Introduction to CFD Questions & Answers  
Question by Student 201427564
Professor, I have a question about $\phi(r)$. In the table, $ \phi(r)=max(0, min(1,r)) $. But last class, you wrote like this. $ {({\phi^{+}}_{i+ \frac {1} {2}})}_{1} = max(0,min(1, 2{r^{+}}_{i})) $ Why did you use $2{r^{+}}_{i}$ rather than ${r^{+}}_{i}$ ?
Both $\phi=\max(0,\min(1,r))$ and $\phi=\max(0,\min(1,2r))$ respect the rule of the positive coefficients and reduce to first order at extrema and are hence valid. But $\phi=\max(0,\min(1,2r))$ is better because it is closer to the second-order stencil. 1.0 point bonus.
Question by Student 201227147
Professor, I have a question about assignment 8 question #2-(b). Last class, you calculated $r _{i} ^{-}$ to find out $ \phi _{i+1/2} ^{-}$ because you used this relationship I guess: $$ \phi _{i+1/2} ^{-} = max(0, min(1,2r _{i} ^{-} ))$$ However, I think $r _{i} ^{-}$ should be $r _{i+1} ^{-}$ because $ \phi _{i+1/2}$ is $ \phi _{i+1/2} ^{-} $. Is it right? or is there some reasons you calculated $r _{i} ^{-}$?
Hm no, if we set $\phi _{i+1/2} ^{-} = \max(0, \min(1,2r _{i+1} ^{-} ))$ then we should define $r_{i+1}^-\equiv\frac{u_i-u_{i+1}}{u_{i+1}-u_{i+2}}$. However, because $r_i^-\equiv\frac{u_i-u_{i+1}}{u_{i+1}-u_{i+2}}$ then we have to set $\phi _{i+1/2} ^{-} = \max(0, \min(1,2r _{i} ^{-} ))$. 1 point bonus.
Question by Student 201227147
I have another question about $ r_{i+1}^-$. In the table, $ r_{i} = \frac{u_{i+1}-u_{i}}{u_{i}-u_{i-1}}$. But you calculated $ r_{i}^-$ with $ r_{i+1}^-\equiv\frac{u_i-u_{i+1}}{u_{i+1}-u_{i+2}} $. Does it mean that the equation in the table is assumed positive upwind (from left($i-1$) to right($i+1$)) so that notations should be from right($i+2$) to left($i$) when I apply it in negative upwind case?(it means $i-1$->$i+2$, $i$->$i+1$, and $i+1$->$i$)$$ $$By the way, Does your answer about previous question mean that there's just difference in notating $r$ and use same equation?
I guess your concerns are due to the fact you are using an older version of tables.pdf.. The tables have changed slightly in the last few days. After downloading the latest version, if you still have some concerns, ask the question again below.
Question by Student 201238707
Professor, I have a question about subsonic inflow's velocity angle. Why those are same? $$\theta^{n+\frac {1} {2}}_1 = \theta^{n+1}_1$$ because we assumed steady-state?
I don't recall this step. You need to put your question into context so I can recall why we did this.
Question by Student 201238707
when we found velocity vector $u^{n+1}_1$ $\&$ $v^{n+1}_1$. we extrapolated $u^{n+\frac {1} {2}}_1$ $\&$ $v^{n+\frac {1} {2}}_1$. And we put $\theta^{n+\frac {1} {2}}_1 = \theta^{n+1}_1$ then used $$\theta^{n+1}_1 = atan(\frac{v^{n+1}_1}{u^{n+1}_1})= atan(\frac{\alpha*v^{n+\frac{1}{2}}_1}{\alpha*v^{n+\frac{1}{2}}_1})$$ but i wonder why those $\theta$ are same?
I don't understand. What other value would you give it?
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