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Two long parallel surfaces, each of emissivity 0.7 are maintained atdifferent temperatures and accordingly have radiation exchange between them. It is desired to reduce 75% of this radiant heat transfer by inserting thin parallel shields of equal emissivity (0.7) on both sides.What would be the number of shields?
- a)1
- b)2
- c)3
- d)4
Correct answer is option 'C'. Can you explain this answer?
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Two long parallel surfaces, each of emissivity 0.7 are maintained atdi...
Ans. (c)
or 0.25 
or n = 3
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Two long parallel surfaces, each of emissivity 0.7 are maintained atdi...
To reduce 75% of the radiant heat transfer between the two long parallel surfaces, thin parallel shields can be inserted on both sides. The shields should have an emissivity of 0.7, which is the same as the emissivity of the surfaces. The question asks for the number of shields required to achieve this reduction.
The number of shields can be determined by considering the reduction in heat transfer achieved by each shield and the overall reduction required.
Determining the Reduction in Heat Transfer by Each Shield:
When a shield is inserted between the two surfaces, it reduces the heat transfer by absorbing and re-emitting a portion of the radiant heat. The reduction in heat transfer achieved by each shield can be calculated using the formula:
Reduction in Heat Transfer = (1 - Shield Emissivity) / (1 + Shield Emissivity)
In this case, the shield emissivity is given as 0.7. Substituting this value into the formula:
Reduction in Heat Transfer = (1 - 0.7) / (1 + 0.7) = 0.3 / 1.7 ≈ 0.176
So, each shield reduces the heat transfer by approximately 17.6%.
Determining the Overall Reduction Required:
To achieve a reduction of 75% in the radiant heat transfer, the overall reduction required can be calculated using the formula:
Overall Reduction = 1 - Desired Heat Transfer / Initial Heat Transfer
Substituting the values:
Overall Reduction = 1 - 0.25 = 0.75
So, the overall reduction required is 75%.
Determining the Number of Shields:
To determine the number of shields required, we need to find the minimum number of shields that can achieve the overall reduction of 75%.
Let's assume 'n' is the number of shields required.
The reduction achieved by 'n' shields can be calculated as:
Reduction achieved by n shields = (Reduction in Heat Transfer)^n
Substituting the values:
(0.176)^n = 0.75
Taking the logarithm on both sides:
n * log(0.176) = log(0.75)
n = log(0.75) / log(0.176) ≈ 2.67
Since the number of shields cannot be a fraction, the next whole number is 3.
Therefore, the number of shields required to achieve a 75% reduction in radiant heat transfer is 3.
The number of shields can be determined by considering the reduction in heat transfer achieved by each shield and the overall reduction required.
Determining the Reduction in Heat Transfer by Each Shield:
When a shield is inserted between the two surfaces, it reduces the heat transfer by absorbing and re-emitting a portion of the radiant heat. The reduction in heat transfer achieved by each shield can be calculated using the formula:
Reduction in Heat Transfer = (1 - Shield Emissivity) / (1 + Shield Emissivity)
In this case, the shield emissivity is given as 0.7. Substituting this value into the formula:
Reduction in Heat Transfer = (1 - 0.7) / (1 + 0.7) = 0.3 / 1.7 ≈ 0.176
So, each shield reduces the heat transfer by approximately 17.6%.
Determining the Overall Reduction Required:
To achieve a reduction of 75% in the radiant heat transfer, the overall reduction required can be calculated using the formula:
Overall Reduction = 1 - Desired Heat Transfer / Initial Heat Transfer
Substituting the values:
Overall Reduction = 1 - 0.25 = 0.75
So, the overall reduction required is 75%.
Determining the Number of Shields:
To determine the number of shields required, we need to find the minimum number of shields that can achieve the overall reduction of 75%.
Let's assume 'n' is the number of shields required.
The reduction achieved by 'n' shields can be calculated as:
Reduction achieved by n shields = (Reduction in Heat Transfer)^n
Substituting the values:
(0.176)^n = 0.75
Taking the logarithm on both sides:
n * log(0.176) = log(0.75)
n = log(0.75) / log(0.176) ≈ 2.67
Since the number of shields cannot be a fraction, the next whole number is 3.
Therefore, the number of shields required to achieve a 75% reduction in radiant heat transfer is 3.
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Two long parallel surfaces, each of emissivity 0.7 are maintained atdifferent temperatures and accordingly have radiation exchange between them. It is desired to reduce 75% of this radiant heat transfer by inserting thin parallel shields of equal emissivity (0.7) on both sides.What would be the number of shields?a)1b)2c)3d)4Correct answer is option 'C'. Can you explain this answer?
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Two long parallel surfaces, each of emissivity 0.7 are maintained atdifferent temperatures and accordingly have radiation exchange between them. It is desired to reduce 75% of this radiant heat transfer by inserting thin parallel shields of equal emissivity (0.7) on both sides.What would be the number of shields?a)1b)2c)3d)4Correct 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 0.7 are maintained atdifferent temperatures and accordingly have radiation exchange between them. It is desired to reduce 75% of this radiant heat transfer by inserting thin parallel shields of equal emissivity (0.7) on both sides.What would be the number of shields?a)1b)2c)3d)4Correct 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 0.7 are maintained atdifferent temperatures and accordingly have radiation exchange between them. It is desired to reduce 75% of this radiant heat transfer by inserting thin parallel shields of equal emissivity (0.7) on both sides.What would be the number of shields?a)1b)2c)3d)4Correct answer is option 'C'. Can you explain this answer?.
Two long parallel surfaces, each of emissivity 0.7 are maintained atdifferent temperatures and accordingly have radiation exchange between them. It is desired to reduce 75% of this radiant heat transfer by inserting thin parallel shields of equal emissivity (0.7) on both sides.What would be the number of shields?a)1b)2c)3d)4Correct 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 0.7 are maintained atdifferent temperatures and accordingly have radiation exchange between them. It is desired to reduce 75% of this radiant heat transfer by inserting thin parallel shields of equal emissivity (0.7) on both sides.What would be the number of shields?a)1b)2c)3d)4Correct 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 0.7 are maintained atdifferent temperatures and accordingly have radiation exchange between them. It is desired to reduce 75% of this radiant heat transfer by inserting thin parallel shields of equal emissivity (0.7) on both sides.What would be the number of shields?a)1b)2c)3d)4Correct answer is option 'C'. Can you explain this answer?.
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