AE320/AE706 Assignment 3 Internal Flow in a Channel with a Circular Bump to be submitted on 06/11/2011
Task (I) Write a Finite volume code based on HLL scheme for computing flow inside a channel with a 10% thick circular arc bump for following three test cases: 1. Subsonic M=0.5 2. Transonic M=0.675 3. Supersonic M=1.4. (II) For subsonic and transonic cases use Riemann Invarient based inflow and outflow boundary conditions. You have your choice about solid wall boundary treatment.
Problem Definitions Consider the 2D domain as shown in figure 1, which corresponds to a rectangular channel of height L and length 3 L. The channel has a circular arc bump (in the middle) of length L and maximum thickness equal to 0.1 L along the bottom wall.
Proceed now as follows 1. Construct an H-grid using 71 points in the axial direction with equal spacing and 31 points in-between top and bottom walls. A quasi-uniform grid distribution is selected by taking ∆x = 3L/70 and dividing each vertical mesh line in 30 uniform steps (see Figure 2). 2. For all cases assume following atmospheric conditions at outlet: pressure and temperature as pa = 101300 Pa and Ta = 288 K respectively. 1
3. In order to match these flow conditions, the total pressure and temperature are fixed at the inlet as: (1) corresponsing to M = 0.5 • po = 120163 Pa • To = 302.4 K (2) corresponsing to M = 0.675 • po = 137458 Pa • To = 302.4 K (3) corresponsing to M = 1.4 • po = 322364 Pa • To = 400.9 K The inlet velocity vector is imposed in the axial direction.
Plot 1. Plot Mach number contours for all the three cases 2. Plot coefficient of pressure cp along the bottom and upper walls.
References for Steger Warming Scheme [1]Steger J. and Warming R., Flux Vector Splitting of the inviscid Gas dynamics equation with application to the finite difference methods, Journal of Computational Physics, Vol 40 (pp 263-293), 1981. [2] C. Hirsch ,Numerical computations of internal nad external flows: vol-1 [3]Anderson Jr, Computational fluid dynamics [4] J. Blazek, Computational Fluid Dynamics: Principle and applications
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Figure 1: Schematic of Computational Domain
Figure 2: Typical computational grid (71 × 31)
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Assignment no 03 Submitted by: Ravi Krishna Roll no: 113014003 Course no: AE 706 Title: Finite Volume Scheme using HLL Riemann solver Procedure applied: 1. A Finite Volume Based code has been developed in which the Flux vector is calculated using HLL Riemann solver technique. 2. A single code has been developed that works both in subsonic and supersonic conditions. 3. For subsonic case, Inlet and Outlet conditions are applies using Riemann Invariants method as described in section 8.4, Inlet/Outlet Boundary of “Computational Fluid Dynamics” by J. Blazek. 4. For supersonic case, given values has been applied at the inlet whereas, at exit the interior values are copied. 5. Cell vertex Finite Volume Method has been used and Wall Boundary is imposed such that normal component of velocity is zero. 6. For time step calculation Courant-Friedrich-Lewis criterion has been used: ∆𝑥 ∆𝑡 = × 𝐶𝑃 𝑢𝑚𝑎𝑥 Where, ∆𝑥 is the grid spacing, CP is the Courant parameter and umax is the maximum value of the local flow speed + local sound speed. The returned timestep is the minimum timestep along all dimensions. 7. For calculation of Flux vector for a particular cell, normal component of Flux and velocity has been obtained at all the edges and then HLL scheme is applied at all the edges individually. 8. In order to save the computation time, for each cell Lower and Right Flux component has been copied from the adjacent cells (Lower and Left cell) with change of sign. Whereas, Right and Upper Flux component has been evaluated for each cell.
i,j+1 D
C
UP i+1,j Right
Left i-1,j
i,j B
A
Down i,j-1
9. The iteration was done till the residual Rss is <= 10-6. 10. The main code “twoDhump.cpp” exports the output in .vtk file format: a. “twoDhump_data.vtk” contains the Mach number profile for all grid points. b. “twoDcp_data.vtk” contains the pressure coefficient over the lower and upper
surface. c. These files can be directly imported into Paraview for plotting and visualization.
11. The results has been obtained and plotted for three cases in the following sections.
Results: Case I: Mach 0.5
Figure 1 Mach number profile for inlet Mach no 0.5 at time = 0.03842 sec Here time of iterations for (Rss<10-6) = 0.03842 sec
Figure 2 Cp profile at upper and lower surface for inlet Mach 0.5
Case II: Mach 0.675
Figure 3Mach number profile for inlet Mach 0.675 at time = 0.0386509 sec Here time of iterations for (Rss<10-6) = 0.0386509 sec
Figure 4 Cp profile over Upper and Lower surfaces for Mach 0.675
Case III: Mach 1.4 For Mach 1.4, three conditions has been plotted at three different times: i.
For time of iterations = 0.01 sec
Figure 5 Mach number profile for inlet Mach 1.4 at time = 0.01sec
Figure 6 Cp profile at Lower and Upper Surface at time = 0.01 sec
ii.
For time of iterations = 0.05 sec
Figure 7 Mach number profile for inlet Mach 1.4 at time = 0.05 sec
Figure 8 Cp profile at Lower and Upper Surface at time = 0.05s
iii.
For Rss<=10-6 giving time of iterations = 0.092 sec
Figure 9 Mach number profile for inlet Mach 1.4 and time = 0.092
Figure 10 Cp profile for Lower and Upper surface for time = 0.092s
Discussion of Results: 1. For Mach 0.5 the disturbances travel in all direction and there is no supersonic flow throughout the domain. 2. For Mach 0.675 the Mach number just reaches a maximum of 0.955 at the minimum throat of the cross section. Hence in this case also the flow remains subsonic throughout the domain. 3. For Mach 1.4, three cases have been considered for three time of iterations. a. For time of iterations = 0.01 sec, a shock wave is formed at the nose of the hump which is reflected from the upper surface. Expansion fan is formed downstream of the hump leading to acceleration of flow to supersonic speed and again a shock wave is formed at the end of the hump. b. For time of iterations = 0.05 sec, the shock before the hump is pushed further upstream. c. For time of iterations = 0.092 sec (for which Rss <= 10-6 condition is satisfied), the shock is pushed upstream at the exit at and a uniform flow develops inside the flow domain.
