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Transcending Reactive vs Real Power Concepts.

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  • Harvey Norris
    As most of us know the REASON that we cannot simply multiply the voltage times the amperage, (VI); to obtain the power expressed in an inductor placed across
    Message 1 of 1 , May 29, 2005
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      As most of us know the REASON that we cannot simply
      multiply the voltage times the amperage, (VI); to
      obtain the power expressed in an inductor placed
      across AC, is that there is a time lag between the
      cause and effect; where voltage being the cause, and
      its amperage being the effect. For the AC inductor,
      because of this time lag, that for the external AC
      source we can speculate that if the current were truly
      90 degrees out of phase with the voltage, (this often
      stated premise is not completely true, it depends
      whether the inductor has a large enough inductive
      reactance X(L)= 2pi*F*L to form a phase angle near 90
      degrees); but if the amperage were near 90 degrees out
      of phase with the voltage, by the time the AC voltage
      input has made a complete pulse in one polarity, and
      thus gone back to its zero crossing point; at that
      moment in time the amperage in the coil will have
      reached its maximum, again using the 90 degree phase
      angle example. Up until this starting point 180
      degrees into the AC cycle, a magnetic field from the
      inductor is established that agrees with its polarity
      input. But in the third quadrant of the AC cycle, the
      voltage input reverses polarity, but the current has
      not yet followed its example, because of the time lag
      the current is going in the opposite direction then
      what its voltage source would cause it to act. In the
      third quadrant then the magnetic field of that
      inductor is then causing a self induction by virtue of
      its magnetic field collapse that establishes a
      polarity OPPOSITE to the polarity of the source. This
      is why back emf, or a backwards working voltage source
      established by mistimed magnetic fields exists. The
      net result of this backward working voltage during
      portions of the AC cycle is that the forward working
      voltage does not conduct it's OHM's law amount of
      amperage, because of this extra effect of AC
      resistance. We call this AC resistance inductive
      reactance. The effects of both the actual resistance
      and its inductive reactance is reffered to as
      impedance, noted as Z where the effects of both
      principles are obtained by summing the squares of both
      quantities, and taking the square root of this result
      shown as the impedance equation Z= sq rt[R^2+
      X(L)^2].

      Now some commentators note that because the coils
      magnetic field establishes its own self induction, we
      are actually dealing with the summation of two power
      sources. They explain that because of the storage of
      energy manifested as a magnetic field, in certain
      portions of the AC cycle the coils collapsing magnetic
      field is establishing a polarity opposite to its
      source, and in terms of the AC source being a
      generator, that generator as its VI apparent power
      input also experiences borrowed and returned energy.
      When the coils magnetic field collapses a portion of
      the power input is returned to the generator. Now the
      real amount of energy expeneded on the coil is just
      its conversion to heat by I^2R heat losses, and the
      rest is just borrowed and returned energy. The
      essential question then becomes does this source also
      act like a banker and charge interest on its borrowed
      and returned assets? This would seem the sensible
      thing to do, but perhaps nature doesnt always model
      mans actions.

