The table below represents a calculation of the energy used by two kettles
to boil the same amount, 1.08L, of water. (If your newsreader doesn't space
it correctly, copy'n'paste it into something that uses a fixed width font,
eg: Notepad)
Start Finish Time Error kW Units Un Err Min Max
RH 00:15 02:55 02:40 00:02 2.750 0.1222 0.0015 0.1207 0.1238
MR 00:15 03:40 03:25 00:02 2.025 0.1153 0.0011 0.1142 0.1164
If the two kettles were 'perfectly' insulated and were 'perfect' convertors
of electrical energy into heat, the units used should be the same, but they
are not.
Discuss
(You should take no more than 15 minutes to complete this question)
... but actually, I've been pondering it longer than that, and am somewhat
baffled.
I ran the experiment after reading this (my current kettle is leaking and
requires replacement):
http://tinyurl.com/2r9u35
... standing in for ...
http://www.russellhobbs.com/buyingguides/kettles/
"2Kw is energy saving as it uses less power to boil the kettle."
As indicated in the mock exam question, if all the heat from the electricity
went into the water, without any being wasted in heating up the environment
through the walls of the kettle, then they should use the same amount of
electricity to boil the same amount of water.
Further, where, as in real life, heat actually is wasted through the walls
of the kettle, then the rate of heat loss at any given moment is determined
by the temperature difference between inside and outside the kettle (the
temperature gradient), and the total lost will therefore be determined
partly by the amount of time the kettle takes to boil - the longer it
takes, the longer it will spend at temperatures where the temperature
gradient is significant, and therefore the more heat should be lost overall.
And this line of reasoning is born out by my experiences camping, trying to
boil water with one of those small canister gas stoves. On a cold windy
day, heat can be lost at such a rate that although the kettle quite quickly
gets hot, it never actually boils until you move it behind some windshield
like a big tree.
Yet the experiment shows, contrary to what I might have reasoned, that the
RH statement is in fact correct, and I can't explain it ...
There aren't many significant differences between the kettles that I can
see, apart from the power rating and element design - the latter might go
some way to explaining things. The older, lower-powered Morphy Richards one
has the 'traditional' exposed element, whereas the newer. higher-powered
Russell Hobbs has a concealed element, and perhaps there is greater wastage
of heat into the environment with that design.
Other than that, they are both columnar in shape (so would probably have a
similar ratio of mass of water to heat against surface area to lose heat
through), both are plastic, though different looking plastic (and both
leak).
> Yet the experiment shows, contrary to what I might have reasoned, that the
> RH statement is in fact correct, and I can't explain it ...
At first thought, if you apply more power (and therefore reach boiling
point quicker) you will have less time for heat transfer (through the
wall and out of the top via steam) to take place, so the total energy
required to reach boiling point will be less.
Geoff
> As others have seen fit to go rabidly off-topic about electricity
> consumption here, here's my ha'p'worth ...
>
> The table below represents a calculation of the energy used by two
> kettles
> to boil the same amount, 1.08L, of water. (If your newsreader doesn't
> space it correctly, copy'n'paste it into something that uses a fixed
> width font, eg: Notepad)
>
> Start Finish Time Error kW Units Un Err Min Max
> RH 00:15 02:55 02:40 00:02 2.750 0.1222 0.0015 0.1207 0.1238
> MR 00:15 03:40 03:25 00:02 2.025 0.1153 0.0011 0.1142 0.1164
>
> If the two kettles were 'perfectly' insulated and were 'perfect'
> convertors of electrical energy into heat, the units used should be the
> same, but they are not.
>
> Discuss
How did you determine the 'finish' time?
If you waited for the auto shut-off to operate, maybe the high-power
kettle has a slower steam sensor. In any event, during the time that the
kettle is boiling, the high-power one will make more steam and use more
power in the process.
