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Controlling my own boundaries
Posted 06.03.2013, 05:41 GMT-5 Heat Transfer & Phase Change 2 Replies
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I am modelling a closed tube of air in 3D with a heating plate at one end of the tube. I have measured the temperature distribution across the heating plate as a function of time and location and would like to implement this in my model. I then use the boundary condition “Temperature” at the end of the tube with my measured profile, which is:
(4.1483*10^(-9)*t^2 -2.1712*10^(-5)*t -3.3232*10^(-2))*x^2 + (1.0432*10^(-8)*t^2 -1.5035*10^(-4)*t -0.11336)*y^2 + (-9.5557*10^(-9)*t^2 -2.3085*10^(-4)*t -1.028*10^(-1))*x + ( -1.0534*10^(-8)*t^2 + 2.3914*10^(-4)*t -0.1234)*y +(6.8938*10^(-10)*t^2 +4.1934*10^(-6)*t -5.4406*10^(-4))*x*y + ( 4.9445*10^(-8)*t^2 + 9.8562*10^(-3)*t + 21.079).
The heating plate is in the z-plane and x and y are both positive .
My problem is now, that this “heating profile” changes depending on which boundary conditions I use the tube walls. E.g. if I use thermal insulation for the tube walls, the temperature is practically constant across the heating plate.
How can I ensure that the heating profile of the plate does not changes at the end of the tube no matter what the other boundaries, in the tube, are?
Thank you for your time
Morten
(4.1483*10^(-9)*t^2 -2.1712*10^(-5)*t -3.3232*10^(-2))*x^2 + (1.0432*10^(-8)*t^2 -1.5035*10^(-4)*t -0.11336)*y^2 + (-9.5557*10^(-9)*t^2 -2.3085*10^(-4)*t -1.028*10^(-1))*x + ( -1.0534*10^(-8)*t^2 + 2.3914*10^(-4)*t -0.1234)*y +(6.8938*10^(-10)*t^2 +4.1934*10^(-6)*t -5.4406*10^(-4))*x*y + ( 4.9445*10^(-8)*t^2 + 9.8562*10^(-3)*t + 21.079).
The heating plate is in the z-plane and x and y are both positive .
My problem is now, that this “heating profile” changes depending on which boundary conditions I use the tube walls. E.g. if I use thermal insulation for the tube walls, the temperature is practically constant across the heating plate.
How can I ensure that the heating profile of the plate does not changes at the end of the tube no matter what the other boundaries, in the tube, are?
Thank you for your time
Morten
2 Replies Last Post 13.03.2013, 11:54 GMT-4