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A 3-m-internal-diameter spherical tank made of 1 -cm-thick stainless steel \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is used to store iced water at \(0^{\circ} \mathrm{C}\). The tank is located outdoors at \(30^{\circ} \mathrm{C}\) and is subjected to winds at \(25 \mathrm{~km} / \mathrm{h}\). Assuming the entire steel tank to be at \(0^{\circ} \mathrm{C}\) and thus its thermal resistance to be negligible, determine (a) the rate of heat transfer to the iced water in the tank and \((b)\) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24-h period. The heat of fusion of water at atmospheric pressure is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\). Disregard any heat transfer by radiation.

Step by step solution

01

Calculate the convective heat transfer coefficient (h)

First, we need to convert the wind speed from km/h to m/s: Wind speed = 25 km/h = 6.94 m/s Now, we can use the following correlation to estimate the convective heat transfer coefficient, h: \(h = 0.0296(D)v^{0.8} = 0.0296(3)(6.94)^{0.8}\) Solving for h, we find: \(h = 26.09 \, \mathrm{W/m^2\cdot K}\)

02

Calculate the rate of heat transfer (Q)

Now we can use this heat transfer coefficient to calculate the rate of heat transfer, denoted as Q: \(Q = hA(T_\text{out} - T_\text{in})\) The surface area of a sphere is given by: \(A = 4 \pi r^2 = 4 \pi (1.5)^2 \) Solving for A, we find: \(A = 28.27 \, \mathrm{m^2}\) And now we can calculate Q: \(Q = (26.09 \, \mathrm{W/m^2\cdot K})(28.27 \, \mathrm{m^2})(30 - 0)\,\mathrm{^\circ C}\) Solving for Q, we find: \(Q = 21986.67 \, \mathrm{W}\) or \(21.99 \, \mathrm{kW}\)

03

Calculate the amount of ice melted in 24 hours

Using the heat of fusion of water, \(h_{if} = 333.7 \, \mathrm{kJ/kg}\), we can find the amount of ice melted in 24 hours: \(M = \frac{Q\Delta t}{h_{if}}\) \(\Delta t\) is the time interval (24 hours in this case), which needs to be converted to seconds: \(\Delta t = 24\, hours = 86400\, seconds\) Now we can compute M: \(M = \frac{(21986.67\, \mathrm{W})(86400\, s)}{333.7 \, \mathrm{kJ/kg}}\) Solving for M, we find: \(M = 5703.2\, kg\) So, the answers are: (a) The rate of heat transfer to the iced water is \(21.99 \, \mathrm{kW}\). (b) The amount of ice at \(0^{\circ}\mathrm{C}\) that melts during a 24-h period is 5703.2 kg.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convective Heat Transfer Coefficient

One fundamental aspect of heat transfer in fluids involves the convective heat transfer coefficient, denoted by the symbol 'h'. This coefficient is a measure of the amount of heat transferred between a solid surface and a fluid per unit area per unit temperature difference. In simpler terms, it quantifies how effectively heat is being carried away by the movement of the fluid (air or liquid) that's in contact with the solid.

Using an analogy, consider 'h' as the efficiency of a worker; the higher the value of 'h', the more efficient the worker is at moving

Rate of Heat Transfer

When dealing with problems in thermodynamics, one often comes across the term 'rate of heat transfer', denoted as Q. It represents the amount of heat energy moving from one system to another over time. Specifically in our case, it is the heat moving from the warmer outdoor air to the cooler iced water inside the tank. A higher value of Q indicates a greater amount of heat being transferred every second. Think of it as water flow; where Q is analogous to the flow rate of water from a tap—the larger the Q, the stronger the flow.

To calculate the rate of heat transfer, we use the formula:

Q Formula

Heat of Fusion

The heat of fusion, symbolized by

h_{if}

, is a crucial property of substances, reflecting the amount of energy required to change a unit mass of a substance from the solid to the liquid phase at constant temperature and pressure. For water, this happens at 0°C and one atmosphere, which is the scenario we have for the iced water in the tank. This property helps us compute how much ice would melt when a certain quantity of heat is supplied to it.

In educational exercises, like the one we're discussing, the heat of fusion allows us to understand the relationship between heat energy and phase change of a substance