Airflow required required in a cabinet to cool 6349 btu per hour?

Answer 1

The specific heat capacity of air is very nearly 1 kilojoule per kilogram per degree celsius, so assume that value.

1 BTU is 1.055 kilojoules, so 6349 BTU is 6698 kilojoules. now we have to decide what temperature rise to allow, you don't mention that. Shall we say 5 deg celsius? Then the heat balance is:

6698 = m x 5 , m = 6698/5 = 1340 kg air per hour.

Density of air is about 1.25 kg per cubic meter, so 1340 kg = 1072 m3 = 38,216 ft3/hour or 10.6 cubic feet per second (1 cubic meter = 35.3 cubic feet).

Remember this is for a 5 deg rise in air temperature. I'm not sure what power you will need to blow that much air, but there will be some rise in air temperature over the fan, before it gets to the cabinet. I can only give you the air flow required, the actual desired temperature of the cabinet is important too and you might need to cool the air flow before it reaches the cabinet.

Answer 2

To cool 6349 BTU per hour, you would need to calculate the required airflow based on the specific heat capacity of the air. The formula is CFM = 6349 / (1.08 * ΔT). Where CFM is the required airflow in cubic feet per minute, 1.08 is the specific heat capacity of air, and ΔT is the temperature difference in °F between the inlet and outlet air in the cabinet.