Computational Aerodynamics Questions & Answers | |
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Hm, I don't understand the question. You need to explain better what you don't understand..
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I'm not sure what you mean. But it is not correct that $A_{i+1/2}U_{i+1}=F_{i+1}$. This is not the Roe average. The Roe average if $\Delta F= A_{i+1/2}\Delta U$. I'll give you 0.5 point for the effort.
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What does “alpha” mean? I didn't use this in class. You need to rephrase your question and use the same symbols as used in class. Or, define clearly a new symbol you are introducing.
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Now I see what you mean. When dealing with a system of equations, the wave speeds are not necessarily $a$ or $u$. Rather, the wave speeds are the eigenvalues. So, when determining $F^\pm_{i+1/2}$, the wave speeds are within $\Lambda^\pm$ (those play the same role as the “$a$” does for a scalar equation). 1.5 point bonus.
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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.
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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.
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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.
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I don't recall this step. You need to put your question into context so I can recall why we did this.
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I don't understand. What other value would you give it?
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True, but $\Gamma=1$, so it doesnt' matter in this case. 1 point bonus.
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This is great! 2.0 points bonus. I would have given you 3 points bonus if you would have typeset it using LATEX.
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Its due on Thursday. I fixed the mistake.
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$\pi$ |