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Chapter 13: Problem 59

Heat transfer by radiation occurs between two large parallel plates, which are maintained at temperatures \(T_{1}\) and \(T_{2}\), with \(T_{1}>T_{2}\). To reduce the rate of heat transfer between the plates, it is proposed that they be separated by a thin shield that has different emissivities on opposite surfaces. In particular, one surface has the emissivity \(\varepsilon_{s}<0.5\), while the opposite surface has an emissivity of \(2 \varepsilon_{s}\). (a) How should the shield be oriented to provide the larger reduction in heat transfer between the plates? That is, should the surface of emissivity \(\varepsilon_{s}\) or that of emissivity \(2 s_{s}\) be oriented toward the plate at \(T_{1}\) ? (b) What orientation will result in the larger value of the shield temperature \(T_{s}\) ?

Short Answer

Expert verified
(a) To provide the largest reduction in heat transfer between the plates, the shield should be oriented with the surface of emissivity \(\varepsilon_s\) towards the plate at \(T_1\). (b) The orientation with the surface of emissivity \(2\varepsilon_s\) towards the plate at \(T_1\) will result in the larger value of the shield temperature \(T_s\).

Step by step solution

01

Calculate the view factors

The view factor is a geometrical parameter that describes the interaction between two surfaces in terms of radiative heat exchange. Since the two plates and the shield are large, we can say that the view factors between the surfaces are 1.
02

Apply the Stefan-Boltzmann Law

The Stefan-Boltzmann Law states that the heat transfer by radiation between two surfaces is proportional to the difference in the fourth power of their temperatures and their surface emissivities. For the two cases, we are asked to consider the orientation of the shield which affects the emissivities between surfaces: 1. First case: Surface \(\varepsilon_s\) towards the plate at \(T_1\) 2. Second case: Surface \(2\varepsilon_s\) towards the plate at \(T_1\) Now, let's compute the heat transfer rates for both cases. 1. First case: Surface \(\varepsilon_s\) towards the plate at \(T_1\) \(Q_1 = \varepsilon_s A_1 \sigma (T_1^4 - T_s^4)\) and \(Q_2 = (2\varepsilon_s) A_2 \sigma (T_s^4 - T_2^4)\) 2. Second case: Surface \(2\varepsilon_s\) towards the plate at \(T_1\) \(Q_1 = (2\varepsilon_s) A_1 \sigma (T_1^4 - T_s^4)\) and \(Q_2 = \varepsilon_s A_2 \sigma (T_s^4 - T_2^4)\)
03

Conservation of heat transfer rate across the shield

The heat transfer rate is conserved across the shield, so we have \(Q_1 = Q_2\). This leads to the following relationship for both cases: 1. First case: \(\varepsilon_s A_1 \sigma (T_1^4 - T_s^4) = (2\varepsilon_s) A_2 \sigma (T_s^4 - T_2^4)\) 2. Second case: \((2\varepsilon_s) A_1 \sigma (T_1^4 - T_s^4) = \varepsilon_s A_2 \sigma (T_s^4 - T_2^4)\) We can now compare these two cases to find the orientation that results in the largest reduction in heat transfer and the shield temperature.
04

Compare heat transfer rates and shield temperatures

We are looking for the orientation of the shield that results in the largest reduction in heat transfer and the shield temperature. From the equations above, we can see that when the surface with lower emissivity (\(\varepsilon_s\)) is oriented towards the plate at \(T_1\), the heat transfer rate is smaller, leading to a larger reduction in heat transfer between the plates. On the other hand, for the shield temperature, since the heat transfer rate is conserved, the orientation that results in the larger value of shield temperature can be derived from the equations as well. It can be seen that when the surface with higher emissivity (\(2\varepsilon_s\)) is oriented towards the plate at \(T_1\), the shield temperature increases. Answer: (a) The shield should be oriented with the surface of emissivity \(\varepsilon_s\) towards the plate at \(T_1\) to provide the largest reduction in heat transfer between the plates. (b) The orientation where the surface with emissivity \(2\varepsilon_s\) is oriented towards the plate at \(T_1\) will result in the larger value of the shield temperature \(T_s\).

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Key Concepts

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

Understanding the Stefan-Boltzmann Law
This core principle of thermal physics states that the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body鈥檚 temperature. Expressed mathematically, the law is given by the equation
\[ Q = \text{emissivity} \times A \times \text{Boltzmann Constant} \times (T^4) \] where
- \(Q\) is the total radiated power,
- \(A\) is the radiating surface area,
- \(T\) is the absolute temperature in Kelvin,
- \(\text{Boltzmann Constant}\) is a fixed number (\(5.67 \times 10^{-8} \text{W/m}^2\text{K}^4\)),
and
- \(\text{emissivity}\) is a measure of a material's ability to emit energy as radiation.
In the exercise, the Stefan-Boltzmann Law is vital to understand the effect of temperature difference and emissivity on the heat transfer rate between surfaces.
The Critical Role of Emissivity in Thermal Radiation
Emissivity is a numerical value ranging from 0 to 1 that is a measure of a material's ability to radiate energy. It is an intrinsic property of materials that indicates how effectively a surface emits infrared radiation, compared to a perfect black body. A perfect black body, which absorbs all incoming light and emits the maximum amount of radiation, has an emissivity of 1, while a perfect reflector would have an emissivity of 0.

In the given exercise, the effectiveness of the shield in reducing heat transfer is influenced by its emissivity on both sides. With \(\varepsilon_s < 0.5\) and the other side being \(2\varepsilon_s\), a lower emissivity facing the hotter plate will lead to a reduced rate of radiative heat transfer, aligning with the Stefan-Boltzmann Law.
How Shield Temperature Affects Heat Transfer
Shield temperature, denoted by \(T_s\), is a significant factor when calculating radiative heat transfer. It's the temperature of the thin barrier or 'shield' placed between bodies at different temperatures to control the rate of thermal radiation passing through. This setup is demonstrated in the exercise by placing a shield with different emissivities between two plates.

The orientation impacting the shield temperature can be understood by recognizing that higher emissivity correlates with greater energy emission. Therefore, positioning the higher emissivity side (\(2\varepsilon_s\)) towards the hot plate (\(T_1\)) results in the shield absorbing more energy, hence raising its temperature. Conversely, a lower emissivity surface absorbs less, which keeps the shield temperature lower, also aligning with the principle of conservation of energy across the shield.

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Most popular questions from this chapter

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?

Derive expressions for the view factor \(F_{12}\) associated with the following arrangements. Express your results in terms of \(A_{1}, A_{2}\) and any appropriate hypothetical surface area, as well as the view factor for coaxial parallel disks (Table 13.2, Figure 13.5). (a) A circular disk and a coaxial, ring-shaped disk. (b) A circular disk and a coaxial, right-circular cone. Sketch the variation of \(F_{12}\) with \(\theta\) for \(0 \leq \theta \leq \pi / 2\), and explain the key features.

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?

A diffuse, gray radiation shield of \(60-\mathrm{mm}\) diameter and emissivities of \(\varepsilon_{2,1}=0.01\) and \(\varepsilon_{2,0}=0.1\) on the inner and outer surfaces, respectively, is concentric with a long tube transporting a hot process flu盲d. The tube surface is black with a diameter of \(20 \mathrm{~mm}\). The region interior to the shield is evacuated. The exterior surface of the shield is exposed to a large room whose walls are at \(17^{\circ} \mathrm{C}\) and experiences convection with air at \(27^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). Determine the operating temperature for the inner tube if the shield temperature is maintained at \(42^{\circ} \mathrm{C}\).

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.
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