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Sagot :
Solution :
Characteristic length = thickness / 2
[tex]$=\frac{0.04}{2}$[/tex]
= 0.02 m
Thermal conductivity for steel is 42.5 W/m.K
[tex]$\text{Biot number} = \frac{\text{convective heat transfer coefficient} \times \text{characteristic length}}{\text{thermal conductivity}}$[/tex]
[tex]$=\frac{30 \times 0.02}{42.5}$[/tex]
= 0.014
Since the Biot number is less than 0.01, the lumped system analysis is applicable.
[tex]$\frac{T-T_{\infty}}{T_0-T_{\infty}} = e^{-b\times t}$[/tex]
Where,
T = temperature after t time
[tex]$T_{\infty}$[/tex] = surrounding temperature
[tex]$T_0$[/tex] = initial temperature
[tex]$b=\frac{\text{heat transfer coefficient}}{\text{density} \times {\text{specific heat } \times \text{characteristic length }}}$[/tex]
t = time
We calculate B:
[tex]$b=\frac{30}{7833 \times 460 \times 0.02}$[/tex]
= 0.000416
Thus, [tex]$\frac{100-50}{500-50}=e^{-0.00416 \times t}$[/tex]
t = 5281.78 second
= 88.02 minutes
Thus the time taken for reaching 100 degree Celsius is 88.02 minutes.
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