Chapter 21, Problem 65P

To determine

The view factor $F_{12}$ for the two geometries, and which geometry has the higher view factor value?

Explanation of Solution

Given:

The diameter of cone base is $D$.

The height of cone is $L$.

The diameter of cylindrical surface is $D$.

The height of cylinder is $L$.

Calculation:

For circular cone and circular disc:

Calculate the areas $A_1,A_2$ and $A_3$.using the relation

    $\begin{array}{c}{A}_1=\frac{\pi D}{2}{\left[L^2+{\left(\frac{D}{2}\right)}^2\right]}^{1/2}\\ =\frac{\sqrt{5}{D}^2\pi }{4}\end{array}$

    $\begin{array}{c}{A}_2=A_3\\ =\frac{\pi }{4}{D}^2\end{array}$

Since radiation leaving $A_2$ is intercepted by $A_1$ and $A_3$ equally, so $F_{21}=F_{23}$

Calculate the view factor $F_{12}$ using the relation (reciprocity theorem).

    $\begin{array}{c}{A}_1{F}_{12}=A_2{F}_{21}\\ F_{12}=\left(\frac{A_2}{A_1}\right)F_{21}\\ =\left(\frac{A_2}{A_1}\right)F_{23}\\ =\frac{F_{23}}{\sqrt{5}}\end{array}$

Calculate the view factor $F_{23}$.using the relation.

Refer Table 21-3 “View factor expressions for some common geometries of finite size (3-D)”.

Obtain the view factor $F_{23}$ as follows:

    $\begin{array}{c}{F}_{23}=\frac{1}{2}\left\{S-{\left[S^2-4{\left(\frac{D_3}{D_2}\right)}^2\right]}^{1/2}\right\}\\ =\frac{1}{2}\left[S-{\left(S^2-4\right)}^{1/2}\right]\\ =\frac{1}{2}\left[S-{\left(S^2-4\right)}^{1/2}\right]\\ =\frac{1}{2}\left[6-{\left(6^2-4\right)}^{1/2}\right]\\ =0.1716\end{array}$

Where

    $\begin{array}{c}{R}_1=R_2\\ =R\\ =\frac{D}{2L}\\ =\frac{1}{2}\end{array}$

And

  $\begin{array}{c}S=1+\frac{1+R_2^2}{R_1^2}\\ =2+\frac{1}{R^2}\\ =2+4\\ =6\end{array}$

Calculate the view factor $F_{12}$ using the relation.

    $\begin{array}{c}{F}_{12}=\frac{F_{23}}{\sqrt{5}}\\ =\frac{0.1716}{\sqrt{5}}\\ =0.0767\end{array}$

For circular surface and disk geometry:

Calculate the view factor $F_{21}$ using the relation (summation rule).

    $\begin{array}{l}{F}_{23}=F_{21}+F_{24}\\ F_{21}=F_{23}-F_{24}\end{array}$

Calculate the view factor $F_{12}$ using the relation (reciprocity theorem).

    $\begin{array}{c}{A}_1{F}_{12}=A_2{F}_{21}\\ F_{12}=\left(\frac{A_2}{A_1}\right)F_{21}\\ =\left(\frac{A_2}{A_1}\right)\left(F_{23}-F_{24}\right)\end{array}$        (I)

Calculate the value of $R$ for surface 3 using the relation.

    $\begin{array}{c}R=\frac{r}{L}\\ =\frac{D/2}{L}\\ =\frac{D/2}{D}\\ =0.5\end{array}$

Refer Table 21-3 “View factor expressions for some common geometries of finite size (3-D)”.

Obtain the view factor $F_{23}$ as follows:

    $\begin{array}{c}{F}_{23}=1+\frac{1-\sqrt{4R^2+1}}{2R^2}\\ =1+\frac{1-\sqrt{4{\left(0.5\right)}^2+1}}{2{\left(0.5\right)}^2}\\ =0.1716\end{array}$

Calculate the value of $R$ for surface 4 using the relation.

    $\begin{array}{c}R=\frac{r}{2L}\\ =\frac{D/2}{2L}\\ =\frac{D/2}{2D}\\ =0.25\end{array}$

Refer Table 21-3 “View factor expressions for some common geometries of finite size (3-D)”.

Obtain the view factor $F_{23}$ as follows:

    $\begin{array}{c}{F}_{23}=1+\frac{1-\sqrt{4R^2+1}}{2R^2}\\ =1+\frac{1-\sqrt{4{\left(0.25\right)}^2+1}}{2{\left(0.25\right)}^2}\\ =0.05573\end{array}$

Calculate view factor $F_{12}$ using equation $\left(1\right)$ and $L=D$.

    $\begin{array}{c}{F}_{12}=\frac{A_1}{A_2}\left(F_{23}-F_{24}\right)\\ =\frac{\pi D^2/4}{\pi DL}\left(F_{23}-F_{24}\right)\\ =\frac{1}{4}\left(0.1716-0.05573\right)\\ =0.0290\end{array}$

Thus, the view factor $F_{12}$ for circular cone and disc is $0.0767$, the view factor $F_{12}$ for circular surface and disc is $0.0290$, the circular cone and disc possess higher view factor value.