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Heat transfer with time derivative heat source
Posted Jun 3, 2015, 12:15 a.m. EDT 3 Replies
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Dear all,
I try to model heat transfer with time derivative heat source Q=dq/dt, with q=Qm*exp(Tau/t)^(Beta)
Qm, Tau, Beta are constant.
Are there anyone know how write dQ/dt in COMSOL?
I have tried either qt or d(Qm*exp(Tau/t)^(Beta),t) but both of them doesn't work.
I got the following error:
Failed to evaluate variable Jacobian.
- Variable: qt
- Geometry: 1
- Domain: 1
- Feature: Time-Dependent Solver 1 (sol1/t1)
Are there anyone know how write dQ/dt in COMSOL?
Thank you for any help.
Agus
I try to model heat transfer with time derivative heat source Q=dq/dt, with q=Qm*exp(Tau/t)^(Beta)
Qm, Tau, Beta are constant.
Are there anyone know how write dQ/dt in COMSOL?
I have tried either qt or d(Qm*exp(Tau/t)^(Beta),t) but both of them doesn't work.
I got the following error:
Failed to evaluate variable Jacobian.
- Variable: qt
- Geometry: 1
- Domain: 1
- Feature: Time-Dependent Solver 1 (sol1/t1)
Are there anyone know how write dQ/dt in COMSOL?
Thank you for any help.
Agus
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3 Replies Last Post Jun 3, 2015, 11:34 a.m. EDT
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Posted:
9 years ago
In addition, when I used Q as d(Qmax*exp(-(Tau/t)^(Beta)),t), I got the error below:
Failed to find consistent initial values.
Last time step is not converged.
- Feature: Time-Dependent Solver 1 (sol1/t1)
Can you help me?
Thanks.
Agus
Failed to find consistent initial values.
Last time step is not converged.
- Feature: Time-Dependent Solver 1 (sol1/t1)
Can you help me?
Thanks.
Agus
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Posted:
9 years ago
Hi
It is of course possible to calculate the derivative in closed form:
dq/dt = -[(Qm·beta·Tau)/t²]*exp(beta·Tau/t)
The problem is that at t = 0 this has a singularity, as well as q itself. Your error message probably is related to that.
It is not my business, but from where such a peculiar function? Usually you see the form exp(t/Tau) where Tau is some characteristic time of the system.
br
Lasse
It is of course possible to calculate the derivative in closed form:
dq/dt = -[(Qm·beta·Tau)/t²]*exp(beta·Tau/t)
The problem is that at t = 0 this has a singularity, as well as q itself. Your error message probably is related to that.
It is not my business, but from where such a peculiar function? Usually you see the form exp(t/Tau) where Tau is some characteristic time of the system.
br
Lasse
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Posted:
9 years ago
Hi Lasse,
Thank you very much for your quick respond.
I agree with you that at t = 0 this has a singularity.
From your answer above, I got new idea to avoid t=0 by changing t to very small values for instance t=0.0001.
It is working properly.
Regarding to the function, it is typical function describing the heat produced by the hydration reaction of cement-based materials. It is also mentioned in the several literature's.
Anyway, thank you very much for your answer. It gave me fresh inspiration to solve it.
Br
Agus
Thank you very much for your quick respond.
I agree with you that at t = 0 this has a singularity.
From your answer above, I got new idea to avoid t=0 by changing t to very small values for instance t=0.0001.
It is working properly.
Regarding to the function, it is typical function describing the heat produced by the hydration reaction of cement-based materials. It is also mentioned in the several literature's.
Anyway, thank you very much for your answer. It gave me fresh inspiration to solve it.
Br
Agus