Author Topic: improvement (Read 6099 times)

Arno · « **on:** December 12, 2005, 07:19:28 PM »

A 100 watt incandesent light bulb gives a certain amount of light and a certain amount of heat. Since it's the light we want and the watts we pay for, improvement comes in the form of a CFL bulb. It gives the same amount of light but less heat, and uses less watts. This is great!

Now going in the other direction- A 100 watt resistant heater gives a certain amount of heat and a certain amount of light. Since it's the heat we want and the watts we pay for, improvement comes in the form of---?

I think this sounds logical, but I'm an old man with no formal education, and don't have a clue. Hope you folks have some answers.

arno

strider3700 · « **Reply #1 on:** December 12, 2005, 01:25:33 PM »

Simple. Pick a metal that glows at a higher temperature to use as the heater element.

Of course electric heating is about 99% efficient so I don't really see the point of making this change.

Experimental · « **Reply #2 on:** December 12, 2005, 01:37:25 PM »

Wow, Arno,

GREAT question, but except for a lot of theary -- I think we are going to wait a long time for a "real", answer !!

I sure hope I,m wrong, as I,m an old guy too and being on a fixed income, sure would like to save some heating money !!!

I,m going to watch -- and my best to you, Bill H....

Laylow · « **Reply #3 on:** December 12, 2005, 02:07:32 PM »

Improvement comes in the form of application. This is the one energy conversion where you have nearly 100% efficiency. If you take it a step further, you would say that some of the conversion is lost to light and possibly noise but... those would eventually turn into heat as well. You can even call your 100W lightbulb a 100% efficient heater.

Now that you have a 100% efficient device you need to apply it in an efficient manner. This means putting the heat where you need it and keeping it there. Putting your heater in an uninsulated attic would be a poor way of applying heat to your home. Putting a hotpad under your butt would be much better.

You can also spend a lot of time contemplating space heater designs. Almost all of them are 1500W heaters so what is the difference from one to the other? A larger heating element will spend more time on than a smaller heating element. It might also last longer. Oil filled radiaters have thermal mass that will hold the heat longer and cause the element to not come on as often. Infrared heaters can throw heat all the way across the room and keep one person warm. Some heaters have thermostats or selectable wattage settings. What's the best? No idea. It all depends on what you are going to do with it.

wooferhound · « **Reply #4 on:** December 12, 2005, 02:39:37 PM »

A 100 watt heater that puts out light.

I made a heater out of a lightbulb and a dogfood can. I put it under the house to keep the pipes from freezing. Also used one in the bathroom to keep the toilet seat from getting so cold that I would stick to it.

http://www.fieldlines.com/story/2004/2/8/183240/2517

Arno · « **Reply #5 on:** December 12, 2005, 03:33:09 PM »

Thanks Laylow,

I'm more confused than ever. Since my 100watt incandesent had so much room for improvement by less heat, less watts and same light, why not my 100watt heater by less light, less watts and same heat? Surely my heater element glows [light]. Is there no way to reduce the glow?

Strider says pick a metal that glows at a higher temperature. I think that's going the wrong way. I don't want a higher temperature I want the same temperature with less watts.

I guess I'm even denser than I thought.

arno

strider3700 · « **Reply #6 on:** December 12, 2005, 03:46:14 PM »

The problem is heat measured in BTU's/hour is simply Watts/hour x 3.413

So if you put in 100 watts you get 341.3 BTU's - conversion losses out of it. Since electric to heat is about 99% efficient you get 338 BTU's out. If you want more heat put in more power. My comment about changing metals would just help you gain a bit of that lost 3 btu's back.

You can change things around so that you use that 341.3 BTU's more efficiently but you can't get more then that. Laws of Physics are a real pain at times.

dinges · « **Reply #7 on:** December 12, 2005, 03:54:06 PM »

A solution looking for a problem?

Conversion of energy into heat is the ONLY ONE that is 100% efficient. All energy will end up as heat sooner or later. Forgot the correct term, something to do with enthalpy and entropy?

Note: conversion is 100% efficient; but like another poster stated, you want the energy to end up where you really want/need it. This is non-trivial.

So, why put a lot of time/effort in improving a heater (2000W?) that puts out perhaps 1W as light (ever tried reading by the light of an electric heater?); it's already 99.9(99?)% efficient.

If you really want to save energy, look at the whole chain: from oil in the ground to electricity to your doorstep. There's where the real inefficiencies are, when it comes to electric heating (this includes heating with incandescents).

