Questions of the Week 05/13/2018

Inspection Education Topics Inspection Question of the Week
21

Congratulations, Brad! He got “Bragging Rights” to the Bonus Question.

There are 4 more questions that are begging to be answered…:smiley:

22

Good job on the bonus question Brad - and the world awaits the main answers!

23

Alright. To help get things going, I’m going to give the answer to question #2

  1. Ordinary glass is opaque to infrared radiation.
    True
    False
    It depends… (explain)

Answer:

It depends on the infrared wavelength. Glass is opaque to long wave infrared (LWIR), which is what most thermal imagers detect, but not short wave infrared (SWIR). If you image an incandescent light bulb with a LWIR imager, you will see the heat radiated from the glass itself. A SWIR imager can actually see the filament inside the light bulb.

The two images that I posted earlier were taken by Scott Gilligan, using his SWIR imager. The first one shows through the bulb to the reflector and you can actually see a reflection of the filament in the reflector. The second image is tuned to only show the filament.

OK so let’s give everyone a second bite at the apple. The folks who have already posted a response get to take a second crack at it. As a hint, there is still one question (other than #2) that nobody has provided an adequate answer to yet - "RAT"s.

24
  1. B
  2. It depends on the infrared wavelength
  3. B, C
  4. F
  5. up [feeling like the coffee still isn’t kicking in, but thanks for the mulligan]
25

OK. Here’s an at-home experiment for question #5 for those who have an imager at home.

Set your Emissivity to 1.00
Set your Reflected Temp to 70°F

Take an ice cube out of the freezer and shoot it with your imager (make sure you get close enough to it). Record the temperature. Now change your Emissivity to .50 and do it again. What happened to your reading?

Now set your Emissivity back to 1.00 take a glass of hot water (100°F+) and shoot it with your imager. Record the temperature. Now change your Emissivity to .50 and shoot it again. What happened to your reading?

When you change the Emissivity value of your imager, you’re telling it what % of incident radiation it is detecting is coming from the object that you are observing and what % is coming from other places. Think about what process your imager is going through when you change emissivity values…

Now somebody answer this:

Decreasing the emittance value of your imager will cause the observed temperature to go ____
Up
Down
It depends… (explain)

Note: the above experiment will not give you an accurate temperature result, but it will demonstrate the process the imager goes through and provide the answer to question #5.

26

My Seek Reveal Pro has 5 emissivity settings, from matte (.97) to gloss (.30). It has no reflected temp setting. That being said, ice cubes at .97 measured 19 deg f, and at .30 measured 30 deg f, (more accurate). Microwaved a cup of water, at .97 measured 165 deg (more accurate) and at .30 measured 271 deg (it was not superheated boiling!)

It depends on accuracy, but the lower the emissivity setting the higher the temperature reading…
Thanks for the lesson.

27

Steve,

Thanks for running the experiment, but unfortunately, you can’t do it with your imager because it provides no control over reflected temperature. I have no idea what it is assuming for reflected temp or what it’s trying to do in the way of calculation.

The four images below represent the experiment on an imager with both emissivity and reflected temperature control.

As you can see with the ice cube (looking at the spot temperature), reducing the emissivity from 1.00 to .50 causes the imager reported temperature to decrease (note: the * on the display indicates that the temperature is outside of the selected range set on the imager)

On the glass of hot water, reducing the emissivity from 1.00 to .50 causes the imager reported temperature to increase.

So, the answer to the question is:

  1. Decreasing the emittance value of your imager will cause the observed temperature to go ____
    Up
    Down
    It depends on whether the reflected temperature is higher or lower than the target object temperature. If the reflected temperature is higher than the object temperature reducing the emissivity value will cause the observed temperature to go down. If the reflected temperature is lower than the object temperature, reducing the emissivity value will cause the observed temperature to go up.

Setting imager emissivity tells the imager how much weight to give to the apparent temperature of the target object vs. the reflected apparent temperature when calculating the observed temperature from the total incident radiation.

For an authoritative reference, you could refer to the “Infraspection Institute Level-II Certified Thermographer Reference Manual” chapter on emittance.

