Disc lifted by pressurized air in a vertical tube

[jmex]

Hello,

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

 

[Lnewqban]

Pressure remains the same after the weight reaches a new height, you just increase the volume of the fluid inside the cylinder.

 

[jmex]

yes, assuming that as well that the pressure remains the same. How will I evaluate the height at which weight will reach?

 

[Baluncore]

If the fluid is a liquid, the hydrostatic pressure will vary with depth in the liquid.
https://en.wikipedia.org/wiki/Vertical_pressure_variation

 

[jmex]

If the fluid is a liquid, the hydrostatic pressure will vary with depth in the liquid.
https://en.wikipedia.org/wiki/Vertical_pressure_variation

The fluid media is air. Visualize blowing air through a pipe and trying to lift a disc. Disc is constrained its motion in vertical direction. While blowing, there will be pressure buildup in the pipe and will keep rising until the force due to pressure is equal to the weight of the disc. Once it is achieved, the disc will start moving up. I am trying to find a relation of the lift of a disc to the force due to the pressure buildup.

 

[Baluncore]

While blowing, there will be pressure buildup in the pipe and will keep rising until the force due to pressure is equal to the weight of the disc. Once it is achieved, the disc will start moving up. I am trying to find a relation of the lift of a disc to the force due to the pressure buildup.

If the pressure is less than needed to support the disc, the disc will gradually sink as air is lost around the edge of the disc, or back through the air supply.
If the pressure is greater than needed to support the disc, the disc will gradually rise as air will flow from the supply supporting the disc.

A weighted close-fitting piston, floating free without friction in a vertical tube, is used as the pressure reference when calibrating gauges. The piston is first pushed up to the top by excess air, then the air supply is turned off. As air escapes around the skirt of the piston, the piston gradually sinks, maintaining a constant pressure throughout the fall. The piston is weighed, and the area of the piston is known, so the pressure can be computed.

If a disc rests on the flat top of a tube, then it will probably lift and slide sideways when the pressure is increased.

 

[erobz]

How will I evaluate the height at which weight will reach?

I think it's a complex problem to actually predict the height for the puck. The flow underneath the puck is diffusing into 3-dimensional space at basically atmospheric pressure. However, pressure distribution in the free air stream, elevation head, and viscous dissipation must make the difference in being able to readily write something down of utility in answering the question (I've tried otherwise, so far unsuccessfully). If anyone has some theory on this, great. I'll probably share my head scratching if the OP is engaged in the meantime.

 

[jmex]

If the pressure is less than needed to support the disc, the disc will gradually sink as air is lost around the edge of the disc, or back through the air supply.
If the pressure is greater than needed to support the disc, the disc will gradually rise as air will flow from the supply supporting the disc.

A weighted close-fitting piston, floating free without friction in a vertical tube, is used as the pressure reference when calibrating gauges. The piston is first pushed up to the top by excess air, then the air supply is turned off. As air escapes around the skirt of the piston, the piston gradually sinks, maintaining a constant pressure throughout the fall. The piston is weighed, and the area of the piston is known, so the pressure can be computed.

If a disc rests on the flat top of a tube, then it will probably lift and slide sideways when the pressure is increased.

Agree with you, but I am looking for a mathematical relation.

I think it's a complex problem to actually predict the height for the puck. The flow underneath the puck is diffusing into 3-dimensional space at basically atmospheric pressure. However, pressure distribution in the free air stream, elevation head, and viscous dissipation must make the difference in being able to readily write something down of utility in answering the question (I've tried otherwise, so far unsuccessfully). If anyone has some theory on this, great. I'll probably share my head scratching if the OP is engaged in the meantime.

Yes, I thought so, I could evaluate using CFD but it takes a long time to evaluate. Also, if there is any variation in any dimension, there would be change in results too. I was hoping to have a relation here that can provide an approximate results.

 

[Baluncore]

Agree with you, but I am looking for a mathematical relation.

Does the disc rise and fall within the tube, or does the disc rest on the end of the tube and lift slightly?

 

[jmex]

end

Does the disc rise and fall within the tube, or does the disc rest on the end of the tube and lift slightly?