So if you really want to make an impact on the world, concentrate your research efforts on that area (or windturbines, or solar, or whatever else you like) instead of putting a lot of effort into marginally improving the electric heater. Electric heating is stone-age technology. Simple, and it works. Like the wheel, it can't be improved upon.

Peter,

The Netherlands.

Laylow · « **Reply #8 on:** December 12, 2005, 04:10:27 PM »

Whenever you try to take one form of energy and convert it to another usable form of energy you experience a loss of usuable energy. The ratio of energy succesfully converted to what we wanted it to compared to the energy that was "lost" is called efficiency. It is expressed as a percentage. Usually, what we are always trying to combat is how much of this energy turns to heat. In the case of an electric heater that is not a problem. You want to know why a heater would not benefit from not putting out any light but in most practices it is not a problem. Even if it wasn't putting out visible light, it would still be putting out infrared light both of which are easily converted to heat whenever they hit something that can absorb it.

As for the "less watts" part of your question, you just need to read up a little. Do a search for energy conversion and pick out an easy page to read. Watts is the term usually used to describe a quantity of electrical energy. 100 watts is equal to about 340 btus or 86,000 calories. Go to this site and look at all the ways that you can convert units of energy. http://www.onlineconversion.com/energy.htm We use all of these units to describe the quantity of different forms of energy but when you do a little studying you will find out that they are really all the same thing.

In the case of your lightbulb; it puts out 100 watts of heat, it uses 100 watts of electricity, but it does not put out 100 watts of light.

You're not dense. You just haven't been exposed to the topic before. When you get to the point where you realize that there is only one kind of energy, the light bulb in your head will light up and you will quickly turn it off so as not to be wasting too much brain power on such an inefficient process.

Laylow · « **Reply #9 on:** December 12, 2005, 04:28:53 PM »

Sorry, I gave you some search terms that maybe don't turn up what you need so readily. Here's a good little page on the nature of energy. http://www.ftexploring.com/energy/enrg-types.htm

Reminds me of an excersize in futility that I had started a while back. Maybe I'll post it here someday and see how many people it infuriates. I was trying to compile an exhaustive list of all of the possible ways to convert energy and examples of the devices that did it. I'll have to find it.

Laylow · « **Reply #10 on:** December 12, 2005, 04:58:35 PM »

When I lived in Okinawa (a tropical island) all of the closets had something similar installed in the back about 12" off the floor. It was a lightbulb in a vented double wall can. I think the idea was to raise the temperature in the closet just enough that if any of the moisture in the air was going to condense it would do it outside somewhere. Or it was just that raising the temperarture slightly would lower the RH enough to have a drying effect.

Arno · « **Reply #11 on:** December 12, 2005, 05:52:22 PM »

OK people, I give up! I've heated with wood for years but it's becoming too big a chore for me. Heating oil, natural gas and propane are going nowhere but up. Electric, although the most expensive way to heat, seems to be going up the slowest, plus it's a lot cleaner. I had just hoped to switch into electric heat a little cheaper. You have lead me to one good idea though, and I will insulate, and then insulate some more!

Experimental, looks like you're right on about the theories and we won't see the results in our time.

Thank you all for the input,

arno

richhagen · « **Reply #12 on:** December 12, 2005, 06:59:10 PM »

About the only way to heat your home using electricity more efficiently than a resistance heating element is with a heat pump, and then only under favorable conditions. If the outside temperature is warm enough, you can move more thermal energy from the outside to the inside of your home than the electrical energy required to move it. A heat pump is a carnot cycle engine, just like an air conditioner, only backwards, so that the hot liquid phase would run through the A-coil inside, and the cold gas phase would run through the coils ouside to soak up some of the available thermal energy. The temperature difference between the colder gas phase and the warmer outside air, and the warmer liquid phase and the inside air has to be great enough, given the efficiency of the design such that more thermal energy is transfered than electrical energy used to operate the heat pump. Rich Hagen

richhagen · « **Reply #13 on:** December 12, 2005, 07:03:45 PM »

Oh, and it also helps if the pump that compresses the working fluid is inside for a heat pump, so that as much of the energy lost due to inefficiencies ends up as heat in your house. There are designs built that are reversible such that they can be used for both heating and air conditioning as well, but they would likely be less efficient than one purpose built for moving heat to the inside of a house. Rich

MountainMan · « **Reply #14 on:** December 12, 2005, 08:41:11 PM »

Just to get the goat of the "conservation is cheaper than RE" crowd that had their way with me a week or so back, I suggest that this thread has uncovered a need for different type of light related conservation.

