Hydro-Motor Simple math
- 1 The System
- 2 Investigation
- 3 Constants
- 4 Math
- 5 Math applied
- 6 System HP
- 7 Pump Math
- 8 Pump HP
- 9 System Efficiency
- 10 System Mechanical advantage
- 11 Math references
- 12 Published REFERENCES
- 13 Quotes by Richard Feynman Nobel Prize Winner
- 14 [By Willie A. Green Jr]
- 15 Summary
- 16 Other High efficiency devices
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.
The information herein may be investigated 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 various loading condition for this model.
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.
.0007 in Pump HP calculation.
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
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
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*4,241 = 1,236.95 ft_lbs.
With the above given; the following is 2 different HP formulas representing work out:
1.) HP=T*N/5252 = 211.8 HP or = 158,002 Watt
2.) HP=(C*N / 12/60) * F / 550 = 211.85 HP or = 158,040 Watt
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 per min. The the GPM = Total Displacement Cubic Inches / GPM Constant = 11304 Cu In / 231 = 48.93 GPM.
With 48.93 GPM and having a pressure of 450 psi we can compute pump HP work in below.
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 calculate forward then you can transpose.
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 efficiencies 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 is a fluid mechanics and volumetric system, not thermal...
For some efficiencies its ok to go over 100% for others, not according to today's understanding...
Consider this formula workout / workout + workin * 100, of which you can never go over 100%
System Mechanical advantage
The ability of a machine to amplify force is called its Mechanical Advantage. In this situation, the force is increased by a factor of 9.42. Mathematically, the mechanical advantage (MA) is equal to the force exerted by the machine (resistance force - Fr) divided by the force input (effort force - Fe). Thus the equation for mechanical advantage is:
MA = Fr / Fe
Fr = 4239
Fe = 450
MA = 4,239 / 450 = 9.42
Mechanical advantage is the ratio of output force divided by input force.
If the output force is bigger than the input force, a machine has a mechanical advantage greater than one.
Reference 1 http://www.gulfcoastairandhyd.com/formulas.htm
Reference 2 Courtesy of Sidener Engineering http://www.sidenereng.com/formulas.htm
Reference 3 MacDizzy© http://www.macdizzy.com/formulas.htm
Reference 4 Engineering fundamentals http://www.efunda.com/home.cfm
Reference 5 Pumps formulas http://www.airlinehyd.com/KnowledgeCenter/Hydraulic/Formulas/PumpH.asp
Engine Volumetric Efficiency Calculator
1. Dale, T.W.,"Spherical Piston Radial Action Engine", U.S. Patent #5,419,288, May 30, 1995.
2. Avallone, E.A. and Baumeister, T. III,"Marks' Standard Handbook for Mechanical Engineers", Ninth edition, McGraw-Hill, New York, 1987.
3. Richards, T.D.,"The Hanes Engine", informational report, copyright 1994, available from the author at P.O. Box 21147, Carson City, NV, 89721.
4. Ashley, S.,"A New Spin on the Rotary Engine", Mechanical Engineering, April 1995, p80-82.
5. Bloch, H.P.,"A Practical Guide to Compressor Technology", McGraw-Hill, New York, 1996.
6. Anon.,"GAUSS Volume I, System and Graphics Manual", Aptech Systems, Inc., Maple Valley, WA, July 18, 1994.
7. Heywood, J.B.,"Internal Combustion Engine Fundamentals", Mcgraw-Hill, New York, 1988.
8. Sliney, H.E. and Dellocorte, C.," The Friction and Wear of Ceramic/Ceramic and Ceramic/Metal Combinations in Sliding Contact", NASA TM-106348, DOE/NASA/50306-3, N94-15769, October 1993.
9. Wikipedia, ,"Scientific Method, Scientific Evidence, & Mechanical Advantage"
Quotes by Richard Feynman Nobel Prize Winner
Science is the belief in the ignorance of experts…
No government has the right to decide on the truth of scientific principles, nor to prescribe in any way the character of the questions investigated.
I learned very early the difference between knowing the name of something and knowing something.
What I cannot create, I do not understand.
"The more you know the more you know you don't know"
"If I understand it, I can create it"
Patent # 6,526,925
By an increase in mechanical advantage and decrease in volumetric mass with the area of force remaining the same the efficiency in any system is increased, there is a direct relationship within any given fluid dynamic or applied mechanics system.
In the above system, no new energy is being created, rather or more amply put this system better utilizes the energy available.
Todate I have 6 each, new prime mover devices, created and realized out of the above math.
[First law of thermodynamics]
The increase in the internal energy of a thermodynamic system is equal to the amount of heat energy added to the system minus the work done by the system on the surroundings.
[Second law of thermodynamics]
Heat cannot of itself pass from a colder to a hotter body. This is equivalent to this scientific statement: The entropy of an isolated system not at equilibrium will tend to increase over time, approaching a maximum value.
[Third law of thermodynamics]
As a system approaches absolute zero of temperature all processes cease and the entropy of the system approaches a minimum value.
[Zeroth law of thermodynamics]
If two thermodynamic systems are in thermal equilibrium with a third, they are also in thermal equilibrium with each other.
Will Green laws of dynamic systems
1"All dynamic systems are a balancing act."
2"Once a system is out of balance there is movement."
3"Movement will continue in a system until a balance is reached in that system"
4"An out of balance movement will pass from one system into another"
5"We can never realize the complete potential from dynamic energy"
Other High efficiency devices
Copyright by Will Green 2006