| |
VK3ABK > TECH 23.02.04 12:00l 82 Lines 3454 Bytes #999 (0) @ WW
BID : 16433_VK3KAY
Read: DB0FHN GUEST OE7FMI
Subj: Visible Colours and Heat.
Path: DB0FHN<DB0RGB<OK0PPL<DB0RES<ON0AR<IK1ZNW<ZL2TZE<VK3KAY
Sent: 040223/0918Z @:VK3KAY.#WEV.VIC.AUS.OC #:16433 [Wendouree] $:16433_VK3KAY
From: VK3ABK@VK3KAY.#WEV.VIC.AUS.OC
To : TECH@WW
Hello again all
The subject of 'black cars' introduced by Ralph, VK2ZRG, has created a
good bit of interest. This extract from a recent bulletin from John, G4YUU,
continues the generally accepted notion that black is the 'colour' for a
hot car!
John writes, in part...
"BTW...Black Gloss reflects sunshine more than any other colour BUT
unfortunately just look at a house that has been painted Black Gloss and in
a very short time it will start peeling due to the absorption of sunshine.."
This, and a comment from Ian, G0TEZ, goes close to a heat transfer effect
that I have been anticipating. Ian commented, with a touch of mystery,
"...as you get further into it, it's not just as simple as black bodies
absorb heat and white ones reflect."
I thought he would have continued with an explanation, but no; not yet. :-)
Then, Pete, G6KUI, sent this comment...
"What everyone seems to forget is that the "colour" of a car is what we
see in the visible spectrum, and what heat is absorbed is dependent on
what is absorbed or not in the infrared spectrum.
Something that looks black in the visible spectrum might not be so in
the in infrared spectrum. Silver/white in the visible may look black in the
infrared and hence absorb more heat energy than black in the visible that
may look white in the infrared." (I think I understand that!) :-)
The missing piece in this discussion is 'EMISSIVITY'. That is, the heat that
is emitted depending on the emitting surface. A perfect emitter (a blackbody)
is given an emissivity of '1' and is the reference for other emitters.
So, when a coloured surface (a car) is absorbing heat energy, it is, at the
same time, emitting heat.
+ +
+ | +
+ | +
+ | +
+ | +
+ | +
| +
| +
| +
|
--------------------------------------------------------------
visible <ÄÄÄÄÄÄ 600øC ÄÄÄÄÄÄÄÄ> heat (IR)?
Energy distribution from a 600øC source, ie. dull red in colour.
The area under the curve represents energy (not to scale) and much more is
as heat extending well into the near infra-red. Visible energy is probably
shown too generously as my graphic has limitations. Light emitted would not
extent beyond 'red' (about 600ø) but Wein's law provides for this. Amplitude
is arbitrary as heat energy would depend on the mass of the source.
So, Pete is worrying about is (a) The colour black.
But 'black' is just the absence of reflection of all 'colours'. But not the
infrared. So, a 'black' surface will still emit energy but we can't see it.
And (b) Pete is getting us confused with what is or isn't visible! But the
point here is that a light colour, (or a "silver/white") will both reflect
'and' have an emissivity of less than 1. That is, it emits less of the
incident heat than a black one. This is where it gets "not just as simple
as black bodies absorb heat and white ones reflect" as G0TEZ tells us.
73. Dick. VK3ABK.
Read previous mail | Read next mail
| |
