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Question 4.3: For the billet in Example 4.2, what will be the temperature ......

For the billet in Example 4.2, what will be the temperature at the center (t_c) after one hour of heating under the same conditions?

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The Bi will remain unchanged.

For \mathrm{Bi}=1.6  \mathrm{Fo}=\frac{7 \times 10^6 \times 3600}{0.3^2}=0.28 \\ Θx = 0.8

Similarly

For Bi = 0.8 Fo = 1.12
Θy = 0.65

and

For Bi = 0.4 Fo = 4.480

Θz = 0.22

\begin{aligned}\frac{t_f-t_c}{t_f-t_i} & =\frac{1400-t_c}{1400-25}=\Theta x \times \Theta y \times \Theta z \\& =0.8 \times 0.65 \times 0.22 \\t_c & =1400-157=1243^{\circ} \mathrm{C}\end{aligned}

What will be the temperature at the center after heating for 30 and 15 min under the same conditions?

Again, Bi will remain unchanged.

\begin{aligned}& \mathrm{Bi}=1.6 \quad \mathrm{Fo}=0.14 \quad \Theta x=0.95 \\& \mathrm{Bi}=0.8 \quad \mathrm{Fo}=0.56 \quad \Theta y=0.76 \\& \mathrm{Bi}=0.4 \quad \mathrm{Fo}=2.24 \quad \Theta z=0.48 \\& \frac{t_f-t_c}{t_f-t_i}= \frac{1400-t_c}{1400-25}=\Theta x \times \Theta y \times \Theta z \\&= 0.95 \times 0.76 \times 0.48=0.615 \\& t_c= 1400-1357 \times 0.3466=899^{\circ} \mathrm{C} \\& \text { For } \mathrm{Bi}= 1.6 \quad \mathrm{Fo}=0.07 \quad \Theta x=1.0 \\& \mathrm{Bi}=0.8 \quad \mathrm{Fo}=0.28 \quad \Theta y=0.82 \\& \mathrm{Bi}=0.4 \quad \mathrm{Fo}=1.12 \quad \Theta z=0.75 \\& \frac{t_f-t_c}{t_f-t_i}=\frac{1400-t_c}{1400-25}=\Theta x \times \Theta y \times \Theta z \\&= 1.0 \times 0.82 \times 0.75=0.615 \\& t_c= 1400-1357 \times 0.615=554^{\circ} \mathrm{C}\end{aligned}

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