Chapter 13: Problem 135
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
Step by step solution
Calculate the heat transfer rate through radiation
Calculate the heat transfer rate through conduction
Calculate the total heat transfer rate
Calculate the amount of ice melted during a 24-hour period
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Stefan-Boltzmann law
\( q = \epsilon \sigma A (T1^4 - T2^4) \)
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
\( q = k A \frac{(T2-T1)}{t} \)
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.