--
Dave
Java Jive wrote:
> Start Finish Time Error kW Units Un Err Min Max
> RH 00:15 02:55 02:40 00:02 2.750 0.1222 0.0015 0.1207 0.1238
> MR 00:15 03:40 03:25 00:02 2.025 0.1153 0.0011 0.1142 0.1164
[snip]
> "2Kw is energy saving as it uses less power to boil the kettle."
[snip]
> Yet the experiment shows, contrary to what I might have reasoned, that the
> RH statement is in fact correct, and I can't explain it ...
Is it me or you getting confused here? Are you talking about a different
experiment to your figures above? Because they show fairly conclusively
that actually the RH used more power... so is the RH actually the
higher-power of the two or is it the low power device? And are RH saying
that their lower- or higher- power device uses less energy?
Looks like you've swapped the initials round....
So if it's actually
MR 00:15 02:55 02:40 00:02 2.750 0.1222 0.0015 0.1207 0.1238
RH 00:15 03:40 03:25 00:02 2.025 0.1153 0.0011 0.1142 0.1164
then, like you, I find it hard to comprehend. I'd say it's the age of
the 2.7kW kettle and improvements in the design of the RH which have
caused the improvement in efficiency. It's clearly not just because it's
a lower-power device.
Geoff
No, the auto shut off on both used to work much better than they do now, to
the point where I nearly always switch them off by hand.
It was a judgement call, and that's why I gave an error figure of 2 sec, if
it had just been the error of me working the stopwatch, I would have allowed
1 sec error.
Incidentally, if anyone's wondering, there's a reason why I started the
stopwatch before the kettles - I did consider growing an extra hand, but I
wanted a result sooner than that ...
"Geoff Winkless" <usenet-at-geoff-dot-dj@[127.0.0.1]> wrote in message
news:4628add7$0$310$4d4e...@read.news.uk.uu.net...
Yes, the RH used more, but, as the table shows, it was the higher-powered of
the two kettles I used.
I hadn't realised that the fact that one of the kettles I happened to have
to hand also happened to be made by the firm whose statement I was
contesting, would be the source of misleading confusion. Sorry for not
being more explicit. To confirm, the two kettles were:
Russell Hobbs: Model 3161-40, 2.5-3.0kW, concealed element, newest model,
probably a little over 5 years old
Morphy Richards: Model 24343, 1850-2200W, exposed element, older model,
probably nearly 15 years old
The statement I was contesting was that all other things being equal, a
lower-powered kettle should consume less electricity overall, my reasoning
being that more heat would be lost to the environment during the slower boil
than with a quicker higher-powered one. But my limited experiment failed to
prove this.
I believe the explanation most probably lies in the difference between the
elements. I hadn't thought about it before today, but it seems clear to me
that an exposed element must be significantly more efficient than a
concealed one, because it will be transferring heat to the water all around
it, except just where it is mounted to the kettle, whereas at least half the
heat coming out of a concealed element would be going downwards and
therefore has the potential to be lost (not having dismantled one, I don't
know what reflective coatings, etc, the manufacturers use to reflect the
heat that is heading downwards back up into the water, but presumably they
must do something like that).
Did you measure the starting water temperature in each case? That has a
bearing on the energy required to raise it to boiling point. Were the
kettles always cold when you started, or was there any residual heat from a
previous boiling? How accurately did you measure your 1.08 litres? (seems an
odd amount to use!)
Did you measure the *actual* electrical energy consumed, or did you simply
base it on the nominal kW ratings of the kettles and the time taken?
--
Cheers,
Roger
______
Email address maintained for newsgroup use only, and not regularly
monitored.. Messages sent to it may not be read for several weeks.
PLEASE REPLY TO NEWSGROUP!
--
Chris Green
Reminds me of my A-level physics practical (OK, so I'm old!) which require
simultaneously releasing two weights, 1m apart, and starting a stopwatch.
The teachers had clearly read the notes because one of them was assigned to
start the watch for you!
Paul DS.
