Calculating time for temperature to change in room

[dmehling]

How does one calculate by how much the temperature of a room would fall over a specific amount of time, if suddenly a door was opened to the outside? For example, suppose the room had a relatively uniform temperature of 75° F, and the outdoor temperature was 45° F. If a single door to the outside was opened, how quickly would the temperature drop? And given that, how would you factor in other things like doors open to other rooms in the same house/building?

 

[Chestermiller]

How does one calculate by how much the temperature of a room would fall over a specific amount of time, if suddenly a door was opened to the outside? For example, suppose the room had a relatively uniform temperature of 75° F, and the outdoor temperature was 45° F. If a single door to the outside was opened, how quickly would the temperature drop? And given that, how would you factor in other things like doors open to other rooms in the same house/building?

This is a solvable problem, but it may not be as simple as you may think. For example, after you open the door, hot air from the room is going to start flowing out through the top part of the opening, while cold air from the outside is going to start coming in through the bottom part of the opening. There will be mixing taking place within the room. All this will be determined by the detailed gas dynamics, including differential mass-, momentum-, and energy balances. Natural convection will be significant. Adding other rooms to the picture is possible also, but the gas dynamics analysis will be more extensive. The air dynamics outside the open doors may also have to be included, at least in close proximity to the house.

Chet

 

[dmehling]

What you have shared is somewhat helpful, but I need more specific information like equations or formulas, so I can make an educated guess as to what the answer would be. I have limited background in physics, so I have no idea what kinds of formulas should be employed for solving this kind of problem.

 

[russ_watters]

Unfortunately, there are no simple formulas for this. The problem is really, really complicated, mostly because of the huge number of variables involved.

 

[sophiecentaur]

What you have shared is somewhat helpful, but I need more specific information like equations or formulas, so I can make an educated guess as to what the answer would be. I have limited background in physics, so I have no idea what kinds of formulas should be employed for solving this kind of problem.

For a start, just stick to the effects on the air in the room. You could base an estimate on the old method of mixtures (from 1960s O Level Physics). Start with a mass M of air in the room at T₁ Take away 10% of it and mix in (well) another M/10 at the lower temperature T₂. Heat lost from the warm air = heat gained by the cold air.
0.9M X SH X (T₁ -T_final) = 0.1M X SH X (T_final-T₂)
SH is specific heat of air.
The SH's and the M's cancel out so
0.9(T₁ -T_final) = 0.1 (T_final-T₂)
Re-arrange that to find what T_final is when there's a 10% change in room air.
0.9T₁ + 0.1 T₂ = T_final
Put in different dilution values 'to taste' and decide on the time involved and you have your answer.

 

[dmehling]

I would certainly imagine it to be a very complex problem if I wanted a fairly exact answer. However if I really tried to simplify things, perhaps I could get a somewhat realistic answer, or range of answers. For example, if I found that the room of a given size would have its temperature drop by 30° over a period of 15 to 30 minutes, I would find that to be a fairly helpful answer.

So if it is possible to solve the problem, as the earlier reply to my question suggested, how would I begin?

 

[dmehling]

For a start, just stick to the effects on the air in the room. You could base an estimate on the old method of mixtures (from 1960s O Level Physics). Start with a mass M of air in the room at T₁ Take away 10% of it and mix in (well) another M/10 at the lower temperature T₂. Heat lost from the warm air = heat gained by the cold air.
0.9M X SH X (T₁ -T_final) = 0.1M X SH X (T_final-T₂)
SH is specific heat of air.
The SH's and the M's cancel out so
0.9(T₁ -T_final) = 0.1 (T_final-T₂)
Re-arrange that to find what T_final is when there's a 10% change in room air.
0.9T₁ + 0.1 T₂ = T_final
Put in different dilution values 'to taste' and decide on the time involved and you have your answer.

Thanks. That should help me get started.

 

[sophiecentaur]

Thanks. That should help me get started.

Very basic but it would give you an idea of the 'feel' of the room if you can estimate, accurately enough, the rate at which the air changes. That is the hard bit, I think.

 

[Chestermiller]

Very basic but it would give you an idea of the 'feel' of the room if you can estimate, accurately enough, the rate at which the air changes. That is the hard bit, I think.

Yes. That's what Russ and I were referring to.

Chet

 

[sophiecentaur]

Yes. That's what Russ and I were referring to.

Chet

But you wouldn't need to have 'perfect' mixing for a subjective effect of temperature. Your body would do a bit of averaging but will be more fussy about the air around your feet and on your face.

I seem to remember, when I was trying to design a Central Heating system, there were (government?) figures available for required radiator sizes, based 1. On U values and areas, through walls etc. but also 2. On temperature difference, room volume and rate of air change. So an estimation method will surely be available via some Heating Website. It's a bit of a sore point for me because my initial calculations were only on U values and the required radiator sizes were laughably small. It got much more realistic when I included the air change factor.
Funnily enough, the plumber who came to do the installation just bought in the best value radiators which were each just a bit bigger than what I had calculated - just by putting his nose inside each room. 'Analogue computing' at its best.

 

[CWatters]

How does one calculate by how much the temperature of a room would fall over a specific amount of time, if suddenly a door was opened to the outside?

Most rooms already have some ventilation. Typically measured in terms of "Air changes per hour". The recommendation for a well sealed house (aka Passive House) is 0.6 per hour. So more than half the air might be changed per hour anyway even with the door closed.

Edit..

http://epb.apogee.net/res/reevair.asp [Broken]

The national average of air change rates, for existing homes, is between one and two per hour, and is dropping with tighter building practices and more stringent building codes. Standard homes built today usually have air change rates from .5 to 1.0

 

[TKJ365]

On a day with 45 degree weather, heat your house to 75 degrees. Get a thermometer, open the door. Wait a specified amount of time. Now you know. Sometimes math is not the easiest way to do something with a common sense solution.

 

[sophiecentaur]

On a day with 45 degree weather, heat your house to 75 degrees. Get a thermometer, open the door. Wait a specified amount of time. Now you know. Sometimes math is not the easiest way to do something with a common sense solution.

Absolutely. Do the experiment - take the temperature every 10 minutes and draw a cooling / heating curve. It may be a bit long winded but you should expect a nice Exponential shaped curve - gradually approaching the outside temperature (if the conditions don't change much).
The Maths will just not give you a useful answer and you would need to check it afterwards, in any case. (And the answer will be wrong, for sure! )