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Consider two large (infinite) parallel planes that are diffuse-gray with temperatures and emissivities of \(T_{1}\), \(\varepsilon_{1}\) and \(T_{2}, \varepsilon_{2}\). Show that the ratio of the radiation transfer rate with multiple shields, \(N\), of emissivity \(\varepsilon_{s}\) to that with no shields, \(N=0\), is $$ \frac{q_{12, N}}{q_{120}}=\frac{\left[1 / \varepsilon_{1}+1 / \varepsilon_{2}-1\right]}{\left[1 / \varepsilon_{1}+1 / s_{2}-1\right]+N\left[2 / s_{s}-1\right]} $$ where \(q_{12, N}\) and \(q_{12,0}\) represent the radiation heat transfer rates for \(N\) shields and no shields, respectively.

Short Answer

Expert verified
The ratio of the radiation transfer rate with multiple shields to that with no shields is given by: \[ \frac{q_{12,N}}{q_{12,0}} = \frac{\left[1 / \varepsilon_{1} + 1 / \varepsilon_{2} - 1\right]}{\left[1 / \varepsilon_{1} + 1 / \varepsilon_{2} - 1\right] + N\left[2 / \varepsilon_s - 1\right]} \]

Step by step solution

01

Understand the radiation heat transfer rate with no shields

When no shields are present, the radiation heat transfer rate between two parallel plates is modeled by the following equation: \[ q_{12,0} = \sigma F_{12}\left( T_1^4 - T_2^4 \right) \] where \(\sigma\) is the Stefan-Boltzmann constant, and \(F_{12}\) is the view factor between the two plates, representing the fraction of radiation leaving surface 1 and reaching surface 2.
02

Find the view factor for no shields

For the case with no shields, the view factor \(F_{12}\) can be found using the following equation: \[ F_{12} = \frac{1}{1/\varepsilon_{1} + 1/\varepsilon_{2} - 1} \]
03

Determine the radiation heat transfer rate with multiple shields

For the scenario with \(N\) shields, each shield introduces an additional layer that impacts the fraction of radiation reaching surface 2. The view factor for this case, \(F_{12,N}\), is given by the following equation: \[ F_{12,N} = \frac{1}{1/\varepsilon_{1} + 1/\varepsilon_{2} - 1 + N(\frac{2}{\varepsilon_s} - 1)} \]
04

Calculate the radiation heat transfer rate for the case with multiple shields

Now that we have the view factor for the case with multiple shields, we can determine the radiation heat transfer rate, \(q_{12,N}\), using the same formula as in Step 1, simply substituting \(F_{12,N}\) for \(F_{12}\): \[ q_{12,N} = \sigma F_{12,N}\left( T_1^4 - T_2^4 \right) \]
05

Determine the ratio of the radiation transfer rates

Finally, we divide the radiation heat transfer rate with multiple shields by the radiation heat transfer rate with no shields to find the ratio: \[ \frac{ q_{12,N} }{ q_{12,0} } = \frac{\sigma F_{12,N}\left( T_1^4 - T_2^4 \right)}{\sigma F_{12,0}\left( T_1^4 - T_2^4 \right)} \] The \(T_1^4 - T_2^4\) terms and the Stefan-Boltzmann constant, \(\sigma\), cancel out, resulting in: \[ \frac{q_{12,N}}{q_{12,0}} = \frac{F_{12,N}}{F_{12,0}} = \frac{\left[1 / \varepsilon_{1} + 1 / \varepsilon_{2} - 1\right]}{\left[1 / \varepsilon_{1} + 1 / \varepsilon_{2} - 1\right] + N\left[2 / \varepsilon_s - 1\right]} \] The result is the expression we were asked to show, as requested in the exercise.

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Key Concepts

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

Diffuse-Gray Surfaces
Diffuse-gray surfaces are an idealized concept used in the study of radiation heat transfer. These surfaces are characterized by their uniform reflectivity and emissivity in all directions and across all wavelengths. In simpler terms, a diffuse-gray surface is a perfect emitter and scatterer of radiation, without preference for direction or wavelength.

When we say a surface is 'gray,' we mean that its emissivity does not change with wavelength, which is not the case for real materials. Consequently, we can represent complex, actual material properties with a single, average value for emissivity, \(\varepsilon\). This simplification allows for more manageable calculations in engineering problems involving radiative heat transfer and is typically quite accurate for engineering estimates.

