how to calculate hvac capacity in a cabinet in an easy way?? 7

Find all the power supplies within the cabinet.
Add up their input wattage ratings.
That's an estimate of the heat you will need to get rid of.

(
Also inspect for devices powered by supplies outside the cabinet, which is less likely but not impossible. If you find any, those supplies' input wattage has to go into the sum also. )




Mike Halloran
Pembroke Pines, FL, USA

"...specific heat of air is 1,005."

There's more thermal mass in the cabinet besides the air. When you cut the cooling for your propose experiment, all the hardware would also have a temperature rise.

On the assumption that the cabinet is ultimately powered from a power line, I'd simply measure the AC power entering the cabinet - and adjust for other power entering or leaving (often minor, sometimes not).

the machine is a tying machine, it has a plc and 3 small VSD and few other things inside the cabinet. It works on demand if there's nothing to tie, waits, if a bale comes, it makes a sequence with the 3 VSD so it's complicated to measure power, it is not continuous.
I can get a data aqisition (dewetron) and measure either total current to the tying machine cabinet and get some kind of service factor and with the manuals of the vsd get the heat dissipation and somehow figure out... or i can measure temperature rise cutting the air to the vortex inside the cabinet with the line in operation, I think practically is a better option but don't know if the calculations are more or less ok...
I was thinking of the specific heat of air because of the operation of the air conditioner

Your proposed thermodynamic method of discovering the needed cooling power is not at all workable. Your method would be acceptable if the air in the cabinet were homogenized. It is not. It will be completely stratified, hot at the top and room temperature at the bottom. It cannot be homogenized unless the enclosure was carefully designed - from conception - to mechanically force homogenization. Adding a fan or fans does not homogenize the air.

The very best way to measure the power is run your cabinet thru a watt-hour meter for a set period of 'n' cycles.

The next best way to get the power dissipation is to run a cycle with the input power fed thru a wattmeter. Look at the time a cycle takes, (cycle time). Then estimate the highest rate the cycles would ever occur (duty cycle). From that you can calculate the average power and that's the exact cooling you need.

Barring this ability you're left with looking in the manuals for the maximum dissipation the VFDs can have and the piddling average power the PLC will use. Again you'll need the cycle period and the duty cycle of those periods. This number will likely be quit s bit higher than reality.

I'd suggest that the most efficient solution would be to take advantage of the stratification by targeting the vortex cooler at cooling the hot air at the top. I'd put the vortex cooler on a thermostat. Make sure it's mounted on the top of the enclosure and that the thermostat is mounted in the enclosure, level with the highest device in the cabinet. I could see this cutting the air consumption down to perhaps 1 percent or less of current usage from your described cabinet inventory.

Keith Cress
kcress -

today i placed a data aquisition and we'll get the duty cycle, maybe tomorrow we'll have a better picture.

thanks.

good link mcgyvr, today i got the data aquisition of the feeder current and now we can get the duty cycle of the machine.

martinrelayer; Don't kid yourself.

That paper is complete BS.

It's the most simpleminded bit of drivel imaginable. It's blindly dependent on surface area of a box assuming the air inside the box is completely homogenized and that cooling of the six surfaces of the box are equal. Both assumptions are invalid.

Go put your hand on your existing enclosure! You will see the bottom half is the same temperature as the room it's in. From somewhere up the side, towards the top, will begin a gradient. The top will be far warmer.

You can use the radiant and convective heat losses thru a horizontal top surface to compute the available cooling thru the top. This will be about 3/4 of the heat removal from the box and the conservative total. If you want more accuracy you will have to measure down from the top of the box to the bottom point of warmth on the sides. Compute the total area of this then run the integral over the vertical height of dT to see the radiant losses verses the surrounding ambient. Then do the same to find the convective losses of the 'warmed' part of the sides.

You will seriously regret following that moronic paper done by imbeciles..

Keith Cress
kcress -

wow that's a statement... i mean i try to get some practical info from whom are supposed to have the experience cooling cabinets (this kind of companies), because costs to experiment with different coolers are high... at least to try a good approximation of what should be the final solution. The task is not to use air to cool cabinets because electrical energy is cheaper than compressed air in our place...
someone handed me this other info, that goes in similar direction... i know this is not accurate, and in the other paper that graph i know should change if the cabinet has natural or forced circulation.

The reason i started this thread is to get other opinions and futher information on this issue. I will always consider all the opinions and if they come with fundament are more valuable to me, like yours.

That's why I thank all of you for your time and hope this thread is useful to someone else.

I promise I will update this.
Regards!

martinrelayer; Glad you comprehended my info above.

As a young engineering tyke I was handed this problem because we put all our products out in the field in enclosures. Some cost thousands of dollars to even get to for service. We had some thermal issues caused by lots of assumptions and from following the illogic of papers like the one referenced above.

I spent about 3 engineering weeks studying the problem with a dozen thermocouples, adjustable heat sources, mock-up panel assemblies, fans of all types, and solar exposures. It was truly educational especially when my boss was a pretty savvy mechanical engineer.

We carried the entire subject clear to sand cast'd aluminum finned back sides.

As for the paper above... you have to remember Hoffman is a sheet metal shop. They don't build electronics into them. I'm sure they've been asked by a million customers about temperature and power dissipation so they cobbled-up that paper obviously not done by a thinking engineer since, as YOU so ably noticed, their graphs don't even reflect major obvious differences in what could be happening inside.

Keep in mind one of my discoveries. You can put all the fans you want inside an enclosure. The air will not circulate - unless - it's mechanically ducted. We also found virtually no heat is transmitted out of the back of the box owing to the inner panel forming an insulating wall and the fact that the back of the box is often flush against the wall behind it.

We quickly learned how to kill two birds with one stone. We drilled fan holes along the bottom of the inner panel and mounted fans blowing into the panel. We sealed the two sides and the bottom of the panel so the air forced in had only the top of the panel to exit. This dependably ducted all the air to cause a uniform circular flow pattern homogenizing the box. Next we did things to force the enclosure to be mounted an inch or two from the wall behind, which brought that whole face into cooling. The back space created an effective thermal chimney.

As for your task, I recognize the mechanical air verse mechanical refrigeration issue. It's a common one. While I completely understand the desire to ditch the vortex cooler I can imagine someone saying "too much air and air is expensive" while, in fact, the problem is only too much or ineffective cooling not really vortex cooling.

I find mechanical refrigeration units to be costly, externally bulky, sometimes noisy, and not particularly reliable. They still have their place certainly. I would at least study the possibility of excess air use via excess vortex cooling, before jumping to a different solution with it's own problems. A line-voltage thermostat controlling an air solenoid to the vortex cooler could be a very cost effective reliable solution.

I'd grab a thermocouple and do some studies. You certainly seem like you have the chops for it. Stick the TC level with the highest thing on the panel and with the unit all sealed up kill the air to the vortex cooler. Monitor the temperature. If it gets too high turn the air back on for a short period. Be a human thermostat. You will soon learn just how much duty cycle is really needed and hence, how much air is truly needed. A thermostat has the nice feature that when the system is inactive for any reason no air will be consumed.

If the air is oil free, as sometimes it is, you can also use it to directly displace the air in the box. With way less consumption than vortex cooling requires. (probably 1/10th) I just did that on a project and it worked very well. It used only the air dumped by the pneumatics. Freeeee air.

Keith Cress
kcress -

Using the inside panel to duct the interior cooling air into a torus is definitely an idea worth stealing. Thanks, Keith.



Mike Halloran
Pembroke Pines, FL, USA