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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

Isothermal furnaces with small apertures approximating a blackbody are frequently used to calibrate heat flux gages, radiation thermometers, and other radiometric devices. In such applications, it is necessary to control power to the furnace such that the variation of temperature and the spectral intensity of the aperture are within desired limits. (a) By considering the Planck spectral distribution, Equation \(12.30\), show that the ratio of the fractional change in the spectral intensity to the fractional change in the temperature of the furnace has the form $$ \frac{d I_{\lambda} / I_{\lambda}}{d T / T}=\frac{C_{2}}{\lambda T} \frac{1}{1-\exp \left(-C_{2} / \lambda T\right)} $$ (b) Using this relation, determine the allowable variation in temperature of the furnace operating at \(2000 \mathrm{~K}\) to ensure that the spectral intensity at \(0.65 \mu \mathrm{m}\) will not vary by more than \(0.5 \%\). What is the allowable variation at \(10 \mu \mathrm{m}\) ?

A horizontal semitransparent plate is uniformly irradiated from above and below, while air at \(T_{c}=300 \mathrm{~K}\) flows over the top and bottom surfaces, providing a uniform convection heat transfer coefficient of \(h=40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The absorptivity of the plate to the irradiation is \(0.40\). Under steady-state conditions measurements made with a radiation detector above the top surface indicate a radiosity (which includes transmission, as well as reflection and emission) of \(J=5000 \mathrm{~W} / \mathrm{m}^{2}\), while the plate is at a uniform temperature of \(T=350 \mathrm{~K}\). Determine the irradiation \(G\) and the emissivity of the plate. Is the plate gray \((\varepsilon=\alpha)\) for the prescribed conditions?

Square plates freshly sprayed with an epoxy paint must be cured at \(140^{\circ} \mathrm{C}\) for an extended period of time. The plates are located in a large enclosure and heated by a bank of infrared lamps. The top surface of each plate has an emissivity of \(\varepsilon=0.8\) and experiences convection with a ventilation airstream that is at \(T_{\infty}=27^{\circ} \mathrm{C}\) and provides a convection coefficient of \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The irradiation from the enclosure walls is estimated to be \(G_{\text {wall }}=450 \mathrm{~W} / \mathrm{m}^{2}\), for which the plate absorptivity is \(\alpha_{\text {wall }}=0.7\). (a) Determine the irradiation that must be provided by the lamps, \(G_{\text {lamp. }}\). The absorptivity of the plate surface for this irradiation is \(\alpha_{\text {Lamp }}=0.6\). (b) For convection coefficients of \(h=15,20\), and \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), plot the lamp irradiation, \(G_{\text {lamp, as a }}\) function of the plate temperature, \(T_{s}\), for \(100 \leq\) \(T_{x} \leq 300^{\circ} \mathrm{C}\). (c) For convection coefficients in the range from 10 to \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and a lamp irradiation of \(G_{\text {lmp }}=\) \(3000 \mathrm{~W} / \mathrm{m}^{2}\), plot the airstream temperature \(T_{x}\) required to maintain the plate at \(T_{x}=140^{\circ} \mathrm{C}\).

A thermocouple whose surface is diffuse and gray with an emissivity of \(0.6\) indicates a temperature of \(180^{\circ} \mathrm{C}\) when used to measure the temperature of a gas flowing through a large duct whose walls have an emissivity of \(0.85\) and a uniform temperature of \(450^{\circ} \mathrm{C}\). (a) If the convection heat transfer coefficient between the thermocouple and the gas stream is \(h=125 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and there are negligible conduction losses from the thermocouple, determine the temperature of the gas. (b) Consider a gas temperature of \(125^{\circ} \mathrm{C}\). Compute and plot the thermocouple measurement error as a function of the convection coefficient for \(10 \leq 5\) \(h \leq 1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What are the implications of your results?

It is not uncommon for the night sky temperature in desert regions to drop to \(-40^{\circ} \mathrm{C}\). If the ambient air temperature is \(20^{\circ} \mathrm{C}\) and the convection coefticient for still air conditions is approximately \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), can a shallow pan of water freeze?
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