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For what value of end temperature difference ratio, is the arithmetic mean temperature difference 5% higher than the log-mean temperature difference?
  • a)
    2.4
  • b)
    2.3
  • c)
    2.2
  • d)
    2.1
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
For what value of end temperature difference ratio, is the arithmetic ...
α 1/ α 2 = 2.2.
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Most Upvoted Answer
For what value of end temperature difference ratio, is the arithmetic ...
Given data:
Arithmetic mean temperature difference = 1.05 × Log-mean temperature difference.
Let the end temperature difference ratio be 'R'.

Formula used:
Arithmetic mean temperature difference = (T1 - t2) / ln(T1 / t2)
Log-mean temperature difference = (T1 - t2) / [ln(T1 / t2) + ln(R)]

Calculation:
Let T1 = 100°C and t2 = 50°C
Arithmetic mean temperature difference = (100 - 50) / ln(100 / 50) = 34.657
Log-mean temperature difference = (100 - 50) / [ln(100 / 50) + ln(R)]
1.05 × Log-mean temperature difference = 1.05 × (100 - 50) / [ln(100 / 50) + ln(R)] = 36.39

Therefore, 36.39 = 34.657 × R^(0.05)
R^(0.05) = 1.0513
R = 2.22

Therefore, the end temperature difference ratio required to make the arithmetic mean temperature difference 5% higher than the log-mean temperature difference is 2.2. Hence, option (c) is the correct answer.
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For what value of end temperature difference ratio, is the arithmetic mean temperature difference 5% higher than the log-mean temperature difference?a)2.4b)2.3c)2.2d)2.1Correct answer is option 'C'. Can you explain this answer?
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