Question

A 5-mm-diameter spherical ball at 50c is covered by a 1-mm-thick plastic insulation (k = 0.13 w/m c). The ball is exposed to a medium at 15c, with a combined convection and radiation heat transfer coefficient of 20 w/m2 c. Determine if the plastic insulation on the ball will help or hurt heat transfer from the bal

Answer 1

r o  <  r c r  the plastic insulation will increase the heat transfer from the ball.

Explanation:

• The diameter of the spherical ball di  =  5  m m
• The temperature of the ball :  T 1  =  50 ∘ C
• The thickness of the plastic insulation:  t  =  1 m m
• The thermal conductivity of the spherical ball:  k  =  0.13 W / m ∘ C
• The temperature of the medium:  T 2   =  15 ∘ C
• The heat transfer coefficient :  h  =  20 W / m^ 2 ⋅ ∘ C

Solution:

The outer radius of the ball is:

r o  =  d i  +  2 t / 2

Substitute the values in above equation,

r o  =  ( 5  +  2  ×  1 ) /  2

r o  =  3.5  m m

The expression to calculate the critical radius of the plastic insulation,

r c r=  2 k /  h

Substitute the values in above equation,

r cr = 2 × 0.13 /  20

r c r  =  0.013  m

r c r  =  13  m m

Hence,  r o  <  r c r  the plastic insulation will increase the heat transfer from the ball.

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