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Linear variation of gradient of a scalar in a cell.

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Old   March 15, 2020, 08:45
Default Linear variation of gradient of a scalar in a cell.
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Mandeep Shetty
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If the temperature (or any scalar) is varying linearly in a cell, will the variation of the gradient of the temperature also be linear?



If we say that temperature, T, is varying linearly in a square plate given by

T = Tp +(x-xp)grad(Tp)

where Tp is the Temperature at the centroid of the square plate, xp is the centroid, x is any point on the plate so (x-xp) is the position (so a vector) of any point on the plate from the centroid, grad(Tp) is the temperature gradient at the centroid.

Is it correct to write,

a) grad(T) = grad(Tp)+(x-xp)grad(grad(Tp))

as (x-xp), even though is a vector, doesn't actually change with x,y,z

OR

b) grad(T) = grad(Tp)+grad[(x-xp)grad(Tp)]

as we cannot consider (x-xp) to be a constant with respect to x, y, z.

Ref: Prof.Hrvoje Jasak Thesis, Error Analysis and Estimation for the Finite Volume Method with Applications to Fluid Flows
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Old   March 15, 2020, 10:53
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Quote:
Originally Posted by granzer View Post
If the temperature (or any scalar) is varying linearly in a cell, will the variation of the gradient of the temperature also be linear?



If we say that temperature, T, is varying linearly in a square plate given by

T = Tp +(x-xp)grad(Tp)

where Tp is the Temperature at the centroid of the square plate, xp is the centroid, x is any point on the plate so (x-xp) is the position (so a vector) of any point on the plate from the centroid, grad(Tp) is the temperature gradient at the centroid.

Is it correct to write,

a) grad(T) = grad(Tp)+(x-xp)grad(grad(Tp))

as (x-xp), even though is a vector, doesn't actually change with x,y,z

OR

b) grad(T) = grad(Tp)+grad[(x-xp)grad(Tp)]

as we cannot consider (x-xp) to be a constant with respect to x, y, z.

Ref: Prof.Hrvoje Jasak Thesis, Error Analysis and Estimation for the Finite Volume Method with Applications to Fluid Flows



Once the temperature is assumed linear, its derivative is a constant in the cell:
T(x)=T0+T1*x -> dT/dx=T1
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Old   March 15, 2020, 13:11
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Originally Posted by FMDenaro View Post
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
grad(T) = grad(Tp)+(x-xp)grad(grad(Tp))?
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Old   March 15, 2020, 13:16
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Thank you! If I assume the gradient of temperature itself to vary lineraly can i write
grad(T) = grad(Tp)+(x-xp)grad(grad(Tp))?

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
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Old   March 15, 2020, 13:26
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Originally Posted by FMDenaro View Post
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.
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Old   March 15, 2020, 13:41
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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
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