Question

Heat Transfer, But really just an integral I can't quite get correct...

Answer 1

So I've gone through this a couple times and haven't quite gotten it...maybe I'm just missing something because it's late...

\[\large Q_o = -k\frac{dT}{dx}\]
we know \(\large k = k_o(1+\alpha x)\) so subbing that in...
\[\large Q_o = -k_o(1+\alpha x)\frac{dT}{dx}\]
Integrate rhs
\[\large Q_o = \int_{}^{} -k_o(1+\alpha x)\frac{dT}{dx}\]
\[\large Q_o=-k_o(\int_{0}^{L} \frac{dT}{dx}+ \alpha \int_{0}^{L} x\frac{dT}{dx})\]
\[\large Q_o = -k_o((T_L - T_o)\int_{0}^{L} \frac{d}{dx} + \alpha(T_L - T_o)\int_{0}^{L} x\frac{d}{dx})\]
\[\large Q_o = -k_o((T_L - T_o)L + \frac{1}{2}\alpha(T_L - T_o)L^2\]

From here, even if I let \(\large \alpha \rightarrow 0\) I still won't arrive at the correct expression given.

*Definitely point out what i did wrong, again it's late and i'm probably just not thinking clearly!

Answer 2

Why are you allowed to only integrate the RHS though?
I did things a little differently... I'm not coming out to the correct answer though :( Grr maybe it'll give you some idea though..

Answer 3

\[\large\rm -k\frac{dT}{dx}=Q_o\]Separating variables,\[\large\rm -dT=\frac{Q_o}{k}dx\]Which is,\[\large\rm -dT=\frac{Q_o}{k_o(1+\alpha x)}dx\]Integrating each side,\[\large\rm \int\limits-dT=\int\limits\frac{Q_o}{k_o(1+\alpha x)}dx\]giving us,\[\large\rm -T(x)=\frac{Q_o}{\alpha k_o}\ln|1+\alpha x|\]You're integrating from 0 to L?
Sorry I'm just trying to focus on the math of the problem XD
I don't really understand the rest of it haha.\[\large\rm T_L-T_o=\frac{-Q_o}{\alpha k_o}\left[\ln(1+L \alpha)-\ln(1+0\alpha)\right]\]

Answer 4

Which leads to,\[\large\rm Q_o=\frac{k_o(T_o-T_L)}{\alpha \ln(1+\alpha L)}\]So ya my way isn't working out either :P Hmm

Answer 5

No that's fine lol...I did that approach too but because I had an \(\large \alpha\) in every term, when it goes to 0 it just wipes out everything so that's why i went this route

Wrong...but still route XD

Answer 6

Grr hmm

Answer 7

I also, just for kicks, when I got to this part:

\[\large \int \frac{dT}{dx}\] wrote it as \(\large \frac{T_L - T_o}{L - 0}\) since it says it's linear...it looked nice...wrong but nice lol

Answer 8

As you can tell, at this point I'm just having fun with it since I know in my sleep deprived state I most likely will NOT be solving this tonight XD

Answer 9

:P

Answer 10

Well first off, I sorta did "the quick way" just to check, I just integrate with \(k\) as a constant:

\[-k\frac{dT}{dx} = Q_0\]\[\int_0^L -k\frac{dT}{dx} dx = \int_0^L Q_0 dx\]\[-k(T_L-T_0) = Q_0(L-0)\]\[Q_0=\frac{-k(T_L-T_0)}{L}\]
So I'm already wrong from the start, since I am getting the answer but with a negative sign that won't go away. Maybe it's a typo?

At any rate, what I did is not too far off but it seems incomplete and ends up getting the same answer as this when plugging in \(\alpha = 0\),

\[Q_0 = \frac{-k_0(T_L-T_0)}{L} -k_0\alpha T_L + \frac{k_0 \alpha}{L}\int_0^L T dx\]

I think that final term there is probably not what they want hanging around, I think this has an interpretation as being the total heat content from 0 to L, but whatever it is when you plug in \(\alpha=0\) it goes away... Kinda sheisty so idk lol.