Yes, in a perfect room, 2KW is 2KW and any 2KW electric heater will heat the room equally.
In the real world though, a fan heater may prove more costly since the warmed air is blown around with some force and is therefore more likely to escape from the space that one wishes to heat.
If a cold room is to be quickly heated, and then kept warm with electric heaters, then it is probably best to bring the room up to temperature with a substantial loading of fan heaters, 6KW is not excessive for a domestic living room.
This should take only a few minutes for an average size room, after which the fan heaters should be turned off and a suitably sized electric radiator used.
Electric heating tends to be expensive, but does have the merit of being quick acting, given a large enough loading.
A large room equiped with say a 5KW woodstove might take hours to become comfortably warmed, remembering that the woodstove might take 30 minutes to produce noticable heat, an hour or more to reach 5KW, and several hours more for the 5KW to fully warm the room.
15KW of electric heat would produce noticable warmth in ONE MINUTE and probably fully warm the room in perhaps 10 minutes.
The 15KW of electric heat would be very costly if run continually, but this would not be needed. If a large room normally heated by a woodstove was to be heated at short notice, then 15KW for 10 minutes, then 10KW for the next 10 minutes, and 5KW for the last 10 minutes would warm the room until the stove was burning well.
Total electricty use would be 5KWH or less than £1 which might be well woth it.
Electric heat is often the best option for short term use, as in temporary or emergency living spaces, or for daily use if for less than an hour a day, or for long hour use if the average load is less than 1KW.
In most circumstances electric heating can not be recomended for large or long hour heating demands, gas, wood, coal, or even oil normally being cheaper for such demands.
Bullshit/snake oil alert re electric heaters
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Little John
So, if I’ve got my understanding of thermodynamics right, this would mean the following two scenarios would produce exactly the same amount of heat.;
1) A fully thermally sealed room has a lead acid battery in it with a known amount of stored electrical energy in it. The battery is connected to a small motor that is turning and is using the electrical energy of the battery. Eventually the battery runs out of energy and the motor stops turning. At the point at which the temperature of the motor matches that of the ambient temperature of the room a measurement is taken of the total thermal energy contained in the room.
2) A fully thermally sealed room has a lead acid battery in it with a known amount of stored electrical energy in it. The battery is connected to an electric resistance heater that is radiating heat and is using the electrical energy of the battery. Eventually the battery runs out of energy and the heater stops radiating heat. At the point at which the temperature of the heater matches that of the ambient temperature of the room a measurement is taken of the total thermal energy contained in the room.
In both cases, the total thermal energy contained in the room will be identical.
So, to take a real-world equivalent, if someone lives in a house that is centrally heated electrically, they may as well leave all of their electrical household appliances on standby during the evening since the heat (and cost) generated from them being on standby will be directly offset by reduced heating demands (and cost) on their thermostatically controlled electrical central heating system. Obviously, I am ignoring the fact that storage heaters consume their energy off peak. But, the principle stands even if the practice does not.
Also, the above principle breaks down in practice, presumably, when the electrical appliances are generating light and noise. The reason being that light and noise can both more easily escape the confines of a room than other forms of energy (through glass windows and by vibration through walls).
However, going back to my two sealed room scenarios, if the battery were connected to a CD player playing loud music; so long as the room was perfectly acoustically insulated, the noise (in the form of vibrating air molecules) would also eventually turn to heat.
1) A fully thermally sealed room has a lead acid battery in it with a known amount of stored electrical energy in it. The battery is connected to a small motor that is turning and is using the electrical energy of the battery. Eventually the battery runs out of energy and the motor stops turning. At the point at which the temperature of the motor matches that of the ambient temperature of the room a measurement is taken of the total thermal energy contained in the room.
2) A fully thermally sealed room has a lead acid battery in it with a known amount of stored electrical energy in it. The battery is connected to an electric resistance heater that is radiating heat and is using the electrical energy of the battery. Eventually the battery runs out of energy and the heater stops radiating heat. At the point at which the temperature of the heater matches that of the ambient temperature of the room a measurement is taken of the total thermal energy contained in the room.
