Basic Math
Up Bucket Friction Gas Ring Physics

 

Basic Math

by Ed Seykota, 1999

Energy of a Ball:

A ball of mass m has velocity v with components x, y and z.
The kinetic energy, Ek = mv/2 where v = x + y + z.

Pressure and Force of a Ball in a Cube:
A ball is in a cube of side s.
The pressure in the cube, P, is Energy/Volume or Ek/s.
As the ball bounces back and forth between the walls
it changes momentum, p = mv, with a frequency n = x/s.
The force on side x, Fx, is momentum change times frequency.
Fx = 2mx * x/2s = mx/s.

 

Some Notes and Derivations

Two Types of Energy
Balls bouncing around, bumping into each other,
with no net velocity in any one direction
have a static energy proportional to temperature.
Balls with a net velocity in one direction
have a dynamic energy, Ek = mv/2.
The Great Magic Disk invites a study
of the conversions between these two forms of energy. See
Units of Measure.


Volume Flux Through an Aperture:
The following derives, using fluid dynamics, equations for the escape velocity of the air through the aperture and the height of the gap.

For Reynolds numbers above about 3 (I estimate the Reynolds number for a levitator much higher), Q, the volume flux through a circular aperture, in terms of C, the orifice coefficient (about .6), A, the area of the aperture, dP, the pressure differential across the aperture and p, the density of the fluid, is, per Vogel, page 187:

Q = C*Pi*A*sqrt(2*dP/p)

Note that for high Reynolds numbers, the volume flux depends on fluid density, the square of area and the suare root of pressure drop; at low Reynolds numbers the volume flux depends on the cube of area, the first power of the pressure drop and m, the dynamic viscosity:

Q = A*dP/3m

Assuming that at high, and not too high, Reynolds numbers, the shape of the orifice is irrelevant, we then have A, the area of the aperature:

A = 2*Pi*Rh*G, and
Q = C*Pi*(2*Pi*Rh*G)*sqrt(2*dP/p)
Q = k * G*sqrt(dP)

Also:

Solving for the gap:
Volume Flux = f(Gap, dPressure)
Gap = f(Volume Flux, Aperture)

Second Order Effects:
Velocity(r) = f(Escape Velocity, Pressure(r))
Total Head = f(Pressure(r), Velocity(r))
Virtual Rim = f(Total Head(r), Ambient Pressure)
Pressure Above Disk = f(Hole Pressure, Pressure(r), Virtual Rim)

 

 

The Math
The following derives, using standard physics and the 1/r theory, an equation for the forces on the disk as a function of pressures and radii. Note: no such derivation exists based on the Bernoulli principle.

Fw = Fu (1 - (Ph/Pa)(Rh/Rd) - 2*(Rh/Rd)(1 - Rh/Rd))


Terms:
Pa - ambient pressure (pascals)
Ph - pressure in hole (pascals)
Px - pressure just past aperture (pascals)
Rd - radius of disk (meters)
Rh - radius of hole (meters)
Fw - down force on disk from weight of disk (newtons)
Fh - down force on disk under the hole (newtons)
Fd - down force on disk under gap hole (nestons)
Fu - up force on underside of disk from ambient pressure (newtons)
G - height of the gap between the top of the disk and the ceiling
Pi - 3.14159

Assumptions:
Air is non-compressible.
Static pressure drops to ambient pressure across the aperture (Pa = Px).
This drop accelerates the molecules to their maximum "escape" velocity.
The molecules, having momentum, keep moving and fan out radially.
The total head opposes the ambient pressure at the periphery of the gap and establishes the virtual radius of the disk. 1/r Law: The static pressure at a point in the gap = Ph*(Rh/r)

Analysis of forces as a function of pressures and radii:

Up force on underside: Fu = Pa*Pi*Rd
Down force under hole: Fh = Ph*Pi*Rh
Down force in a thin ring in gap: dFr = Px*(Rh/r) * 2*Pi*r*dr
Rearranging: Fr = 2*Pi*Px*Rh
Down force under entire gap: 2*Pi*Px*Rh(Rd - Rh)

Now from force balance: Fw + Fh + Fd = Fu
So: Fw = Fu - Fh - Fd

Substituting:
Fw = Pa*Pi*Rd#178; - Ph*Pi*Rh - 2*Pi*Px*Rh(Rd - Rh)
Fw = 2*Pi* (Pa*Rd#178; - Ph*Rh - 2*Px*Rh(Rd - Rh))
Fw = Fu (1 - (Ph/Pa)(Rh/Rd) - 2*(Px/Pa)(Rh/Rd)(1 - Rh/Rd))

Fw = Fu (1 - (Ph/Pa)(Rh/Rd) - 2*(Rh/Rd)(1 - Rh/Rd))

Reasonablility Tests:

For a levitator with a very, very small hole:
Rh << Rd
Fw = Fu (1 - 0 - 0)) = Fu
The levitator lifts a weight equal to the ambient force on the bottom of the disk.

For a disk with no overhang and a vacuum in the hole:
Rh = Rd and Ph = 0
Fw = Fu (1 - 0 - 0)
The levitator lifts a weight equal to the ambient force on the bottom of the disk.

For a disk with no overhang and a half vacuum in the hole:
Rh = Rd and Ph = Pa/2
Fw = Fu (1 - 1/2 - 0)
The levitator lifts a weight equal to half the ambient force on the bottom of the disk.

For a disk with no overhang:
Rh = Rd
Ph > Pa
Fw = Fu (1 - Ph/Pa - 0)
Fw = Fu (1 - Ph/Pa) < 0
The levitator lifts a negative weight. Thatis, it only balances by pushing down on a lighter-that-air disk.

For a disk with the motor off:
Ph = Pa
Fw = Fu (1 - (Rh/Rd) - 2*(Rh/Rd(1 - Rh/Rd))
Fw = Fu (1 - (Rh/Rd) - 2*(Rh/Rd) + 2*(Rh/Rd)))
Fw = Fu (1 - 2*(Rh/Rd) + (Rh/Rd)))
Fw = Fu (1 - (Rh/Rd))
The levitator lifts a weight somewhere between zero and the ambient force on the bottom of the disk, depending on the ratio of the radii of the hole and disk. At this zero air velocity limit, the equilibrium gap is also zero, so the disk acts like a suction cup.

For a typical disk:
Rh = Rd/4 and Ph = 2*Pa
Fw = Fu (1 - (2)(1/4) 2*(1/4)(1 - 1/4))
Fw = Fu (1 - 1/8 - 3/8) = Fu/2
The levitator lifts a weight equal to half the ambient force on the bottom of the disk.

If you don't agree with these derivations, and/or have suggestions and comments, I'd like to hear from you. Good debates make for good science.

Units of Measure

Quantities Dimensions SI Units
Length, Distance L Meter (m)
Area, Surface L Square Meter (m)
Volume L Cubic Meter (m)
Time T Second (s)
Velocity, Speed L/T Meters per Second (m/s)
Acceleration L/T Meters per Second Squared (m/s)
Mass M Kilogram (kg)
Force ML/T Newton (N) or kg-m/s
Density M/L Kilogram per Cubic Meter (kg/L)
Work ML /T Joule (J) or Newton-Meter (N-m)
Power ML/T Watt (W) or Joule per Second (J/s)
Pressure, Shear Stress M/LT Pascal (Pa or N-s/m)
Dynamic Viscosity M/LT Pascal Second (Pa-s or N-s/m)
Kinematic Viscosity L/T Squate Meter per Second (m/s)


 

Dear Professor Brad Snyder (UNR) 784-6939: For our conversation, scheduled for Monday May 10 at 11:00 AM, you asked for clear diagrams defining the problems. I hope the following is clear. If anything is unclear or ambiguous, please let me know at 832-7377. Non disclosure request: please keep these matters confidential until the end of 1999. Yours truly, Ed Seykota

 

Definitions, Symbols and Units:

States

Symbol

Units

Notes

Mass

M

Kg

Air: d ~ 1.2 kg/m3

Momentum

Mo

Kg-m/s

Mo = M * U

Thermal Energy

Te

Kg-m2/s2

Te = PV = NRT

Rates and Flows

 

 

 

Mass Flux

M-dot

Kg/s

 M-dot = dM/dt

Momentum Flux

Mo-dot

Kg-m/s2

 Mo-dot = dMo/dt

Thermal Energy Flux

Te-dot

Kg-m2/s3

 Te-dot = dTe/dt

 

Problems:

Problem #1: Air escapes through a round orifice in a solid plate.

Given:

0 < P < 5 atm (gauge); P0 = 0 atm.; T = 70 F; T0 = 50 F; D = 1/16".

Find the flows at the exit:

Find M-dot, Mo-dot, Te-dot

 

Problem #2: A circular plate over the orifice creates a ring-shaped "valve" at the center and changes the flows.

Given:

0 < P < 5 atm; P0 = 0 atm.; T = 70 F; T0 = 50 F; D = 1/16", h ~ .001 m.

Find the flows at the exit:

Find M-dot; Mo-dot; and Te-dot.

 

Problem #3: Air leaving the valve area continues to experience skin-friction drag.

Given:

0 < P < 5 atm; P0 = 0 atm.; T = 70 F; T0 = 50 F; D = 1/16", E0 ~ .0000003 m.

Find drag from skin friction:

Derive U (velocity), p (pressure) and d (density) from the satates as functions of M, Mo, and Te.

Find drag as a function of U, p and d.

 

Problem #4: Difference Equations:

To do: Create control volumes from concentric rings of thickness dR, radiating out from the orifice between the plates, like nested bicycle tires. (The rings in this panel are shown on their sides.) During the traverse across dR, within a control volume, Mo and Te change slightly as a function of drag and pressure gradient across dR. For control volume[r], write equations for these small changes.

