[ARCHIVED THREAD] - Heating With A Fan (Page 1 of 2)
Posted: 2/1/2010 2:57:48 PM EDT
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You have two closed (airtight) and insulated rooms. In the first room you put a 1500W fan. In the second room you put an electric space heater. How much power will the space heater have to draw to heat that room at the same rate as the room with the fan? |
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Quoted: 1500 watt fan? Horsecrap!! Whats the wattage on the space heater? 1500W fan, yea it's pretty big but I've always wanted to move a lot of air! The wattage on the space heater is the question. What wattage would be required to heat the room the same as the fan? |
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1500 watt fan? Horsecrap!! Whats the wattage on the space heater? 1500W fan, yea it's pretty big but I've always wanted to move a lot of air! The wattage on the space heater is the question. What wattage would be required to heat the room the same as the fan? 1500 watts to move air is different then using resistance to heat a wire. Unless you hooked the heater up to 220 or above, all you will ever get is 1500 watts to heat with as that is all a 120V outlet can provide. [12.5 amps] The fan isn't going to do any heating whatsover, [other then the heat the motor sheds] all it can do is transfer the heat from something else. |
| Assuming the motor on the fan is 90% efficient, about a 150 watt heater...or a large light bulb. Air is like a fluid and you will never change kinetic to heat energy that way. Have the fan blades rub on something to produce friction and you may have hope of getting some heat out of it. |
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Quoted: Assuming the motor on the fan is 90% efficient, about a 150 watt heater...or a large light bulb. Air is like a fluid and you will never change kinetic to heat energy that way. Have the fan blades rub on something to produce friction and you may have hope of getting some heat out of it. Interesting point. ***Also note that this is the first one of these threads that I have done not knowing the answer ahead of time. I have my own theories but no written proofs to back them up. |
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I propose a similar experiment:
Two identical, well insulated rooms, no windows, door (s) seal tight. Room A contains one electric kitchen stove, oven door open, set on "BROIL" the electric element draws 3000 watts.. . Room B contains 10 horizontal chest freezers, each consuming 300 watts average. , After a number of hours, which room is warmer, the one with the oven or the one with the freezers? . edit, made numbers simpler, ( same ratio) |
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you Sir are correct.
1350watts will be transformed in the rotational work of moving the fan blades and the other 150watts will be heat loss. But.... What about the heat generated by the friction between the air and the blades? That is well beyond my knowledge of physics and too many variables to know for sure but if i had to guess i would say somewhere in the neighborhood of a 200 watt heater. |
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Quoted: Quoted: So were do the btu's come from?The moving air from the blades adds work to the room. 1 BTU = 1054.8 Joules The fan is using 1500 Watts = 1500 Joules / Second So the fan is using 1.422 BTU / Second = 5119.45 BTU / Hour That energy can't go anywhere because the room is sealed. It has to be converted to heat (I think).  |
| Once upon a time 30y ago I could solve that but don't remember how anymore.There are other variables you would need to know such as size and weight and number of fan blades involved and the rpm of the motor so as to determine the amount of energy used to move air as opposed to producing heat.One reason you forget all that stuff is when you realize you can't heat a room with a fan. |
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Quoted: I propose a similar experiment: Two identical, well insulated rooms, no windows, door (s) seal tight. Room A contains one electric kitchen stove, oven door open, set on "BROIL" the electric element draws 3000 watts.. . Room B contains 10 horizontal chest freezers, each consuming 300 watts average. , After a number of hours, which room is warmer, the one with the oven or the one with the freezers? . edit, made numbers simpler, ( same ratio)  |
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Quoted: Once upon a time 30y ago I could solve that but don't remember how anymore.There are other variables you would need to know such as size and weight and number of fan blades involved and the rpm of the motor so as to determine the amount of energy used to move air as opposed to producing heat.One reason you forget all that stuff is when you realize you can't heat a room with a fan. Says who?  |
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It wouldn't matter about perception however the humidity level,ambient temp and the pitch of the fan blades would.The energy transfer from evaporative cooling would affect the heat exchange into the room. BTW I did not intend to offend the OP just that physics was really my bag in the 70's and I liked problems such as these. |
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Quoted: Have you factored in the perceived air temp caused by the evaporative effects of that much air movement on your skin, er sompin'? I mean wind chill . . .  Humidity in the room is at 100% so there will be no evaporation. Edit: I also don't really care about perceived temperatures for the sake of this question. |
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Quoted: It wouldn't matter about perception however the humidity level,ambient temp and the pitch of the fan blades would.The energy transfer from evaporative cooling would affect the heat exchange into the room. BTW I did not intend to offend the OP just that physics was really my bag in the 70's and I liked problems such as these. No problem.  |
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Quoted: Thermodynamics question. Fan would add heat to an adiabatic room. I'm in my fourth college Thermodynamics class right now and the nuances of this problem are still not that obvious. That's what makes it a good question, lots of room to discuss what's going to happen! |
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1500watts = 1500watts
Fan heated room temp increase would come from resistive heating in the motor coils, friction in the bearings, friction between the fan blades and the air, and friction between the air and itself, the air and the walls, and everything it is blowing by in the room. 1500w = 1500w always. Adiabatic or not, the room will get hotter. If you don't make the adiabatic assumption, the fan heated room won't get as hot because of the increased convection and associated heat loss through the walls, floor, and ceiling. Here's one that was posed to me: If a human puts out 1000BTU of heat, and is put in an adiabatic room, what will the final temperature of the room be? |
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you Sir are correct. 1350watts will be transformed in the rotational work of moving the fan blades and the other 150watts will be heat loss. But.... What about the heat generated by the friction between the air and the blades? That is well beyond my knowledge of physics and too many variables to know for sure but if i had to guess i would say somewhere in the neighborhood of a 200 watt heater. The kinetic heating of the blades would near zero. Most of the energy used by the fan is to keep a pressure difference between the front and the back of the blades. There would be some heating but not in the amount of energy used to maintain a pressure difference between each side of the fan blades. |
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Quoted: 1500watts = 1500watts Fan heated room temp increase would come from resistive heating in the motor coils, friction in the bearings, friction between the fan blades and the air, and friction between the air and itself, the air and the walls, and everything it is blowing by in the room. 1500w = 1500w always. Adiabatic or not, the room will get hotter. If you don't make the adiabatic assumption, the fan heated room won't get as hot because of the increased convection and associated heat loss through the walls, floor, and ceiling. Here's one that was posed to me: If a human puts out 1000BTU of heat, and is put in an adiabatic room, what will the final temperature of the room be? I've had to solve those before. |
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Assuming the motor on the fan is 90% efficient, about a 150 watt heater...or a large light bulb. Air is like a fluid and you will never change kinetic to heat energy that way. Have the fan blades rub on something to produce friction and you may have hope of getting some heat out of it. Interesting point. ***Also note that this is the first one of these threads that I have done not knowing the answer ahead of time. I have my own theories but no written proofs to back them up. No, it's not. The air will come to a halt due to friction. The 1500 watts of work you put into the air will wind up as heat in the end. The same is true of lights. |
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So were do the btu's come from?The moving air from the blades adds work to the room. 1 BTU = 1054.8 Joules The fan is using 1500 Watts = 1500 Joules / Second So the fan is using 1.422 BTU / Second = 5119.45 BTU / Hour That energy can't go anywhere because the room is sealed. It has to be converted to heat (I think).
Q in = Q out, for steady state, Q dot in = Q dot out. Thermodynamics, not just a good idea, it's the law. |
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This thread is full of epic fail, in the case of the 10 freezer room, if the doors were closed the freezer room would be warmer. Heat is just kinetic energy, it makes no difference where the movement comes from. This. Not only do you have the energy put into the room from the compressor running the freezer, you also have the heat removed from the freezers. But of course, when the freezers reach temp, they all stop so the measurement would have to be made before then. |
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So were do the btu's come from?The moving air from the blades adds work to the room. 1 BTU = 1054.8 Joules The fan is using 1500 Watts = 1500 Joules / Second So the fan is using 1.422 BTU / Second = 5119.45 BTU / Hour That energy can't go anywhere because the room is sealed. It has to be converted to heat (I think).
Q in = Q out, for steady state, Q dot in = Q dot out. Thermodynamics, not just a good idea, it's the law. Fuckin A it is
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1500 watt fan? Horsecrap!! Whats the wattage on the space heater? 1500W fan, yea it's pretty big but I've always wanted to move a lot of air! The wattage on the space heater is the question. What wattage would be required to heat the room the same as the fan? 1500 watts to move air is different then using resistance to heat a wire. Unless you hooked the heater up to 220 or above, all you will ever get is 1500 watts to heat with as that is all a 120V outlet can provide. [12.5 amps] The fan isn't going to do any heating whatsover, [other then the heat the motor sheds] all it can do is transfer the heat from something else. You fail at physics. |
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1500watts = 1500watts Fan heated room temp increase would come from resistive heating in the motor coils, friction in the bearings, friction between the fan blades and the air, and friction between the air and itself, the air and the walls, and everything it is blowing by in the room. 1500w = 1500w always. Adiabatic or not, the room will get hotter. If you don't make the adiabatic assumption, the fan heated room won't get as hot because of the increased convection and associated heat loss through the walls, floor, and ceiling. Here's one that was posed to me: If a human puts out 1000BTU of heat, and is put in an adiabatic room, what will the final temperature of the room be? I've had to solve those before. Doesen't nemo neo come and relieve you from copper top duty before you have to answer the question? |
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Heat is work is energy. Adding 1500 Watts to a room, regardless of the method, will heat the perfectly insulated room regardless of the mode. Now 1500 Watts of power applied to a heat pump, communicating heat from another source INTO the room will be much more effective at increasing the temperature. Since the Watt is a unit of power, energy is just that power over time. The integral of power is energy. Or just a force over a distance. |
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1500 watt fan? Horsecrap!! Whats the wattage on the space heater? 1500W fan, yea it's pretty big but I've always wanted to move a lot of air! The wattage on the space heater is the question. What wattage would be required to heat the room the same as the fan? 1500 watts to move air is different then using resistance to heat a wire. Unless you hooked the heater up to 220 or above, all you will ever get is 1500 watts to heat with as that is all a 120V outlet can provide. [12.5 amps] The fan isn't going to do any heating whatsover, [other then the heat the motor sheds] all it can do is transfer the heat from something else. You fail at physics. I don't have a lifetime to wait for the room to heat up. There is no such things as a perfectly insulated room. |
| Imagine the rooms are "black boxes." Into black box 1 you've got an extension cord in which you're adding 1500 watts of energy. How much energy do you have to pump through the extension cord going to black box 2 to equal the amount of energy you're adding to black box 1? In other words, it's a pretty easy question. |