PSU in airtight enclosure

[Jack Benson]

Hello,

we are developing a product, and i need to put the PCB + 12v PSU in a waterproof airtight enclosure.

The PCB generates virtually no heat, but the PSU does.

The power supply will either be:

Meanwell EPS-65-12

https://www.tme.eu/Document/89f564d5574113cd0d06547e02713c61/EPS-65-SPEC.PDF

Meanwell EPS-120-12

https://www.tme.eu/Document/7810c130ad988901beb80e012e1c3214/EPS-120-spec.pdf

the PSU's are approx. 100mm * 50 mm * 30 mm high.

I am hoping to be able to use an enclosure with the internal measures: 130 mm * 130 mm * 40 mm

the product will be not be used constantly.

normally, 10/15 minutes at a time. it is possible that it could be used for up to an hour.

any insight would be welcome.

thank-you

 

[3DDave]

It is less costly to buy an enclosure and a power supply and put it together and test to see what happens than to play 200 questions to coax out what the actual use is going to be.

If the PCB produces no heat, it doesn't need a 120 Watt power supply that specifies forced air cooling.

 

[Jack Benson]

I get what you are saying, but 5 questions will get me on the right track.

the PCB gives power to a pump that is not in the waterproof enclosure - which draws the current.

 

[che12345]

Mr. Jack Benson (Industrial)(OP)28 Jul 23 07:36
"...we are developing a product, and i need to put the PCB + 12v PSU in a waterproof airtight enclosure. The PCB generates virtually no heat, but the PSU does...the PSU's are approx. 100mm * 50 mm * 30 mm high....I am hoping to be able to use an enclosure with the internal measures: 130 mm * 130 mm * 40 mm."
1. Assuming your chosen enclosure is good for your PSU. The issue is whether the enclosure will be able to maintain the temperature within the allowed of PSU.
Basic data : a) total heat generated by PSUs in (W)
b) temperature difference [inside temp allowed - outside environmental temp] in K.
c) the deposition of the enclosure i.e. base/wall mounted , area of the cooling surfaces.
d) see VDE0660 part500 for further detail.
2. As your enclosure is rather small, for simplicity, consider the cooling surface to be 3 W/Km².
Che Kuan Yau (Singapore)

 

[LittleInch]

Insight?

Depends on what level of cooling your enclosure is going to get.

Still air - this will rapidly heat up to over 80-90 C in minutes, far less than one hour, but that's at full power output.
Air doesn't take much to heat it up.
What's the enclosure made of? Metal is best for heat transfer, but plastic oof any sort will act like an insulator

Note that the power output drops off after 50C.

Your heat output is ~10-12% of the output. So you need to first define the output.

Will this be in air or water? Water all the time - no issues.

Air most or all of the time in still air - you will overheat your PSU and your PCB.

You will ned to bolt on some sort of heat sink device, but will probably find this difficult to seal.

So lots of missing information and lots of things to think about for you

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[FacEngrPE]

You can use vendors software to estimate heat rise.
Example

https://www.saginawcontrol.com/resources/thermal-calculator/

 

[IRstuff]

Waterproofing and airtightness do not preclude cooling; you can put fins in the outside of the box and place the internal heat sources in intimate contact with the inside wall where the fins are attached.

Note that airtightness is a challenging problem; you'd need seals and surfaces stiff enough to compress the seals without bending.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529 Entire Forum list

http://www.eng-tips.com/forumlist.cfm

 

[Jack Benson]

65 watts PSU @ 12% loss = 7.8 watts of heat

the volume of the box is 150 * 150 * 60 mm = 0.00135 m3

if 1 m3 of air = 1.222 kg, then 0.00135 m3 of air weights 1.6497g

i used this specific heat calculator to work out the delta T

https://www.omnicalculator.com/physics/specific-heat

is there any chance this is correct?

i divided 7.8 watts / 60 / 60 = 0.002166 Wh

does this mean every second the temp would rise 4.698 C?

i probably have made lots of mistakes

but i want to learn

 

[LittleInch]

Your box has suddenly got bigger? ( was 130 x 130 x 40... ) You have ignored the volume of the PSU itself, so air mass will be lower.

You are assuming absolute insulation and no heating of the box material or the PSU itself which will be many times more heat capacity than air.

So in theory, yes it would rise by 4.6 deg per second, but all these other effects mean that that is in fact meaningless.

However a rise of 1 or 0.5 Deg per second is possible so within a minute, the PSU will reach 50C and start to de-rate according to the charts provided.

As temp increases though, heat output from the box increases so maybe 3 - 5 minutes to get to 50C starting at 20.

But there are so many variables here we don't know that any one could mean that the temp rise is a lot faster or a lot slower or may reach steady state before getting to 50C.
Things like box material and thickness. Metal will transmit heat many times more than PVC or "plastic"
Actual power going out
Is box exposed to water which cools it or still air?
Is there a external heat sink?
Is the box insulated?



Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Jack Benson]

everything you write is correct.

I was focusing on the maths i.e. making sure that I understood the calculations and could correctly do the calculations.

yes - I made a number of errors. 130 * 130 * 40 was the approximate internal measure.

I mistakenly based my calculations on the approximate external dimensions 150 * 150 * 60 which is incorrect.

i also did not take into account that the box would not be empty.