      So we have two aspects, and this is where the third
      aspect comes into being, where the concept of phase
      angles comes into play. We have the apparent power
      input which is just the voltage times the amperage
      across the load,or (VI) of the generator as the
      apparent power input. Then we have the real power
      input which is undeniable because we can measure the
      heat release as the I^2R quantity. Because we have a
      load containing reactance, or a reactive load, the
      difference between the real and apparent power inputs
      is merely the % of borrowed and returned energies vs
      real energy output. But we also have a method to
      compute what this heat release should be by methods of
      trigonometry. We can draw out the AC voltage and
      amperage AC cycles, and find out the % portions where
      they are in agreement, where this portion will be
      expressed as a cosine angle. The real power input
      therefore becomes VI* cos( acting phase angle of the
      inductor), and we conclude that this also equals I^2R
      as the heat loss. Recall that in trigonometry we have
      a y coordinant, [expressed as X(L)in phase angles]:
      where y/radius is the sine quantity of the angle, and
      conversely x/radius is the cosine. (In trigonmetry we
      allow the radius of the circle to be the unit 1 to
      simplify these calculations.) To find the phase angle
      of an inductor we let the inductive reactance to be
      the positive y quantity, and the actual resistance R
      to be the x quantity. But a little more trig is need
      to find the actual phase angle which can be calculated
      with any calculator having trig functions. The tangent
      of the angle is expressed as y/x. So if we have a coil
      where X(L) and R is both known, the tangent is easily
      known; it is just X(L)/R, which is also the
      theoretical Q factor of the coil. But what we NEED to
      find out is the inverse of this, which will tell us
      the radians, which can further be reduced to an angle
      by multiplying by 360 degrees/ 2pi radians, which is
      the conversion factor to degrees. Suppose I have an
      inductive reactance 5 times the resistance; what is
      the phase angle? We know the tangent is merely 5, but
      this tells us nothing. Instead what we NEED to find
      out is the answer to the question; the tangent of what
      amount of radians will give an answer of 5. Thus we
      use the inverse function, called the co-tangent, or
      tan^-1 to determine the answer. And tan^-1 (5) = an
      answer in radians that can be converted to degrees,
      and this is the method for finding the phase angle.
      Now if X(L)is very large vs the R quantity of
      resistance, X((L) forms the y axis, R forms the x
      axis, and the hypotenus, (ordinarily the radius
      quantity in trig functions) becomes the impedance Z.
      We already know by the Pythagorean theorem that the
      sum of the squares of the right angle triangle sides
      is equal to the square of the hypotenus, and by then
      taking the square root of both sides of the equation
      we reach the same equation for the impedance Z. And if
      X(L)>> R then X(L)~= Z. The length of the radius and
      its y reflection are almost equal, but the x
      reflection as the additive vector is very small
      compared to the other two quatites by definition since
      we have initially made the condition X(L) is far
      greater then R. The x reflection is also the cosine of
      the phase angle, and it is this no. as a percentage
      that is used for the real power calculations. Thus
      with this condition the apparent power will be far
      greater then the real power. These things are all
      rudimentary for the electrical engineer.

      Now we come to electrical resonance, where X(C) forms
      the negative Y axis, the opposite reactances in series
      can cancel, leaving R as the predominant factor for
      conduction, and in this condition then we suspect that
      now the apparent power will be the same thing as the
      real power, and the amperage has now come closely in
      phase with the impressed voltage. However because of a
      factor known as "internal capacity" we may not always
      arrive at a conduction that equals ohms law
      requirements. The depends on the construction of the
      inductor we are resonating. However a VERY
      misunderstood thing can happen here. We might assume
      that since now the apparent power equals the real
      power, which also occurs in DC circuits after the
      magnetic field is established, that % wise the
      generator is no longer having a significant portion of
      "borrowed and returned" energy allocations. We have
      cancelled the reactive state of the circuit, so have
      we not also cancelled the "borrowed and returned"
      field energies? The generator would seem to indicate
      so, because the apparent power equals the real power,
      and we are not seeing differences between those
      definitions. However in actuality the opposite thing
      has occured, the borrowed and returned field energies
      will have gone up q times, even though the generator
      does not seem to indicate this % wise in its
      allocations. A simple proof of this is the definition
      of stored energy for the L and C components. The
      energy stored in C is defined as [CV^2]/2, and for L
      it is [LI^2]/2. In series resonance each L and C
      component developes an internal voltage rise against
      each other, in opposite directions with respect to
      each other, so that on the outside of the circuit, the
      generator only sees this net cancellation. For the C
      case, since the V term is squared, and we are
      obtaining series resonant internal voltage rise, the
      stored energy has gone up Q times, where the acting Q
      factor is the ratio of the inside voltage rise to the
      outside voltage source. We then have increased the
      borrowed and returned energies, but now a different
      relationship exists. Formerly the energy was borrowed
      and returned to the generator, but now the borrowed
      and returned energies only exist between the L and C
      values... It is literally as if the expanded electric
      and magnetic field energies were being obtained in
      resonance for free... But can we use this fact to
      power another load? That isnt so easy a proposition.
      If we divert the expanded voltages to another circuit,
      this in turn destroys the properties of the resonance
      itself, so that the actual resonant rise of voltage is
      wiped out.

