two power resistors as element

[JohanWagener]

Please comment on my calculations

If I take two 10W resistors of 1k each and put them in parallel I would get
a 500ohm resistor right? Now if I apply 220V over it there would be about
0.44A flowing through it. That is power of about 98.6W. How long will it
take to boil 500ml of water?

 

[Sir Charles W. Shults III]

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 - http://home.cfl.rr.com/aichip

 

[nospam]

"JohanWagener" <rebeldog@webmail.co.za> wrote:

Please comment on my calculations

If I take two 10W resistors of 1k each and put them in parallel I would get
a 500ohm resistor right? Now if I apply 220V over it there would be about
0.44A flowing through it. That is power of about 98.6W. How long will it
take to boil 500ml of water?

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.

 

[Sir Charles W. Shults III]

"Dmytry Lavrov" <dmytrylavrov@fsmail.net> wrote in message
news:3F086F5A.449E@fsmail.net...

Sir Charles W. Shults III wrote:

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 - http://home.cfl.rr.com/aichip
Just put your fingers into water and die.

;-)

Who, exactly, is this comment directed toward?

Cheers!

Chip Shults
My robotics, space and CGI web page - http://home.cfl.rr.com/aichip

 

[Dmytry Lavrov]

JohanWagener wrote:

Please comment on my calculations

If I take two 10W resistors of 1k each and put them in parallel I would get
a 500ohm resistor right? Now if I apply 220V over it there would be about
0.44A flowing through it. That is power of about 98.6W. How long will it
take to boil 500ml of water?

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

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

 

[Clifton T. Sharp Jr.]

JohanWagener wrote:

If I take two 10W resistors of 1k each and put them in parallel I would get
a 500ohm resistor right? Now if I apply 220V over it there would be about
0.44A flowing through it. That is power of about 98.6W. How long will it
take to boil 500ml of water?

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

--
The function of an asshole is to emit quantities of crap. Spammers do
a very good job of that. However, I do object to my inbox being a
spammer's toilet bowl. -- Walter Dnes

 

[Pete Culf]

"Clifton T. Sharp Jr." <clifto@clifto.com> wrote in message
news:3F087A3F.268FB135@clifto.com...

JohanWagener wrote:
If I take two 10W resistors of 1k each and put them in parallel I would
get
a 500ohm resistor right? Now if I apply 220V over it there would be
about
0.44A flowing through it. That is power of about 98.6W. How long will it
take to boil 500ml of water?

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

At what altitude?

 

[Gareth]

JohanWagener wrote:

Please comment on my calculations

If I take two 10W resistors of 1k each and put them in parallel I would get
a 500ohm resistor right? Now if I apply 220V over it there would be about
0.44A flowing through it. That is power of about 98.6W.

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

How long will it take to boil 500ml of water?

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?

-----------------------------------------------------------------------
To reply to me directly:

Replace the text after the@symbol with: totalise DOT co DOT uk

 

[Mark (UK)]

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 Mahringer wrote:

Hi Johan

JohanWagener wrote:

Please comment on my calculations

If I take two 10W resistors of 1k each and put them in parallel I would get
a 500ohm resistor right? Now if I apply 220V over it there would be about
0.44A flowing through it. That is power of about 98.6W. How long will it
take to boil 500ml of water?

The resisitors won't survive this longer than a few seconds, even if
cooled in water. Also the low electric resistance of water will
cause some problems .... kids, don't try that at home!

HTH
Wolfgang

 

[Clifton T. Sharp Jr.]

Pete Culf wrote:

"Clifton T. Sharp Jr." <clifto@clifto.com> wrote...
JohanWagener wrote:
If I take two 10W resistors of 1k each and put them in parallel I would
get
a 500ohm resistor right? Now if I apply 220V over it there would be
about
0.44A flowing through it. That is power of about 98.6W. How long will it
take to boil 500ml of water?

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

At what altitude?

Anywhere above the water line.

--
The function of an asshole is to emit quantities of crap. Spammers do
a very good job of that. However, I do object to my inbox being a
spammer's toilet bowl. -- Walter Dnes

 

[Pauline Aston]

"JohanWagener" <rebeldog@webmail.co.za> wrote in message
news:be9l1a$1ci$1@ctb-nnrp2.saix.net...

Please comment on my calculations

If I take two 10W resistors of 1k each and put them in parallel I would
get
a 500ohm resistor right? Now if I apply 220V over it there would be about
0.44A flowing through it. That is power of about 98.6W. How long will it
take to boil 500ml of water?

