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Chapter 12: Problem 1

Consider an opaque horizontal plate that is well insulated on its back side. The irradiation on the plate is \(2500 \mathrm{~W} / \mathrm{m}^{2}\), of which \(500 \mathrm{~W} / \mathrm{m}^{2}\) is reflected. The plate is at \(227^{\circ} \mathrm{C}\) and has an emissive power of \(1200 \mathrm{~W} / \mathrm{m}^{2}\). Air at \(127^{\circ} \mathrm{C}\) flows over the plate with a heat transfer convection coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the emissivity, absorptivity, and radiosity of the plate. What is the net heat transfer rate per unit area?

Short Answer

Expert verified
In conclusion, the plate has an emissivity (蔚) of 0.889, absorptivity (伪) of 0.889, and radiosity (J) of 1700 W/m虏. The net heat transfer rate per unit area is 2725 W/m虏.

Step by step solution

01

Plug in the values into the emissivity formula: \[ 蔚 = \frac{1200}{(5.67 \times 10^{-8})(500^4)} \]

Calculate the emissivity: \( 蔚 \approx 0.889 \) #Step 2: Calculate Absorptivity# Absorptivity (伪) can be calculated using the energy balance equation at the surface: \( 伪G = E + (1 - R_{fraction})G \) where G is the irradiation and \(R_{fraction}\) is the fraction of reflected energy from the plate. Here, \( R_{fraction} = R/G \).
02

Substitute the given values into the energy balance equation: \( 伪G = 1200 + (1 - 500/2500)2500 \)

Calculate the absorptivity: \( 伪 \approx 0.889 \) #Step 3: Calculate Radiosity# Radiosity (J) is the sum of emitted energy (E) and reflected energy (R): \( J = E + R \)
03

Substitute the given values into the radiosity equation: \( J = 1200 + 500 \)

Calculate the radiosity: \( J \approx 1700 \: W/m^2 \) Now let's find the net heat transfer rate per unit area. #Step 4: Calculate the radiative heat transfer rate# The radiative heat transfer rate per unit area (q_rad) can be found using the equation: \( q_{rad} = 伪G - E \)
04

Substitute the values into the radiative heat transfer rate equation: \( q_{rad} = 0.889 \cdot 2500 - 1200 \)

Calculate the radiative heat transfer rate: \( q_{rad} \approx 1225 \: W/m^2 \) #Step 5: Calculate the convective heat transfer rate# The convective heat transfer rate per unit area (q_conv) can be found using the equation: \( q_{conv} = h (T_{plate} - T_{air}) \)
05

Substitute the values into the convective heat transfer rate equation: \( q_{conv} = 15 (500 - 400) \)

Calculate the convective heat transfer rate: \( q_{conv} \approx 1500 \: W/m^2 \) #Step 6: Calculate the net heat transfer rate per unit area# The net heat transfer rate per unit area (q_net) is the sum of the radiative and convective heat transfer rates: \( q_{net} = q_{rad} + q_{conv} \)
06

Substitute the calculated values into the net heat transfer rate equation: \( q_{net} = 1225 + 1500 \)

Calculate the net heat transfer rate per unit area: \( q_{net} \approx 2725 \: W/m^2 \) In conclusion, the plate has an emissivity (蔚) of 0.889, absorptivity (伪) of 0.889, and radiosity (J) of 1700 W/m虏. The net heat transfer rate per unit area is 2725 W/m虏.