For the purpose of textbook exercises and practical engineering calculations, assuming surfaces to be diffuse and gray helps in applying the Stefan-Boltzmann law. This law is only strictly accurate for ideal blackbodies, but by treating surfaces as diffuse-gray, we can make use of the law by adjusting it with an emissivity factor. Thus, such surfaces serve as an essential model in thermodynamics and heat transfer courses.
Stefan-Boltzmann Constant
The Stefan-Boltzmann constant, denoted as \(\sigma\), is a physical constant that plays a crucial role in the domain of thermal radiation. It appears in the famous Stefan-Boltzmann law, which states that the total energy radiated per unit surface area of a black body per unit time is directly proportional to the fourth power of the black body's thermodynamic temperature.

The precise value of the Stefan-Boltzmann constant is \(5.670374419 \times 10^{-8} \text{ W m}^{-2} \text{K}^{-4}\). What this means is that each square meter of a blackbody's surface, at a temperature of one Kelvin, radiates \(5.670374419 \times 10^{-8} \text{ watts}\).

In our textbook example, the constant is used to compare the heat transfer rates with and without shields. The beautiful simplicity of the Stefan-Boltzmann law, when combined with the constant, allows us to estimate the thermal radiation from bodies that are close to ideal black bodies or, as in our case, from diffuse-gray surfaces by using their emissivity.
View Factor
The view factor, also known as the configuration factor or shape factor, is a dimensionless quantity in thermal radiation that represents the fraction of the radiation leaving one surface that directly reaches another specified surface. View factors depend solely on the geometry of the involved surfaces.

To calculate the heat transfer between surfaces, especially in cases without intervening media, the view factor is crucial. The value of a view factor lies between 0 and 1, inclusive. A view factor of 1 means that all the radiation from one surface directly strikes the other surface, while 0 implies none reaches it.

In the context of our textbook problem, the presence of shields alters the view factor, and hence, changes the radiation heat transfer rate between the two planes. View factors come into play in this problem to indicate how adding shields affects the fraction of radiation between the two initial large parallel planes. Understanding how view factors change with different configurations is essential for predicting the efficacy of thermal shields and similar barriers.

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Most popular questions from this chapter

Consider a circular furnace that is \(0.3 \mathrm{~m}\) long and \(0.3 \mathrm{~m}\) in diameter. The two ends have diffuse, gray surfaces that are maintained at 400 and \(500 \mathrm{~K}\) with emissivities of \(0.4\) and \(0.5\), respectively. The lateral surface is also diffuse and gray with an emissivity of \(0.8\) and a temperature of \(800 \mathrm{~K}\). Determine the net radiative heat transfer from each of the surfaces.

Most architects know that the ceiling of an ice-skating rink must have a high reflectivity. Otherwise, condensation may occur on the ceiling, and water may drip onto the ice, causing bumps on the skating surface. Condensation will occur on the ceiling when its surface temperature drops below the dew point of the rink air. Your assignment is to perform an analysis to determine the effect of the ceiling emissivity on the ceiling temperature, and hence the propensity for condensation. The rink has a diameter of \(D=50 \mathrm{~m}\) and a height of \(L=10 \mathrm{~m}\), and the temperatures of the ice and walls are \(-5^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively. The rink air temperature is \(15^{\circ} \mathrm{C}\), and a convection coefficient of \(5 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) characterizes conditions on the ceiling surface. The thickness and thermal conductivity of the ceiling insulation are \(0.3 \mathrm{~m}\) and \(0.035 \mathrm{~W} / \mathrm{m}-\mathrm{K}\), respectively, and the temperature of the outdoor air is \(-5^{\circ} \mathrm{C}\). Assume that the ceiling is a diffuse-gray surface and that the Walls and ice may be approximated as blackbodies. (a) Consider a flat ceiling having an emissivity of \(0.05\) (highly reflective panels) or \(0.94\) (painted panels). Perform an energy balance on the ceiling to calculate the corresponding values of the ceiling temperature. If the relative humidity of the rink air is \(70 \%\), will condensation occur for either or both of the emissivities? (b) For each of the emissivities, calculate and plot the ceiling temperature as a function of the insulation thickness for \(0.1 \leq r \leq 1 \mathrm{~m}\). Identify conditions for which condensation will occur on the ceiling.

At the bottom of a very large vacuum chamber whose walls are at \(300 \mathrm{~K}\), a black panel \(0.1 \mathrm{~m}\) in diameter is maintained at \(77 \mathrm{~K}\). To reduce the heat gain to this panel, a radiation shield of the same diameter \(D\) and an emissivity of \(0.05\) is placed very close to the panel. Calculate the net heat gain to the panel.

Consider two large, diffuse, gray, parallel surfaces separated by a small distance. If the surface emissivities are \(0.8\), what emissivity should a thin radiation shield have to reduce the radiation heat transfer rate between the two surfaces by a factor of 10 ?

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?

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