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The Seykota Levitator Model
© 1999 by Ed Seykota
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Models
The author constructs models that explain the behavior of the
Bernoulli Levitator. The model results, based on Radial Momentum
correlate nicely with experimental measurements. This figure shows a
simulation of the pressure (red), air velocity (blue) and other
parameters along the gap between the levitator table and the disk, from
the central orifice to the edge of the disk. The characteristic 1/r dip
in the pressure just outside the orifice correlates with the location of
the cavitation ring that appears when using water as the fluid. |
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The levitator model is a numeric
(simulation) model. In this case, there is a central valve and also
an array of nested control volumes, each in the shape of a ring,
concentrically arranged between the plate and the table. Mass flows from
the valve at the center (ring #0) into the first ring (ring #1). From
there, it continues to flow into the larger rings.
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Array of nested control volumes, concentrically arranged. |
Each ring has three fundamental physical states and two computational
properties. Together, these are sufficient to define the system and also
to carry out the numeric simulation.
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Fundamental Physical States |
Density
Pressure
Velocity |
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Computational Properties |
Number_of_Rings
Delta Time |
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Derived Properties |
Delta_Radius = Disk_Radius / Number_Of_Rings
Radius = Delta_Radius * Ring_Number
Face_Area = 2 * pi * Radius * Gap_Height
Volume = Face_Area * Delta_Radius
Mass = Volume * Density
Momentum = Mass * Velocity
Energy = Pressure * Volume
Pressure Gradient = Pressure[n+1] - Pressure[n-1]
Temperature = Energy / (Mass * R) |
| Note that the particular choice of fundamental physical
states makes them independent of volume. Mass, kinetic energy, thermal
energy and momentum all depend on volume and are therefore
computation-dependent. |
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Section of a ring
Each ring has three fundamental physical states: density, pressure and
temperature. Mass flows from the inner rings to the outer rings and the
mass carries momentum and thermal energy with it.
The simulation proceeds by computing the flows to and from the rings,
and the changes in states within the rings, then incrementing the states
over a very small delta time, re-computing the flows and changes, and so
forth. |
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The Basic Laws
The model observes the basic laws of fluid dynamics: conservation of
mass, conservation of momentum and conservation of energy. |
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Conservation Equations
The three conservation equations are conservation of mass, conservation
of energy and conservation of momentum. I use the following notational
conventions: (1) Variables are of the form, Var[Ring, Time], where Ring is
the ring number and Time is either [t] for the present, or [t-1] for one
dT, or delta time, earlier. (2) The flux of ring[n] is the flux leaving
ring[n] and flowing into ring[n+1]. |
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Conservation of Mass
Mass is the accumulation of the net mass flux.
Mass[n, t] = Mass[n, t-1] - Net_Mass_Flux[n, t-1] * dT
The net mass flux equals the mass flux in minus the mass flux out.
Net_Mass_Flux[n, t] = Mass_Flux[n-1, t] - Mass_Flux[n, t] |
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Conservation of Energy
The total energy (TE) in a ring is the sum of the kinetic energy (KE)
plus the thermal energy (TE).
TE[n, t] = KE[n, t] + TE[n, t] |
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Kinetic Energy
The kinetic energy is a function of the net forward motion of the
fluid, flowing from the center of the disk toward the perimeter.
KE[n, t] = 1/2 * Mass[n, t] * Velocity[n, t]2
Kinetic energy is the accumulation of the net KE flux plus the energy
conversion (see below).
KE[n, t] = KE[n, t-1] + (KE_Flux[n, t-1] + Energy_Conversion[n, t-1]) *
dT
KE_Flux[n, t] = KE_Flux[n-1, t] - KE_Flux[n, t] |
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Thermal Energy
The thermal energy is a function of the random static motion of the
molecules in the fluid.
TE[n, t] = Pressure[n, t] * Volume[n, t]
Also, by PV = NRT,
TE[n, t] = Mass[n, t] * R * Temperature[n, t]
Thermal energy is the accumulation of the net TE flux minus the energy
conversion (see below).
TE[n, t] = TE[n, t-1] (TE_Flux[n, t-1] - Energy_Conversion[n, t-1]) *
dT
TE_Flux[n, t] = TE_Flux[n-1, t] - TE_Flux[n, t] |
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Energy: Conversion Between Forms
Total energy must remain constant, by conservation of energy. Energy
can, however, change between thermal and kinetic forms. The energy
conversion from thermal energy to kinetic energy results from the force
from the pressure head across a ring, acting on the fluid in the ring,
over a delta distance.
