A0201-01 Thanks sandia24 . Nice work
A0201-03 I was setting up the variable list for Problem 4 on the power output ad input see if this is reasonable--Q's on
clarification or suggestions to team drop box or pm if your are an ionizationx member
_________________________________________________
_________________________________________________
________________________
MAIN DISCUSSION FOLLOWS BELOW VARIABLE LIST AND VALUE RANGE
VARIABLE VALUE SOURCE/// REFERENCE
V1 = velocity of magnetic field movement per second 50-90 ips In 2019 Handout in Bremen Conference
N1 = number of twists per unit length of non-magnetic spiral divider per unit length 0.3 - 1.2 Estimate for M4steel considering thickness
and core diameter.
N3 = value of magnetic field strength ? TBD by calc. and type of EPG
F1 = value of the frequency pulsing alignment coils for dyne-axis of magnetic field 60 Hz/sec Mains frequency in US 50 some parts of EU--- wiki
N4= number of coils per tier 1 - 58 Don Gabel, photogrammetry and SEPG022,
N5 = number of turns in each coil 200 - 12000 Estimates using packing fractions, winding depth,
length photogrammetry as secondary verification
P1 = effective cross-section winding factor: random, hexagonal or precision winding 0.78- 0.906 Wiki refs circle packing theory
N6 = number of core sectors enclosed by pick-up coil 3- 4 Don Gabel images of various EPG's
N7= number of tiers 1 - 7 Birth of New Technology 1994 or 1995 ed
A1= cross-sectional area of tubing uses in EPG tier ( in inches0 0.218- 0.254 The Copper Handbook
Power Input Variables
)
W1 = watts required for initiation of flow ( Initial inertial load) Rheological, mass density and volume consideration TBC
W2= steady state power load for mag-media circulation see appendix TBC
W3 = dyne-axis load see appendix TBC
Known values
N1 known
N2 known
N3 calculation to be completed
N4 known
N5 known
N6 known
F1 known
P1 known
V1 known
Stated design output was 220 VAC @ 300 amps ( per seminar notes)
A0202-01 Let's try another attempt at N3. At one of the conferences in 2019 ( SMC 2019 Bremen Ohio), it was proposed that the Transformer EMF
equation might be used in the mathematical model of the Meyer EPG series regarding the flux density problem.
Through photogrammetry the maximum number of turns , number of coils, diameter and volume of the core
magnetized slurry/gas can be determined. Since the output power, velocity, and frequency are known with some precision
It may be possible to arrange the transformer EMF equation to obtain a Beta Max for the flux density!!
Another observation was made at the 2019 Bremen Conference that the larger the core volume, the lower value of the magnetic
saturation could be and still maintain the same power output. This is because the total power output for the device is dependent
in part upon the total amount of flux present in the magnetic core.
A0202-02 Correct, if the other design factors such as the number of coils, number of winds and same velocity of the magnetic
gas or slurry are maintained, the limitations of the maximum level of magnetic saturation of the EFH series ferrofluids
can be mitigated. Basically scale up the volume and the magnetic saturation can be lower and still provide the
design power output. While the 400 Hz mil-spec converters are still an option for the magnetic drives I think
you may want to just keep it simple so that operating frequency matches the 50 or 60 Hz standards for output for
residential use.
A0202-03 thnx to thorzpwr
A0203=01 ok, you forgot to hide your location ,metadata I'll fix it zo the mib's don't get ya....lol
A0205-01 Since output data is only available for the 6Tmaggas EPG and for the velocity of the magnetic medium, 'I think a problem approach might be
to determine the flux in the 6 multi-tier system as if EFH-1 was present and then scale down to the magnetic pump system and volume of EFH-1
at the stated velocity and use a ,calculated flux density to determine output characteristics of the magnetic pump device in terms of output.
The sizing of the bus bars, the parallel arrangement of the pick-up coils and the breakdown voltage of the insulation might put some upper
limits to how it was being operated and limits to the possible voltagexs and amps produced
Design output 220 VAC at 300 amps = 66,000 watts
A020601 So now let's assign values to some of the variables
---------------------------------------------------------------------------------------------------------------
Cross sectional area is calculated as follows:
1. determine the diameter of the tubing 0.5" obtained by photogrammetry 0.5 outside diameter
also confirmed by actual measurement by Don Gabel . (see notebook photos)
2. determine the range of. possible internal diameters Common types of pipe K L and M that have the
same outside diameter but thickness of inner diameter and wall thickness vary.
Stan Meyer may have used pre-coiled air conditioning or water supply tubing. for ease of construction.
