Subtopic: Transient heat transfer in simple geometry
Question Statement:
A spherical fruit of radius $r = 25\text{ mm}$ with thermal conductivity $k = 0.5\text{ W/(m}\cdot\text{K)}$ is hydrocooled in chilled water with a convective heat transfer coefficient $h = 65.0\text{ W/(m}^2\cdot\text{K)}$. Taking characteristic length as $L_c = r/3$, the Biot number ($Bi$) of the fruit is:
Answer Key:
1.083
Step-by-Step Solution & Explanation
For a sphere, characteristic dimension $L_c = \frac{V}{A} = \frac{\frac{4}{3}\pi r^3}{4\pi r^2} = \frac{r}{3} = \frac{0.025}{3} = 0.00833\text{ m}$.
Biot number:
$Bi = \frac{h L_c}{k} = \frac{65.0 \times 0.00833}{0.5} = 1.083$
Since $Bi > 0.1$, internal thermal gradients are significant and lumped capacity is not valid.