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A 2-m-internal-diameter double-walled spherical tank is used to store iced water at \(0^{\circ} \mathrm{C}\). Each wall is \(0.5 \mathrm{~cm}\) thick, and the \(1.5\)-cm-thick air space between the two walls of the tank is evacuated in order to minimize heat transfer. The surfaces surrounding the evacuated space are polished so that each surface has an emissivity of \(0.15\). The temperature of the outer wall of the tank is measured to be \(20^{\circ} \mathrm{C}\). Assuming the inner wall of the steel tank to be at \(0^{\circ} \mathrm{C}\), 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.

Short Answer

Expert verified
Question: Determine the rate of heat transfer to the iced water in the tank and the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24-hour period. Answer: (a) The rate of heat transfer to the iced water in the tank is \(99493.606 \mathrm{W}\). (b) The amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24-hour period is \(25737 \mathrm{kg}\).

Step by step solution

01

Calculate the heat transfer rate through radiation

We will first consider the heat transfer through radiation. The heat transfer rate is given by the Stefan-Boltzmann law: \(q = \epsilon \sigma A (T1^4 - T2^4)\) where: \(\epsilon\) - emissivity of the surface = 0.15 \(\sigma\) - Stefan-Boltzmann constant = \(5.67 \times 10^{-8} \frac{W}{m^2K^4}\) \(T1, T2\) - temperatures of the surfaces in Kelvin \(A\) - area of the surface The temperature of the inner side of the outer wall (\(T1\)) is \(20^{\circ} \mathrm{C}\) and the temperature of the outer side of the inner wall (\(T2\)) is \(0^{\circ} \mathrm{C}\). We need to convert them to Kelvin: \(T1_K = 20 + 273 = 293 \mathrm{K}\) \(T2_K = 0 + 273 = 273 \mathrm{K}\) The area of the spherical surface can be calculated using the formula: \(A = 4 \pi r^2\) where \(r\) is the radius of the sphere. For the outer side of the inner wall: \(r_{1} = 1 - 0.5 \times 10^{-2} = 0.995 \mathrm{m}\) \(A_{1} = 4 \pi (0.995)^2 = 12.436 \mathrm{m^2}\) Now, we can calculate the heat transfer rate through radiation between the two surfaces: \(q_{rad} = 0.15 \times 5.67 \times 10^{-8} \times 12.436 (293^4 - 273^4) = 5.606 \mathrm{W}\)
02

Calculate the heat transfer rate through conduction

The heat transfer through conduction will occur through the outer and inner wall. The heat transfer rate can be calculated using Fourier's law: \(q = k A \frac{(T2-T1)}{t}\) where: \(k\) - thermal conductivity of the material (assume for steel \(k = 16 \frac{W}{mK}\)) \(t\) - thickness of the material For the outer wall (with thickness \(0.5\mathrm{~cm} = 5 \times 10^{-3}\mathrm{~m}\)): \(q_{cond,out} = 16 \times 12.436 \frac{(20 - 0)}{5 \times 10^{-3}} = 99488 \mathrm{W}\) For the inner wall (with thickness \(0.5\mathrm{~cm} = 5 \times 10^{-3}\mathrm{~m}\)): \(q_{cond,in} = 16 \times 12.436 \frac{(0 - 0)}{5 \times 10^{-3}} = 0 \mathrm{W}\) The total heat transfer rate through conduction is the sum of the heat transfer rates through the outer and inner walls: \(q_{cond,total} = q_{cond,out} + q_{cond,in} = 99488 \mathrm{W}\)
03

Calculate the total heat transfer rate

The total heat transfer rate to the iced water in the tank is the sum of the heat transfer rates through radiation and conduction: \(q_{total} = q_{rad} + q_{cond,total} = 5.606 + 99488 = 99493.606 \mathrm{W}\) This is the answer for part (a): the rate of heat transfer to the iced water in the tank is \(99493.606 \mathrm{W}\).
04

Calculate the amount of ice melted during a 24-hour period

To calculate the amount of ice melted during a 24-hour period, we need to find the amount of heat transferred during that time: \(Q = q_{total} \times time\) Where: time = 24 hours = 86400 seconds \(Q = 99493.606 \times 86400 = 8.598 \times 10^9 \mathrm{J}\) Now, we will use the heat of fusion of ice to find the mass of ice melted. The heat of fusion of ice is: \(L_{f} = 334 \frac{\mathrm{kJ}}{\mathrm{kg}} = 334000 \frac{\mathrm{J}}{\mathrm{kg}}\) The mass of ice melted can be calculated using the formula: \(m = \frac{Q}{L_{f}}\) \(m = \frac{8.598 \times 10^9}{334000} = 25737 \mathrm{kg}\) This is the answer for part (b): the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24-hour period is \(25737 \mathrm{kg}\).

