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Re: BTU removal

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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.

Good Luck
William

Follow Ups

Re: Re: BTU removal Damle Shrirang 02:10:17 02/17/03 (0)

Subject: Re: Re: BTU removal

Comments:: : 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. : Good Luck : William :

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