non-contact thermomter opinions

[stu]

I am looking at buying one of these to measure stuff around the house before
and after insulation etc and was wondering if anyone had one and what they
thought of it.
http://www1.jaycar.com.au/productView.asp?ID=QM7222

also does anyone know if it will measure the temperature of glass?
thanks
Stuart

 

[[email protected]]

A really nice little unit, albeit fixed emissivity, is available at
Radio Shack, of all places. Look for stock number 22-325. I see that
Rat Shack is currently listing it for $50 but it regularly is on sale
for $29 and sometimes $19.95. I bought several at the $29 price point
after opening up my first one and finding "Raytek" on the circuit
board. Raytek is the big name in non-contact temperature measurements.

Harbor Freight sells one of these too.

No. Glass requires an instrument tuned to a different wavelength of
infrared than these units. Under certain conditions, a general
purpose unit can be used on glass...

Raytek's assume a 0.95 emissivity. Glass is 0.88. Not much different,
but you might aim at a piece of tape on the glass to improve accuracy.
Shiny surfaces (eg foil-faced foamboard) are more difficult.

Nick

 

[m Ransley]

It will measure glass it will measure shiney surfaces , my cheap
RadioShack unit does, but it is of limited use to finding leaks, best is
a blower door test to determine air exchanges per day,and find leaks,
and an IR photo before and after , you will have to find someone
providing these services.

 

[[email protected]]

It will measure glass it will measure shiney surfaces...

Wrong, esp for a shiny surface, which it will read low.

Nick

 

[stu]

Wrong, esp for a shiny surface, which it will read low.

Nick

Fistly thanks for the replies guys
I just have a few more questions
I was looking at this one because it reads to .1 of a deg and was hoping
that it would be a repeatable measurement, even though the absolute accuracy
is 2% or 2 deg. (not inculding any emissivity error, if i understand what
has been said here)
Would it be fair to say that it can be used to compare the temp of "like"
surfaces or the same surface at Different times of the day and have a
accurate comparison(even if the absolute temp is wrong)?
How bad is the error likely to be on surfaces with different emissivity?

Stuart

 

[[email protected]]

Would it be fair to say that it can be used to compare the temp of "like"
surfaces or the same surface at Different times of the day and have a
accurate comparison(even if the absolute temp is wrong)?
Sure.

How bad is the error likely to be on surfaces with different emissivity?

A 100 F surface with e = 0.95 would radiate 0.95s(460+100)^4 Btu/h-ft^2,
where s is the Stefan-Boltzman constant, vs 0.05s(460+T)^4 for a T (F) foil
surface with e = 0.05. If these are equal, T = (0.95/0.05x560^4)^(1/4)-460
= 709 F, hot enough for pizza...

Questions? Ask m.ransley

Nick

 

[stu]

Sure.

well thats pretty much all i need

A 100 F surface with e = 0.95 would radiate 0.95s(460+100)^4 Btu/h-ft^2,

5297 ???(i think, but the formula i have seen for the Stefan-Boltzman
constant gives the answer in Watts m-2? have i got something ass about?)

where s is the Stefan-Boltzman constant, vs 0.05s(460+T)^4 for a T (F) foil
surface with e = 0.05. 278
If these are equal, T = (0.95/0.05x560^4)^(1/4)-460
= 709 F, hot enough for pizza...

ok now you lost me
i was thinking more like
T^4=E/(.95*s)
T^4=278(.95*.0000000567)
the way i work it out if i was standing infront of a black body the foil
would read -191.7F
of course i find this a little hard to believe

while we are looking at formula I don't fully understand, is there a simple
formula for working out how many W in total are coming through a square
meter of wall for different temperature differences.

last question promise...... does anyone know the conversion for R values,
because australian R values seem to be about 1/10th of the US

 

[[email protected]]

5297 ???(i think, but the formula i have seen for the Stefan-Boltzman
constant gives the answer in Watts m-2? have i got something ass about?)

The metric version is P = 5.67x10^-8(273+T)^4 W/m^2, with T in C.

while we are looking at formula I don't fully understand, is there a simple
formula for working out how many W in total are coming through a square
meter of wall for different temperature differences.

Multiply the wall conductance by the temperature difference.

last question promise...... does anyone know the conversion for R values,
because australian R values seem to be about 1/10th of the US

It's closer to 5.68.

Nick