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.
"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).
Java Jive wrote: > 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.
Java Jive wrote: > 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?
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.
> If you waited for the auto shut-off to operate [...]
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 ...
> 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.
> 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 [...]
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).
But you're not just heating the water - you're also heating the kettle. What is the thermal capacity of the two kettles? The one that took longer is probably much heavier.
> 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.
> "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).
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!
> 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
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.
> 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 ...
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!
Java Jive wrote: > 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.
> "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).
It would be interesting to find their efficiencies as well. Here is a useful guide
(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.
I don't think so, that's why I mentioned that they were of similar design and materials. Otherwise I would expect one that was nearer to spherical in shape to lose less heat to the environment than a columnar one, and one made of metal to lose more than one made of plastic.
> But you're not just heating the water - you're also heating the > kettle. What is the thermal capacity of the two kettles? The one that > took longer is probably much heavier.
If I could have done all that with otherwise identical kettles, I would have, (nice little O-level Physics experiment) but the kettles are a different design of element to start with, so there's no way that I could ever compare like with like with the only difference being the power rating, so any greater accuracy gained would have been insignificant compared with the variation between the kettles.
Also, I haven't got a suitable thermometer ...
"Rob Horton" <yahoo@mr_horton.com> wrote in message
> 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?
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?
I think the difference is probably explained by the poorer 'coupling' of heat from the element into the water in the newer model. It must be difficult to improve on the coupling from the old, exposed element, surrounded as it is on all sides by water which is in intimate contact with it.
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.
In message <GYadnUTNxINdbbXbnZ2dnUVZ8qXin...@eclipse.net.uk>, Java Jive <j...@evij.com> wrote
>> 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
>> 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
>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. How did you measure the power? -- Alan news2006 {at} amac {dot} f2s {dot} com
Alan <junk_re...@amac.f2s.com> wrote: > In message <4628d022$0$514$bed64...@news.gradwell.net>, > tinn...@isbd.co.uk wrote > >Java Jive <j...@evij.com> wrote:
> >> 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
> >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. How did you measure the power?
The inductance of the coiled element will have neglegible effect at 50Hz and can safely be ignored.
Rob Horton <yahoo@mr_horton.com> wrote: > 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.
On Fri, 20 Apr 2007 17:04:11 +0100, Rob Horton <yahoo@mr_horton.com> wrote: > 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.
On Fri, 20 Apr 2007 18:53:33 +0100, Alan <junk_re...@amac.f2s.com> wrote: >>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.
On Fri, 20 Apr 2007 12:38:03 +0100, Java Jive <j...@evij.com> wrote: > If the two kettles were 'perfectly' insulated and were 'perfect' convertors > of electrical energy into heat
