Time for a cold volume of air to reach a higher air temperature

In summary: Why?In summary, the cold air inside of the cube will take a long time to equalize to the ambient air temperature, depending on how the cube is arranged. If the cube is turned so that the open side is facing upwards, then the process will be faster.
  • #1
Bhope69199
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TL;DR Summary
What equation do I use to work out the time it takes for a 1000m3 volume of air at 285K to reach ambient air temp (290K at 1 atm).
I have a cube with a volume of 1000m3 at an initial temp of 290K. The bottom side (10m by 10m) is open to the ambient air. I put this cube into a huge fridge and cool the whole volume by 5K. I close the open side by placing a cover on it. This cube has now got a volume of air at a temperature of 285K inside. I then take the cube out of the fridge and suspend it in mid air with the face with the cover facing down. The ambient air that it is placed in is much larger than the volume of 1000m3, is at 1 atm and is at a temp of 290K. I then remove the cover opening the bottom side to the ambient surrounding air at 290K.

1. What I would like to know is what equation I can use to calculate how long the volume in the cube will take to equal 290K once the cover is removed. I am assuming (is the assumption right?) that as soon as the cover is removed due to the cold air being more dense than the ambient air it will "fall" (or be displaced), with ambient air replacing it. What equation can I use to show the length of time it takes for the volume of air to reach 290K. We can neglect any heat transfer through the other closed sides of the volume.

2. If this cold volume of air does "fall" how long will it take for it to equalise to the ambient air temp?

3. How do the equations change if I turn the cube so the open face is facing upwards?

If there is further information needed please let me know and I will add.
 
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  • #2
The 285 K will warm towards 290 K at a decreasing exponential rate, but will never quite get there.

Also, as the 1000 m3 cube of air gets closer to 290 K, conservation of energy will reduce the ambient air towards 289.99 K.
 
  • #3
What equations would I use to calculate the time taken?

Baluncore said:
conservation of energy will reduce the ambient air towards 289.99 K.
Would this still be the case if the ambient air reservoir was much much larger than the volume?
 
  • #4
With only the bottom of the box open, the cold air will immediately begin to flow downwards, being replaced by ambient air. The cold air will not warm, it will be mixed with, and diffuse into, ambient air somewhere below the box.

With only the top of the box open, the system will be stable.
To break that situation, the inside wall of the box could be heated by radiant heat entering through the missing top panel. The air in contact with the inside of the wall, will be warmed by conduction, until it begins to rise up the wall. That rising air will be replaced by air sinking in the middle of the box. That is a closed circulation within the open box.
How will it end. Since the air is being heated from a radiant source, via the internal wall of the box, it might heat above ambient, in which case it will flow up, and out of the box, to be replaced by ambient.
 
  • #5
Baluncore said:
the cold air will immediately begin to flow downwards, being replaced by ambient air
Will this replacement be instant?
 
  • #6
Bhope69199 said:
Will this replacement be instant?
It will not be instant. The air flow may reach a rate of maybe one metre per second. For a ten-metre cube, the process could take from half to a couple of minutes. I expect the middle part of the open base will fall first, as ambient air can move in from the sides.

When presented with a parameterised problem like this, we often find that the wrong question has been asked, and that there is a simpler answer. Why not explain what this situation is related to.
 
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  • #7
Baluncore said:
Why not explain what this situation is related to.
I wanted to know how long it would take (and how I could calculate it) for a volume of air that is cold to equal the ambient air temperature, whether that is by air displacement or by heating due to the higher temperature ambient air it is surrounded by.
 
  • #8
Bhope69199 said:
I wanted to know how long it would take (and how I could calculate it) for a volume of air that is cold to equal the ambient air temperature, whether that is by air displacement or by heating due to the higher temperature ambient air it is surrounded by.
Why ?
 
  • #9
I was looking at my cold cup of coffee and wondered how I would calculate how long it took to cool down. Then thought about turning it upside down and how the volume would be displaced with ambient air pretty much instantly, and whether this would be the case if the coffee cup was filled with cold air. Would the cold air be displaced immediately and what equation I could use to calculate the time it takes. And then thought it would be interesting to know the time taken in the configuration where the cup was just left on the table. (i.e not flipped upside down). I searched online and couldn't find any answers (perhaps the question wasn't worded correctly) so thought I would ask here.
 
  • #10
Bhope69199 said:
I wanted to know how long it would take (and how I could calculate it) for a volume of air that is cold to equal the ambient air temperature, whether that is by air displacement or by heating due to the higher temperature ambient air it is surrounded by.
There is no easy way to calculate it. It requires a computational fluid dynamic analysis (a complex computer simulation).
 
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  • #11
russ_watters said:
It requires a computational fluid dynamic analysis
I searched the thread for the word "convection" and couldn't find it. Convection will be the main mechanism for heat transfer (going back to School Physics).
 
  • #12
I can visualize the streamline pattern much more easily (at least qualitatively) if the cube is replaced by a cylinder 1 m in diameter and 1 m high. Any problem with that?
 

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