# Linear variation of gradient of a scalar in a cell.

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March 15, 2020, 10:53
#2
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Filippo Maria Denaro
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Quote:

Once the temperature is assumed linear, its derivative is a constant in the cell:
T(x)=T0+T1*x -> dT/dx=T1

March 15, 2020, 13:11
#3
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Mandeep Shetty
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Quote:
 Originally Posted by FMDenaro Once the temperature is assumed linear, its derivative is a constant in the cell: T(x)=T0+T1*x -> dT/dx=T1
Thank you! If I assume the gradient of temperature itself to vary lineraly can i write

March 15, 2020, 13:16
#4
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Filippo Maria Denaro
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Quote:

Well, be careful that grad is also a vector operator. Work on the components dT/dx and dT/dy. But this way you are assuming a quadratic law for the temperature

March 15, 2020, 13:26
#5
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Mandeep Shetty
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Quote:
 Originally Posted by FMDenaro Well, be careful that grad is also a vector operator. Work on the components dT/dx and dT/dy. But this way you are assuming a quadratic law for the temperature
I am trying to use a second-order FVM method. I was told that using the second-order FVM method means all the variables (which I am starting to understand just means the primitive variables and not the gradient of the said variables) would be assumed to vary linearly within the cell. So as I am already assuming the temperature to vary linearly, and so I cannot assume the grad(T) to vary linearly again.

March 15, 2020, 13:41
#6
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Filippo Maria Denaro
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Quote:
 Originally Posted by granzer I am trying to use a second-order FVM method. I was told that using the second-order FVM method means all the variables (which I am starting to understand just means the primitive variables and not the gradient of the said variables) would be assumed to vary linearly within the cell. So as I am already assuming the temperature to vary linearly, and so I cannot assume the grad(T) to vary linearly again.

Indeed in a FVM you have a second order accuracy using a linear assumption for the function. The derivative is constant on each of the face so that you can evaluate their difference