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I have been doing a lot of derivations recently involving heat transfer.

I was attempting to derive heat accumulation in a differential element based upon inflow and outflow as well as thermodynamic principles.

I came up with:

dQ_control_volume_dot = Q_dot_in - Q_dot_out

And by the fundamental theorem of calculus:

$mC\frac{dT}{dt}=Q_{dot}(x)-Q_{dot}(x+dx)=dx\frac{\partial Q_{dot}}{\partial x}$

Now I have to figure out what partial(Q,x) is, so I write:

$\frac{\partial Q_{dot}}{\partial x}=-kA\frac{dT}{dx}$ using upon Fourier's law

And so I arrive at:

$mC\frac{dT}{dt}=-kAdT$

Now remember, you're looking at a thermodynamic statement set equal to Fourier's law.

Now, let's think qualitatively about this term "dT"

The dT on the RHS is a spatial dT, but the dT on the left is the change in energy induced in the control volume over time by by heat flow.

So, one dT is spatial, and one is related to time.

Now, I come to you and say, those dT can't be set equal to each other, they're completely different terms. One is spatial, one is temporal.

You could respond, they're both infinitely small, and have the same units, so as long as they have the same sign they're the same.

Now, can we verify that they have the same sign? I'm not sure...

Am I completely botching the problem setup and do I have the fundamental equation setup incorrect, or can someone rectify my confusion?

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  • $\begingroup$ Please use MathJax on this site. $\endgroup$ – Danu Feb 6 '16 at 20:15
  • $\begingroup$ Have you learned about partial derivatives and partial differentiation yet? $\endgroup$ – Chet Miller Feb 6 '16 at 20:26
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The derivation is not correct. The mass within the control volume is $\rho A\Delta x$. The rate of energy accumulation within the control volume is $\rho A\Delta xC\frac{\partial T}{\partial t}$. So the heat balance should be:$$\rho A\Delta xC\frac{\partial T}{\partial t}=Q_x-Q_{x+\Delta x}$$Dividing by $\Delta x$ and taking the limit as $\Delta x$ approaches zero gives:$$\rho AC\frac{\partial T}{\partial t}=-\frac{\partial Q}{\partial x}$$But, $Q=-kA\frac{\partial T}{\partial x}$. Substituting this into the differential heat balance equation yields:$$\rho C\frac{\partial T}{\partial t}=k\frac{\partial^2 T}{\partial x^2}$$This is the desired equation.

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