Answer:
t = 2244.3 sec
Explanation:
calculate the thermal diffusivity
[tex]\alpha = \frac{k}{\rho c}[/tex]
[tex]= \frac{50}{7800\times 480} = 1.34 \times 10^{-5} m^2/s[/tex]
Temperature at 28 mm distance after t time = = 50 degree C
we know that
[tex]\frac[ T_{28} - T_s}{T_i -T_s} = erf(\frac{x}{2\sqrt{at}})[/tex]
[tex]\frac{ 50 -25}{300-25} = erf [\frac{28\times 10^{-3}}{2\sqrt{1.34\times 10^{-5}\times t}}][/tex]
[tex]0.909 = erf{\frac{3.8245}{\sqrt{t}}}[/tex]
from gaussian error function table , similarity variable w calculated as
erf w = 0.909
it is lie between erf w = 0.9008 and erf w = 0.11246 so by interpolation we have
w = 0.08073
[tex]erf 0.08073 = erf[\frac{3.8245}{\sqrt{t}}][/tex]
[tex]0.08073 = \frac{3.8245}{\sqrt{t}}[/tex]
solving fot t we get
t = 2244.3 sec
