[Amps] Power-off tube cooling

[Mike Hyder --N4NT--]

Here's how I've looked at it (remember I'm not an engineer) :

The big heat generators in a tube are the anode and the filament / heater.
Keep in mind that the cooling air flow is not designed to cool these
generators, but to cool the enclosure and particularly the connecting rods
(pins) which pass through the enclosure.  For my thinking, I will consider
the anode of a 3-500 tube because that is the largest tube I've ever used.

>From the anode, heat travels by conduction through the rod which supports
it.  The temperature along that rod is a gradient from the temperature of
the anode proper down to the temperature we achieve with our cooling air
flow past the anode cooling fins and glass seal.  If we suddenly stop our
cooling, then that temperature gradient will change because heat at the
external end of the rod is no longer being removed by the air flow.  Thus
while the glass envelope is cooling, the rod is becoming hotter where it
passes through the glass.  This is because the thermal energy in the anode
is still conducting through that rod.

Sometimes my thinking in cases like this is helped by what is called a
'reductio ad absurdum.'  Imagine a solid sphere about the size of a golf
ball heated to 2000º supported by a wire two inches long connected to the
perfect heat sink that brings the temperature down to 0º.  The temperature
gradient along this wire will be from 2000º at one end to 0º at the other.
Now we remove the ability of our heat sink to sink heat.  The ball is still
2000º so the gradient along the supporting rod now changes as the supported
end's temperature rises until the temperature in the rod reaches
equilibrium: 2000º at both ends.

The same logic holds for the filaments in any tube, I would think.

73, Mike N4NT

----- Original Message ----- 
From: <RFlabnotes at aol.com>
To: <amps at contesting.com>
Sent: Thursday, January 15, 2004 10:55 AM
Subject: [Amps] Power-off tube cooling


> I am curious about the origin of this mythology. It would be nice to see
some
>  factual data, or, at least, logic to support it.
>
> My simple-minded physical reasoning says that when the power is all
removed,
> there can be no additional heat-rise. So, at least the filament for a
short
> time is no better or worse off than during steady-state standby. You might
then
> be tempted to argue that thermal stresses induced by the cooling down
period
> will go to work, possibly causing dangerous differential stresses. If
true, the
> forced air will actually accelerate such processes. Letting everything
slowly
> equilibriate back to ambient would clearly be less stressful of everything
in
> the thermal 'circuit'. What good is accomp[lished by forcing it quickly
back
> down to ambient (of course, in the process, it will then spend a few
minutes
> less time at its operating temperature - is that what this is all about).
>
> Is there really some good physics to rely on here, or is this just another
> piece of folklore that won't go away?
>
> Eric von Valtier K8LV