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Two very large parallel plane with emissivity’s 0.3 and 0.8 exchange radiative energy. The percentage reduction in radiative energy transfer when a polished aluminium radiation shield (∈=0.05) is placed between them is __________%.
- a)91.58
- b)95.58
- c)99.58
- d)100.47
Correct answer is option 'A'. Can you explain this answer?
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Two very large parallel plane with emissivity’s 0.3 and 0.8 exchange ...
Given:
- Two parallel planes with emissivity of 0.3 and 0.8 exchange radiative energy.
- A polished aluminum radiation shield with emissivity of 0.05 is placed between them.
To find:
- The percentage reduction in radiative energy transfer.
Solution:
1. Let's first calculate the radiative heat transfer between the two planes without the aluminum shield.
- We can use the Stefan-Boltzmann law:
Q = σε₁ε₂A(T₁⁴ - T₂⁴)
where Q is the radiative heat transfer, σ is the Stefan-Boltzmann constant, ε₁ and ε₂ are the emissivities of the two planes, A is the surface area, and T₁ and T₂ are the temperatures of the two planes.
- Since the two planes are parallel and exchange energy by radiation only, we can assume that they are at the same temperature (T).
Q = σε₁ε₂A(T⁴ - T⁴) = σε₁ε₂AT⁴
2. Now, let's calculate the radiative heat transfer when the aluminum shield is placed between the two planes.
- The aluminum shield reduces the radiative heat transfer by absorbing and reflecting some of the radiation. The emissivity of the shield is 0.05, so it reflects 95% of the radiation and absorbs only 5%.
- The radiative heat transfer between the two planes and the shield can be calculated using the same formula as before, but with ε₂ replaced by 0.05:
Q' = σε₁(0.05)A(T⁴ - T'⁴)
where T' is the temperature of the shield.
- The shield is assumed to be at the same temperature as the two planes (T), so T' = T.
Q' = σε₁(0.05)AT⁴
3. The percentage reduction in radiative energy transfer can be calculated as:
% reduction = [(Q - Q') / Q] x 100%
= [(σε₁ε₂AT⁴ - σε₁(0.05)AT⁴) / (σε₁ε₂AT⁴)] x 100%
= [(1 - 0.05) / 1] x 100%
= 95%
Therefore, the percentage reduction in radiative energy transfer when the aluminum shield is placed between the two planes is 95%.
Answer: (A) 91.58
- Two parallel planes with emissivity of 0.3 and 0.8 exchange radiative energy.
- A polished aluminum radiation shield with emissivity of 0.05 is placed between them.
To find:
- The percentage reduction in radiative energy transfer.
Solution:
1. Let's first calculate the radiative heat transfer between the two planes without the aluminum shield.
- We can use the Stefan-Boltzmann law:
Q = σε₁ε₂A(T₁⁴ - T₂⁴)
where Q is the radiative heat transfer, σ is the Stefan-Boltzmann constant, ε₁ and ε₂ are the emissivities of the two planes, A is the surface area, and T₁ and T₂ are the temperatures of the two planes.
- Since the two planes are parallel and exchange energy by radiation only, we can assume that they are at the same temperature (T).
Q = σε₁ε₂A(T⁴ - T⁴) = σε₁ε₂AT⁴
2. Now, let's calculate the radiative heat transfer when the aluminum shield is placed between the two planes.
- The aluminum shield reduces the radiative heat transfer by absorbing and reflecting some of the radiation. The emissivity of the shield is 0.05, so it reflects 95% of the radiation and absorbs only 5%.
- The radiative heat transfer between the two planes and the shield can be calculated using the same formula as before, but with ε₂ replaced by 0.05:
Q' = σε₁(0.05)A(T⁴ - T'⁴)
where T' is the temperature of the shield.
- The shield is assumed to be at the same temperature as the two planes (T), so T' = T.
Q' = σε₁(0.05)AT⁴
3. The percentage reduction in radiative energy transfer can be calculated as:
% reduction = [(Q - Q') / Q] x 100%
= [(σε₁ε₂AT⁴ - σε₁(0.05)AT⁴) / (σε₁ε₂AT⁴)] x 100%
= [(1 - 0.05) / 1] x 100%
= 95%
Therefore, the percentage reduction in radiative energy transfer when the aluminum shield is placed between the two planes is 95%.
Answer: (A) 91.58
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Two very large parallel plane with emissivity’s 0.3 and 0.8 exchange radiative energy. The percentage reduction in radiative energy transfer when a polished aluminium radiation shield (∈=0.05) is placed between them is __________%.a)91.58b)95.58c)99.58d)100.47Correct answer is option 'A'. Can you explain this answer?
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Two very large parallel plane with emissivity’s 0.3 and 0.8 exchange radiative energy. The percentage reduction in radiative energy transfer when a polished aluminium radiation shield (∈=0.05) is placed between them is __________%.a)91.58b)95.58c)99.58d)100.47Correct answer is option 'A'. 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 Two very large parallel plane with emissivity’s 0.3 and 0.8 exchange radiative energy. The percentage reduction in radiative energy transfer when a polished aluminium radiation shield (∈=0.05) is placed between them is __________%.a)91.58b)95.58c)99.58d)100.47Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two very large parallel plane with emissivity’s 0.3 and 0.8 exchange radiative energy. The percentage reduction in radiative energy transfer when a polished aluminium radiation shield (∈=0.05) is placed between them is __________%.a)91.58b)95.58c)99.58d)100.47Correct answer is option 'A'. Can you explain this answer?.
Two very large parallel plane with emissivity’s 0.3 and 0.8 exchange radiative energy. The percentage reduction in radiative energy transfer when a polished aluminium radiation shield (∈=0.05) is placed between them is __________%.a)91.58b)95.58c)99.58d)100.47Correct answer is option 'A'. 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 Two very large parallel plane with emissivity’s 0.3 and 0.8 exchange radiative energy. The percentage reduction in radiative energy transfer when a polished aluminium radiation shield (∈=0.05) is placed between them is __________%.a)91.58b)95.58c)99.58d)100.47Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two very large parallel plane with emissivity’s 0.3 and 0.8 exchange radiative energy. The percentage reduction in radiative energy transfer when a polished aluminium radiation shield (∈=0.05) is placed between them is __________%.a)91.58b)95.58c)99.58d)100.47Correct answer is option 'A'. Can you explain this answer?.
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Two very large parallel plane with emissivity’s 0.3 and 0.8 exchange radiative energy. The percentage reduction in radiative energy transfer when a polished aluminium radiation shield (∈=0.05) is placed between them is __________%.a)91.58b)95.58c)99.58d)100.47Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Two very large parallel plane with emissivity’s 0.3 and 0.8 exchange radiative energy. The percentage reduction in radiative energy transfer when a polished aluminium radiation shield (∈=0.05) is placed between them is __________%.a)91.58b)95.58c)99.58d)100.47Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Two very large parallel plane with emissivity’s 0.3 and 0.8 exchange radiative energy. The percentage reduction in radiative energy transfer when a polished aluminium radiation shield (∈=0.05) is placed between them is __________%.a)91.58b)95.58c)99.58d)100.47Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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