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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 = mv²/2 where v² = x² + y² + z².
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/s³.
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 = mx²/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 = mv²/2.
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*A²*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 = A³*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)²*sqrt(2*dP/p)
Q = k * G²*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 - 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*Rd²
Down force under hole: Fh = Ph*Pi*Rh²
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*Rh² - 2*Pi*Px*Rh(Rd - Rh)
Fw = 2*Pi* (Pa*Rd#178; - Ph*Rh² - 2*Px*Rh(Rd - Rh))
Fw = Fu (1 - (Ph/Pa)(Rh/Rd)² - 2*(Px/Pa)(Rh/Rd)(1 - Rh/Rd))
Fw = Fu (1 - (Ph/Pa)(Rh/Rd)² - 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 - 2*(Rh/Rd(1 - Rh/Rd))
Fw = Fu (1 - (Rh/Rd)² - 2*(Rh/Rd) + 2*(Rh/Rd))²)
Fw = Fu (1 - 2*(Rh/Rd) + (Rh/Rd))²)
Fw = Fu (1 - (Rh/Rd))²
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 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 |
L² |
Square Meter (m²) |
| Volume |
L³ |
Cubic Meter (m³) |
| Time |
T |
Second (s) |
| Velocity, Speed |
L/T |
Meters per Second (m/s) |
| Acceleration |
L/T² |
Meters per Second Squared (m/s²) |
| Mass |
M |
Kilogram (kg) |
| Force |
ML/T² |
Newton (N) or kg-m/s² |
| Density |
M/L³ |
Kilogram per Cubic Meter (kg/L³) |
| Work |
ML² /T² |
Joule (J) or Newton-Meter (N-m) |
| Power |
ML²/T³ |
Watt (W) or Joule per Second (J/s) |
| Pressure, Shear Stress |
M/LT² |
Pascal (Pa or N-s/m²) |
| Dynamic Viscosity |
M/LT |
Pascal Second (Pa-s or N-s/m²) |
| Kinematic Viscosity |
L²/T |
Squate Meter per Second (m²/s) |
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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:
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States |
Symbol |
Units |
Notes |
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Mass |
M |
Kg |
Air: d ~ 1.2 kg/m3 |
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Momentum |
Mo |
Kg-m/s |
Mo = M * U |
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Thermal Energy |
Te |
Kg-m2/s2 |
Te = PV = NRT |
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Rates and Flows |
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Mass Flux |
M-dot |
Kg/s |
M-dot = dM/dt |
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Momentum Flux |
Mo-dot |
Kg-m/s2 |
Mo-dot = dMo/dt |
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Thermal Energy Flux |
Te-dot |
Kg-m2/s3 |
Te-dot = dTe/dt |
Problems:
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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 |
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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. |
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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. |
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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 … |
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… 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. |
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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 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.
v² = x² + y² + z²; 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
* vx²/ s3; (by substitution).
ET = P * V; Thermal energy equals pressure times volume.
ET = m * vx²/ s3 * s3
= m * vx²; 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 = mv²/2; 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.
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Units of Measure
A good way to check your computations is to carry along units of
measure, and balance them as you go.
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Quantities |
Dimensions |
SI Units |
|
Length, Distance |
L |
Meter (m) |
|
Area, Surface |
L² |
Square Meter (m²) |
|
Volume |
L³ |
Cubic Meter (m³) |
|
Time |
T |
Second (s) |
|
Velocity, Speed |
L/T |
Meters per Second (m/s) |
|
Acceleration |
L/T² |
Meters per Second Squared (m/s²) |
|
Mass |
M |
Kilogram (kg) |
|
Force |
ML/T² |
Newton (N) or kg-m/s² |
|
Density |
M/L³ |
Kilogram per Cubic Meter (kg/L³) |
|
Work |
ML² /T² |
Joule (J) or Newton-Meter (N-m) |
|
Power |
ML²/T³ |
Watt (W) or Joule per Second (J/s) |
|
Pressure, Shear Stress |
M/LT² |
Pascal (Pa or N-s/m²) |
|
Dynamic Viscosity |
M/LT |
Pascal Second (Pa-s or N-s/m²) |
|
Kinematic Viscosity |
L²/T |
Squate Meter per Second (m²/s) |
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