Does a Fan Produce More Heat or Just Blow the Air Around?

tony17112acst · Post by **tony17112acst** » Wed. Jan. 15, 2014 4:19 pm

Isayre/Larry: A fan may be put on a car's radiator for efficiency OR longevity ...it doesn't matter! What matters is that they put it there to transfer heat from the medium better.

But could you please address my last post (next to last) instead of just stating:

"Just remember that the various fundamental laws of thermodynamics can not and will not be violated."
"Science by consensus is a dangerous practice."
or "The laws of thermodynamics state in the most rudimentary terms that no matter what your whims, wishes, or desires or group therapy sessions (like this forum)..."

Please address my question with a coil electric heater with discretely limited BTU's and heat transfer before it goes up the stack. I thought it was a very good question. Please answer: " Will more heat transfer to the room by pointing a fan on the stove that contains this 50,000 BTU electric heater? "

That's a "yes" or "no" answer please.

Post by **Lightning** » Wed. Jan. 15, 2014 4:20 pm

lsayre wrote:Now riddle me this: Why is it that across the board, no matter what fuel is involved, the efficiencies of boilers seem to lag a bit with regard to the efficiencies achieved via furnaces or mere stoves. If radiators increase efficiency, wouldn't boilers have the highest efficiencies of the bunch?

It depends on surface area for heat transfer and their design.. You are always going to have exhaust that is carrying heat out with it unless you set up secondary heat exchangers to squeeze out the few remaining Btus in the exhaust. Then it would be capable of closer to 100% heat transfer efficiency. Water is a much better conductor than air which is why boilers don't have as much surface area for heat transfer. Look at how small gas and propane boilers can be. My sister has a tiny propane boiler that is wall mounted and is about 2ft x 2ft by 1ft wide and heats around 2500 square foot house.

tony17112acst · Post by **tony17112acst** » Wed. Jan. 15, 2014 4:30 pm

This is one of my favorite Simpsons moments:
https://www.youtube.com/watch?v=vWxZm8WjlI8

lsayre · Post by **lsayre** » Wed. Jan. 15, 2014 5:17 pm

Doesn't EdenPure try to argue that their 100% efficient electrical resistance heaters can produce more heat than anyone else's 100% efficient electrical resistance heaters?

Post by **Lightning** » Wed. Jan. 15, 2014 5:20 pm

lsayre wrote:Doesn't EdenPure try to argue that their 100% efficient electrical resistance heaters can produce more heat than anyone else's 100% efficient electrical resistance heaters?

I don't know how that would be possible

tony17112acst · Post by **tony17112acst** » Wed. Jan. 15, 2014 5:23 pm

Larry/Isayre

Please address my question with a coil electric heater with discretely limited BTU's and heat transfer before it goes up the stack. I thought it was a very good question. Please answer: " Will more heat transfer to the room by pointing a fan on the stove that contains this 50,000 BTU electric heater? "

That's a "yes" or "no" answer please.

lsayre · Post by **lsayre** » Wed. Jan. 15, 2014 7:17 pm

tony17112acst wrote:Larry/Isayre

Please address my question with a coil electric heater with discretely limited BTU's and heat transfer before it goes up the stack. I thought it was a very good question. Please answer: " Will more heat transfer to the room by pointing a fan on the stove that contains this 50,000 BTU electric heater? "

That's a "yes" or "no" answer please.

Can the resistance of a resistor change if air (or a heat sink, or a water jacket, etc...) is applied to cool it? The answer to this is apparently yes. The magnitude and direction of the change are dependent upon the material involved. To quote Sting, it depends.

A few potentially helpful relationships are in order (pick the ones that suite you):
1 Watt = 1 Joule per second
Watts x 3.412= BTU/h
Watts = V*A
power P in watts (W) is equal to the energy E in joules (J), divided by the time period t in seconds (s):
P(W) = E(J) / t(s)
P = I x V
and also:
P = I^2 x R (Joules first law)
power P in watts (W) is equal to the energy E in joules (J), divided by the time period t in seconds (s):
P(W) = E(J) / t(s)

By definition a current of one ampere flowing through a wire with one ohm of resistance heats the wire at a rate of one watt, which by definition = 3.412 BTU's of heat.

2 amperes through the same wire would therefore heat it at a rate of 4 watts, or 13.65 BTU's.