 

[Baluncore]

A disc resting on the end of the tube will be unstable. The mathematics and the numerical model must also demonstrate that instability.

You must fix the disc to remain horizontal and centred over the tube before you will start getting stable results.

You can get an initial estimate of the gap, based on the following. The pressure outside the tube will be atmospheric. The pressure inside the tube will be higher by the weight/area of the disc. The pressure drop across the lip of the tube can be assumed constant. Bernoulli will give you the velocity of airflow through the gap, based on that pressure difference. The height of the disc, above the tube lip, will be determined by the flow rate of the available air.

 

[jmex]

A disc resting on the end of the tube will be unstable. The mathematics and the numerical model must also demonstrate that instability.

You must fix the disc to remain horizontal and centred over the tube before you will start getting stable results.

You can get an initial estimate of the gap, based on the following. The pressure outside the tube will be atmospheric. The pressure inside the tube will be higher by the weight/area of the disc. The pressure drop across the lip of the tube can be assumed constant. Bernoulli will give you the velocity of airflow through the gap, based on that pressure difference. The height of the disc, above the tube lip, will be determined by the flow rate of the available air.

yes, it will have constrained to move only in vertical direction. It will not be able to move in horizontal direction.
Okay, using Bernoulli, I will get the velocity, still how do I find the gap using mass flow rate?

 

[Baluncore]

Mass flow rate and density give a volume flow rate. Find the area needed for the flow. The width of the channel, is the circumference of the tube in contact with the disc. The height of the gap is area / width.

 

[erobz]

Mass flow rate and density give a volume flow rate. Find the area needed for the flow. The width of the channel, is the circumference of the tube in contact with the disc. The height of the gap is area / width.

If you apply Bernoulli's to a fluid jet for an incompressible flow (uniformly distributed velocity - neglecting small elevation head for a gas jet) we find that $v_{\text{inlet}} = v_{\text{outlet}} = v$, that just implies that $A_{\text{inlet}} = A_{\text{outlet}}$. I don't think this is useful, as the height is going to be fixed independent of $Q$ by that method? What am I missing.

I expect there to be a definite dependence of puck height, on flow rate $Q$. I would expect the puck to accelerate, reach some point, and oscillate a bit. I think the drag force acting on the puck is decreasing as the flow expands into space. That could account for the behavior if the expansion of the flow field could be modeled.

 

[Baluncore]

What am I missing.

I don't know. I am not an expert in air-pucks or hovercraft. I model the end of the tube as a knife-edge below the flow, opposed by the flat disc above the flow. The pressure is dropped across the knife-edge. The pressure is reduced by the weight/area of the disc, with that reduction in PE, becoming an increase in KE, as the fluid moves into and through the gap.

I would ignore any oscillation or instability, by holding the disc level, damped, and centred over the tube. I would reduce the disc diameter to that of the tube diameter, and ignore any drag force on the disc.

 

[Baluncore]

Consider flow of 1 m3 of air with density rho = 1.204 kg/m3
dP is pressure due to weight/area of disc
PE = dP * 1 m3 ; joules = Pa * m3
KE = 0.5 * rho * v^2
PE = KE
dP = 0.5 * rho * v^2
2 * dP / rho = v^2
v = Sqrt( 2 * dP / rho )
gap * circ * v = flow ; where circ is circumference of tube edge
gap = flow / ( circ * v )

 

[erobz]

Consider flow of 1 m3 of air with density rho = 1.204 kg/m3
dP is pressure due to weight/area of disc
PE = dP * 1 m3 ; joules = Pa * m3
KE = 0.5 * rho * v^2
PE = KE
dP = 0.5 * rho * v^2
2 * dP / rho = v^2
v = Sqrt( 2 * dP / rho )
gap * circ * v = flow ; where circ is circumference of tube edge
gap = flow / ( circ * v )

Is velocity "v" the flow /area (approximately) in this (uniformly distributed flow)?

 

[Baluncore]

Is velocity "v" the flow /area (approximately) in this (uniformly distributed flow)?