      A little over a year ago I set up some circuits where
      the resonances were obtained from a 3 phase alternator
      functioning @ 480 hz. These resonances were special in
      that they were designed to be maximum energy transfer
      circuits, generally defined as a circuit that drops
      the open circuit voltage of the source in half, but on
      the only side of the coin, the maximum amount of power
      to load is obtained from that generator. Between these
      3 AC phases, set up as Delta Series Resonances,
      (DSR's), the voltage rise q factor of 5 was noted.
      What this means is that 5 times more amperage is
      procurred then what exists in the reactive state, and
      also the internal rise of voltage was 5 times that of
      the source. Now between these three 120 degrees of
      voltage rises, the AC was turned into DC by means of
      full wave rectifications, which because of the poly
      phased inputs, thus fills up the DC ripple without use
      of a filter capacity. The load for these DC currents
      was a 3/8ths inch width of a ferrite block, normally
      considered an insulator. Now ferrite is actually
      considered a class of semiconductor, in that it looses
      resistance with heat. The ferrite in this case
      gradually grew hot, its resistance decreased
      phenomenally going from 30,000 ohms to around 7 ohms,
      until it began to glow on a corner, the temp around
      900 degree fahrenheit. Meanwhile the outside coils,
      where this internal current was obtained from on the
      rectifications also experienced a drop of current from
      what existed in its resonant state. The addition of
      this low ohmic load between the resonances had caused
      its amperage input to drop, in fact it drops low
      enough so that only half of the current that the coils
      would consume in its reactive state was noted. We
      might say that X(L) of the coils appeared to double.
      What this also means is that once again we should be
      dealing with apparent vs real power input ratios. Now
      the inside load, being DC has no power factor
      arguments associated with it, but the outside coils
      do, because they are AC reactances, placed into
      resonance, and then reduced from resonance by addition
      of the interphasal load. This load has become a 2nd
      generation maximum power transfer principle, where the
      internal impedance or reactance of the outside coil
      systems have been matched by an equal resistive load.
      In terms of this Z(int) of the source has been given
      an equal R(load) in ohmic values, the requirement for
      maximum energy transfer. The outside spiral coils
      themselves, constructed as maxium energy tranfer
      resonances, (METR components), use the same principle
      in the R(int) of the alternator stator coils equals
      R(load) of the coils resistance R value.

      Now in the cited circumstance, the stator lines each
      serving two phases contains ~ 2.0 -2.4 Amps, and the
      division into phase currents should reduce this by
      about 1.7 times the stator line currents. The outer
      phase currents derived from these stator line currents
      are somewhat inbalanced, and the pic I took of the
      power input shows 1.4, .85 and 1 amp currents on this
      outer triangle, a sum of 3.25 Amps. On the DC current
      procurred from the outside misphased AC currents I
      arrive at a sum of 3 DC amps, obtained from this
      outside AC current oscillation. Both the alternator
      input voltage and the ferrite DC voltage are about
      equalized at 17 volts. As such no resonant rise of
      voltage is taking place, the voltage input is about
      the same as the voltage output to the load. Now if we
      average the phases currents to ~ 1.1 amps, with 17
      volts across them, the apparent power input is 17* 1.1
      = 18.7 watts per phase, or the generator is inputing
      an apparent power of 56.1 watts. But the REAL power of
      the inside triangle load being shown by the 900 degree
      heat of the ferrite is also 17VDC*3A = 51 watts. The
      coils themselves that lay on the delivery lines to the
      ferrite have a combined 3 watts heat loss, close to
      what the apparent power shows itself as, where the
      combined loads of 51 + 3 watts almost equals the
      apparent power input of 56 watts.

      Here is where the kicker comes in; the apparent power
      in is yeilding almost the same amount as real power
      out!. We surmise that since no resonant rise of
      voltage is occuring on the outer triangle coils, that
      we should treat this as a apparent power input, and
      not a real one. Now we have about a 84 degree phase
      angle, made as tan^-!(10), from a situation that
      started out with a phase angle dictated by the
      tan^-1(5), from the coil systems having a q of 5.
      After the load was added the coils appeared to have
      about double the impedance on their conductions, which
      is where the 10 figure in tan^-1 estimations comes in.
      If this were true, the mechanical energy required to
      turn the alternator should only reflect a real power
      load of only 10 watts, by the conventional phase angle
      analysis, but yet we have a bonafide real power load
      of 5 times this amount!

      Have we indeed turned an apparent power input into a
      real one? Perhaps the apparent vs real power arguments
      do not apply for this special case. What I would like
      to be able to do is power the alternator with a geared
      up bicycle gearing rig, and drive the alternator by
      bicycled leg power. For only a 10 watt power
      requirement to make 900 degrees, this seems possible
      with a human energy input. Then I could stick in real
      resistive power loads, and compare the respective
      drags for both situations. In any case the above
      arguments either show that the current electrical
      phase angle theory is inadequate to explain the above
      situation, or conversely we have used the "free" field
      oscillations inherent in resonance to create a
      situation of overunity. The calculations also show
      that ~ 6 times the energy transfer occurs in the LC
      field energy transfers then do the actual real power
      being inputed as I^2R heat losses on the DSR's
      themselves, which are just under 1 ohms resistance.

      Sincerely
      Harvey D Norris






      Tesla Research Group; Pioneering the Applications of Interphasal Resonances http://groups.yahoo.com/group/teslafy/
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