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Determine the steady-state temperatures of two radiation shields placed in the evacuated space between two infinite planes at temperatures of 600 and \(325 \mathrm{~K}\). All the surfaces are diffuse and gray with emissivities of \(0.7\).

Short Answer

Expert verified
The steady-state temperatures for the two radiation shields are T鈧 = 462.5 K and T鈧 = 137.5 K.

Step by step solution

01

Identify the known parameters

In this exercise, we know: - Temperature of Plane 1: T鈧 = 600 K - Temperature of Plane 2: T鈧 = 325 K - Emissivity of all surfaces: 蔚 = 0.7
02

Understand the geometry

We have two shields placed between the infinite planes. Let's denote the temperature of each shield as: - Shield 1: T鈧 - Shield 2: T鈧 Note that these temperatures might not be the same. We need to find T鈧 and T鈧.
03

Define the radiative resistance

The radiative resistance R between two surfaces is given by: \[ R=\frac{1}{ 蔚_1A_1 F_{1鈫2} }+\frac{1- 蔚_2}{ 蔚_2A_2 } \] Here, 蔚鈧 and 蔚鈧 are emissivities, A鈧 and A鈧 are the areas (infinite for our case), and F鈧佲啋鈧 is the radiative heat exchange view factor. Since all surfaces have the same emissivity, we can simplify this relation by considering just the view factors: \[ R=\frac{1}{ 蔚 (F_{1鈫2}+F_{2鈫1}) } \]
04

List the view factors

For infinite planes, all view factors between the planes and the shields are equal. Let's denote this value as F. Then we can use: \[ R=\frac{1}{ 蔚 \cdot 2F } = \frac{1}{ 1.4F } \]
05

Apply the resistance network concept

The overall heat transfer resistance R_total between the infinite planes is given by the sum of individual resistances between the planes and the shields: \[ R_{total} = R_{1-3} + R_{3-4} + R_{4-2} \] Plugging the values for radiative resistance, we can write: \[ R_{total} = \frac{1}{1.4F} + \frac{1}{1.4F} + \frac{1}{1.4F} \] \[ R_{total} = \frac{3}{1.4F} \]
06

Find the heat transfer

Now, the radiative heat transfer Q between the infinite planes can be found using the following equation: \[ Q = \frac{ T鈧 - T鈧 }{ R_{total} } \] Replace the value of R_total: \[ Q = \frac{ T鈧 - T鈧 }{ (3 / 1.4F) } \]
07

Solve for shield temperatures using the heat transfer value

With the heat transfer value, we can write the equations for the temperatures of the shields. For shield 1, we can use: \[ Q = \frac{ T鈧 - T鈧 }{ (1 / 1.4F) } \Rightarrow T鈧 = T鈧-1.4 F Q \] Similarly, for shield 2, we can use: \[ Q = \frac{ T鈧 - T鈧 }{ (3 / 1.4F) } \Rightarrow T鈧 = T鈧+1.4 F Q \]
08

Consider that heat transfer value is the same between all surfaces

As we progress through each stage, the radiative heat transfer remains the same. Therefore: \[ \frac{T鈧-T鈧剗{(\frac{1}{1.4F})} = Q \] \[ T鈧-T鈧 = 1.4 F Q \] Now we have two equations for shield temperatures, and we can solve for T鈧 and T鈧 using these relationships: \[ T鈧 = T鈧-1.4 F Q \] \[ T鈧 = T鈧+1.4 F Q \] Add both equations to get: \[ T鈧 = 2T鈧+2.8 F Q \Rightarrow T鈧 = \frac{T鈧-T鈧倉{2} = \frac{600K - 325K}{2} = 137.5K\] Now substitute T鈧 in the temperature equation of shield 1: \[ T鈧 = T鈧-1.4 F Q = 600K - 1.4 F Q = 462.5K \] Now, the steady-state temperatures for the shields are T鈧 = 462.5 K and T鈧 = 137.5 K.

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Key Concepts

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

Radiative Heat Transfer
Radiative heat transfer is a mode of energy exchange between two bodies or surfaces via electromagnetic waves, usually in the form of infrared radiation. This phenomenon is crucial in understanding how energy is transferred in systems such as our textbook example with infinite planes and radiation shields. Unlike conduction or convection, radiation does not require a medium; it can occur through a vacuum, which is why it's particularly interesting for applications like space vehicles or satellite design.

Interestingly, the energy transmitted by radiation is proportional to the fourth power of the absolute temperature of the body, as described by the Stefan-Boltzmann law. The law is expressed as: \[ Q = \text{Stefan-Boltzmann constant} \times \text{emissivity} \times \text{area} \times (T^{4} - T_{s}^{4}) \] where \( Q \) is the rate of heat transfer, \( T \) and \( T_{s} \) are the temperatures of the bodies, and the emissivity represents the efficiency of the surface in emitting thermal energy.
Thermal Resistance Network
The concept of a thermal resistance network simplifies the understanding of heat transfer processes through an analogy with electrical circuits. Just as electrical resistance slows down the flow of electrons, thermal resistance hinders the flow of heat. In our textbook problem, the infinite planes and radiation shields can be modeled using a series of thermal resistances.

