A sphere has a radius of 0.100 m. Its surface can be treated as a grey surface with emissivity 0.20 and a uniform temperature of 19 degrees C. It is placed in a room where the room surfaces are at an average temperature of 18 degrees C. The air in the room is at 20 degrees C. Determine the total heat transfer rate to/from the sphere if its convective heat transfer coefficient is 3.0 W/m^2K. (The surface area of a sphere is 4πr^2.

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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A sphere has a radius of 0.100 m. Its surface can be treated as a grey
surface with emissivity 0.20 and a uniform temperature of 19 degrees C. It is
placed in a room where the room surfaces are at an average
temperature of 18 degrees C. The air in the room is at 20 degrees C. Determine the total
heat transfer rate to/from the sphere if its convective heat transfer
coefficient is 3.0 W/m^2K. (The surface area of a sphere is 4πr^2.)

The answer is 0.236W

Equations to help:

Heat Transfer by convection:

Q = Ahc(Tf-Ts)

Heat Transfer by radiation:

Q = Aεσ(T^4(1)-T^4(2))

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Follow-up Question

You have stated that the convective area is πr^2. Please explain this. πr^2 is the area for a circle, not a sphere. It should be 4πr^2 as this is equal to the surface area of the sphere as you have done correctly in the radiative area. 

A conv= 4πr^2*3*293-292 = 0.377

A rad = 4πr^2*0.2*5.67x10^-8*(292^4-291^4) = 0.144

I have tried to multiply the convective area by 4πr^2 instead of πr^2 but it is giving me an incorrect total heat transfer rate of 0.377+0.144 = 0.521

the correct answer is supposed to be 0.236

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Follow-up Question

Why did you multiply the convective area by πr^2 and not 4πr^2 ?  The surface area of a sphere is 4πr^2. 

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