Disc lifted by pressurized air in a vertical tube

[Baluncore]

The flow as I have drawn is arbitrarily shaped and devoid of a force that manifestly decreases with increasing height by empirical evidence.

Internal pressure provides the force, that decreases with increasing puck height, for the specified constant flow.

There are ways of computing the air flow through a rectangular orifice for a specified pressure drop. That pressure supports the disc, through a negative feedback loop. The control variable in the experiment is the fixed air flow.
https://toolbox.tlv.com/global/AU/calculator/air-flow-rate-through-orifice.html

 

[erobz]

There are ways of computing the air flow through a rectangular orifice for a specified pressure drop. That pressure supports the disc, through a negative feedback loop. The control variable in the experiment is the fixed air flow.
https://toolbox.tlv.com/global/AU/calculator/air-flow-rate-through-orifice.html

That employs the first law with a compressible flow and has viscous head loss built into it via the coefficient of discharge. Its empirical.

Also, in the rotameter you talk about earlier. The drag balances the weight is the exact phrasing for the description of the device in my textbook.

What internal pressure? Do you me the approximately atmospheric pressure of the incompressible fluid jet which is supposedly valid for subsonic flows. Putting a vane (this puck) in its way and deflecting it causes it to lose its bulk kinetic energy how? There are clearly babies in bath water here that need rescuing.

 

[Baluncore]

What internal pressure? Do you me the approximately atmospheric pressure of the incompressible fluid jet which is supposedly valid for subsonic flows.

There is a pressure drop associated with the increase in air velocity. The exit pressure is atmospheric pressure, the input pressure is NOT approximately atmospheric pressure, it is always greater than atmospheric pressure.

A wind turbine slows the air to extract kinetic energy. That is different to the disc supported above the tube, where the air accelerates into the orifice. The kinetic energy then heats the atmosphere outside the tube with the turbulence.

 

[erobz]

There is a pressure drop associated with the increase in air velocity. The exit pressure is atmospheric pressure, the input pressure is NOT approximately atmospheric pressure, it is always greater than atmospheric pressure.

I think we need to talk magnitudes here.

 

[Baluncore]

I think we need to talk magnitudes here.

Magnitude is very important.

The disc, above the end of the tube, does not fit any simple classical model used in text books. To transfer simplifying assumptions from education, to real world situations is a mistake.

There is one model I see, an analogy to the disc above the end of a tube.
When the safety valve on a steam engine blows continuously. The excess flow of steam is determined by the energy released by the fire in the firebox, which sets the steam production flow rate. The flow of steam through the gap between the weighted safety valve and the seat, drops steam from boiler pressure to atmospheric pressure. No condensation is involved. The pressure ratio is very high at about 15:1, so I expect the steam will be supersonic at some point in the valve. Turbulence in the steam outside the valve would explain the noise.
How much noise is made when a disc is supported above the end of a tube, where the pressure ratio is closer to 100500:100000 = 1.005 ?

 

[jmex]

Hi, I was away for some time, haven't read all the comments, will go through it and will definitely try to look into how we can implement here. Thank you everyone for all the inputs and I really appreciate your time.

 

[erobz]

The disc, above the end of the tube, does not fit any simple classical model used in text books. To transfer simplifying assumptions from education, to real world situations is a mistake.

We can't trust the textbook... Do you say this as an expert in the field of fluid dynamics or not? Published works in the field? You just said PE=KE in the beginning of all this- now you say - "To transfer simplifying assumptions from education to real world situations is a mistake." I don't appreciate being toyed with. To be clear, when I asked if this is just some "bull spit for engineers" I expected much more than the "textbooks are all wrong", but PE = KE is fine...

 

[erobz]

Look at this, how can it be said that the dominant effect is integrating the static pressure across the inlet of the control volume? It doesn't even get honorable mention here. Really that different than a weighted puck deflecting the flow? Show me how.

That being said, I still think it is insufficient to explain the equilibrium height of the puck because it fails to account for viscous loss. We see no mention of pressure other than “because the pressure is constant” etc…, we get a constant force $F$. In order to get an equilibrium height, we need to capture how to diminish $F$ vs $h$. Decreasing a pressure that is already assumed nil, wont help there. In the case of water, we could use elevation head as the energy bank and get away with it...maybe. However, with air the elevation head of the jet would be tiny.

