Answer:
t = 59.37 s
Explanation:
Given data:
thermal diffusivity [tex]= \alpha = \frac{k}{\rho c_p} =0.40\times 10^{-0.5}[/tex]
theraml conductivity = k = 22 W/m.K
h = 300 W/ m^2.K
[tex]T_i[/tex] = 25 degree C = 298 k
[tex]T_o[/tex] = 60 degree C = 333 k
[tex]T_{\infty} [/tex]= 75 degree C = 348 L
diameter d = 0.1 m
characteristics length Lc = r/3 = = 0.0166
[tex]Bi = \frac{hLc}{K} = \frac{300\times 0.0166}{22} = 0.226[/tex]
[tex]\tau = \frac{\alpha t}{lc^2} = \frac{0.4\times 10^{-5}\times t}{0.0166^2}[/tex]
[tex]\tau = 0.036 t[/tex]
[tex]\frac{T_o -T_{\infty}}{T_i -T_{\infty}} = Ae^[\lambda^2 \tau}[/tex]
at Bi = 0.226
Ai = 0.982
[tex]\lambda = 0.876[/tex]
[tex]\frac{333348}{298-348} = 0.982e^{-0.879^2 0.036t}[/tex]
[tex]0.3 = 0.982 e^{-0.2t}[/tex]
[tex]0.305 = e^{-0.2t}[/tex]
-1.187 = - 0.02t
t = 59.37 s