Rather than just converting over to CFL, I think everybody needs to cover every square inch of their interior with front surfaced mirrors, so that all that light that is bouncing around the room will get reflected and have another chance to land on whatever you intended to light with your CFL.

You could probably get by with half as much CFL wattage just by painting everything white. Probably still want mirrors on the ceiling though.

Intended as humor, but no doubt it will spawn a subthread about the reflective quality of different types of paint...

...I'm gonna go build another mill now.

xox,

jp

wooferhound · « **Reply #15 on:** December 12, 2005, 10:04:33 PM »

The problem is that

you can't buy a 100 watt heater

I did'nt need more

Therefore the invention of the lightbulb heater

and yes eventually all of the light will be converted to heat by radiating onto the walls. But I don't want warm walls and ceiling. I want warm air rising off of the floor.

nothing to lose · « **Reply #16 on:** December 12, 2005, 10:38:04 PM »

Yep, I normally use a 100watt light bulb in the pump house to keep the pipes there from freezing. Gee I wish I had used it the other night. We finally got into the teens (when I did not expect it), house froze up but nothing broke, pump house froze up and I lost all the pipes, fittings, and a couple valves that were all PVC

As for more effecient electric heaters. Perhaps the question is not getting more heat for the same power, but more work for the same power?

For instance, I can use a 1500watt space heater AND CFL lighting using more power in my shop (room of junk trailer house). I probably also want a fan moving the heat around the room?

However what if I used 5 of those 300watt halogen floor lamps I am stuck with and have no use for? I would still have 1500watts of heat, but all the light (and more) I want and not use extra power for lighting. The walls are already dark panneling and they should absorb light/heat. By directing the lights various places or sitting them in seperate areas of the room I don't need the fan either.

I actually have wood heat in the shop trailer now, but to work till I get the place warmed up I may need some short term heat in the back room. Making stands for those lamp heads so I can direct the light and heat where I want it may be better power wise than using any normal 1500watt electric heaters. Also perhaps when only wanting to work 30 minutes or so on a small part, I would need light and heat is nice, no reason to fire up the wood burner for less than an hours work time.

I did think about getting a ventless propane gas heater for that one room, but then I figure I can make short term electric heat even using my inverter for power, and any longer term heating needs are wood heat anyway. So back to the lighting for heating ideas again.

nothing to lose · « **Reply #17 on:** December 12, 2005, 10:56:03 PM »

Although that sort of works to a point, here we are trying to heat for less power not make more light. So that is working backwards in this thread. Actually you would want to paint the walls black and absorb as much light as posible which will become heat and warm the walls. Of course then the room will seem darker though with the same amount of light.

Painting walls white to reflect light rather than absorb it does brighten the room a bit. The mirror tricks don't really work that well like in the movies though. True in the MUMMY they did light an entire underground cavern with 1 sun beam and a bunch of mirrors, but in my bathroom a quartz flashlight shining randomly around the room made as much light as shinning it into the mirror and bouncing it elsewhere

Although perhaps the light was spreadout more in the room, I think the room was actually dimmer where the light was.

scottsAI · « **Reply #18 on:** December 12, 2005, 11:14:49 PM »

Richhagen, on the right track,

The most efficient is a geo-heat pump.

Using ground source water, pond etc, is warmer than the cold air outside. Makes for a much more efficient system. System is reversible at very high efficiency seer 13, reports of 19. Problem is the cost of getting the water. If you have a pond then is not bad, if you must drill a well or dig a big hole to get enough surface area to use as a heat source or sink, then $$.

Have fun,

Scott.

wildbill hickup · « **Reply #19 on:** December 13, 2005, 05:08:38 AM »

The insulation thing is the key. In addition to that is going around your house and finding any cold drafts find a way to stop them. Case in point(not avery scientific one). A friend has a small cabin. Year before last(in winter so he could feel the cold spots) he carefully checked and filled every little crack and crevice he could find, anyplace he could find a cold draft(even just slight) he tried his best to 'shut it off', he also noted to me that he didn't find many(it seemed to him). Howevery the following year his fuel bill was reduced by 1/3. The cabin was already well insulated the savings came from just filling the cracks. I realize the temps might have not been exactly the same(all winters here in Northern Vermont are about the same COLD) from year to year and there are other discrepencies, but it still points out the savings. The only way to increase the effiency of a near 100% heater is increase the effiency of the surroundings.

By the way one that I'm trying this year(to reduce loss through window glass is sticking bubble wrap on window panes that are not used (to see outside) but you still want to get light from. It's supposed to increase R value by a factor of 1. Not much but every little bit counts. Just an idea I got from some solar site.

