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Click here for the results of the 2 series of cases, completely summarized in the last two worksheet of the Excel workbook.

Convergenc level

As shown in the workbook, the contour of energy dissipation rate changed little judging from eyeball examination between different convergence levels, but the calculated performance parameters still showed some variability as the convergence level changes. However, they stablized from e-5 to e-6.

Boundary conditions

A larger varibility was shown in the parameters when compared between symmetry and wall boundary conditions, while little differences were observed in the energy dissipation contours with bare eyes. However, the variability are still within an acceptable ranging in terms of a varieity of other uncertainties in the practice of design and operation.

Convergence rate

The iteration steps needed And the iteration steps need for each cases at the three different convergence levels:


Conclusions



The above table suggests the case with wall boundary condition converges a lot faster

More comments
Also note that in all these cases the solutions were obtained by directly iterating towards e-6 with second order scheme, instead of the normal practice: iterating with first order to obtain an good initial guess and then with second order for more accurate values.

Conclusions

  • e-6 is a residual level where the solutions are stablized.
  • Using
  • Energy dissipation rate contour and quantitative analysis show dependence of performance parameters as a function of h/b ratio, which suggest it worth further investigation.
  • Results are very sensitive to convergence levels.
  • Symmetric boundary condition makes it difficult to converge.
  • Symmetric boundary condition and wall boundary condition may have similar results at good convergence levels (residual below e-6). We can use wall boundary condition to replace the symmetry boundary condition to ensure accuracy of the results.

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  • is justifiable.
  • Wall boundary condition makes iteration converge a lot faster.
  • A residual of e-6 could be obtained by directly iterate toward e-6 with second order scheme.