The flow is the total volume that exits the tube, per second, through the orifice.
v is the velocity of the fluid.
gap = flow / ( circ * v )
circ * gap * v = flow
orifice_area = ( circ * gap )
v = flow / orifice_area

There are two areas.

1. The area of the disc that opposes the internal tube pressure.
dP = mass * g / area_of_disc

2. The area of the orifice that is gap high, by circ long.

I do expect criticism of my crude initial model. I also expect that criticism to be accompanied by an improved numerical model, that is more realistic.

 

[erobz]

I do expect criticism of my crude initial model. I also expect that criticism to be accompanied by an improved numerical model, that is more realistic.

If you try to invoke Bernoulli's and continuity I get $A_{inlet} = A_{orifice}$.

$\implies h=D/4$.

The is an obvious flaw with that if you experiment. I stuck small foam ball (about the size of a ping pong ball) and levitated it using low setting and high setting on my wife's hair dryer. Low setting, $h$ very close to the inlet, slightly oscillating. On high setting it climbs to $H$ for my equipment it was on the order of $h \approx 1 \text{cm}$ to $H \approx 10 \text{cm}$. I don't have an anemometer, nor did I measure and other parameters.

I'm not trying to pick an argument, but it's just not correct, or I'm doing a poor job of interpreting the model. The fluid mechanics is fundamentally more complex and not easily able to be distilled theoretically. I have some examples of some textbook case that I consider in the vicinity of the question, but significantly less difficult of the task at hand.

I'm not going to make a numerical model, the OP already knows that is best done with existing CFD software. I will share the textbook sample problem to see if its malleable in this way.

 

[Baluncore]

The is an obvious flaw with that if you experiment. I stuck small foam ball (about the size of a ping pong ball) and levitated it using low setting and high setting on my wife's hair dryer.

The gap, h, needs to be significantly smaller than the diameter of the tube.
The flow needs to be fixed, independent of dP, while the weight of the ball is changed.

 

[jmex]

I believe the approach @Baluncore mentioned is correct of PE=KE. Still figuring out a way to find the lift iteratively. As the disc weight lifts up, its potential energy increases. This will increase the Force in the downward direction resulting more pressure energy required to lift it. Now it will keep iterating until the pressure energy required is equal to kinetic energy (flow which is converted into pressure energy) provided to the system.
Correct me if I am wrong.

 

[jmex]

If you try to invoke Bernoulli's and continuity I get Ainlet=Aorifice.

⟹h=D/4.

This is a usual practice in Valve industry for a minimum valve lift required for least pressure drop by comparing two cross-section area. Here it would be
pi*(D^2)/4 = pi*D*h
hence h = D/4

I'm not going to make a numerical model, the OP already knows that is best done with existing CFD software. I will share the textbook sample problem to see if its malleable in this way.

It will be very helpful if you can share the sample problem similar to this model. I have done workout in CFD software but is quite time consuming. I am working on a mathematical model that can find the lift approximately.

 

[Baluncore]

'===========================================================================
' height of disc = gap above tube end, for a specified disc weight and flow
'===========================================================================
' Tube and disc diameter =  50.0 mm        Gap is in millimetres.
' mass      dP     vel     flow
'  kg        Pa     m/s    1 ml/s  10 ml/s  100 ml/s   1 l/s   10 l/s  100 l/s
'  0.001      5.    2.88   0.0022   0.0221   0.2210   2.2102  22.1021 221.0207
'  0.002      9.    3.84   0.0017   0.0166   0.1657   1.6574  16.5742 165.7422
'  0.003     16.    5.12   0.0012   0.0124   0.1243   1.2429  12.4289 124.2891
'  0.006     28.    6.83   0.0009   0.0093   0.0932   0.9320   9.3204  93.2037
'  0.010     50.    9.11   0.0007   0.0070   0.0699   0.6989   6.9893  69.8929
'  0.018     89.   12.15   0.0005   0.0052   0.0524   0.5241   5.2412  52.4123
'  0.032    158.   16.20   0.0004   0.0039   0.0393   0.3930   3.9304  39.3037
'  0.056    281.   21.60   0.0003   0.0029   0.0295   0.2947   2.9474  29.4736
'  0.100    499.   28.80   0.0002   0.0022   0.0221   0.2210   2.2102  22.1021
'  0.178    888.   38.41   0.0002   0.0017   0.0166   0.1657   1.6574  16.5742
'  0.316   1579.   51.22   0.0001   0.0012   0.0124   0.1243   1.2429  12.4289
'  0.562   2809.   68.30   0.0001   0.0009   0.0093   0.0932   0.9320   9.3204
'  1.000   4994.   91.09   0.0001   0.0007   0.0070   0.0699   0.6989   6.9893
'  1.778   8882.  121.46   0.0001   0.0005   0.0052   0.0524   0.5241   5.2412
'  3.162  15794.  161.97   0.0000   0.0004   0.0039   0.0393   0.3930   3.9304
'  5.623  28086.  216.00   0.0000   0.0003   0.0029   0.0295   0.2947   2.9474
' 10.000  49945.  288.04   0.0000   0.0002   0.0022   0.0221   0.2210   2.2102
' 17.783  88816.  343.00   0.0000   0.0002   0.0019   0.0186   0.1856   1.8560
' 31.623 157940.  343.00   0.0000   0.0002   0.0019   0.0186   0.1856   1.8560
' 56.234 280861.  343.00   0.0000   0.0002   0.0019   0.0186   0.1856   1.8560
'100.000 499449.  343.00   0.0000   0.0002   0.0019   0.0186   0.1856   1.8560
'
'===========================================================================
' based entirely on energy equivalence, between pressure and velocity
' ignores changes in density of the air
' ignores adiabatic cooling of air released
' ignores all drag and viscosity effects
' assumes stable disc, fixed in position and orientation, only gap changes
' assumes flow in tube is not restricted
' assumes maximum velocity is the speed of sound
'
'---------------------------------------------------------------------------
' dP = mass * g / area ; pressure step at knife edge, end of tube, in Pa
' consider flow of 1 m3 of air with density rho
' PE = dP * 1 m3 ; joules = Pa * m3
' KE = 0.5 * rho * v^2 ; joules
' PE = KE ; conservation of energy
' dP = 0.5 * rho * v^2
' 2 * dP / rho = v^2
' v = Sqrt( 2 * dP / rho ) ; limit 343 m/s
' gap * circ * v = flow ; where circ is circumference of tube edge
' gap = flow/1000 / circ / v ; gap in metres, flow in litre/sec
'
'===========================================================================
' input parameters
Dim As Double mass = 0.10   ' mass of disc in kg
Dim As Double diam = 0.05   ' diameter of disc and tube end in metres
Dim As Double flow = 1.000  ' flow in litre/sec

' constants and standards
Dim As Double Pi = 4 * Atn( 1 )
Dim As Double g = 9.80665   ' acceleration due to gravity
Dim As Double rho = 1.204   ' density of air, in kg/m3
Dim As Double sos = 343     ' speed of sound at sea level, in m/s 

' precompute
Dim As Double area = Pi * ( diam / 2 )^ 2   ' of tube and disc
Dim As Double circ = Pi * diam  ' circumference of contact line

'---------------------------------------------------------------------------
Print Using " Tube and disc diameter =####.# mm        Gap is in millimetres."; diam * 1000
Print " mass      dP     vel     flow"
Print "  kg        Pa     m/s    1 ml/s  10 ml/s  100 ml/s   1 l/s   10 l/s  100 l/s"

' compute and print data
Dim As Double dP, velo, gap
For j As Double = -3 To 2 Step 0.25 ' mass in kg
    mass = 10^j
    dP = mass * g / area  ' pressure step, in pascals = N/m2
    velo = Sqr( 2 * dP / rho )
    If velo > sos Then velo = sos
    Print Using "###.### ######. ####.## "; mass; dP; velo;
    For i As Integer = -3 To 2  ' 1 gram to 100 kg
        flow = 10^i
        gap = flow /1000 / velo / circ
        Print Using "###.#### "; gap * 1000; ' gap in mm 
    Next i
    Print
Next j

'===========================================================================

 

[erobz]

I feel like this problem is worth discussing some things. Alot of these problems are obviously simplified versions of reality. That being said, we have expansion for the flow at atmospheric pressure. I don't think this is unique to a windmill. The air jet out flow out of a leaf blower for example expands like this without any interrupting media, just toss some flour into the flow. So the control volume is around the blades, it shears the support structure. The force of thrust becomes an external force acting on the control volume at that point.

