Question 4.2: A rectangular billet of steel measures 150 × 300 × 600 mm wi......
A rectangular billet of steel measures 150 × 300 × 600 mm with an initial temperature of 25°C. It is placed in a furnace at 1400°C.
What will be the temperature of the center of ingot after 2 hours?
Coefficient of heat transmission h = 150 w/m²°C
Coefficient of thermal diffusivity a = 7 × 10^{−6} m²/sec
Coefficient of thermal conductivity λ = 28 w/m°C
The billet can be considered as having formed by the intersection of three infinite plates — 1, 2, and 3.
The dimensionless temperature Q at any point in a finite rectangular billet is equal to the product of dimensionless temperatures of the three infinite plates forming the given rectangle.
Let t_c be the temperature at the center (located at origin O)
\frac{t_f-t_c}{t_f-t_i}=\frac{t_f-t_{x=0}}{t_f-t_i} \times \frac{t_f-t_{y=0}}{t_f-t_i} \times \frac{t_f-t_{z=0}}{t_f-t_i} (i)
t_i = Initial temperature = 25°C
t_f = Furnace temperature = 1400°C
For a plate having thickness (2δx) = 600 mm
\begin{aligned}& \mathrm{Bi}=\frac{h \delta x}{\lambda}=\frac{150 \times 0.3}{28}=1.6 \\& \mathrm{Fo}=\frac{a \tau}{\delta x^2}=\frac{7 \times 10^{-6} \times 2 \times 3600}{0.3^2}=0.56 \end{aligned}From chart for Bi = 1.6 and Fo = 0.5
Θx = 0.66 (ii)
For a plate of thickness (2δy) = 300 mm
\begin{aligned}& \mathrm{Bi}=\frac{150 \times 0.15}{28}=0.8 \\& \mathrm{Fo}=\frac{7 \times 10^{-6} \times 2 \times 3600}{0.15^2}=2.24\end{aligned}From chart for Bi = 0.8 and Fo = 2.24
Θy = 0.28 (iii)
For a plate of thickness (2δz) = 150 mm
\begin{aligned}& \mathrm{Bi}=\frac{150 \times 0.075}{28}=0.4 \\& \mathrm{Fo}=\frac{7 \times 10^{-6} \times 2 \times 3600}{0.075^2}=8.842\end{aligned}From chart for Bi = 0.4 and Fo = 8.842
Θz = 0.055 (iv)
\begin{aligned}\frac{t_f-t_c}{t_f-t_i} & =\frac{1400-t_c}{1400-25}=0.66 \times 0.28 \times 0.055=0.0101 \\t_c & =1400-12=1386^{\circ} \mathrm{C}\end{aligned}