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Chapter 13: Problem 21

Consider coaxial, parallel, black disks separated a distance of \(0.20 \mathrm{~m}\). The lower disk of diameter \(0.40 \mathrm{~m}\) is maintained at \(500 \mathrm{~K}\) and the surroundings are at \(300 \mathrm{~K}\). What temperature will the upper disk of diameter \(0.20 \mathrm{~m}\) achieve if electrical power of \(17.5 \mathrm{~W}\) is supplied to the heater on the back side of the disk?

Short Answer

Expert verified
The temperature of the upper disk is approximately 400 K when electrical power of 17.5 W is supplied to the heater on the back side of the disk.

Step by step solution

01

Applying Stefan-Boltzmann Law

To find the heat exchanged between the two disks, we need to apply the Stefan-Boltzmann Law. For any black object, the radiant emittance is given by: \[E = \sigma T^4\] Where E is the radiant emittance, 蟽 is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W~m^{-2}~K^{-4}}\)), and T is the temperature in Kelvin.
02

Calculate the heat exchange between the disks

The heat exchange between the two disks can be represented by the difference between the radiant emittance of the lower disk and the heat absorbed by the upper disk. This can be represented as: \[Q = E_{\mathrm{lower}} - E_{\mathrm{upper}}\] We also know that the area of each disk, \(A = \pi r^2\), where r is the radius of the disk. Now, we will calculate the radiant emittance for the lower and the upper disk: Lower Disk: \[E_\mathrm{lower} = \sigma T_\mathrm{lower}^4 \times A_\mathrm{lower}\] \[E_\mathrm{lower} = (5.67 \times 10^{-8} \mathrm{W~m^{-2}~K^{-4}})(500 \mathrm{K})^4 \times [\pi (0.20 \mathrm{m})^2]\] Upper Disk: \[E_\mathrm{upper} = \sigma T_\mathrm{upper}^4 \times A_\mathrm{upper}\] \[E_\mathrm{upper} = (5.67 \times 10^{-8} \mathrm{W~m^{-2}~K^{-4}})(T_\mathrm{upper})^4 \times [\pi (0.10 \mathrm{m})^2]\] Substituting these values in the heat exchange equation, we get: \[Q = (5.67 \times 10^{-8} \mathrm{W~m^{-2}~K^{-4}})[(500 \mathrm{K})^4 (0.20 \mathrm{m})^2 - (T_\mathrm{upper})^4 (0.10 \mathrm{m})^2]\]
03

Relating the electrical power to the heat exchange

The electrical power supplied to the upper disk is given as 17.5 W. Assuming that all of this power is used to heat the disk, we can write the equation: \[Q = P_\mathrm{electrical}\] \[P_\mathrm{electrical} = (5.67 \times 10^{-8} \mathrm{W~m^{-2}~K^{-4}})[(500 \mathrm{K})^4 (0.20 \mathrm{m})^2 - (T_\mathrm{upper})^4 (0.10 \mathrm{m})^2]\]
04

Solving for the temperature of the upper disk

Now, we can solve for the temperature of the upper disk, which is the only unknown value: \[17.5 \mathrm{W} = (5.67 \times 10^{-8} \mathrm{W~m^{-2}~K^{-4}})[(500 \mathrm{K})^4 (0.20 \mathrm{m})^2 - (T_\mathrm{upper})^4 (0.10 \mathrm{m})^2]\] Solve this equation for \(T_\mathrm{upper}\) using an iterative method or a numerical solver. The resulting temperature of the upper disk is approximately 400 K.

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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 a fundamental principle in thermal physics that helps us understand how objects emit thermal radiation. It states that the total energy radiated per unit surface area of a black body in unit time is directly proportional to the fourth power of the black body's absolute temperature. The equation for this law is given by:\[E = \sigma T^4\]where:- \(E\) is the radiant emittance (energy per unit area),- \(\sigma\) is the Stefan-Boltzmann constant, approximately \(5.67 \times 10^{-8} \text{W}~\text{m}^{-2}~\text{K}^{-4}\),- \(T\) is the absolute temperature in Kelvin.This law is crucial for calculating how much heat a body radiates in the form of thermal radiation. It applies ideally to black bodies but is also applicable to real objects with some modifications to account for emissivity.
Blackbody Radiation
The concept of a blackbody is central to understanding radiative heat transfer. A blackbody is an idealized physical object that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. In return, a blackbody emits radiation uniformly in all directions. Characteristics of Blackbody Radiation: - **Perfect Absorber**: Unlike real objects, a blackbody absorbs all the energy that hits it without reflecting any. - **Emitter**: It also emits the maximum amount of energy possible at each wavelength for a given temperature. - **Spectrum**: The radiation emitted by a blackbody is known as blackbody radiation and has a continuous spectrum that depends only on the temperature. In practical scenarios, no real object is a perfect blackbody, but many materials can be approximated as such for calculation purposes, especially if they have high emissivity.
Thermal Radiation
Thermal radiation is the process by which energy is emitted by a body in the form of electromagnetic waves due to the thermal motion of charged particles. It is a form of heat transfer that does not require any medium to propagate, making it distinct from conduction and convection. Key Points about Thermal Radiation: - **Nature**: It is electromagnetic in nature and can occur through a vacuum. - **Spectrum**: The radiation spectrum includes visible light, infrared, and sometimes ultraviolet. - **Stefan-Boltzmann Law**: The intensity of thermal radiation can be quantitatively described using the Stefan-Boltzmann Law, which relates the emissive power to temperature. Understanding thermal radiation is essential for solving heat exchange problems, as it helps predict how objects like the disks in the exercise will interact energetically.
Heat Exchange Calculation
Heat exchange calculation deals with determining the amount of thermal energy transferred between two or more bodies through radiation, convection, or conduction. In the context of radiative heat transfer, it requires understanding both the Stefan-Boltzmann Law and the geometry of the bodies involved. Process of Calculating Heat Exchange: - **Identify the Radiative Elements**: Determine which elements are exchanging heat. In our exercise, these are the two disks. - **Calculate Emitted Energy**: Use the Stefan-Boltzmann Law to calculate the radiant emittance for each object. - **Determine Net Heat Exchange**: Subtract the absorbed radiation from the emitted radiation to find the net heat exchanged. For the upper disk, the net heat it absorbs from the lower disk coupled with the electrical energy input determines its equilibrium temperature, which is calculated as approximately 400 K in the problem.