Wildbill

willib · « **Reply #20 on:** December 13, 2005, 07:09:11 AM »

Peter , if thats true , 100 % efficiency for heating...

then why did i recieve so much "heat" when trying to heat water with a wind gen. ??

http://www.fieldlines.com/story/2005/11/11/18646/621

wooferhound · « **Reply #21 on:** December 13, 2005, 07:29:58 AM »

I used to work at a Theatrical Lighting Rental company. They had incandescent lighting fixtures that wnt up to 2000watts. They all have reflectors that focus the light out of the front of the fixture at about a 30 degree angle.

Everybody would put a 1000 watt light under their desk pointed up toward their bodys. This would be Local Radient Heat and worked quite well even in nonheated areas. I own a small lighting system that uses 120watt floodlights. On some shows I spend the cold night in the van with an extension cord running into th building and several of the floodlights pointed at me for heat.

Arno · « **Reply #22 on:** December 13, 2005, 12:06:23 PM »

No one seems to have noticed that I stressed the 'SAME' amount of heat from my 100% electric heater, not more, and hoped for a way to get it with less watts , since that's the way the "electric bosses" make me pay them, but perhaps it makes no difference either way since our heaters are already perfect.

Thanks again for the imput

arno

Laylow · « **Reply #23 on:** December 13, 2005, 12:58:34 PM »

You can't get 100 watts of heat for less than 100 watts of electricity. You use 100 watts of electricity and you get 100 watts of heat. You can't ask for more. You're not allowed to have 101% efficiency. Any electric heater that you can buy turns every single watt of electricity into heat. Every single 100 watt heater is going to put out 100 watts of heat. You can not get 100 watts of heat with less than 100 watts of electricity. The answer is not going to change Arno.

dinges · « **Reply #24 on:** December 13, 2005, 01:03:56 PM »

Willib,

You didn't receive heat, just a bit of 'constructive criticism' :-) My statement of 'back to the drawing board' may have been interpreted as a bit harsh, but it's what I say to myself when I have one of those great ideas that doesn't work in the end (according to calcs). Or would you rather that we kept you in the dark, with you doing lots of work on a system that doesn't work in the end? (or at least, not as expected). As they say over here 'soft doctors make stinking wounds'.

My statement about energy conversion (100%) is true. What was in the other thread was also all true, even your thinking wasn't flawed; it should be very well possible to heat water that way, but not to heat a lot of water to a high temperature.

BTW, if you really do want to heat water, do what mr. Joule did: stir the water; that way, at least you don't have the electrical conversion losses, so a bit more of efficiency.

No need to repeat the other thread, but it's quite an eye-opener when you realize how much energy it takes to heat water.

So, to recap: the ONLY energy conversion that is 100% efficient is that of (whatever form of energy) into heat. The other way around (heat into electricity/light/etc.) is less than 100%.

Take e.g. wiring losses in an electrical heater: the connecting wires have a bit of resistance, and thus dissipate electrical power, which is converted to heat! (Just not where you want it, though).

Peter,

The Netherlands.

Arno · « **Reply #25 on:** December 13, 2005, 05:54:17 PM »

Yo Laylow,

I use to affect my teachers the same way, they would get frustrated and tell me 8x8=64 and the answer won't change. You can relax now and know that you have gotten through to me. Thanks again for the patience.

I still think the answer is 72 arno

willib · « **Reply #26 on:** December 13, 2005, 07:02:44 PM »

no , i prefer you give it to me strait .

What electrical conversion losses are you talking about ??

you just said that converting electricity to heat is 100 % efficient??

not being argumentative i just dont understand what you mean??

alcul8r · « **Reply #27 on:** December 14, 2005, 05:43:29 AM »

Willibur,

Electricity is quality energy and can be used for lots of applications. Just using it for heat is normally considered a waste unless you have lots of excess power.

When you use heater coils to heat water from your windmill, lots of the electricity goes toward heating up the alternator. The same current flowing through the alternator is flowing through the heat coils, so if their inductance (resistance if DC) is the same you are using half of your energy to heat the alternator.

If the resistance in the heater is significantly larger you may be blocking much of the current which could be produced. Because power is I^2 R, you are losing power.

If the other way around and your heater has low resistance, the same formula, I^2 R, says most of the heat is being produced in your alternator. Remember the same current which flows through the heat coils flows through the alternator coils.

