Question 4.5: steel ball having radius 4.0 mm is heated to 900°C throughou......
steel ball having radius 4.0 mm is heated to 900°C throughout and then dropped in an oil bath at 50°C. For the ball Thermal diffusivity α = 0.053 cm²/sec
Specific heat c = 0.5 J/gm°C Density ρ = 7.8 gm/cm³ Heat transfer coefficient (average, long term) h = 0.2 w/cm²°C
Determine
1. The time regimes of cooling
2. The temperature gradient in each regime
The cooling regimes in the ball will depend on the ratio
\tau=\frac{r_o^2}{\alpha}=\frac{0.4^2}{0.053} \sim 3.0 \ \mathrm{sec}The early regime will be when
\tau \ll \frac{r_o^2}{\alpha} \text { i.e., } \tau \ll 3.0 \ \mathrm{sec}Let \tau = 0.1 sec
In early regime it is presumed that the surface temperature of the ball becomes equal to the oil temperature, i.e., 50°C immediately on submersion. The center temperature is 900°C.
At a depth of 0.1 cm from the surface the temperature will be
substituting
\begin{aligned}\frac{t_{0.1}-50}{900-50} & =\operatorname{erf} \frac{0.1}{2 \sqrt{0.053 \times 0.1}}=\operatorname{erf} 0.687 \\& =0.65 \\t_{0.1} & =0.65(850)+50 \sim 600^{\circ} \mathrm{C} \\\text { Similarly } t_{0.2} & =850^{\circ} \mathrm{C} \\t_{0.3} & =900^{\circ} \mathrm{C} \\t_{0.4} & =900^{\circ} \mathrm{C}\end{aligned}This shows that the cooling transient has not reached the center.
The transient regime is defined as
Let \tau = 2.5 sec
Similar calculation shows that in the transient regime
\begin{aligned}& t_{0.1}=173^{\circ} \mathrm{C} \\& t_{0.2}=297^{\circ} \mathrm{C} \\& t_{0.3}=400^{\circ} \mathrm{C} \\& t_{0.4}=526^{\circ} \mathrm{C}\end{aligned}This shows that the center has started cooling.
For long term, i.e., steady state
Let \tau = 15 sec
Volume of the ball
V=\frac{4}{3} \pi r^3=0.267 \mathrm{~cm}^3Surface area A = 4 π r² = 2.01 cm²
We will calculate the temperature at the center after 15 sec, i.e., (18 − 3) sec after the transient period of 3 sec when the center has reached about 500°C.
t_{\infty}=50^{\circ} \mathrm{C} t_1=500^{\circ} \mathrm{C}The equation to be applied for longtime cooling is
\begin{aligned}\frac{t_{0.1}-t_{\infty}}{t_i-t_{\infty}} & =\exp -\left[\frac{h A}{\rho c V}\left(\tau-\tau_c\right)\right] \\\frac{t_{0.4}-50}{500-50} & =\exp -\left[\frac{0.2 \times 2.01}{7.8 \times 0.5 \times 0.267} 15\right]=\exp -5.80 \\& =3.03 \times 10^{-3} \\t_{0.4} & =3.03 \times 10^{-3} \times 450+50 \\& =51.3^{\circ} \mathrm{C}\end{aligned}showing that the center has cooled.
If we consider
\tau = 10 sec t = 60°C and
\tau = 5 sec t = 110°C