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Author: Subject: I want to cool down my room with ice
vmelkon
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[*] posted on 14-7-2016 at 05:24
I want to cool down my room with ice


What's wrong with my calculation?
I am ending up with 5.05 g of ice @ -20 °C is needed and that is obviously wrong.

Room width (m) = 5
Room length (m) = 6
Room height (m) = 2.5

Volume of air (m3) = 75
TOTAL (cm3) = 75000000

Air temperature (°C) = 27
Heat capacity (J/(g*K)) = 0.001006
Air density (kg/m3) =1.2041
Air density (g/m3) = 1204.1
Air density (g/cm3) =0.0012041

Mass of air (g) = 90307.5

I want to take the temperature down to (°C) = 20
Quantity of energy that must be removed (J) = 635.945415

Ice temperature (°C) = -20
Heat capacity of ice (J/(g*K)) = 2.11
Heat capacity of water (J/(g*K)) = 4.1813

Mass of water needed (g) = 5.05




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Fulmen
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[*] posted on 14-7-2016 at 05:45


You need to
a) Account for all the mass in the room, not just the air.
b) calculate the heat flux going into the room.




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[*] posted on 14-7-2016 at 06:26


(C) Account for air exchange in the climate controlled area- Unless you're not planning on breathing in there, and like high humidity...



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vmelkon
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[*] posted on 14-7-2016 at 07:00


Accounting for the junk in my room would be difficult.
I would estimate that the temperature would go up by ~1 °C per hour if my initial room temp is 20 and outside, it is 27. (Of course, it would not be linear).

Also, I did not include into my calculation the phase change of the ice to water, so instead of 5.05 g of ice, it would be some lower value.

Thanks guys.




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[*] posted on 14-7-2016 at 08:18


Your math is off by a factor of 1000.

The specific heat of air at 20C is 1.005 kJ/kg·K, which is also 1.005 J/g·K. Watch your unit analysis.

So the real answer is about 5kg, not 5g.

Of course, as mentioned above, the latent heat of objects in the room will reheat the air fairly quickly since their combined surface area is much greater than that of the ice.

You should recalculate with the Hf. It is significant at 0.333 kJ/g.




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[*] posted on 14-7-2016 at 15:03


You also forgot to consider the enthalpy of fusion of water, assuming your ice will melt.

Never mind, you did mention that. Your assertion that the room heats up by 1 degree per hour is not enough information, we need to know one of the following variables (or we need a second equation): (1) the heat capacity of the room (2) the thermal conductivity of the walls. The surface area of your room is 115 m^2; fiberglass insulation has an R value of around 0.5 m2·K/(W·in); and I'll assume you're house is built with 2x4 studs so you have 4 inches of insulation. That means the resistance of the walls is (4 inches) x (0.5 m2·K/(W·in)) / (115 m^2) = 0.01739 K/W; the conductivity is 57.5 W/K.

Since calculating how much ice you have to melt to reach 20 C requires solving a differential equation, and making further assumptions (about how quickly ice melts, which in turn depends on how quickly air moves past the ice and what form the ice is in, and all intermediate shapes the ice takes on as it melts), I will make it a lot simpler by figuring out how much ice you have to melt per second to keep the room cool.

It's 27 C outside, 20 C inside. So with a thermal conductivity of 57.5 W/K, the walls leak 402.5 W. Now we need to know how much energy is released when H2O goes from -20 to 20 C.

Enthalpy of fusion: 333.55 J/g
Heat capacity of ice (J/(g*K)) = 2.11
2.11 x 20 = 42.2 J/g
Heat capacity of water (J/(g*K)) = 4.1813
4.1813 x 20 = 83.626 J/g
All three steps together: 2.11+42.2+83.626 = 127.9 J/g


402.5 W / (127.9 J/g) = 3.147 g/s.

Therefore, if you melt 3.147 g/s of ice you should approach the desired temperature assuming that your room is constructed only of 4 inches of fiberglass insulation.

