Answer:
t= 17 .1 s
Explanation:
Given that
Ti= 25 C
T∞= 75°C
T= 60 ºC
K=22 W/(m·K),
[tex]h=300 W/m^2.k[/tex]
[tex]\alpha = 9\times 10^{-5}\ m^2/s[/tex]
d= 0.1 m
So for sphere
Lc= d/6 = 0.0166 m
We know that
[tex]Bi=\dfrac{hL_c}{K}[/tex]
[tex]Bi=\dfrac{300\times 0.0166}{22}[/tex]
Bi = 0.22
[tex]Fo=\dfrac{\alpha t}{L_c^2}[/tex]
[tex]Fo=\dfrac{9\times 10^{-5}\times t}{0.0166^2}[/tex]
Fo = 0.32 t
Lets take
θo= Ti - T∞ =25 - 75 = 50 °C
θ = T-T∞ = 60 -75 = 15 °C
We know that
[tex]\theta =\theta _oe^{{-Bi.Fo}}[/tex]
[tex]15 =50e^{{-0.22\times 0.32t}}[/tex]
by solving this t= 17 .1 s