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Heat transfer occurs in three fundamental ways: conduction, convection, and radiation. While conduction and convection require a medium (solid, liquid, or gas) to transfer heat, thermal radiation is the transfer of energy through electromagnetic waves and does not require any medium. This makes radiation the only mode of heat transfer that can occur through a vacuum, such as the heat from the Sun reaching the Earth.
Thermal radiation is emitted by all bodies with a temperature above absolute zero. The amount and nature of this radiation depend on the body's temperature and surface properties.
To understand radiation quantitatively, we introduce the concept of a blackbody. A blackbody is an idealized surface that absorbs all incident radiation, regardless of wavelength or angle, and emits the maximum possible radiation at every wavelength for a given temperature. Real surfaces emit less radiation and are called gray bodies.
Studying blackbody radiation provides the foundation for understanding real-world radiative heat transfer. The laws governing blackbody radiation-Stefan-Boltzmann, Wien's displacement, and Kirchhoff's laws-are essential tools for engineers to analyze and design systems involving thermal radiation.
The Stefan-Boltzmann law quantifies the total energy radiated per unit surface area of a blackbody across all wavelengths. It states that this energy is proportional to the fourth power of the absolute temperature of the body.
Mathematically, for a perfect blackbody:
For real surfaces, which are not perfect emitters, the emissive power is reduced by a factor called emissivity (\( \varepsilon \)), a dimensionless number between 0 and 1. Thus, the emissive power of a real surface is:
Why the fourth power? The \( T^4 \) dependence arises from the physics of electromagnetic radiation and quantum mechanics, indicating that even small increases in temperature cause a large increase in radiated energy.
Emissivity depends on material, surface finish, temperature, and wavelength. For example, a polished metal surface may have emissivity around 0.1, while a black painted surface can have emissivity close to 0.95.
While Stefan-Boltzmann law gives the total energy radiated, the Wien's displacement law tells us the wavelength at which the radiation intensity is maximum for a blackbody at a given temperature. This peak wavelength shifts to shorter wavelengths as temperature increases.
Mathematically:
This law is extremely useful in engineering and science to estimate the temperature of an object by measuring the wavelength of its peak emission, such as in pyrometry or astrophysics.
Kirchhoff's law states that for a body in thermal equilibrium, the emissivity (\( \varepsilon_\lambda \)) at a particular wavelength equals its absorptivity (\( \alpha_\lambda \)) at that wavelength:
This law implies that a good absorber at a particular wavelength is also a good emitter at that wavelength. It is fundamental in understanding radiative heat exchange and designing surfaces for thermal control.
For example, a surface that absorbs 90% of incident radiation at a certain wavelength will also emit 90% of the radiation at that wavelength when heated.
Step 1: Identify given data:
Step 2: Use Stefan-Boltzmann law:
\[ E = \varepsilon \sigma T^4 = 1 \times 5.67 \times 10^{-8} \times (1000)^4 \]
Calculate \( T^4 = 1000^4 = 10^{12} \)
\[ E = 5.67 \times 10^{-8} \times 10^{12} = 5.67 \times 10^{4} \text{ W/m}^2 \]
Step 3: Calculate total power radiated:
\[ Q = E \times A = 5.67 \times 10^{4} \times 0.5 = 2.835 \times 10^{4} \text{ W} \]
Answer: The blackbody surface radiates 28,350 W of power.
Step 1: Given temperature \( T = 1500 \) K, Wien's constant \( b = 2.897 \times 10^{-3} \) m·K.
Step 2: Apply Wien's displacement law:
\[ \lambda_{max} = \frac{b}{T} = \frac{2.897 \times 10^{-3}}{1500} = 1.931 \times 10^{-6} \text{ m} \]
Step 3: Convert to micrometers (1 μm = \(10^{-6}\) m):
\[ \lambda_{max} = 1.931 \, \mu m \]
Answer: The peak wavelength is approximately 1.93 μm.
Step 1: Given absorptivity \( \alpha = 0.8 \), radiated power \( E = 400 \) W/m², temperature \( T = 600 \) K.
Step 2: According to Kirchhoff's law, emissivity equals absorptivity at thermal equilibrium:
\[ \varepsilon = \alpha = 0.8 \]
Step 3: Check if the radiated power matches Stefan-Boltzmann law:
\[ E = \varepsilon \sigma T^4 = 0.8 \times 5.67 \times 10^{-8} \times (600)^4 \]
Calculate \( 600^4 = 1.296 \times 10^{11} \)
\[ E = 0.8 \times 5.67 \times 10^{-8} \times 1.296 \times 10^{11} = 0.8 \times 7350 = 5880 \text{ W/m}^2 \]
The given radiated power (400 W/m²) is much less than calculated, indicating the measured value might be at a specific wavelength or under different conditions. However, emissivity from Kirchhoff's law remains 0.8.
Answer: Emissivity of the surface is 0.8.
Step 1: Given data:
Step 2: Calculate \( T_1^4 \) and \( T_2^4 \):
\[ T_1^4 = 800^4 = (8 \times 10^2)^4 = 4.096 \times 10^{11} \]
\[ T_2^4 = 600^4 = 1.296 \times 10^{11} \]
Step 3: Calculate denominator terms:
\[ \frac{1 - \varepsilon_1}{\varepsilon_1 A_1} = \frac{1 - 0.9}{0.9 \times 2} = \frac{0.1}{1.8} = 0.0556 \]
\[ \frac{1}{A_1 F_{12}} = \frac{1}{2 \times 1} = 0.5 \]
\[ \frac{1 - \varepsilon_2}{\varepsilon_2 A_2} = \frac{1 - 0.7}{0.7 \times 2} = \frac{0.3}{1.4} = 0.2143 \]
Step 4: Sum denominator:
\[ 0.0556 + 0.5 + 0.2143 = 0.7699 \]
Step 5: Calculate net heat transfer:
\[ Q = \sigma A_1 (T_1^4 - T_2^4) \times \frac{1}{0.7699} \]
\[ Q = 5.67 \times 10^{-8} \times 2 \times (4.096 \times 10^{11} - 1.296 \times 10^{11}) \times \frac{1}{0.7699} \]
\[ Q = 5.67 \times 10^{-8} \times 2 \times 2.8 \times 10^{11} \times 1.299 \]
\[ Q = 5.67 \times 10^{-8} \times 2 \times 2.8 \times 10^{11} \times 1.299 \]
Calculate stepwise:
\[ 5.67 \times 10^{-8} \times 2 = 1.134 \times 10^{-7} \]
\[ 1.134 \times 10^{-7} \times 2.8 \times 10^{11} = 1.134 \times 2.8 \times 10^{4} = 3.175 \times 10^{4} \]
\[ 3.175 \times 10^{4} \times 1.299 = 41260 \text{ W} \]
Answer: The net radiative heat transfer between the plates is approximately 41,260 W.
Step 1: Given peak wavelength \( \lambda_{max} = 2.5 \, \mu m = 2.5 \times 10^{-6} \) m.
Step 2: Use Wien's displacement law:
\[ T = \frac{b}{\lambda_{max}} = \frac{2.897 \times 10^{-3}}{2.5 \times 10^{-6}} = 1158.8 \text{ K} \]
Answer: The estimated temperature of the furnace wall is approximately 1159 K.
When to use: In all problems involving Stefan-Boltzmann and Wien's laws.
When to use: When exact emissivity data is not provided in exam problems.
When to use: While verifying answers in radiative heat transfer problems.
When to use: During formula application and problem solving.
When to use: To save time during competitive exams.
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