Solar radiation of falls on a grey opaque surface at steady state. The surface has a temperature of and emissivity of 0.8. Find radiosity from the surface?Correct answer is 'A'. Can you explain this answer?

Question

Answer

J = E+G_r

E = εσT4

E = 0.8 x 5.67 x 10-8 x (50 + 273)4

E = 493.72 W/m2

Given G = 1200 W/m2

But as surface is at steady state

α = ε = 0.8

∴ ρ = 1 -α = 0.2

∴ G_r = ρG

G_r = 0.2 x 1200 = 240 W/m²

Therefore

J = E+G_r

J = 493.72 + 240

J = 733.72 W/m²

Answer

Understanding the Problem
In this scenario, we are examining a grey opaque surface receiving solar radiation at steady state. The surface has a defined temperature and an emissivity of 0.8. Our goal is to determine the radiosity of this surface.
Key Concepts
- Radiosity (J): This is the total energy leaving a surface per unit area, which includes both emitted and reflected energy.
- Emissivity (ε): This is a measure of how effectively a surface emits thermal radiation relative to a perfect black body. For this case, ε is given as 0.8.
Formulating the Radiosity
The formula for radiosity can be expressed as:
J = ε * σ * T^4 + (1 - ε) * G
Where:
- J = Radiosity
- ε = Emissivity of the surface (0.8)
- σ = Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/m²K⁴)
- T = Temperature of the surface in Kelvin
- G = Irradiation or solar radiation incident on the surface
Calculating the Components
1. Emitted Radiation: This is calculated using the emissivity and the temperature of the surface.
2. Reflected Radiation: This is determined by the proportion of solar radiation that is not absorbed due to emissivity.
Final Calculation
To find the final value of radiosity, simply plug in the values for temperature, emissivity, and solar radiation into the formula. The final output will yield the radiosity of the grey opaque surface.
Conclusion
In summary, the total radiosity J combines both emitted and reflected components, accounting for the surface's emissivity. By understanding the contribution from both thermal emission and solar radiation absorption, one can accurately determine the radiosity from the surface in question.

Answer

J = E+G_r

E = εσT4

E = 0.8 x 5.67 x 10-8 x (50 + 273)4

E = 493.72 W/m2

Given G = 1200 W/m2

But as surface is at steady state

α = ε = 0.8

∴ ρ = 1 -α = 0.2

∴ G_r = ρG

G_r = 0.2 x 1200 = 240 W/m²

Therefore

J = E+G_r

J = 493.72 + 240

J = 733.72 W/m²

Similar Chemical Engineering Doubts

For an opaque plane surface at steady state the irradiation, radiosity and emissive power are respectively 20, 12 and . What is the emissivity of the surface?

The spectral emissivity of an opaque surface at 800 K is approximated asBlack body radiation function values for a black surface at 800 K are as followsFind the total emissivity of the surface?

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