Thermodynamic Breakdown of a Fan
- Thread starter Layman
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Well said!
Thanks.
He won't understand all of it but it was worth a try.
I fear you are right.:shrug:
I don't even think that would be the right equation to use to solve the problem or a similar problem. It is not a homework problem, but the reason I address that specific problem was to see if a closed room with a ceiling fan would end up having some ridiculous temperature afterwards.
It takes in no account of how much energy would be generated from it being transferred from the fan. I would rather assume that the temperature of the human body is negligible. I am not concerned about how cold it feels really, just how hot it actually would be according to thermodynamics. A fan rotating at 500 rpm is probably a bit too much, and a normal fan may actually go more like 120 rpm on high. Then I don't see anything in the equation that addresses the real problem about how that kinetic energy would increase the heat of the room.
It doesn't have to blow directly on you - in a closed room, the fan will make ALL the air start moving, even just a little, which in turn helps you FEEL cooler. It doesn't take much movement to accomplish this.
That may be true, but that is not the point I am trying to make. It is very common for people to leave fans going when out of a room for extended periods of time. If the fan actually made it hotter by adding kinetic energy to the room that transfers to thermal energy, then they would be hit by a wall of hot air when going into the room. The problem is that I don't think it actually makes the real room temperature rise due to that, because it goes against everyday experience. I could leave a fan running in a room and come back to expect it to be cooler instead of expecting to enter a broiler room. A ceiling fan would add a lot of energy to room, and I don't think there is a temperature increase that is significant enough to show that. If I left a fan running in a room, I would expect the complete opposite to happen. I would expect the room to be cooler than normal. Then I don't have substantial gust of winds inside of my house.
The thing is, most fans are reasonably energy efficient, and thus do not give off a lot of heat (electrically speaking, waste energy is discharged primarily as heat from friction and resistance in the electrical components).
Take a match and light it. It gives off heat, right? Yet it doesn't raise the temperature of the room (at least, not in any statistically significant way).
Why is this? Simple - it doesn't give off ENOUGH heat to do so.
Now, take, say, ten thousand matches and light them (don't do it for real, you will probably burn your house down)
You will notice an increase in temperature... still not likely to be much.
It doesn't help matters any that very few houses are "airtight" (in fact, I don't believe any are unless specifically designed to be that way)
I was talking about heat gained from the transfer of the kinetic energy made by the blades of the fan itself. The kinetic energy of the fan turning would be transferred to heat. It doesn't take much of this kind of friction to start a fire, and that is how fires are created with little tools in the wilderness. Then I am not so much concerned with the heat generated from the mechanisms of the fan itself. The fan would continually be dumping kinetic energy to a closed system. That would in turn mean that a lot of heat would be created.
The heat from the blades themselves would be transferred to the air that would continually have friction with the air. I don't think the temperature increase of a room would be able to account for this as the temperature would get lower. The blades of the fan would not get hot, and they would actually be colder as well. The energy of the kinetic energy would have to be lost. Kinetic energy would be lost, and nothing would actually be getting hotter.
Either way, it will just be a bunch of hand-waving on both sides of the argument, because I don't think anyone actually even knows what those values would even turn out to be here. If you had a small fan that was put in a jar that was air tight even, I wouldn't expect it to make it hotter. If you could, people could have coffee mugs that ran on batteries to stir the beverage keeping it warm on the go. Then I don't think that would even work, even though it seems like descriptions of thermodynamics implies that it would.
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One time I was replacing my processor in my computer, and I failed to put the fan on the right way. I thought that if the fan was blowing up off of the processor it would make it cooler, because it wouldn't be blowing the hot air inside the computer on it. Then I found that this ended up breaking the computer. The main processor is actually cooler if it has a fan blowing the heated air inside of the computer directly on it. It makes me think that the processor wouldn't just feel colder by having the hot air blow on it, but it would actually be at a cooler temperature. Having the fan blow the hot air by it to another fan blowing the air out of the computer wouldn't work. The CPU would be cooler having the fan blow on it, rather than blowing the air out of the system altogether.
Thermodynamically speaking the heat generated by the blades of the fan is negligable. The small amount of heat will be transfered to the outside.
If you had a room that was PERFECTLY insulated and you turned on the fan in that room the room would definitely heat up.
The CPU has a large heat exchanger that removes heat from the CPU by conduction. The heat is removed from the heat exchanger by convection and to remove the most heat from the heat exchanger you need a large mass flow of air. The largest mass flow is accomplished by blowing the fan directly onto the heat exchanger. The air in the computer is hotter than the outside air (typically) but it is very much cooler than the high temperature of the CPU so there is effective heat exchange and the CPU is cooled..
Yes, but by how much?
I thought it had a heat sink, and it increases the surface area that heat can be transferred from.
Yes that is exactly what it does - I refered the heat sink as a heat exchanger both terms are correct.
exchemist
Valued Senior Member
The power consumption of a typical ceiling fan is comparable with that of a filament light bulb, in the range 60-100W. So that will be the heating effect, as all the power consumed in its operation is turned to heat. What effect this has on the temperature of the room obviously depends on the size of the room. To put it in perspective, 100W is about one tenth of a one bar electric heater. So it is not a lot.
Using VERY rough numbers:
Heat from the fan motor:
Fan power - 150 watts
Efficiency - 80%
Watts lost and converted to heat - 30 watts
Room Size 15ft x 10ft x 7ft
Insulation - near perfect
Temperature increase is ~ 1 deg F / hour
Heat from the friction of the blades is many magnitudes less than from the motor - it is negligible and can be ignored.
In any real situation there will not be an increase in temperature from a ceiling fan. If you have ever been in a fan room at an industrial facility that is a different matter, the fan room requires cooling.
No, that's wrong: all of the wattage is converted to heat. Fan efficiency only tells you how much isn't IMMEDIATELY converted to heat and spends a few seconds being airflow energy.
In that case it would be 5 deg F/ hr for a perfectly insulated room. So in a normally insulated room you should detect an increase in temperature after a couple of hours. If I remember I am going to run an experiemnt when I get home.
Edit to add: Russ thanks for reminding me that energy doesn't just dissapear in this particular universe.
Perhaps I should engage the old noggin for a few minutes prior to jumping into calculating...Common issue. The more fun argument in my office is WHERE, exactly, the fan heating happens in an HVAC system.
Er... what? How do you figure on that?
Take the AERATRON E503 model fan - it draws 2.4 watts on a 120 volt circuit, giving a total draw of 20mA on low speed, and draws 14.5 watts at on a 120 volt circuit for a maximum draw of 121mA on its highest setting. It is energy star certified to have an electrical efficiency of at least 80% even at maximum draw. If it converted 100% of that energy to heat, a simplistic heat output would be:
BTU/h = (V * I) * 3.412
BTU/h = (120V * .121A) * 3.412
BTU/h = 49.54
This means it can raise roughly one square foot per hour of air by 49.54 degrees Celsius... (assuming I have done my math and research right on this) or, in a 500 square foot room, it'll raise the temperature .099 degrees celsius per hour.
Cubic feet, not square feet. You are high by a factor of 8 (typical residential).
Why do I get the feeling that if you worked a problem that way on a physics test it would be wrong? I really don't think the wattage of the power supply could be directly related to the amount of heat a fan would give off due to the transfer of it's kinetic energy.
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