BTW, the input power of the PSU is usually not the same as its output power. That would mean 100% efficiency.
BTW, the input power of the PSU is usually not the same as its output power. That would mean 100% efficiency.
Buy a "Kill-a-watt" meter to gauge electric input (the power supply rating is NOT the input). It can assess the power over a period of time as well, since a typical computer will jump in power usage due to various loads taken place. You can then do the calculations. Good device to measure electricity costs per device as well. Sells for about $30.
the question is "how much".
energy balance equation
Q-W=U (assume closed system)
Q=Qin-Qout
W=0 (no work)
Qin= power from the PSU + sun light (just for the hell of it), 500W+100W (100W from sunlight, assumed)
Qout=1w/s (assumed)
U=Cv*deltaT
starting at 25 degree celcius, after 1 hour, the room temp would be
500W+100W-1W/s*60s/1min*60-0=Cv(Tf-25+273) (look up the Cv @ 25 degree Celcius, Temp in Kelvin I believe, correct me if I'm wrong)
so you get 240W net input, divide that by Cv and subtract initial temperature.
See, you assume 1 W of heat output, thereby automatically dictating the results. I'm sure the real number will be higher than this. However, this value can be found exactly using energy conservation equations, assuming the materials of construction of the room (and surrounding walls/rooms) is known. It's fairly complex, but it can be done by making some pretty simple, accurate assumptions.
Plus, you can't really have a Watt/second... A Watt is a Joule/second, so it's already a rate. I think that's why there's an obvious error in your calculations - 500 W input - 1 W output = 499 W accumulation. The other problem is that, as the room heats up, the heat loss from that room will increase, since the heat loss is always proportional to the temperature gradient (Fourier's law of heat conduction and typical convection boundary conditions). Using your model, the room would heat up infinitely, since the heat input is always greater than the heat output. Clearly, this can't be true.
Originally posted by: CycloWizard
See, you assume 1 W of heat output, thereby automatically dictating the results. I'm sure the real number will be higher than this. However, this value can be found exactly using energy conservation equations, assuming the materials of construction of the room (and surrounding walls/rooms) is known. It's fairly complex, but it can be done by making some pretty simple, accurate assumptions.
Plus, you can't really have a Watt/second... A Watt is a Joule/second, so it's already a rate. I think that's why there's an obvious error in your calculations - 500 W input - 1 W output = 499 W accumulation. The other problem is that, as the room heats up, the heat loss from that room will increase, since the heat loss is always proportional to the temperature gradient (Fourier's law of heat conduction and typical convection boundary conditions). Using your model, the room would heat up infinitely, since the heat input is always greater than the heat output. Clearly, this can't be true.
Engineering = approximation that works. There is no point of figuring out the heat transfer through the insulation when the OP just wants a ballpark figure of how much the room will heat up. You can probably look up the material data somewhere (matweb? haven't touched that with fiberglass insulation stuff).
you can have watt per second, sure. I just don't have my thermo book with me (4th year civil engineering, we don't really do thermo stuff except a section in the FE/EIT exam). You'd be surprised how close it'll be for the normal room temperature taking into account of temp gradient. Cv is a function of temperature.
it'd probably be wiser to assume sunlight as W/s also.
No, you can't have Watts per second. You can have Joules per second - an energy rate of transfer, also known as the Watt. The entire question is whether or not the heat transfer rate through the walls, ceiling, and floor compare to the rate of generation by the computer, period. If it's less, then the room will heat up. If you assume this transfer rate, you've completely assumed away the problem. Doctoral candidate in chemical engineering here - this is what I do.
Yes, my apologies, it should've been joules per second.
delta U = m*Cv*delta T
Given good enough assumptions, we can figure out how much it heats up in a period of time without very significant errors. I don't think it's necessary to take everything into account just to calculate how many degrees a room will heat up (i.e. how much the value for the specific heat air changes as a function of temperature, etc) in a relatively short period of time, valid assumptions (mine were pretty bad, I apologize) are more than enough for these type of calcs.
What will you be researching for your doctorate? Heat transfer? Chemical engineers have always given me the impression that they design refineries and such, and I don't know much about ChemE at all.
delta U = m*Cv*delta T
Given good enough assumptions, we can figure out how much it heats up in a period of time without very significant errors. I don't think it's necessary to take everything into account just to calculate how many degrees a room will heat up (i.e. how much the value for the specific heat air changes as a function of temperature, etc) in a relatively short period of time, valid assumptions (mine were pretty bad, I apologize) are more than enough for these type of calcs.
What will you be researching for your doctorate? Heat transfer? Chemical engineers have always given me the impression that they design refineries and such, and I don't know much about ChemE at all.
To really know the extent of room heating, you need to consider the steady state temperature. This occurs as time goes to infinity. At short times, clearly the small amount of energy (in Joules) output by the system will be very small relative to the heat capacity of all the air in the room, unless the room is very small, of course.
The traditional chemical engineer has designed chemical plants, such as refineries. This includes heat exchangers, chemical/nuclear reactors, modeling fluid and solid flows (even though solids don't really 'flow' per se), separation units (distillation columns, gas-liquid absorbers, liquid-gas strippers, liquid-liquid extractors, and others). My research is in a completely unrelated field (developing a viscoelastic mechanical model for how the eye focuses), but I'm still responsible for knowing all of the basics on a pretty deep level just as if I were doing research in that area - I might have to teach classes on it someday, so... I've taken at least nine courses that relate to heat transfer both in a thermodynamic (independent of time) and transport (time-dependent) fashion. Basically, we have to know lots of stuff that no one else wants to learn about, which is why we get paid the big bucks (unless you're stupid enough to go to grad school and get paid minimum wage or so).