Assume you have a 1000 Watt windmill. Keeping in mind that at low wind speeds you'll get very little power and in high wind speeds you'll be furling out of the wind. A lot of your best winds are in the spring and fall, when you may not need extra heat. How much usable heat are you going to get from the mill?

The electricity could probably be used for much better purposes by concentrating on year-round needs.

Rex

dinges · « **Reply #28 on:** December 14, 2005, 07:02:25 AM »

I couldn't have said it better...

All electricity generated by a windmill is converted to heat; of the 1000W, perhaps 200W in the stator coils (a waste, if you want to heat water), transitions (wire-wire connectors, screws, etc.), wiring (heating up, consuming power --> less power for your hot water), and, at the end of the chain, your actual heating element.

Like I said: all electricity is converted to heat, but not all is converted to heat in the place you want it (the water, not your wiring or stator coils).

BTW, Willi, you have made calculations yourself about the losses in your bike stator coils? So it should be already clear to you that not all the energy you generate ends up at its intended place? That's the whole point (at least part of it) to calculating and minimizing internal resistance of your coils. Or by wiring it in delta, so internal wiring resistance becomes R/1.73, instead of R*1.73 in star.

Hope this is clear enough. Never mind about you being argumentative. I can take it :-)

Peter,

The Netherlands.

nothing to lose · « **Reply #29 on:** December 14, 2005, 05:20:57 PM »

Yes, that is correct. So the only way to improve on it would be get more work from it. Same power (watts) doing more work.

An example I think of this would be using a stirling engine to power a fan to move the heated air. It take 1,500 watts of power to make 1,500 watts of heat still. But if you sit a stirling engine on top of that heater in the room to be heated, then there are no losses or power used to move that air. All heat stays in the room. The tempature difference between the heater and air is what runs the engine to turn the fan.

Take it a step farther and you have 1,500 watts of LIGHT which equals 1,500 watts of heat, and a fan moving the heated air. Still only 1,500 watts of power for heat, but you now have Light and fans in use for free.

1,500 watts of electric heat is going to take 1,500 watts of power, BUT how much work will that 1,500 watts of heat do? If the heat is made and used in the same room, then any loses just go back into heating the room which is what you first wanted to do anyway, so therfore there really is no loss.

Not too serious here, but this IS ONE way to get more heat for less power

Use 1,500 watts of electric to boil water, run a small model steam engine on that boiled water, connect a small generator to that steam engine to make some power.

OK, there is NO loss in the engine or generator because any loss is heat and you want that heat to begin with. So all heat stays in the room, therefor there are no losses.

Any work done by the steam engine then is free work, so if it makes 200watts of power on 1,500watts of electric heat, then you actually only consumed 1,300 watts of power for 1,500 watts of heat.

So use the power grids 1,500watts to boil water and get 1,500watts of heat into your room while running the steam engine and producing 200watts of power which you also use for heating. You now got 17,000 watts of heat for only 1,500 watts of grid power.

That would and does work that way, how much power you could generate from a steam engine with 1,500 watts of electrically heated water I don't know, could be more or less. But since all you wanted in the first place is the heat anyway you got it and anything extra is free.

I like working model steam engines and have given this alot of thought since I am building one myself. How to fire the boiler and keep everything safe? Electric boiler should be as easy to control and safe as your hot water tank, just a bit different.

And as always, all heat stays in the room. Well at least as much as any other heat source would keep heat in the room.

nothing to lose · « **Reply #30 on:** December 14, 2005, 05:26:12 PM »

Whoops,,

" producing 200watts of power which you also use for heating. You now got 17,000 watts of heat for only 1,500 watts of grid power."

That should be 1,700watts of heat, not 17,000

nothing to lose · « **Reply #31 on:** December 14, 2005, 05:32:07 PM »

"I own a small lighting system that uses 120watt floodlights. On some shows I spend the cold night in the van with an extension cord running into th building and several of the floodlights pointed at me for heat."

I bet you stay nice and warm, and have no problem reading in bed either

More work, same power.

willib · « **Reply #32 on:** December 14, 2005, 09:21:27 PM »

Yes , thanks for making it clear, i just wasnt sure what losses you were talking about..

i have to dissagree with your calculations on the delta resistance though , still not being agrumentative mind you.lol

look at it this way,

you have three resistances in standard delta configuration, right.?

rhe resistance between any two points is X in parallel with 2X..

product over sum yealds (x*2X)/(X+2X) = 2X^2/X(1+2)= 2X/3 where X = your phase resistance..

News:

Author Topic: improvement (Read 6099 times)