The enclosure is planned to be plastic so it does not conduct electricity - but that means that it does not conduct heat well which introduces other problems.

since my initial post I have found that the walls of waterproof enclosures are approx. 2/3 mm thick.

the box will be exposed to still air.

at the moment i have not factored an external heat sink.

the box is not insulated - other than the material it is made from.


it occurred to me that if I could find a waterproof PSU that could be outside the enclosure then it would remove the problem:

https://www.lumenalights.com/shop/product/12v-waterproof-transformer-various-wattages/

I would need to try to source this at a sensible price from China.


or maybe this:

https://www.tme.eu/sk/en/details/xlg-75-12-a/led-power-supplies/mean-well/

does an LED power supply mean it is unsuitable for powering for example a pump?

[URL unfurl="true"]https://res.cloudinary.com/engineering-com/image/upload/v1690637562/tips/Meanwell_LED_transformer_nruhmg.webp[/url]



based on what i have written, do you think that removing the PSU from the enclosure is a sensible way forward?


thank-you for taking the time to point out the factors that i need to take into account.

 

[LittleInch]

I think it's a bad move putting a PSU inside a small box when its design is intended for essentially an open environment, so using a suitably protected PSU able to withstand water would seem to be a much better idea (IP 67 or better).

I don't know anything about small power supplies so can't help you, but a motor is different to a LED so they might not be compatible.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[7anoter4]

Try this

https://www.heatsinkcalculator.com/blog/how-to-calculate-the-temperature-rise-in-a-sealed-enclosure/

 

[IRstuff]

There are lots of options
> molding of the enclosure to include penetrating cooling fins
> thermally conductive coatings -- the below are what I found with a simple search, but you'd need to find something that does the waterproofing

https://www.surfacetechnology.co.uk/surface-coatings/thermally-conductive-conformal-coating/

https://ctherm.com/resources/tech-l...tive-pvdf-graphene-nanoplatelet-gnp-coatings/

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529 Entire Forum list

http://www.eng-tips.com/forumlist.cfm

 

[che12345]

Jack Benson (Industrial)(OP)28 Jul 23 07:36
"....we are developing a product, and i need to put the PCB + 12v PSU in a waterproof airtight enclosure....."
I look at it with numerous assumptions, for your consideration:
1. a) using 65W PSU, Eff 0.9. Heat generated = 6.5W. Note: it is loaded for only 1h on with 23h off. Take it that this is equivalent to say 0.5 x 6.5W =3.25W.
) Inside temp allowed is 40^oC, ambient air cooling temperature is 30^oC. That is dt=10K.
c) metallic enclosure, the thermal co-efficient k taken 4W/K.m².
2. Now consider what is the metallic enclosure free cooling surface area needed.
A= Q/k(dt).......in m². A=3.25W/4W x 10K = 0.082.....m².
3. Next consider the free cooling surface of the enclosure which depends on the mounting. Your intended size (130x130x40) mm is too small to dissipate the heat generated.
4. Suggested solution:
a) use a bigger metallic enclosure,
b) weld cooling fins on all available external free cooling surfaces,
c) attach heating items directly on the enclosure surface,
d) paint the enclosure inside and outside (including fins) matt black.
Che Kuan Yau (Singapore)

 

[7anoter4]

Jack Benson said:
65 watts PSU @ 12% loss = 7.8 watts of heat
the volume of the box is 150 * 150 * 60 mm
We don't know still, what is the air temperature -it is 40oC, I suppose-and what has to be maximum inner air temperature-50oC, I suppose.
However, calculating following [see attached link] we get inner air temperature of 64oC if the box material is steel, and 64.7 if the box material is p.v.c. 2 mm.
Changing the dimensions to 300*300*100 we get 50.5.

 

[Jack Benson]

i really appreciate all your replies.

i am learning a lot.

the machine will be normally at room temperature - so 18-24C. That will be the temperature of the air outside the waterproof enclosure (but still within the enclosure of the device that is expected to be 500 * 340 * 150 mm) - but as it stands there is no forced airflow outside the box. It is possible that the room temperature can be higher in summer if the building insulation is not great.

The main device case is currently made of 1.5 mm steel. It is not air tight.

It is possible in the future, we will make an IPX4 plastic external case so the main case is also waterproof. There would still be an internal waterproof plastic box for the electric parts as the device is a water heater and i want to make sure that if there is a leak the electric components are protected.

are you saying that if the box was 300*300*100, based on 7.8 watts of heat, the max temperature it would reach is 50.5C. Is that the same for both PVC or steel case?

I assume the dimensions of the box are not important - just the overall volume?

do you have a worked example of how the calculation is done?

thank-you

 

[7anoter4]

Do you mean, if I adapted such a box to any PSU some time. Never.
I am a retired power station physical design engineer and all my life I did only design.
I designed also, technological processes for power and communication cables, transformers,and other. But I never did anything with my hand directly.

 

[Jack Benson]

I would like to see how you make the calculations so I understand the maths

 

[7anoter4]

I will try to pass the calculation in excel

 

[Jack Benson]

thank-you all so much for your input.

In the short term, the product we are currently selling doesn't need the PCB to be in an airtight enclosure.

The product we are working on will be placed in a location where there will be water sprays.

given the complications of making a small airtight box for the PCB, i will focus on making the outer case airtight and check the appropriate regulations