Computational Aerodynamics Questions & Answers  




I'm glad to hear that. Because your post may help others, I'll give you 2 points bonus boost.




I corrected it.




Both the integral form and the differential form can be used in CFD. But we can derive the integral form by integrating the differential form over a volume.. We'll get to this at one point. Interesting question: I'll give 2 points bonus boost.




The following is always correct: $$ \frac { \partial ( {\frac {1} {2}} {\phi^2} ) } { \partial \xi} = {\phi} {\frac {\partial \phi} {\partial \xi}} $$ where $\phi$ is any property and $\xi$ can be $x$, $y$, $t$, or any coordinate. It doesn't matter if $\phi$ is $v_x$ or $t$, the above is a mathematical transformation, not a physical one. Not a bad question, I'll give you 1.5 points bonus.




This is a good question. First note that $\Gamma$ must be chosen such that it respects the dependencies we specified during the derivation. By definition, $$ \Gamma\equiv\frac{\partial \tau}{\partial t} $$ Thus, because $\tau=\tau(t)$, $\Gamma$ can not depend on $x$ or $y$. $\Gamma$ could depend on $t$ thus. We here choose it to be simply $\Gamma=1$ because that is the simplest expression that satisfies the above dependency.. I'll give you 2 points bonus boost.




Good question. I have pondered this as well when I first encountered the generalized coordinates.. The definition of $\Omega$ is simply ${x_{\xi}}{y_{\eta}} + y_{\xi}x_{\eta}$, not the cell area. But it is very close to the cell area thus. In the next assignment, I will ask you to compare the cell area with $\Omega$ and assess the differences. I'll give you 2 points bonus.




We can set $\Gamma$ to any value as long as it respects the dependency we originally specified. That is, $\Gamma=\partial\tau/\partial t$ should be such that $\tau=\tau(t)$. Thus, we can choose $\Gamma$ to be $1$ or $t$ or $t^2$ but not $x$ or $tx$, etc. Here, it is perfectly fine to choose $\Gamma=1$. One advantage of choosing $\Gamma=1$ is that $$ \frac{\partial}{\partial\tau}\left(\frac{\partial\tau}{\partial t}\right) = \frac{\partial}{\partial\tau}\left(\Gamma\right) = \frac{\partial}{\partial\tau}\left(1\right) = 0 $$ If $\Gamma$ would have been chosen as $t$ or $t^2$, then the latter wouldn't be 0, and this would lead to $\Phi_1\ne 0$, and this would prevent us to write the Euler equations in generalized coordinates in conservative form. I'll give you 2 points bonus boost.




Warp is a binary file (executable), so it can not be opened in pluma. With pluma you can open text files like .wrp or .txt or .c etc. Please don't do assignment #4 now: the tgz package is not complete yet. I'll explain what to do in assignment 4 tomorrow.




Hmm, I am not sure what your first question is.. As for the second, 'NO' does not make the mesh equally 1, it simply needs to be used when the block is 1D (when there are no segments along a certain dimension). I'll give you 1 point bonus boost.




First, the command is Join(), not Joint(), and is case sensitive. So, if you use JOIN or join it will not work: it has to be spelled Join. For the third mesh, you can not use Join(is,js,ie,je,j,GG..) because the GG segment grabs the spacing from the mesh along the j coordinate. So in this case, he will set the first grid spacing (between j=js and j=js+1) to the distance between the point (i,js1) and the point (i,js). However, there is no point (i,js1) because it is outside of the domain. Good question: 2 points bonus boost.




You are not wrong.. Rather, I forgot to include some term. It should have been written: $$ \rm js1=round(0.5 * (jejs))+js; $$ and $$ \rm is1=round(0.3*(ieis)/2)+is; $$ Good observation: 2 points bonus boost.




That's right: I fixed it. I'll give you 1 point bonus boost because your typesetting was hard to read.



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