Stumped! What do you think?

Hi Everyone;

Happy New Year to all.

A job I did several months ago revealed some apparent shrinkage in a closed-cell foam ceiling. This is not the same residence for which I posted those dramatic images. But it's in the same neighborhood and was sprayed by the same individual. The foam had been in place for about one year.

The anomalies detected in this particular home were limited to a small number of non-contiguous bays in the ceiling. The home owner has since had occcasion to remove quite a bit of his tongue-in-groove ceiling lumber. He found dramatic foam separations that my imager had not even hinted at! My span was 3F.

Under what conditions would an imager, held at a ~90-degree angle to the ceiling, miss thermal anomalies in the middle section of several contiguous bays, yet detect thermal anomalies in the two outside bays?

Was there an 10C delta for several hours (4) before you scanned? Were the anomalies you found near exterior walls?

Hi Jeff;

Thanks for your reply. We had a good delta-t for the scan at 8AM. The home sits under trees so there's not much sun until late in the morning. And I have to assume the home owner was truthful in saying the heat had been on all night. But you're right, the anomalies were mostly near the exterior walls.

Lorna, on these jobs, is the foam trimmed flush with the framing, or is it sprayed in at less than full depth so that there's an air gap between the foam and the ceiling finish?

Ms. Fear,

I am assuming that when you say “tongue-in-grove ceiling lumber” you are referring to a Wood Product and not some composite laminate  (plastic / manmade material based).

If a wood product , you should  take into consideration the K and R value of the species being used and adjust your Thermal Differential dwell time to achieve a usable Δ-T accordingly.

As an example if the ceiling is Western Red Cedar it would have a K value of approx. 0.74 BTU in. per square ft.h degrees F and a R value: 1.35 per inch of thickness.

Douglas Fir would have a K value of approx. 0.74 BTU in. per square ft.h degrees F

and a R value: 0.94  per inch of thickness.

Or in other words it would require approx. 31% more time, of a applicable temperature differential, to detect a possible Thermal Anomaly to show up on the exterior surface of a Red Cedar ceiling then it would on a ceiling, of the same thickness, constructed of Douglas Fir.

Now if you add the additional R-Value of the Foam in question you can see the problems you can run into. Don’t think it was your equipment, angle of dangle or any of that stuff, my guess would be not a sufficient Thermal Differential and exposure time to achieve the desired results.

Hope this helps, not intended to be offensive in any way, just predicated on some past experiences.

Hi Guys;

Thanks for the feedback and questions.

David, the bays were not completely filled.

Al, sounds like I need the Level II class. I am not aware of how to make changes based on K factors!

Ms. Fear,

Think you missed the point; please notice that the K-Value of the two species is exactly the same, the Thinking Thermally Variable was the R-Value between the two types of wood.

If you look closely of the attachment I posted you can clearly see the difference between a Solid 2x4 and a Composite ½ inch wood product and the Defused Thermal Image produced on the Rear of the subject matter.

The Diffusion is caused by the Cellulose Structure of the objects in question, the CelluloseStructure and R-Value, of the species being inspected is going to affect the Thermal Patterns you will see, also note the low ∆-T’s involved.

The 103.6 ºF point on the Front of my example is reduced to a 90.5º surface temperature on the opposing surface.

To my knowledge no IR equipment on the market will allow you to enter a K-Value differential (could be wrong on that).

While I agree that you can not have too much Classroom Training, your own experiences, to date, have shown you that Boots On The Ground exposure with Real Life Scenarios far exceeds a week of Power Point Presentations, in some hotels conference room and a Level Something piece of paper.

In my opinion, reviewing some of your post, you are on a course to success, offer your services to some Not For Profit Organizations like Habitat For Humanity, or your local Weatherization Program, shoot all you can and Keep Asking the Contractors questions on your findings.

Lots more interesting than a hotel conference room and you can take the dollars you saved and invest it into a higher grade of Thermal Camera.

Keep up the good work, cheers,

Al,

I held off saying anything after your first post, because I assumed you had a typo.  Your most recent post however confirms that you meant to use a common value for K in combination with two different values for R.  Given that R is essentially 1/k, this does not make sense to me.  Would you explain further, please.

Thanks

Jack

Lorna, I don't know how well you'll be able to image a tongue-and-groove ceiling with an airspace behind it to the foam. There will probably be enough air movement through the board gaps and in the air space that your thermal signatures will at least be blurred. It's the same problem as you would have imaging a brick exterior wall with an air gap behind it. You mention that the areas that showed were near exterior walls, and this might mean that the extra thermal bridging there was what you were seeing, at least partially.

Jack,

Not trying to be disrespectful in any way, you are correct the K-Value of the species I listed is the same, and I might add not my calculations but the ASTM published test results.

To my knowledge the ASTM K-Value Test are based on Thermal Conductivity, while the R-Value Test is based on Thermal Resistance.

Therefore, in accordance with the ASTM Test results, both the species, I originally used as an example have the Same Conductivity but a Different Thermal Resistance in a given, corresponding, material thickness.

Again not my numbers but the data from various species test results;

Characteristics Of Douglas Fir

Density(oven-dry): 0.51 g/cm3

Specific Gravity (oven dry): 0.50

Durability: Rated as moderately resistant to decay.

