# Thread: Cooling Time Calculation For a Volume of Air

1. Hi,

I am currently working in the industrial refrigeration industry and have been tasked with finding out how to calculate the time it takes reduce the temperature in a cold store by 1°C. I thought this would be relatively simple for me, but I am really struggling with it. Basically, the question arises because in order to protect against the cold store from pulling a vacuum by reducing its temperature rapidly, pressure relieving valves are installed within the cold store box. Manufacturers of these relief valves require this 'time' figure in there calculations for the number of releif valves required.

The example I am currently working on is as follows:

Cold store running at -20°C, with 6 x air coolers all having a cooling capacity of 88kW (total = 528kW), each providing a volume air flow rate of 24m3/s. The total volume of the cold store is approximately 60,000m3. Assuming that the room is at ambient (20°C) on the start of the temperature pull down. Can anyone help me here? It would be much appreciated. If there is any additional info I need to provide please let me know. Thanks in advance

2.

3. Disclaimer: this is not really my area of expertise. That said,
The density of air at sea level is 1.2kg/cubic meter.
http://en.wikipedia.org/wiki/Density_of_air
The heat capacity of air is approximately 1 kj/kg per degree C.
http://www.engineeringtoolbox.com/ai...ity-d_705.html
The cooler capacity is 528 kw, and a watt is a joule/second so that is 528 kj/sec.

So we have

Now, if your cold store is partly filled with something besides air, you would have to reduce that proportionally to the volume occupied.

4. The only thing I would add is that the cooling capacity of 88 kW for the cooling coils is rated at a specific air temperature. If it's rated at 20C then the cooling rate will be reduced as the air gets cooler (because the temperature difference between the air and the coolant is reduced). If it's rated at -20C then the opposite would be true. So depending on the details of the capacity rating you might cool down faster or slower than the constant cooling rate calculation would indicate. I recommend you take a careful look at the operating manual.

5. Thanks a lot guys, much appreciated. I'll spend a bit of time today going through the replies and getting my head around them. Thanks again

6. This is from the Wikipedia article
http://en.wikipedia.org/wiki/Specifi...asic_equations
The equation relating heat energy to specific heat capacity, where the unit quantity is in terms of mass is:

ΔQ = mcΔT

where ΔQ is the heat energy put into or taken out of the substance, m is the mass of the substance, c is the specific heat capacity, and ΔT is the temperature differential where the initial temperature of the reaction is subtracted from the final temperature.
In your case we are looking for the time Δt to reach a change in temperature ΔT where ΔT=1. ΔQ/Δt is the cooling capacity.

The equation becomes.
ΔQ/Δt = mcΔT/Δt
Δt= mc/(ΔQ/Δt).
You weren't given the mass, just the volume, so we had to multiply by the density. So to get the answer we had to multiply the volume by heat capacity c and density and divide by the cooling capacity.

William used the density at a lower temperature and that's why he got a different answer. It is more conservative to use 20C for the purposes of sizing the relief valves.

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