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Cooling Equipment in a Cabinet 2

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Worker7

Mechanical
Dec 4, 2016
18
I need to do a thermal analysis to make sure that the electronic equipment items in a 6 foot tall metal cabinet are not getting too hot when they are all running. The cabinet has solid walls on two sides, a door on the rear and a top. There is no front door. The equipment is placed into the cabinet from the front. So I am going to assume the input air is all room temperature. Currently I do not know if there is any type of fan being used to cool the equipment. Please tell me what equation I should use. Do I apply the equation to each equipment item in the cabinet or to the cabinet as a whole? I am not savvy when it comes to conversions, so I would like the explanation to be as simple as possible. I don't necessarily need to fully understand all the math, I just have to get the project done quickly to keep on schedule. I do not currently have data on any of the equipment item, but will be getting it soon. I am just trying to get a head start on the project, so I can just plug in values once I get the data sheets. I do not think I need to consider radiation given off by the metal cabinet itself, unless that is something very easy to do. I do think that they are expecting me to be looking at the convection within the cabinet and to determine if the air inside the cabinet will be hotter than the acceptable operating temperature of the most critical piece of equipment. They will also probably want me to size a fan if one is required. Thank you for your help.
 
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Power in = power out for steady state, so divide the power lost in the cabinet by the thermal capacity of air for whatever temperature delta you can accept. That will give you the volume of air required to take away the power at the temperature rise allowed. This assumes that all the air passes over all the equipment; if there is any blockage items can reach very high temperatures.

This is a crude estimate. Good luck.
 
Thanks 3DDave. So I plan to use 1.005 for the thermal capacity of air because the cabinet is at room temperature approx.
What units will the volume of air be in?
 
Worker7,

I have been working on a similar problem. You need Cooling Techniques for Electronics Equipment, by Dave Steinberg. A fan, ventilating or circulating, makes a huge difference. You need to know.

--
JHG
 
All of the individual components you are mounting will have been designed to work in certain ambient conditions, with their own internal fans and heat sinks. You need to figure out what temperature of air they are counting on and then ensure that you can constantly provide that to them. If you are in a large warehouse then having sufficient air circulation will not be a problem. If your cabinet is in a small room you may need to put in a dedicated HVAC system.

As 3DDave said, all of the power your system uses will be put off into the air as heat so, using the heat capacity of air, your air supply temp, total system power, and your max acceptable air temperature rise dictated by the components you can figure out how many CFM of air you need to move in and out of the room and size your HVAC accordingly.
 
Would calculating how many BTUs/hr are produced, you can design something according to that. Just make sure without any data as far as the equipment inside said box. Use a multi speed system or VRV inverter system. You can oversize and leave it on auto it will adjust to return air temp or where you have your temp sensor.


John H
Pacific Coast HVAC
 
Worker7,

I just re-read your comments.

The heat capacity of your air is not relevant at steady state.

On my problem, I was told the electric power going into the cabinet. As noted by 3DDave, all of this must be dissipated outside, somehow. With the reference I noted above, I have the means to work out [Δ]T at the outside face of the enclosure. I can work out [Δ]T through the wall of the enclosure. I can work out [Δ]T from the inside air to the wall of the enclosure. I can work out [Δ]T from the components to the inside air. For a given ambient temperature, I can estimate the temperature the components are running at. If this is too much, you need to reduce the [Δ]Ts. Fans and heat sinks are an obvious solution. Heat sinking components directly to the outside wall is a solution. Heat pipes and thermo-electric coolers are fancier and more expensive, but they may be necessary.

--
JHG
 
drawoh,

The cabinet is supposedly missing one side, the front, so it isn't a closed box.
 
3DDave,

I am trying to interpret what he is saying.

Worker7 said:
The cabinet has solid walls on two sides, a door on the rear and a top. There is no front door. The equipment is placed into the cabinet from the front.

There is no front door, or there is no front enclosure?

If this was my design and it was open, I would worry about each individual module. The cabinet would be nothing more than a mount.

--
JHG
 
Drawoh,

my favorite "I don't necessarily need to fully understand ..." As long as he doesn't neither do I.
 
Sorry 3DDave. There is no front door. Only a rear door. Since this is a cabinet that is in the lab, I am thinking that instead of doing some sort of calculations all I really need to do is take some temperature readings where the air exists the fans in the equipment items. If the air temperature is lower than the maximum operating temperature on the data sheet, it should be OK. Do you agree?
 
Your testing approach is probably reasonable if the boxes are all working at maximum load (or some approximation of max real load etc.) for some period before & while you do the measurements.

Posting guidelines faq731-376 (probably not aimed specifically at you)
What is Engineering anyway: faq1088-1484
 
Worker7,

Is your cabinet enclosed, or open on one side?

If you dump heat to air, the air is cooler than your components. I suggest an infra-red thermometer. You can get one at Home Depot. You can measure the outside temperature of your components. Mine has a reliable class[ ]2 laser, which makes an excellent cat toy.

--
JHG
 
"There is no front door. The equipment is placed into the cabinet from the front. " was not misleading at all. Best of luck.
 
Everyone, thanks for all your help. I really appreciate it.
 
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