In both cases, the total thermal energy contained in the room will be identical.
So, to take a real-world equivalent, if someone lives in a house that is centrally heated electrically, they may as well leave all of their electrical household appliances on standby during the evening since the heat (and cost) generated from them being on standby will be directly offset by reduced heating demands (and cost) on their thermostatically controlled electrical central heating system. Obviously, I am ignoring the fact that storage heaters consume their energy off peak. But, the principle stands even if the practice does not.
Also, the above principle breaks down in practice, presumably, when the electrical appliances are generating light and noise. The reason being that light and noise can both more easily escape the confines of a room than other forms of energy (through glass windows and by vibration through walls).
However, going back to my two sealed room scenarios, if the battery were connected to a CD player playing loud music; so long as the room was perfectly acoustically insulated, the noise (in the form of vibrating air molecules) would also eventually turn to heat.
Last edited by Little John on 13 Jun 2013, 23:21, edited 1 time in total.
Are we talking about heating the room, or the people in it? As well as my nearby halogen heater, that heated me quite a lot, and made the rest of the room bearable to walk around, I used a fan heater in my workshop, standing on the bench. It kept the bits of me that I needed while working warm. I suppose turning an electric blanket into clothes would be an even more efficient option, but a bit dangerous!
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UndercoverElephant
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Yes, I think. Although it only works in the winter, when the heating is on. In the summer, leaving the appliances on standby really does just waste energy/money. Even worse is the situation where the problem is heat rather than cold, and money is being spent on air conditioning while the appliances are left on standby.stevecook172001 wrote:So, if I’ve got my understanding of thermodynamics right, this would mean the following two scenarios would produce exactly the same amount of heat.;
1) A fully thermally sealed room has a lead acid battery in it with a known amount of stored electrical energy in it. The battery is connected to a small motor that is turning and is using the electrical energy of the battery. Eventually the battery runs out of energy and the motor stops turning. At the point at which the temperature of the motor matches that of the ambient temperature of the room a measurement is taken of the total thermal energy contained in the room.
2) A fully thermally sealed room has a lead acid battery in it with a known amount of stored electrical energy in it. The battery is connected to an electric resistance heater that is radiating heat and is using the electrical energy of the battery. Eventually the battery runs out of energy and the heater stops radiating heat. At the point at which the temperature of the heater matches that of the ambient temperature of the room a measurement is taken of the total thermal energy contained in the room.
In both cases, the total thermal energy contained in the room will be identical.
So, to take a real-world equivalent, if someone lives in a house that is centrally heated electrically, they may as well leave all of their electrical household appliances on standby during the evening since the heat (and cost) generated from them being on standby will be directly offset by reduced heating demands (and cost) on their thermostatically controlled electrical central heating system. Obviously, I am ignoring the fact that storage heaters consume their energy off peak. But, the principle stands even if the practice does not.
Yes, although I suspect the amount of energy escaping as light and noise is usually minimal. It would have to be very bright light, or noise on the scale of a serious loudspeaker.Also, the above principle breaks down in practice, presumably, when the electrical appliances are generating light and noise. The reason being that light and noise can both more easily escape the confines of a room than other forms of energy (through glass windows and by vibration through walls).
Yes, I think.However, going back to my two sealed room scenarios, if the battery were connected to a CD player playing loud music; so long as the room was perfectly acoustically insulated, the noise (in the form of vibrating air molecules) would also eventually turn to heat.
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kenneal - lagger
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Going back to the lighting, the problem with using a light bulb to heat your room is that two to three times the amount of heat that the bulb puts into the room is put into the atmosphere and wasted at the power station because of the inefficiency of the electrical generation system. Your gas central heating system will put 90% of the energy it uses into the room while the light bulb only puts about 30%.
