two power resistors as element

That will depend on the mass of your pot, the specific heat of the material, the heat loss based on whether it is insulated or not... plus the mass and specific heat of the water. Best determined empirically.

Cheers!

Chip Shults My robotics, space and CGI web page -

The specific heat capacity of water is 4.19 J/g C or 4.19 J/ml C.

Starting at 22C you need 4.19 * 500 * (100-22) J of energy = 16.3 kJ.

So at 98W it will take about 28 minutes.

You would need to factor in losses, say an average 10W and specifc heat of the heater and container, say equivilent to another 50ml of water.

The answer is then nearer 34 minutes.

Now you know why kettles come with 3kW and not 100W elements.

material,

Who, exactly, is this comment directed toward?

Cheers!

Chip Shults My robotics, space and CGI web page -

0.5*water_k*80/100 (m*k*dT/power)

water_k is about 4800 ,if memory serves.(joules/kg*c)

Hiya!

I preffered Pauls suggestion.

Take 15kgs of Plutonium, and squash together real hard. That'll boil plenty of water for you :-)

Yours, Mark.

Wolfgang Mahr>

It will never boil. 98.6W pumped into two 10W resistors will destroy the resistors quickly.

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get

about

At what altitude?

I make that 96.8W, but apart from that small typing error I agree with your calculations

I assume this is related to your previous post about building a kettle?

I do not think a kettle is a good project; the combination of mains voltages and water is the obvious danger, but there may be others such as toxic chemicals (e.g lead from solder) getting into the water, or producing (explosive) hydrogen gas due to electrolysis of the water.

Why not just buy a kettle?

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Anywhere above the water line.

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get

96.8W is equivallent to dissipation at a rate of 96.8 joules per second. 500mL (cc) of water requires 80 times 500 calories to reach 100C from 20C or 40000 calories. Taking 1 calorie as 4.18 joules then you need 167200 joules which equates to 1727 seconds or 28.8 minutes. Losses would extend this time, possibly to infinity without good insulation.

In order to actually boil the water you need another 2240.5J per cc or 1.12 MJ to supply the latent heat for vapourisation.

All that after half a bottle of wine!

Best wishes,

Pauline Aston.

>

Perhaps not. If they are wirewound resistors on a hollow core, they may not get hot enough to fail until the water boils away.

Domestic water heater elements will fail if not submerged, this is the same.

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Doubtful those resistors will last long enough to finish electrocuting you. At 10w dissipation each they'll run hot enough to burn your fingers not to mention how you're going to power them safely while submerged, not to further mention that their maximum rating temperature is likely less than your anticipated cup of tea (assuming you could get them to dissipate the heat effectively while being powered safely). Just nothing right with this particular project as it appears to be inquiring about.

What a bunch of pesimists.

I just dug a Eurohm 2.5W 1R8 wirewround resistor out of the junk box. Not one of those nice green ones - it is physically smaller.

It is currently 1/2 inch deep in a plastic beaker of water and passing

3.7A. That is 10 x rated not 5 and it is just sitting there bubbling a bit. I wouldn't like to drink the water and it isn't getting any hotter than 62 C.

My guess is it would eventually fail from corrosion of the leads, few seconds - pahhh.

What would be wrong with using a resistor made for the job and cheap to boot ?? I refer of course to the "Electric jug element". In fact the whole idea seems very much like re-invention of the wheel !!

-- Regards ............... Rheilly Phoull

material,

and

The resistor ratings are in still air. He has them submerged in water. Although not a good idea, the water cools the resistors.

Al

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will

the

No. Except for the fact that the resistors are probably not rated to work under water (hence corrosion etc., becomes likely), there is no reason for the resistors to fail. The point is that a resistors temperature depends on the power input, and the rate of dissipation by conduction/radiation. The resistors failure 'curve', depends primarily on the temperature that it is working at. Thinking of a resistor as having a specific 'rating', that applies all the time, is a dangerous, and incorrect way of working. A typical power resistor, will be given a rating figure, but this will carry with it a whole series of specifications. Normally the spec will be something like 10W, in 'free air', at 25C, with a temperature rise of 100C. Now the actual rating in use, will change if any of these factors is altered (so it is typical to happily run 10W resistors dissipating 20W, by the simple expedient of forced air cooling). Similarly, if your box, does not provide 'free air', with an airflow obstruction, then the rating has to be derated. The actual 'life' of the resistor will depend (ignoring the other effects like corrosion mentioned), on the temperature of the element. Now some wirewound designs (such as the Arcol HS family), are rated to operate with the element at up to 200C. Assuming the resistor chosen is rated to operate with the element this hot, then water cooling the resistor, keeps it's element from getting above it's rated maximum temperature, and life from this point of view, should be acceptable. Some power applications, use multiple resistors, operating in flourine liquid coolants (though this is now very expensive), and 'overrate' the resistors by often as much as 800% in this setup. Really though, the chances are that the resistor will undergo significant decay, because of both chemical, and electrolytic effects, and will also result in the 'potential' for electrical connection to the liquid, which make this a solution, rather like using a sledgehammer, to try to catch a fly (very dangerous for anyone else in the room...). Small waterproof heating elements are readily available from industrial suppliers, or by dismantling cheap kettles (including things like car designs), which are complete sealed assemblies, far better suited to this job....

Best Wishes

Well for carbon resistors the max temp they _can_ go to is white hot. Bit beyond the specs, but they can work for a few seconds at such temps. Yes, I know from experience :) IIRC that was at about 500w diss in a lil 1/3w R. You should use UV filtering for safety.

OTOH some of the green ceramic ones can run the elements upto IIRC about 350C - and thats within specs.

BTW, why not use a low V transformer and pass the current through the metal case of the kettle? Seems like a better idea to me. You'd have a lovely clean light kettle then - and a base weighting kilo-kilos.

Regards, NT

I never considered submerging the resistors. It wasn't a given, and I can't think of a decent method of insulating the entirety of the resistor leads that would be electrically safe and prevent corrosion failure (other than heating a vessel).

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oil

Yes. This is very dependant on the resistor construction involved. Some 'old' designs, have the wire exposed on the core. With these, the water will directly cool perhaps half the wires surface. Designs which have a ceramic body, have the wire wound round a former (normally glassfibre), then the ceramic body over this. This increases massively the thermal resistance to the case, but with the gain, of protecting the wire (both from you, and you from it...). Then there are aluminium bodied designs, which use a similar approach, but normally with a heatsink compound between the wire and the casing (these are smaller for a given power 'rating' than the ceramic designs, but require a heatsink. I'd expect water cooling to work, with a resistor with the wire on the surface. Possibly work, but to a lesser extent with the aluminium cased designs, then to work far less well with the other designs. I'd not expect any design to get much beyond perhaps 10:1 (you were pushing 26:1 with your experiment...).

Best Wishes