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Example 13 - Shock Tube

Example 13 - Shock Tube

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Example 13 - Shock Tube

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rad_ex_13_shock-tube

Summary


This famous experiment is interesting for observing the shock-wave propagation. Moreover, this case uses the representation of perfect gas and compares the different formulations: The ALE uses Lagrangian or Eulerian and Smooth Particle Hydrodynamics (SPH).

The first part of the study deals with the modeling description of perfect gas with the hydrodynamic viscous fluid law 6. The purpose is to test the different formulations:

Lagrangian (mesh points coincident to material points)
Eulerian (mesh points fixed)

For the Eulerian formulation, different scale factors on time step are also tested. Furthermore, the SPH formulation is also tested; which does not use mesh, but rather particles distributed uniformly over the volume.

The propagation of the gas in the tube can be studied in an analytical manner. The gas is separated into different parts characterizing the expansion wave, the shock front and the contact surface. The simulation results are compared with the analytical solution for velocity, density and pressure.

 

Title

Shock tube

 

rad_ex_13.1

Number

13.1

Brief Description

The transitory response of a perfect gas in a long tube separated into two parts using a diaphragm is studied. The problem is well-known as the Riemann problem. The numerical results based on the SPH method and the finite element method with the Lagrangian and Eulerian formulations, are compared to the analytical solution.

Keywords

Brick elements
Lagrangian and Eulerian formulations
SPH modeling and hexagonal net
Scale factor for time step
Hydrodynamic viscous fluid law (/MAT/LAW6) and perfect gas modeling

RADIOSS Options

ALE boundary conditions (/ALE/BCS)
ALE material formulation (/ALE/MAT)
SPH symmetry conditions (/SPHBCS)

Compared to / Validation Method

Analytical solution

Input File

Eulerian_formulation: <install_directory>/demos/hwsolvers/radioss/13_Shock_tube/Eulerian_formulation/TACEUL*

Lagrangian_formulation: <install_directory>/demos/hwsolvers/radioss/13_Shock_tube/Lagrangian_formulation/TACLAG*

SPH_hexagonal-net: <install_directory>/demos/hwsolvers/radioss/13_Shock_tube/SPH_formulation/TUBSPH*

Technical / Theoretical Level

Advanced

Overview


Aim of the Problem

The shock tube problem is one of the standard problems in gas dynamics. It is a very interesting test since the exact solution is known and can be compared with the simulation results. The Smooth Particle Hydrodynamics (SPH) method, as well as the Finite Element method using the Eulerian and Lagrangian formulations served in the numerical models.

Physical Problem Description

A shock tube consists of a long tube filled with the same gas in two different physical states. The tube is divided into two parts, separated by a diaphragm. The initial state is defined by the values for density, pressure and velocity, as shown in Figures 1 and 2. All the viscous effects are negligible along the tube sides; it is also assumed that there is no motion in the beginning.

rad_ex_fig_13-1

Fig 1: Sketch of the shock tube.

rad_ex_fig_13-2

Fig 2: Initial states with discontinuities.

The initial state at time t = 0 consists of two constant states 1 and 4 with p4 > p1, density4 > density1, and V1 = V4 = 0 (table).

Table 1: Initial conditions in the shock tube.

 

High pressure side (4)

Low pressure side (1)

Pressure p

500000 Pa

20000 Pa

Velocity u

0 m/s

0 m/s

Density density

5.7487 kg/mm3

0.22995 kg/mm3

Temperature T

303 K

303 K

Just after the membrane is removed, a compression shock runs into the low pressure region, while a rarefaction (decompression) wave moves into the high pressure part of the tube. Furthermore, a contact discontinuity usually occurs.

Analysis, Assumptions and Modeling Description


Perfect Gas Modeling with RADIOSS

The hydrodynamic viscous fluid law 6 is used to describe compressed gas.

The general equation describing pressure is:

ex_13_equation

with   ex_13_equation2

where, p is the pressure, Ci are the hydrodynamic constants, En is the internal energy per initial volume, and density is the density.

Perfect gas is modeled by setting all coefficients C0, C1, C2 and C3 to zero.

Also:

C4 = C5 = Y - 1

Where, Y is the gas constant.

Then the initial internal energy, per initial volume is calculated from initial pressure:

ex_13_iie

Under the assumption Y = CST = 1.4 (valid for low temperature range), the hydrodynamic constants C4 and C5 are equal to 0.4.

Gas pressure is described by:

ex_13_gas_pressure

Parameters of material law 6 are provided in Table 2.

Table 2: Material properties of gas in law 6.

 

High pressure side (4)

Low pressure side (1)

Initial volumetric energy density (E0)

1.25x106 J/m3

5x104 J/m3

C4 and C5

0.4

0.4

Density density

5.7487 kg/mm3

0.22995 kg/mm3

 

Analytical Approach

The shock tube problem has an analytical solution of time before the shock hits the extremity of the tube [1].

rad_ex_fig_13-3

Fig 3: Schematic shock tube problem with pressure distribution for pre- and post-diaphragm removal.

