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Starting Equations (Ref. 1)
Derive Vector Equations
Combining the above, the vector form of the viscous portion of the momentum equation,
, is found.
Focusing on the last three terms.
Putting it back together.
The following geometry relation will be assumed.
Next, taking the vector triple product
and rearranging it
and then inserting it in the
equation leads to
Next, the velocity difference in the η direction is given by
is along the axis.
Assuming U is in the η plane, then
Dot products are independent of the η index, therefore the following results when differencing in η direction.
is the area vector for a given cell side (η in this case) and vol is the volume of the cell.
Next, the relationship that the area vectors of the cell walls must add to zero will be used.
Substituting this into
3D Thin Layer Set
Rearrange the 3D viscous momentum equations.
Next, the cross product and dot terms are dropped, leaving only the thin-layer set.
The viscous Jacobians are derived from the 3D thin layer set.
To help derive the viscous Jacobian the following viscous solution vector is defined.
The viscous Jacobian in the η direction is then defined as follows.
The derivative of the viscous Flux is then split into a part that explicitly contains the velocity and explicitly contains the speed of sound.
Since the transpose of a scaler is equal to the scaler, the second matrix multiplication set can be transposed.
Finally the derivative of the viscous flux due to the explicit velocity terms is,
must be placed in terms of the viscous solution vector.
By inspection it can be seen that:
Therefore the derivative of the viscous flux with respect to the viscous solution vector when taking into account that
Next, dealing with
Putting it all together.
The eigenvalues for this are:
Diagonal Viscous Jacobian
Speed of sound part.
1) Krist, S. L., Biedron, R. T., and Rumsey, C. L., "CFL3D User's Manual (Version 5.0)," NASA TM 1998-208444, June 1998.