The solutions derives from application of the momentum equation to the control volume cv(this I don't have a book worked solution for).

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

My final result for reference is:

$$ F_T= \rho v_i^2 A_i \left( 1 - \frac{A_i}{A_o} \right)$$

Here is another somewhat related question I feel. The problem is its enclosed in pipe. I feel like the approach your problem is an amalgamation of each problem. I can upload the result to this if interested.

I think we have to be very precise with assumptions. Notice in either of these that Bernoulli's would not hold. The effects to be explored in these problems are implicitly attributed to loss, I have extreme doubt that any application of Bernoulli's make them appear. If you try to apply Bernoulli's to 6.77 you get a direct contradiction to the problem statement.

So potential energy gained by the mass is lost by the flow...ok maybe, but how. What are the assumptions? I see Bernoulli's being called into action in post 11.

From my textbook, for subsonic incompressible flow the pressure in the jet is constant (atmospheric). The elevation head here in air, negligible. Hence $v_i=v_o$ ,and $h = \frac{D}{4}$. That feels like garbage in = garbage out.

Under these assumptions the force of thrust from the remains constant, and the puck keeps accelerating if it is sufficient to cause acceleration. clearly, not what is observed. I think the flow is expanding and the force decreasing as in the first problem. the flow is slowing down because of heat generated via inelastic collisions with still gas surrounding the flow.

Or everything is a ghost of assumptions past. Also, a strong possibility. An easy mistake to make no? I'm certainly left with more questions than answers as usual... I'd like to take some time to try and untangle it.

I think it's mainly heat generation that is responsible for the effect here. The flows energy is being shared with the surroundings, its expanding, and the forces doing work on the puck are diminishing with distances from inlet. I also don't believe the total kinetic energy of the flow is lost to surroundings at the position of equilibrium (or oscillation about it).

Specialists in fluid mechanics (I'm just a hobbyist in everything)? @Chestermiller, @boneh3ad , @pasmith , others @jmex ? I'm always excited to take a beating.

EDIT: all my latex turned to unicode...why does it do that sometimes?

 

[erobz]

Here is the model I wish to entertain:

Isolate the control volume:

So we have the force of drag acting downward $F_D$, Weight of air in control volume $W_{cv}$, uniformly distributed (approximately atmospheric) $P_{atm}$ acting across the inlet and outlet cross sectional areas. The immediate goal is to find the drag force. Also velocities are assumed uniformly distributed across the inlet/outlet. Begin with (1):


$$ \sum \mathbf{F} = \frac{d}{dt} \int_{cv} \rho \mathbf{v} ~ dV\llap{-} + \int_{cs} \rho \mathbf{v} \left( \mathbf{V} \cdot d \mathbf{A} \right) \tag{1}$$


$$ -F_D - W_{cv} -P_{atm} A(y) + P_{atm}A_i = \frac{d}{dt} \int_{cv} \rho \mathbf{v} ~ dV\llap{-} + \int_{cs} \rho \mathbf{v} \left( \mathbf{V} \cdot d \mathbf{A} \right) $$

I neglect the difference in the pressure terms and the weight of the control volume. I am left with:

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

On to the momentum side of the equation:

Since the mass inside the control volume is taken to be very small, I expect the rate of momentum accumulation within the control volume to be small, hence:

$$ -F_D = \cancel{\frac{d}{dt} \int_{cv} \rho \mathbf{v} ~ dV\llap{-}}^{\approx 0} + \int_{cs} \rho \mathbf{v} \left( \mathbf{V} \cdot d \mathbf{A} \right) $$

$$ -F_D = \int_{cs} \rho \mathbf{v} \left( \mathbf{V} \cdot d \mathbf{A} \right) $$