If electrical resistance changes (and we have concluded that it can in a system where the resistive element is subject to cooling) then amperes (current is measured in these) must change also, since V = IR , whereby V is an implied constant here, so therefore the BTU's change as a consequence, as can be seen above.

But don't ever forget "system equilibrium" or that you can't have your cake and eat it too (the laws of thermodynamics).

In the end we are still talking in circles. I think I'll go out a buy an EdenPure heater now.

tony17112acst · Post by **tony17112acst** » Wed. Jan. 15, 2014 8:29 pm

I'm glad you finally agree with me/us.

lsayre · Post by **lsayre** » Wed. Jan. 15, 2014 8:35 pm

tony17112acst wrote:I'm glad you finally agree with me/us.

Does a fan produce more heat or just blow the air (previously heated air that is) around?

PS: Did you know that James Prescott Joule discovered the first law of thermodynamics?

KLook · Post by **KLook** » Wed. Jan. 15, 2014 9:09 pm

YoU BeEn To ColaRaDo lAtelY?

Yous sounds lake yous bean smocking yous brakefust!

KeVIn

Coalfire · Post by **Coalfire** » Wed. Jan. 15, 2014 9:35 pm

Whatever the answer just remember to get more heat out of this electric heater it must have a MPD followed by two barometric dampers for maximum efficiency

Eric

tony17112acst · Post by **tony17112acst** » Wed. Jan. 15, 2014 10:52 pm

I think you two are the only ones left here. I'm out. Turn out the lights ...lock up.
-Tony

lsayre · Post by **lsayre** » Thu. Jan. 16, 2014 4:01 am

The answer was staring me in the face earlier and I didn't see it. Until now.

If the resistance and therefore the amperes are changed in an electrical system due to the heating of the resistor element to where (using my example above) the Watts generated increase from 3.412 (1 BTU) to 13.65 (4 BTU's), then your electric bill (which is paid for directly by the measure of the Watts the system has generated/consumed) has just increased by a factor of 4 also. If you lower the resistance to current flow and thereby generate additional Watts/BTU's/heat you must simultaneously increase the current flow by a square factor. In a resistance heater (or toaster or range top or oven or whatever) the resistance to current flow is directly correlated (I.E. = to) to Watts generated = BTU's. Lessened resistance means a square factor increase in the current flowing into the system. Likewise coal burned = Watts = BTU's. The energy in an electrical system is in the current flow (and if you increase it you must pay for it accordingly). The energy in a coal stove is in the coal (and if you burn it you must pay for it accordingly). There is no difference. If you increase the energy output of a system by altering it's resistance to energy flow (be it a coal stove with a fan blowing across its surface accomplishing this, or an electrical circuit accomplishing this via heating [a toaster for example] the resistor element and then adding more heat output by lowering the resistance and increasing the current flow, you simultaneously are increasing the energy input/consumption and the cost to the end user accordingly. The total energy of the system is conserved (the first law of thermodynamics) because mass = energy. But the total energy output of the system (measured in Watts/BTU's) is increased, so to sustain the increased energy output the energy input/consumption of the system increases likewise. For a resistance circuit generating Watts (BTU's, or heat) this consumption of energy increase is a square function of the current flow. Perhaps the increase in coal consumption goes up as a square function for a coal stove with lowered surface resistance as well. ???

My final answer is therefore as it has been from the beginning: A fan does not produce more heat. Burning more coal or generating more current flow produces more heat.

Post by **Lightning** » Thu. Jan. 16, 2014 6:46 am

Larry, pardon me for saying so but I think you being a bit stubborn at this point lol. I have to agree,, the fan doesn't produce any heat. Hell the stove doesn't produce any heat either. It's merely a heat transfer station. The only thing producing heat is the coal.

The elevated volume of air moving across the stove's surface is causing better heat transfer efficiency as shown by the drop in stack temperature. I believe this was the original poster's intention.

You are toying with us...

titleist1 · Post by **titleist1** » Thu. Jan. 16, 2014 6:49 am

lsayre wrote:My final answer is therefore as it has been from the beginning: A fan does not produce more heat. Burning more coal or generating more current flow produces more heat.

Yep I think everybody has agreed on that just like they agree that a circulator pump on a boiler doesn't produce more heat. But I don't see any boiler guys turning them off as a result of this 10 page discussion.