Heat loss in a control panel

Hello, I am trying to figure out how to effectively cool a control cabinet which has a VFD in it. And I don't want to do this with an online
heat transfer program. I would like to understand the equations which consider input heat and air flow.
So, here's the data: 1. I have a vented enclosure with an internal volume of 3.72 square meters. 2. There is a VFD in the enclosure which produces 1500W of waste heat. 3. The enclosure has a fan capable of moving 100 cfm (0.472 cubic meters per second) 4. The ambient temperature outside the enclosure will be 32 C.
Assuming that there is no heat transfer through the walls of the enclosure, and there is no heat source other than the VFD (no sunlight or nearby machinery), what will be the approximate temperature inside the enclosure? Please help me out and explain your answer.
thanks, Andy
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Oops. Let me start over. The volume should be 50 cubic feet, or 1.42 cubic meters. And the air flow of 100 cfm is equal to 0.047 cubic meters per second.
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wrote:

The cabinet inside temperature will be 144.2 degress F. Yes, this is aproximate, so don't zing me about having something after the decimal point. I can't tell you the information that can help you understand the equation, only that this is an equation given to me from a VFD manufacturer(Danfoss) for a forced ventilation cooling method. I have since misplaced the application note.
Temprise(Deg F) = wattsloss * 3.16 / fan CFM
In your case, the Temprise(Deg F) = 1500 Watts * 3.16 / 100 CFM = 47.4 Deg F. Therefore the internal temperature will be 36 Deg C + the 47.4 Deg F temperature rise = 144.2 Deg F or 62.3 Deg C. Way more than the typical 122 Deg F(or 50 Deg C) max on a VFD. You will need more CFM to cool this enough.
FYI, for nonventilated enclosures the formula is Temprise(Deg F) wattsloss/0.244/area(Sq. FT) in steel. Stainless needs to be derated, but I don't remember the derating value. Area is the surface area of the enclosure in square feet that is exposed to allow cooling. Sides, top, bottom and door.
IIRC, PDL Electronics had a more technical appnote on sizing drive cabinets.
Hope this helps, Jeff
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Jeff. Thanks. Yes, it helps. I derived an equation yesterday based on some earlier recommendations, and I came up with heat rise will be equal to: T = 0.95*V/F, where T is temperature rise in Celsius, V is volume of enclosure, and F is flow rate of fan. My answer came out to 140.9F, so I assume we are doing the same thing. And yes, I realize that there are more exact ways of doing this. I just want to get a ball park estimate. If I use this equation to size my fan, I am considering the worst case scenario, so I am not likely to underestimate the cooling.
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On Tue, 20 Nov 2007 06:30:46 -0800, Andrew wrote:

The experienced mechanical engineers I know compute this sort of thing by putting enough light bulbs into the enclosure to generate substantial heat, then measuring the resulting temperature rise.
Your temperature rise inside the cabinet is going to be determined mostly by the speed at which you get air exchange through the vents, and that's going to be anywhere from much slower than normal because of all the crap you have in the cabinet, to much faster than normal because incidental forced convection through the vents.
Of course, your temperature rise inside the cabinet is going to be _much_ less if you put fans on the vents.
Me, I'd start collecting light bulbs...
--
Tim Wescott
Control systems and communications consulting
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Andy,
Try looking at the space time of the air. You have an internal volume which is hopefully in cubic meters, not square meters, but lets let it be 3.72. You move 0.472 cubic meters per second, meaning that about every 7 or 8 seconds you exchange air entirely. This would mean that the volume is about 8s * 1500 W = 12000J warmer than the exterior. Get the specific heat of air in joules/m^3*C. Multiply the spedivid heat by 3.72 meters. Divide the 12000j by the specific heat of the volume to get the temperature rise. This is going to assume perfect mixing (which you don't have), and a few other things. It would represent the minimum temperature rise you need to worry about.
What is also important is that some places will see a much greater temperature rise due to stagnant conditions. Some air will also just whistle on through without warming much at all. Applying baffles (properly) may redirect air to where it will do the most good.
Next, do just what Tim said with the lightbulbs.
Michael
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On Wed, 21 Nov 2007 06:42:34 +0000, Herman Family wrote:

It wasn't clear from what the OP said whether the fans would force air through the external vents, or merely stir the air inside the cabinet. In the former case, they'll do a lot less than the latter (although stirring the air inside a space is often invaluable, too).
--
Tim Wescott
Control systems and communications consulting
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On Nov 20, 11:42 pm, "Herman Family"

Michael and Tim, Thanks. Michael's method is how I was trying to approach it, using energy. I figured that there were X kg of air in the cabinet, and the air stayed in the cabinet for T seconds, during which time it was exposed to 1500W, or 1500 Joules/sec. Based on the Cp of air, the air temperature would rise a certain amount of degrees within T seconds. But I have to say that the light bulb idea is excellent. I might try it if for no other reason than to check my heat calculations.
thanks, Andy
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