Next, with the uniformly distributed velocities we evaluate the integrals over the control surfaces:

$$ -F_D = \int_{outlet} \rho \mathbf{v}^2(y) d A(y) - \int_{inlet} \rho \mathbf{v}^2 d A_i $$

Applying continuity $Q = v_{inlet} A_{inlet} = v(y) A(y)$:

$$-F_D = \rho Q^2 \left( \frac{1}{A(y)} - \frac{1}{A_i} \right)$$

$$ \implies F_D = \rho Q^2 \left( \frac{1}{A_i} - \frac{1}{A(y)} \right) $$

The area of the annulus is given by:

$$ A(y) = \pi ( ky + r_i)^2 - A_p $$

Now apply this to drag force to the puck:

$$ F_D - W = m \ddot y $$

$$ \rho Q^2 \left( \frac{1}{A_i} - \frac{1}{A(y)} \right) - W = m\ddot y $$

$$ \rho Q^2 \left( \frac{1}{A_i} - \frac{1}{\pi ( ky + r_i)^2 - A_p} \right) - W = m\ddot y $$

From here you get the steady state solution for $y$ by setting $\ddot y = 0$ and solving.

Now, what seems to be sensible (to me) is this model for the expanding flow area as energy is robbed from it (flow work from drag) in the form of heat.

How can a sensible $k$ be found from first principles seems to be the remaining thorn.

Thoughts?

EDITS: corrected some of the mathematics pertaining to the evaluation of the control surface integrals as well as some careless subscript usage.

 

[boneh3ad]

I can't seem to write in LaTeX today, so unfortunately, I can't dig too deeply into this or check work, but @erobz is pointed in the right direction.

• You can enforce mass continuity based on the height of the puck capping the pipe to relate velocities/flow rates.
• You can relate pressures to velocities via the Bernoulli equation (if you assume incompressible and viscous losses are negligible).
• You can use conservation of momentum and an intelligently-drawn control volume to relate the force of the fluid to the weight of the puck.

That will give you three equations for three unknowns and let you solve for $h$ (the levitation height) as a function of #p_{gage}#, the pressure inside the pipe referenced to atmosphere.

 

[erobz]

I can't seem to write in LaTeX today, so unfortunately, I can't dig too deeply into this or check work, but @erobz is pointed in the right direction.

• You can enforce mass continuity based on the height of the puck capping the pipe to relate velocities/flow rates.
• You can relate pressures to velocities via the Bernoulli equation (if you assume incompressible and viscous losses are negligible).
• You can use conservation of momentum and an intelligently-drawn control volume to relate the force of the fluid to the weight of the puck.

That will give you three equations for three unknowns and let you solve for $h$ (the levitation height) as a function of #p_{gage}#, the pressure inside the pipe referenced to atmosphere.

Thanks for the reply. I think I’m having trouble understanding the Bernoulli bit. I’m interpreting my intro fluids textbook as saying an incompressible fluid jet at subsonic flow has everywhere atmospheric pressure. They give an example, the one with the windmill in a post above and the pressure, elevation are constant along a streamline. However, I find that if we were to ignore viscous effects there is no reason for the flow to expand like it does i.e. without the viscosity the turbine would be unable to extract work from the flow. The flow would have no motive to expand as far as I can see mathematically. Indeed, Bernoulli's is not satisfied with the examples from the text above. What is actually going on?

 

[erobz]

Trying to clearly convey my grievances. I think this problem at hand is basically this problem. Turn it vertical, replace the turbine extracting work from the flow with the puck being held in place by the force of drag.

Bernoulli's becomes problematic:

$$ \frac{ p_1 }{\gamma} + z_1 + \frac{v_1^2}{2g} = \frac{ p_2 }{\gamma} + z_2 + \frac{v_2^2}{2g} $$

If we follow the letter of the problem trying to apply this:

$p_1 = p_2 = p_{atm}$
$z_1 = z_2 = z$

$$ \frac{ p_{atm} }{\gamma} + z + \frac{v_1^2}{2g} = \frac{ p_{atm} }{\gamma} + z + \frac{v_2^2}{2g} $$

$\implies v_1 = v_2$

A direct contradiction of the problem statement. A head loss term is apparent, the work that is being extracted from the flow depends on it...is basically it?