Keep in mind that changing emissivity does not alter the image itself in any way. Emissivity only matters when you need to determine actual temperatures. It is not necessary for qualitative thermography. This is why Jim Seffrin advises Home Inspectors not to include temperatures in their reports as they can be very inaccurate if the inspector does not take these and other factors into account.

Note that the objective of this experiment was to illustrate the effect of changes to the emissivity settling, not to produce an accurate reading of the temperature of the ice cube and water glass. The values that we used for emissivity and reflected temperature were not the correct values to yield an accurate temperature measurement.

IR.jpg

IR.jpg

IR.jpg

IR_12900.jpg

IR_12898.jpg

28

Bumpity…bump…bump. :smiley:

29

Very informative - thank you. The books are a bonus, but the learning keeps me coming back.

30

Still waiting for someone to put up all 5 right answers. No more clues.

31

Thanks, Steve, the Awards Committee hopes that is what it is about.

We appreciate ALL those who take the time to participate. :smiley:

32

I’m heading for Colorado in the wee hours of the morning so rather than leave folks hanging, the answers are below.

I wanted to make this one a bit challenging. I didn’t intend to make it insurmountable. Thanks to all who participated.

Chuck

Answers:

  1. A solid object that is opaque to infrared radiation and is a good reflector of infrared radiation will be a _____ emitter of infrared radiation.
    A. strong
    B. poor (For opaque objects reflectivity and emissivity are complementary and combined always equal 1.0. The more reflective an object is, the less emissive it is.)
    C. unpredictable based on information provided… (explain)
  2. Ordinary glass is opaque to infrared radiation.
    True
    False
    It depends on the infrared wavelength. Glass is opaque to long wave infrared (LWIR), which is what most thermal imagers detect, but not short wave infrared (SWIR). If you image an incandescent light bulb with a LWIR imager, you will see the heat radiated from the glass itself. A SWIR imager can actually see the filament inside the light bulb.
  3. The emissivity of an object may be affected by (identify all that are FALSE).
    A. the surface smoothness of the object
    B. the shape of the object
    C. the angle at which the object is viewed
    D. the temperature of the object
    E. the color of the object
    F. the IR wavelength
  4. If you plug the appropriate emissivity of an object into your infrared imager, it will give you an accurate representation of the actual temperature of that object.
    True
    False (there are multiple other parameters to set and factors to take into account in order to be able to accurately determine temperature)
  5. Decreasing the emittance value of your imager will cause the observed temperature to go ____
    Up
    Down
    It depends on whether the reflected temperature is higher or lower than the target object temperature. If the reflected temperature is higher than the object temperature reducing the emissivity value will cause the observed temperature to go down. If the reflected temperature is lower than the object temperature, reducing the emissivity value will cause the observed temperature to go up.

Bonus question (the prize is bragging rights): How do Low-E windows for hot climates differ from Low-E windows for cold climates and why?

In very hot climates, the Low-E coating should be on surface #2 (reducing the ability of the outer glass to radiate heat into the structure across the space between the panes). In very cold climates, it goes on surface #3 (reducing the ability of the inner pane to radiate heat out of the structure across the space between the panes).

Surfaces are numbered from the outside in: The outside surface of outside pane is #1; The inside surface of outside pane is #2; The outside surface of the inside pane is #3; the inside surface of the inside pane is #4

33

Excellent questions and a true learning opportunity, thank you for your time. Enjoy your trip to CO.

34

GREAT job as usual Chuck!
Thank you :smiley:
:cool:

35

Ditto!! :smiley:

36

Thanks Chuck :slight_smile:

37

Thanks Chuck!

38

Good job Chuck. Don’t eat the Colorado brownies, but do stop by every brewery you get the opportunity to.

39

Thanks guys. I usually hit at least one brewery when I go. Hails was pretty crazy. Looks like more to come.

40

Answers (without looking at responses)

1 - B
2 - Depends. If the glass is in contact with a hot item, the imager will detect heat on the opposite side of the glass at the point of contact.
3 - DEF
4 - False
5 - Up

A little more than halfway through Level I. This was a very educational QOTW. Thanks!