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ramachandrasg
14-07-2010, 01:01 PM
Hi,

How to arrive cooling load for clean room.

For example: If clean room requires for certain air changes (100% FResh air) is say 10,000 CFM and room to be maintained at 73 Deg F. and outside air temperature is 96 deg F DB and 62 Deg F WB.
Pls help how to arrive cooling load in TR for this??

tmm
14-07-2010, 04:14 PM
To determine the load you can use the following:

[Enthalpy Outside Air- Enthalpy Inside Air] x Specific Gravity of Air x Vent Rate.

As we use the SI system the conversions are as follows:

10,000 CFM = 6990.108 meters cubed/hour
96F = 35.6C
62F = 16.67C
74F = 22.78C [Since you have not stated RH = 50%]
Spec Gravity of Air = 1.2
Enthalpy Ouside Air [CIBSE Guide] 46.901
Enthalpy Inside Air [CIBSE Guide] 44.869

From formula above:

[46.910-44.869] x 1.2 x 6990.108 = 17120.173 KJ/h

Converting to Kw = 4.756.

tmm

frank
14-07-2010, 06:42 PM
Hi,

How to arrive cooling load for clean room.

For example: If clean room requires for certain air changes (100% FResh air) is say 10,000 CFM and room to be maintained at 73 Deg F. and outside air temperature is 96 deg F DB and 62 Deg F WB.
Pls help how to arrive cooling load in TR for this??
The total cooling load will need to be calculated to deal with building fabric gains, occupancy, lighting and equipment gains in addition to the ventilation load.

Without further information from you it will not be possible to advise you correctly.

I make the 100% fresh air load to be 106kW with a dt of 13K

10000cfm = 4.719m3/s