Computational Aerodynamics Questions & Answers  
Question by Student 201227106
Professor, I installed VMware and tried to run LINUX virtual machine. but it has an error with the message. I think my laptop has problem with hardware. Please recomment with my problem.
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I think this is because VMware on your laptop can not run a 64 bit Linux OS for some reason. So, download the 32 bit iso image of Ubuntu Mate 16.04.2 and install it (after clicking on Ubuntu MATE 16.04.2 LTS, click on 32 bit). Let me know if this fixed the problem.
Question by Student 201227106
Thank you, professor. I could install that program with your help.
I'm glad to hear that. Because your post may help others, I'll give you 2 points bonus boost.
Question by Student 201127151
Professor, I just would like to inform you that my student ID isn't listed in Introduction to CFD Scores. So could you check it again, please?
I corrected it.
Question by Student 201238707
Professor, I learnt in other class that there are two methods to solve the fluid problem. I wonder why we dealt with just differential form of fluid motion. Is it because Equation in integral form is not useful to CFD?
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.
Question by Student 201427564
Dear Professor, I can't understand why these two equations $ \frac { \partial ( {\frac {1} {2}} {{v}_{x}^2} ) } { \partial t} = {{v}_{x}} {\frac {\partial {v}_{x}} {\partial t}}$ , $ \frac { \partial ( {\frac {1} {2}} {{v}_{x}}^2 ) } { \partial y} = {{v}_{x}} {\frac {\partial {v}_{x}} {\partial y}} $ are reasonable. Because $ {v}_{x} $ is velocity of x-direction, but t and y are not x.
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.
Question by Student 201527110
Professor, I don't understand how $\Gamma,(\frac{\partial \tau}{\partial t})$ can be chosen. You set $\Gamma=1$ because $\Gamma$ is constant when we trying to get $\Phi_1=0$ but, actually I don't understand how it really can be like this.
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.
Question by Student 201127151
Professor, I have a question about $\Omega$. You said that ${\Omega}={x_{\xi}}{y_{\eta}} + y_{\xi}x_{\eta}$ and it means the area of a cell. I understood that $\Omega$ represents area because its dimension is $[m^2]$. But I don't understand that it is the area of a cell. Is it just the definition, or can we demonstrate it?
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.
Question by Student 201427142
Professor, I still cannot understand about $\Gamma$ in quastion by student #201527110 $$ \Phi_1 = U \frac{\partial}{\partial\tau}\left(\Omega\Gamma\right) = U \frac{\partial}{\partial\tau}\left(\frac{\partial x}{\partial\xi}\frac{\partial y}{\partial\eta}\frac{\partial\tau}{\partial t}-\frac{\partial y}{\partial\xi}\frac{\partial x}{\partial\eta}\frac{\partial\tau}{\partial t} \right) $$ at here $$\frac{\partial}{\partial\tau}\left(\frac{\partial x}{\partial\xi}\frac{\partial y}{\partial\eta}\frac{\partial\tau}{\partial t}\right) = \left(\frac{\partial}{\partial\tau}\left(\frac{\partial x}{\partial\xi}\right)\right)\frac{\partial y}{\partial\eta}\frac{\partial\tau}{\partial t}+\frac{\partial x}{\partial\xi}\left(\frac{\partial}{\partial\tau}\left(\frac{\partial y}{\partial\eta}\right)\right)\frac{\partial\tau}{\partial t}+\frac{\partial x}{\partial\xi}\frac{\partial y}{\partial\eta}\left(\frac{\partial}{\partial\tau}\left(\frac{\partial\tau}{\partial t}\right)\right)$$ and I understand 1st and 2nd in right term. But I think that the 3rd term($\left(\frac{\partial}{\partial\tau}\left(\frac{\partial\tau}{\partial t}\right)\right)$) cannot be omitted. Because $\frac{\partial\tau}{\partial t}$ is still depending on $\tau$. How can we assume $\Gamma = 1$??
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.
Question by Student 201427564
Professor, When I try to open warp file through 'Pluma', this error occured. How should I do?
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.
Question by Student 201427564
Professor, I wonder about difference between 'EE' and 'NO'. 'E' makes mesh equally space and 'NO' also. Can we adjust distance with 'E' like below? Block(is,js,ie,je, EE,0.5,2,2, NO); Also, is it right that 'NO' only makes mesh equally 1? (like default) I know this is very basic question but I want to know clearly. Thank you.
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.
Question by Student 201427102
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for middle block, you use "BLOCK(    ,GG,   );"
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for middle block, you use "JOINT(    ,GG,   );"
3.jpg  ./download/file.php?id=3369&sid=49a5e2e4c1277e1af69f8bccaeb3534d  ./download/file.php?id=3369&t=1&sid=49a5e2e4c1277e1af69f8bccaeb3534d
However, this case, for middle block, you use "JOINT(   ,EE,   );" Why I shoud use 'EE' rather than 'GG'??
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,js-1) and the point (i,js). However, there is no point (i,js-1) because it is outside of the domain. Good question: 2 points bonus boost.
Question by Student 201227147
Professor, I have a question about the today's class. During the class, to set $js1$ which is the midpoint of $js$ and $je$, you wrote as follow: $$js1 = round(0.5*(je-js));$$ but if I set $js = 40$ and $je = 100$, $js1$ becomes 30 which is not located between $js$ and $je$. Similarly, you wrote $is_1$ as follow: $$is1 = round(((ie-is)-0.7*(ie-is))/2);$$ but if I set $is = 40$ and $ie = 100$, $is1$ becomes 9 which is also not located between $is$ and $ie$. Could you tell me what I'm thinking wrong?
You are not wrong.. Rather, I forgot to include some term. It should have been written: $$ \rm js1=round(0.5 * (je-js))+js; $$ and $$ \rm is1=round(0.3*(ie-is)/2)+is; $$ Good observation: 2 points bonus boost.
Question by Student 201227106
Professor, I think Assignment 3 has a problem. In question #2, Q is consist of omega*U, but I think Q is consist of omega*capital gama*U.
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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