So it pays, environmentally and monetarily, to use the most efficient appliance in the most efficient way. Use your central heating for heating the house and the most efficient lighting to light it. This holds for any electrical appliance unless you are generating electricity from PV. Even then the cost, both environmental and monetary, of generating electricity comes into it.
Regarding the heating, the method of distribution of the heat by the various appliances will affect the efficiency. A fan heater will distribute the heated air in a more horizontal direction that a convector heater, thus spreading it better, so a low level fan heater will put the heat into the lower parts of the room before the heated air rises to the ceiling. The convector will heat the air adjacent to it which will then rise to the ceiling. The convector then has to heat the whole volume of air in the room before it gets to your feet.
While that comparatively hot convected air is at the ceiling there will be a much higher heat loss out of the room than with the air warmed by the fan heater which starts off a lower temperature and loses much of its heat to its surroundings while still at a low level. The fan heater has the disadvantage of rapidly moving the air so the recipient of the heat literally suffers from wind chill and requires a higher temperature to keep warm. While a convector moves the air more slowly there is still a wind chill effect at work, mainly at foot level.
In a well insulated room a radiant fire would be most effective as it doesn't move the air in the room. The insulation of the room would enable the inner surface of the wall to warm quickly and reradiate the radiation from the fire around the room warming the contents, including the occupant. If the insulation were not good the occupant would have to point the heater at himself to gain heat as any radiation from the fire hitting a cold surface would be absorbed by that surface to warm itself.
Again only about 30% of the gross energy used for an electric fire would heat the room whereas 75% or more of the energy that a wood burning stove uses would go to heat the room.
Once again this shows that we should insulate before we think about providing heating to a house so that we only supply the minimum amount of energy to gain the comfort that we need.
The Lagger!!
So it pays, environmentally and monetarily, to use the most efficient appliance in the most efficient way. Use your central heating for heating the house and the most efficient lighting to light it. This holds for any electrical appliance unless you are generating electricity from PV. Even then the cost, both environmental and monetary, of generating electricity comes into it.
Regarding the heating, the method of distribution of the heat by the various appliances will affect the efficiency. A fan heater will distribute the heated air in a more horizontal direction that a convector heater, thus spreading it better, so a low level fan heater will put the heat into the lower parts of the room before the heated air rises to the ceiling. The convector will heat the air adjacent to it which will then rise to the ceiling. The convector then has to heat the whole volume of air in the room before it gets to your feet.
While that comparatively hot convected air is at the ceiling there will be a much higher heat loss out of the room than with the air warmed by the fan heater which starts off a lower temperature and loses much of its heat to its surroundings while still at a low level. The fan heater has the disadvantage of rapidly moving the air so the recipient of the heat literally suffers from wind chill and requires a higher temperature to keep warm. While a convector moves the air more slowly there is still a wind chill effect at work, mainly at foot level.
In a well insulated room a radiant fire would be most effective as it doesn't move the air in the room. The insulation of the room would enable the inner surface of the wall to warm quickly and reradiate the radiation from the fire around the room warming the contents, including the occupant. If the insulation were not good the occupant would have to point the heater at himself to gain heat as any radiation from the fire hitting a cold surface would be absorbed by that surface to warm itself.
Again only about 30% of the gross energy used for an electric fire would heat the room whereas 75% or more of the energy that a wood burning stove uses would go to heat the room.
Once again this shows that we should insulate before we think about providing heating to a house so that we only supply the minimum amount of energy to gain the comfort that we need.
The Lagger!!
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woodburner
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The fan heater may prove more costly to run as the moving air reduces the boundary layer at the walls. This means the walls will get warmer than with a convector, but I doubt this effect will be as great as the self deception people indulge in to convince themselves what ever they believe is the reality.adam2 wrote:Yes, in a perfect room, 2KW is 2KW and any 2KW electric heater will heat the room equally.
