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For the FROMM scheme, what is the flux limiter ψ(r) equal to?(a) 1-\(\frac{r}{2}\)(b) 1+\(\frac{r}{2}\)(c) \(\frac{1-r}{2}\)(d) \(\frac{1+r}{2}\)I have been asked this question during an interview.I need to ask this question from Convection-Diffusion Problems in portion Convection-Diffusion Problems of Computational Fluid Dynamics |
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Answer» CORRECT answer is (d) \(\frac{1+r}{2}\) Best EXPLANATION: To FIND the flux limiter, Φf=Φc+\(\frac{1}{2}\) Ψ(r)(ΦD-Φc ) For the FROMM scheme, Φf=Φc+\(\frac{1}{4}\)(ΦD-ΦU) Comparing both, Ψ(r)(ΦD-Φc)=\(\frac{1}{2}\)(ΦD-ΦU) Ψ(r)=\(\frac{1}{2}\frac{(\phi_D-\phi_U)}{(\phi_D-\phi_c)}\) Ψ(r)=\(\frac{1}{2}\frac{(\phi_D-\phi_c+\phi_c-\phi_U)}{(\phi_D-\phi_c)}\) But, \(\frac{(\phi_c-\phi_U)}{(\phi_D-\phi_c)}=r\) Therefore, \(\PSI(r)=\frac{1}{2}(1+r)\). |
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