A very useful free reference is The Copper Tubing Handbook fermi which provides the specifications and measurements
for copper tubing and pipe.
You can google The Copper Tubing Handbook for the pdf or just click on this link:
https://pbar.fnal.gov/organizationalchart/Leveling/2004%20water%20cage%20work/Cutubehandbook.pdf
One observation concerning the publicly available EPG images, it that there do not seem to be joints on the spiralled sections
themselves although the connecting copper pipes to the pumps or other means of moving the slurry or gas are straight.
I believe Stan Meyerswas practical and tried to keep things simple, so I believe he just used piping that was already coiled when purchased.
So, now let's use the above reference to get a range of possible values for the cross-sections of the copper tubing and pipes
commonly available.. Copper pipe has three basic wall thicknesses: Type K, Type L and Type M So even though the outside
diameter may remain the same, a THICKER wall means a SMALLER cross-section inside the tube
So here's the values of cross-sectional area for different copper tubing and pipe in square inches:
Type K 0.218 Type L 0.233 Type M 0.254 So the cross-sectional area for coiled copper pipe is between 0.218 and 0.254 square inches
Since the 6 tier system is not available for examination at this time, there is a degree of imprecision for the cross-sectional area value
Because the cross-sectional area is used in volume calculations and in the calculation of total magnetic flux for these systems, the estimates
of system performance depend upon the type of tubing used in the construction
--------------------------------------------------------------------------------------------------------------------------------------------------------
Length of tubing carrying magnetic slurry/gas
Since the EPGs are of a general circular design, the formula C = D x Pi or stated -- Circumference of a circle equals the
diameter of the circle multiplied by Pi (approximately 3.1416)
Now, if you are trying to find the total length of tubing used in an EPG which is a spiral, for example(In this case exactly
3 loops, then thinkof this as 3 circles each with a different diameter and circumference
The outer loop is longer than the middle loop which is in turn larger the the innermost ring of loop.
So roughly speaking, let's say you had an EPG like the Magnetic Drive (Red Pump) System and that by examination
or photogrammetryand it was determined that diameter was 17 inches.
If you are using 1/2 inch tubing in the construction, what would be the diameter of the middle loop?
The radius of the middle loop is moved in by 1/2 inch because of the width of the outer loop or to put it another
way, the diameter of the middle loop would be 16 inches measured across its outside By a similar reasoning, the innermost loop
is or about 15 inches in diameter.
So the length the spiral is approximately ( 15 + 16 + 17) times Pi. Now Stan Meyer for reasons of type of pump used (B-500 had input and output
connections at right angles)then some portions of the spiral had four loops instead of three so adjustments will have to be
made for this added length. The total length of is important because this is used in the calculations
for the Volume of gas or ferrofluid being used and also in the calculations for inductance and the number windings for the
coils as well as the length of wire required for making the windings
--------------------------------------------------------------------------------------------------------------------------------------------------------
Coils and length of wire need for project and per coil
End View "Tube" length
A formula for a single wind around a single circular core O diameter of wire times 1
1.A formula for multiple winds around a singular tubular core of length L O diameter of wire x N number of windings or wraps
2 A formula for multiple winds around two adjacent tubular cores of length L OO diameter of wire x N number of windings or wraps
.3 A formula for multiple winds around three adjacent tubular cores of length OOO diameter of wire x N number of windings or wraps
4 General Formula for multiple winds around multiple tubes OOOO... diameter of wire x N number of windings or wraps
So the length of the tube determines the total number of wraps possible independent of the number of adjacent tubes
(close wraps no spacing between wraps on tube
Formula Length of tube (think inductor core) equals the number of wraps times the width or diameter of the wire L= N times W or L/ divided by W = N
ay of determining the number of wraps that can fit on a given length of tube or core
-------------------------------------------------------------------------------------------------------------------------------------------------------
Now for the fun part determining the Length of Wire needed for one wrap around multiple adjacent cores
Formula for 1 core O L = Diameter of core times Pi
Formula for 2 adjacent cores OO L = (Diameter of core times Pi) PLUS 2D <--- for the wire that bridges the "notch" between the adjacent tubes (top and bottom)
