Seykota Model
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The Seykota Levitator Model

1999 by Ed Seykota

 

 

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.

 

 

 

 

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.

 

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.

 

Fundamental Physical States

Density

Pressure

Velocity

Computational Properties

Number_of_Rings

Delta Time

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.

 

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.

 

 

The Basic Laws

 

The model observes the basic laws of fluid dynamics: conservation of mass, conservation of momentum and conservation of energy.

 

 

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].

 

 

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]

 

 

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]

 

 

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]

 

 

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]

 

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.

 

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.

 

 

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.

 

 

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.

 

 

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.  

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.

  

 

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.

 

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.

  

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.

 

  

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.

 

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.

 

  

Actual

The actual device draws more air as the radius of the disk increases.

 

Model

The model also draws more air as the radius of the disk increases.

 

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.

 

  

Actual

At lower pressures, the actual device draws less air as the pressure increases.

 

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.

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:

Property

Equations

Notes

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

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.

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.

 

Derivation of Force on Plate

Notes

1

dF/dt = pi * P0r0 * v

all constants

2

F = pi * P0r0 * v * t

simple integration

3

F = pi * P0r0 * (R- r0)

from r0 to R

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.

 

Derivation of Drag

Notes

1

D = v * A * x / h

drag = velocity * area * dynamic viscosity / gap

2

a = v * A/M * x / h

divide by mass to get acceleration

3

a = v * x / (h2 * d)

divide by density

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.

 

Derivation of Force on Plate

Notes

1

dF/dt = pi * Pvr0 * v

v is not constant

2

dF/dt = pi * Pvr0 * [v0 - a * t]

Substituting for v

3

F = pi * Pvr0 * [v0 * t - a * 1/2 * t2]

integrating

4

F = pi * Pvr0 * [v02 / a - 1/2 * v02 / a]

substituting for t

5

F = 1/2 * pi * Pvr0 * v02 / a

rearranging

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.

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.