Owen Duffy wrote:
(Dieter Kiel) wrote in news:1ihnz3n.wxm0z7nvxaivN%
:

http://www.w8ji.com/vswr_reactive_power.htm



I assume Dieter that this is your recommendation of the article.

That article uses the term 'reactive power' in a non-conventional way,
though the term is a well known one (ie has a conventional meaning).

Conventional use is that the term 'apparent power' is applied to the
product of RMS voltage and current flowing into a two terminal load, and
the units are VoltAmps (VA), not Watts as used in the article.

Reactive power is the reactive component of apparent power, and expressed
in units of 'VoltAmpsReactive' (VAR).

'Real power' is the real component of apparent power and expressed in
units of Watts (W).

The relationship is that
apparentpower = (realpower^2 + reactive power^2)^0.5 .

This is all basic lumped component AC circuit theory, and holds at RF.

True, but we still aren't there.

It's very misleading to quote "VAR powers" in the kilowatt range,
because the only power available to melt the feedline is 100W from the
transmitter. There is no magnification of real power.

The high value of "VAR power" is a theoretical result of the large RF
currents in the system. These result from an antenna feedpoint impedance
that has a very low resistive part and is almost entirely reactive. The
large RF currents are a genuine physical phenomenon, as also are the
high voltages a quarter-wavelength back along the feedline (if the
feedline is long enough, of course)... but if there were no losses in
the feedline, these would have no further effect. In spite of the wild
values of impedance, current, voltage, VSWR etc, if there were no losses
in the feedline then all of the RF power would still reach the antenna.

In a real feedline, the effect of the high currents is to divert almost
all of the available RF power away from the antenna and into the
feedline's own resistive losses - skin-effect losses in the copper
conductor, and dielectric losses in the plastic. Both of these result in
heating and softening of the plastic, which makes the dielectric loss
even higher. This tends to divert even more of the available power into
the weak spots, where the plastic finally melts.

But there is still only 100W available to do the damage.

No argument about the final conclusion - it ain't gonna work - but I
don't care for the explanation. There's no problem with "VAR power" for
anyone who already has a firm grip on the concepts, but it is not a good
way to explain those concepts to a newcomer.


--

73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek