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

[Czech Martin]

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. Like this:

.subckt tank out in
x1 p1 0 in 0 quadrat <-taking the square of input real voltage
x2 p1 p ideal_diode  <- ideal diode preveting heat backward flow
r1 p 1 1k            <- diffusion network
c1 1 0 10m
r2 1 2 1k
c2 2 0 10m
r3 2 3 1k
c3 3 0 10m
rst 3 st 3k          <- loss network
vst st 0 7           <- outside temperature
e1 out 0 2 0 2.5     <- scaling by 2.5 to give correct temperature reading  (as voltage)
.ends

I'm sorry if I have caused any confusion.


m.c.

-----Original Message-----
From: ChristianH [mailto:chris at scp.de]
Sent: Dienstag, 11. Februar 2003 15:34
To: synth-diy at dropmix.xs4all.nl
Subject: Re: [sdiy] help needed: thermal equations to RC networks
analogy


On Tue, 11 Feb 2003 14:58:07 +0100 Czech Martin wrote:

> I believe that a heat source plus thermal medium can be described
> in electrical analogy:
> 
> temperature -> voltage
> heat stream -> current
> heat source -> voltage source
> thermal isolation -> resistance
> heat capacity -> capacitor
> 
> It's only a believe, but I'd like to know instead!
> 
> So ,e.g. a resistor firing a volume of air can be described
> as a voltage source (voltage ~ total heat power),
> and a RC network: a string of resistors, and from nodes
> in between capacitors to ground.
> If the whole is to take place in a package one needs some 
> resistor to model heat losses to the outside world.
> 
> Problem:the voltage source will finally give voltage
> to all nodes (i.e. heat will distribute).
> But if the voltage is switched of, the voltage source
> will draw current back from the network.
> This would mean that a resistor that is not burning
> any power input will take some heat current back
> from the air, actually cooling it. ouch!

I think this is where the analogy is limping a bit. An electrical
resistor generating heat has a heat capacitance as well. When it stops
running electrical current through itself, it doesn't stop emitting heat.

It would do so if it could be immediately put back to room temperature
(i.e. thermal ground), and in this case it would indeed drain heat from
the surrounding air. So much for the theory part...

Christian