Delta_Energy_Thermal_To_Kinetic = Force[n, t] * dS
Force[n, t] = Pressure_Head[n, t] / Face_Area[n]
Pressure_Head[n, t] = (Pressure[n-1, t] - Presure[n+1, t]) / 2
dS = Velocity[n, t] * dT
Note: The change in energy from thermal to kinetic results in the same
magnitude change in thermal and kinetic energies. Therefore, the total
energy remains constant. |
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Conservation of Momentum
The fluid in a ring will, by conservation of momentum, continue to flow
at the same velocity and leave that ring and flow into the next ring.
Note: As a confirmation of the derivation of Delta_Energy above, we can
also arrive at the same result for dV, the delta velocity, by considering
Delta_Momentum. Delta_Momentum results from the net force acting on the
fluid in the disk, acting over a delta time.
(1) Derive delta velocity from delta momentum
Delta_Momentum = Force * dT
Now since velocity = momentum / mass,
dV = Force / Mass * dT
(2) Derive delta velocity from delta energy
Delta_Energy = Force * dS = Force * Velocity * dT
Delta_Energy = 1/2 * Mass * (Velocity + dV)2 - 1/2 * Mass *
Velocity2
Now since dV2 vanishes for very small dT,
Delta Energy = 1/2 * Mass * (2 * Velocity * dV)
Therefore,
1/2 * Mass * (2 * Velocity * dV) = Force * Velocity * dT
dV = Force / Mass * dT
Thus, we arrive at the same result for dV, by considering (1) delta
momentum and (2) delta energy. |
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Friction Factor
Fluid passing through the gap between the table and the disk
experiences a friction force opposing the direction of flow. This is
generally a negligible effect and is not necessary to explain the overall
lift mechanism. It is, however, important in determining the equilibrium
position of the disk.
I was not able to find definitive literature to compute friction for
the levitator configuration. I tried various estimates and found that they
gave qualitatively reasonable results. |
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Central Valve (ring #0)
Air from a central plenum at pressure, P and temperature, T exits
through a circular orifice of diameter, D and into the gap between the
table and the disk of height, h. The valve, then is a ring of height = h
and circumference = pi * D. For constant density, simple energy balance
gives the escape velocity in terms of the pressure drop across the valve:
Velocity = sqrt (2 * Pressure_Head / Density)
For compressible fluid, the density does not stay constant and it also
exhibits the "vena contracta" effect, so the exit velocity is
about 2/3 of the above formula. |
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Qualitative Verification of
the Model
The first test of model verification is to confirmation that it
behaves reasonably. The following runs were for a disk with radius = .02
meters, an orifice of .00159 meters, a gap of .0002 meters and a motive
pressure of 1.5 atmospheres. The horizontal axis shows the distance from
the orifice to the circumference.
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Run #1: Radial Momentum
Pressure initially falls as it passes through the valve. Then, as the
air fans out due to Radial Momentum, pressure continues to fall. This
pressure gradient further accelerates the air and velocity increases. Drag
acceleration from friction converts some momentum to heat. Eventually, the
pressure must rise to meet ambient pressure at the circumference of the
disk. As pressure starts to rise against ambient pressure and low-momentum
air ahead, it experiences additional counter-acceleration, which further
reduces velocity and momentum. This positive-feedback results in a rapid
rise of pressure, sometimes knows as hydraulic jump. After the jump, the
pressure decreases toward the circumference, per normal flow through a
duct. The low-pressure Radial Momentum process at the center contributes
all the lift. Lift has little to do with velocity. |
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Run #2: Linear Momentum
In run #2, everything is the same as in run #1, except that a parallel
channel constrains the flow. Thus, the flow has linear momentum and no
Radial Momentum. Since the air cannot expand radially, there is no Radial
Momentum expansion effect and the pressure near center does not drop below
ambient. The gradual decrease in pressure against drag from friction is
typical of flow through parallel pipes and ducts. The pressure between the
plate and the table is always above ambient. There is no possibility of
lift, no matter how fast the flow of air. |
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Quantitative Verification of the Model
Gathering direct measurement of the pressure and mass flux in the
tiny space between the disk and the table is impractical. Nevertheless,
it is possible to collect some fairly revealing data for inferential
comparison. For example, we can take a flux/pressure profile. By placing
instrumentation in-line between the pump and the device, we can measure
the central orifice pressure and the mass flux rate. Then, by varying
the pump pressure we can make graphs of how mass flux varies with
pressure under various conditions.