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Key Concepts

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

Emissivity
Emissivity is a measure of a surface's ability to emit thermal radiation compared to a perfect black body at the same temperature.
It is expressed as a value between 0 and 1, with 1 indicating a perfect emitter. For example, if a surface has an emissivity of 0.889, it means that the surface emits 88.9% of the radiation a black body would emit at the same temperature.
High emissivity surfaces are efficient at releasing absorbed energy as thermal radiation.
  • Used in heat transfer calculations to assess radiative emission efficiency.
  • Essential for determining the heat loss or gain of a material in different environmental conditions.
In practical applications, knowing the emissivity helps design energy-efficient systems and can assist in temperature measurement using infrared thermometers.
Absorptivity
Absorptivity denotes how well a material can absorb incident radiation. It is also a number between 0 and 1, indicating the fraction of irradiance the surface absorbs.
If a surface has an absorptivity of 0.889, it absorbs 88.9% of the incident radiation, with the remaining 11.1% being reflected.
Absorptivity is crucial when considering how much energy a surface captures from its surroundings.
  • Important for solar energy applications, determining efficiency of solar panels.
  • Influences thermal comfort and energy consumption in buildings.
Understanding absorptivity is crucial in designing systems for efficient energy transfer, such as in heating or cooling systems.
Radiosity
Radiosity is the total radiation leaving a surface, which includes both emitted and reflected radiation.
It provides a comprehensive picture of the radiation interaction between a surface and its surroundings. In the given example, the radiosity is calculated to be 1700 W/m虏.
Radiosity considers all forms of radiation-related energy transfer from a surface:
  • Includes components of both thermal emission and reflection.
  • Depends on both the emissivity and the incident radiation.
This concept is vital in radiative heat transfer calculations, helping to accurately model thermal exchanges in engineering applications.
Convective Heat Transfer
Convective heat transfer occurs when heat is carried away from a surface by a fluid (liquid or gas) flowing over it. It involves both conduction and fluid movement principles.
The rate of convective heat transfer can be influenced by factors such as the temperature difference between the surface and the fluid, the fluid's speed, and its properties.
This process is described by the equation:
  • \[ q_{conv} = h (T_{surface} - T_{fluid}) \]
  • Where \( h \) is the convection heat transfer coefficient, \( T_{surface} \) is the temperature of the surface, and \( T_{fluid} \) is the temperature of the fluid.
Understanding convective heat transfer is essential for designing heating and cooling systems and analyzing thermal interactions in a fluid environment.
Radiative Heat Transfer
Radiative heat transfer involves the emission or absorption of electromagnetic radiation, particularly in the infrared spectrum.
Unlike conduction and convection, it does not require a medium and can occur in a vacuum. It depends on the temperature and surface characteristics of the emitting and absorbing bodies.
For radiative heat transfer calculations, consider:
  • Emissivity and absorptivity values of surfaces.
  • Any temperature differences between bodies involved.
Radiative heat transfer is a key factor in a wide range of applications, from understanding planetary climates to designing efficient thermal management systems in engineering.

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Most popular questions from this chapter

One scheme for extending the operation of gas turbine blades to higher temperatures involves applying a ceramic coating to the surfaces of blades fabricated from a superalloy such as inconel. To assess the reliability of such coatings, an apparatus has been developed for testing samples under laboratory conditions. The sample is placed at the bottom of a large vacuum chamber whose walls are cryogenically cooled and which is equipped with a radiation detector at the top surface. The detector has a surface area of \(A_{d}=10^{-5} \mathrm{~m}^{2}\), is located at a distance of \(L_{\text {sl }}=1 \mathrm{~m}\) from the sample, and views radiation originating from a portion of the ceramic surface having an area of \(\Delta A_{c}=10^{-4} \mathrm{~m}^{2}\). An electric heater attached to the bottom of the sample dissipates a uniform heat flux, \(q_{b}^{\prime \prime}\), which is transferred upward through the sample. The bottom of the heater and sides of the sample are well insulated. Consider conditions for which a ceramic coating of thickness \(L_{c}=0.5 \mathrm{~mm}\) and thermal conductivity \(k_{c}=\) \(6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) has been sprayed on a metal substrate of thickness \(L_{s}=8 \mathrm{~mm}\) and thermal conductivity \(k_{s}=\) \(25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The opaque surface of the ceramic may be approximated as diffuse and gray, with a total, hemispherical emissivity of \(\varepsilon_{c}=0.8\). (a) Consider steady-state conditions for which the bottom surface of the substrate is maintained at \(T_{1}=1500 \mathrm{~K}\), while the chamber walls (including the surface of the radiation detector) are maintained at \(T_{w}=90 \mathrm{~K}\). Assuming negligible thermal contact resistance at the ceramic- substrate interface, determine the ceramic top surface temperature \(T_{2}\) and the heat flux \(q_{b}^{\prime \prime}\). (b) For the prescribed conditions, what is the rate at which radiation emitted by the ceramic is intercepted by the detector?

