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Two long parallel surfaces each of emissivity 07 are maintained at different temperature and accordingly have radiation heat exchange between them. It is desirable to reduce 75% of this radiant heat transfer by inserting thin parallel shields of equal emissivity on both sides. The number of shields should be
- a)one
- b)two
- c)three
- d)four
Correct answer is option 'C'. Can you explain this answer?
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Two long parallel surfaces each of emissivity 07 are maintained at dif...
where N = Number of radiation shield
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Two long parallel surfaces each of emissivity 07 are maintained at dif...
Explanation:
Given:
- Two long parallel surfaces with emissivity 0.7
- Desired reduction in radiant heat transfer: 75%
- Thin parallel shields with equal emissivity on both sides
- Number of shields required: ?
Solution:
Step 1: Calculate the initial radiant heat transfer
- Let's assume the initial radiant heat transfer between the two surfaces is Q.
- The radiant heat transfer between two surfaces can be calculated using the Stefan-Boltzmann Law:
Q = εσA(T1^4 - T2^4)
where ε is the emissivity, σ is the Stefan-Boltzmann constant, A is the surface area, T1 is the temperature of the first surface, and T2 is the temperature of the second surface.
- Since the emissivity of both surfaces is 0.7, we can simplify the equation as:
Q = 0.7σA(T1^4 - T2^4)
Step 2: Calculate the desired reduction in radiant heat transfer
- The desired reduction in radiant heat transfer is 75%.
- Therefore, the radiant heat transfer after inserting the shields should be 25% of the initial radiant heat transfer.
- Let's assume the radiant heat transfer after inserting the shields is Q'.
- Hence, Q' = 0.25Q
Step 3: Calculate the number of shields required
- Let's assume the number of shields required is n.
- When n shields are inserted between the two surfaces, the radiant heat transfer between each shield is reduced by a factor of 0.7 (emissivity of the shields).
- Therefore, the radiant heat transfer between the shields can be calculated as:
Q'' = (0.7)^nQ
- Since there are two sides to each shield, the total reduction in radiant heat transfer is:
Q''' = (0.7)^{2n}Q
- The radiant heat transfer after inserting the shields is given by:
Q' = Q - Q'''
- Substituting the values of Q' and Q''', we get:
0.25Q = Q - (0.7)^{2n}Q
- Simplifying the equation, we get:
(0.7)^{2n} = 0.75
- Taking the logarithm on both sides of the equation, we get:
2n log(0.7) = log(0.75)
n = log(0.75) / (2 log(0.7))
n ≈ 2.45
Conclusion:
- The number of shields required to reduce 75% of the radiant heat transfer is approximately 2.
- Therefore, the correct answer is option 'C' (two shields).
Given:
- Two long parallel surfaces with emissivity 0.7
- Desired reduction in radiant heat transfer: 75%
- Thin parallel shields with equal emissivity on both sides
- Number of shields required: ?
Solution:
Step 1: Calculate the initial radiant heat transfer
- Let's assume the initial radiant heat transfer between the two surfaces is Q.
- The radiant heat transfer between two surfaces can be calculated using the Stefan-Boltzmann Law:
Q = εσA(T1^4 - T2^4)
where ε is the emissivity, σ is the Stefan-Boltzmann constant, A is the surface area, T1 is the temperature of the first surface, and T2 is the temperature of the second surface.
- Since the emissivity of both surfaces is 0.7, we can simplify the equation as:
Q = 0.7σA(T1^4 - T2^4)
Step 2: Calculate the desired reduction in radiant heat transfer
- The desired reduction in radiant heat transfer is 75%.
- Therefore, the radiant heat transfer after inserting the shields should be 25% of the initial radiant heat transfer.
- Let's assume the radiant heat transfer after inserting the shields is Q'.
- Hence, Q' = 0.25Q
Step 3: Calculate the number of shields required
- Let's assume the number of shields required is n.
- When n shields are inserted between the two surfaces, the radiant heat transfer between each shield is reduced by a factor of 0.7 (emissivity of the shields).
- Therefore, the radiant heat transfer between the shields can be calculated as:
Q'' = (0.7)^nQ
- Since there are two sides to each shield, the total reduction in radiant heat transfer is:
Q''' = (0.7)^{2n}Q
- The radiant heat transfer after inserting the shields is given by:
Q' = Q - Q'''
- Substituting the values of Q' and Q''', we get:
0.25Q = Q - (0.7)^{2n}Q
- Simplifying the equation, we get:
(0.7)^{2n} = 0.75
- Taking the logarithm on both sides of the equation, we get:
2n log(0.7) = log(0.75)
n = log(0.75) / (2 log(0.7))
n ≈ 2.45
Conclusion:
- The number of shields required to reduce 75% of the radiant heat transfer is approximately 2.
- Therefore, the correct answer is option 'C' (two shields).
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Two long parallel surfaces each of emissivity 07 are maintained at different temperature andaccordingly have radiation heat exchange between them. It is desirable to reduce 75% of this radiant heat transfer by inserting thin parallel shields of equal emissivity on both sides. The number of shields should bea)one b)twoc)three d)fourCorrect answer is option 'C'. Can you explain this answer?
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Two long parallel surfaces each of emissivity 07 are maintained at different temperature andaccordingly have radiation heat exchange between them. It is desirable to reduce 75% of this radiant heat transfer by inserting thin parallel shields of equal emissivity on both sides. The number of shields should bea)one b)twoc)three d)fourCorrect 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 Two long parallel surfaces each of emissivity 07 are maintained at different temperature andaccordingly have radiation heat exchange between them. It is desirable to reduce 75% of this radiant heat transfer by inserting thin parallel shields of equal emissivity on both sides. The number of shields should bea)one b)twoc)three d)fourCorrect 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 Two long parallel surfaces each of emissivity 07 are maintained at different temperature andaccordingly have radiation heat exchange between them. It is desirable to reduce 75% of this radiant heat transfer by inserting thin parallel shields of equal emissivity on both sides. The number of shields should bea)one b)twoc)three d)fourCorrect answer is option 'C'. Can you explain this answer?.
Two long parallel surfaces each of emissivity 07 are maintained at different temperature andaccordingly have radiation heat exchange between them. It is desirable to reduce 75% of this radiant heat transfer by inserting thin parallel shields of equal emissivity on both sides. The number of shields should bea)one b)twoc)three d)fourCorrect 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 Two long parallel surfaces each of emissivity 07 are maintained at different temperature andaccordingly have radiation heat exchange between them. It is desirable to reduce 75% of this radiant heat transfer by inserting thin parallel shields of equal emissivity on both sides. The number of shields should bea)one b)twoc)three d)fourCorrect 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 Two long parallel surfaces each of emissivity 07 are maintained at different temperature andaccordingly have radiation heat exchange between them. It is desirable to reduce 75% of this radiant heat transfer by inserting thin parallel shields of equal emissivity on both sides. The number of shields should bea)one b)twoc)three d)fourCorrect answer is option 'C'. Can you explain this answer?.
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