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

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.

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Đan Thu 3 years 2021-07-14T07:13:15+00:00 1 Answers 99 views 0

Answers ( )

  1. diemthu
    -1
    2021-07-14T07:14:32+00:00
    Reply

    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  

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