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.
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6 comments:
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!
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.
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?
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. :{
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!!
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 !
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