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.



>

 

[Sam Goldwasser]

"Clifton T. Sharp Jr." <clifto@clifto.com> writes:

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

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.

--- sam | Sci.Electronics.Repair FAQ Home Page: http://www.repairfaq.org/
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Important: The email address in this message header may no longer work. To
contact me, please use the Feedback Form at repairfaq.org. Thanks.

 

[G. Skiffington]

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.

 

[nospam]

Wolfgang Mahringer <wolfgang.mahringer@sbg.at> wrote:

If I take two 10W resistors of 1k each and put them in parallel I would get
a 500ohm resistor right? Now if I apply 220V over it there would be about
0.44A flowing through it. That is power of about 98.6W. How long will it

The resisitors won't survive this longer than a few seconds, even if
cooled in water. Also the low electric resistance of water will
cause some problems .... kids, don't try that at home!

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.

 

[Rheilly Phoull]

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

"Michael Hofmann" <westbound@gmx.net> wrote in message
news:3F091851.60904@gmx.net...

Sir Charles W. Shults III wrote:
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.

... plus the ambient air pressure

SCNR,
Michael

 

[Al]

In article <3F087A3F.268FB135@clifto.com>,
"Clifton T. Sharp Jr." <clifto@clifto.com> wrote:

JohanWagener wrote:
If I take two 10W resistors of 1k each and put them in parallel I would get
a 500ohm resistor right? Now if I apply 220V over it there would be about
0.44A flowing through it. That is power of about 98.6W. How long will it
take to boil 500ml of water?

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

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

Al

--

Reverse address to reply.

 

[Roger Hamlett]

"Richard Crowley" <rcrowley7@xprt.net> wrote in message
news:vgir6hkijsudca@corp.supernews.com...

"Al" <ten.knilhtrae@yajfa.remove> wrote in message
news:ten.knilhtrae-104E7D.08403107072003@news.bellatlantic.net...
In article <3F087A3F.268FB135@clifto.com>,
"Clifton T. Sharp Jr." <clifto@clifto.com> wrote:

JohanWagener wrote:
If I take two 10W resistors of 1k each and put them in parallel I
would get
a 500ohm resistor right? Now if I apply 220V over it there would be
about
0.44A flowing through it. That is power of about 98.6W. How long
will
it
take to boil 500ml of water?

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

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

So it will take a bit longer for the resistors to fail.
But still unlikely to last long enough to boil water.
(Except maybe a few drops on the surface).
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

 

[N. Thornton]

"G. Skiffington" <skiffiREDUCEng@nbnet.nb.ca> wrote in message news:<12D3B984.7079@nbnet.nb.ca>...

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.

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

 

[Clifton T. Sharp Jr.]

Sam Goldwasser wrote:

"Clifton T. Sharp Jr." <clifto@clifto.com> writes:
It will never boil. 98.6W pumped into two 10W resistors will destroy the
resistors quickly.

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.

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).

--
The function of an asshole is to emit quantities of crap. Spammers do
a very good job of that. However, I do object to my inbox being a
spammer's toilet bowl. -- Walter Dnes

 

[Roger Hamlett]

"nospam" <nospam@nospam.invalid> wrote in message
news:arvvgvk2na19ntkqa0di7leovv3rc4sr25@4ax.com...

Wolfgang Mahringer <wolfgang.mahringer@sbg.at> wrote:

Despite of what the others write: your resistors might even survive:
the water conducts heat much better than the air for which the
resistoir was designed.

That's not really true.

You have 2 thermal resistances:
i) Resistance wire (or body) to case
ii) Resistance case to ambient.

Resistance i) is _always_ the same, and mostly larger than resistance
ii)
Only resistance ii) is influenced (greatly reduced) by the water
"cooling".

And if the water heats up (presuming the resistors survive that long,
which i doubt, btw) the resistance ii) rises, because of the lesser
temp difference between resistor case and the fluid.

Hey you are right.

I took my 2.5W 1R8 Eurohm resistor (body measures 11mm x 4.5mm diameter)
and ran it at 65W submerged in a bit of corn oil.

After 15 minutes the oil was at 150 C the resistor was discoloured and a
bit of the body coating flaked off leading to localised heating of the
oil
which made smokey bubbles.

So the moral is even with oil cooling compact wire wound resistors don't
much like running with a 2500% overload.
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