Google Answers Logo
View Question
 
Q: RADIATION HEAT TRANSFER ( Answered,   1 Comment )
Question  
Subject: RADIATION HEAT TRANSFER
Category: Science > Technology
Asked by: jasonwjs-ga
List Price: $50.00
Posted: 24 Sep 2002 01:52 PDT
Expires: 24 Oct 2002 01:52 PDT
Question ID: 68346
1.A specific flat surface orbiting the earth is coated with black nickel.
  Assume the surface is a diffuse emitter with the plate is insulated on 
  one surface and the other surface is exposed to solar radiation.The solar
  irration is 1350w/m^2.Find equilibrium Temperature?
Answer  
Subject: Re: RADIATION HEAT TRANSFER
Answered By: voyager-ga on 24 Sep 2002 05:47 PDT
 
Hi jasonwjs,

this is by no means an easy question you are asking! First I will have
to make a few assumptions for the calculation to be possible at all:

- Temperature in deep space usually only reaches 3C above absolute
zero. However, you are asking about an object in earth orbit. If I
assume that it is actually orbiting the earth, (planetary and) solar
radiation would heat up an object that perfectly absorbs and emits to
280 K (7 C) (or about 5 C if planetary radiation and shielding
effects are taken into account) - which is what I will assume as your
environmental temperature. (
http://www.faqs.org/faqs/astronomy/faq/part4/section-14.html )
- I will assume that your black nickel has an emissivity of 0.06 and
an absorbtivity of 0.93 of the incoming radiation (
http://www.efdlant.navfac.navy.mil/criteria/documents/MH/1003_13A.PDF
)
- I will also assume that one surface is hit by the radiation at a 90
angle.

To calculate an equilibrium temperature we will have to use the
Stefan-Boltzmann Law:

Radiation Power ( * absporptivity ) = emissivity * Stefan's Constant *
Radiation Area (Temperature of Radiator^4 - Temperature of
surroundings^4)

In your case we assumed emissivity as 0.06, absorptivity 0.93 ,
Stefan's Constant is 5.6703*10^-8 watt/((m^2)(K^4)), Radiation Area is
assumed to be 1 m^2 for this calculation and Temperature of
surroundings is 280K. To achieve equilibrium we also have to assume
that the emission of energy is equal to the solar radiation: 1350
watt/m^2.

This brings us to the following equation:

1350 W/m^2 * 0,93 = 0.06 * 5.6703*10^-8 W/((m^2)(K^4)) * 1 m^2 * ( T^4
- 6146560000 K^4 )

If we bring T to the left side and all the other parts to the right,
the equation looks like this:

1350 W/m^2 * 0.93 = 0.06 * 5.6703*10^-8 W*m^2/K^4 * T^4 - 0.06 *
6146560000 * 5.6703*10^-8 W*m^2

T^4 = ((0.93 * 1350 + 0.66 * 6146560000 * 5.6703*10^-8)/0.06 *
5.6703*10^-8) K^4

T^4 = (0.93 * 1350/(0.06 * 5.6703*10^-8) + 6146560000) K^4

T =~ 783 K 

However, be aware that if this was a homework question or similar,
that you teacher might have assumed that the temperature in space is
2.7 K due to the famous background radiation only, without taking
planetary and solar radiation into account for the calculation of
surrounding temperature (it is very difficult to say what the
surrounding temperature in a certain area of space really is). If you
insert approximately 3 K into the above equation the result would be

T =~ 779

The resulting temperatures seem to be very high - that is due to the
black nickel coating being a nearly ideal absorber of solar radiation
while being a much poorer emitter.

I have also found an article describing the equilibrium temperature of
ballistic missiles ( http://www.ucsusa.org/security/CM_apA-E.pdf )
where the scientists took the temperature of the surrounding out of
the equation and instead integrated albedo effects into the equation.
However, it is not so easy to use their equation because it requires a
spherical rotating body.

If you assume that your body is orbiting with one of its thin sides
towards earth and in some distance to it, you could assume that the
radiation from the planet is fairly small and can be ignored. In such
a case you would just have to assume the surrounding temperature as
zero and do the calculation with that in mind:

T^4 = 1350 * 0.93/(0.06 * 5.6703*10^-8) K^4

T =~ 779 K

Please request a clarification if you have any questions about the
answer!

Additional Resources

Stefan-Boltzmann Law 
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c3

Is Space Hot or Cold?
http://spider.ipac.caltech.edu/staff/waw/mad/mad5.html

Search Terms

equilibrium temperature space diffuse
://www.google.com/search?sourceid=navclient&q=equilibrium+temperature+space+diffuse

temperature of space
://www.google.com/search?sourceid=navclient&q=temperature+of+space

absorption radiation "black nickel"
://www.google.com/search?sourceid=navclient&q=absorption+radiation+%22black+nickel%22

emissivity nickel
://www.google.com/search?sourceid=navclient&q=emissivity+nickel
Comments  
Subject: Re: RADIATION HEAT TRANSFER
From: bphillips-ga on 29 Aug 2004 21:35 PDT
 
The answer which has been posted is appropriate to the question but
ignores the bigger picture.  In assuming an area of 1m2, voyager-ga
has actually established an equation with different units on either
side, the LHS is in W/m2 and the RHS in W.

The bigger picture is to add to the LHS, the area of the body which is
absorbing energy, and change the area on the RHS to the area which is
emitting energy.  In this example, they would be one and the same,
hence use of 1m2 gives the right numerical answer, but for most real
life cases, the two areas are different, with the emitting area larger
than the absorbing area.

Important Disclaimer: Answers and comments provided on Google Answers are general information, and are not intended to substitute for informed professional medical, psychiatric, psychological, tax, legal, investment, accounting, or other professional advice. Google does not endorse, and expressly disclaims liability for any product, manufacturer, distributor, service or service provider mentioned or any opinion expressed in answers or comments. Please read carefully the Google Answers Terms of Service.

If you feel that you have found inappropriate content, please let us know by emailing us at answers-support@google.com with the question ID listed above. Thank you.
Search Google Answers for
Google Answers  


Google Home - Answers FAQ - Terms of Service - Privacy Policy