On Mon, Aug 16, 2010 at 08:58:06PM +0000, Paul Decker wrote:
>
> "...the anode dissipation happens primarily by radiation,
> not conduction..."
>
> Doesn't radiation heat the air molecules around the tube, which in turn heats
> the glass and metal parts of the tube?
Depends on how much IR the glass transmits. If it's perfectly transparent
to IR, then the air gets hot from the IR, but the envelope doesn't. If it's
perfectly opaque to IR, then the envelope gets hot from the IR, but the air
doesn't. IR transmission factors between 0 and 1 give a linear combination
of the two.
The above is much simplified.
If the envelope is IR-transparent, then
o the air gets hot from the IR (though not very), and heats the
envelope (though not much): air is pretty transparent to IR.
o the heat conducted from the elements to the pins heats the envelope;
o other things I haven't thought of may heat the envelope.
If the envelope is IR-opaque, then
o the envelope gets hot from the IR and heats the air;
o the heat conducted from the elements to the pins heats the envelope;
o other things I haven't thought of may heat the envelope.
--
Mike Andrews, W5EGO
mikea@mikea.ath.cx
Tired old sysadmin
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