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HOW MANY "CFM'S" I NEED TO OCTAIN AN AIR FLOW OF 30 MILES PER HOUR IN A (6' WIDE BY 8' HIGH BY 16' LONG CHAMBER) ?

- AXEL VAZQUEZ

AXEL VAZQUEZ wrote:HOW MANY "CFM'S" I NEED TO OCTAIN AN AIR FLOW OF 30 MILES PER HOUR IN A (6' WIDE BY 8' HIGH BY 16' LONG CHAMBER) ?

You don't specify in which direction. I will assume flow is in the 16' direction so the cross section of the flow is 6' x 8' = 48 sq ft.

30 mph x 22 fps/15 mph = 44 ft/sec.

44 ft/sec across 48 sq ft = 2112 cubic feet per minute. This ignores the fact there will be slight stagnation at the walls, causing a higher velocity in the center and lower velocity around the edge.

You are talking serious fans plus straightening louvers to create this.

- Guest

Anonymous wrote:AXEL VAZQUEZ wrote:30 mph x 22 fps/15 mph = 44 ft/sec.

44 ft/sec across 48 sq ft = 2112 cubic feet per minute.

Big Oops:

44 ft/sec across 48 sq ft = 2112 cubic feet per SECOND!!!

That's 126.720 cubic feet per minute, sorry about that.

- Guest

Anonymous wrote:Anonymous wrote:AXEL VAZQUEZ wrote:30 mph x 22 fps/15 mph = 44 ft/sec.

44 ft/sec across 48 sq ft = 2112 cubic feet per minute.

Big Oops:

44 ft/sec across 48 sq ft = 2112 cubic feet per SECOND!!!

That's 126.720 cubic feet per minute, sorry about that.

Hi Guest !

Could you explain little bit in detail how/where do you get 22fps/15mph from? I am kind of confuse at this step. Thanks

- dinhj2002

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