Boiling Eggs
An ordinary egg can be approximated as a 5-cm-diameter sphere (Fig). The egg is initially at a uniform temperature of 5^{\circ} C and is dropped into boiling water at 95^{\circ} C . Taking the convection heat transfer coefficient to be h = 1200 W / m ^{2} \cdot{ }^{\circ} C , determine how long it will take for the center of the egg to reach 70^{\circ} C .
SOLUTION An egg is cooked in boiling water. The cooking time of the egg is to be determined.
Assumptions 1 The egg is spherical in shape with a radius of r_{0} = 2.5 cm . 2 Heat conduction in the egg is one-dimensional because of thermal symmetry about the midpoint. 3 The thermal properties of the egg and the heat transfer coefficient are constant. 4 The Fourier number is \tau > 0.2 so that the one-term approximate solutions are applicable.
Properties The water content of eggs is about 74 percent, and thus the thermal conductivity and diffusivity of eggs can be approximated by those of water at the average temperature of
(5+70) / 2 = 37.5^{\circ} C ; k = 0.627 W / m \cdot{ }^{\circ} C and \alpha = k / \rho C_{p} = 0.151 \times 10^{-6} m ^{2} / s (Table A-9).
Analysis The temperature within the egg varies with radial distance as well as time, and the temperature at a specified location at a given time can be determined from the Heisler charts or the one-term solutions. Here we will use the latter to demonstrate their use. The Biot number for this problem is
Bi = \frac{h r_{0}}{k} = \frac{\left(1200 W / m ^{2} \cdot{ }^{\circ} C \right)(0.025 m )}{0.627 W / m \cdot{ }^{\circ} C } = 47.8
which is much greater than 0.1, and thus the lumped system analysis is not applicable. The coefficients \lambda_{1} and A_{1} for a sphere corresponding to this Bi are, from Table 4–1,
\lambda_{1} = 3.0753, \quad A_{1} = 1.9958
Substituting these and other values into Eq. \theta_{0, \text { sph }} = \frac{T_{o} – T_{\infty}}{T_{i} – T_{\infty}} = A_{1} e^{-\lambda_{1}^{2} \tau} and solving for \tau gives
\frac{T_{o} – T_{\infty}}{T_{i} – T_{\infty}} = A_{1} e^{- \lambda_{1}^{2} \tau} \longrightarrow \frac{70 – 95}{5 – 95}= 1.9958 e^{-(3.0753)^{2} \tau} \longrightarrow \tau = 0.209
which is greater than 0.2, and thus the one-term solution is applicable with an error of less than 2 percent. Then the cooking time is determined from the definition of the Fourier number to be
t = \frac{\tau r_{o}^{2}}{\alpha} = \frac{(0.209)(0.025 m )^{2}}{0.151 \times 10^{-6} m ^{2} / s } = 865 s \approx 14.4 min
Therefore, it will take about 15 min for the center of the egg to be heated from 5^{\circ} C \text { to } 70^{\circ} C.
Discussion Note that the Biot number in lumped system analysis was defined differently as Bi = h L_{c} / k = h(r / 3) / k , However, either definition can be used in determining the applicability of the lumped system analysis unless Bi \approx 0.1.
