A 4-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 ball
Answers
Answer and Explanation:
Given data
The diameter of the spherical ball: di=5mmdi=5mm
The temperature of the ball : T1=50∘CT1=50∘C
The thickness of the plastic insulation: t=1mmt=1mm
The thermal conductivity of the spherical ball: k=0.13W/m∘Ck=0.13W/m∘C
The temperature of the medium: T2=15∘CT2=15∘C
The heat transfer coefficient : h=20W/m2⋅∘Ch=20W/m2⋅∘C
The outer radius of the ball is:
ro=di+2t2ro=di+2t2
Substitute the values in above equation,
ro=(5+2×1)2ro=3.5mmro=(5+2×1)2ro=3.5mm
The expression to calculate the critical radius of the plastic insulation,
rcr=2khrcr=2kh
Substitute the values in above equation,
rcr=2×0.1320rcr=0.013mrcr=13mmrcr=2×0.1320rcr=0.013mrcr=13mm
Thus the critical radius of insulation is 13 mm.
The rate of heat transfer is maximum when the critical radius of insulation is equal to the outer radius of the sphere and minimum value when the outer radius is less than to the critical radius.
Hence, ro<rcrro<rcr the plastic insulation will increase the heat transfer from the ball.