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Solar radiation of 1000 W/m2 is incident on a grey opaque surface with emissivity of 0.4 and emissive power of 400 W/m2. The radiosity of the surface will be:
- a)940 W/m2
- b)850 W/m2
- c)760 W/m2
- d)670 W/m2
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Solar radiation of 1000 W/m2 is incident on a grey opaque surface with...
In this question emissive power(E) = 400 W/m2.
G = 1000W/m2
For, opaque surface, ζ = 0
α + ρ + ζ = 1
α = 0.4
ρ = 1-α = 1 - ε = 1 - 0.4 = 0.6
Radiosity(J) = εE + ρG = 0.4 × 400 + 0.6 × 1000 = 760W/m2
Most Upvoted Answer
Solar radiation of 1000 W/m2 is incident on a grey opaque surface with...
Given data:
- Solar radiation incident on the surface: 1000 W/m2
- Emissivity of the surface: 0.4
- Emissive power of the surface: 400 W/m2
Formula:
The radiosity (J) of a surface is given by the sum of the reflected solar radiation (R) and the emitted thermal radiation (E) from the surface.
J = R + E
Calculation:
1. Reflected solar radiation (R):
The reflected solar radiation is given by the product of the solar radiation incident on the surface and the reflectivity (ρ) of the surface.
Since the surface is opaque and grey, we assume it has a reflectivity of 1 - emissivity = 1 - 0.4 = 0.6.
R = ρ * Solar radiation
= 0.6 * 1000
= 600 W/m2
2. Emitted thermal radiation (E):
The emitted thermal radiation is given by the product of the emissive power (ε) and the emissivity (σ) of the surface.
Since the surface is opaque and grey, we assume it has an emissivity (σ) of 0.4.
E = ε * Emissivity
= 400 * 0.4
= 160 W/m2
3. Radiosity (J):
The radiosity is the sum of the reflected solar radiation and the emitted thermal radiation.
J = R + E
= 600 + 160
= 760 W/m2
Therefore, the radiosity of the surface is 760 W/m2, which corresponds to option C.
- Solar radiation incident on the surface: 1000 W/m2
- Emissivity of the surface: 0.4
- Emissive power of the surface: 400 W/m2
Formula:
The radiosity (J) of a surface is given by the sum of the reflected solar radiation (R) and the emitted thermal radiation (E) from the surface.
J = R + E
Calculation:
1. Reflected solar radiation (R):
The reflected solar radiation is given by the product of the solar radiation incident on the surface and the reflectivity (ρ) of the surface.
Since the surface is opaque and grey, we assume it has a reflectivity of 1 - emissivity = 1 - 0.4 = 0.6.
R = ρ * Solar radiation
= 0.6 * 1000
= 600 W/m2
2. Emitted thermal radiation (E):
The emitted thermal radiation is given by the product of the emissive power (ε) and the emissivity (σ) of the surface.
Since the surface is opaque and grey, we assume it has an emissivity (σ) of 0.4.
E = ε * Emissivity
= 400 * 0.4
= 160 W/m2
3. Radiosity (J):
The radiosity is the sum of the reflected solar radiation and the emitted thermal radiation.
J = R + E
= 600 + 160
= 760 W/m2
Therefore, the radiosity of the surface is 760 W/m2, which corresponds to option C.
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Solar radiation of 1000 W/m2 is incident on a grey opaque surface with emissivity of 0.4 and emissive power of 400 W/m2. The radiosity of the surface will be:a)940 W/m2b)850 W/m2c)760 W/m2d)670 W/m2Correct answer is option 'C'. Can you explain this answer?
Question Description
Solar radiation of 1000 W/m2 is incident on a grey opaque surface with emissivity of 0.4 and emissive power of 400 W/m2. The radiosity of the surface will be:a)940 W/m2b)850 W/m2c)760 W/m2d)670 W/m2Correct answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Solar radiation of 1000 W/m2 is incident on a grey opaque surface with emissivity of 0.4 and emissive power of 400 W/m2. The radiosity of the surface will be:a)940 W/m2b)850 W/m2c)760 W/m2d)670 W/m2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Solar radiation of 1000 W/m2 is incident on a grey opaque surface with emissivity of 0.4 and emissive power of 400 W/m2. The radiosity of the surface will be:a)940 W/m2b)850 W/m2c)760 W/m2d)670 W/m2Correct answer is option 'C'. Can you explain this answer?.
Solar radiation of 1000 W/m2 is incident on a grey opaque surface with emissivity of 0.4 and emissive power of 400 W/m2. The radiosity of the surface will be:a)940 W/m2b)850 W/m2c)760 W/m2d)670 W/m2Correct answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Solar radiation of 1000 W/m2 is incident on a grey opaque surface with emissivity of 0.4 and emissive power of 400 W/m2. The radiosity of the surface will be:a)940 W/m2b)850 W/m2c)760 W/m2d)670 W/m2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Solar radiation of 1000 W/m2 is incident on a grey opaque surface with emissivity of 0.4 and emissive power of 400 W/m2. The radiosity of the surface will be:a)940 W/m2b)850 W/m2c)760 W/m2d)670 W/m2Correct answer is option 'C'. Can you explain this answer?.
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