Also, there is no "negative feedback loop" that is of notable significance. If you are claiming that this force application to the jet is different from steady state of the puck deflecting flow, I would like to see how the selective explanation is justified.

 

[erobz]

I'm going to switch to water (incompressible with some significant density) and account for elevation head to account for the expansion of the flow.

I assume the authors of my textbook are correct and that the pressure of the jet is atmospheric for less sonic (local speed of sound in the fluid comprising the jet ~ 1400 m/s). My control volume looks like the following:

The flow just before it is deflected radially outward under the puck is assumed uniformly distributed. It does not lose significant kinetic energy in the turn because the elevation change is assumed negligible in comparison with the height of the column.

Applying Bernoulli's between the inlet and outlet of the control volume (ignoring viscosity):

$$\frac{P_i}{\rho g} + z_{i} + \frac{v_i^2}{2g} = \frac{P_o}{\rho g} + z_{o} + \frac{v_o^2}{2g} \tag{1}$$

With
$P_i = P(z) = P_{atm}$
$z_i = 0$
$v_i = \frac{Q}{A_i}$
$z_o = z$
$v(z) = \frac{Q}{A(z)}$

$$ \implies \frac{1}{A(z)} = \sqrt{\frac{1}{A_i^2} - \frac{2g}{Q^2} z} \tag{2}$$

Applying the momentum equation accounting for momentum accumulation in the vertical column and control volume weight:

$$ \sum \mathbf{F} = \frac{d}{dt} \int_{cv} \rho \mathbf{v} ~ dV\llap{-} + \int_{cs} \rho \mathbf{v} \left( \mathbf{V} \cdot d \mathbf{A} \right) $$

$$ \overbrace{-F}^{\text{reaction from puck}} - \overbrace{g\int \rho A(z)dz}^{\text{weight of cv}} = \overbrace{\rho Q \frac{dz}{dt}}^{\text{cv accumulation}} - \overbrace{ \rho \frac{Q^2}{A_i} }^{ \text{net momentum efflux in} ~z ~\text{direction} } \tag{3} $$

You solve for $F$ and it gets plugged into Newtons Second for the puck uzing $A(z)$:

$$F - W_p = m_p \ddot z \tag{4} $$


The steady date solution should now fall out of setting $\dot z = 0$ and $\ddot z = 0$ inside of (4).

It look useful to me, and internally consistent with the textbook to a very reasonable degree.

With air as the incompressible fluid we should be able to ignore weight (density $\rho$ is a factor scattered through) $W_{cv}$ and accumulation term in (3). However, I cannot ignore elevation head maintaining consistency, if you ignore it the baby again goes out with the bath water and you get a constant force $F$ for the solution...death of reality. Viscous effects have to be accounted for in air or the equations fall apart at the seams.

 

[erobz]

Just to put forth the formula for constructive criticism:

$$ m \ddot z +\rho Q \dot z - \frac{\rho Q^2}{A_i} + \rho g \int A(z) ~dz+ W_p = 0 $$

For steady state this reduces to:

$$- \frac{\rho Q^2}{A_i} + \rho g \int A(z) ~dz+ W_p = 0 $$

Evaluate the integral:

$$ \rho g \int A(z) ~dz = \frac{\rho Q^2}{A_i} - \rho Q^2 \sqrt{ \frac{1}{A_i^2} - \frac{2gz}{Q^2}}$$

Then you substitute and solve for $Q$ as a function of equilibrium height of the puck $z_{eq}$ and you end with quadratic in $Q^2$:

$$ Q^4 - 2gz_{eq} A_i^2 Q^2 - \left( \frac{A_i W_p}{\rho}\right)^2= 0 $$


Using the quadratic formula I find:

$$ Q = \sqrt{ g z_{eq} A_i^2 + \sqrt{(g z_{eq} A_i^2)^2 + \left( \frac{A_i W_p}{\rho}\right)^2 } } $$


So you sub in parameters $A_i, \rho, W_p$, and desired equilibrium height of the puck $z_{eq}$ and you compute the required volumetric flowrate $Q$ to maintain equilibrium.

Here are the numerical results for elevating a standard hockey puck 15 cm with a something close to a garden hose. The units are handled by the program, so I haven't fumbled it.