Continued

continuation.

There is (1) a pressure gradient from the inside to the outside and (2) some skin friction across the top and the bottom. These act to change the momentum and thermal energy of the air passing through.

Very small section of a ring showing direction of mass flux. Width (w) is actually curved slightly and continues around as the circumference of the ring.

 

Mathematics

Fluid velocity - Fluid flux - Total energy - Radial Action Theory - Thermal Energy and Pressure - Kinetic Energy and Pressure -  Units of Measure

Fluid velocity

The air at the entrance to the gap sees a clear way ahead and expands into it such that it loses thermal (pressure) energy and gains radial (velocity) energy. To simplify the math, I assume that right at the entrance to the gap: (1) the pressure instantly drops right to ambient pressure and (2) the air instantly achieves its maximum "escape velocity." These assumptions are best for Reynolds numbers above 3 and I feel the Levitator qualifies. A more rigorous examination would show these assumptions are not exactly accurate although the discrepancies would not seem to effect the overall Theory of Radial Action.

1/2 mvr2 = dP * V0; kinetic energy gain from thermal energy; dP = P0 - PA.

vr2 = 2 * dP / D; velocity squared equals twice pressure drop over density;

vr = sqrt(2 * dP / D); by taking square root.

For a pressure drop of 25 newtons/cm2 and a density of .01 gram/cm3, (seat-of-pants estimate)

vr = sqrt(2 * (25 newtons/cm2 ) / (.01 gram/ cm3) = sqrt(5000 m2/sec2);

vr = 70 m/sec = 200 ft/sec = 120 miles per hour, approx

Fluid flux

Q = C * Pi * A * vr;

Q is the volume flux through a circular aperture. C is the orifice coefficient, about .6. A is the area of the aperture. Pi is about 3.14156. vr is the fluid velocity.

Total energy

Er = eT + eK; the energy of a fluid equals the thermal plus kinetic energy.

Radial Action Theory: the pressure of a radially expanding fluid is inverse to the radius.

At the entrance to the gap, the volume of a unit of air:

V0 = h * pi * ((r0 + dr)2 - r02); volume equals the gap times a small delta radius.

V0 = 2 * h * pi * r0 * dr; approximate, for very small dr.

As the unit of air radiates to radius, r1, it becomes a larger and larger ring.

V1 = 2 * h * pi * r1 * dr

V1 / V0 = r1 / r0; the volume is proportional to the radius.

P1 = P0 * V0 * V1; the pressure is inverse to the volume.

P1 = P0 * r0 / r1; the pressure is inverse to the radius.

Thermal Energy and Pressure

A molecule has velocity along three axes.

v = x + y + z; the composite velocity equals the vector sum of the components.

Balls bouncing around in a box, bumping into each other and into the walls, with no net velocity in any one direction, have a thermal energy or heat, proportional to that thermal motion. The collisions with the walls exert pressure on the walls. Upon colliding with the walls, the balls either gain or lose energy depending on the temperature of the box. In this way, molecules in a box stay in thermal equilibrium with the box. From examination of force from a molecule bouncing back and forth between walls:

nx = vx/(2*s); the frequency of a hit equals the velocity divided by twice the side length.

Fx = px * nx; the force on a side equals momentum change times frequency.

Fx = [m * vx] * [vx/(2*s)]; ( by substitution).

P = F / A; pressure equals force divided by area.

P = [m * vx] * [vx/(2*s)] / s3 = m * vx/ s3; (by substitution).

ET = P * V; Thermal energy equals pressure times volume.

ET = m * vx/ s3 * s3 = m * vx; thermal energy equals mass times the square of velocity.

Kinetic Energy and Pressure

Air in motion has momentum and kinetic energy relative to objects at rest. As molecules in the flow stream collide with an object they exchange momentum and energy with the object.

Ek = mv/2; kinetic energy equals one half times mass times velocity squared.

px = m * vx; the momentum of the molecule is its mass times its x-velocity.

 

 

 

 Units of Measure 

A good way to check your computations is to carry along units of measure, and balance them as you go.

 

 

Quantities

Dimensions

SI Units

Length, Distance

L

Meter (m)

Area, Surface

L

Square Meter (m)

Volume

L

Cubic Meter (m)

Time

T

Second (s)

Velocity, Speed

L/T

Meters per Second (m/s)

Acceleration

L/T

Meters per Second Squared (m/s)

Mass

M

Kilogram (kg)

Force

ML/T

Newton (N) or kg-m/s

Density

M/L

Kilogram per Cubic Meter (kg/L)

Work

ML /T

Joule (J) or Newton-Meter (N-m)

Power

ML/T

Watt (W) or Joule per Second (J/s)

Pressure, Shear Stress

M/LT

Pascal (Pa or N-s/m)

Dynamic Viscosity

M/LT

Pascal Second (Pa-s or N-s/m)

Kinematic Viscosity

L/T

Squate Meter per Second (m/s)