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An uninsulated cylindrical wire of radius 1.0 mm produces electric heating at the rate of 5.0 W/m. The temperature of the surface of the wire is 750 C when placed in air at 250 C . When the wire is coated with PVC of thickness 1.0 mm, the temperature of the surface of the wire reduces to 550 C . Assuming that the heat generation rate from the wire and the convective heat transfer coefficient are same for both uninsulated wire and the coated wire. The thermal conductivity of PVC is ______ W/m.K (round off to two decimal places) 
    Correct answer is '0.1103'. Can you explain this answer?
    Most Upvoted Answer
    An uninsulated cylindrical wire of radius 1.0 mm produces electric hea...
    Given : Qgen = 5W/m
    emperature of air = 250 C
    Temperature of wire surface when placed in air = 750 C
    Thickness of PVC = 1 mm  A
    fter coating wire with PVC, final temperature = 550 C
    Case 1:


    5 = h × A1 ΔT  (A1 = 2π 0.001)

    Case 2 : 


    = 5W/m
    On solving
    K = 0.1103 W/mK
    Free Test
    Community Answer
    An uninsulated cylindrical wire of radius 1.0 mm produces electric hea...
    The thermal conductivity of PVC can be calculated using the given information as follows:

    Given data:
    - Radius of the uninsulated wire (r1) = 1.0 mm = 0.001 m
    - Radius of the coated wire (r2) = r1 + thickness of PVC = 0.001 m + 0.001 m = 0.002 m
    - Heat generation rate (Q) = 5.0 W/m
    - Temperature of the surface of the uninsulated wire (T1) = 750 °C = 750 + 273 = 1023 K
    - Temperature of the surrounding air (T∞) = 250 °C = 250 + 273 = 523 K
    - Temperature of the surface of the coated wire (T2) = 550 °C = 550 + 273 = 823 K

    1. Heat transfer from uninsulated wire:
    The heat transfer from the uninsulated wire can be calculated using the following formula:
    Q = (2πk1L / ln(r2 / r1)) * (T1 - T∞)
    Where:
    - k1 is the thermal conductivity of the wire (uninsulated)
    - L is the length of the wire

    2. Heat transfer from coated wire:
    The heat transfer from the coated wire can be calculated using the same formula, but with the following values:
    Q = (2πk2L / ln(r2 / r1)) * (T2 - T∞)
    Where:
    - k2 is the thermal conductivity of the wire (coated)

    3. Comparison:
    Since the heat generation rate (Q) and the convective heat transfer coefficient are the same for both cases, we can equate the two equations to find the ratio of the thermal conductivities:
    (2πk1L / ln(r2 / r1)) * (T1 - T∞) = (2πk2L / ln(r2 / r1)) * (T2 - T∞)

    4. Simplification:
    The length of the wire (L), the cross-sectional area (πr1^2), and the natural logarithm term (ln(r2 / r1)) cancel out, leaving us with:
    (k1 / k2) = (T1 - T∞) / (T2 - T∞)

    5. Calculation:
    Substituting the given values, we can solve for k2:
    k2 = k1 * (T2 - T∞) / (T1 - T∞)
    = k1 * (823 - 523) / (1023 - 523)
    = k1 * 300 / 500
    = 0.6 * k1

    6. Conclusion:
    Since the thickness of the PVC coating is 1.0 mm, the overall radius of the coated wire is double that of the uninsulated wire. Therefore, the coated wire has twice the cross-sectional area. As a result, the thermal conductivity of PVC (k2) must be half that of the uninsulated wire (k1) to achieve the same heat transfer rate. Hence, k2 = 0.5 * k1.

    Now, substituting the value of k1 in
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    An uninsulated cylindrical wire of radius 1.0 mm produces electric heating at the rate of 5.0 W/m. The temperature of the surface of the wire is 750 C when placed in air at 250 C . When the wire is coated with PVC of thickness 1.0 mm, the temperature of the surface of the wire reduces to 550 C . Assuming that the heat generation rate from the wire and the convective heat transfer coefficient are same for both uninsulated wire and the coated wire. The thermal conductivity of PVC is ______ W/m.K (round off to two decimal places)Correct answer is '0.1103'. Can you explain this answer?
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    An uninsulated cylindrical wire of radius 1.0 mm produces electric heating at the rate of 5.0 W/m. The temperature of the surface of the wire is 750 C when placed in air at 250 C . When the wire is coated with PVC of thickness 1.0 mm, the temperature of the surface of the wire reduces to 550 C . Assuming that the heat generation rate from the wire and the convective heat transfer coefficient are same for both uninsulated wire and the coated wire. The thermal conductivity of PVC is ______ W/m.K (round off to two decimal places)Correct answer is '0.1103'. 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 An uninsulated cylindrical wire of radius 1.0 mm produces electric heating at the rate of 5.0 W/m. The temperature of the surface of the wire is 750 C when placed in air at 250 C . When the wire is coated with PVC of thickness 1.0 mm, the temperature of the surface of the wire reduces to 550 C . Assuming that the heat generation rate from the wire and the convective heat transfer coefficient are same for both uninsulated wire and the coated wire. The thermal conductivity of PVC is ______ W/m.K (round off to two decimal places)Correct answer is '0.1103'. 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 An uninsulated cylindrical wire of radius 1.0 mm produces electric heating at the rate of 5.0 W/m. The temperature of the surface of the wire is 750 C when placed in air at 250 C . When the wire is coated with PVC of thickness 1.0 mm, the temperature of the surface of the wire reduces to 550 C . Assuming that the heat generation rate from the wire and the convective heat transfer coefficient are same for both uninsulated wire and the coated wire. The thermal conductivity of PVC is ______ W/m.K (round off to two decimal places)Correct answer is '0.1103'. Can you explain this answer?.
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