Disc lifted by pressurized air in a vertical tube

[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.