Geometry: \( \{(-3,-1) \times (-1,1)\} \cup \{(-1,1)\times (-0.1,0.1) \} \cup \{(1,3)\times (-1,1)\} \)
Boundary Conditions: The slip boundary condition (\(\boldsymbol{m} \cdot \boldsymbol{n} = 0 \)) is enforced on the triangle and on the top and bottom sides.
Discretization: The mesh was composed of uniform quadrilaterals (squares in this case) with 863,297 DoFs.
Final time: \(t = 50\). CFL = 0.9
Initial State: \( (\rho_L, \boldsymbol{v}_L, p_L) = (0.7, \boldsymbol{0}, 0.6) \) and \( (\rho_R, \boldsymbol{v}_R, p_R) = (0.125, \boldsymbol{0}, 0.1) \). The diaphram is at \( \{x = -1\} \).
Equation of state: van der Waals: \[ p(\rho, e) = (\gamma-1) \frac{\rho e + a\rho^2}{1 - b\rho} - a \rho^2 \]
Parameters: \(\gamma = 1.4\), \(a = 0.3215\), \(b = 0.1\).
Software: A modified version of Ryujin
An analytic solution to the Euler equations can be derived by assuming the flow is isentropic (constant entropy) and divergence free (\(\nabla \cdot \boldsymbol{v} = 0)\). An ansantz solution is prescribed to the velocity and the derivation follows. A detailed derivation can be found here: (Put a link here).
Geometry: \( [-0.5,-0.5]^2 \)
Boundary Conditions: The slip boundary condition (\(\boldsymbol{m} \cdot \boldsymbol{n} = 0 \)) is enforced on all sides.
Discretization: The mesh was composed of uniform quadrilaterals (squares in this case) with ??? DoFs.
Final time: \(t = ???\). CFL = ???
Initial State: The initial states are given with reference to the cardinal directions: north-west (NW), north-east (NE), south-west (SW), and south-east (SE). \[ (\rho_{\text{NW}}, u_{\text{NW}}, v_{\text{NW}}, p_{\text{NW}}) = ... \] \[ (\rho_{\text{NE}}, u_{\text{NE}}, v_{\text{NE}}, p_{\text{NE}}) = ... \] \[ (\rho_{\text{SW}}, u_{\text{SW}}, v_{\text{SW}}, p_{\text{SW}}) = ... \] \[ (\rho_{\text{SE}}, u_{\text{SE}}, v_{\text{SE}}, p_{\text{SE}}) = ... \] The center point is \( (0.25,0.25) \).
Equation of state:
Parameters:
Software: A modified version of Ryujin
Coming soon...
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The setup for this problem involves a shock moving with Mach ??? to the right which collides with a triangular obstacle. We follow the same setup given in Toro, Casto, Lee (2015). The problem is motivated from real physical experiments (CITE).
Geometry: \( \{(-0.65,0.5) \times (-0.5,0.5)\} \setminus K \) where \(K\) is the triangle formed by the vertices: \( (-0.2,0.0) \), \( (0.1, -1/6) \), \( (0.1, 1/6) \).
Boundary Conditions: The slip boundary condition (\(\boldsymbol{m} \cdot \boldsymbol{n} = 0 \)) is enforced on all sides. No boundary conditions were applied to the left or right sides.
Discretization: The mesh is unstructured and composed of triangles with 1,222,183 DoFs. The mesh is generated using Triangle.
Final time: \(t = 0.0022\). CFL = 0.33
Initial State: \( (\rho_L, \boldsymbol{v}_L, p_L) = (1.82039, 148.597, 0, 185145) \) and \( (\rho_R, \boldsymbol{v}_R, p_R) = (1.225, 0, 0, 101325) \). The diaphram is at \( \{x = -0.55\} \).
Equation of state: van der Waals: \[ p(\rho, e) = (\gamma-1) \frac{\rho e + a\rho^2}{1 - b\rho} - a \rho^2 \]
Parameters: \(\gamma = c_p/c_V = 864.7/577.8\), \(a = 0.14\), \(b = 3.258 \times 10^{-5}\).
Software: In-house Fortran code.
This problem models a shock exiting a "tube". Similar to the Shock Hitting Triangle problem, we start with an initial state and a Mach number and compute the state behind the shock using the Rankine-Hugoniot conditions. As the shock leaves the tube, we observe the effect of diffraction around the corner. The setup for this problem can greatly vary see CITE CITE CITE. We follow the setup described in CITE.
Geometry: The computation domain is: \( [0,2]^2 \setminus ((0,0.5) \times (0,1)) \).
Boundary Conditions: Dirichlet conditions are enforced on the far left and far right sides of the domain and slip boundary conditions (\(\boldsymbol{m} \cdot \boldsymbol{n} = 0 \)) are enforced on all other sides.
Discretization: The mesh was composed of uniform quadrilaterals (squares in this case) with DoFs.
Final time: \(t = 0.01\). CFL =
Initial State: \( (\rho_L, \boldsymbol{v}_L, p_L) = (???, ???, 0, ???) \) and \( (\rho_R, \boldsymbol{v}_R, p_R) = (1, 0, 0, 1) \). The diaphram is at \( \{x = 0.5\} \).
Equation of state: Jones-Wilkins-Lee: \[ p(\rho, e) = A \Big(1 - \frac{\omega}{R_1}\cdot \frac{\rho}{\rho_0} \Big) \exp\Big(-R_1 \frac{\rho_0}{\rho}\Big) + B \Big(1 - \frac{\omega}{R_2} \cdot \frac{\rho}{\rho_0} \Big) \exp\Big(-R_2 \frac{\rho_0}{\rho}\Big) + \rho \omega e \]
Parameters: \(A = \), \(B = \), \(R_1 = \), \(R_2 = \), \(\omega = \), and \(\rho_0 = \).
Software: A modified version of Ryujin
Coming soon...