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Heat dissipation in IP 67 Enclosure

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structure59

structure59

Structural
Joined
Nov 23, 2008
Messages
17
Location
IN
Hi,
I have a 750mmDx600mmWx450mmH IP 67 enclosure of SS 304 and it has one 5KVA online UPS plus 4SMPS inside.Total power input is 3000Watts.Willthe heat dissipation be a problem? The enclosure will be used in open air.
 
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Assuming your efficiencies are better than 85%

This appears to be outside your area of expertise; you should hire an ME with thermal design experience.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
 
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From practical experience, yes you have a serious problem. You will need either active cooling or forced ventilation.

If you are outdoors then you need to consider heat gain from the environment - ambient air temperature, and the effects of insolation (solar gain) unless you're mounting it indoors within an air-conditioned space. If indoors then you may get away with a much larger enclosure and rely on passive heat dissipation.
 
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Dear Mr. structure59 (Structural)
Q1. " ...I have a 750mmDx600mmWx450mmH IP 67 enclosure of SS 304 and it has one 5KVA online UPS plus 4SMPS inside. Total power input is 3000Watts. Will the heat dissipation be a problem? The enclosure will be used in open air ".
A1. Please tell us a little more so that we have some data to base on:
a) what is the max ambient temperature of the location?
b) what is the max temperature the equipments are designed for proper operation (e.g. 30,40... deg C)?
C) i) are all 6-surfaces (i.e. 4-sides + top + bottom are having >100mm clearance for cooling?,
ii) or, only 5-surfaces ( i.e. 3-sides + top + bottom having >100mm clearance for cooling?,
iii) or, only 4-surfaces ( i.e. 3-sides + top having >100mm clearance for cooling? Etc.
d) whether the SS 304 enclosure is painted or bear?
e) re-evaluate the [total power lose = heat generated by the equipments]. NOT the [total power input], FYI (e.g. a 5kAV UPS say with 95% efficiency, would generate 250W of heat. The 4SMPS would generate ??? W. Take into consideration whether are ALL the equipments are at [full-load] at ALL time.
f) BTW The internal temperature would > the ambient temperature if without force cooling or refrigeration.
Che Kuan Yau (Singapore).
 
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Trick of the trade:
Go to a website for an enclosure company that also sells industrial air conditioners and find their wizard program for picking the right size Air Conditioner. Load in your ambient conditions, exposed surfaces and heat rejection, then put in your enclosure size, it will return the size of AC unit you will need, if any. If it does come back saying you need an AC, keep increasing the size of the box until it tells you that you don't!

Warning, it's an eye opener. I once was asked to put a 75HP VFD into a sealed SS box and no AC, in an area that had a maximum ambient of 30C. The box ended up 90" H x 96" W x 36" D (approx. 2300 x 2400 x 900mm).


" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden
 
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If we take ABB Switchgear Manual-11th revised edition vol.4.chapter 4.4.1 Temperature rise in enclosed switch boards:
the average air temperature rise inside the enclosure it is Δϑ=Pveff/α/Am where Pveff=power dissipation
Am=dissipation surface α=heat transfer coefficient
α=3 W/m^2/Ko on top and 4.5 W/m^2/Ko on lateral surface
and if the top surface is 1.26 m^2 and 5.4675 m^2 lateral, the temperature rise is 3000/6.7275=446 oC!!!
If maximum temperature has to be 50oC and ambient air temperature could be 40oC then Pveff=67.3 W only.
A load factor may be used according to IEC 60439-1

 
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Dear Mr. 7anoter4

7.1. ".. ABB Switchgear Manual-11th ... Temperature rise in enclosed switch boards:
the average air temperature rise inside the enclosure it is Δϑ=Pveff/α/Am where Pveff=power dissipation
Am=dissipation surface α=heat transfer coefficient.
C.1. the formula " Δϑ=Pveff/α/Am" is confusing. The original text is Δϑ = Pv eff/(α x Am).
7.2. α=3 W/m^2/Ko on top and 4.5 W/m^2/Ko on lateral surface.
C.2 α=3 W/m^2/Ko is confusing. The original text is α=3 W/(m^2 . K ) ....
7.3. and if the top surface is 1.26 m^2 and 5.4675 m^2 lateral, the temperature rise is 3000/6.7275=446 oC!!!
C3. Total confusion.
The enclosure is 750mmDx600mmWx450mmH . Δϑ = 50C-40C = 10 K .
a) top only 0.75 x 0.6 = 0.45 m2. (α x Am) = 3W x 0.45 m2 x 10K = 13.5W.
b) 2 sides only: (0.45x 0.75x 2)= 0.67 m2. (α x Am) = 4.5W x 0.67 m2 x 10K = 30.15W
c) 1 front only 0.6x.45 = 0.27 m2]. (α x Am) = 4.5W x 0.27 m2 x 10K = 12.15W.
d) total (α x Am) = (a+b+c) = 55.8W. with Δϑ = 50C-40C = 10 K . Where Pv eff= 55.8W.
7.4. If maximum temperature has to be 50oC and ambient air temperature could be 40oC then Pveff=67.3 W only. A load factor may be used according to IEC 60439-1.
C4. Agreed. The difference in the final value could be due to the back and bottom surface were not included in the heat dissipation.
Che Kuan Yau (Singapore)
 
