Tuesday, April 11, 2006

HW #4 - P #3 - Convection and Radiation Heat Transfer

A 5cm diameter spherical ball whose surface is maintained at a temperature of 70oC is suspended in the middle of a room at 20oC. If the convection heat transfer coefficient is 15 W/m2-oC and the emissivity of the surface is 0.8, determine the total rate of heat transfer from the ball.

6 comments:

Anonymous said...

My answer is off by 2.6 Watts and I have no idea why! I used

Qtotal = Qconv + Qrad
Qconv = hA(Ts - Tf)
Qrad = E(sigma)A(Ts)^4

And all the units cancel correctly to give Watts. The thing is that I had used Celsius in Qconv and I converted Ts to Kelvin for Qrad so that the Stefan-Boltzmann units cancelled - would this be incorrect? How could I convert either constant h or sigma to be the other temperature scale, if it is possible? Or maybe I'm missing something completely unrelated :P

Thank you for your time!

Dr. B said...

Lisa 12:36
I think you did everything correctly. It is OK to use T in degC in Newton's Law of Cooling (Qconv) because it is a change in temperature and a change in T of 1 degC is the same as a change of 1 degK.

I don't know why you are off by 2.6 W. Did you use A = 4 pi R^2 = pi D^2 ? That is the only thing I can think of.

I hope this is helpful.

Anonymous said...

I did the same way as Lisa with A = 4 pi R^2 = pi D^2. My answer came out to be off by 2.6 Watts as well. Could something else be wrong somewhere?

Dr. B said...

Kim 5:22 PM
The only other thing that is tricky is Qrad.
Qrad = psilon*sigma*A*(Thot^4-Tcold^4)
This is NOT the same as:
Qrad = epsilon*sigma*A*(Thot-Tcold)^4

That is my last idea. You are going to have to give me lots of intermediate numbers if you want more help. :{

Anonymous said...

Oh man! That was my problem! I was raising Thot - Tcold entirely to the fourth power rather than separately. Sleep is a good thing. Thank you for your help!!

Dr. B said...

Lisa 12:14 AM
It is my pleasure to help.
It is common to make a similar mistake with kinetic energy.

v2^2 - v1^2 is NOT the same as (v2-v1)^2

I hope this exchange on the blog helps you avoid this pitfall in the future !