Finishing: Paints, stains, varnishes, oils and waxes all work well

Flame spread rating: 90(class III)

Smoke developed classification: 70

K value(12% mc): 0.74 BTU in.per square ft.h degrees F

R value: 0.94 / in. of thickness.

Characteristics Of Ponderosa Pine

Acoustic properties: Cedar tends to dampen sound transmission

Density(oven-dry): 22 lb/ft3

Specific Gravity (oven dry): 0.35

Durability: Durable species

Fasteners: Corrosion resistant only (stainless steel, aluminum, hot-dipped galvanized, brass, etc.)

Finishing: Paints, stains, varnishes, oils and waxes all work well

Flame spread rating: 69(class II)

Smoke developed classification: 98

K value(12% mc): 0.74 BTU in.per square ft.h degrees F

R value: 1.35 in. of thickness.

Characteristics Of Western Red Cedar

Acoustic properties: Cedar tends to dampen sound transmission

Density(oven-dry): 22 lb/ft3

Specific Gravity (oven dry): 0.35

Durability: Durable species

Fasteners: Corrosion resistant only (stainless steel, aluminum, hot-dipped galvanized, brass, etc.)

Finishing: Paints, stains, varnishes, oils and waxes all work well

Flame spread rating: 69(class II)

Smoke developed classification: 98

K value(12% mc): 0.74 BTU in.per square ft.h degrees F

R value: 1.35 in. of thickness.

And I reiterate, using the R-Value to calculate the amount of time I need to expose a Temperature Differential my subject surfaces too has always produced repeatable results for me in the climate zones I have worked in.

As Mr. Snell has stated in prior post “Al in not always correct” just trying to share what has worked for me.

Cheers,

OK, so I guess adjusting for the K value is the same as using the emissivity of the specific wood type?

Don't accuse me of being a scientist, but I think what Al is talking about is using the R, U, or K value of the material you are imaging to determine the timeframe for a thermal signature originating within an assembly to become evident on the surface. I don't think it has anything to do with e, which in this case would be >.90 so basically a non-issue in a normal home inspection scenario, unless maybe there's an over-fired woodstove in the room or something else that's screamin' hot. I suppose you could use Al's math to predict when you have to make your inspection, but in most cases, if you can get a relatively stable delta T for a few hours or more, it will show up if it's going to. It sounds like you did that, and my opinion is still that the air gap is blurring the signature or even eliminating it. I would sure like to have the is2 files to play with.

Or maybe I have this totally wrong.

Al,

Thanks for your response, and nothing disrespectful about it.  I trust you feel the same about my comments.  

I do not have the ASTM test references you cite, and certainly don't have the test results you are reporting.  Going from general considerations, k and R values are generally cited for the material alone, that is they do not incorporate any consideration for the heat transfer coefficients between the surface of the material and the surrounding air. (so-called "film coefficients" usually abbreviated as h).  Using the general concept of k, thermal conductivity, in units of (BTU in)/( ft2 F h) and R, thermal resistance, in units of (ft2 F h)/BTU per inch of thickness, we can see that R = 1/k.  This works for k=0.74 and R=1.35.  It obviously does not work for k=0.74 and R=0.94.  (1/R for R=0.94 would be a k=1.06)  I am sure this is obvious to you, but I wanted to be clear for everyone.  All I can conclude is that you have some bad data.  

I am not disagreeing with you in principle about the significance of R or k on the time it takes for a change to show up, although I think there are other factors that also come into play, such as the density and specific heat of the material.  

David,

You have it right.  Scientist.  There, I said it.  

Jack

Jack,

Elated to see that we have buried the the preferable hatchet, never considered any of your post other then a Professional Query, regardless of what our counterparts across the pond may think.

That being said, can you see what we have created by Ms. Fears lasts Post?

The Bottom Line is that is the Experience Curve of the Thermographer that counts!

Hanging your hat on ASTM Factors, taken with a 40º ∆t, in a  Hot to Cold environment, in a Laboratory Climatic Controlled Chamber (NFRC standards included) ,  and then  Homogenizing the results  into a One Size Fits All scenario, for all the DOE climate zones, just doesn’t cut it for the varied geographic areas WE are expected to work with.

As an X-Employee of the United States Department Of Energy, and having to Whiteness such Certification Test, at DOE Approved Facilities, approving such Building Materials and Fenestration systems, I can assure you that, in my opinion, the System Is Seriously Flawed!

Or in another words I do not disagree with any of your or other posters comments (Mr. Midland's), concerning this thread,  just adding some real life experiences like the example below.

I know that I will get arguments that I can Not accurately pick up a Thermally Anomaly on a Power Line, (totally off subject), causing (RFI) Radio Frequency Interference in violation to FCC minimum standards from a ¼ Mile away.

I will guarantee you, if you know what you are looking for, it is possible (i.e. is the Experience Curve of the Thermographer that counts!).

While Ms. Fears assumption that “using the emissivity of the specific wood type” is part of the equation, in my real time experience, it is not the resolution to the entire question at hand.

Calculating your Thermal differential dwell time, in my experience, is your best bet.

Cheers-N-Beers or the beverage of your choice,

Al, I'll say what everyone's thinking...

"English Please!"

I value your experience and contributions but I simply cannot absorb the information you're trying to convey! Maybe, in the future, try explaining yourself using less numbers and abbreviations. Those tend to break the flow and impede the idea you're trying to convey from getting to our brains.