In the real world though, a fan heater may prove more costly since the warmed air is blown around with some force and is therefore more likely to escape from the space that one wishes to heat.
If a cold room is to be quickly heated, and then kept warm with electric heaters, then it is probably best to bring the room up to temperature with a substantial loading of fan heaters, 6KW is not excessive for a domestic living room.
This should take only a few minutes for an average size room, after which the fan heaters should be turned off and a suitably sized electric radiator used.
Electric heating tends to be expensive, but does have the merit of being quick acting, given a large enough loading.
A large room equiped with say a 5KW woodstove might take hours to become comfortably warmed, remembering that the woodstove might take 30 minutes to produce noticable heat, an hour or more to reach 5KW, and several hours more for the 5KW to fully warm the room.
15KW of electric heat would produce noticable warmth in ONE MINUTE and probably fully warm the room in perhaps 10 minutes.
The 15KW of electric heat would be very costly if run continually, but this would not be needed. If a large room normally heated by a woodstove was to be heated at short notice, then 15KW for 10 minutes, then 10KW for the next 10 minutes, and 5KW for the last 10 minutes would warm the room until the stove was burning well.
Total electricty use would be 5KWH or less than £1 which might be well woth it.
Electric heat is often the best option for short term use, as in temporary or emergency living spaces, or for daily use if for less than an hour a day, or for long hour use if the average load is less than 1KW.
In most circumstances electric heating can not be recomended for large or long hour heating demands, gas, wood, coal, or even oil normally being cheaper for such demands.
We have a woodburner for heating, and like JohnB, we can feel warm when the air temperature is low, just from the radiant heat. This changes a lot when we have a cold north-easterly and it blows through the gaps, but we need ventilation.
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adam2
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Yesstevecook172001 wrote:So, if I’ve got my understanding of thermodynamics right, this would mean the following two scenarios would produce exactly the same amount of heat.;
1) A fully thermally sealed room has a lead acid battery in it with a known amount of stored electrical energy in it. The battery is connected to a small motor that is turning and is using the electrical energy of the battery. Eventually the battery runs out of energy and the motor stops turning. At the point at which the temperature of the motor matches that of the ambient temperature of the room a measurement is taken of the total thermal energy contained in the room.
2) A fully thermally sealed room has a lead acid battery in it with a known amount of stored electrical energy in it. The battery is connected to an electric resistance heater that is radiating heat and is using the electrical energy of the battery. Eventually the battery runs out of energy and the heater stops radiating heat. At the point at which the temperature of the heater matches that of the ambient temperature of the room a measurement is taken of the total thermal energy contained in the room.
In both cases, the total thermal energy contained in the room will be identical.
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adam2
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In a space heated with peak rate electricity, then there is nothing to be gained by not leaving a TV for example on standby.
The energy "wasted" by the TV will appear as heat and reduce the electricity used by the heating system.
However only a minority of houses are heated by peak rate electricity, and then probably not 24/7, and certainly not heated all year. There is therefore a saving by not leaving appliances on or on standby needlessly whenever the heating is not required.
Energy wasted by low efficiency lighting still tends to add to costs even in an electricly heated room. Lamps are normally high up and the wasted energy makes the ceiling hotter. No ceiling is perfectly insulated and therefore some of the energy wasted by the lamps is gone forever rather than warming the lower part of the room in which persons sit or stand.
Energy wasted by a low efficiency lamp at low level, in an electricly heated room is not truly wasted as it does help warm the room and reduce the energy used by the electric heater.
I would still avoid such lamps though as they are undeniably a waste at times when the heating is not needed.
Recessed halogen downlights waste energy even when turned off ! due to heated air escaping via the holes in which they are fitted.
The energy "wasted" by the TV will appear as heat and reduce the electricity used by the heating system.
However only a minority of houses are heated by peak rate electricity, and then probably not 24/7, and certainly not heated all year. There is therefore a saving by not leaving appliances on or on standby needlessly whenever the heating is not required.
