Question 4.4: A steel ingot measuring 100 × 250 × 500 mm is at an initial ......
A steel ingot measuring 100 × 250 × 500 mm is at an initial temperature of 30°C. It is placed in a furnace at a temperature of 1250°C for a period of 1 hr.
Determine the temperature at the center and at the center of faces of the ingot at the end of the heating period.
For steel λ = 26 w/m°C a = 6.5 × 10^{-6} m²/sec, ρ = 7.83 × 10³ kg/m³
Heat transfer coefficient to the ingot α = 200 w/m²°C.
The ingot is shown in Figure 4.7. The origin is chosen at the center.
The dimensionless temperature at any point Θ in the ingot will be equal to the product of the dimensionless temperatures Θx, Θy, and Θz of the constituent slabs at that point, i.e., (for center x = y = z = 0),
Θ_O = Θ_{x=0} × Θ_{y=0}=0 × Θ_{z=0}
Here, the thicknesses of the constituent plates are 100, 250, and 500 mm.
First calculate Bi, Fo, and Θ for the center and surface of the three constituent slabs
\begin{aligned}\delta x & =0.250 \ \mathrm{~m} \quad T=60 \ \mathrm{~min}=3600 \ \mathrm{sec} \\ a & =6.5 \times 10^{-6} \ \mathrm{~m}^2 / \mathrm{sec} \quad \lambda=26 \ \mathrm{w} / \mathrm{m}^{\circ} \mathrm{C} \\ h & =200 \ \mathrm{w} / \mathrm{m}^{\circ} \mathrm{C} \\ \mathrm{Fo} & =\frac{a \tau}{\delta x^2}=\frac{6.5 \times 10^{-6} \times 3600}{0.25^2}=0.375 \\ \mathrm{Bi} & =\frac{h \delta x}{\lambda}=\frac{200 \times 0.25}{28}=1.92\end{aligned}Similarly
For δy = 0.05 m
Fo = 9.36 Bi = 0.3846
And
For δz = 0.125 m
Fo = 1.5, Bi = 0.96
Referring to Figure 4.3 and Figure 4.4, the center and surface dimensionless temperatures Θ are
\begin{array}{rcc}\quad \delta m & \text { Ocenter } & \text { Өsurface } \\\delta x=0.250 & 0.64 & 0.34 \\\delta y=0.05 & 0.05 & 0.035 \\ \delta z=0.125 & 0.42 & 0.48\end{array}The dimensionless temperature at the center is
Θc = Θ_x \times Θ_y \times Θ_z \\ =0.64 \times 0.05 \times 0.42 = 0.01344 \\ t_c = t_f – (t_f – t_a) \times Θc \\ = 1250 – (1250 – 30 ) \times 0.01344 \\ = 1086°CFor the same Fo and Bi numbers the values of Θx = dx, Θy = dy, and Θz = dz are obtained by referring to the appropriate figure.
In Figure 4.7 P is a point at the center of the face, x = y = 0 and z = dz. It can be seen that there will be six such points, one on the center of each face, and these six points will form three pairs.
The center of each face, e.g., point P is on the face of one
slab and on the axis of the other two slabs. Hence, Θ for each such point will be obtained by the multiplication of
e.g., \Theta_{\text {center } x}, \Theta_{\text {center } y} \text { and } \mathrm{Q}_{\text {surface } z}
i.e., \Theta_P=\Theta_{\text {center } x} \times \Theta_{\text {center } y} \times \Theta_{\text {surface } z} \\ \Theta_{P x}=0.64 \times 0.05 \times 0.48=0.01536 \\ t_P=1250-1220 \times 0.01536 \\ = 1231°C
Similarly
\begin{aligned}& \Theta_{P y}=1241^{\circ} \mathrm{C} \\& \Theta_{P z}=1238^{\circ} \mathrm{C}\end{aligned}
