×
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.

#### Jobs

(OP)
Good Day to All,

I'm a bit puzzled by a phenomenon that I'm seeing during a set of CFD simulations.

As an example, Say you have an enclosure with an internal heat source that is thermally connected to this enclosure.

There is a solar load beating down on the enclosure. Simulation shows, when the object is static, the Heat source reaches 180C.
Then, with the object moving at 70 mph, the temperature due to the forced convection reduces to 150C.

Then, I decide to run the simulation without the solar load. Same scenarios, results at stationary are 165C. So a DT 15C reduction due to no solar loading, this is understandable.

What I'm really puzzled about is if I then run at 70 mph (No Solar Load), I get almost the same value as 70mph with Solar Load = 150C.

Does anyone have any suggestions as to why this maybe, as my thinking is that you should see a reduction in both instances without solar loading.

70 mph wind creates ffiction and turbulence. Friction and turbulence create heat.

(OP)
Kind of makes sense, but would it create that much heat

Also, when I compare both 70mph results (with an with out loading), there is no difference

"... if I then run at 70 mph (No Solar Load), I get almost the same value as 70mph with Solar Load = 150C." You almost get the same value which means to me that both values are not identical which would make sense. I would expect, with no solar load at 70 mph wind, the temperature would be less than with 70 mph with solar load.

(OP)
Hi,

Correct, there is about 0.4 Degrees C difference

But, when I look at the value of stationary, we get 180C with Solar Load, and 165C with out solar load, so that's a DT of 15C.

Then, at 70mph, we get results of 150C with Solar Load, and 149.6C with out solar load. Would it not be in the order of a similar value to the stationary results, as you could interoperate that the solar loading applies a 15c DT load?

70 mph is FORCED convection, compared to NATURAL convection.

But, you do not say what the actual POWER level is nor how much area is actually loaded. This results in a loaded question with no obvious answers.
> What area?
> What ambient temperature?
> What power?

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

70mph wind is a pretty serious forced convection. I can't imagine a scenario where a body would have its temperature significantly raised by solar radiation while also subject to a hurricaine-class wind.

What this says is that a 70 mph wind is sufficient cooling to almost completely eliminate any temperature rise from solar heat. Remember that the solar radiation is applied to the exact same surface as the wind, not to opposite sides.

Do a quick reality check on relative heat transfer rates.

Assume an outer skin temperature of 150°C.
Calculate heat transfer rate for forced convection with Q = h A (150°C- Tambient)
Now calcualte radiant heat transfer rate with Q = σ A F ((150°C)4 - (Tambient)4)

My guess is that the forced convection is much higher than the radiation heat transfer.

Yuor sense of direction on where results should tend to is probably correct, except that you may be using a speed which is far too high to make a difference whether you have solar loading or not, limited by internal heat generation. Try some lower speed and maybe you'll see what you expect.

Agree. WhT THE HECK IS YOUR INTERNAL SIMULATED HEAT LOAD IN WATTS?

Try 0 mph wind.
2 mph, 5 mph, and 10 mph.

Look at your units too! Seriously, temperatures and heat loads in metric, and winds in miles per hour? Sure it doesn't think you have meters/hour? Wind speed should be meters/second, but just perhaps knots. (Nautical miles per hours)

There's should, and there's reality. While military requirements documents are supposed to be in SI units, I've seen a single specification use meters, kilometers, yards, nautical miles per hour, and feet, all in the same document. Even in the case of a military standard test, there's often the use of 40 mph instead of the SI equivalent, simply because that's what it traditionally was.

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

#### 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!