Doorway time factor (Load Calc)

Hello, I am trying to make a vba excel application to make load calculation to cold rooms.

I am following ashrae 2006 load calculation indications, but i do not get to match infiltration load with other soft load calculation like TRENTON or CALC-RITE (are they the same company?). Neither with the ashrae example. I think it is because I missunderstand the doorway time factor or the Gosney&Olama expresion although it is so easy to apply the ecuations...

May somebody help me and try to explain me about how to apply the doorway time factor corretcly?

Thank you.

In cold room temperature low than ambient. If there are not equalizing valve than pressure in cold room will be smaller. When the door opened warm air come to cold room from difference pressure. Air velocity depends on difference pressure...if you know air velocity, door dimensions and opened time you can calculate heat. Try to think on this direction.

Aik, Even if there is equalizing valve the hot air will inlet and cold air will outlet. This is due to air densities differences. Approximately on the middle up of the door will pass hot air and the middle down will pass cold air.

Sandro, you are right :)

Hi, thank you for your answer.

If I apply the ASHRAE expresion to a beef room (17x8x4,8m LxWxH) with a door (2,1mx2,5m WxH), 15ºC infiltration air 55%RH, temp.box 3ºC, 2 hours open door so that I asset door time factor aprox. 0,083, no curtains so that door flow factor=1, I will obtain 3,43Kw. (All this using ashrae expresion, chapter 13, 2006)

If i introduce these datas in the TRENTON soft for instance, I obtain 21Kw infiltration load what it seems to be not ver agree with the ashrae expression. This is what i meant on how to apply these factors because I am very surprised to obtain such disagreement.

I have rechecked this assesment and I did not find mistakes.

Does somebody compare these calculation methods?

Try this sortware from IRC.http://www.irc.wisc.edu/?/file&id=247

Hi, I have check the link you showed, and i think there is something wrong with this application too.

I pass all the SI units to IP units, and the application indicates an infiltration load of 10W!! what it is too low.

I am supposing a 2 hours open door, in a 24h period, what it is 5min/h opened.
Attachment 6772

Hi, as I remember, the door infiltration is calculated via the formula 70/sqrt(V). This gives you the amount of hot air that enters the room if the door is opened for one hour. (in number of room volumes).
For example, a 100 cubic meter room, will have the air change factor of 7. It means that if you keep the door opened for 1 hour, you will have 700 cubic meters of hot air enetering the room.

Quote:

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Originally Posted by Romanicus

Hi, as I remember, the door infiltration is calculated via the formula 70/sqrt(V). This gives you the amount of hot air that enters the room if the door is opened for one hour. (in number of room volumes).
For example, a 100 cubic meter room, will have the air change factor of 7. It means that if you keep the door opened for 1 hour, you will have 700 cubic meters of hot air enetering the room.

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Romanicus,

I think we are debating the use of mathematical model that depends of several parameters not a STOCHASTIC equation.