Anyone know how flame detection works in (Potterton) boilers?

[[email protected]]

I've got a 1998 Potterton Profile boiler. It has a single electrode
which provides the sparks and also does the flame detection. I've
read somewhere on the Internet that you can do flame detection by
measuring the current that passes through a flame between two
electrodes. Rectification of an ac current removes the possibility of
false detection due to sooting because of some ionisation effect.

Anyone got any detailed explanations or circuits?

Thanks,

Paul

 

[Homer J Simpson]

I've got a 1998 Potterton Profile boiler. It has a single electrode
which provides the sparks and also does the flame detection. I've
read somewhere on the Internet that you can do flame detection by
measuring the current that passes through a flame between two
electrodes. Rectification of an ac current removes the possibility of
false detection due to sooting because of some ionisation effect.

Anyone got any detailed explanations or circuits?

Like http://www.babcock.com/fps/scanners_sunspot.html ??

Honeywell always preferred the UV detectors BTW.

 

[Lionel]

I've got a 1998 Potterton Profile boiler. It has a single electrode
which provides the sparks and also does the flame detection. I've
read somewhere on the Internet that you can do flame detection by
measuring the current that passes through a flame between two
electrodes.
Wrong.

Rectification of an ac current removes the possibility of
false detection due to sooting because of some ionisation effect.

Wrong again.

Anyone got any detailed explanations or circuits?

--
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[Lionel]

On Fri, 01 Jun 2007 02:45:22 +0000, Lionel <[email protected]>

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From: Lionel <[email protected]>

^^^^
**** off, Steve "Bowtard" Young.

Organization: Kookologists
Subject: Re: Anyone know how flame detection works in (Potterton) boilers?
Date: Fri, 01 Jun 2007 02:45:22 +0000
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Wrong again.


--
\___ Proud Cog #1 in the AUK Hate Machine
_(AUK)====:: Do *you* think that you have the Right Stuff?
/='='='='-, Apply TODAY by addressing a gratuitously cruel
(O+O+O+O+O) flame to: "Uncle Fester", C/O soc.singles & AUK.
~^^^^^^^^^~~~^~^^~'~~^'^~~~"~~'"~^~'"~~^~"~'~^'^~^~^^~^~"~^~"'~'"~^~~

--
\___ Proud Cog #1 in the AUK Hate Machine
_(AUK)====:: Do *you* think that you have the Right Stuff?
/='='='='-, Apply TODAY by addressing a gratuitously cruel
(O+O+O+O+O) flame to: "Uncle Fester", C/O soc.singles & AUK.
~^^^^^^^^^~~~^~^^~'~~^'^~~~"~~'"~^~'"~~^~"~'~^'^~^~^^~^~"~^~"'~'"~^~~

 

[Jamie]

squashstring wrote:




Wrong again.

I could be wrong how ever, It maybe possible that lest
current or hardly no current is detected from the HV xformer
when the burner is ignited and pushing flames between the to
electrodes. The arc wouldn't be able to make and thus you
could detect this..
Just and idea.

 

[Spehro Pefhany]

Flames are electrically conductive. After an electrode is HV pulsed a
few times to ignite a flame, that same electrode can be used at
low-level to verify that the flame is present. Lots of furnaces and
gas ranges work this way.

John

If the path got gummed up with soot or moisture it would inevitably
end rather badly (a really big bang), so that's not how they work. The
flame conducts asymmetrically.


Best regards,
Spehro Pefhany

 

[Jamie]

Did I say anything that's not true? Did I say the conduction was
symmetric?

John

Well John, You and others have cleared that up with me, which also
gives me an idea for a problem we have at work that may now be able
to offer a simpler solution.
You must remember, not every one appreciates one's help how ever,
there are always a few that do and those are the one's that count.

Have a good day..

 

[Glen Walpert]

On Fri, 01 Jun 2007 19:19:39 -0500, Spehro Pefhany

If the path got gummed up with soot or moisture it would inevitably
end rather badly (a really big bang), so that's not how they work. The
flame conducts asymmetrically.

That may be the standard explanation, but I don't think the flame is
actually capable of conducting asymmetrically; how could that possibly
work? The asymmetricric conduction is probably the result of one
electrode being hotter than the other due to position or size, and
thus emitting electrons more easily than the other one - so lower
voltage drop with fixed current or more current with the same voltage
when current is from the colder to the hotter electrode (hotter
electrode negative).

With no voltage applied to two electrodes in a flame positioned so
that one is hotter than the other a small thermocouple voltage will
result also, even if both electrodes are the same material, but this
might be harder to detect. With the larger thermal voltage gradient
along the hotter electrode, and the more negative voltage at the hot
end for both electrodes (always true for all metals), this effect
would tend to add to the effect of lower electron emission work for
the hotter electrode in the case of a small applied AC voltage.

The effect could probably be enhanced with different electrode
materials, and they could then possibly both be at the same tip
temperature and still produce the differential conduction effect.

(A good example where the myth of thermocouple voltages being produced
"by the junction" instead of by the temperature gradient along the
length of the conductors cannot explain the effect.)

So both of these effects will contribute to asymmetrical conduction,
but I can't offhand think of any way the flame itself could fail to
conduct equally in either direction. Perhaps the conduction of both
heat and electricity in the flame is primarily by electron (or ion)
transport, similar to metals, in which case the flame might develop a
voltage gradient between the hot and less hot parts of the flame as in
thermocouple wires, but I think that effect requires the electrons to
be in a conduction band vice free ions and it still would be a DC
voltage rather than differential conduction (though it might look like
it).

Interesting effect, never heard of it before, and my explanation might
well be wrong . Contrary info welcome.

Glen