilibrium?

 

[printereater]

I would like to discuss this with anyone that is still listening? They try to invoke Bernoulli's, and say that the pressure is lower in the moving air by dint of it simply being in motion. They are acting as if energy is conserved when Bernoulli's is called into action, but it clearly isn't. Work is being done on the air by the fluid jet, there is an external force acting on the air. The fluid jet is grabbing air (via friction), pulling it down at the interface with the jet as it falls. It's not as if the air had a certain energy and spontaneously changed its velocity, demanding a change in pressure because nothing else is able to change in the equation (no remaining energy reservoirs to trade energy with). The energy of in the air necessary to transition to higher velocity comes from friction with the fluid jet.

The way I see it, we do not need a pressure gradient to describe this force that manifests. Air being dragged by the fluid jet, sticks to the ball and is deflected, the same as the water jet. The force would manifest as a reaction to its rate of momentum change, just as the water does, albeit much smaller (the object must be quite close to the fluid jet for this to initiate).

What am I missing?

I am quite confused too. This is one of the few documents I could find related to my experiment and it keeps talking about it as if it's the pressure gradient that keeps the sphere at equilibrium.

"the external pressure will continue keeping this situation"

That was why I was wondering if the air pressure needs to be considered in the free body diagram. But after reading your explanation, I think yours makes more sense

 

[printereater]

If this is the case. Is it wrong for me to explain the effect the way they explained in this article?

"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"

@Delta2 What is your opinion on this?

 

[Delta2]

@Delta2 What is your opinion on this?

Yes this seems to be a qualitative explanation of the transient state, when the ball is attracted to the stream of water. According to this once we reach equilibrium the "external" atmospheric pressure (in one side of the ball, and one side of the water stream) helps to maintain the equilibrium.

 

[Delta2]

I am not sure I understand very well @erobz post #127

 

[Delta2]

Another way to see this is that though we are at equilibrium we say $$\sum F=\dot m \vec{v_o}-\dot m\vec{v_i}\neq 0$$ though since we are at equilibrium we should set $$\sum F =0$$.

It seems to me that regarding the ball, the force from the atmospheric pressure minus the force from the water at the water interface equals the term $$\dot m \vec{v_o}-\dot m \vec{v_i}$$ so that if we transfer this term to the other side of the equation we can include it inside the forces and write $$\sum F'=0$$ which seems like more an equilibrium of forces condition.

 

[Delta2]

if you ask me the full math behind my conclusion "Force from atmospheric pressure on one side of the ball-(minus) Force from water pressure on the other side of the ball=$\dot m\vec{v_o}-\dot m\vec{v_i}$"" I am not able to provide them.

No this , certainly does not imply that "Force from atmospheric pressure=$\dot m \vec{v_o}$" and "Force from water pressure=$\dot m\vec{v_i}$".

 

[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),
$$\frac{\dot{m}^{2}\sin(\theta)\cos(\beta )}{\rho A_{o}\sin(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}=\frac{\dot{m}^{2}}{\rho }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+W$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\tan(\beta )}=\frac{\dot{m}^{2}\rho A_{o}}{\rho }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+W\rho A_{o}$$
$$\frac{1}{\tan(\beta )}=\frac{\dot{m}^{2}\rho A_{o}}{\dot{m}^{2}\rho\sin(\theta) }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+\frac{W\rho A_{o}}{\dot{m}^{2}\sin(\theta)}$$
$$\tan(\beta )=\frac{1}{\frac{\ A_{o}}{\sin(\theta) }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+\frac{W\rho A_{o}}{\dot{m}^{2}\sin(\theta)}}$$
$$\beta =\arctan[\frac{1}{\frac{\ A_{o}}{\sin(\theta) }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+\frac{W\rho A_{o}}{\dot{m}^{2}\sin(\theta)}}]$$

@erobz @Delta2 Am I not allowed to use this formula now?

 

[erobz]

@erobz @Delta2 Am I not allowed to use this formula now?

The only debate that is happening is what is responsible for the force. The analysis is fine.

 

[erobz]

Can you provide the mass, diameter or the sphere, angle $\beta$, and angle $\theta$, in that experiment I'd like to do some calculation surrounding pressure gradients in the film (flow), via centripetal acceleration.

 

[printereater]

The diameter of the sphere is 10cm. I do not have the accurate measurements of $\theta$ and $\beta$ yet though, sorry:(

 

[printereater]

$$ \sum \vec{F} = \dot m \vec {v_o} - \dot m \vec {v_i}$$

Can you show me how to derive this from RTT please

 

[erobz]

Can you show me how to derive this from RTT please

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.

 

[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?