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The intensity of radiation in normal direction of a diffused surface at 600°C with emissivity of 0.15 (in W/m2sr is__________ . (Round off to two decimal places)
Correct answer is '1573'. Can you explain this answer?
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The intensity of radiation in normal direction of a diffused surface ...
T = 600oC = 873 K
E = πin
⇒ in = E/π
= εEb/π
= εσT4/π
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The intensity of radiation in normal direction of a diffused surface ...
Given:
- Temperature (T) = 600°C = 600 + 273.15 = 873.15 K
- Emissivity (ε) = 0.15
To Find:
- Intensity of radiation in the normal direction of the diffused surface
Solution:
Stefan-Boltzmann Law:
The Stefan-Boltzmann law states that the total radiant heat energy emitted per unit surface area of a black body is directly proportional to the fourth power of its absolute temperature (in Kelvin). Mathematically, it can be expressed as:
E = σ * ε * T^4
Where,
- E is the radiant heat energy emitted per unit area (in W/m^2)
- σ is the Stefan-Boltzmann constant (5.67 × 10^-8 W/m^2K^4)
- ε is the emissivity of the surface (dimensionless)
- T is the absolute temperature (in K)
Calculating the Intensity:
Substituting the given values into the Stefan-Boltzmann law equation:
E = σ * ε * T^4
= (5.67 × 10^-8) * 0.15 * (873.15)^4
≈ 1572.83 W/m^2
Rounding off to two decimal places, the intensity of radiation in the normal direction of the diffused surface at 600°C with an emissivity of 0.15 is approximately 1573 W/m^2.
Therefore, the correct answer is 1573.
Note:
The intensity of radiation is calculated assuming that the surface is a diffused surface, which means it is not a perfect black body. The emissivity value (ε) represents the ability of the surface to emit radiation compared to a perfect black body. A perfect black body has an emissivity of 1.
- Temperature (T) = 600°C = 600 + 273.15 = 873.15 K
- Emissivity (ε) = 0.15
To Find:
- Intensity of radiation in the normal direction of the diffused surface
Solution:
Stefan-Boltzmann Law:
The Stefan-Boltzmann law states that the total radiant heat energy emitted per unit surface area of a black body is directly proportional to the fourth power of its absolute temperature (in Kelvin). Mathematically, it can be expressed as:
E = σ * ε * T^4
Where,
- E is the radiant heat energy emitted per unit area (in W/m^2)
- σ is the Stefan-Boltzmann constant (5.67 × 10^-8 W/m^2K^4)
- ε is the emissivity of the surface (dimensionless)
- T is the absolute temperature (in K)
Calculating the Intensity:
Substituting the given values into the Stefan-Boltzmann law equation:
E = σ * ε * T^4
= (5.67 × 10^-8) * 0.15 * (873.15)^4
≈ 1572.83 W/m^2
Rounding off to two decimal places, the intensity of radiation in the normal direction of the diffused surface at 600°C with an emissivity of 0.15 is approximately 1573 W/m^2.
Therefore, the correct answer is 1573.
Note:
The intensity of radiation is calculated assuming that the surface is a diffused surface, which means it is not a perfect black body. The emissivity value (ε) represents the ability of the surface to emit radiation compared to a perfect black body. A perfect black body has an emissivity of 1.
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The intensity of radiation in normal direction of a diffused surface at 600°C with emissivity of 0.15 (in W/m2sr is__________ . (Round off to two decimal places)Correct answer is '1573'. Can you explain this answer?
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