Question 4.3: Determine the view factor from a sphere of radius R1 to (a) ...

From Table, we use the relevant view factors.
(a) Sphere to Coaxial Square Plate Figure shows that

F_{1-2}\mid _{to     complete     square}=4F_{1-2}\mid_{ to \frac{1}{4}  of   a    sphere } =4\times \frac{1}{4\pi }\tan ^{-1}\left(\frac{1}{l_{1}^{\ast 2}+l_{2}^{\ast 2}+l_{1}^{\ast 2}l_{2}^{\ast 2}} \right) ^{1/2}   Table

where l_{1}^{\ast }= l/R_{2},      l_{2}^{\ast }= l/R_{2}.
(b) Sphere to Coaxial Disk

F_{1-2}\mid _{disk}=\frac{1}{2}\left[1-\frac{1}{(1+R_{2}^{\ast 2})^{1/2}} \right]      Table

where

(i) For finite R_{2} :
Now using the numerical values, we have for the finite R_{2},     l_{1}^{\ast} = 1,     l_{2}^{\ast} = 1, and R_{2}^{\ast } = 1 , and the results are

(a) F_{1-2}\mid _{square    plate }=4\times \frac{1}{4\pi } tan ^{-1}\left(\frac{1}{1+1+1} \right) ^{1/2}=0.1668

(b)  F_{1-2}\mid _{disk }=\frac{1}{2}\left[1-\frac{1}{(1+1)^{1/2}} \right] =0.1464

As expected, the view factor for the square plate is slightly larger.

(ii) For R_{2}\longrightarrow \infty :
Here we have l_{1}^{\ast }=l_{2}^{\ast }=0,R_{2}^{\ast }\longrightarrow \infty  , and noting that tan^{−1}(\infty ) = \pi/2 , we have

(a) F_{1-2}\mid _{square   plate }=4\times \frac{1}{4\pi } tan ^{-1}\left(\frac{1}{0+0+0} \right) ^{1/2}=\frac{1}{2}

(b)  F_{1-2}\mid _{disk }=\frac{1}{2}\left[1-\frac{1}{\left(1+\infty\right)^{1/2} }\right]=\frac{1}{2}