A small disk \(5 \mathrm{~mm}\) in diameter is positioned at the center of an isothermal, hemispherical enclosure. The disk is diffuse and gray with an emissivity of \(0.7\) and is maintained at \(900 \mathrm{~K}\). The hemispherical enclosure, maintained at \(300 \mathrm{~K}\), has a radius of \(100 \mathrm{~mm}\) and an emissivity of \(0.85\). Calculate the radiant power leaving an aperture of diameter \(2 \mathrm{~mm}\) located on the enclosure as shown.

The absorber plate of a solar collector may be coated with an opaque material for which the spectral, directional absorptivity is characterized by relations of the form $$ \begin{array}{ll} \alpha_{\lambda, \dot{\theta}}(\lambda, \theta)=\alpha_{1} \cos \theta & \lambda<\lambda_{c} \\ \alpha_{\lambda, \theta}(\lambda, \theta)=\alpha_{2} & \lambda>\lambda_{c} \end{array} $$ The zenith angle \(\theta\) is formed by the sun's rays and the plate normal, and \(\alpha_{1}\) and \(\alpha_{2}\) are constants. (a) Obtain an expression for the total, hemispherical absorptivity, \(\alpha_{S}\), of the plate to solar radiation incident at \(\theta=45^{\circ}\). Evaluate \(\alpha_{5}\) for \(\alpha_{1}=0.93, \alpha_{2}=\) \(0.25\), and a cut-off wavelength of \(\lambda_{c}=2 \mu \mathrm{m}\). (b) Obtain an expression for the total, hemispherical emissivity \(\varepsilon\) of the plate. Evaluate \(\varepsilon\) for a plate temperature of \(T_{p}=60^{\circ} \mathrm{C}\) and the prescribed values of \(\alpha_{1}, \alpha_{2}\), and \(\lambda_{c}\). (c) For a solar flux of \(q_{s}^{\prime \prime}=1000 \mathrm{~W} / \mathrm{m}^{2}\) incident at \(\theta=45^{\circ}\) and the prescribed values of \(\alpha_{1}, \alpha_{2}, \lambda_{c}\), and \(T_{p}\), what is the net radiant heat flux, \(q_{\text {net }}^{\prime \prime}\), to the plate? (d) Using the prescribed conditions and the Radiation/ Band Emission Factor option in the Tools section of \(I H T\) to evaluate \(F_{\left(0 \rightarrow \lambda_{j}\right)}\), explore the effect of \(\lambda_{c}\) on \(\alpha_{S}, \varepsilon\), and \(q_{\text {net }}^{N}\) for the wavelength range \(0.7 \leq \lambda_{c} \leq 5 \mu \mathrm{m}\).

Four diffuse surfaces having the spectral characteristics shown are at \(300 \mathrm{~K}\) and are exposed to solar radiation. Which of the surfaces may be approximated as being gray?

A deep cavity of \(50-\mathrm{mm}\) diameter approximates a blackbody and is maintained at \(250^{\circ} \mathrm{C}\) while exposed to solar irradiation of \(800 \mathrm{~W} / \mathrm{m}^{2}\) and surroundings and ambient air at \(25^{\circ} \mathrm{C}\). A thin window of spectral transmissivity and reflectivity \(0.9\) and 0 , respectively, for the spectral range \(0.2\) to \(4 \mu \mathrm{m}\) is placed over the cavity opening. In the spectral range beyond \(4 \mu \mathrm{m}\), the window behaves as an opaque, diffuse, gray body of emissivity \(0.95\). Assuming that the convection coefficient on the upper surface of the window is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\),
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