The traditional chemical engineer has designed chemical plants, such as refineries. This includes heat exchangers, chemical/nuclear reactors, modeling fluid and solid flows (even though solids don't really 'flow' per se), separation units (distillation columns, gas-liquid absorbers, liquid-gas strippers, liquid-liquid extractors, and others). My research is in a completely unrelated field (developing a viscoelastic mechanical model for how the eye focuses), but I'm still responsible for knowing all of the basics on a pretty deep level just as if I were doing research in that area - I might have to teach classes on it someday, so... I've taken at least nine courses that relate to heat transfer both in a thermodynamic (independent of time) and transport (time-dependent) fashion. Basically, we have to know lots of stuff that no one else wants to learn about, which is why we get paid the big bucks (unless you're stupid enough to go to grad school and get paid minimum wage or so
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Mombull jumdo I have a 510 watts pc power and cooling P/S it does not even come close to using that peak power it has a 650 watts let alone the 510 watts its rated at . my system draws a max ox 400watts at peak load and thats the very most heat that it well have to dissipatate. You stated plainly the heat output would be equal to the rated power supply in watts.
Here is your post so don't tell me I missed the OP question OK!
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CycloWizard
Diamond Member
Posts: 4753
Joined: 08/26/2004
What's the maximum power output of your power supply? That's the amount of heat that your computer can add to your room. When dissipated over the entire amount of air present, the effect on temperature is pretty negligible unless you're in a closet.
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well, you CAN have watts per second...if you're talking about, say, changes in heat dissipation as a function of time
Yeah, it's dumb to assume a rate of heat loss, since that rate will change as the room temperature changes, which is the whole essence of the problem.
Learn English. The answer to all of your rambling insanity is in the very post you quoted. Go to www.merriam-webster.com and look up definitions for 'maximum', 'peak', and any other word that you don't understand. Good Lord!
MY system will not even get close to 650 peak watts of my power/supply.What part of that don't you understand?
I understand that you're an idiot who paid too much for a power supply he can't use. Any other genius questions?
A P/S that is rated over system specs is not a bad idea for system stability. But you go ahead and use a P/S thats rated at your system power usage . To each his own.
May be he bought a power supply that can deliver more than he needs for some reason! Whatever that reason, he is right that the maximum power of the supply may not be the accurate figure to use for the calculation. Rather, the power used should be. The difference may be minimal or negligible depending on the exact figures. But, he has a point!
What is the point of this argument anyway? You made some good points addressing the question of the OP. Why are you hammering this guy now?
Let me know when you learn what a first approximation is in your engineering classes. Then maybe you'll understand why I made this simple suggestion.
Because he was being an ass, repeatedly. Besides, if you already know the exact peak power draw of your system, why would you be reading this thread?
Need to think a little outside the box on this one also. You can figure out how much heat your room loses due to the materials in the walls, ceiling, floor, doors and windows.
As an example:
When the computer is off to when it is fully heated up you may (for numbers sake) get a change in temp of 10degrees (deltaT) but lets say you find out your computer is putting out 15Btu's of heat per hour when the room it is in is losing 2btus/hr.
See what im getting at? just taking the ambient temp in the room first off would be a good way to go but it wouldnt be fully accurate. Sorry to bring it to another level with a mention of heat loss calcs of the room
As an example:
When the computer is off to when it is fully heated up you may (for numbers sake) get a change in temp of 10degrees (deltaT) but lets say you find out your computer is putting out 15Btu's of heat per hour when the room it is in is losing 2btus/hr.
See what im getting at? just taking the ambient temp in the room first off would be a good way to go but it wouldnt be fully accurate. Sorry to bring it to another level with a mention of heat loss calcs of the room
Originally posted by: GZDonner
Need to think a little outside the box on this one also. You can figure out how much heat your room loses due to the materials in the walls, ceiling, floor, doors and windows.
As an example:
When the computer is off to when it is fully heated up you may (for numbers sake) get a change in temp of 10degrees (deltaT) but lets say you find out your computer is putting out 15Btu's of heat per hour when the room it is in is losing 2btus/hr.
See what im getting at? just taking the ambient temp in the room first off would be a good way to go but it wouldnt be fully accurate. Sorry to bring it to another level with a mention of heat loss calcs of the room
After equilibrium (after the average room temperature stabilizes), the amount of power (energy rate) generated by all the energy sources in the room and outside (Sun) will be equal to the total power lost by the room.
The two numbers will not be equal at first right after you turn on the PC. At first, there is more power coming in than going out! That is why the room temperature goes up. But, that is temporary. Otherwise, the room temperature would keep going up.
So, the example you are talking about, where the computer is putting out 15BTUs per hour while the room is losing only 2BTUs per hour, results in the room temperature going up. As the room temperature goes up, the room loses heat faster (more power leaving the room) because of a greater thermal gradient between the room and outside. Eventually, the two numbers become equal and the temperature remains constant (at least on average).
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