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Chapter 7: Problem 128

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.

Short Answer

Expert verified
Question: Determine the (a) rate of heat transfer to the iced water and (b) amount of ice at 0°C that melts during a 24-hour period, given a wind speed of 25 km/h and a sphere of ice of 3 m in diameter with the outside temperature of 30°C. Answer: (a) The rate of heat transfer to the iced water is 21.99 kW. (b) The amount of ice at 0°C that melts during a 24-hour period is 5703.2 kg.

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

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Most popular questions from this chapter

Solar radiation is incident on the glass cover of a solar collector at a rate of \(700 \mathrm{~W} / \mathrm{m}^{2}\). The glass transmits 88 percent of the incident radiation and has an emissivity of \(0.90\). The entire hot water needs of a family in summer can be met by two collectors \(1.2 \mathrm{~m}\) high and \(1 \mathrm{~m}\) wide. The two collectors are attached to each other on one side so that they appear like a single collector \(1.2 \mathrm{~m} \times 2 \mathrm{~m}\) in size. The temperature of the glass cover is measured to be \(35^{\circ} \mathrm{C}\) on a day when the surrounding air temperature is \(25^{\circ} \mathrm{C}\) and the wind is blowing at \(30 \mathrm{~km} / \mathrm{h}\). The effective sky temperature for radiation exchange between the glass cover and the open sky is \(-40^{\circ} \mathrm{C}\). Water enters the tubes attached to the absorber plate at a rate of \(1 \mathrm{~kg} / \mathrm{min}\). Assuming the back surface of the absorber plate to be heavily insulated and the only heat loss to occur through the glass cover, determine \((a)\) the total rate of heat loss from the collector, \((b)\) the collector efficiency, which is the ratio of the amount of heat transferred to the water to the solar energy incident on the collector, and \((c)\) the temperature rise of water as it flows through the collector.

A \(0.2 \mathrm{~m} \times 0.2 \mathrm{~m}\) street sign surface has an absorptivity of \(0.6\) and an emissivity of \(0.7\), while the street sign is subjected to a cross flow wind at \(20^{\circ} \mathrm{C}\) with a velocity of \(1 \mathrm{~m} / \mathrm{s}\). Solar radiation is incident on the street sign at a rate of \(1100 \mathrm{~W} / \mathrm{m}^{2}\), and the surrounding temperature is \(20^{\circ} \mathrm{C}\). Determine the surface temperature of the street sign. Evaluate the air properties at \(30^{\circ} \mathrm{C}\). Treat the sign surface as a vertical plate in cross flow.

Air at \(15^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) flows over a \(0.3\)-m-wide plate at \(65^{\circ} \mathrm{C}\) at a velocity of \(3.0 \mathrm{~m} / \mathrm{s}\). Compute the following quantities at \(x=x_{\mathrm{cr}}\) : (a) Hydrodynamic boundary layer thickness, \(\mathrm{m}\) (b) Local friction coefficient (c) Average friction coefficient (d) Total drag force due to friction, \(\mathrm{N}\) (e) Local convection heat transfer coefficient, \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (f) Average convection heat transfer coefficient, W/m² \(\cdot \mathrm{K}\) (g) Rate of convective heat transfer, W

A 10 -cm-diameter, 30-cm-high cylindrical bottle contains cold water at \(3^{\circ} \mathrm{C}\). The bottle is placed in windy air at \(27^{\circ} \mathrm{C}\). The water temperature is measured to be \(11^{\circ} \mathrm{C}\) after \(45 \mathrm{~min}\) of cooling. Disregarding radiation effects and heat transfer from the top and bottom surfaces, estimate the average wind velocity.

A \(20 \mathrm{~mm} \times 20 \mathrm{~mm}\) silicon chip is mounted such that the edges are flush in a substrate. The substrate provides an unheated starting length of \(20 \mathrm{~mm}\) that acts as turbulator. Airflow at \(25^{\circ} \mathrm{C}(1 \mathrm{~atm})\) with a velocity of \(25 \mathrm{~m} / \mathrm{s}\) is used to cool the upper surface of the chip. If the maximum surface temperature of the chip cannot exceed \(75^{\circ} \mathrm{C}\), determine the maximum allowable power dissipation on the chip surface.
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