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Interesting argument between friends |
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| Nov12-12, 10:44 PM | #18 |
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Interesting argument between friendsi doubt more than a couple of watts will make it out. pretty much all of the power the computer and electronics draws will end up heating the room just as well as a space heater of the same amount of power. |
| Nov12-12, 10:52 PM | #19 |
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| Nov12-12, 11:25 PM | #20 |
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So we stray from the point. In the defense of my point, I argued that a heater have as much resistance as possible. This would ideally lead to large amounts of inelastic collisions with the lattice which would generate an irreversible thermal process "heat". When an engineer designs a heater they want the conversion from electrical power to "heat" to be as high as possible.
However when an engineer designs a computer they want their conversion from electrical power to "heat" to be as small as possible. Ideally, a computer engineer would wish to design a computer which dissipates no heat (which is impossible). I believe I remember from Solid State Physics that for silicon there is a way to calculate how much energy is used to "power" a device and how much is used to heat it. If I recall it was very simple, something like if the electron requires 6eV to overcome the band gap and its given 7eV then the electron has a final kinetic energy of 1eV which is expended as thermal radiation (the leftover KE creates collisions in the lattice → heat). Regardless of this specific example, I believe that if you think about it microscopically rather than microscopically, you see that a lot of the energy is going towards different processes that AREN'T heat. In the long run all energy WILL turn into heat (entropy ftw), however I'm talking about amounts and time scales on the order of "will this device heat my apartment more than my heater" in which case I don't think it will. |
| Nov13-12, 05:30 AM | #21 |
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| Nov13-12, 05:31 AM | #22 |
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And again: you still haven't thought of such a process, have you? |
| Nov13-12, 09:32 AM | #23 |
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Sorry I haven't read all the replies in this thread as I have limited time today, so apologies I end up repeating things that have already been said, but I'll add my two cents worth anyway.
The simple answer is that unless you have a device that is either, 1. Converting electrical energy in to stored energy (such as chemical or gravitational potential). For example a battery charger. or 2. Exporting power by design. Say for example I had a 2kW motor inside the house, but exporting that power through a drive-shaft to outside in the yard where it was driving something. then the vast majority of electrical energy consumed by any appliance (TV, Computer, Vacuum cleaner etc) *will* be converted to heat within the house. Take the TV example given by the OP. A TV consuming say 250 watts will have all but at most a few watts converted to heat. No energy is exported or stored so therefore it must all be converted to heat + light + sound and RFI within the house (mostly heat). Though the energy involved in the light and sound will be small, most of it will be absorbed and converted to thermal energy in the walls and furnishings anyway. Typically less than one or two watts would escape the house as light or sound, and a negligible amount would escape as RFI. In summary, 99% of the electrical energy consumed by the TV would normally be converted to heat within the house! |
| Nov13-12, 09:57 AM | #24 |
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Tell your friend he is a fool, you are exactly right. Power of a heater is designed to only heat a house/room etc. Power from a PC power source is to power a PC, heat is just an undesired side effect.
Is like he is saying a tank made of solid Uranium and a race car made of carbon fiber have the same engine, so they have to go the same speed. That's very wrong. Tell your friend to put on a coat and stop arguing. |
| Nov13-12, 10:25 AM | #25 |
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I think most of you are overestimating the efficiency of the electronics used in our homes. I tend to agree with Russ. Even if I didn't know Russ is an HVAC engineer I would still agree with him.
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| Nov13-12, 10:40 AM | #26 |
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As I mention above, the only real exceptions to this are devices that either store or export power by design, a battery charger for example. |
| Nov13-12, 10:47 AM | #27 |
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| Nov13-12, 10:51 AM | #28 |
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I don't really like to do argument from authority, but in this case someone challenged whether I have ever tested this. The fact of the matter is that I have.Data centers are easy: what causes debates in my company is fan energy. But not how much heat is added; WHERE it is added (dissipated). |
| Nov13-12, 11:26 AM | #29 |
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Ima guess on the fan blades, |
| Nov13-12, 11:27 AM | #30 |
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| Nov13-12, 11:48 AM | #31 |
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All agree that due to the inefficiency of the fan, a decent fraction immediately becomes heat. But what about the rest? My position is that it is converted everywhere you see a pressure loss, in proportion to the loss: across coils and dampers, due to friction in ducts, etc. |
| Nov13-12, 11:52 AM | #32 |
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In an ideal case with a computer actively consuming 1000W and a heater actively consuming 1000W the heat produced would be fairly close. Most power consumed by a computer is converted to heat. The exceptions I can think of are a minor amount stored in capacitors, spinning of disc drives/fans, LEDs and speakers, power that leaves to go to external to the computer (Ethernet for example), and stray radiation. In all of those cases the end result will be some additional heat generation (friction from the disc drives for example), but some is converted to mechanical, stored, or radiated energy.
Where the thought experiment falls short in reality is a computer with a 1000W power supply will rarely if ever actually draws 1000W. |
| Nov13-12, 12:17 PM | #33 |
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I am not a power engineer. Please explain from my drawing what is the power drawn from the battery. Assuming capacitor C is ideal and lossless, assuming the on channel resistance of the MOSFET is 0Ω so no ohmic loss.
When IN is low, Q1 turns on, voltage from the battery is connect to the C and charge it up as shown in Loop1. When IN switch to high, Q1 turn off and Q2 turn on, the C is discharged by Q2 through Loop2. Q1, Q2 and C are inside an isolated enclosed place in the dashed box in . Battery and the IN driver is outside of the box. 1) What is the power input to the box when IN is pulsing to charge and discharge the cap continuously? 2) If there is real power input, how is heat generated inside the box as there is no resistance? 3) If there is no power input, do we expect the battery to last forever in this ideal case? |
| Nov13-12, 12:43 PM | #34 |
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the issue is, yung, that when that cap discharges, the energy stored in the cap ends up somewhere. where do you think it goes? better clip on some heat sinks to the MOSFETs. |
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| conservation of ener, electrical energy, heat, statistical mechanic, thermodynamics |
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