1200% efficient hydraulic motor?

Question

I am seeking, someone to verify the math on my hydraulic motor?

In this fluid system, this motor is 14” diameter, 8 cylinders, has 2” dia. pistons, with a 7” dia. crank shaft 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.

Constants
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
1Horse Power =746 Watts
Efficiency = eff.

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
(Constant 231 Cu. In. Per Gallon)

Hydraulic Horsepower = Pressure (psi) × gpm / 1714 HP = (psi × gpm) / 1714
(Constant 1714)
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

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
Torque = T = Force applied x lever arm in “ft ” or Ta*psi = F lbs.”
F = Ta*P = 9.42477*450 = 4,241 lbs Force.
T = ft-lbs = (crankshaft radius / 12)*4239 = (3.5/12) =. 2916*4,239 = 1,236.7 ft-lbs.
With the above given; the following is 2 different HP formulas for work out:
1.) HP=T*N/5252 = 211.98 HP or = 158,002 Watts
2.) HP=(C*N / 12/60) * F / 550 =211.97 HP or = 158,040 Watts

We now compute the input pump fluid in gallons per minute for
the eight cylinders.

Displacement = A*d*8 = 12.5663701 cu. in. = total displacement per revolution for all eight cylinders.

Gallons = d*N/231 = 48.959888551 gpm = input pump fluid flow.

Then G = 48.96 is used for GPM on the second page.



With 48.96 GPM and a pressure of 450 psi we can compute pump HP work in.

The following are 2 Hydro HP formulas representing work in.
1.) PUMP HP=GPM*P / 1714 * eff. of 80% = 48.96 * 450 /1,714 =12.84HP * 1.2 = 15.43 HP or = 11,507.03 W
2.) PUMP HP=GPM*P*.0007 = 48.96 * 450 * .0007 = 15.42 HP or = 11,503.32W

Efficiency work out / work in * 100 = 211.98HP / 15.43 HP * 100 = 1,373.82%
The efficiency of a machine, which is defined as the ratio of the work output to the work input, is always less than one, since 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

I put everything in an excel sheet and checked the formulad over and over...

The pump is turing the motor not the other way around...

A hydraulic fluid pump with an electric motor drives the above system.

If you dont understand the math please dont assume, ask me another question about the math...

Math references
These formulas are courtesy of Gulf Coast Air & Hydraulics Lion Hydraulics
Reference 1 http://www.gulfcoastairandhyd.com/formulas.htm
Reference 2 Courtesy of Sidener Engineering http://www.sidenereng.com/formulas.htm
Reference 3 MacDizzy© 1993-1999 http://www.macdizzy.com/formulas.htm
Reference 4 Efficiency formulas http://www.kent.k12.wa.us/staff/trobinso/physicspages/PhysicsOf/Wade/physics.htm http://www.reliance.com/prodserv/motgen/b7087_5/b7087_5_3.htm http://www.zoeller.com/zep/techbrief/JF1article.htm
Reference 5 Pump HP formulas see http://www.hfsindustrial.com/formulas/formulas.htm
Reference 6 HP and volumetric efficiency http://www.prestage.com/carmath/formulas.asp
http://www.octinc.com/dwnlds/Wtrmthfo.pdf
All math and formulas may be referenced at this site http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Published REFERENCES
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.

Did you come up with about 1,373.82% efficent???

You are assuming tha I dont understand the geometry of my own system...
I will try to find a way to download a drawing...
this will better help you see the system

There are no assumptions... This is the way that the motor is designed... It is designd to allow fluid to come in contact with each cylinder over 180^ of travle... that is how the volume of fluid running through the motor is calculated...Actually there are 4 cylinder in use but that makes the HP go up...

EVERYONE please keep in mind this is a rotory device, Not reciprocating...

Answer 1

A representative from Exxon-Mobil Oil will be contacing you later today.
Please make yourself and your plans available for showing to our engineers.
We have your name, number and address.
Thanks,
Management team

Answer 2

There's a logical inconsistency in the value of the displacement of the piston rods (.5 inch) and the radius of the crankshaft (3.5 inch). You're assuming that the full force F operates through the 22 inch circumference even though the total motion of all 8 pistons can't be more than 8*.5 = 4 inches at full force.

To be more precise, using the assumption that 3 pistons are acting at once, the maximum linear motion per cycle at full force would be (8/3)*.5 = 1.33 inch.
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The factor by which you're overestimating output power is:

22/1.333 = 16.5

If you divide this by the efficiency factor of the input motor (1.2):

16.5/1.2 = 13.75 = 1375 %

which is where your erroneous efficiency percent comes from.

So check to make sure of the radius at which the piston rods are attached - maybe the 7 inch diameter refers to some other component or wheel on the crankshaft.

With respect to Mesper's answer: The assumption that the motor runs at 900 RPM is not an issue that I can see. Both the power of the pump motor and the load on the hydraulic motor are unspecified and are presumably selected to make the system work as specified (except for the problem I just pointed out).

Answer 3

The answer to why you are getting an efficiency of well beyond 100% is actually easy. Your making a classical mistake. The problem is to do with the fact that you have worked out what the output of the pump would be if driven at 900rpm. You have then worked out how much power would be needed to drive the hydraulic motor at 900rpm and then assumed that you can divide these to get the efficiency. Unfortunately, the assumption that the motor actually could turn the pump at 900rpm is the problem.

Imagine the pump is the gearbox end of a car. You can calculate that to get the car to go at 150mph you need the engine to be turning at 5000rpm. If you now put a 25cc lawnmower engine in the car turning at 5000rpm will the car travel at 150mph? The obvious answer is no.

So where did you go wrong? The answer is that you assumed the hydraulic motor had enough power to actually turn the pump at 900rpm. In practice it doesn't. You would need a motor that could turn at 900rpm but also could generate enough power at these revs to keep the pump turning. Your pump motor can't because it doesn't have enough power and would simply stall or run at a slower speed to the point where the pump output matched the motor power input (minus losses).

If this wasn't the case. you've cured the enercy crisis- you could run cars on really small motors and use your system to run itself by feeding the pump back to the hydraulic motor with power to spare.

Answer 4

Hehehe. This kind of bullshit will make you crazy (believe me, I know ☺) We both know that there's some kind of 'bug' somewhere in your calculations, either in the excell equations or in the way your data is being entered (double check that all the units are correct). You just have to keep going back over it (without getting depressed or pissed off ☺) until you find it.

That, or your eyes will pop open at 0330 and the first words out of your mouth will be. "Motherf*cker!!!! That's what it is!!!"

It's things such as this that develop character in young engineers ☺


Doug

Answer 5

Re-do the math. I think you've got the point in the wrong place or something. 100% efficient would be absolutely amazing (if not impossible)considering the friction and hose routing and fluid concerns.
For your next question, include the math.

Answer 6

Err.. i have calculated using a conventional calculator and the efficiency is not 1200%. I have counterchecked your calculation, and seems to me the formula is ok. Perhaps, you may consider to use conventional calculator to perform the algorithm. Please take extra care on the Dimension eg, all change to Minute change to Second, Meter change to Feet, and etc. For convenient purposes, you may standardize all your unit to S.I. unit, eg. Meter, Second, Kg, Newton, and so on.

For me, i have change all to S.I. before performing calculation . This get my job easier

Answer 7

Funny how that seems too good to be true, ain't it, now?

Answer 8

Ok, Where is it.