Question

Heat transfer question

Provide a sketch and state your assumptions as you determine the shape factor F12 for the following: Z-0.4 m 2,-0.2m Y-0.3 m X-0.5 m

Answer 1

Solution

The superposition rule states that the view factor from a surface i to a surface j is equal to the sum of the view factors from surface i to the parts of surface j.

$F_{1\rightarrow 2, 3} = F_{12} + F_{13}$

Applying this rule to our problem,

3 0 4

$F_{2\rightarrow 1, 4} =F_{23}= F_{24} + F_{21}$

$\Rightarrow F_{21}=F_{23}- F_{24}$

By reciprocity theorem,

$A_1F_{12}=A_2F_{21}$

$\Rightarrow F_{12}=\frac{A_2F_{21}}{A_1}$

Assumptions

1. All surfaces are diffuse

2. Planes formed by A1+A4 is perpendicular to A2

Let us assume that there was no missing surface perpendicular to surface 2 as shown in the figure below,

Surface 3 is a combination of surface 1 and surface 4.

Z=0.4m 3 TOTAL SURFACE 2 7=0.3m x=0.5 m

X = 0.5 m, Y = 0.3 m and Z = Z_a = 0.4 m

For the figure above,

$\frac{Z}{X}=\frac{0.4}{0.5}=0.8$$\frac{Y}{X}=\frac{0.3}{0.5}=0.6$

0.5 YIX = 0.02 Z Y X 0.05 0.4 0.1 0.2 0.3 0.4 0.6 0.2 1.0 1.5 0.1 2.0 4 10 20 0 0.1 0.2 0.4 4 6 8 10 N 0.6 0.8 1 ZIX

From the given graph,shape factor is,

$F_{23}=0.25$

Now let us assume that there was only missing surface perpendicular to surface 2, as shown in the figure below,

Z=0.2 m MISSING SURFACE 2 Y=0.3 m X2o.5m

X = 0.5 m, Y = 0.3 m and Z = Z_a-Z_b = 0.4 - 0.2 = 0.2 m

For the figure above,

$\frac{Z}{X}=\frac{0.2}{0.5}=0.4$   $\frac{Y}{X}=\frac{0.3}{0.5}=0.6$

0.5 YIX = 0.02 Z Y X 0.05 0.4 0.1 0.2 0.3 0.4 0.6 0.2 1.0 1.5 0.1 2.0 4 10 20 0 0.1 0.2 0.4 4 6 8 10 N 0.6 0.8 1 ZIX

From the given graph, shape factor is,

$F_{24}=0.19$

From the formula we derived initially,

$F_{12}=\frac{A_2F_{21}}{A_1}=\frac{0.5*0.3*0.06}{0.2*0.5}=0.09$