EDITs: trying to fix up algebra typos and sloppy work

 

[boneh3ad]

I got a bit hasty in my early reply in addition to suffering from the $\LaTeX$ bug. It happens when I try to do control volumes while on Zoom calls. Ultimately, @erobz is on the right track in #30.

If you use that control volume (though I would draw the outlet to be even with the inner edge of the pipe), you can do an analysis using conservation of mass and momentum. Define the inlet size as station 1 and the outlet under the disc as station 2. Then conservation of mass, assuming steady and incompressible flow, states that
$$\boxed{V_1 A_1 = V_2 A_2.}$$
You know that $A_1 = \pi D^2/4$ and $A_2 = \pi D h$, where $h$ is the height the cap is lifted. That lets you relate the inlet and outlet velocities. You could start with the integral form of conservation of mass here, but that's a bit much for this case.

Conservation of momentum is a bit trickier, but is how we relate flow quantities to forces. Assuming steady flow and neglecting viscosity and gravity (which seems reasonable if you draw the small CV),
$$\iint_{\mathcal{A}}\rho\vec{V}(\vec{V}\cdot\hat{n})dA = -\iint_{\mathcal{A}} p\hat{n}dA +\vec{F}_{ext}.$$
You can simplify that and you'll find that the $x$-momentum equation is satisfied by default, so all that's left is the $y$ direction. There, you'll have some pressure at the top of the pipe (station 1), $p_1$, atmospheric pressure at the outlets, $p_{atm}$, and the weight of the cap, $W$, which points down. Simplifying the momentum equation in the $y$ direction yields
$$\boxed{-\rho v_1^2 A_1 = p_{atm} A_1 - p_1 A_1 + W.}$$

If the goal is to get the height the cap lifts, $h$, as a function of $p_1$ (or more usefully, $p_g = p_1 - p_{atm}$), then we currently have 2 equations for 3 variables. Enter the Bernoulli equation (again neglecting gravity).
$$\boxed{p_1 + \frac{1}{2}\rho v_1^2 = p_{atm} + \frac{1}{2}\rho v_2^2.}$$

That should allow you to solve for $h$ as a function of $D$, the pipe diameter, $p_g$, the gage pressure inside the pipe, and $W$, the weight of the "cap."

This time I chicken-scratched this during a different meeting, so someone check my work.

 

[erobz]

This time I chicken-scratched this during a different meeting, so someone check my work.

I think there I have a theoretical problem relating pressure $p_1$ to $z$. According to my text in incompressible flow of a free jet $p_1$ is atmospheric for subsonic flow velocity. In all the problems relating to incompressible jets I've encountered (this is very far from comprehensive) the pressure is taken as atmospheric. with the explicitly stated caveat that otherwise for supersonic flow velocity jet outlet can be above atmospheric.

Please see post no.34 (an excerpt from my textbook).

Is this only true for liquid incompressible flows? Until now I assumed the assumption of incompressible flow meant incompressible flow. Which is why for air I'm unsatisfied, and just deicide to switch to water to remove the requirement for pressure inconstancy in the flow.

What I did to fix this this was by using elevation head to give the necessary motive force for the change in velocity (expanding the area). It seems to give reasonable result. see post no's 39,40 if you get a chance.

I feel like there is a good bit of hopping about going on with assumptions that I'm unsure of and would like untangle.

 

[boneh3ad]

The way you have the CV drawn, the elevation head will be minimal. This is especially true if the fluid medium is air.

Otherwise, you are correct that an inviscid flow through a pipe with no external forces would have no pressure drop recorded. However, we have an external force here in that the weight of that "cap" is pressing down on the fluid. If you account for that, you will get a pressure drop. That was the $F_{ext}$ term in what I wrote.

 

[erobz]

The way you have the CV drawn, the elevation head will be minimal. This is especially true if the fluid medium is air.

I figured that and that is why I flipped the script in 39 re-drawing the control volume assuming a sizeable water jet to get a fix on this...I'm trying to isolate the difference between air and water and how large $z$ is. I get a reasonable result in 39 and 40?

If it were air lifting the puck in post 39, and the puck was at 15 cm, like I did for the sample calculation it doesn't involve pressure head, and it would lose elevation head as a bank because the density is so low. I think the solution would fall apart there. The problem is that it could be the same puck...not necessarily, but I hope you get my point.

