Re: Kapton emissivity in thermal IR

[Nutz]

Here's a piece of copperclad with a bit of 3 mil Kapton tape stuck to
it,

ftp://jjlarkin.lmi.net/Cool1.JPG

(no, my hand's not that short and pudgy... that's just the angle)

and here's the same scene in the thermal IR:

ftp://jjlarkin.lmi.net/Cool2.jpg


The copper is a near-perfect mirror at these wavelengths, so you can
see the reflection of my hand. But the kapton is almost black, so it
ignores my hand and displays the actual temperature of the copper.

The ceiling is warmer than my desk, which explains the rest.

Neat: Kapton is also OK in ultra-high vacuum.


John

A question - is it that the Kapton emissivity is low for IR or that the
Kapton reflectivity at IR is low?



That camaera would sure make a handy fault finding tool for boards.

 

[[email protected]]

The emissivity is high and the reflectivity is low. The sum of E + R
always equals 1. You can judge E and R by waving your hand over some
material and looking at its temp with an imager or an IR thermometer;
if it has low emissivity = high reflectivity, you'll see the
reflection of your warm hand change the apparent temperature of the
stuff. Which is why IR temperature measurements can be so tricky: you
can seriously burn your hand touching a piece of copper or brass that
reads room temp on an IR: the IR is seeing the reflection of the room,
not the temp of the metal.




It sure does, like when a power supply is caved in and you don't know
why. I found a shorted cap like that recently. Too bad they're so
expensive.

John- Hide quoted text -

- Show quoted text -

The emissivity is high and the reflectivity is low. The sum of E + R
always equals 1.

Is this true? Reflectivity I think I understand. There will be an
angular dependence to the reflectivity. Emissivity I'm a bit vague
about....ratio of energy radiated to black body radiator at same
temperature. http://en.wikipedia.org/wiki/Emissivity
Is there any angular dependence?

George Herold

 

[Rich Grise]

The emissivity is high and the reflectivity is low. The sum of E + R
always equals 1. You can judge E and R by waving your hand over some
material and looking at its temp with an imager or an IR thermometer;
if it has low emissivity = high reflectivity, you'll see the
reflection of your warm hand change the apparent temperature of the
stuff. Which is why IR temperature measurements can be so tricky: you
can seriously burn your hand touching a piece of copper or brass that
reads room temp on an IR: the IR is seeing the reflection of the room,
not the temp of the metal.

I once used a smear of wet cigarette ashes to get an IR thermograph
reading from aluminum.

Cheers!
Rich

 

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