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Stan’s Legacy The Stanley Meyer Archive
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Energy balance of the loop

What goes in, what comes out, and the ratio — stated addition, nothing hidden?

The formula
r=PoutPd+Pp,Pd=I12·Rd·Dd100
r
Output over input
Pout
Pickup mean power, W
Pd
Drive copper loss, W
Pp
Pump electrical power, W
I1
Drive pulse current, A
Rd
Drive coil resistance, Ω
Dd
Drive duty, %
LaTeX 
r = \frac{Pout}{Pd + Pp}, \qquad Pd = I1^{2} \cdot Rd \cdot \frac{Dd}{100}

Work it out

µW

The loaded-output page's mean power into the load — mean, over pulses and gaps, not the peak.

A

The current actually in the drive coil during a pulse — the current-rise page's figure, not the supply's rating.

Ω

The drive winding's DC resistance, warm.

%

The share of the time the drive pulse is on.

W

What the pump draws from the wall. The estate's Little Giant B-500 is rated about 40 W.

W

Flow times pressure drop round the loop — what the medium actually takes from the pump. Reported as information; it is inside the pump's figure, not added to it.

or see a worked example with typical numbers

Method

  1. The drive coil's mean copper loss is the pulse current squared, times the coil's resistance, times the fraction of the time the pulse is on.
  2. The power in is that loss plus the pump's electrical draw. The hydraulic power is not added: it is part of what the pump draws, and is reported only as the pump's efficiency, hydraulic over electrical.
  3. The ratio is the pickup's mean power into its load over the power in. It is stated as a number; what it means is for the reader.

Assumptions

  • The drive's loss is its copper loss only. Switching losses in the driver, the supply's own inefficiency and the energy dumped from the coil at the end of each pulse (½LI², which a freewheel diode turns to heat) are all additional inputs the wall pays for and this page does not count — so the input is, if anything, understated.
  • The pump's electrical power is what the meter reads, not the motor's nameplate; the two differ, and the nameplate is the larger.
  • The pickup power is the mean into the load, taken from the loaded-output page. Peak power is larger by the inverse of the duty and is not a power the load can use.
  • Nothing here is Meyer's. The estate material gives the pump and the speeds and makes a claim of output; this page is addition over the other pages' figures, and the archive takes no position on the ratio beyond insisting that every term of it be visible.