Help using coolpack for cold room

[W.Miller]

Hi I have a room being stored in a supermarket that is 3 x 2.5 x 2 and it will be storing meats. The ambient store temp is 20c. How can I utilise coolpack to make this calculation. I also would like to know what Tcool(h) means. Any help would be useful, this is for a college work and I just want to know if I using this program correctly. Thanks.

[Brian_UK]

Have you tried the HELP files of Coolpack?

[W.Miller]

They don't tell you anything. I may have to do this manually.

[SteinarN]

Tcool [h] means the time for the cool down of the warm products loaded into the cool room. "T" says it is time, "[h]" says the time is in hours.

[W.Miller]

Tcool [h] means the time for the cool down of the warm products loaded into the cool room. "T" says it is time, "[h]" says the time is in hours.

Ok, thanks.

[nike123]

They don't tell you anything. I may have to do this manually.

This is from Help:2773

I think that description is good enough for average person.;)

[W.Miller]

Sorry to be a pain, but does that mean if I wanted to cool meat down to -18 from 15c would I change the Tcool to 1 hour. Is that normal industry standard.

[nike123]

Sorry to be a pain, but does that mean if I wanted to cool meat down to -18 from 15c would I change the Tcool to 1 hour. Is that normal industry standard.

Yes, that mean that, and not, that is not industry standard.
For illustration purpose:
http://www.meatupdate.csiro.au/data/Meat_93_24.pdf

[SteinarN]

Sorry to be a pain, but does that mean if I wanted to cool meat down to -18 from 15c would I change the Tcool to 1 hour. Is that normal industry standard.

I doubt 1 hour is industry standard.

The actual cooling/freezing time is highly dependent of the air flow, the size of the meat pieces and whether it is packed or hanging free.

If it is not a fast cooling/freezing room I would have set the Tcool to 24 hours and the set the quantity of products to load every day some 20 to 30 percent higher than the real quantity loaded into the room every day. Or alternatively set the real quantity and set a Tcool of 15 to 18 hours. This would give a reasonable spare capasity for defrosting.

[W.Miller]

Ok after using cool pack I came up with these answers.

(uncooked meat)
Room size = 3 x 2.5 x 2
Troom = 0c
Rh = 85%
Ambient = 23c
Rh = 65%
Acf = 9
goods = 1000g
Tin(c) = 17
tcool = 15
Persons = 1
Work type = heavy
Lights 100w/m2

Now it says I need a maximum of 7.8 kw for the room stated.
All I would like to know is 7.8kw enough for this space?

[nike123]

Ok after using cool pack I came up with these answers.

(uncooked meat)
Room size = 3 x 2.5 x 2
Troom = 0c
Rh = 85%
Ambient = 23c
Rh = 65%
Acf = 9
goods = 1000g
Tin(c) = 17
tcool = 15
Persons = 1
Work type = heavy
Lights 100w/m2

Now it says I need a maximum of 7.8 kw for the room stated.
All I would like to know is 7.8kw enough for this space?

For 1000g of goods in 15 hours, more than enough.;)

[W.Miller]

Excellent I will add that calculation to my work and hopefully get a pass mark.

[SteinarN]

What did you get as an average capasity requirement?

[W.Miller]

I think it was around 3.7 kw.

[SteinarN]

I would have used the average capasity requirement as this gives a reasonable spare capasity.

The max capasity is the capasity needed to avoid the cool room temp increasing temporarily if the daily meat is put into the room all at once. This capasity requirement is also subjected to a number of aproximations in the coolpack program. A somewhat different number can be as correct as the coolpack number.

[W.Miller]

I would have used the average capasity requirement as this gives a reasonable spare capasity.

The max capasity is the capasity needed to avoid the cool room temp increasing temporarily if the daily meat is put into the room at once. This capasity requirement is also subjected to a number of aproximations in the coolpack program. A somewhat different number can be as correct as the coolpack number.

Thanks I will take your advice.

[W.Miller]

I would have used the average capasity requirement as this gives a reasonable spare capasity.

The max capasity is the capasity needed to avoid the cool room temp increasing temporarily if the daily meat is put into the room at once. This capasity requirement is also subjected to a number of aproximations in the coolpack program. A somewhat different number can be as correct as the coolpack number.

When I think about it if I have to size and select refrigeration plant for this. So working of the maximum capacity is probably the best option. For example if I size it at 3.6 kw and the temp does rises unexpected, the system will not be able to cope with the extra demand. If you catch my drift.