Choose Your Own Temperature Kettle

My local tea shoppe in the last sale had an Upton tea kettle 1L Model AK16. It is a blow torch at 1500w. It needs to be 1/3 full for all temperature ranges from 160 to 212. I use my KamJove since I always boil water. I'll never go back to a kettle on the stove.

Jim

Michael Plant wrote:

Those units are Home-Tek HT355 (white) and 358B (brushed stainless). They're 3kW (!) kettles. Unfortunately, the reason their heating elements can pump out 3kW is that they're 240V UK models.

I emailed the company ( snipped-for-privacy@hometekinternational.co.uk) in June, and they replied that their kettles are currently only for the UK market, and they had no plans to make a lower voltage version, though they'd forward my request to the manufacturer for consideration. I haven't heard (nor did I expect to) anything since then. (The increasingly unfavorable exchange rate is probably not making this more likely to become available here.)

N.

For 3KW, it would take an optimistic 25A to make one that operates on

You should be able to operate the thing as-is on 120V, and it should put out about 1500W and pull under 15A if you do so. The thermostatic cutoff will be fine if it's mechanical. It will take more than twice as long to heat, though, with half the power.

However, there's no reason you can't get a 15A 240V electrical outlet installed in your kitchen. An electrician will do it for a pretty reasonable fee (most of the cost is the labour to route the cables) and it will allow you to use all kinds of European appliances in your kitchen. Heating appliances won't care about the different line frequency and universal motors won't care either. Shaded-pole motors (like in my Thai coconut grinding machine) run 20% slower.

--scott

This is true for a simple heating circuit with a mechanical thermostat as I have done it with the Kamjove kettles I imported from China, basicaly you just get half of the rated wattage when using it on 120v. However this particular unit has an active electronic thermostat/controller with LED display in order to maintain temperature control. These active electronics are almost certainly voltage sensitive.

This would be the olny option if you wanted to use this unit.

I also emailed the company yesterday and they still maintain that they have no plans to make a 120v version.

Mike

I brought a Kamjove back from China with me this time around. The thing wouldn't work because of the different current it runs on; we are on 110 and China is on 220. Anyway, I bought a transformer to convert the current so it works now. It's rather funny to see this small tea boiler running from a 23 pound transformer.

Actually, the Kamjove I have is pretty awesome. It boils water in about 3 minutes.

Actually the Kamjove units with mechanical thermostats will work just fine on 120v, all you have to do is change the plug. The water will heat slower but it works fine.

Mike

If you all can hang in there a few more months, a new one will be coming out in the market that combines form and function to offer a very uniquely beautiful vessel for getting your water exactly how you want it. I'll be sure to alert you all to when it is out.

SMB

Actually, it wouldn't at all. I would have had to totally rewire the thing and we tried several things.

What is the wattage of the Kamjove? Price of transformer? Does the transformer draw much current (amperes) from the 110v connection? What size circuit breaker would I need?

This is intriguing; can you share more details at this time?

N.

Well, the wattage mine uses is about 1500W, so I can say it boils water starting from cold in less than 5 minutes. heh.

As for your other questions, I have not much of an idea. I wasted way to much money on a transformer, that's all I am going to say. You can search "step down transformers" on google.

If it's an element rated for 1500W on 240V, it will actually produce

You can work out the current demand with P=IV. A 750W element on 120V will draw 6.25 amperes.

--scott

Well, not exactly. Halving the voltage also halves the current, thus quartering the drawn power; P=v^2/R=I^2R. However, elements in a device like that often run at dull-red heat, at which temperature the resistance is much higher than at room temperature. (Hence the burnout surge on turning incandescent lamps on.) Without knowing the particular element, it's hard to guess how actual power at half voltage will compare with (power at rated voltage)/4.

-DM