If you go to a LAN party, the room it’s in will be quite warm if you don’t open windows and turn on several fans. Businesses with large server rooms have to install extra cooling systems in them to deal with all the heat.
quote:
Originally posted by El Marko
In reality (and I think in thermodynamics as well) the work extracted by these various appliances in the form of mechanical or electrical energy is not converted efficiently into heat as it is with a heater.
scr4:
I’m sorry but you are wrong. Energy is always conserved. If a 100W light bulb is 5% efficient, that means 5W worth of light is emitted in addition to 95W heat.
El Marko is correct, but you are misinterpreting what he is saying. In the same breath you seem to be claiming that something can be 5% efficient and 100% efficient at the same time.
Energy is ultimately conserved, yes. But the efficency with which one form of energy is converted to another (heat to mechanical, light to mechanical, stored magnetic to electrical, nuclear to chemical) is never 100% efficient. So if you are talking strictly about temperature, then in the real world your 300W computer is never going to heat as efficiently as a 300W heater.
Even if you idealize it to a perfect room where energy never escapes and allowed to achieve a steady state, then again it isn’t necessarily the case that energy outputs between two devices will achieve the same resultant temperature. Energy can simultaneously exist in mulitple states in that room, it isn’t required to ultimately turn into kinetic motion of particles or equilibrium of radiation.
Take for example a 1000W machine that is combining component chemicals (sulfur and such) to make matches. Some of that work done by the machine becomes stored chemical energy bound up in the final matchhead fuel and will never convert to energy used to heat the room unless the match is struck. This device is not 100% efficient in converting that 1000W into thermal energy.
On the plus side, Trigonal, this is really more of a heat transfer question than a thermodynamic question.
I’ve sized refrigeration systems for small computer rooms before, and what works for me is adding up the nameplate wattage numbers on all of the equipment in the room, and then taking 50% of this value as the heat load. The 50% recognizes that the equipment probably won’t be utilizing all of the watts it is rated for, or may not be running all of the time, etc. In real life, I’m not going to get an exact answer, and even if did, I’m still limited to using refrigeration equipment that is sized in 1/2 ton increments at best.
If the cooling of the equipment is critical, then a more conservative approach would be required, but I have not had any problems with this method yet.
Real world, but steady state. If this helps, if a family heats with electric resistance heat can the parents legitimatly complain about little Johny leaving the lamps on due to power consumption? Or put another way, if one is using electric resistance heat, can you just use electricity willy-nilly since that will just heat the room anyway and cost exactly the same (disregarding wear and tear)?
No. A “5% efficient light bulb” is a bulb that turns 5% of the consumed energy into visible light, and the rest into heat. If you consider it an electrical heater, it’s a 100% efficient heater. (assuming no windows)
That’s right, because processes usually produce waste heat. If you look at it as a heater then it’s 100% efficient.
That’s what I said too. If the energy is accumulated in the room as something other than heat, it won’t be 100% efficient. But I think you’ll be hard pressed to come up with any household appliance which, in a steady-state usage, is less than 100% efficient as a heater. The OP mentioned computers, fridge, light bulb, etc. Those do not store energy.
If you believe that an electric space heater isn’t 100% efficient, then your understanding of basic thermo is completely inadequate.
Way too many people who don’t know enough thermo are posting to this thread.
Any device that generates only heat, be it a computer, a refrigerator, etc., is for all practical purposes 100% efficient. Two such devices consuming the same power (actual consumption, not rating) are both generating the same amount of heat.
All of the energy that goes into “losses” like turning motors ends up as waste heat. And heat is heat. It doesn’t matter that it once turned a motor.
In addition, not only is light escaping thru windows not a significant issue, windows are usually a net gain in terms of light. (Just ask my plants.) They are a loss of heat, but then you are losing heat, which is the form of energy under discussion.
Electric heat pumps work better than electric heaters because they use an external source of (low level) heat. I.e., you have escaped the “closed system” limitation of thermo (as far as your house is concerned).
