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g Steel plates (AISI 1010) of 4 cm thickness initially at a uniform temperature of 500 deg C are cooled by air at 50 deg C with a convection coefficient of 30 W-m2-K-1. Estimate the time it will take for their midplane temperature to reach 100 deg C.

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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