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Greetings: An accurate answer to you question would require much more information to design an efficient air flow system, however let's use a back of the envelope calculation to get an order of magnitude estimate. Your question is similar to heating and air-conditioning problems except that the air is usually modified external to the box rather than in the box although some gymnasium heaters are suspended in the room volume. The volume of your box is very large and the air flow velocity is reasonably slow so air compressibility is not a large factor. The design of the room shape could help to make air flow uniform through out the box (room?). I will use a standard fan ventilation engineering type estimate. Air horsepower is based on the fundamental fluid-horsepower equation, neglecting compressibility: hp (air) = p times Q divided by 33,000 where hp is the power required, p is the total pressure drop (pounds/square foot) required through the fan and Q is the air flow (cubic feet per minute). Mechanical efficiency (ME) is the ratio of air horse power to horsepower input (electrical, mechanical etc.). In your question you do not mention the box length or width so I will assume it is a square box 660 ft on a side (10 acres) with a 30 ft high ceiling (43560 square feet/acre times 10 acres times 30 feet high = 13,068,000 cubic feet). We can place a 330 foot long by 30 ft high structure across one half of the floor of the box to mount a bank of fans (F). <-----660 ft ------> -------------------- I <--- F <--- I I <--- F <--- I I <--- F <--- I I <--- F <--- I I <--- F <--- I I I I I I Air Flow I I ----------> I I I --------------------- The 12 ft /sec maximum air velocity that you require determines the maximum Q through the fans Q = 30 (ft) times 330 (ft) times 12 (ft/sec) times 60 (sec/min) Q = 7,128,000 cubic feet per minute. Standard ventilation engineering practice for a fan blowing into an large open volume uses 1/4 pound per square foot as the pressure drop through the fans. This gives an air horse power hp (air) = 1/4 times 7,128,000 divided by 33,000 = 54 horse power It would help to curve the corners of the box and to place large air flow vanes through out the box to keep laminar air flow through out the volume. This is typical practice in low speed wind tunnels. Also the distribution of the fans can be adjusted for laminar air flow Now comes the hard part of the problem. What mechanical hp is required to generate our required air flow and 54 hp (air)? I looked up the largest fan that I have in our labs engineering catalogs and I found a 60 inch diameter fan from Peerless Electric (Model VA60C-42). There probably are larger fans available but I am not aware of them. Information and design software can be obtained from Peerless at the following URL: http://www.industry.net/peerless-winsmith The VA60C-42 fan duct is 60 inches in diameter, and the fan rotates at 1160 RPM , providing 148,100 CFM of air flow and requires 126.1 BHP. Thus the hp (air) of the fan is: hp = 1/4 times 148,100 divided by 33,000 = 1.12 hp (air) The mechanical efficiency ME = 1.12 hp divided by 126.1 BHP = about 1% Thus you would require 7,128,000 CFM /148,100 CFM per fan = 48 fans 48 fans times 126.1 BHP/fan = 6,053 BHP. As a check if we multiply the hp air of the Peerless fans we get 48 fans times 1.12 hp per fan = 53.76 hp compared to 54 hp in our first calculation. Thus the real fans are well characterized by our engineerring model equation! The air once moving in the box might reduce the pressure drop through the fans, however this increased efficiency would be offset by the poor flow and change in momentum of the air mass as the direction of the air flow changes at the corners. However, the Peerless fan tables indicate that fan BHP is quite insensitive to pressure changes of several octives providing us with a generous margin for error.. Regards your Mad Scientist Adrian Popa

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