Energy wasted by low efficiency lighting still tends to add to costs even in an electricly heated room. Lamps are normally high up and the wasted energy makes the ceiling hotter. No ceiling is perfectly insulated and therefore some of the energy wasted by the lamps is gone forever rather than warming the lower part of the room in which persons sit or stand.
Energy wasted by a low efficiency lamp at low level, in an electricly heated room is not truly wasted as it does help warm the room and reduce the energy used by the electric heater.
I would still avoid such lamps though as they are undeniably a waste at times when the heating is not needed.
Recessed halogen downlights waste energy even when turned off ! due to heated air escaping via the holes in which they are fitted.
"Installers and owners of emergency diesels must assume that they will have to run for a week or more"
Also you don't want to be using electricity to heat your home if you live in a gas heated home (as more than 80% of the UK does) because it's much more expensive and carbon intensive.JohnB wrote:The problem with heat from lighting is that as hot air rises it stays above head height close to the ceiling. Draughts tend to come in close to the floor, so lights don't heat the parts of the room where the people are. I found this when I lived in my van, with a ceiling that just touches my head. I could be sitting down feeling cold, but when I stood up my head would be hot and my feet cold!
I had a temperature gauge that showed the temperature at seat height and at shoulder height when standing, and there was usually a big difference when I had any heating on.
Any savings you see for energy saving bulbs should take the heat replacement effect into account and be the net saving.
ETA: if I'd read to the end of the thread I would see that Adam covers off everything nicely as usual
Interesting. I generally advise people in smaller, better insulated rooms who need a secondary heating source to go with an oil filled convector because it's worth trying to warm the air in the room. For larger, badly insulated rooms I suggest radiant heaters because there's no point trying to warm the air, just warm yourself.kenneal - lagger wrote:In a well insulated room a radiant fire would be most effective as it doesn't move the air in the room. The insulation of the room would enable the inner surface of the wall to warm quickly and reradiate the radiation from the fire around the room warming the contents, including the occupant. If the insulation were not good the occupant would have to point the heater at himself to gain heat as any radiation from the fire hitting a cold surface would be absorbed by that surface to warm itself.
Again only about 30% of the gross energy used for an electric fire would heat the room whereas 75% or more of the energy that a wood burning stove uses would go to heat the room.
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mr brightside
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I'm used to considering all losses in electric motors as 'dead losses', due to them being lost at the shaft; which is the bit i'm concerned with on a day to day basis. I'd like to comment on your thought experiment if i may...stevecook172001 wrote:The point is that all the energy gets lost to heat, in the end. Even the energy that was not initially lost to heat because it turned a motor. Even that energy ends up as heat in the room. It can't simply disappear.
I don't know anything about Thermodynamics any more, and what i did know was mostly related to Steam Tables and left my adolescent mind about 3mins after passing the exam; so i won't be commenting on the subject or attempting to re-explain it to you!stevecook172001 wrote:So, if I’ve got my understanding of thermodynamics right, this would mean the following two scenarios would produce exactly the same amount of heat.;
1) A fully thermally sealed room has a lead acid battery in it with a known amount of stored electrical energy in it. The battery is connected to a small motor that is turning and is using the electrical energy of the battery. Eventually the battery runs out of energy and the motor stops turning. At the point at which the temperature of the motor matches that of the ambient temperature of the room a measurement is taken of the total thermal energy contained in the room.
2) A fully thermally sealed room has a lead acid battery in it with a known amount of stored electrical energy in it. The battery is connected to an electric resistance heater that is radiating heat and is using the electrical energy of the battery. Eventually the battery runs out of energy and the heater stops radiating heat. At the point at which the temperature of the heater matches that of the ambient temperature of the room a measurement is taken of the total thermal energy contained in the room.
