Computational Aerodynamics Questions & Answers  
Question by Student 201427116
Professor, I have a question about WENO. We used below terms in WENO: $$ {\bar w_0} = \frac{\gamma_{0}}{{(\epsilon+\beta_0)}^2}, {\bar w_1} = \frac{\gamma_{1}}{{(\epsilon+\beta_1)}^2} $$ There is $\epsilon$ in denominator but what is it for? and how can I decide the value of $\epsilon$ ? $$ $$ Another question is about $\gamma_0$ and $\gamma_1$. With Taylor series expansion, we found $\gamma_0=\frac{1}{3}$ and $\gamma_1=\frac{2}{3}$ for 3rd order accuracy. With 3rd order accuracy, that is, WENO3, does $\gamma_0$ and $\gamma_1$ have fixed value of $\frac{1}{3}$ and $\frac{2}{3}$, respectively? Or are they also variants?
06.13.18
The user-defined constant $\epsilon$ is included to prevent a division by zero. Set it to a very small value. As for $\gamma_0$ and $\gamma_1$, they are fixed constants: don't change them.
Question by Student 201327132
Professor I have a question about bdry condition. In my note, One property must come from out of the domain at subsonic outflow bdry condition. So We choose Pressure. And We use $P_1^{n+1}=P_\inf$. Other bdry condition case that subsonic inflow, Two properties must be obtained from outside domain. So we choose Temperature and Pressure. And we use Stagnation Temperature and Pressure for time level n+1. Why we use difference methods to obtain pressure?(Stagnation pressure and Freestream pressure) Thank you.
For the inflow BC, the stagnation pressure can be assumed equal to the one in the freestream because the flow along a streamline is isentropic. But such is not the case for the outflow BC. What if there is a shock somewhere within the domain? Then, the entropy rises and the stagnation pressure will go down and not be equal to the one in the freestream. However, for external flows around a body, the pressure will eventually become equal to the freestream pressure even if shocks are present (as long as the BC is far away from the body). Hence why it's better at the outflow BC to choose to fix pressure rather than stagnation pressure.
Question by Student 201227138
Professor, I think I have bad luck bonus in your coruse, Viscous Flow. I will appreciate if you check my bad luck bonus.
07.02.18
OK, will look into this now.
Question by Prasanna
Professor, I am a bit confused about question #5 of Assignment 4. I have to find $$\frac{\partial F_3}{\partial U_4}=\frac{\partial (\rho u^2 +P)}{\partial U_4}=\frac{\partial (\rho u^2)}{\partial U_4}+\frac{\partial P}{\partial U_4}$$ and the difficult part seems to be in determining the $\frac{\partial P}{\partial U_4}$ term. The alternative method you taught involves using the chain rule, for example, $$\frac{\partial F_3}{\partial U_4}=\frac{\partial F_3}{\partial \rho_1} \frac{\partial \rho_1}{\partial U_4}+\frac{\partial F_3}{\partial \rho_2} \frac{\partial \rho_2}{\partial U_4}+\frac{\partial F_3}{\partial u} \frac{\partial u}{\partial U_4} +\frac{\partial F_3}{\partial \phi} \frac{\partial \phi}{\partial U_4} $$ where $F_3=F_3(\rho_1,\rho_2,u,\phi)$ and $\phi$ is some variable. For this problem, $\phi$ has to be a function of $U_1,U_2,U_3,U_4$ and also a function of P such that I can evaluate $\frac{\partial \phi}{\partial U_4}$. But then if I could express P in terms of $\phi$ which is a function of $U_1,U_2,U_3,U_4$, I would use the first method which you taught to evaluate the flux jacobian terms instead. I would like your comment regarding this.
04.15.19
You don't necessarily need to express $\phi$ as a function of $U$ to determine $\partial \phi/\partial U$ in the same way as you don't need to express $F$ as a function of $U$ to obtain $\partial F/\partial U$.
04.16.19
Question by Van Tien
Professor, in Assignment 5, Question #3, for the extrapolation, I am confused to use the 1D-Lagrange interpolation or 2D-Lagrange interpolation. In the case of 1D-Lagrange interpolation, I am not sure the polynomial function is based on x or y coordinate. In the case of 2D-Lagrange interpolation, I think I need more information from the other nodes.
05.08.19
Use a 1D extrapolation polynomial. 2D is too time consuming to compute.
Question by Student 201983196
Professor, in Assignment7, Question#2, How do I calculate this equation, $\frac {\mid{A}\mid (Z_L,Z_R)}{2}(U(Z_R)-U(Z_L))$? Is $\frac {\mid{A}\mid (Z_L,Z_R)}{2}$ 2X1 matrix? But $(U(Z_R)-U(Z_L))$ also is 2X1 matrix? I don't know how to calculate $F_(i+\frac{1}{2})=\frac{F(Z_L)+F(Z_R)}{2}+\frac {\mid{A}\mid (Z_L,Z_R)}{2}(U(Z_R)-U(Z_L))$
05.27.19
No, $|A|$ is a $2 \times 2$ matrix determined from an average state function of $Z_L$ and $Z_R$.
Question by Prasanna
Professor, for Assignment #7, Question #3, are the answers posted in the following order: $u_L$,$u_R$ in decimal digits, $u_R$ in fraction, $f(u_L)$ in decimal digits, $f(u_L)$ in fraction respectively?
I updated the answers to make them more clear.
Question by Student 201983196
Professor, In Assignment#7, Question#2 (b), I use 2nd order polynomial about node(4,5,6) and node(5,6,7). and then using optimal weight, calculate $u_R$. but my solution is wrong. I don't know how to solve this Question#2(b).
05.28.19
Hm, I see a problem in your approach. You shouldn't be finding a polynomial when determining the flux with a TVD minmod2 limiter.
Question by Student 201627128
Professor, in class when you explained how to find WENO3, you found a highest degree polynomial through the data points. Using a similar approach I was able to find $u_L$ equal to 4.5 as in the solutions, however, when I apply the same strategy to find $u_R$, I get 4.25 instead, which does not match the solution. I tried to do it in reconstruction evolution and again found 4.5 for $u_L$ but this time $u_R$ becomes 5. Is there a separate approach to find $u_R$?
05.29.19
I don't understand why using reconstruction-evolution would give you a different answer. You need to find $u_{\rm R}$ by interpolating $u$. Once $u$ is interpolated and $u_{\rm L}$ and $u_{\rm R}$ are found, then apply reconstruction evolution.
Question by Prasanna
Professor, for assignment #7, question #3 (c), I did my calculation as follows: I get a slightly different answer from my calculation. However, the answer is the same as the one posted only if I choose the wavespeed $a=u$ and not $a=u/2$.
For $f$ and $u$ given, there is only one wave speed not two.
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