Every last Watt of electrical energy that comes into your house can only leave via a small number of ways. As heat thru walls, windows, drafts, etc. As light thru windows. (A quite small effect.) As vibrational energy (which most houses don’t generally produce to a measurable degree). And you can “carry” energy out. Stuff some energy into something (a hot rock, battery, flywheel, etc.). Take it outside. Again, not usually worth calculating. And then there’s really unworthy-of-mention methods, like fusing atoms and such.
In the particular case of a computer. Everything except the light from LEDs on the front (or other weird stuff case modders use) ends up as heat inside the house. And even the overwhelming majority of LED light gets absorbed by something and turned into heat. The energy stored in capacitors or disk spinning will eventually turn into heat. (Unless you throw the computer out the window quickly while the caps are still charged and disk spinning.)
Repeat: Space heaters, computers, DVD players, etc. are for all practical purposes 100% efficient at converting electricity into heat.
I have come to the conclusion that NO ONE on this forum knows what he/she is talking about.
Move this thread to the IMHO forum.
Discuss nothing simply if it can be confused to the point where it is too complex and wonderful to understand.
Why do you say that? ftg, among several others, knows exactly what he’s talking about.
Why do you say that? ftg, among several others, knows exactly what he’s talking about.
ftg:“Every last Watt of electrical energy that comes into your house can only leave via a small number of ways…”
Yes, and my point being that it isn’t ultimately required to end up as heat. That is the point I feel that many are missing.
If my room has a bowling-alley-pin-racking machine, then it is less than 100% efficient as a heat machine because some of the energy becomes stored as potential energy in the erect bowling pins.
I could have a machine dissociating water molecules, where energy isn’t recoverable unless they again react.
I could have a particle accelerator in my room where some energy is lost as it gets lost in the mass-defects of the higher order particles it builds.
I have a machine that turns a ratcheted wheel with a spring, storing the energy in the spring tension never to be released as heat.
And so on…
Only as long as nothing KNOCKS THE PINS OVER. Even in bowling alleys, that situation only occurs when the lane is shut down for the evening. Common appliances such as toasters, light bulbs and refrigerators don’t store energy. The OP does not mention defibrillators, battery chargers, stellarators or rope-twining machines.
Putting aside linear accelerators, classical mechanics and electrochemical reactions, lets look at plain old incadescent light bulbs and whether or not they are equivalent to electric furnaces.
It depends.
If the light fixture is positioned in such a way as it elevates the temperature of an exterior wall, floor, or ceiling, then it’s not 100% like a furnace(math will follow). Otherwise, ignoring emitted electromagnetic energy, it’s the same.
The math(All English Measures, BTW):
Here in the US home insulation is rated by it thermal resistance, or R rating. For our purposes we’re going to assume that our single room house is 10ft by 10ft by 10ft. Furthermore, the R factor for the walls will be 13, the ceiling 22 and the floor 22 as well. This corresponsed to 3.5 inches of fiberglass insulation in the walls, six inches in the floor and ceiling with dryway and wood flooring added in. Additionally, I’ll assume your parent’s furnace keeps the house at a steady 68F, it’s 32F outside, and the Earth temperature is 50F. Just to cover all the obvious stuff we’ll finally assume an infiltration rate of 0.5 air exchanges per hour. That means 500 cu. ft. of air in the room is replaced with air from outside. I’ve been lead to believe that’s good.
Note on the calculations: R is the thermal resistance. It’s additive, if you put two R11 pieces of 3.5 inch fiberglass insulation together it’s R22. We don’t use R in calculations, we use U=1/R. U is the coefficient of thermal transmission. Insulation sort of adds like capacitors, go figure. Anyway, the Units for U are BTU/hour per square foot per degree difference(°F). So, to compute the amount of enery(in BTUs/Hour) that pass through an insulator, It is U*(Area)*(Temperature difference). For the infiltration, it takes 0.018 BTU to raise the temperature of 1 cu. ft. of air 1 degree F.
