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Old July 21st 05, 04:24 AM
Richard Clark
 
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On Wed, 20 Jul 2005 21:13:42 -0500, (Richard
Harrison) wrote:

Richard Clark wrote:
"The very first formula from the Standard Handbook for Electrical
Engineers, Section 11. Power Transmission, Electrical Calculsations:
I = P/E (1) "

Unless a d-c transmission is specified, that`s wrong.


Hi Richard,

You may disagree with the contents from this tome, but the citation is
accurate.

P=EI cos theta, where theta is the angle between E & I. For 0-deg. coa
theta=1.


The section preceding distribution is Section 10, Power Plants, which
devotes Para. 722 Inherent regulation; Para. 724 The Load Factor;
Para. 725 Load Fluctuation; Para. 726 The Power-Factor.
Sections10-795 through 836 is given over to AC switching.

And of course casting back towards the beginning of the Handbook we
find ourselves at Section 7, Alternating Current Generators which
covers Power-Factor, Slip, Stalling torque.... The treatment later in
the section covers representative efficiencies in Polyphase equipment
up to 1000HP well into the 90's of percent. All may note that this is
an accuracy that is still 10 times better than the "Can you solve
this" math offered here.

Returning to Section 11, Power Transmission; coverage there includes
Power-Factor Correction. Para. 95, in part:
The per cent, quadrature current required for unity power-factor
[which renders the formula responded to as accurate] at the
receiver may be determined from Fig. 14, or may be calculated
for a three-phase system as follows.... [which I won't go into
but covers power-factors from 60% to 95%].

There is still no requirement to fill the need for a vector of
direction.

This relates, in part, to my service in the Power Industry designing
power grid control systems and testers (SCADA).

73's
Richard Clark, KB7QHC
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Old July 21st 05, 04:41 AM
Richard Harrison
 
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Ian White, GM3SEK wrote:
"That statement bears no physical relationship to how this instrument
actually works---."

We`ve been through detailled explanations of how a Bird works. Cecil did
not need to do another. The wattmeter takes actual samples of the
voltages and currents at any single point on the coax. These are
representative of the powers which are moving toward the load and away
from the load. Careful calibration allows indicarions in watts.

An electric current through a speedometer is calibrated to indicate
miles per hour. It works. So does the Bird Wattmeter.

Best regards, Richard Harrison, KB5WZI

  #125   Report Post  
Old July 21st 05, 05:43 AM
Richard Harrison
 
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Richard Clark wrote:
"Where does he (Terman) equate phase with direction?"

From page 90 of Terman`s 1955 edition:
Bottom of the page; "Transmission Line with Short-circuited Load.
Where the load end of the line is short-circuited, that is ZL=0,
reference to Eq.(4-14) shows the reflection coeficient has a value of
-1.0 on an angle of 0-deg. = +1.0 on an angle of 180-deg. As in the
open-circuited case, the reflected wave has an amplitude equal to the
amplitude of the incident wave. However, the reflection takes place with
reversal in phase of voltage and without change in phase of the
current."

I`m a lousy typist but tried to make an exact copy of part of the page.
I assume you agree the incident and reflected waves travel in opposite
directions in Terman`s example.

Best regards, Richard Harrison, KB5WZI



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Old July 21st 05, 07:24 AM
Richard Harrison
 
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Richard Clark wrote:
"Again, what is the vector of direction for the light bulb?"

Electromagnetic waves include light and heat whicjh have extremely short
wavelengths. The light bulb may not be a perfect point source but the
waves travel away from the source with the velocity of light and consist
of electric and magnetic fields that are at right angles to each other
and also at right angles to the direction of travel. Wave energy is
divided 50-50 between the electric and magnetic fields.

Many frequencies (colors) make up the radiation from a light bulb. Much
more heat is radiated than visible light.

In a radio wave the essential properties are frequency, intensity,
direction of travel, and plane of polarization, For the constituents of
light bulb radiation, it is the same.

300 million m/sec is the velocity and this equals the product of
frequency X wavelength. Emissions of a light bulb are of extremely high
frequency but of extremely short wavelenggth too.

All points on a wavefront are equidistant from the source and emerged
simultaneouslly so they share the same phase.. From a point source light
bulb we would be in the far field.

The field is transverse. The power flow (J.D. Kraus` words), or Poynting
vector, is entirely radial.

Best regards, Richard Harrison, KB5WZI

  #128   Report Post  
Old July 21st 05, 07:39 AM
Richard Harrison
 
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Richard Clark wrote:
"To obtain a complete cancellation it requires identical powers with
identical but opposing phases."

Yes. That is why closely spaced balanced transmission lines have no
significant radiation.

Best regards, Richard Harrison, KB5WZI

  #129   Report Post  
Old July 21st 05, 08:02 AM
Ian White G/GM3SEK
 
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Cecil Moore wrote:
Ian White G/GM3SEK wrote:

Cecil Moore wrote:
The Bird is indirectly measuring [(E^for) x (H^for)] as forward
power and [(E^ref) x (H^ref)] as reflected power.

That statement bears no physical relationship to how the instrument
actually works (and "indirectly" won't get you off the hook either).


I was hoping someone would assert such. E^for is proportional to
Vfor which is what the Bird samples. H^for is proportional to
Ifor which is what the Bird samples. Within a 50 ohm environment
that yields forward power. Same for reflected power.


The Bird does not generate a vector cross product. There is nothing
inside the instrument that's capable of doing such a thing.

The hardware displays readings of detected RF voltages - not power. The
forward/reflected power calibration on the meter scale is an external
calculation, based on transmission line theory.

You know exactly how instruments like the Bird work, because at various
times you have posted accurate descriptions here. Your enthusiasm for
your pet theory is making you distort the truth.


--
73 from Ian G/GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek
  #130   Report Post  
Old July 21st 05, 08:57 AM
Richard Clark
 
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On Thu, 21 Jul 2005 01:24:36 -0500, (Richard
Harrison) wrote:

Many frequencies (colors) make up the radiation from a light bulb. Much
more heat is radiated than visible light.


Hi Richard,

Actually that is quite wrong. IR is not heat, it is radiation. Heat,
actually phonons, constitutes something less that 10% of the
conversion of electrical power in a light bulb. Lest we take off on
the tangent of IR bulbs being used for heating, it is the load of that
IR radiation (directed upon a dissipative surface) that renders
phonons, otherwise IR is radiated in exactly the same manner as any
radiation. There are any number of simple, practical tests to confirm
this. For one, IR passes through most glass without heating it. You
have to go out of your way to obtain IR blocking glass (which doesn't
even absorb that much either). There are some IR wavelengths that go
right through water, and others that are entirely absorbed.

However, this is not about heat, nor IR, nor even the loss of a
principle vector property, its angle notation, or even the whole
absence of the vector property from the solution to wave interference
powers. Rather, it is about the facade of complete cancellation

Entirely ignoring all these other trivial details, that cancellation
is incomplete in and of necessity for real or imagined initial
conditions. This is revealed in any mathematical solution, and
certainly by examination.

73's
Richard Clark, KB7QHC
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