|
|
|
Comment |
Sorted by
Date Written |
wow (Reply to this comment)
by shopaholic_man
my first science lesson from epinions! I liked the laymans explanations, you lost me in the formulas though! Nicely done!
|
Dec 15 '06
10:06 am PST
|
|
Hi (Reply to this comment)
by jeavinl
Excellent job... of reminding me of my mechanical engineer husband and the many convoluted conversations he tries to have with me. Yes, and, of course of explaining the principles of heat transfer. :)
Jeanne
|
Dec 14 '06
10:40 am PST
|
|
Re: Re: Genuinely Geeky (Reply to this comment)
by tch7
Yes, I suppose that adds some clarity to the matter.
|
Dec 12 '06
3:04 pm PST
|
|
Re: Genuinely Geeky (Reply to this comment)
by zorrothepiking
I made the following changes:
Where k is the thermal conductivity, A is the cross-sectional area, and t is the thickness. Using this to calculate the total resistance to heat transfer, we can derive the following:
R = t / ( k * A )
q = Temp_Diff / R
In some texts, you'll also see R calculated without the area term. Using this system, the heat rate would equal the temperature differential times the area divided by Resistance. For what we are doing here, though, it is fine to use a composite Resistance value that represents all material and geometric considerations.
This, of course, all makes several simplifying assumptions, including that our system is acting at full efficiency.
Thanks for pointing that out to me. It was very unclear. Do you think what I did above is more effective? This way, I can ignore topic of heat flux entirely.
Glad to know I'm not the only geek. :)
|
Dec 12 '06
1:24 pm PST
|
|
Re: Genuinely Geeky (Reply to this comment)
by zorrothepiking
Close, but no cigar. Heat flux is equal to the temperature difference over the resistance. Heat rate/flow is only proportional to the temperature diffential over the resistance.
The only difference is the cross-sectional area, which can be integrated into Resistance equation. In the paper, I wrote this:
Maybe I was a little deceptive earlier by leaving this out, but it's kinda inconsequential unless you're intending to go out and measure all this, since the area could be integrated into the resistance value.
It's true that you generally just the t/k to calculate R, but this should really be considered R" (Resistance to Flux) instead of R (Resistance to Transfer). Afterall, flow (not flux) being equivalent to the driving force divided by the resistance is a universal dynamic principle.
Point take, though... I ought to either use flux resistance and flux rate or transfer resistance and transfer rate. It's hard enough to keep is straight without switching back and forth. So to most of you reading this now and wondering what we're talking about, I've already made the change.
|
Dec 12 '06
1:08 pm PST
|
|
Genuinely Geeky (Reply to this comment)
by tch7
Still short of a temperature gradient diagram and sample calculations, but I suppose I can forgive you for that. It's just a shame that you can't get equations to display worth a lick on Epinions...
In the spirit of geekhood:
"So like I stated earlier, heat rate (or heat flux) equals the temperature difference divided by the resistance."
Close, but no cigar. Heat flux is equal to the temperature difference over the resistance. Heat rate/flow is only proportional to the temperature diffential over the resistance.
|
Dec 12 '06
12:28 pm PST
|
|
Update (Reply to this comment)
by zorrothepiking
Updated 12/12 to include math. :)
|
Dec 12 '06
7:54 am PST
|
|
Background (Reply to this comment)
by zorrothepiking
I've had physics and chemistry, but only as part of my Mechanical Engineering Major. Between those and Heat Transfer 1 and 2, I picked up most of the basics. I was trying to gear it towards people without a really strong background in engineering, thermodynamics, or math. I just wanted it to be interesting and a very approachable essay.
And about the spacesuit, it'd probably keep you nice and toasty without running up your electricity costs. Then again, I bet it'll do nothing for your figure. :)
|
Dec 12 '06
5:13 am PST
|
|
I'm kind of cold right now... (Reply to this comment)
by joyfulgirl91
So you think a space suit is the answer? Maybe just around the house, instead of running the heater?
Jessica
|
Dec 11 '06
4:10 pm PST
|
|
God I love Physics... (Reply to this comment)
by gatorgirlie
This brought back the three years I spent in Engineering.
You did a great job with the explanations -- they were pin-point perfect and easy enough to understand even for those without a physics background.
:)
Jen
|
Dec 11 '06
1:06 pm PST
|
|
Were you a physics or chemistry major? (Reply to this comment)
by bettega
I was!
Congrats on the nice piece. Well written; you are probably well educated!
Happy holidays
Bettega
|
Dec 11 '06
12:59 pm PST
|
|
|
|
© 1999-2009 Shopping.com, Inc.