Basic radiation problem

[franck]

Hi all,

I am wondering if someone has the analytical solution (or point me to a reference book) of the following thermal problem:

I have a beam of cross section (Ac), conductivity (lambda), length (L), exposed area to space (As) and finally with an emissivity of 0.8. One end of the beam is fixed to 300K for instance and the rest of the beam is free to radiate to space (0K). I would like to know the temperature distribution along the bar.

Any comments greatly welcome!

Best regards,

Franck

 

[pmover]

i would think that your heat xsfer textbook contains sufficient technical information to aid you in solving your problem. i would not be surprised if you found a sample problem nearly identical to what you described.

of course, this does depend upon you having taken a heat transfer course . . . are you a student?

from a previous posting, there is a heat transfer textbook at:

http://web.mit.edu/lienhard/

http://www/ahtt.html

And this site is very good.

http://www.onesmartclick.com/engineering/heat-transfer.html

good luck!
-pmover

 

[franck]

I am not a student but unfortunately I am working 6000 kms from my home place where all my technical documents are located.

I am just checking if someone would have this info handy before doing the calculation myself.

Thanks for your response.

Best regards,

Franck

 

[corus]

At a guess you need to solve the differential k.d^2T/dx^2=Sx0.8x(Tk^4-273^4)/L with the boundary condition that T=300 at x=0 and
-k.dt/dx=S*0.8*(Tk^4-273^4) at x=L. For an exam question I'd simply say that the solution is trivial.

corus

 

[onebug]

you can assume 1D heat transfer problem, T(x). take an element with length dx and apply energy balance {q(x) = q(x+dx) + dq(radiation)}. apply Taylor series along with Fourier's law and you should arrive at the equation you want to solve. you might have to solve this numerically since radiation effect is non-linear and hence your derived equation. if your cross-sectional area is uniform, this will simplify your problem, but still a non-linear problem.

 

[chicopee]

franck-is there any convective heat loss from that beam in addition to the radiative heat loss that you mentioned?

 

[franck]

Hi Chicopee,

Fortunately this is for a space application so no convection. I am ended up solving this simple problem with MathCad since the solution is not really trivial.

Best regards,

Franck

 

[hacksaw]

radiation problems are never simple, with or without mathcad...

 

[chicopee]

This is what I came up with:
assumptions-one dimensional steady state flow
finite length and uniform cross section beam
T(0)=300,T'(0)=300,T'(L)=0
T"(X)= (s*As*e/k*Ax)*(T^4-T#^4)
T(X)=300+300X-1/2*X^2*T#^4+
1/30*(s*As*e/k*Ax)*X^6
s and e are the radiation properties; X-the variable beam length; Ax-cross sectional area of beam;As-surface area of beam; T(x)-is the temperature gradient along beam; T'(0)- was estimatedand probably incorrect;T#-ambient temperature. I did not have a chance to plot this equation so try it out and tell me of the outcome.

 

[franck]

Thanks Chicopee for trying to find an analytical solution to this problem! Unfortunately, your equation does not work. First, you forgot (s*As*e/k*Ax) before T#^4 but this is not a big deal. The most important issue is when integration two times T(X)^4 you are finding X^6. Basically, if you are deriving 2 times your equation you will not find your differential equation anymore. Do not spend too much time on it since I am not sure if there is an analytical solution (or maybe there is a fuchsienne function but I do not want to go there...)

 

[chicopee]

I did not get back to your reply sooner due to other commitments. Reviewing my work this is what I got:
Second order, 4th degree DE: T''(X)+M^2T^4+M^2T#^4
whereby M^2= s*Ps*e/kAx
One of the mistakes in my first reply was w/ As instead
of Ps(being the surface perimeter of the slender beam).
The solution that I gave you was based on Laplace Transform and one of the problem was to determine the I.C.
for T'(0). So far I can not get a solution to this non-linear D.E. but I think that you may try this approximation method:T(Xn)=T(Xn-1)-2*(T(Xn-1)/N+T#^4)^(1/4) on a spread sheet.
Note: N=s*e*Ps*Dx^2/k*Ax; Dx is the incremental delta x.
By the way is Ts=300 in Celsius or Farhenheit?
and T#=??? -if assumed at absolute zero above D.E. simplifies to two terms.

 

[hacksaw]

chico,

that is a tough problem period and a spread sheet calc only works for limited cases. try it is is a real learning experiece...

 

[chicopee]

Hacksaw,my solution to the DE of 5/8/05 is thru Laplace transformation. I first needed IC's for T(0) and T'(0) both of which I got. T(0) was easy but T'(0) had to be estimated thru the Relaxation method. My solution to the Laplace transform is as follows:
T(X)=T(0)+T'(0)*X + (2^-1)*M^2*X^6
where T(0) is the beam base temperature
T'(0)=dT/dX @X=0 and is a negative value
X = length measurement from beam base
M=s*Ps*e/k*Ax
also, the term involving the ambient temperature
was neglected being too low in value.
I ran a set of values other than that suggested by Franck on the above formula thru Excel and the shape of the curve was almost linear but still had a lazy S appearance.