I can't get my head around the two scenarios resulting in the same temperature yeild. Are you saying that motor noise, housing vibration, and forces generated by it and acting on the floor in startup and, to a lesser extent, during running (as per Newton's 3rd Law) all end up as heat? For example, if you put load cells under the feet you'd find that one side was pressing down harder than the other.
If that is what you are saying then don't take my post as an insult, i accept your argument. Having said that though if we could set the thought experiment up practically, i'd bet you a pint of premium Pale Ale that the motor would yeild less heat; even if it's only 0.2% or something.
Persistence of habitat, is the fundamental basis of persistence of a species.
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Little John
In practical or theoretical terms, assuming a precisely equivalent consumption of electricity, the motor simply has to give up the same heat as the radiator, eventually. However, I do accept that the way in which the motor might give up its heat may not be as desirable as the way the heater gives it up. In other words, given that the heater is designed to give up its heat in the most efficient way possible, we can assume that the motor will probably hold onto its heat in its body parts for longer and so will give up it's heat less quickly. However, that being said, it will still eventually give up that heat. That is to say, all things, including temperature gradients, tend towards equilibrium and, eventually, the temperature of the heater and/or the motor will match that of the environment in which they reside.mr brightside wrote:I'm used to considering all losses in electric motors as 'dead losses', due to them being lost at the shaft; which is the bit i'm concerned with on a day to day basis. I'd like to comment on your thought experiment if i may...stevecook172001 wrote:The point is that all the energy gets lost to heat, in the end. Even the energy that was not initially lost to heat because it turned a motor. Even that energy ends up as heat in the room. It can't simply disappear.
I don't know anything about Thermodynamics any more, and what i did know was mostly related to Steam Tables and left my adolescent mind about 3mins after passing the exam; so i won't be commenting on the subject or attempting to re-explain it to you!stevecook172001 wrote:So, if I’ve got my understanding of thermodynamics right, this would mean the following two scenarios would produce exactly the same amount of heat.;
1) A fully thermally sealed room has a lead acid battery in it with a known amount of stored electrical energy in it. The battery is connected to a small motor that is turning and is using the electrical energy of the battery. Eventually the battery runs out of energy and the motor stops turning. At the point at which the temperature of the motor matches that of the ambient temperature of the room a measurement is taken of the total thermal energy contained in the room.
2) A fully thermally sealed room has a lead acid battery in it with a known amount of stored electrical energy in it. The battery is connected to an electric resistance heater that is radiating heat and is using the electrical energy of the battery. Eventually the battery runs out of energy and the heater stops radiating heat. At the point at which the temperature of the heater matches that of the ambient temperature of the room a measurement is taken of the total thermal energy contained in the room.
I can't get my head around the two scenarios resulting in the same temperature yeild. Are you saying that motor noise, housing vibration, and forces generated by it and acting on the floor in startup and, to a lesser extent, during running (as per Newton's 3rd Law) all end up as heat? For example, if you put load cells under the feet you'd find that one side was pressing down harder than the other.
If that is what you are saying then don't take my post as an insult, i accept your argument. Having said that though if we could set the thought experiment up practically, i'd bet you a pint of premium Pale Ale that the motor would yeild less heat; even if it's only 0.2% or something.
Where I am happy to qualify my post with a concession is with regards to vibration. Obviously, any vibration escaping from the room as a result of the action of the motor that is not matched by an equal escape of energy from the room containing the heater will count as a real energy loss and so cause a discrepancy in the final thermal measurement between the two rooms.
Last edited by Little John on 15 Jun 2013, 17:14, edited 1 time in total.
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kenneal - lagger
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Little John
Actually, that's something that has always genuinely confused me.kenneal - lagger wrote:Even noise ends up as heat as both are just a vibration of molecules.
If heat is merely extreme vibration of a medium due to it being highly energised, this implies that heat can only be transmitted across space via a medium. In the case of here on earth, that would be via air or water or some such medium.
In which case, how come heat from the sun can travel through the empty mediumless reaches of space and cause me to feel it here on the surface of the earth.