The difficulty in this approach is that since the disk tends to rise
until it restricts its own air supply, the equilibrium position of the
disk is dependent on the roughness of the disk and on the way friction
determines drag. Unfortunately, precious little seems to be known about
friction for air when velocity, density, pressure and Reynolds Number
are widely dynamic. On the principle that a reasonable estimate of a
factor is preferable to ignoring it entirely, I have included an
estimate based on the Moody diagram. Perhaps due to the tenuous nature
of this estimate, the model gives only crude numerical predictions.
Nevertheless, the model does generate outputs that all move in the
right direction in response to changes in inputs.
Weight Tests
By suspending various weights beneath the disk, the flux/pressure
profiles indicate that increasing the weight also increases the flux. In
effect, the weight pulls the gap open and so the levitator draws more
air. In addition, the importance of weight as a determinant of flux is
more pronounced at low pressures. The model shows the same behavior on
both counts.
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Actual
The actual device draws more air as the weight of the disk increases.
In addition, this effect is more pronounced at lower pressure and narrows
at higher pressures. |
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Model
The model also draws more air as the weight of the disk increases. The
model effect is also more pronounced at lower pressures and narrows at
higher pressures. |
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Radius Tests
By suspending disks of various radii, the flux/pressure profiles
indicate that increasing the disk radius also increases the flux. The
model shows the same behavior.
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Actual
The actual device draws more air as the radius of the disk increases. |
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Model
The model also draws more air as the radius of the disk increases. |
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Negative Resistance Effect
Some of the flux/pressure profiles show a curious result. The flux
sometimes decreases as more pressure is applied. This indicates negative
incremental resistance to pressure. The model shows the same effect
although the model shows it for a disk of about one fifth as large as
the corresponding actual device. This inaccuracy may have to do with
formulations of equations for the valve or for air friction. While
negative resistance is a curious artifact of the levitator it does not
appear to be central to its operation. Nor does it seem to matter to the
use of the model to as a demonstration that lift is a function of radial
Momentum, not of air velocity. Bernoulli's Principle simply does not
apply as an explanation for lift.
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Actual
At lower pressures, the actual device draws less air as the pressure
increases. |
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Model
At lower pressures, the model also draws less air as the pressure
increases. |
Theory of Radial Momentum
The theory of Radial Momentum: a fluid with Radial Momentum loses pressure.
For example, consider the levitation table.
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The key geometry of the theory of Radial Momentum is the expanding ring
of air with cross section dA = 2 * pi * r * dr. Since dr = v * dt, we
have:
dA/dt = 2 * pi * r * v
This is the key equation for the theory of Radial Momentum. The time
rate of change of area the area of the ring equals two times pi times the
radius times the radial velocity of the air. |
A more complete derivation of the equations that describe the
levitator:
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Property |
Equations |
Notes |
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Area |
A = pi * r2
dA/dt = 2 * pi * r * dr
dA/dt = 2 * pi * r * v |
Area of a plate inside the circumference of a point on the disk |
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Pressure |
P = P0 * r0 / r
dP/dt = - P0r0 * dr / r2
dP/dt = - P0r0 * v / r2 |
Pressure at a point on the disk inverse to area of a ring, hence falls
off with radius. |
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Force |
F=PA
dF/dt = d(PA)/dt
dF/dt = PdA/dt + AdP/dt
dF/dt = pi * P0r0 * v |
Force on the plate at a point on the disk. |
The theory of Radial Momentum seems to predict the levitator behavior at both
low and high pressure operation.
Case I: Low pressure
In this case, the gap is relatively wide and the levitator draws all the
airflow the pump can deliver. The well pressure is relatively low. P0,
the pressure just past the valve is the same as the pressure in the well.
Friction is about zero and the velocity is constant to the edge of the physical
disk. The time for a ring of air to traverse the plate, t = (R- r0)/v.
The force on the plate of the air in the well, F0 = pi * r0^2
* P0.
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Derivation of Force on Plate |
Notes |
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1 |
dF/dt = pi * P0r0 * v |
all constants |
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2 |
F = pi * P0r0 * v * t |
simple integration |
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3 |
F = pi * P0r0 * (R- r0) |
from r0 to R |
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4 |
F = F0 * (R/ r0 -1) |
substituting F0 |
Note that force (4) is independent of velocity and the size of the gap. Force
depends on the ratio of the disk radius to the orifice radius.