(i) Record the power rating of the heater element;
(ii) Measure and record the temperature of 500 mL of water and pour it
into a kettle;
(iii) Switch on the power supply to the kettle and start timing how long
it takes the kettle to bring the water to boil;
(iv) Switch off the power supply when the water boils, and record the
time it took for the water to come to boil;
To calculate the efficiency of the kettle you need to find how much
energy the water absorbed to bring it to boiling point.
Use the formula
Q = mc (Tf - Ti)
where
m = mass of water
c = specific heat of water
Tf = final temperature
Ti = initial temperature
Then divide this value by the time it took to bring the water to boiling
and you get the power consumed in boiling the water. Finally you divide
this value by the power rating of the element to give the efficiency of
the kettle.
The following example is based on a kitchen kettle with an element
rating of 2,200 watts:
m = 0.5 kg, c = 4186 J/kg oC, Tf = 100oC, Ti = 22oC.
So Q = 0.5 x 4186 x (100 - 22) = 163,254 Joules
The time taken to bring the water to boil was 94 seconds.
Therefore the power consumed to boil the water = 163,254 / 94 = 1,737 Watts
To find the efficiency of the kettle divide the power used to boil the
water by the power output of the element and multiply by 100 to give a
percentage value, i.e. (1,737 / 2,200) x 100 = 79%
The efficiency of the kitchen kettle is 79%, or 21% of the power output
is wasted.
"Dave W" <dave...@yahoo.co.uk> wrote in message
news:1177076809....@q75g2000hsh.googlegroups.com...
Also, I haven't got a suitable thermometer ...
"Rob Horton" <yahoo@mr_horton.com> wrote in message
news:q4ydnXQtLqzpebXb...@pipex.net...
>
> It would be interesting to find their efficiencies as well. Here is a
> useful guide [...]
The one that used most energy (the RH) had made a cup of tea about 15
minutes before, though I let the tap run for a while (to get ground
temperature water) and flushed both several times beforehand to get rid of
loose limescale and to try and get them to the same starting temperature.
> How accurately did you measure your 1.08 litres? (seems an
> odd amount to use!)
A juice bottle brim full, I'm not sure exactly about it being 1.08L, but I
am sure to an acceptable accuracy that both kettles had the same amount of
water.
> Did you measure the *actual* electrical energy consumed, or did you simply
> base it on the nominal kW ratings of the kettles and the time taken?
The latter.
In other words, I bet there was much more residual heat in the newer element
and its enclosure after you'd switched it off, hence the lower apparent
efficiency.
Some of this will be due to the higher power input raising it to a higher
temperature; some (most, I think) will be due to the relatively poor
transport of heat energy into the water.
Steve
>> Did you measure the *actual* electrical energy consumed, or did you simply
>> base it on the nominal kW ratings of the kettles and the time taken?
>
>The latter.
>
The stated nominal KW rating could be out by +/- 10%, or more.
The manufactures make these kettles for more than one market so may take
the nominal mains voltage in the country of manufacture and this will
affect the nominal power they declare.
--
Alan
news2006 {at} amac {dot} f2s {dot} com
One element may have been 'printed' as a method of manufacture and be
purely resistive and the other element may be wound (a coil) and be
inductive. How did you measure the power?
The inductance of the coiled element will have neglegible effect at 50Hz
and can safely be ignored.
Peter
> The following example is based on a kitchen kettle with an element
> rating of 2,200 watts:
> m = 0.5 kg, c = 4186 J/kg oC, Tf = 100oC, Ti = 22oC.
> So Q = 0.5 x 4186 x (100 - 22) = 163,254 Joules
> The time taken to bring the water to boil was 94 seconds.
> Therefore the power consumed to boil the water = 163,254 / 94 = 1,737 Watts
>
> To find the efficiency of the kettle divide the power used to boil the
> water by the power output of the element and multiply by 100 to give a
> percentage value, i.e. (1,737 / 2,200) x 100 = 79%
> The efficiency of the kitchen kettle is 79%, or 21% of the power output
> is wasted.
Then try the experiment with the same amount of water but using a
microwave oven. I'd expect the heating process to be more efficient, but
the efficiency of the magnetron would be lower.
