Question

heat transfer

Consider the picture shown below. The very long cylindrical rod (surface 1) acts as a black surface 1 m in diameter which is maintained at 1000 K. The rod is surrounded by a very thin cylindrical shield with inner surface 3-1 and outer surface 3-2. The diameter of this shield is 2 m; and is so thin that the inner and outer surface areas are essentially the same. The shield has an emissivity of 0.3 on both sides. The shield is protecting a third cylindrical surface (surface 2) with an inner diameter of 3 m, temperature of 200 K and emissivity of 0.5. Gray inner surface, D2 = 3m, T2 = 200 K Ex=0.5 Black surface, A,- D = 1 m, T = 1000 K Shield with gray inner and outer surfaces D3 = 2 m, £3,2 and 3,1 = 0.3 a. What is the view factor between the inner rod (surface 1) and the inside of the shield (surface 3-1)? No tables or charts required! b. What is the view factor between the outside of the shield (surface 3-2) and the inside of the outer cylinder (surface 2)? No tables or charts required!

Answer 1

Solution- (a) We have following equations to solve-

F11 + F13 =1

A1F13 = A3F31

Now, as from the figure, F11= 0 ,

So,

F13 = 1

$F_{31}=\frac{A_{1}F_{13}}{A_{3}}$

$F_{31}=\frac{A_{1}}{A_{3}}=\frac{\Pi *1*L}{\Pi *2*L}=0.5$

Ans: F31=0.5

(b) For surface 2-3 we have

  $F_{32}+F_{33}=1$

$A_{2}F_{23}=A_{3}F_{32}$

Now, from fig we have  

Then,

$F_{23}=\frac{A_{3}F_{32}}{A_{2}}$

$F_{23}=\frac{A_{3}}{A_{2}}=\frac{\Pi *2*L}{\Pi *3*L}=0.67$

Ans F23= 0.67