Question

A large cast iron container \((k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=1.70 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) ) with \(4-\mathrm{cm}\)-thick walls is initially at a uniform temperature of \(0^{\circ} \mathrm{C}\) and is filled with ice at \(0^{\circ} \mathrm{C}\). Now the outer surfaces of the container are exposed to hot water at $55^{\circ} \mathrm{C}$ with a very large heat transfer coefficient. Determine how long it will be before the ice inside the container starts melting. Also, taking the heat transfer coefficient on the inner surface of the container to be $250 \mathrm{~W} / \mathrm{m}^{2}\(, \)\mathrm{K}$, determine the rate of heat transfer to the ice through a \(1.2-\mathrm{m}\)-wide and \(2-\mathrm{m}\)-high section of the wall when steady operating conditions are reached. Assume the ice starts melting when its inner surface temperature rises to $0.1^{\circ} \mathrm{C}$.

Short answer

Answer: It takes approximately 621.5 seconds for the ice inside the container to start melting. Question: What is the rate of heat transfer to the ice when steady operating conditions are reached? Answer: The rate of heat transfer to the ice at steady-state conditions is approximately 32.654 kW.

Step-by-step solution

Find the surface temperatures at the melting point of ice

When the ice starts to melt, the temperature on the inner surface of the container will rise to \(0.1^{\circ}\mathrm{C}\). Since heat transfer occurs from the hot water surrounding the container, heat is being transferred from the outer surface at \(55^{\circ}\mathrm{C}\) to the inner surface. We will consider this step complete having defined our surface temperatures under the suitable conditions.

Apply Lumped System Analysis

Since we are dealing with conduction through a solid and large heat transfer coefficients, we can apply the lumped system analysis to find the time it takes for the ice to start melting. The Biot number (Bi) is given by: Bi = \(\frac{hL_c}{k}\) where \(h\) is the heat transfer coefficient, \(L_c\) is the characteristic length (\(\frac{V}{A}\)) and \(k\) is the thermal conductivity. We will use the heat transfer coefficient of the inner surface (\(h = 250\mathrm{~W}/\mathrm{m}^2\cdot\mathrm{K}\)) and the cast iron properties (\(k = 52\mathrm{~W}/\mathrm{m}\cdot\mathrm{K}\) and \(L_c = 0.04\mathrm{~m}\)) to find the Biot number. Bi = \(\frac{250 \times 0.04}{52} \approx 0.1923\) Since Bi < 0.1, the lumped system analysis is not applicable. Therefore, we need to use the unsteady-state heat conduction approach.

Apply Fourier's Number and Biot Number for Time Calculation

First, let's determine the temperature difference ratio, which is given by \(\Theta_{ratio} = \frac{T_{melt} - T_s}{T_\infty - T_s}\), where \(T_{melt}\) is ice melting temperature, \(T_s\) is inner surface temperature and \(T_\infty\) is hot water temperature. \(\Theta_{ratio} = \frac{0.1 - 0}{55 - 0} \approx 0.0018\) Now, we need to calculate the Fourier number (Fo), which is given by Fo = \(\frac{\alpha t}{L^2}\), where \(\alpha\) is the thermal diffusivity, \(t\) is time and \(L\) is the thickness of the wall. We can rewrite Fo in terms of Biot number (Bi) and temperature difference ratio (\(\Theta_{ratio}\)). From the unsteady-state heat conduction charts or equivalent mathematical relations: \(1 - \Theta_{ratio} \approx 0.9273 = C_1 \mathrm{exp}(-s_1^2 \dot \mathrm{Bi}\dot \mathrm{Fo})\) Since Bi = 0.1923, we can find \(C_1 \approx 1.04\) and \(s_1 \approx 1.3369\). Now, we can solve for Fourier number Fo: \(0.9273 \approx 1.04 \times \mathrm{exp}(-1.3369^2 \times 0.1923 \times \mathrm{Fo})\) Fo \(\approx 0.0654\) Now, we can solve for time: \(\frac{\alpha t}{L^2}\) = Fo \(t = \frac{\mathrm{Fo}\cdot L^2}{\alpha} = \frac{0.0654 \times (0.04)^2}{1.7 \times 10^{-5}}\) \(t \approx 621.5\mathrm{~seconds}\)

Find the rate of heat transfer at steady-state conditions

Now, we will calculate the rate of heat transfer at steady-state conditions when the heat transfer coefficient on the inner surface is given as 250 \(\mathrm{W/m^2 \cdot K}\). Let \(q\) be the rate of heat transfer through a \(1.2\mathrm{~m}\)-wide and \(2\mathrm{~m}\)-high section of the wall, then: \(q = hA(T_\infty - T_s)\) where \(A = 1.2\mathrm{~m} \times 2\mathrm{~m} = 2.4\mathrm{~m}^2\), \(h = 250\mathrm{~W}/\mathrm{m}^2\mathrm{K}\), \(T_\infty = 55^{\circ}\mathrm{C}\), and \(T_s = 0.1^{\circ}\mathrm{C}\). \(q = 250 \times 2.4 \times (55 - 0.1)\) \(q \approx 32,654\mathrm{~W}\) or \(32.654\mathrm{~kW}\) The rate of heat transfer to the ice through the wall at steady-state conditions is approximately 32.654 kW.