--
Immunity is better than innoculation.
Peter
> The following example is based on a kitchen kettle with an element
> rating of 2,200 watts:
> m = 0.5 kg, c = 4186 J/kg oC, Tf = 100oC, Ti = 22oC.
> So Q = 0.5 x 4186 x (100 - 22) = 163,254 Joules
Assuming the boiling point of water is exactly 100 degrees C. It is rather
dependant on atmospheric pressure (weather and height above sea level) at
the time.
> To find the efficiency of the kettle divide the power used to boil the
> water by the power output of the element and multiply by 100 to give a
> percentage value, i.e. (1,737 / 2,200) x 100 = 79%
The output power of the element is not constant and certainly is very
unlikely to be what's written on the label.
The resistance of the element changes with temperature (it is lower when
colder) and there is no guarantee what the supply voltage is at any point
in time.
So your calculation of efficiency is liable to be somewhat erroneous.
The real question is... is it more interesting than what is on the telly?
Obviously it would seem so.
>>I think that may be the clue - concealed element versus exposed
>>element. The exposed element probably loses less heat to the outside
>>world than the concealed element does.
>
> One element may have been 'printed' as a method of manufacture and be
> purely resistive and the other element may be wound (a coil) and be
> inductive.
So what? Reactive power doesn't have any heating effect. One kind of
assumes the power figure the manufacturer prints on the label assumes a
power factor of 1.
> How did you measure the power?
He didn't. He guessed, which makes the whole discussion somewhat moot.
> If the two kettles were 'perfectly' insulated and were 'perfect' convertors
> of electrical energy into heat
Where else do the energy go if not into heat?
>The inductance of the coiled element will have neglegible effect at 50Hz
>and can safely be ignored.
It depends on the measurment method. Some meters can be
extremely inaccurate for inductive loads.
Nick.
What's the inductance of a coiled element? A few mH, I'd expect.
Peter
>I ran the experiment after reading this (my current kettle is leaking and
>requires replacement):
> http://tinyurl.com/2r9u35
>... standing in for ...
> http://www.russellhobbs.com/buyingguides/kettles/
>
>"2Kw is energy saving as it uses less power to boil the kettle."
This would appear to have been written by advertising copy writers who
don't know the difference between power and energy. Probably the same
ones that try to flog hi-fi amplifiers by quoting "RMS Watts".
As it stands, it's gobbledygook - are they trying to say that a 2kW
kettle uses less *energy* to boil the same amount of water, and if so,
less than what? Of course a 2kW appliance could be said to be
"energy-saving" compared with a higher powered one because it will use
energy at a lower rate - that's what a lower power means - and will
use less of it during a given amount of time, but it wouldn't surprise
me if the writer simply didn't realise that the constant factor is the
amount of water, and a lower powered kettle would have to run for
longer to heat it to boiling point.
Rod.
Usually based on 230V, as that is the harmonised voltage for the EU
(plus many other countries who were 220-240v also harmonised), though in
reality in the UK it would still be around 240v, the voltage in my house
it 234v.
--
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Mobile Offers: www.southeastbirmingham.co.uk/mob
Win Amazon Vouchers:
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"Roderick Stewart" <escap...@removethisbit.beeb.net> wrote in message
news:0rgi231uh9d70v976...@4ax.com...
As no-one has proposed otherwise, it seems that those knowledgeable about
Physics here agree with my basic suggestion that, <emphasis on> all other
things being equal <emphasis off>, due to greater heat losses over the
longer boiling time, a low-powered kettle should take more energy overall to
heat a given amount of water than a high-powered one, and that the failure
of my admittedly limited experiment to prove this is due to other
differences between the two kettles, most probably the element design.
Wrt to the latter, I've been casting an eye over various models in various
shops, and didn't find a single one with the putatively more efficient
exposed element, seemingly they're all now with the putatively less
efficient concealed element!
In these days of concerns about global warming, such is 'progress' ...
"Java Jive" <ja...@evij.com> wrote in message
news:Pfydnf0vuLaLNbXb...@eclipse.net.uk...
