User talk:Will2green
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Simple math that proves Overunity
Contents |
[edit] The System
In a fluid system, this rotary motor is 14” diameter, 8 cylinders, has 2” dia. pistons, with a 7” dia. crank and 0.5” displacement, there are 3 active cylinders. In this example we are using the constants 450-psi cylinder pressure, rotating at 900 rpm. A hydraulic fluid pump with an electric motor drives the above system.
A pump and motor were chosen to give me the ability, to control both the flow and pressure on the input side. Thereby more accurately calculate, the torque and speed on the output side.
This technology is better than existing technologies for the following two reasons
1) Greater leverage within a more confined space.
2) Reduced mass used in linear to rotary motion translation.
[edit] Investigation
The information herein may be investigeted by use of the following technique.
The Scientific method, which is a body of techniques for investigating phenomena and acquiring new knowledge, as well as for correcting and integrating previous knowledge. It is based on observable, empirical, measurable evidence, and subject to laws of reasoning. All such evidence is collectively called scientific evidence and is used as a means of structural cohesion to new information..
FEA Software offers a wide spectrum of powerful tools to help engineers who are familiar with design validation concepts to perform virtual testing and analysis of parts and assemblies. Engineers who need more specific design analysis capabilities can use FEA Software to predict the physical behavior of practically any part or assembly under any loading condition.
FEA Software was used in the concept validation stage and in testing verious loading condition for this model.
[edit] Constants
[1]
Displacement = disp = 0.5 inches = d
Piston diameter = 2”
Active Cylinders = 3
Star-wheel / Crankshaft = 7” diameter
Pressure = PSI = 450 PSI
Speed = N = 900 RPM’s
1 HorsePower = 746 Watts
5252 = constant (33,000 divided by 3.14 x 2 = 5252)
1 HP = 33,000 ft·pound-force·min−1 = exactly 0.74569987158227022 kW
1 Gallon = 231 Cu. In.
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[edit] Math
Fluid Pressure (PSI) = Force (pounds) / Area ( sq. in.).........P = PSI = F/A
Cylinder Area (sq. In.) = 3.1416 × Radius (inch)(sq.)...........A = Pi × R(sq.)
Cylinder Force (pounds) = Pressure (psi) × Area (sq. in.).....F = P × A
Cylinder Speed (ft./sec.) = (231 × gpm) / (12 × 60 × Area)....v = (0.3208 × gpm) / A
Fluid Motor Speed (rpm) = (231 × gpm) / disp. (cu. in.).........n = (231 × gpm) / d
Pump Output Flow (gpm) = (Speed (rpm) × disp. (cu. in.)) / 231 gpm = (n ×d) / 231
Torque = Force applied x lever arm in “ft” or Ta*psi = F lbs.”
1.) PUMP HP=GPM*P / 1714
2.) PUMP HP=GPM*P*.0007
1.) HP=F*N/5252
2.) HP=(C*N / 12/60) * F / 550
Efficiency=work out / work in * 100
Where D is Displacement, L is Lever radius and E is efficiency D/L=E
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[edit] Math applied
A = Cylinder Head Area = pi*(r^2) = 3.14*(1*1) = 3.14159 sq. in.
Ta = Total Area = A*active cylinders = 3.14*3” = 9.42477 sq. in.
P = Pressure = 450psi
N = RPM = 900
C = Circumference of Star wheel = 7 * 3.14159 = 21.99
L = Star-wheel Crankshaft Radius / 12 inches = 3.5/12 = .29 ft
D = Distance in 1 Min = (C*N) /12 = (21.99*900) /12 = 19,791in/12 = 1,649.25 ft
F = Ta*P = 9.42477*450 = 4,241 lbs Force.
T = Torque in ft_lbs = (crankshaft radius / 12)*4241 = (3.5/12) =. 2916*4,2341 = 1,236.95 ft_lbs.
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[edit] System HP
With the above given; the following is 2 different HP formulas representing work out:
1.) HP=F*N/5252 = 211.8 HP or = 158,002 Watt
2.) HP=(C*N / 12/60) * F / 550 = 211.85 HP or = 158,040 Watt
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[edit] Pump Math
[Hydraulic Formulas][6]
With the above, we then calculate pump fluid flow & HP
Based on a volume of 12.56 cu in per rev., Then 12.56 * 900 rpm = 11,304 cu. in.
Converting that into gal per min divide 11,304 cu. in./ 231 = 48.93 gal per min.
The total displacement per cylinder = Cylinder Area* Displacement = 3.14*.5=1.57 cu in
The total displacement per rev = Cubic Inches* 8 Cylinders = 3.14*0.5*8 = 12.56 displacement per rev.
Total Displacement at speed = Displacement per rev* RPM = 12.56 * 900 rpm = 11,304 total displaced cubic inches. GPM = Total Displacement Cubic Inches / GPM Constant = 11304 Cu In / 231 = 48.93 GPM.
With 48.93 GPM and a pressure of 450 psi we can compute pump HP work in below.
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[edit] Pump HP
The following are 2 Pump HP formulas representing work in.
1.) PUMP HP=GPM*P / 1714 = 48.93 * 450 /1,714 =12.84HP or = 9,578.64 Watts
2.) PUMP HP=GPM*P*.0007 = 48.93 * 450 * .0007 = 15.41 HP or = 11,495.86 Watts
An electric motor and pump were chosen for this example to control the input with calculable knowns. If you can calulate forward then you can transpose.
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[edit] System Efficiency
Efficiency work out / work in * 100 = 211.8HP / 15.41 HP * 100 = 1,374.43%
The Mechanical_efficiency, which is defined as the ratio of the work out to the work in, some of the input is invariably wasted in overcoming friction. The element of time does not enter into the computation of work; the time rate of doing work is called power.
There are so many types of efficiences that each must be considered to understand efficiency...
Algorithmic efficiency, Energy efficiency, Electrical efficiency, Fuel efficiency, Lighting efficiency, Mechanical efficiency, Productive efficiency, Thermal efficiency, System efficiency & Volumetric efficiency,
In the above system, no new energy is being created, but rather or more amply put, this system better utilizes the energy available.
This system it is a fluid mechanics and volumetric system not thermal...
For some efficiencies its ok to go over 100% for others not, according to our understanding...
Consider this formula workout/workout+workin * 100, of which you may never go over 100%
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