Posted By<" ">William on October 25, 2002 at 17:37:23:
In Reply to: BTU removal posted byJames Partain on October 23, 2002 at 15:14:30:
: If I have ambient air temperature of 80°F, 80 percent relative humidity and I want to maintain a maximum of 105°F in an enclosure while removing 1783019 BTU's per hour produced by an electric motor, how many CFM's do I need? Enclosure is 3,032 cubic feet ID.
: James Partain
Here's a formula that will give you a rough order of magnitude.
CFM= kW/temp rise x 3000 (a constant)
Using your numbers:
CFM= 522.2kW/25 degree rise x 3000= 62,664 CFM
Assuming zero losses.
A blower capable of this CFM at 2" s.p. would be about 60" dia. Rotating at 550 rpm, with a 40 hp motor.
So say your room is 10'w x 10'h x 30'long, the air flow traveling lengthwise would be 620feet per minute (approx) a wind tunnel almost.
I maintain that your heat load is less than you have stated. Arriving at your actual heat load you will realize your CFM needs will be lower.
< "> Subject: Re: Re: BTU removal
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