this is the short verison left out some of the more techincal stuff
Home Electrical Savings?
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I know the nuts and bolts of this one from chemistry class, if that helps: http://en.wikipedia.org/wiki/Avogadro's_law
Heat is pumped out, then it seeps back in. You pump it out, it seeps back in, etc, etc....
The more times you pump, the more you pay.
The more times you pump, the more you pay.
It has to do with Maxwell-Boltzmann distributions. When you take a gas and compress it, the molecules don't actually speed up, so the temperature doesn't really rise, but in a zillionth of a second, the increased density of collisions creates a new speed distribution which simulates a higher temperature, and is in fact actually a higher temperature. The the heat can leak from the new higher temperature into the outside air, which is warmer than the indoor air.
It's really quite simple.
It's really quite simple.
Ok, I must have been out dropping acid for these lectures (hey ekpolk, would I have been a scientist instead?
):
http://en.wikipedia.org/wiki/Ideal_gas_law
http://en.wikipedia.org/wiki/Universal_gas_constant
That kind of makes more sense now. Umm, kind of. This might make a good hobby.
http://en.wikipedia.org/wiki/Ideal_gas_law
http://en.wikipedia.org/wiki/Universal_gas_constant
That kind of makes more sense now. Umm, kind of. This might make a good hobby.
Originally Posted By: oilyriser
It has to do with Maxwell-Boltzmann distributions. When you take a gas and compress it, the molecules don't actually speed up, so the temperature doesn't really rise, but in a zillionth of a second, the increased density of collisions creates a new speed distribution which simulates a higher temperature, and is in fact actually a higher temperature. The the heat can leak from the new higher temperature into the outside air, which is warmer than the indoor air.
It's really quite simple.
CliffsNotes! Hooray!
It has to do with Maxwell-Boltzmann distributions. When you take a gas and compress it, the molecules don't actually speed up, so the temperature doesn't really rise, but in a zillionth of a second, the increased density of collisions creates a new speed distribution which simulates a higher temperature, and is in fact actually a higher temperature. The the heat can leak from the new higher temperature into the outside air, which is warmer than the indoor air.
It's really quite simple.
CliffsNotes! Hooray!
Ha ha! I used to be really good at this: http://www.jezzball.net/
Quote:
no julian, the refriderant is used for the heat transfer, under a low pressure it absorbs heat and under a high pressure it gives off heat, the indoor part of the system, in the cooling mode is under a low pressure which allows the refriderant to pick up heat from the house and dehumidify mostly because the temp of the indoor coil is lower than the due point of the entering air
Sounds good to me.
Quote:
So is it more accurate to say heat from outside is exchanging with the refrigerant, which is exchanging with the indoor heat?
This probably should have been worded: "heat from outside is exchanging (heat) with the refrigerant..." if that makes more sense.
no julian, the refriderant is used for the heat transfer, under a low pressure it absorbs heat and under a high pressure it gives off heat, the indoor part of the system, in the cooling mode is under a low pressure which allows the refriderant to pick up heat from the house and dehumidify mostly because the temp of the indoor coil is lower than the due point of the entering air
Sounds good to me.
Quote:
So is it more accurate to say heat from outside is exchanging with the refrigerant, which is exchanging with the indoor heat?
This probably should have been worded: "heat from outside is exchanging (heat) with the refrigerant..." if that makes more sense.
Originally Posted By: oilyriser
It has to do with Maxwell-Boltzmann distributions.
Quote:
If you increase the temperature of the sample, the distribution changes as show below. It becomes stretched out and has a lower peak, but the area under the curve remains the same.
http://www.bustertests.co.uk/studies/kinetics-collision-theory-maxwell-boltzmann-distribution.php
I haven't thought this through - with respect to how Maxwell-Boltzmann distributions work in AC units - but at this point I'm not seeing the connection.
It has to do with Maxwell-Boltzmann distributions.
Quote:
If you increase the temperature of the sample, the distribution changes as show below. It becomes stretched out and has a lower peak, but the area under the curve remains the same.
http://www.bustertests.co.uk/studies/kinetics-collision-theory-maxwell-boltzmann-distribution.php
I haven't thought this through - with respect to how Maxwell-Boltzmann distributions work in AC units - but at this point I'm not seeing the connection.
I was just trying to confuse the issue. When you squeeze the gas, the molecules do actually speed up (instead what I said earlier) since you are pushing them with a piston in one direction. The shape of the graph stays the same, but since work is being done on all the molecules (eventually, through collisions), their energy must increase. So, the curve gets stretched to the right. There are now more high energy particles, and tiny thermal vibrations (phonons) move into the material of the container, and transfer energy from the gas.
It's very interesting that elastic bands also heat up and cool off as you stretch them and let them relax.
It's very interesting that elastic bands also heat up and cool off as you stretch them and let them relax.
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