Coffee Urn Heating Circuit Help

[Helmanfrow]

Our coffee urn recently failed when the 130°C thermal limit switch started tripping at a much lower temperature.
I ordered a new switch and bypassed the broken one temporarily in the meantime.
I figure the fuse, the GFCI and the fact that there's always water in the urn provide a reasonable degree of safety.

This anecdote is tangential to my question, but I opened with it to explain my sudden interest in coffee urn heating circuits.

What's actually puzzling me is something else:

You can see from the service manual's schematic (marked up by me) that thermal switch T1 controls the high-wattage element.
What I don't understand is why, when the switch is open, the circuit to the element is not completed by way of wiring through the lamp, KWL (green highlight).

I thought that perhaps the schematic was wrong so I quickly made my own as a sanity check... and ended up with an identical drawing.

Why, when the switch is open, does element HP not energize through the lamp wiring?

Can someone please walk me through this?

[Helmanfrow]

I'm also wondering why the fuse and thermal limit are on the neutral and not the line.

[james_s]

The heating element does energize through the lamp, but the current required to light the lamp is negligible, it isn't enough to cause the element to warm measurably.

The reason the fuse is on the neutral is likely that it made mechanical layout easier. With a closed device that has no user accessible conductors it makes no difference which side those are on.

[BradC]

What I don't understand is why, when the switch is open, the circuit to the element is not completed by way of wiring through the lamp, KWL (green highlight).

It does, but the power will be limited to bugger all. The KWL lamp drops near enough to all the voltage, so the element will almost just be a piece of wire. If it wasn't in water I suspect you wouldn't even notice a temperature rise in it.

Edit : beaten

[RJSV]

While reading the main idea, here, I did notice a statement that has safety concerns, that your GFI device will protect against any hazard... So I'm wondering if a smokey overheat situation would even be able to trip that breaker ?  Not an expert, but I'd bet real money that an 'overheat' and potential fire could happen, while still having no 'ground fault...at least until a fire gets a roaring start.
   Ditto that, for having more caution even if there is a 'fuse'.  By trying things, some combination of conditions could still, uh, result in loss of house and lives.  (My computer has 'backups', maybe, but I don't wish to test that.)

[james_s]

What are you talking about? A GFCI has nothing to do with fires or overheating, it is to protect the user from shock should they come into contact with a live conductor and they themselves provide a leakage path to ground. The built in thermal fuse provides protection against overheat and will open the circuit, and if there is an overcurrent situation the branch circuit breaker will protect against that fault and de-energize the circuit.

[Helmanfrow]

What are you talking about? A GFCI has nothing to do with fires or overheating, it is to protect the user from shock should they come into contact with a live conductor and they themselves provide a leakage path to ground. The built in thermal fuse provides protection against overheat and will open the circuit, and if there is an overcurrent situation the branch circuit breaker will protect against that fault and de-energize the circuit.

If the jacket on a live or neutral wire were to melt and make contact with the chassis the GFCI would trip.
That said, the wires inside the device appear to be of the high-temperature variety so I figure their jackets can resist a good deal of heat.

My main line of defense are the functioning High Power thermostat and the fact that the urn is always full of water. Also the fuse.

[Helmanfrow]

The heating element does energize through the lamp, but the current required to light the lamp is negligible, it isn't enough to cause the element to warm measurably.

So the all the voltage would drop across the lamp as BradC said?
Why doesn't the element draw the current it needs?
Back to Khan Accademy for me...

The reason the fuse is on the neutral is likely that it made mechanical layout easier. With a closed device that has no user accessible conductors it makes no difference which side those are on.

Right, that's what I figured. Just making sure.

[themadhippy]

Why doesn't the element draw the current it needs?

Think of it  not as a a lamp,but a resistor in series with the element

[james_s]

So the all the voltage would drop across the lamp as BradC said?
Why doesn't the element draw the current it needs?
Back to Khan Accademy for me...

Look at it this way, assuming the kettle is 120V 1200W the element will have a resistance of 12 ohms. Now for the sake of simplicity let's pretend the lamp is a 1 watt 120V incandescent bulb, when illuminated that will have a resistance of 14.4k. Since they're in series with 120V across the pair you can easily use ohms law to calculate the voltage drop across each. Now that you have the voltage drop, you can use Ohms law again to calculate the power dissipated by the lamp and by the element. Kirchoff's law dictates that the current through both will be identical since they're in series.

I'd still encourage you to do the calculations yourself but what you'll find is that the 14.4k resistance of the lamp dominates the 12 ohms of the heating element and limits the total current, thus the vast majority of the voltage is dropped across the lamp and the drop across the element is tiny. Power is a product of voltage and current, so if the voltage drop is negligible then the power diispated is also negligible.

[Helmanfrow]

Look at it this way, assuming the kettle is 120V 1200W the element will have a resistance of 12 ohms. Now for the sake of simplicity let's pretend the lamp is a 1 watt 120V incandescent bulb, when illuminated that will have a resistance of 14.4k. Since they're in series with 120V across the pair you can easily use ohms law to calculate the voltage drop across each. Now that you have the voltage drop, you can use Ohms law again to calculate the power dissipated by the lamp and by the element. Kirchoff's law dictates that the current through both will be identical since they're in series.

I'd still encourage you to do the calculations yourself but what you'll find is that the 14.4k resistance of the lamp dominates the 12 ohms of the heating element and limits the total current, thus the vast majority of the voltage is dropped across the lamp and the drop across the element is tiny. Power is a product of voltage and current, so if the voltage drop is negligible then the power diispated is also negligible.

Resistors in series. Voltage drop. Current the same everywhere in the circuit. Got it!

🙏

[Helmanfrow]

If nobody minds I'm going to hijack my own thread:

As it turns out, the replacement parts from the manufacturer are priced through the roof and into the lower atmosphere.
They're charging about CAD$46 (plus shipping) for what ought to be a $5 part at retail.

I've scoured the interwebs and discovered that there are many similar components for a few dollars but but none that meet all the original specs.

The original is designated KSD301-R-G 130.
KSD301 appears to be a generic designation for this form of thermal limit switches.
130 probably refers to the set temperature of 130°C.
The R-G suffix appears in catalogs from at least three manufacturers, I think it has to do with the manual reset switch and ceramic body or perhaps the threaded terminal lugs.

None of the models available for sale have threaded terminals, so I'd have to either tap my own holes or crimp new connectors onto the wires. I don't think I'd use solder in this application.

My questions are as follows:
Is it okay to use a bakelite part instead of the original ceramic?
Is it okay to increase or decrease the set temperature?

Or should I just bite the bullet and by the original part for 1/4 the cost of a new urn...