[sdiy] help needed: thermal equations to RC networks analogy

[Czech Martin]

Thanx to all for your input.
I think I've understood my error:
I should model the heat source as electrical current source,
with parallel heat capacitance (electrical capacitance)
and heat resistance to the air volume (electrical resistor).
Thus, if the electrical current source is off, it will not
short anything, so wrong heat backward transfer is not
possible. The heat output of the heat source should be proportional
to U^2/R or I^2*R (electrical power).

All the other issues (radiation, nonlinearity) are certainly true.
Because there is two fans in the tank and isloation arround,
this could be diminished. Experiment will show.

Anyway, I have finished this aviary (birdcage) which
kept me busy for 3.5 weekends now. It has about 1m^3 volume,
a lot of work...
Next I can measure static equilibrium and step response
of the tank.

Not that I *really* need that, but I think it is a nice exercise.
I built the first heat servo system in great haste
and I have made many compromises just to get something working.

That is one reason why I never published the circuit or data.
Because people would think from that that I must be absolutely clueless
which is only true to some extend ;->

m.c.




-----Original Message-----
From: Magnus Danielson [mailto:cfmd at swipnet.se]
Sent: Dienstag, 11. Februar 2003 20:36
To: Czech Martin
Cc: chris at scp.de; synth-diy at dropmix.xs4all.nl
Subject: Re: [sdiy] help needed: thermal equations to RC networks
analogy


From: "Czech Martin" <Martin.Czech at Micronas.com>
Subject: RE: [sdiy] help needed: thermal equations to RC networks analogy
Date: Tue, 11 Feb 2003 16:30:18 +0100

> Perhaps I should be more precise:
> 
> I want to model a heat source (resistor) inside 
> a box of air. The box has some losses to the outside.
> My assumption was that the heat capacitance of
> the resistor is not important compared to 
> the delay of heat transfer through the air.
> The point you made is still valid, though.
> 
> I modelled the resistor as voltage source, the voltage
> beeing proportional to the square of the resistor voltage.
> Then I connected a "heat pipe", of resistors and capacitors
> to model heat transfer. At the end I included a loss
> resistor to the outside. Now i have included an ideal
> diode to prevent backwards heat transfer into the
> resistor when it is switched off.

Why? It's not there in real life!

If your resistor does not receives any current to burn up, it will first cool
of, but may also heat up due to other heat sources.

The generate effect could be modeled as a heat-current, right? Loading of the
resistor's body mass is that of a capacitor shunted with a resistor. A few RC
links down you touch the surface and then into the free air.

This is just from the top of my head, but I could make a serious attempt for
an analogy if you really need it. I'm sure I have a few heat-models around
here somewhere.

Cheers,
Magnus - don't know if I should bother answering emails like these... seems so
useless and thankless at times...