 

[boneh3ad]

If you just rearrange all the three boxed equations I wrote so that you get $h = f(D,p_g,W)$, it's a pretty compact, closed form solution that seems to make sense. I just didn't want to write it out because I didn't go back and double check everything or test it to make sure all the behavior made sense. It requires no actual units, either. Leaving these sorts of things in terms of variables until the end is always preferable.

 

[erobz]

If you just rearrange all the three boxed equations I wrote so that you get $h = f(D,p_g,W)$, it's a pretty compact, closed form solution that seems to make sense. I just didn't want to write it out because I didn't go back and double check everything or test it to make sure all the behavior made sense. It requires no actual units, either. Leaving these sorts of things in terms of variables until the end is always preferable.

I agree there, but mine in 40 is closed form too? And different... Thats my issue here.

$$ Q = \sqrt{ g z_{eq} A_i^2 + \sqrt{(g z_{eq} A_i^2)^2 + \left( \frac{A_i W_p}{\rho}\right)^2 } } $$

 

[boneh3ad]

Why are you solving for $Q$? Wasn't the goal to find $h$ as a function of pressure in the pipe?

 

[erobz]

Why are you solving for $Q$? Wasn't the goal to find $h$ as a function of pressure in the pipe?

I think the goal was to find how high the puck would levitate above the end of the pipe for a particular mass flowrate. I chose to find $Q$ because it was easier to envision the problem. I put a light ball on my wife's hair dryer. it levitates at different height for the high and low settings. If I desired a bowling ball to do the same, I'm going to need one heck of a hair dryer, that much is apparent.

 

[erobz]

Getting $z(Q)$ seem theoretically less sound to me. I could blow on a bowling ball with my mouth and it will smash my face...even though I'm full of hot air! There are an infinite number of $Q$'s that don't necessarily do squat for $z_{eq}$. I feel I would be less likely to choose one of them, by picking $z$ instead.

 

[boneh3ad]

I just looked at my solution and clearly I screwed something up because it predicts flow and a lifting of the cap when there is zero pressure in the pipe. Ha! Looking for where I likely made a typo, probably won't get to it until later.

 

[Baluncore]

The height of a stationary ACV over a hard surface, is determined by the flow, Q , and the shape of the skirt, which determines the discharge coefficient.
Directed air-jets, below the craft, are designed to reduce the discharge passing below the skirt. Those jets lower the discharge coefficient to about 0.53, which represents a significant improvement over a simple discharge.
The mass of an ACV is supported by the increased pressure below the ACV. That is no different to the disc hovering above the end of a tube. Each, self-adjusts the height, to discharge the flow being provided.

Liang Yun, Alan Bliault - Theory and Design of Air Cushion Craft - Arnold_ Wiley (2000). Section 2.5: "Flow rate coefficient method". Page 72.

 

[erobz]

You don't get coefficients of discharge without viscous effects. They quantize entropy generation in the flow.

 

[Baluncore]

You don't get coefficients of discharge without viscous effects.

I don't care. The method works.

 

[erobz]

I don't care. The method works.

That’s fine (for a hovercraft), but you said over and over, ignore viscous effects. You also unintentionally invoke viscous effects in your model of PE=KE. Repeatedly I said we shouldn’t ignore them because there seems to a fundamental flaw in that line of reasoning. Also, the theory you present is different as flow is initially directed downward through the hovercraft, practically stagnated, then expanded outward laterally as it exits the skirt. It’s not the same. The viscous effects are the only reason that works. The puck is example is not going to stagnate the entire flow like that. The pumps/fans need viscosity in the system to generate that pressure.

 

[erobz]

I just looked at my solution and clearly I screwed something up because it predicts flow and a lifting of the cap when there is zero pressure in the pipe. Ha! Looking for where I likely made a typo, probably won't get to it until later.

When I solve the system of equations you suggest I get for $z$ direction:

$$ - W + p_1 A_1 = - \frac{ \rho Q^2 }{A_1} $$

I'm using $p_1$ as gauge pressure ( $p_{atm} = 0 ~\text{gauge}$ ).

After I sub in Bernoulli's as you've done I get:

$$ Q = \sqrt{ \frac{W}{\frac{4 \rho}{\pi D^2} + \frac{\rho}{8 \pi z_{eq}^2} }} $$

If its algebraically correct, its certainly unrealistic for larger $z_{eq}$ as $Q$ is approaching a constant.

Another concern is if the pressure is significant to the solution here for a given puck, what is the reason it appears insignificant to the solution I propose in 39 and 40. It seems inconsistent.

I think with air, viscous effects dominate; with water, elevation head dominates. Why the pressure would be significant in one, but not the other incompressible flow (treatment) escapes me. Perhaps for the air incompressible assumption is significant stretch? I don't think its right to half-heartedly accept the assumptions and expect to get a consistent result in the framework.

EDITS: reworked the algebra.

 

[boneh3ad]

Yeah I already told you I'm pretty sure there was a sign error somewhere. If I get the chance, I'll go back and find the error.

Either way, I still don't understand the obsession of finding $Q$.

 

[erobz]

Yeah I already told you I'm pretty sure there was a sign error somewhere. If I get the chance, I'll go back and find the error.

No pressure. pun intended. I was just showing you what I got in your rendition for verification purposes.

Either way, I still don't understand the obsession of finding $Q$.

If there is some equation that writes $z$ as a function of $Q$. Imagine a bowling ball on a hairdryer. If is select Q = 0, I better get $z= 0$. However if I select $Q = 0.1 \text{m}^3/ \text{s}$ I also better get for $z = 0$. There is a critical flow which will elevates the bowling ball. If I pick a random $Q$ and calculate $z$ I expect to get some nonsense in the equation if I've not exceeded the critical flowrate. I don't think picking $Q$ as independent variable is as robust as finding the $Q$ which elevates the weight at a selected $z_{eq}$. There is probably a determinant to worry about with finding $z$ as a function of $Q$ where you get non-real results for some values of $Q$. Why bother with that.

 

[erobz]

In the meantime, let me examine some limits:

$$ Q = \sqrt{ \frac{W}{\frac{4 \rho}{\pi D^2} + \frac{\rho}{8 \pi z_{eq}^2} }} $$

If the equilibrium height of the puck in the flow goes to zero $z_{eq} \to 0$ that sends $Q \to 0$ with it. That is not expected behavior. there is a critical ( non-zero ) $Q$ to just lift the puck.


If the equilibrium height of the puck in the flow goes large $z_{eq} \to \infty$ that sends $Q$ to a non zero constant. That is also not expected behavior in reality.

I think this result seems to fly in the face of empirical evidence.

 

[erobz]

As a recap I feel there are two competing theories (and possibly yet unmodeled - viscous effects). Neither perfect, but one better than the other IMO. This one checks more boxes for me using the flow elevation head (assumes static pressure is negligible). It makes sense with $W=0, z = 0$ etc...

$$ Q = \sqrt{ g z_{eq} A_i^2 + \sqrt{(g z_{eq} A_i^2)^2 + \left( \frac{A_i W_p}{\rho}\right)^2 } } \tag{1} $$

When we try to use pressure ignoring elevation head, we get:

$$ Q = \sqrt{ \frac{W}{\frac{4 \rho}{\pi D^2} + \frac{\rho}{8 \pi z_{eq}^2} }} \tag{2} $$

Am I repeatedly making an algebraic mistake to get this? I can tell (2) is behaviorally a poor result. I have graphical /numerical results that confirm.

I don't hope this goes quietly into the night without verification because I am probably going to miss learning something if it does. Isn't that what we are here for?

P.S. sorry for mixing and matching units (I'll edit if anyone is peeved), but readers get the point I hope.

 

[Ivan Nikiforov]

Hello,

View attachment 352909
I have a circular pipe where there is an inbuild pressure and I have kept a circular weight on it. I would like to know what would be the lift of the weight under certain pressure.
I know the relation that the minimum weight required to keep the pipe closed is F = P*A. (this F will be the required weight to keep the pipe closed.) As there will be increase in pressure, the weight will lift. How do i calculate what would be the lift in case of increase in pressure.
(Assuming that the weight is constrained to vertical direction only)

Thanks

Hello! It seems to me that the following needs to be taken into account in order to perform the calculation. The process you are describing consists of two processes: 1) The process when the load has not yet been lifted (determined by the static pressure of the system); 2) The process when the load is lifted and air flows through the formed free flow section (determined by dynamic pressure). In essence, this is similar to an artillery piece - the projectile in the barrel obeys the laws of internal ballistics, and the projectile that leaves the barrel obeys the laws of external ballistics.