The total resistance to radiative heat transfer in a system with multiple layers鈥攁s with our shields鈥攃an be calculated by summing individual resistances in series. This is analogous to electrical resistances in a simple circuit. By representing heat transfer pathways as resistances, we can employ Ohm's law, originally from electrical engineering, to ascertain the steady-state temperatures across the network. The equation \( Q = \frac{T_1 - T_2}{R_{total}} \) used in our exercise is based on an analogy with Ohm's law where heat transfer is equated to the potential difference, and radiative resistance is seen as the impedance to the heat flux.
Emissivity of Surfaces
Emissivity is a measure of a surface's ability to emit thermal radiation and is a dimensionless value ranging from 0 to 1. A surface with an emissivity of 1 is called a black body and is an ideal emitter, whereas a real object has emissivity less than 1, as in our shields example where the emissivity is 0.7.

The amount of heat that a surface radiates depends on its emissivity, and when calculating radiative heat exchange, the emissivity of the surfaces involved plays a critical role. In the problem we've analyzed, we assume diffuse and gray surfaces, meaning emissivity does not change with the direction and wavelength of the radiation. These simple assumptions are particularly helpful when dealing with complex systems, as they allow us to focus on the underlying principles without getting lost in the specifics of material properties or geometric intricacies.

Incorporating the emissivity into the radiative resistance formulas allows us to more accurately model the temperature distribution between radiating bodies. The step-by-step solution provided systematically applies this concept to derive the steady-state temperatures for the radiation shields situated between two infinite planes.

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Most popular questions from this chapter

A flue gas at 1 -atm total pressure and a temperature of \(1400 \mathrm{~K}\) contains \(\mathrm{CO}_{2}\) and water vapor at partial pressures of \(0.05\) and \(0.10 \mathrm{~atm}\), respectively. If the gas flows through a long flue of \(1-\mathrm{m}\) diameter and \(400 \mathrm{~K}\) surface temperature, determine the net radiative heat flux from the gas to the surface. Blackbody behavior may be assumed for the surface.

The arrangement shown is to be used to calibrate a heat flux gage. The gage has a black surface that is \(10 \mathrm{~mm}\) in diameter and is maintained at \(17^{\circ} \mathrm{C}\) by means of a water-cooled backing plate. The heater, \(200 \mathrm{~mm}\) in diameter, has a black surface that is maintained at \(800 \mathrm{~K}\) and is located \(0.5 \mathrm{~m}\) from the gage. The surroundings and the air are at \(27^{\circ} \mathrm{C}\) and the convection heat transfer coefficient between the gage and the air is \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the net radiation exchange between the heater and the gage. (b) Determine the net transfer of radiation to the gage per unit area of the gage. (c) What is the net heat transfer rate to the gage per unit area of the gage? (d) If the gage is constructed according to the description of Problem 3.107, what heat flux will it indicate?

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.

A flat-bottomed hole \(6 \mathrm{~mm}\) in diameter is bored to a depth of \(24 \mathrm{~mm}\) in a diffuse, gray material having an emissivity of \(0.8\) and a uniform temperature of \(1000 \mathrm{~K} .\) (a) Determine the radiant power leaving the opening of the cavity. (b) The effective emissivity \(\varepsilon_{e}\) of a cavity is defined as the ratio of the radiant power leaving the cavity to that from a blackbody having the area of the cavity opening and a temperature of the inner surfaces of the cavity. Calculate the effective emissivity of the cavity described above. (c) If the depth of the hole were increased, would \(\varepsilon_{e}\) increase or decrease? What is the limit of \(s_{\epsilon}\) as the depth increases?

A wall-mounted natural gas heater uses combustion on a porous catalytic pad to maintain a ceramic plate of emissivity \(\varepsilon_{c}=0.95\) at a uniform temperature of \(T_{c}=1000 \mathrm{~K}\). The ceramic plate is separated from a glass plate by an air gap of thickness \(L=50 \mathrm{~mm}\). The surface of the glass is diffuse, and its spectral transmissivity and absorptivity may be approximated as \(\tau_{\lambda}=0\) and \(\alpha_{\lambda}=1\) for \(0 \leq \lambda \leq 0.4 \mu \mathrm{m}, \tau_{\lambda}=1\) and \(\alpha_{\lambda}=0\) for \(0.4<\lambda \leq 1.6 \mu \mathrm{m}\), and \(\tau_{\lambda}=0\) and \(\alpha_{\lambda}=0.9\) for \(\lambda>1.6 \mu \mathrm{m}\). The exterior sarface of the glass is exposed to quiescent ambient air and large surroundings for which \(T_{m}=T_{a x}=300 \mathrm{~K}\). The height and width of the heater are \(H=W=2 \mathrm{~m}\). (a) What is the total transmissjvity of the glass to irradiation from the ceramic plate? Can the glass be approximated as opaque and gray? (b) For the prescribed conditions, evaluate the glass temperature, \(T_{R}\), and the rate of heat transfer from the heater, \(q_{k}\). (c) A fan may be used to control the convection coefficient \(h_{n}\) at the exterior surface of the glass. Compute and plot \(T_{g}\) and \(q_{\mathrm{b}}\) as a function of \(h_{e}\) for \(10 \leq h_{0} \leq 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

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