Mechanical Engineering Exam > Mechanical Engineering Questions > Solar radiation of 1200 W/m² falls perpendic...
Start Learning for Free
Solar radiation of 1200 W/m² falls perpendicularly on a gray opaque surface of emissivity 0.5. If the surface temperature is 50°C and surface emissive power is 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?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
Solar radiation of 1200 W/m² falls perpendicularly on a gray opaque s...
The radiosity of a surface is the total amount of radiation leaving the surface per unit area, including both emitted and reflected radiation. It is given by the equation:
Radiosity = Emissive Power + Reflective Power
1. Emissive Power:
Emissive power is the amount of radiation emitted by the surface. In this case, the emissive power is given as 600 W/m².
2. Reflective Power:
The reflective power is the amount of radiation reflected by the surface. Since the surface is opaque, it does not transmit any radiation, so all the incident radiation is either absorbed or reflected. The surface is gray, which means it reflects a portion of the incident radiation. The reflectivity of a gray surface is equal to 1 - emissivity. In this case, the emissivity is given as 0.5, so the reflectivity is 1 - 0.5 = 0.5.
The reflective power can be calculated as:
Reflective Power = Reflectivity * Incident Radiation
3. Incident Radiation:
The incident radiation is the solar radiation falling perpendicularly on the surface. It is given as 1200 W/m².
Calculating the Reflective Power:
Reflective Power = 0.5 * 1200 = 600 W/m²
Calculating the Radiosity:
Radiosity = Emissive Power + Reflective Power
Radiosity = 600 W/m² + 600 W/m²
Radiosity = 1200 W/m²
Therefore, the radiosity of the surface is 1200 W/m², which is option C.
Radiosity = Emissive Power + Reflective Power
1. Emissive Power:
Emissive power is the amount of radiation emitted by the surface. In this case, the emissive power is given as 600 W/m².
2. Reflective Power:
The reflective power is the amount of radiation reflected by the surface. Since the surface is opaque, it does not transmit any radiation, so all the incident radiation is either absorbed or reflected. The surface is gray, which means it reflects a portion of the incident radiation. The reflectivity of a gray surface is equal to 1 - emissivity. In this case, the emissivity is given as 0.5, so the reflectivity is 1 - 0.5 = 0.5.
The reflective power can be calculated as:
Reflective Power = Reflectivity * Incident Radiation
3. Incident Radiation:
The incident radiation is the solar radiation falling perpendicularly on the surface. It is given as 1200 W/m².
Calculating the Reflective Power:
Reflective Power = 0.5 * 1200 = 600 W/m²
Calculating the Radiosity:
Radiosity = Emissive Power + Reflective Power
Radiosity = 600 W/m² + 600 W/m²
Radiosity = 1200 W/m²
Therefore, the radiosity of the surface is 1200 W/m², which is option C.
Attention Mechanical Engineering Students!
To make sure you are not studying endlessly, EduRev has designed Mechanical Engineering study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Mechanical Engineering.
|
Explore Courses for Mechanical Engineering exam
|
|
Top Courses for Mechanical EngineeringView all
Solar radiation of 1200 W/m² falls perpendicularly on a gray opaque surface of emissivity 0.5. If the surface temperature is 50°C and surface emissive power is 600 W/m², the radiosity of that surface will bea)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 gray opaque surface of emissivity 0.5. If the surface temperature is 50°C and surface emissive power is 600 W/m², the radiosity of that surface will bea)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 gray opaque surface of emissivity 0.5. If the surface temperature is 50°C and surface emissive power is 600 W/m², the radiosity of that surface will bea)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 gray opaque surface of emissivity 0.5. If the surface temperature is 50°C and surface emissive power is 600 W/m², the radiosity of that surface will bea)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 gray opaque surface of emissivity 0.5. If the surface temperature is 50°C and surface emissive power is 600 W/m², the radiosity of that surface will bea)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 gray opaque surface of emissivity 0.5. If the surface temperature is 50°C and surface emissive power is 600 W/m², the radiosity of that surface will bea)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 gray opaque surface of emissivity 0.5. If the surface temperature is 50°C and surface emissive power is 600 W/m², the radiosity of that surface will bea)600 W/m²b)1000 W/m²c)1200 W/m²d)1800 W/m²Correct answer is option 'C'. Can you explain this answer?.
Solutions for Solar radiation of 1200 W/m² falls perpendicularly on a gray opaque surface of emissivity 0.5. If the surface temperature is 50°C and surface emissive power is 600 W/m², the radiosity of that surface will bea)600 W/m²b)1000 W/m²c)1200 W/m²d)1800 W/m²Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free.
Here you can find the meaning of Solar radiation of 1200 W/m² falls perpendicularly on a gray opaque surface of emissivity 0.5. If the surface temperature is 50°C and surface emissive power is 600 W/m², the radiosity of that surface will bea)600 W/m²b)1000 W/m²c)1200 W/m²d)1800 W/m²Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Solar radiation of 1200 W/m² falls perpendicularly on a gray opaque surface of emissivity 0.5. If the surface temperature is 50°C and surface emissive power is 600 W/m², the radiosity of that surface will bea)600 W/m²b)1000 W/m²c)1200 W/m²d)1800 W/m²Correct answer is option 'C'. Can you explain this answer?, a detailed solution for Solar radiation of 1200 W/m² falls perpendicularly on a gray opaque surface of emissivity 0.5. If the surface temperature is 50°C and surface emissive power is 600 W/m², the radiosity of that surface will bea)600 W/m²b)1000 W/m²c)1200 W/m²d)1800 W/m²Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of Solar radiation of 1200 W/m² falls perpendicularly on a gray opaque surface of emissivity 0.5. If the surface temperature is 50°C and surface emissive power is 600 W/m², the radiosity of that surface will bea)600 W/m²b)1000 W/m²c)1200 W/m²d)1800 W/m²Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Solar radiation of 1200 W/m² falls perpendicularly on a gray opaque surface of emissivity 0.5. If the surface temperature is 50°C and surface emissive power is 600 W/m², the radiosity of that surface will bea)600 W/m²b)1000 W/m²c)1200 W/m²d)1800 W/m²Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
|
|
Explore Courses for Mechanical Engineering exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup