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Solar radiation of 1200 W/m2 falls perpendicularly on a grey opaquesurface of emissivity 0.5. If the surface temperature is 50°C and surfaceemissive power 600 W/m2, the radiosity of that surface will be:
- a)600 W/m2
- b)1000 W/m2
- c)1200 W/m2
- d)1800 W/m2
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Solar radiation of 1200 W/m2 falls perpendicularly on a grey opaquesur...
We know that radiosity is the total amount of radiation leaving the surface, given as sum of emissive power
of surface and amount reflected solar radiation, so we have
Most Upvoted Answer
Solar radiation of 1200 W/m2 falls perpendicularly on a grey opaquesur...
Calculation of Radiosity
Given:
Solar radiation = 1200 W/m2
Emissivity of surface = 0.5
Surface temperature = 50°C
Surface emissive power = 600 W/m2
Formula:
Radiosity = Solar radiation absorbed + Surface emissive power
Calculation:
Solar radiation absorbed = (1 - emissivity) x Solar radiation
= (1 - 0.5) x 1200
= 600 W/m2
Surface emissive power = 600 W/m2
Radiosity = 600 + 600
= 1200 W/m2
Therefore, the radiosity of the grey opaque surface is 1200 W/m2.
Given:
Solar radiation = 1200 W/m2
Emissivity of surface = 0.5
Surface temperature = 50°C
Surface emissive power = 600 W/m2
Formula:
Radiosity = Solar radiation absorbed + Surface emissive power
Calculation:
Solar radiation absorbed = (1 - emissivity) x Solar radiation
= (1 - 0.5) x 1200
= 600 W/m2
Surface emissive power = 600 W/m2
Radiosity = 600 + 600
= 1200 W/m2
Therefore, the radiosity of the grey opaque surface is 1200 W/m2.
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Solar radiation of 1200 W/m2 falls perpendicularly on a grey opaquesurface of emissivity 0.5. If the surface temperature is 50°C and surfaceemissive power 600 W/m2, the radiosity of that surface will be:a)600 W/m2b)1000 W/m2c)1200 W/m2d)1800 W/m2Correct answer is option 'C'. Can you explain this answer?
Question Description
Solar radiation of 1200 W/m2 falls perpendicularly on a grey opaquesurface of emissivity 0.5. If the surface temperature is 50°C and surfaceemissive power 600 W/m2, the radiosity of that surface will be:a)600 W/m2b)1000 W/m2c)1200 W/m2d)1800 W/m2Correct answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Solar radiation of 1200 W/m2 falls perpendicularly on a grey opaquesurface of emissivity 0.5. If the surface temperature is 50°C and surfaceemissive power 600 W/m2, the radiosity of that surface will be:a)600 W/m2b)1000 W/m2c)1200 W/m2d)1800 W/m2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Solar radiation of 1200 W/m2 falls perpendicularly on a grey opaquesurface of emissivity 0.5. If the surface temperature is 50°C and surfaceemissive power 600 W/m2, the radiosity of that surface will be:a)600 W/m2b)1000 W/m2c)1200 W/m2d)1800 W/m2Correct answer is option 'C'. Can you explain this answer?.
Solar radiation of 1200 W/m2 falls perpendicularly on a grey opaquesurface of emissivity 0.5. If the surface temperature is 50°C and surfaceemissive power 600 W/m2, the radiosity of that surface will be:a)600 W/m2b)1000 W/m2c)1200 W/m2d)1800 W/m2Correct answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Solar radiation of 1200 W/m2 falls perpendicularly on a grey opaquesurface of emissivity 0.5. If the surface temperature is 50°C and surfaceemissive power 600 W/m2, the radiosity of that surface will be:a)600 W/m2b)1000 W/m2c)1200 W/m2d)1800 W/m2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Solar radiation of 1200 W/m2 falls perpendicularly on a grey opaquesurface of emissivity 0.5. If the surface temperature is 50°C and surfaceemissive power 600 W/m2, the radiosity of that surface will be:a)600 W/m2b)1000 W/m2c)1200 W/m2d)1800 W/m2Correct answer is option 'C'. Can you explain this answer?.
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