Sig Turbomachinery MRF Library
Contents
1 Continuity Equation
2 Momentum Equation
2.1 Some mathematical useful identities
The first time derivation of any vector formulated in the absolute and relative frame of reference, according to [1], where is the angular frequency between these two coordinate systems. Any quantity with an ' is related to the relative frame of reference.
If is a vector operator, then it is valid
Therefor the total time derivation of any vector can be split off into a local term and a convective term. In the absolute frame of reference it gives with the absolute velocity
And in the relative frame of reference it gives with the relative velocity
If we insert eq. 2-4 into the 1, one can write
With the relationship of eq. 14 this yields to
Additional, the following identities are valid with any vector and a scalar [2]
and [3]
and [4]
2.2 Derivation
We are now deriving the momentum equation in the relative coordinate system but formulated with the absolute velocity.
If we use eq. 6, then it can be easily shown, that the following statement is valid
If we use the solution from eq. 9, and apply them to eq. 7, the following can be shown
Furthermore, if we apply in eq. 5, this gives us
Now we add the first term on the right hand side of eq. 11 to eq. 12 and apply furthermore eq. 11. Then it follows
Consider, that the origin of the absolute and relative frame of reference or coincident. Then it is valid to say, that the position vector is equal in both systems . But, be aware, the components of these vectors are not equal, because they are defined in different base systems. If we now use in eq. 1, it gives usthe well known relation between the absolute and relative velocity.
If we now use the solution from eq. 14 and apply this to the first term on the right hand side of eq. 11, the following is valid
If we now apply eq. 8 to the term A2 of eq. 15, we get
If you consider for example and in cartesian coordinates and apply the cross product component-by-component, then the result from eq. 16 maybe become a little bit more clearer. With eq. 16, eq. 15 becomes
Now with eq. 17 and eq. 7 term A1 from eq. 13 writes to
If we now insert eq. 18 into eq. 13, we get
The momentum equation of an inertial coordinate system is equal to
If we now apply eq. 19 to eq. 20 we get the momentum equation in a relative coordinate system, see for example [1]
Furthermore it can be shown that the tensor 2. rank from the dyadic product of is not symmetric, if you write it down component-by-component in cartesian coordinates. So that it must be valid to write
3 Energy Equation
TODO.
4 Shear Stress Tensor
4.1 Mathematical Expressions
The product rule is also valid with the Nabla-Operator and arbitrary products and . The arrows above indicates the appliance of the Nabla-Operator.
For a cross product of two vectors it is valid
In general one writes for a tensor :
And for a tensor one can also write
Furthermore for a tensor it is valid [2]:
and also
Therefore the following indentities are valid, starting with eq. 34:
With eq. 39 for the tensor the following is valid:
If eq. 40 gets transposed and then eq. 33 is applied to the result, one yields:
4.2 Transformation in a relative frame of reference
For a newtonian fluid the shear stress tensor can be expressed in the inertial frame of reference with as
Where in three-dimensional the number of dimensions . Furthermore the volume viscosity vanishes after Stokes' hypotheses.
As already expressed in eq. 16, the last term on the right hand side is equal to
The same procedure is also applied to the following expression
For the last two terms on the right hand side, one can write:
If one now applies the product rule from eq. 30, one yields the following:
As is a single vector and not a vector field, the gradient of a single vector is always zero. If one now apply eq. 36 and eq. 34 at each term of the right hand side of eq. 54, one yields with the metric tensor :
With the help of eq. 41 it follows:
As the metric tensor is symmetric, one can write . Thus from eq. 56 it follows
With the help of eq. 57 it follows for eq. 52:
Finally with eq. 51 and eq. 58 the shear stress tensor for a newtonian fluid in the relative frame of reference can be expressed as
5 References
- ↑ FLUENT 6.3 User's Guide; "Equations for a Rotating Reference Frame"; Chap. 10.2.2
- ↑ Michael H. Vavra, "Aero-thermodynamics and flow in turbomachines", 1960; Appendix B1