Question

For heat transfer purposes, an egg can be considered to be a 5.5-cm-diameter sphere having the properties of water. An egg that is initially at 4.3°C is dropped into boiling water at 100°C. The heat transfer coefficient at the surface of the egg is estimated to be 800 W/m2⋅K. If the egg is considered cooked when its center temperature reaches 74°C, determine how long the egg should be kept in the boiling water. Solve this problem using the analytical one-term approximation method. The thermal conductivity and diffusivity of the eggs can be approximated by those of water at room temperature to be k = 0.607 W/m·°C and α = 0.146×10−6 m2/s.

Answer 1

Answer:

The time taken is   $t = 40007 sec$  

Explanation:

From the question we are told that

   The diameter of the egg is $d_e = 5.5cm = \frac{5.5}{100} = 5.5*10^{-2}m$

    The initial temperature of egg the $T_e = 4.3^{o}C$

     The temperature of the boiling water $T_b = 100^oC$

    The heat transfer coefficient is  $H = 800 W/m^2 \cdot K$

    The  final temperature is $T_e_f = 74^oC$

     The  thermal  conductivity of water is $k = 0.607 W/m^oC$

     The diffusivity of the egg $\alpha = 0.146 * 10^{-6} m^2 /s$

Using one term approximation

We have the

            $\frac{T_e_f - T_b}{T_e - T_b} = Ae^{-\lambda ^2 \tau}$

The radius is  $r = \frac{5.5*10^ {-2}}{2} =2.75*10^{-2}m$     Note that this radius is approximation to that of  a real egg

    Now we need to obtain the Biot number which help indicate the value of $A \ and \ \lambda$ to use in the above equation

     The Biot number is mathematically represented as

               $Bi = \frac{H r}{k}$

Substituting values  

               $Bi = \frac{800 * 2.75 *10^{-2}}{0.607}$

                    $= 36.24$

So for this value  which greater than 0.1 the  coefficient $\lambda_1 \ and \ A_1$ is  

        $\lambda = 3.06632$

        $A = 1.9942$

Substituting this into equation 1 we have

          $\frac{74- 100}{4.3 - 100} = 1.9942 e^{-(3.0632^2) \tau}$

          $0.2717= 1.9942 e^{-(3.0632^2) \tau}$

          $0.2717= 1.9942 e^{-9.383 \tau}$

           $0.13624 = e^{-9.383 \tau}$

Taking natural log of both sides

           $-1.993 = -9.383\ \tau$

          $\tau = 0.2124$

    The time required for the egg to be cooked is  mathematically represented as

          $t = \frac{\tau r^2}{\alpha }$

substituting value  is  

         $= \frac{0.2124 * 2.75 *10^{-2}}{0.146 *10^{-6}}$

         $t = 40007 sec$