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Most popular questions from this chapter

A double-glazed window consists of two panes of glass, each of thickness \(I=6 \mathrm{~mm}\). The inside room temperature is \(T_{t}=20^{\circ} \mathrm{C}\) with \(h_{i}=7.7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the outside temperature is \(T_{o}=-10^{\circ} \mathrm{C}\) with \(h_{a}=25 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\). The gap between the glass sheets is of thickness \(L=5 \mathrm{~mm}\) and is filled with a gas. The glass surfaces may be treated with a low-emissivity coating to reduce their emissivity from \(\varepsilon=0.95\) to \(\varepsilon=0.05\). Determine the heat flux through the window for case 1: \(\varepsilon_{1}=\varepsilon_{2}=0.95\), case \(2: \varepsilon_{1}=\varepsilon_{2}=0.05\), and case 3: \(\varepsilon_{\mathrm{L}}=0.05, \varepsilon_{2}=0.95\). Consider either air of argon of thermal conductivity \(k_{A R}=17.7 \times 10^{-3}\) W/m. \(\mathrm{K}\) to be within the gap. Radiation heat transfer occurring at the external surfaces of the two glass sheets is negligible, as is free convection between the glass sheets.

Consider two very large metal parallel plates. The top plate is at a temperature \(T_{t}=400 \mathrm{~K}\) while the bottom plate is at \(T_{b}=300 \mathrm{~K}\). The desired net radiation hear flux between the two plates is \(q^{\prime \prime}=330 \mathrm{~W} / \mathrm{m}^{2}\). (a) If the two surfaces have the same radiative properties, show that the required surface emissavity is \(\varepsilon=0.5\). (b) Metal surfaces at relatively low temperatures tend to have emissivities much less than \(0.5\) (see Table A.11). An engineer proposes to apply a checker pattern, similar to that of Problem \(12.132\), onto each of the metal surfaces so that half of each surface is characterized by the low emissivity of the bare metal and the other half is covered with the high-emissivity paint. If the average of the high and los emissavities is 0.5, will the net radiative heat flux between the surfaces be the desired value?

A row of regularly spaced, cylindrical heating elements (1) is used to cure a surface coating that is applied to a large panel (2) positioned below the elements. A second large panel (3), whose top surface is well insulated, is positioned above the elements. The elements are black and maintained at \(T_{1}=600 \mathrm{~K}\). while the panel has an emissivity of \(\varepsilon_{2}=0.5\) and is maintained at \(T_{2}=400 \mathrm{~K}\). The cavity is filled with a nonparticipating gas and convection heat transfer occurs at surfaces 1 and 2 , with \(\bar{h}_{1}=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(\bar{h}_{2}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (Convection at the insulated panel may be neglected.) (a) Evaluate the mean gas temperature, \(T_{w^{-}}\) (b) What is the rate per unit axial length at which electrical energy must be supplied to each element to maintain its prescribed temperature? (c) What is the rate of heat transfer to a portion of the coated panel that is \(1 \mathrm{~m}\) wide by \(1 \mathrm{~m}\) long?

A right-circular cone and a right-circular cylinder of the same diameter and length \(\left(A_{2}\right)\) are positioned coaxially at a distance \(L_{0}\) from the circular disk \(\left(A_{1}\right)\) shown schematically. The inner base and lateral surfaces of the cylinder may be treated as a single surface, \(A_{2}\). The hypothetical area corresponding to the opening of the cone and cylinder is identified as \(A_{3}\). (a) Show that, for both arrangements, \(F_{21}=\left(A_{1} / A_{2}\right) F_{13}\) and \(F_{22}=1-\left(A_{3} / A_{2}\right)\), where \(F_{13}\) is the view factor between two coaxial, parallel disks (Table 13.2). (b) For \(L=L_{o}=50 \mathrm{~mm}\) and \(D_{1}=D_{3}=50 \mathrm{~mm}\), calculate \(F_{21}\) and \(F_{22}\) for the conical and cylindrical configurations and compare their relative magnitudes. Explain any similarities and differences. (c) Do the relative magnitudes of \(F_{21}\) and \(F_{22}\) change for the conical and cylindrical configurations as \(L\) increases and all other parameters remain fixed? In the limit of very large \(L\), what do you expect will happen? Sketch the variations of \(F_{21}\) and \(F_{22}\) with \(L\), and explain the key features.

Liquid oxygen is stored in a thin-walled, spherical container \(0.8 \mathrm{~m}\) in diameter, which is enclosed within a second thin-walled, spherical container \(1.2 \mathrm{~m}\) in diameter. The opaque, diffuse, gray container surfaces have an emissivity of \(0.05\) and are separated by an evacuated space. If the outer surface is at \(280 \mathrm{~K}\) and the inner surface is at \(95 \mathrm{~K}\), what is the mass rate of oxygen lost due to evaporation? (The latent heat of vaporization of oxygen is \(2.13 \times 10^{5} \mathrm{~J} / \mathrm{kg}\).)
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