Case II: High pressure
In this case, the gap [h] is relatively small and does not draw all the
airflow the pump can deliver. The well pressure therefore can be relatively
high. Air escapes through the valve with escape velocity given by conservation
of energy. V = sqrt(2 Pv / d), where [Pv] is the pressure drop across the valve
and [d] is the density. Velocity then falls due to viscous drag, D = velocity *
area * viscosity / gap. To simplify the derivation, I assume average velocity
and constant acceleration.
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Derivation of Drag |
Notes |
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1 |
D = v * A * x / h |
drag = velocity * area * dynamic viscosity / gap |
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2 |
a = v * A/M * x / h |
divide by mass to get acceleration |
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3 |
a = v * x / (h2 * d) |
divide by density |
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4 |
h2 = v * x / (a * d) |
Rearrange |
Note that gap is squared since both drag and mass are inversely proportional
to gap. Typical values for (4) are v = 50 m/s, x = 1.8 * 10-5 Pa s, a
= 105 m/s, d = 1.2 kg/m3 and h = 8.66 * 10-5 m.
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Derivation of Force on Plate |
Notes |
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1 |
dF/dt = pi * Pvr0 * v |
v is not constant |
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2 |
dF/dt = pi * Pvr0 * [v0 - a * t] |
Substituting for v |
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3 |
F = pi * Pvr0 * [v0 * t - a * 1/2 * t2] |
integrating |
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4 |
F = pi * Pvr0 * [v02 / a -
1/2 * v02 / a] |
substituting for t |
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5 |
F = 1/2 * pi * Pvr0 * v02 /
a |
rearranging |
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6 |
F = 1/2 * pi * Pvr0 * v02 *
d * h2 / (v * x) |
substituting for a |
In (6), force is proportional to the square of the gap. As such, the device
acts as a spring, suspending the plate at an equilibrium position. With a pump
that can deliver high flux at high pressure, the induced force on a large plate
may be substantial and the plate may behave like a suction cup. Typical values
for (5) for a home garage laboratory are Pv = 100000 kg/m-s2, r0
= 1.5*10-3 m, v0 = 100 m/s, d = 1.2 kg/m3, h = 10-4
m., v= 50 m/s, x = 1.8*10-5 Pa s and F = 22.5 kg-m/s2.
Simulation Model
A simulation model shows time behavior of the model.
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Simulation of the constant friction case, showing the behavior of a
small ring of air, versus time, starting at the point that it escapes
through the valve and starts expanding radially. Radial Velocity falls due
to friction. As the area of the ring grows, its pressure decreases.
Variables are on different scales, as indicated after each variable name.
The simulation follows Euler's method of recursive solution of a system of
integral equations. |
Experimental Verification
Direct measurement of pressures and velocities in the gap between the plate
and the table is physically difficult. Indirect experimental verification,
however, is possible. The theory of Radial Momentum predicts that the levitating
force on the plate will increase with gap size, similar to the behavior of a
restorative spring. Therefore, increasing the weight of the plate would pull the
plate away from the table, increasing the gap size and inducing higher airflow.
Application of Bernoulli's Principle (a smaller gap induces a faster air
velocity and a higher force) indicates a negative spring; the plate would simply
latch up tight against the table regardless of plate weight
The feed pump used to provide air for the experiments has a 3-gallon holding
tank and is rated at 4.0 scfm at 40 psi and 2.9 scfm at 90 psi. The pump is
therefore less efficient at high pressure and increasing the demand for airflow
results in lowering the holding tank pressure.
Actual experimentation verifies that increasing the weight of the plate
reduces tank pressure. The pump maintained a tank pressure of 57 psi with a
plate that weighed about one half ounce. When an additional 14 ounces was
suspended under the plate, the tank pressure fell to 49 psi. This demonstrates
that the levitator responds to increased disk weight by drawing more airflow to
balance the weight. This confirms the Radial Momentum theory and contradicts the
explanation based on application of Bernoulli's Principle.
[flow experiment results can go here]
Air Table with Other Channel Shapes
Plates with various raised channels to entrain airflow were tested. All of
the plates that allowed Radial Momentum adhered to the table. Only the parallel
channel plate, which does not allow Radial Momentum, did not adhere to the
table. This confirms that Radial Momentum is necessary for pressure decrease.
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