>
> The table below represents a calculation of the energy used by two kettles
> to boil the same amount, 1.08L, of water [...]
Just in case there's any doubt that a low-powered kettle DOES use more
energy, you can apply the principle of 'reductio ad absurdum'.
Below boiling point, the rate of heat loss of the kettle is proportional
to its temperature above the surroundings. The hotter it is, the faster
will be the heat loss. Imagine the situation where the heat input
(approximately constant, being supplied by the electricity) balances the
rising heat loss. If the water has not yet boiled, it never will. But it
will keep on using energy until you switch it off.
Of course, it might have nothing to do with the laws of physics. It may
be simply that you are watching it.
>
>
>
Ian.
--
> the voltage in my house it 234v.
It might have been at the time you read it, but it can vary wildly.
Here is what mine did yesterday:
http://home.clara.co.uk/orac/mains/apc0420.gif
and then you can get horrible things like:
http://home.clara.co.uk/orac/mains/apc0129.gif
A neater, more elegant way of saying what I was trying to say with the
camping stove example.
> Of course, it might have nothing to do with the laws of physics. It may
> be simply that you are watching it.
:-)
I'm only guessing, but I think you must have a decent UPS to hand (if
not in line with the mains feed to your PC)?
--
Regards, John.
Please remove the "ohggcyht" before replying.
The address has been munged to reject Spam-bots.
>>> the voltage in my house it 234v.
>
>> It might have been at the time you read it, but it can vary wildly.
>> Here is what mine did yesterday:
>> http://home.clara.co.uk/orac/mains/apc0420.gif
>> and then you can get horrible things like:
>> http://home.clara.co.uk/orac/mains/apc0129.gif
>
> I'm only guessing, but I think you must have a decent UPS to hand (if
> not in line with the mains feed to your PC)?
The clue's in the file name :-) An APC Smart-UPS 700iNET. Saved the day many
a time from the best that Western Power Distribution can serve up, which is
not always very good.
I think it may read a bit high overall, but the relative change is probably
reasonably accurate (I fiddled with the calibration constants a while back
in the course of experimentation with the control protocol and I'm not sure
what they all were originally, so I had to guess a bit :-) ).
>Wrt to the latter, I've been casting an eye over various models in various
>shops, and didn't find a single one with the putatively more efficient
>exposed element, seemingly they're all now with the putatively less
>efficient concealed element!
>
>In these days of concerns about global warming, such is 'progress' ...
It makes you wonder how much of the rest of what we are told about
global warming, maximising efficiency, saving the planet etc is
actually true, and how much is simply being repeated by those who
don't understand it and daren't question it because it is the
prevailing dogma and they wouldn't understand the answer anyway.
Rod.
Saving energy means boiling only the water you want. With the
traditional exposed element most people boiled 2 or 3 times the water
they needed for one cup of tea/coffee.
Very good point. Most concealed element kettles will safely boil just a
cupful of water, whereas exposed elements need to be fully covered.
Steve
>>It makes you wonder how much of the rest of what we are told about
>>global warming, maximising efficiency, saving the planet etc is
>>actually true,
>
>Saving energy means boiling only the water you want. With the
>traditional exposed element most people boiled 2 or 3 times the water
>they needed for one cup of tea/coffee.
True, but the original posting referred to an advertisement that
appeared to be claiming that a 2kW kettle was more energy efficient
(more efficient than what is not stated), the implication being that
it is the fact that it is a 2kW one that makes it more efficient,
which I think we've all agreed is nonsense. The shape of the element
will of course have an effect, but I saw no mention of this in the
posting or the advertisement.
Rod.
The instruction's for the RH state a min of 1 cupful, but how big is a cup?
The mugs I use are about 300mL. There's no minimum mark on the graduated
window, but the very bottom of it represents 200mL. Shall we compromise at
250mL? That's the amount represented by the minimum mark on the MR.
And I'm pretty sure that I've seen minimum figures for concealed elements
higher than all the above, but can't now find them online, so maybe it was
in a store.
