Math verification on 1200% efficient motor?

Question

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%

I am seeking help... not superfluous babble

The work is only don on 3 ea pistons over 180 deg.
as they rotate new pistons are activated ant the others are deactivated...

This is a rotory device therefore there is always 3 active pistons... As one becomes deactivated another is activated...

Answer 1

The calculation of the input power looks correct, but in the output power calculation 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.

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.

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Edit:

I don't have a perfectly clear picture but it seems reasonable to assume that at any point in time each piston is either engaged (pushing) on the (7 inch) wheel or not. The maximum distance a point on the circumference of the wheel can move during the period a piston is engaged is .5 inch, but since the circumference distance is 22 inches per revolution, it's a mathematical certainty that there are time intervals during that cycle when no pistons are doing work on it.

During those periods the instantaneous torque is zero. The nominal torque is the time average over one revolution which is going to be substantially less than the maximum instantaneous torque, because it includes that dead time when the shaft is turning but no pistons are pushing on it. In your calculation you're assuming the torque is the maximum all through the revolution.

The simplest way to estimate the *average* torque is just divide the energy output in foot pounds per cycle from the pistons and divide by 2pi:

T = 8 * Force * (1/24 ft)/(2pi)

This assumes (and guarantees) no energy loss from the pistons to the shaft which is probably not the case. There could be substantial energy lost to heat (especially if there are metal parts banging into other parts moving at slightly different speeds), but I'm not sure how it could be calculated theoretically.

Answer 2

I've read through it a couple times and the number for your torque doesn't seem right. I'm no expert on these types of engines though to really know what a typical number looks like. However, you cannot have over 100% efficiency, you'd be breaking the laws of physics at that point. For a gas burning engine you should be looking for a number between 5 and 10%. Rocket engines run in the 70% regions, but they are using simple fuels (hydrogen oxygen). They are about as good as it gets. If you flip your work in and work out the number actually seems right. Make sure you have the distinction right from your text.

Answer 3

I'm not a mech eng., but I am always uncomfortable with units other than SI for computation of engineering quantities., especially because you use "Watts" which is an SI with FPS system. Maybe you have your conversion factors right, but I'd suggest converting everything to a consistent set of units and solving again.

Answer 4

no such thing as 1200% efficient motor. what are you building, a warp engine? maybe they can be 900% efficient, but not 1200%