Time needed to cool air temperature

[PaulLag]

Hi there hope everybody is fine.

Please, I’d like to clarify a doubt concerning the cooling time to lower the temperature of air in a room from Ta to Tb.
I have looked in past posts but without finding an answer to my questions, so that I would be grateful if you could help me.

My argumentation would be as follows:

I can estimate the capacity needed once I know the volume of the room, the dispersions, the lighting, the presence of heat sources in the room.
This capacity would determine a selection of a specific air conditioner, having the calculated capacity increased by a safety factor.

My first question is:
How can I relate the capacity calculated to the time needed to cool down the temperature ?
If the capacity doubles, would the time needed to reduce temperature halve ?

The second question is:
How can I relate the air flow of selected air conditioner to the time of cooling ?
I assume the bigger is the air flow, the lower the cooling time, since air flow affects capacity.
How are air flow and cooling time related ? I assume it won't happen that the double the airflow, the half the cooling time.

Will then air flow be somehow related to the air changes, and therefore is there any relation between time needed to cool down and air changes ?

The third question is:
Can anybody suggest me a specific literature to refer to concerning this subject ?


Many thanks !

 

[GT-EGR]

You have to look at it as an energy balance in the room.

The first part that you’ve already done is determine the rate of heat input into the room, and then you selected an AC unit that matched that. So this handles all external heat sources, but not the warmup or cool down of the air itself - it more assumes the air is already at your desired temperature.

On a peak cooling day, your safety factor will be the only thing actually lowering the room temperature, the rest of the cooling that your AC unit provides will just be completely reversed by the heat gain in the space.

So if you take that safety factor BTU/h, and then compare it to the amount of heat you are trying to reduce the air temperature by, and also account for the quantity of air in the room, you’ll end up with how many hours. The air temp and energy in the space is the part of this question you are looking for - the specific heat (Cp) of air (0.24 BTU per pound per degree Fahrenheit). It takes 0.24 BTU of heat to change the temperature of one pound of air by one degree F.

So you calculate how much air mass you have, what temp you start at, what temp you want to get to, covert it to BTUs, then divide that by your extra safety factor BTU/h - and that should give you your answer.

 

[chicopee]

Calculations to cool air temperature alone is an easy calculation but what is not as easy is to figure out is the heat input to the air during the cooling period from furniture and all its furnishing, walls, floor and ceiling

 

[MintJulep]

I guess you somehow didn't find these in you search.

thread403-343325

thread403-232882

 

[IRstuff]

A big miss in your list for heat sources "in" the room, is the "room" itself, i.e., its walls, which radiate and convect heat from their other sides.

The other issue is question of flow, since you cannot arbitrarily increase the air flow (think hurricane) and the room must equalize its pressure, so portions of what air input into the room is exhausted to the outside.

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[IRstuff]

One other consideration is how hot the walls are. You could potentially have cold air in the room but it might still feel hot if they're radiating or convecting large amounts of heat into the room, particularly since most A/Cs have a non-100% duty cycle, since their control is bang-bang, rather than analog.

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[PaulLag]

Hi

Thanks for all your answer.
Thanks for the links, very useful.

One question:

I tried to apply formula in a practical case.

I want to calculate now the capacity needed to reduce temperature

From

t = V*⌂T/(49.6*qnet)

I get

qnet = V*⌂T/(49.6*t)


Let’s consider an insulated room having a volume of 735m³
Let’s then assume

TD = 40K

(that is I need to cool air by 40K)

I get following results

min kW
60 9.9
120 4,9
180 3,3
240 2,5
300 2


The progression seem reasonable, meaning, more time, less capacity I need.
At the same time, if I increase the volume or TD, I’d need more capacity.
But I have the feeling these capacities seem to be low to reduce the temperature difference
I mean: these results refer more to a common domestic split.
My feeling is that more capacity is needed.

Please, can somebody help me to understand the mistake in my argumentation ?

Thanks in advance !

 

[IRstuff]

This "room" is in the desert somewhere? Assuming an endpoint of 22°C, that would put the ambient at a minimum of 62°C, or possibly 55°C, assuming some solar load.

What is the ceiling height? Assuming it's 8 ft, the room area is 300 m^2, so that typically requires 4.6 tons of A/C for residential; but since the room is so big, it's probably filled with people and equipment, so maybe more like 6.5 tons

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[MintJulep]

You've solved for excess capacity.

Your system still needs a base capacity to meet the steady state load.

 

[GT-EGR]

Without checking your equations (not familiar with metric or your approach) I assume those KW capacities are what is needed just to cool the air volume. Now you need to add that to the room cooling loads (internal heat gains - lights, equipment, envelope, etc) - that will give you the full capacity needed to cool the air down while offsetting the additional heat gains of the room

 

[PaulLag]

Hello and many thanks to your answers.

@ITstuff
No, the room is not in the desert.
I need to cool down temperature from summer ambient, said 30°C to -10°C
The ceiling height is 5m
The basement area is 147m²

@MintJulep
Can I understand that:
When I find the capacity needed to balance the power, temperature will stay constant.
What I have calculated is the capacity needed to cool down the air temperature, haven’t I ?

@GT-EGR
Thanks.


One question: does anybody know if temperature I am considering refer to only sensible cooling ?
In case of latent cooling do I need to add some additional capacity ?

I would like to thank you for your patience.
If you can recommend me some literature to refer to, I would be grateful


Thanks again

 

[IRstuff]

No commercial A/C will get to -10C, since that's below freezing; you are talking about a refrigerator, not an air conditioner.

My freezer section is programmed only to -21C

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[MintJulep]

What is 49.6 in your equation?

 

[PaulLag]

Hi

@MintJulep

I get this from the post you suggested me

thread403-343325: Calculating the time to heat up a room

from dbill74 contribution

 

[Zesti]

When there is a large temperature difference between inside and outside of the room, you will need quite a bit of insulation.
And you have to make sure there are no conducting elements between the inside and the outside because this will cause condensation on the outside. Or even condensation within the structure.