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Hoffman has an app available from their website which allows you to design wrt heat loss from an enclosure. (per jraef's comment above.)

"I have not failed. I've just found 10,000 ways that won't work." Thomas Alva Edison (1847-1931)

 
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Dear Mr.che12345
Thank you for your comments and I agree with most of them.
First Pveff/α/Am≡Pv eff/(α x Am). However, the original it is as you said.
Second, in the following sentence "and if the top surface is 1.26 m^2 and 5.4675 m^2 lateral the temperature rise is 3000/6.7275=446 oC"
the 1.26 and 5.4675 are not m^2 but W/oK so W/(W/oK)=oC
The actual top area is 0.6*0.75=0.45m^2[I took it as 0.42-my mistake!] and the lateral area=2*(0.6+0.75)*0.45=1.215.
Then 3*0.45=1.35 W/oK and 1.215*4.5=5.4675 W/oK
However, using the other formulae from the same manual but volume 1-chapter 1.2.5- and considering the radiation also, the correct heat dissipation from SS 304 box wall at 55dgr.C to ambient of 40 dgr.C will be 520W.It could be some conduction from the bottom but I neglected it so far.
In my opinion the 5 kVA UPS efficiency it is not less than 85% and the rectifier efficiency not less than 93% then the maximum losses it could be 1100 W.
If the equipment from the box could withstand 85oC and the prisoner air may be 55oC maximum the IP67 box wall may be 55oC and the total possible evacuated losses will be 520 W [not more].
 
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Dear Mr. 7anoter4

1. I did another calculation based on published IEC Technical Report, arrived at 40W, which is lower than 55.8W; based on the ABB Switchgear Manual. It is far from " ...55dgr.C to ambient of 40 dgr.C will be 520W ... "
2. BTW a) temperature difference adeg C - bdeg C = c K (not cdeg K)
b) thermodynamic temperature T0 = 273.15 K (not degK, where K is the SI unit name for kelvin, symbol K)

3. @structure59 (Structural) 25 Jul 20 06:45
We have not received any reply from you. It is a very important subject in electrical engineering. Is there any thing that we can help?
Che Kuan Yau (Singapore)
 
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Dear Mr. 7anoter4
Corrigendum
My sincere apology for the typographical error. The value should be 50W ; instead of 40W.
Che Kuan Yau (Singapore)
 
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Which is that IEC technical report?
 
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My rough clacs show that a box of that size will have an external surface temperature about 50C above ambient to dissipate 450W.

Assuming:
85% eff power supply providing 3000W​
Ambient about 25C​
All enclosure surfaces have the same temperature​
No solar load​
Natural convection only (no wind)​

The math is easy for external surface temperature which is why I do it this way. Components inside the enclosure can be much hotter, especially if they are not thermally coupled to the enclosure.
 
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My estimation will be as below:
Power to be dissipated = 3000 x 0.15 =450 W
Area for dissipation by radiation and convection for an ambient of 10-40 C = top 0.45 m2 Four sides= (0.75+0.6)x 2 x 0.45 =1.22 m2.Bottom area neglected,being resting on ground. Total area =1.67m2
Power dissipation= 450/1.67=269 W/m2
In oil filled transformers,45 C rise is noticed on tank or cooler surface with 600W/m2 dissipation. The rise varies as (W/m2) 0.8
Rise= 45(269/600)0.8=24C .Inside being in air, temperature rise inside may not be uniform .So an enclosure temperature rise of 30C can be expected
 
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Thanks to all participants for such fruitful discussions.
The ambient may be as high as 40 degree Celsius on some summer days.
Looking at to serious heat accumulation possibilities I am evaluating possibilities of installing heatsinks on 4 sides.

Sorry for getting late in thanking you all.
 
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