I guess that's a big advantage of the columnar design. The disadvantage
might be that with a more average amount of water in it, the surface area of
water touching the walls of the kettle might be greater than with a design
that approximated more closely to a sphere. But there again, quite a lot of
heat is lost from the surface as steam, and a spherical design would have a
greater surface area when half full.
I am not sufficiently expert in this field to say which design would be
better for an average family household, but I suspect the columnar design is
preferable for those like myself who are usually just making one or two cups
at a time.
I have now also discovered these, which are more encouraging ...
http://tinyurl.com/2vw7x4
... standing in for ...
http://www.russellhobbs.com/products/index.aspx?path=/products/kettles/energy_saving_thermal_kettle
... and ...
http://tinyurl.com/2bc5gu
... standing in for ...
"Alan" <junk_...@amac.f2s.com> wrote in message
news:xSLv4OAt...@amac.f2s.com...
The element powers are probably only approximate and will change with
temperature as teh kettle heats up.
"Java Jive" <ja...@evij.com> wrote in message
news:Pfydnf0vuLaLNbXb...@eclipse.net.uk...
> As others have seen fit to go rabidly off-topic about electricity
> consumption here, here's my ha'p'worth ...
>
> The table below represents a calculation of the energy used by two kettles
> to boil the same amount, 1.08L, of water. (If your newsreader doesn't
> space
> it correctly, copy'n'paste it into something that uses a fixed width font,
> eg: Notepad)
>
> Start Finish Time Error kW Units Un Err Min Max
> RH 00:15 02:55 02:40 00:02 2.750 0.1222 0.0015 0.1207 0.1238
> MR 00:15 03:40 03:25 00:02 2.025 0.1153 0.0011 0.1142 0.1164
>
> If the two kettles were 'perfectly' insulated and were 'perfect'
> convertors
> of electrical energy into heat, the units used should be the same, but
> they
> are not.
>
> Discuss
>
> (You should take no more than 15 minutes to complete this question)
>
> ... but actually, I've been pondering it longer than that, and am somewhat
> baffled.
>
> I ran the experiment after reading this (my current kettle is leaking and
> requires replacement):
> http://tinyurl.com/2r9u35
> ... standing in for ...
> http://www.russellhobbs.com/buyingguides/kettles/
>
> "2Kw is energy saving as it uses less power to boil the kettle."
>
> As indicated in the mock exam question, if all the heat from the
> electricity
> went into the water, without any being wasted in heating up the
> environment
> through the walls of the kettle, then they should use the same amount of
> electricity to boil the same amount of water.
>
> Further, where, as in real life, heat actually is wasted through the walls
> of the kettle, then the rate of heat loss at any given moment is
> determined
> by the temperature difference between inside and outside the kettle (the
> temperature gradient), and the total lost will therefore be determined
> partly by the amount of time the kettle takes to boil - the longer it
> takes, the longer it will spend at temperatures where the temperature
> gradient is significant, and therefore the more heat should be lost
> overall.
>
> And this line of reasoning is born out by my experiences camping, trying
> to
> boil water with one of those small canister gas stoves. On a cold windy
> day, heat can be lost at such a rate that although the kettle quite
> quickly
> gets hot, it never actually boils until you move it behind some windshield
> like a big tree.
>
> Yet the experiment shows, contrary to what I might have reasoned, that the
> RH statement is in fact correct, and I can't explain it ...
>
> There aren't many significant differences between the kettles that I can
> see, apart from the power rating and element design - the latter might
> go
> some way to explaining things. The older, lower-powered Morphy Richards
> one
> has the 'traditional' exposed element, whereas the newer. higher-powered
> Russell Hobbs has a concealed element, and perhaps there is greater
> wastage
> of heat into the environment with that design.
>
> Other than that, they are both columnar in shape (so would probably have a
> similar ratio of mass of water to heat against surface area to lose heat
> through), both are plastic, though different looking plastic (and both
> leak).
>
>