Formula foe 3 adjacent cores OOO L = (Diameter of core times Pi) PLUS 4D <--- to account for the length needed to bridge 2 notches between the adjacent tubes (top and bottom)
Formula for 4 adjacent cores OOOO L = Diameter of core times Pi) PLUS 6D <--- To account for the length needed to bridge 3 notches between the adjacent tubes (top and bottom)
In summary, we now can calculate the length of a single wrap of wire around multiple adjacent cores and if we multiply that by the number of wraps or turns that can be wrapped on a given linear length of core
it is helpfun in deciding amount of wire needed
General Formula for Single Layer 1 wrap or turn around multiple adjacent tubes
L length equals ( Diameter of core or tube) plus ( ( N or number of cores minus 1) times 2)
So now is possible to calculate the number of winds or wraps (single layer0 around an EPG if we know the diameter of the outermost core of a spiralled EPG, the number of "loops" in the spiral, the outside
diameter of the core tubing and the gauge, diameter or width of the wire used to wrap the core
t
So lets give a quick try for the multitier 6TmaggasEPG
1 tier is about 17 inches in diameter. Since the line drawing of the 7 tier system and photographs show the drain/connecting tubes are 180 degrees apart so its possible to keep the number of loops for a tier to
be 2.5 3.5 or 4.5 loops or if the connecting tubes are all exact integers of loops the connecting tube could be all on one side. Or the direction of the flow could be counterclockwise one one tier and clockwise in the other tier. So based on the line drawing lets say that that each tier has 3.5 loops
Length of core for 1 tier [ ( 15+16+17)]times Pi ] plus( 1/2 times 14 times Pi) = 150.78 + 29.99 = 172.77 inches 6 tiers 1036 inches
172.77 inches divided by .025 inches per turn (22 gauge wire by photogrammetry = maximum 6910 turns per tier
6 times 6910 = 41,460 turns or if you use exactly 3 loops per tier 150.78 times 6 = 904 inches 904 divided by 0.025 = about 36,191 turns 6 tiers 906 inches
Image a n inductor with between 36 and 41 thousand turns of wire and between 75 and 86 feet long !! depending on method of construction
Design parameters Metric
The design output is 220 volts at 300 amp draw 66,000 watts (Watts)
(W) 220 times 300 amp draw = 66,000 watts
The cross-sectional area of the core is between 0.218 and 0.254 square in
or (A) = 1.406 to 1.634 sq cm or 0.0001406 0.0001634 square meters
F (frequency) is 60 cycle/ second AC
V (voltage) is 220 volts AC output
K Constant = 4.44
Solving of Bm =BetaMax
Basic equation
V = voltage
F = supply frequency
N = number of turns
A = cross sectional area in square meters
B = peak magnetic flux density in Weber / meter squared or T tesla
K = 4.44
V = 4.44 x F x N x A x B or rearranging this
B = divided by( 4.44 x f x N x a )
so let's try plugging in a few figures for a six tier device
V = voltage 220 VAC
F= 60 hertz per second in the US
N= 11,873
A = 0. 000468 sq m area 3 channels of pipe x 0.242 sq inches divided by conversion factor 1550 = 000468 square metres
4.44 = constant
Bmax = 220/ 1480 or 0.1486 Wb/M squared or Tesla for the 5/8" six tier system4.44 times F*N * BetaMax * A
Rearranging: BetaMax = V divided by ( 4.44 x F x N x A)
220 divided by( 4.44 times 60 Hz/sec frequency times 36191 x .218 A sq inches = .0001046
now to work on units. with a different diameter..
V = 220 VAC...
F = supply frequency
N = number of turns
A = cross sectional area in square meters
B = peak magnetic flux density in Wb / meter squared or T tesla
K = constant
V = 4.44 x F x N x A x Bmax, or rearranging this
Bmax = V divided by( 4.44 * F * N * A )
so let's try plugging in a few figures for a six tier device with a 5/8" OD copper spiral
V = voltage 220 VAC
F= 60 hertz per second in the U
N = 11,87
A = 0. 000468 sq m area 3 channels of pipe x 0.242 sq inches divided by conversion factor 1550 = 000468 square metres
B = BetaMax
K = 4.44 ( constant )
Thus Bmax = 220/ 1480 or 0.1486 Wb/M squared or Tesla for the 5/8" six tier system
[ b]Next Topic Multiple layer coils
In terms of construction if the cross sectional area is changed because of using a larger diameter tubing but keeping N number of turns the same and the length of the
spiralled coils is the same and other factors the same (same desired output) t because the output is related to the amount of flux of the core, the larger the core in terms
of cross section (and volume) means that a lower Beta value in the core of the upsized EPG can still result in the desired power output. Basically if more power is
needed the large core can allow for a lower amount of flux to be used if there is a limit to magnetic saturation for the slurry or mag-gas matrix.
This is more useful to calculate wire requirements for the Mechanical Pump EPG .
Since it's possible to estimate the thickness of the coils, the length of the original coils,
the gauge of the wire and velocity of the ferrofluid 50 ips and using a flux value estimate
a power output for the Mechanical Pump EPG.