As for Bert's comment, I am assuming that OP has a heat recovery ventilator that is approaching 100% efficiently (the one in my house is pretty damn good, over 90% I think). That way he can breathe comfortably without losing heat :)

[Edited on 14-7-2016 by DoctorOfPhilosophy]

[Edited on 14-7-2016 by DoctorOfPhilosophy]

[Edited on 14-7-2016 by DoctorOfPhilosophy]

[Edited on 14-7-2016 by DoctorOfPhilosophy]
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vmelkon
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[*] posted on 17-7-2016 at 16:08


Quote: Originally posted by Praxichys  
Your math is off by a factor of 1000.

The specific heat of air at 20C is 1.005 kJ/kg·K, which is also 1.005 J/g·K. Watch your unit analysis.

So the real answer is about 5kg, not 5g.

Of course, as mentioned above, the latent heat of objects in the room will reheat the air fairly quickly since their combined surface area is much greater than that of the ice.

You should recalculate with the Hf. It is significant at 0.333 kJ/g.


Thanks dude. I'm not sure why I had used 0.001005 J/g·K.




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[*] posted on 17-7-2016 at 16:58


Quote: Originally posted by DoctorOfPhilosophy  
Therefore, if you melt 3.147 g/s of ice you should approach the desired temperature assuming that your room is constructed only of 4 inches of fiberglass insulation.


Thanks, it is interesting to see the units / equation in action.

Having to melt 3.147 g/s of ice means that much ice going from -20 °C to +20 °C in 1 second, which is not realistic.

Yes, I would be walking in and out of my room, so there would be significant leakage of air from the next room. My PC is in my room and so am I. My PC monitor is about 100 W. I think the PC is 300 W on Idle and maybe 400 W if gaming.
Maybe my body output 100 W. My LED bulb outputs 10 W.

Yup, it gets complicated.




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[*] posted on 17-7-2016 at 17:41


As an aside, I once attended an end-of-year function at a school. There were some 500 students in the assembly hall plus staff and parents. It was a hot day and there was no air conditioning.

A pretty good attempt was made to control the temperature using about ten 20kg blocks of ice hanging in cloth bags in front of ordinary fans. I was quite close to one of these and was reasonably comfortable. I am sure that others in the room were not. But the idea does have merit.




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[*] posted on 18-7-2016 at 03:29


If the air is dry then a fan blowing over a damp cloth will cool the air, the cloth can 'wick up' water from a reservoir.
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[*] posted on 6-8-2016 at 05:13


Quote: Originally posted by Sulaiman  
If the air is dry then a fan blowing over a damp cloth will cool the air, the cloth can 'wick up' water from a reservoir.


I you live in dry states like Arizona, this method is very effective and competitive to air conditioning.

A simple experiment, spin a damp wash cloth in dry air and feel the significant cooling effect.

See, for example, http://www.newair.com/kb/the-secrets-to-keeping-cool-with-ev...

A secondary advantage, you have also adjusted your excessively dry air.

Note, not mentioned in some promotional material is what the total effect of such products could be in more humid environments. Yes, the temperature in the room may still drop, but the natural ability of your body to cool itself (likewise through evaporative cooling) is impaired as the humidity increases. Bottom line, you may not actually feel (or be) cooler.

[Edit] The article link above does cite an interesting idea of using ones central air conditioner to provide some reduced level of cooling with dehumidifying. Then, use a portable evaporative cooler to 'cool' select areas. This dual strategy can save energy, and is more compatible with more green/solar powered homes, where the energy reserve may be limited.

[Edited on 6-8-2016 by AJKOER]
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[*] posted on 6-8-2016 at 05:34


I remember stopping along the road somewhere in the desert southwest where there was a funny arrangement along the side of some small gift store and wondered what the heck was that and concluded it was some kind of water evaporation cooling device. Maybe as a bonus the humidity would be good for your skin.
Here's one for a car.
https://upload.wikimedia.org/wikipedia/commons/a/af/1949_Hud...
https://en.wikipedia.org/wiki/Evaporative_cooler
https://en.wikipedia.org/wiki/Car_cooler

[Edited on 6-8-2016 by Morgan]
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[*] posted on 10-8-2016 at 06:06


Interesting.
I had heard that ancient egyptians would hang a wet cloth on the window. If the wind blows inside, you get cooled air.




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