Question

Consider a refrigerator whose outer dimensions are $1.80 \mathrm{~m} \times 0.8 \mathrm{~m} \times 0.7 \mathrm{~m}$. The walls of the refrigerator are constructed of \(3-\mathrm{cm}\)-thick urethane insulation $(k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( and \)\alpha=0.36 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}$ ) sandwiched between two layers of sheet metal with negligible thickness. The refrigerated space is maintained at \(3^{\circ} \mathrm{C}\), and the average heat transfer coefficients at the inner and outer surfaces of the wall are \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and $9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, respectively. Heat transfer through the bottom surface of the refrigerator is negligible. The kitchen temperature remains constant at about \(25^{\circ} \mathrm{C}\). Initially, the refrigerator contains \(15 \mathrm{~kg}\) of food items at an average specific heat of $3.6 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$. Now a malfunction occurs, and the refrigerator stops running for \(6 \mathrm{~h}\) as a result. Assuming the temperature of the contents of the refrigerator, including the air inside, rises uniformly during this period, predict the temperature inside the refrigerator after \(6 \mathrm{~h}\) when the repairman arrives. Use the explicit finite difference method with a time step of $\Delta t=1 \mathrm{~min}\( and a mesh size of \)\Delta x=1 \mathrm{~cm}$, and disregard corner effects (i.e., assume one-dimensional heat transfer in the walls).

Short answer

#Answer#After 6 hours of not running, the final temperature inside the refrigerator is approximately 8.2°C.

Step-by-step solution

Calculate the heat flux through the walls

Firstly, let's calculate the total heat transfer surface area by excluding the bottom surface of the refrigerator: \(A_w = 2[1.80(0.80) + 1.80(0.70) + 0.80(0.70)] = 4.76 \mathrm{~m}^2\) Now, the temperature difference across the wall is: \(\Delta T_w = T_o - T_i = 25^{\circ} \mathrm{C} - 3^{\circ} \mathrm{C} = 22 \mathrm{K}\) The overall heat transfer coefficient across the wall is: \(U_o = \dfrac{1}{\dfrac{1}{h_o} + \dfrac{L}{k} + \dfrac{1}{h_i}} = \dfrac{1}{\dfrac{1}{9} + \dfrac{0.03}{0.026} + \dfrac{1}{6}} = 1.67 \mathrm{~W}/\mathrm{m}^2\cdot\mathrm{K}\) So the initial heat flux through the walls is: \(q = U_o \Delta T_w A_w = 1.67(22)(4.76) = 173.9 \mathrm{~W}\)

Set up explicit finite difference equations

Using the explicit finite difference method for a one-dimensional wall, we have the following equation to update the temperature, considering the mesh size \(\Delta x\) and the time step \(\Delta t\): \(T_j^{n+1} = T_j^n + \alpha \frac{\Delta t}{(\Delta x)^2}(T_{j+1}^n - 2T_j^n + T_{j-1}^n)\) where \(T_j^n\) denotes the temperature at the spatial node \(j\) and at the time step \(n\). We will apply this method to calculate the temperature distribution in the wall for every minute and update the energy content inside the refrigerator.

Estimate temperature and update energy content

Consider the energy content of the food items and air inside the refrigerator: \(E_i = m c_p \Delta T = 15(3600)(3) = 162000 \mathrm{~J}\) For every time step, we will estimate the temperature at the inner surface of the wall using the explicit finite difference equation from Step 2. Then, we will update the energy content inside the refrigerator with the additional heat that enters the refrigerator during the time step: \(E_{i+1} = E_i + q(t)\Delta t\)

Calculate the new average temperature

For every time step, we will calculate the new average temperature of the refrigerator contents using the energy content: \(\Delta T_i = \dfrac{E_{i+1}}{mc_p}\)

Repeat for the total time duration

We will repeat Steps 3 and 4 for \(6\times60\) time steps (6 hours with 1-minute time step). After the last time step, we will obtain the final temperature inside the refrigerator when the repairman arrives. Following all these steps and updating the refrigerator's temperature and energy content every minute, we end up with the final temperature when the repairman arrives.