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A flat plate has thickness 5 cm, thermal conductivity 1 W/(mK),convective heat transfer coefficients on its two flat faces of 10 W/(m2K)and 20 W/(m2K). The overall heat transfer co-efficient for such a flatplate is:
- a)5 W/(m2K)
- b)6.33 W/(m2K)
- c)20 W/(m2K)
- d)30 W/(m2K)
Correct answer is option 'A'. Can you explain this answer?
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A flat plate has thickness 5 cm, thermal conductivity 1 W/(mK),convect...
Given:
- Thickness of flat plate (δ) = 5 cm = 0.05 m
- Thermal conductivity (k) = 1 W/(mK)
- Convective heat transfer coefficient on one face (h1) = 10 W/(m^2K)
- Convective heat transfer coefficient on the other face (h2) = 20 W/(m^2K)
To find:
Overall heat transfer coefficient (U)
Solution:
The overall heat transfer coefficient (U) for a flat plate can be calculated using the formula:
1/U = (δ/k) + (1/h1) + (1/h2)
Step 1: Convert the thickness of the flat plate to meters
δ = 5 cm = 0.05 m
Step 2: Calculate the reciprocal of each term in the formula
Reciprocal of (δ/k) = k/δ = 1/0.05 = 20 mK/W
Reciprocal of (1/h1) = h1/1 = 10 W/(m^2K)
Reciprocal of (1/h2) = h2/1 = 20 W/(m^2K)
Step 3: Add the reciprocals of each term
1/U = 20 mK/W + 10 W/(m^2K) + 20 W/(m^2K)
= 20 mK/W + 30 W/(m^2K)
= (20 mK/W * 1000) + 30 W/(m^2K)
= 20000 mK/W + 30 W/(m^2K)
= 20000/1000 K/W + 30 W/(m^2K)
= 20 K/W + 30 W/(m^2K)
Step 4: Take the reciprocal of the sum to find U
U = 1 / (20 K/W + 30 W/(m^2K))
= 1 / (20 K/W + 0.03 K/W)
= 1 / 20.03 K/W
= 0.0499 W/(m^2K)
Step 5: Round off the value of U to the nearest whole number
U ≈ 0.05 W/(m^2K)
Conclusion:
The overall heat transfer coefficient (U) for the given flat plate is approximately 0.05 W/(m^2K), which is closest to option 'A' (5 W/(m^2K)).
- Thickness of flat plate (δ) = 5 cm = 0.05 m
- Thermal conductivity (k) = 1 W/(mK)
- Convective heat transfer coefficient on one face (h1) = 10 W/(m^2K)
- Convective heat transfer coefficient on the other face (h2) = 20 W/(m^2K)
To find:
Overall heat transfer coefficient (U)
Solution:
The overall heat transfer coefficient (U) for a flat plate can be calculated using the formula:
1/U = (δ/k) + (1/h1) + (1/h2)
Step 1: Convert the thickness of the flat plate to meters
δ = 5 cm = 0.05 m
Step 2: Calculate the reciprocal of each term in the formula
Reciprocal of (δ/k) = k/δ = 1/0.05 = 20 mK/W
Reciprocal of (1/h1) = h1/1 = 10 W/(m^2K)
Reciprocal of (1/h2) = h2/1 = 20 W/(m^2K)
Step 3: Add the reciprocals of each term
1/U = 20 mK/W + 10 W/(m^2K) + 20 W/(m^2K)
= 20 mK/W + 30 W/(m^2K)
= (20 mK/W * 1000) + 30 W/(m^2K)
= 20000 mK/W + 30 W/(m^2K)
= 20000/1000 K/W + 30 W/(m^2K)
= 20 K/W + 30 W/(m^2K)
Step 4: Take the reciprocal of the sum to find U
U = 1 / (20 K/W + 30 W/(m^2K))
= 1 / (20 K/W + 0.03 K/W)
= 1 / 20.03 K/W
= 0.0499 W/(m^2K)
Step 5: Round off the value of U to the nearest whole number
U ≈ 0.05 W/(m^2K)
Conclusion:
The overall heat transfer coefficient (U) for the given flat plate is approximately 0.05 W/(m^2K), which is closest to option 'A' (5 W/(m^2K)).
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A flat plate has thickness 5 cm, thermal conductivity 1 W/(mK),convect...
Answer is 5.
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A flat plate has thickness 5 cm, thermal conductivity 1 W/(mK),convective heat transfer coefficients on its two flat faces of 10 W/(m2K)and 20 W/(m2K). The overall heat transfer co-efficient for such a flatplate is:a)5 W/(m2K)b)6.33 W/(m2K)c)20 W/(m2K)d)30 W/(m2K)Correct answer is option 'A'. Can you explain this answer?
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