Respuesta :
Solution :
The dimensionless conduction heat rate, [tex]$q_{ss}^*$[/tex]
[tex]$q_{ss}^*=\frac{q\times L_c}{K A_s(T_1-T_2)}$[/tex] ...........(1)
where [tex]$L_c$[/tex] = characteristic length
[tex]$=\left(\frac{A}{4\pi} \right)^{1/2}. \sqrt{\frac{D^2}{4}}$[/tex]
A is surface area
q = heat transfer
[tex]$q=Sk(T_1-T_2)$[/tex] ..................(2)
where, S = conductor shape factor
Now substituting (2) in (1),
[tex]$q_{ss}^* = \frac{Sk(T_1-T_2)L_c}{kA(T_1-T_2)}$[/tex]
[tex]$q_{ss}^* = \frac{S \times L_c}{A}$[/tex]
[tex]$q_{ss}^* = \frac{S \times D/2}{\pi D^2}$[/tex]
[tex]$q_{ss}^* = \frac{S \times D}{2\pi D^2}$[/tex] ...................(3)
For a sphere, we know S = 2πD
Putting this in (3),
[tex]$q_{ss}^* = \frac{2 \pi D \times D}{2\pi D^2}$[/tex]
[tex]$q_{ss}^* = \frac{2 \pi D^2}{2\pi D^2}$[/tex]
[tex]$q_{ss}^* = 1$[/tex]
Therefore, the dimensionless heat conduction rate for a sphere is 1.