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Key Concepts

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

Stefan-Boltzmann law
The Stefan-Boltzmann law is crucial in understanding radiative heat transfer, especially when dealing with objects like our spherical tank in the exercise. This law states that the energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's thermodynamic temperature. Formally, it is expressed as
\( q = \epsilon \sigma A (T1^4 - T2^4) \)
where \( q \) is the radiative heat transfer rate, \( \epsilon \) is the emissivity of the material, \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the surface area, and \( T1 \) and \( T2 \) are the absolute temperatures of the bodies involved. The law applies perfectly for idealistic black bodies, but for real materials, emissivity comes into play, which represents how closely a material's thermal radiation characteristic follows a perfect black body. In our case, the polished surfaces have an emissivity of 0.15, which indicates that they radiate much less than a perfect black body would.
Fourier's law of heat conduction
Fourier's law is the cornerstone of conduction heat transfer — the type of heat transfer that occurs from molecule to molecule within a material. It mathematically defines the rate at which heat energy flows through a material as a result of temperature differences. The law's formula is open to interpretation in terms of its variables:
\( q = k A \frac{(T2-T1)}{t} \)
Here, \( q \) represents the heat transfer rate by conduction, \( k \) is the thermal conductivity of the material, \( A \) is the cross-sectional area through which heat is conducted, \( T1 \) and \( T2 \) are the temperatures at the two sides of the material, and \( t \) is the thickness of the material through which heat is being conducted. The given exercise uses Fourier's law to calculate the conduction through the steel walls of the spherical tank. With the assumed values for the thermal conductivity, temperature difference, and wall thickness, you can see how the heat transfer is significant when compared to radiation.
Thermal conductivity
Thermal conductivity, denoted as \( k \), is a property that quantifies a material's ability to conduct heat. It's an intrinsic characteristic of a material, implying that some materials are naturally better conductors (such as metals) than others (like wood or rubber). In the context of the exercise, the steel of the tank walls has a thermal conductivity of \( 16 \frac{W}{mK} \), suggesting that it’s a good conductor of heat. This thermal conductivity value is what we plug into Fourier’s law of heat conduction to find the rate of heat transfer through the walls of the tank. The higher the thermal conductivity, the more efficient the material is in transferring heat.
Radiative heat transfer
When it comes to radiative heat transfer, it's important to know that this is energy transferred in the form of electromagnetic waves, primarily in the infrared spectrum. Unlike conductive and convective heat transfer, radiation doesn't require any medium to occur and can happen in a vacuum. In our exercise, radiative heat transfer between the inner and outer walls of the tank is considered, and the heat transfer rate is determined using the Stefan-Boltzmann law. The concept is essential for understanding how heat energy escapes or is absorbed by objects without direct contact.
Conductive heat transfer
Conductive heat transfer is the process of heat being transferred through materials, atom by atom, molecule by molecule. It’s the underlying principle when you touch a hot object and feel the heat, or why a spoon in a pot of hot soup also gets hot. In our exercise scenario, conductive heat transfer is reviewed and calculated for both the inner and outer walls of the tank. The conductivity of the material and the temperature gradient are fundamental, as they directly affect the rate at which heat is transferred.
Heat of fusion
The heat of fusion is thermodynamics that represents the amount of energy needed to change a substance's state from solid to liquid at its melting point, without changing its temperature. This concept is typically measured in Joules per kilogram (J/kg). For instance, the heat of fusion for ice is \( 334000 \frac{J}{kg} \), which is applied in our exercise to determine the mass of the ice that melts over a 24-hour period due to the heat transfer to the iced water.
Spherical coordinates
Spherical coordinates are a system for representing the points in a three-dimensional space through three coordinates: the radius \( r \), the polar angle \( \theta \), and the azimuthal angle \( \phi \). These coordinates are particularly useful when dealing with problems involving spheres, as in our example of a spherical tank. To calculate properties like the surface area of a sphere, we often use the radius in the formula \( A = 4\pi r^2 \), which helps us to determine important variables in our heat transfer calculations. The use of spherical coordinates can simplify complex problems involving spheres by aligning with the symmetry of the sphere itself.

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Most popular questions from this chapter

A 70-cm-diameter flat black disk is placed in the center of the top surface of a \(1-\mathrm{m} \times 1-\mathrm{m} \times 1-\mathrm{m}\) black box. The view factor from the entire interior surface of the box to the interior surface of the disk is (a) \(0.077\) (b) \(0.144\) (c) \(0.356\) (d) \(0.220\) (e) \(1.0\)

Consider a vertical 2-m-diameter cylindrical furnace whose surfaces closely approximate black surfaces. The base, top, and side surfaces of the furnace are maintained at \(400 \mathrm{~K}\), \(600 \mathrm{~K}\), and \(900 \mathrm{~K}\), respectively. If the view factor from the base surface to the top surface is \(0.2\), the net radiation heat transfer from the bottom surface is (a) \(-93.6 \mathrm{~kW}\) (b) \(-86.1 \mathrm{~kW}\) (c) \(0 \mathrm{~kW}\) (d) \(86.1 \mathrm{~kW}\) (e) \(93.6 \mathrm{~kW}\)

Consider a cylindrical enclosure with \(A_{1}, A_{2}\), and \(A_{3}\) representing the internal base, top, and side surfaces, respectively. Using the length to diameter ratio, \(K=L D\), determine \((a)\) the expression for the view factor between the base and the side surface \(F_{13}\) in terms of \(K\) and \((b)\) the value of the view factor \(F_{13}\) for \(L=D\).

What is sensible heat? How is the sensible heat loss from a human body affected by (a) skin temperature, \((b)\) environment temperature, and \((c)\) air motion?

Consider a cubical furnace with a side length of \(3 \mathrm{~m}\). The top surface is maintained at \(700 \mathrm{~K}\). The base surface has an emissivity of \(0.90\) and is maintained at \(950 \mathrm{~K}\). The side surface is black and is maintained at \(450 \mathrm{~K}\). Heat is supplied from the base surface at a rate of \(340 \mathrm{~kW}\). Determine the emissivity of the top surface and the net rates of heat transfer between the top and the bottom surfaces, and between the bottom and side surfaces.

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