3 replies to this topic

[breizh]

Hi ,
The best way to get an answer is to show what you have done for us to help . I don't think you will find people doing your homework . For sure someone will help .

Breizh

[breizh]

Hi ,

Let you perform an heat balance with several hypothesis :

R<<<<L , Thermal conductivity lambda =cte , No heat transfer gradient on R , only on X
h= superficial heat transfer coefficient , Temp at the surface is constant and equal to t .
gradient of temperauture along x ( only) .

t =t(x)

ta= ambiant temp

Heat balance : Qx -Qx+dx - Losses =0


losses= h*2*pi*R*dx ( t-ta)

Qx= -lambda*pi*R2^2*dt/dx >>>> dQ=-lambda *pi*R^2*d2t/dx2*dx

after simplifications

dQ=losses >>>> d2t/dx2= 2*h/(lambda*R) *(t-ta)

to solve this equation perform a variable change and solve it for conditions : t0=200C and tl=200 C with ta =25C

T= t-ta and w= (2*h/lamda*R) ^.5 then d2T=dx2= w^2*T >>>>>>>>>solution of this equation is : T=a1* sh(w*x) +a2*cosh (w*x)

or T = ( Tl*sh(w*x) +T0*sh(w*(L-x) ) /Sh(w*x) with Tl= (tl-ta) & T0=(to-ta)

ultimately replace the T by (t-ta) and w by its value .

Hope this helps.




Breizh

[latexman]

Another way to ensure good responses is to have a title that summarizes what you are doing/looking for, like "Centerline Temperatures of Cylinder - Sides @ 25^o C and Ends @ 200^o C".