Introduction to CFD Questions & Answers  
Question by Student 201227125
Professor, at assignment 2-Question #3, x and y has dimension that is length(unit mm). In this case, dose $\xi$ and $\eta$ also have dimension or not?
No, $\xi$ and $\eta$ don't have dimensions. But their derivatives do of course. So $\xi_x$, $\eta_x$, etc will have dimensions (1/m or 1/mm). 1 point bonus.
Question by Student 201227148
Sir, I can not get my CFD password. I entered my email address. But I can not log in.
Try again now.
Question by Student 201427116
Professor, We learned how to find wave speed from Flux Jacobian on last class. After finding all elements of Jacobian , you showed the way to extract wave speeds from Jacobian. In this process, you found 3 eigenvalues of Jacobian which are expressed by $$ {\phi_1}, {\phi_2}, and {\phi_3}, respectively. $$ And said, these eigenvalues represent each wave speeds. $$ {} $$ But I cannot understand the relation between eigenvalues of Jacobian and wave speed. How do I confirm that these eigenvalues should be wave speed? What relationship is involved in this process?
Consider a system of equations as follows: $$ \frac{\partial U}{\partial t}+ A \frac{\partial U}{\partial x}=0 $$ Recall that the eigenvalues are such that $A=L^{-1}\Lambda L$: $$ \frac{\partial U}{\partial t} + L^{-1} \Lambda L \frac{\partial U}{\partial x}=0 $$ Multiply by $L$: $$ L\frac{\partial U}{\partial t} + \Lambda L \frac{\partial U}{\partial x}=0 $$ Say that a vector $W$ exists such that $L=\partial W/\partial U$. Further, because $L=L(U)$, it follows that $W=W(U)$. Thus, we can say $$ \frac{\partial W}{\partial t}=L\frac{\partial U}{\partial t} ~~{\rm and}~~ \frac{\partial W}{\partial x}=L\frac{\partial U}{\partial x} $$ Substitute the RHS of the latter 2 equations in the former: $$ \frac{\partial W}{\partial t} + \Lambda \frac{\partial W}{\partial x}=0 $$ Because $\Lambda=[\phi_1,~\phi_2,~\phi_3]^{\rm D}$ is diagonal, the latter is simply a list of advection equations each with a wave speed $\phi_1$, $\phi_2$, $\phi_3$. Good question: 2 points bonus.
Question by Student 201327132
Dear professor. We derived Euler equations in genrealized coordinate. First we start $$\frac{\partial U}{\partial \tau}+ \xi_x \frac{\partial F_x }{\partial \xi} + \eta_x \frac{\partial F_x }{\partial \eta} + \xi_y \frac{\partial F_y }{\partial \xi} + \eta_y \frac{\partial F_y }{\partial \eta} =0$$ Second we multiply by $\Omega$. But I think, we don't need to multiply $\Omega$. Because here, $$\frac{\partial U}{\partial \tau}+ \frac{\partial F_x \xi_x }{\partial \xi} - F_x \frac{\partial \xi_x }{\partial \xi} + \frac{\partial F_x \eta_x }{\partial \eta} -F_x \frac{\partial \eta_x }{\partial \eta} +\frac{\partial F_y \xi_y }{\partial \xi} -F_y \frac{\partial \xi_y }{\partial \xi} + \frac{\partial F_y \eta_y }{\partial \eta} - F_y \frac{\partial \eta_y }{\partial \eta} =0$$ And, $$-F_x( \frac{\partial \xi_x}{\partial \xi} -\frac{\partial \eta_x}{\partial \eta})= -F_x( \frac{\partial}{\partial x} \frac{\partial \xi}{\partial \xi} - \frac{\partial}{\partial x} \frac{\partial \eta}{\partial \eta})=0 $$ I think, this is more simple equation. $$\frac{\partial U}{\partial \tau}+ \frac{\partial F_x \xi_x }{\partial \xi}+ \frac{\partial F_x \eta_x }{\partial \eta} +\frac{\partial F_y \xi_y }{\partial \xi}+ \frac{\partial F_y \eta_y }{\partial \eta} =0$$ I want to know why we use $\Omega$. Thank you.
The problem with your logic is that $\partial \xi_x / \partial \xi$ is not zero because $\xi_x$ can vary along the the $\xi$ coordinate. Similarly, $\partial \eta_x / \partial \eta$ is also not zero because $\eta_x$ can vary along $\eta$. Good question, 2 points bonus.
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