Question

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

Answer 1

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,

\(\psi(r)=\frac{1}{2}(1+r)\).