HOW MANY "CFM'S" I NEED TO OCTAIN A AIR FLOW OF 30

[AXEL VAZQUEZ]

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) ?

[Guest]

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]

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.

[dinhj2002]

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

[Guest]

If you convert x miles per hour to feet per sec, you have

x mph * 5280 ft/1 mi * 1 hr/3600 s = 5280x/3600 ft/sec

Any such conversion involves the fraction 5280/3600, which simplifies to 22/15 if you remove common factors. It is exact.