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A 5 mm diameter spherical ball at 50°C is covered by 1 mm thick plastic insulation (k = 0.13 W/mK). The ball is exposed to a medium at 15°C with heat transfer coefficient of 20 W/m2°C. Heat loss from the spherical ball due to further plastic insulation will
  • a)
    Always increase
  • b)
    First increase then decrease
  • c)
    Always decrease
  • d)
    First decrease then increase
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A 5 mm diameter spherical ball at 50°C is covered by 1 mm thick plast...
Heat Loss from Insulated Spherical Ball

Given:
Diameter of the ball (d) = 5 mm
Temperature of the ball (T1) = 50°C
Thickness of the insulation (δ) = 1 mm
Thermal conductivity of the insulation (k) = 0.13 W/mK
Temperature of the medium (T2) = 15°C
Heat transfer coefficient of the medium (h) = 20 W/m2°C

To find: Effect of further plastic insulation on heat loss from the spherical ball.

Solution:

1. Calculation of Heat Transfer Coefficient:
The heat transfer coefficient (h) for spherical objects is given by the relation:

h = 2k/[(1-ε)×d]

where ε is the emissivity of the surface of the ball and is assumed to be 1 for a perfect black body.

Substituting the given values, we get:

h = 2×0.13/[(1-1)×0.005] = 52 W/m2°C

2. Calculation of Heat Loss Without Additional Insulation:
The heat loss from the spherical ball without any additional insulation can be calculated using the following relation:

Q = 4πk(d/2+δ)×(T1-T2)/[1/(h×d)+1/(4k(d/2+δ))]

Substituting the given values, we get:

Q = 4π×0.13×(0.005/2+0.001)×(50-15)/[1/(20×0.005)+1/(4×0.13×(0.005/2+0.001))] = 0.254 W

3. Effect of Further Plastic Insulation:
The heat loss from the spherical ball with additional insulation of thickness δ1 can be calculated using the same relation with δ replaced by δ+δ1.

As the thickness of insulation increases, the heat loss initially increases due to an increase in the thermal resistance of the insulation layer. However, after reaching a certain thickness, the heat loss starts to decrease due to a decrease in the temperature difference between the surface of the ball and the medium.

Therefore, the correct answer is option 'B' - First increase then decrease.
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Community Answer
A 5 mm diameter spherical ball at 50°C is covered by 1 mm thick plast...
R = 2.5 mm
rinsulation = 3.5 mm
For sphere,
Rc = 2K/h = (2 x 0.13)/20 = 13 mm
rinsulation < />c
∴ Due to plastic insulation, heat loss first increases then decreases.
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A 5 mm diameter spherical ball at 50°C is covered by 1 mm thick plastic insulation (k = 0.13 W/mK). The ball is exposed to a medium at 15°C with heat transfer coefficient of 20 W/m2°C. Heat loss from the spherical ball due to further plastic insulation willa)Always increaseb)First increase then decreasec)Always decreased)First decrease then increaseCorrect answer is option 'B'. Can you explain this answer?
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