For Our Hypothetical room, we have
100 sq ft of ceiling with R22 and temp diff of 36 (68-32).
100 sq ft of flooring with R22 and temp diff of 18 (68-50).
400 sq ft of walls with R13 and temp diff of 36.
Computing the flow of enegy out we get
Ceiling: ~164 BTU/Hour
Floor: ~82 BTU/Hour
Walls: ~1108 BTU/Hour
Computing exchange energy
Exchange: 10000.018.05*36=324BTU/Hour
For a total of: 1678 BTU/Hour to maintain 68F when it’s 32F outside.
A 100W bulb produces 341.2 BTU/Hour.
So it provides 20% of the heating for the room(Note: If the bulb is the only heat source, the room would stay at about 38F).
Ok, that was a lot of worthless stuff, but it sets the stage for what follows:
If you’re talking about light fixtures on the ceiling, they definitely do heat up the ceiling directly above the figure considerably. I do not have a fancy thermocoupler like good-ol QED has, but just as a rough guess lets assume a light fixture raises the temperature of the ceiling to 140F in a 2 sq ft area. From all that went on above, we can say that’s equivalent to:
(140-32)*2/22=9.8 BTU/Hour that lost into the attic.
In enclosed fixtures, the number is much higher. My son’s enclosed,recessed 250W bathroom fixture spills about 75% of it’s thermal energy into the attic(Steel enclosure with vents, and thick glass).
Your parents do have a point if it’s a recessed light, but a lamp by the bed is the same(if not better from a thermal radiation standpoint as mentioned way, way above) as the electric house furnace.
My apologies to scr4, ftg, and Squink.
They said, what I was trying to convey, and failed to get across.
NotMrKnowItAll has made a simply posed question complex and wonderful and long reading.
“Beware of the Cog”
Of course the refrigerator doesn’t run all of the time, does it? When it stops the kinetic energy is turned into heat by the friction in the machine. And then it delivers its heat to the lower temperature environment. Over the long haul, all of the energy delivered to the refigerator comes out as heat except the kinetic energy existing at the moment and that is but an insignificant fraction of the total energy delivered to the machine over a period of time.
Sorry, but I don’t buy this. The fact that some of the heat is lost through the ceiling does not change the fact that all of the energy is converted to heat. As you pointed out, all of the heat doesn’t make it into the room, but that’s not the bulbs fault. Furthermore, you are assuming that the electric house furnace is perfect, but of course it will have heat loss through the furnace cabinet, ducts, etc., which are analogous to the bulb’s heat lost through the ceiling.
Sure, that’s all true, but I was answering the question about whether or not the parents have a legitimate beef with the lights being left on. In reflecting on this, I regret I was so conservative with things(forgot about ducts). It seems that depending on where a light is, it would be better at delivering heat to a room than the central heat. But, again, enclosed fixtures do channel most of the heat to the attic, so those don’t deliver parent-appreciable heating value. Might be designed as a Air-Conditioning friendly fixture.
Of course, if the furnace is within your heated part of the house, you wouldn’t be losing that heat…
Look again at the OP and consider how much has been introduced to confuse the issue and confound the uninformed.
My oh my! Haven’t we gotten a long way out into the outfield?
The answer is still 1500 watts is 1500 watt input which goes to raising the temperature of the room PERIOD!
There are side benefits if there are appliances in use in the place of resistance heaters. e.g. a refrigerator to keep food cold or a stove to cook the food, etc. The fact that the input power is put to use in no way affects the end result.
“Beware of the Cog”
You guys crack me up. The answer to this question is:
It is more costly to heat a room this way. You can, however, reduce the size of the heater needed based on the appliances that will be running anyway, it’s just not a 1 to 1 relationship.
Well, more costly, in that you lose lightbulb heat out the ceiling rather than getting it as room heat, but, on the other hand, if you do all of your heating with toasters, you get the same hot room and a lot of toast.
It’s like co-generation for the do-it-yourselfer.