×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Are you an
Engineering professional?
Join Eng-Tips Forums!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

#### Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

# VFD Enclosure Passive Cooling

## VFD Enclosure Passive Cooling

(OP)
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/grou...)
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-13...)

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?

### RE: VFD Enclosure Passive Cooling

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.

### RE: VFD Enclosure Passive Cooling

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.

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

### RE: VFD Enclosure Passive Cooling

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....

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

### RE: VFD Enclosure Passive Cooling

(OP)
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.

### RE: VFD Enclosure Passive Cooling

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".

#### Quote (mechE17)

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.

### RE: VFD Enclosure Passive Cooling

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.

### RE: VFD Enclosure Passive Cooling

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.

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

### RE: VFD Enclosure Passive Cooling

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

### RE: VFD Enclosure Passive Cooling

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

### RE: VFD Enclosure Passive Cooling

(OP)
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.

### RE: VFD Enclosure Passive Cooling

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

Pierre

### RE: VFD Enclosure Passive Cooling

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/m2 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 ft2

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/ft2-oF

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.

### RE: VFD Enclosure Passive Cooling

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.

### RE: VFD Enclosure Passive Cooling

#### Quote:

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)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

#### Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

#### Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Close Box

# Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...

Register now while it's still free!