VFD Enclosure Passive Cooling 1

[mechE17]

Hi everyone. I am designing stainless steel, NEMA 4X electrical enclosures for VFDs and to reduce costs, the goal is to use passive cooling only and dissipate heat through the enclosure itself. I've attempted to calculate the required surface area to dissipate heat through three different methods and received three different results.
1. Rule of thumb from Rockwell Automation which states A = (Watts dissipation)*8 (

https://literature.rockwellautomation.com/idc/groups/literature/documents/td/sp200-td001_-en-e.pdf)

2. Heat transfer conduction through a uniform material: Q = -k*A(T2-T1)/L, using thermal conductivity of SS for k = 16.2 W/(m*K).
3. Simple heat transfer equation: (Watts dissipation) = (heat transfer coeff)*(A)*(ΔT), with a heat transfer coeff = 5.5 W/(m2*K) for stainless steel. (

http://www.vfds.org/vfd-cooling-and-ventilation-136857.html)

For a small VFD generating 18 W of heat, method 1 says 92880 mm², method 2 says 55.6 mm², and method 3 says 163636 mm².

Is one of these methods best, or is there another equation to consider?

 

[3DDave]

Method 2 works if there is a heat sink on the other side that can keep the temperature of the box surface at whatever you put as the limit. In real use, there is a boundary of air that will heat up and insulate the box surface. If the outer surface was in water, with high thermal conductivity, it would likely be enough, but not with air.

You will rarely regret having more area than required and will soon regret if it is far too little. In between you may get 1 year instead of 5 years due to cooking the capacitors or some other trade-off.

 

[LittleInch]

Same input data?

No 2 is simply the square area of steel assuming a constant temp. Steel transmits heat very well, air much less so.



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[LittleInch]

Lots depends on how the outside air is. Inside an AC controlled container at say 25C and with a bit of circulating air and no 1 should be fine as it's based on 40C ambient air.

Expjsed to really hit still air outside and its a different story.

But 18 W heat rejection?? My laptop puts out more than that....

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[mechE17]

Thanks for the quick replies. Installations will be inside, not exposed to sunlight. I'm using the same input data for the methods with heat transfer equations, with ΔT = 20°C assuming 20°C outside the enclosure and a max temp of 40°C inside (this is the max from the VFD vendor).

The 18W heat rejection is the given value for the smallest drive, I just took it as an example. #2 makes the most sense to me and was my first instinct, but a required area of 55.6 mm² isn't realistic compared to the other answers, which are so much larger.

 

[MintJulep]

Your equation (2) is for conduction through the metal only, and does not consider the convection needed on both surfaces.

The source for equation (3) shows a lack of understanding of the fundamentals. "The better heat conduction of the enclosure the more heat dissipated."

It might be safe to assume that the VFD is designed with adequate cooling, and therefore the cooling features on your enclosure need to be "about the same".

max temp of 40°C inside (this is the max from the VFD vendor).

You probably don't want to use the VFD's max allowable condition as your design condition.

 

[georgeverghese]

Eqn 3 is familiar to me, 5.5w/m2/degK is approx equal to the natural convection htc you get from a hot metal surface (bit on the high side of 3 to 4w/m2/degK). So, for this 18w dissipation, you've got a dt=20degC for the total surface area of 0.16m2. Assuming air temp is say 25degC, then metal surface temp = 45degC. At 45degC, you could neglect the additional heat dissipation by radiation. In summer, surface temp would be somewhat higher. Am assuming you have no other comparable heat transfer resistances acting in series with natural convection in this case.

 

[LittleInch]

The Rockwell data says they use 40C outside with a DT of 15C.

Why no fan in the enclosure?

This is 18W. It is a very low number.

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[Compositepro]

With only natural convection, only the surfaces above the elevation of the heat source in the enclosure will transfer heat.

 

[IRstuff]

Method 2 is incomplete; you need to convect the heat into the atmosphere, not just get from one side of the enclosure to the other.

TTFN (ta ta for now)
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[EdStainless]

We never trusted critical equipment to passive cooling.
At the very least we used a set of fins cut into the wall of the enclosure with fins inside and out.
And fans on both sides as well.
If we really cared the enclosure was air conditioned.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed

 

[mechE17]

Apologies for the delayed response. When I add convection to method #2 by adding Newton's Law of Cooling to the equation, it barely changes the area required and it's still significantly lower than the other methods. Is this too simplified of equations to simulate the situation? Conduction equation to get the heat through the metal, then convection to get the heat into the environment.

Q = [-k*A(T2-T1)/L]+[h*A*(T2-T1)], and estimate h = 12 W/(m²*K), then solve for A.

 

[pierreick]

Hi,
It's one or the other: heat transferred by conduction = heat transferred by convection.

Pierre

 

[Snickster]

As others may have mentioned, the 2nd equation just considers conduction heat transfer through the stainless steel metal and does not include the inside and outside air film convection heat transfer coefficients. So this equation is not correct. The correct way to calculate the heat transfer across a surface is to include the inside and outside air film heat convection transfer coefficients. The 3rd method appears to include the convection film coefficients but I think the value stated for U of 5.5 w/m[sup]2[/sup] K may be a little high.

I work in american units so I have values of inside and outside convection heat transfer resitance R both at:

0.68 deg. F per Btu/hr per ft[sup]2[/sup]

And the resistance of the metal as negligible.

So total thermal resistance is 1.36 and U overall heat transfer coefficient is 1/R = 0.735 Btu/hr/ft[sup]2[/sup]-[sup]o[/sup]F

And heat transfer Q = U A dT

I would check by the method 1 which is a rule of thumb method by the manufacturer and then check by this heat transfer equation and use the most conservative value.

For typical electrical equipment the maximum temperature allowed is 104 F which will then be the maximum inside cabinet design temperature and the cabinet external design temperature would be the maximum possible space temperature.

Typically a space with electrical equipment installed is at least ventilated with outside air to remove heat or air conditioned with HVAC units.

 

[georgeverghese]

There is a significant air gap between the VFD electronics and the enclosure; and this hasnt been taken into account. So at 45degC metal cabinet wall temp, the electronics will be much hotter.

 

[IRstuff]

stated for U of 5.5 w/m2 K may be a little high.

1/(0.68 deg. F per Btu/hr per ft2) = 8.35 W/m^2-K

In any case, if your spec limit for the VFD is 45 degC, then a 40 degC enclosure temp is inappropriate, since you'll have at least a 20 degC delta between the VFD and the interior surface of the enclosure

TTFN (ta ta for now)
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