Timeline for Couple a PDE and ODE in NDSolve
Current License: CC BY-SA 3.0
19 events
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| Mar 20, 2020 at 21:28 | comment | added | Young | @user21 you can use it... I was just curious | |
| Mar 20, 2020 at 19:45 | comment | added | user21 | @Young, unfortunately, that is not possible. I''ll not use this then. That's not a problem. You may be interested to hear that we how also have a HeatTransfer tutorial | |
| Mar 20, 2020 at 18:20 | comment | added | Young | @user21 yes, but can I (Garrett Young) get credit. | |
| Mar 10, 2020 at 9:10 | comment | added | user21 | Would you mind if we used this answer in the documentation? | |
| Aug 23, 2016 at 1:02 | history | bounty ended | user21 | ||
| Aug 8, 2016 at 8:33 | vote | accept | Pinti | ||
| Aug 4, 2016 at 23:54 | history | edited | Young | CC BY-SA 3.0 |
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| Aug 4, 2016 at 3:48 | comment | added | xzczd♦ |
You're right. (Seems that I'm a little tired…) Still, I think there should be a k before D[T[t, x], x, x]. (This mistake is inherited from OP's code, of course.)
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| Aug 4, 2016 at 2:50 | comment | added | Young | @xzczd I respectfully disagree ... the boundary condition is being applied to a 0-D space and and Q=cp m dT ... cpR here is a relative term for mass and cp ... so cpR dT/dt = q = -k(T0-T) | |
| Aug 4, 2016 at 2:27 | comment | added | xzczd♦ | Wait, after revisting my PDE text book, I'm afraid the implementation of Newton's law of cooling is still wrong. The correct relationship between ￥$Q$ and $q$ and $T$ in Newton's law of cooling should be $\frac{\text{dQ}}{\text{dt}}=q=-k \frac{\partial T}{\partial x}$. (Fourier's law) i.e. it is corresponding to a Robin boundary condition in x-direction. | |
| Aug 4, 2016 at 1:35 | history | edited | Young | CC BY-SA 3.0 |
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| Aug 4, 2016 at 1:29 | comment | added | xzczd♦ |
Oh, you're right, then I think mR is not mass ratio but heat capacity ratio?
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| Aug 4, 2016 at 1:14 | comment | added | Young |
@xzczd Said another way, T0 is only valid at x=0
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| Aug 4, 2016 at 1:02 | comment | added | Young | @xzczd The x is ignored because the the reservoir response is 0-D | |
| Aug 4, 2016 at 1:01 | comment | added | xzczd♦ |
The reservoir is just attached to the left end, are you sure the equation mR D[T0[t, x], t] == -h (T0[t, x] - T[t, x]) is correct?
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| Aug 3, 2016 at 23:41 | history | edited | Young | CC BY-SA 3.0 |
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| Aug 3, 2016 at 23:18 | history | edited | Young | CC BY-SA 3.0 |
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| Aug 3, 2016 at 23:06 | history | edited | Young | CC BY-SA 3.0 |
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| Aug 3, 2016 at 19:23 | history | answered | Young | CC BY-SA 3.0 |