The problem I am having the momentum and continuity constraints are fine but insufficient. To predict that expansion of the flow we need the head loss in the flow at equilibrium as a function of the variable $y$ to sub into that momentum equation yet. That seems beyond intro fluid mechanics level analysis.

I hope I'm making sense. Maybe the OP @jmex is satisfied, but I'm looking to clarify some ideas/concepts on this. I think Its a deeper problem than it first appears, am I wrong?

 

[Baluncore]

Turn it vertical, replace the turbine extracting work from the flow with the puck being held in place by the force of drag.

I do not agree. The turbine is turned by lift on the blades, not by drag as air moves through between the blades. The solidity of the turbine is fixed, the gap below the puck is variable.

The puck is held in place by static air pressure. Any drag is parallel to the lower surface of the puck. That drag is radial, so self-cancels. The drag is perpendicular to the static pressure against the puck. A specified mass of air flows through a rectangular orifice below the puck. The pressure drop through that orifice, supports the mass of the puck.

The flow coefficient, Cf, through the orifice will be moderated by the sharpness of the edges of the orifice. The bottom of the puck forms one of those edges at the orifice exit. The tube wall forms the other. Worst case sharp edges, Cf = 0.66 will vary to Cf = 0.95 for a smoothly-rounded end, thick-walled tube, with an oversized disc. For the rectangular orifice in this case, I would guess at Cf = 0.8

 

[erobz]

I do not agree. The turbine is turned by lift on the blades, not by drag as air moves through between the blades. The solidity of the turbine is fixed, the gap below the puck is variable.

We'll perhaps its more delicate in defining what is causing the loss in flow energy than I stated as "drag", but the fact of the matter there is no lift without viscosity, so I don't think it resolves the issue (I used the term "viscous effects" in #27 several times ). The end result is flow energy is being converted to heat.

What do you make of problem 6.8 in post 24 if you are dissatisfied with the wind turbine? I think it i also a very close relative to the task at hand here as well.

The puck is held in place by static air pressure. Any drag is parallel to the lower surface of the puck. That drag is radial, so self-cancels. The drag is perpendicular to the static pressure against the puck. A specified mass of air flows through a rectangular orifice below the puck. The pressure drop through that orifice, supports the mass of the puck.

I'm having doubts about static pressure due to two ideas:

1) The pressure in a subsonic fluid jet is atmospheric pressure.
2) Due to (1) its found that in impacting incompressible fluid jet on a vane for example $v_1=v_2=v$ constant velocity. When those problems are solved, the pressure distribution is not integrated over the vane in calculating the reaction force on the vane. Its negligible in comparison to the thrust. The momentum equation (eq1) in post 24 with incompressible flow, and the result of the force falls out. I don't see how this is differs in nature if you are wanting to approach it like this:

If you apply Bernoulli's for incompressible the flow still retains it KE, but changes direction. I think its problematic saying PE = KE. Forget the intermediary (pressure) you can just say $mgh = KE$. It feels good...until you try to justify the fluid mechanics.

The problem is it's not helpful to get the height of equilibrium. The flow as I have drawn is arbitrarily shaped and devoid of a force that manifestly decreases with increasing height by empirical evidence. I have an itch that I cannot scratch with the models. If it's wrong, then it is surely subtle! I don't think it can be inconsequentially swatted away. Saying that the pressure inside the tube supports the weight is not at all clear to me framed within the standard approach. In impinging jet problems in atmosphere the pressure distributions over inlet/outlets are shown practically nil. I can find $F$ without their consideration at all. My textbook does so repeatedly.


I still wait for some explanation that is internally consistent. If the models are bull spit they feed to engineers to pass them through the intro fluids course then my only hope is for someone to show me... use my presented equations against me. I won't to cry for my inequities (I don't have any Ivy League degree to defend).

 

[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 ?