Evolution of the flow pattern is illustrated in Fig 3. When the diaphragm bursts, discontinuity between the two initial states breaks into leftward and rightward moving waves, separated by a contact surface.

Each wave pattern is composed of a contact discontinuity in the middle and a shock or a rarefaction wave on the left and the right sides separating the uniform state solution. The shock wave moves at a supersonic speed into the low pressure side. A one-dimensional problem is considered.

rad_ex_fig_13-4

Fig 4: Diagram of the shock, expansion waves and contact surface.

There are four distinct zones marked 1, 2, 3 and 4 in Fig 4. Zone 1 is the low pressure gas which is not disturbed by the shock wave. Zone 2 (divided in 2 and 2' by the contact surface) contains the gas immediately behind the shock traveling at a constant speed. The contact surface across which the density and the temperature are discontinuous lies within this zone. The zone between the head and the tail of the expansion fan is noted as Zone 3. In this zone, the flow properties gradually change since the expansion process is isentropic. Zone 4 denotes the undisturbed high pressure gas.

Equations in Zone 2 are obtained using the normal shock relations. Pressure and the velocity are constant in Zones 2 and 2’.

The ratio of the specific heat constant of gas Y is fixed at 1.4. It is assumed that the value does not change under the temperature effect, which is valid for the low temperature range.

The analytical solution to the Riemann problem is indicated at t=0.4 ms. A solution is given according to the distinct zones and continuity must be checked. Evolution in Zones 2 and 3 is dependent on the constant conditions of Zone 1 and 4. The analytical equations use pressure, velocity, density, temperature, speed of sound through gas and a specific gas constant. Equations in Zone 2 are obtained using normal shock relations and the gas velocity in Zone 2 is constant throughout. The shock wave and the surface contact speeds make it possible to define the position of the zone limits.

Zone 1 – Zone 4

 

Zone 4

Zone 1

Pressure p

p4 = 500000 Pa

p1 = 20000 Pa

Velocity u

u4 = 0 m/s

u1 = 0 m/s

Density density

density4 = 5.7487 kg/mm3

density1 = 0.22995 kg/mm3

Temperature T

T4 = 303 K

T1 = 303 K

Speed of sound through gas:

ex_13_speed_of_sound

Specific gas constant:

ex_13_gas-constant

 

 

High pressure side (4)

Low pressure side (1)

a

a4 = 348.95 m/s

a1 = 348.95 m/s

R

287.049 J/(kg.K)

 

 

Zone 2

 

Analytical solution

Results at t = 0.4 ms

Pressure p

rad_ex_13_pressure

p2 = 80941.1 Pa

Velocity u

ex_13_velocity

u2 = 399.628 m/s

Density density

density2 = density2RT2

density2 = 0.5786 kg/mm3

Temperature T

ex_13_temp

T2 = 487.308 K

Shock wave speed:

ex_13_shockwave

Therefore, x2/1 = Vs * 0.4 + 500 = 765.266 mm

 

Zone 2'

 

Analytical solution

Results at t = 0.4 ms

Pressure p

p2 = p2'

p2' = 80941.1 Pa

Velocity u

u2 = u2'

u2' = 399.628 m/s

Density density

density2' = density3(x4/3)

density2' = 1.5657 kg/mm3

Temperature T

p2' = r2'RT2'

T2' = 180.096 K

Surface contact speed:  Vc - u2

Therefore, x2/2' = symbol_u2 * 0.4 + 500 = 559.85 mm

 

Zone 3

Zone 3 is defined as:

ex_13_zone3-region

where, x = 500 + X

 

ex_13_zone3A

 

 

Analytical solution

Results at t = 0.4 ms

Pressure p

ex_13_zone3-pressure

ex_13_zone3-pressure2

Velocity u

ex_13_zone3-velocity

u3 = 290.792 + 2.0833 X

Density density

ex_13_zone3-density

ex_13_zone3-density2

Temperature T

ex_13_zone3-temp

clip0843

 

Continuity verifications:

ex_13_continuity-verif

 

Finite Element Modeling with Lagrangian and Eulerian Formulations

Gas is modeled by 200 ALE bricks with solid property type 14 (general solid).

The model consists of regular mesh and elements, the size of which is 5 mm x 5 mm x 5 mm.

rad_ex_fig_13-5

Fig 5: Mesh used for Lagrangian and Eulerian approaches.

In the Lagrangian formulation, the mesh points remain coincident with the material points and the elements deform with the material. Since element accuracy and time step degrade with element distortion, the quality of the results decreases in large deformations.

In the Eulerian formulation, the coordinates of the element nodes are fixed. The nodes remain coincident with special points. Since elements are not changed by the deformation material, no degradation in accuracy occurs in large deformations.

The Lagrangian approach provides more accurate results than the Eulerian approach, due to taking into account the solved equations number.

For the ALE boundary conditions (/ALE/BCS), constraints are applied on:

Material velocity
Grid velocity

The nodes on extremities have material velocities fixed in X and Z directions. The other nodes have material and velocities fixed in X, Y and Z directions.

The ALE materials have to be declared Eulerian or Lagrangian with /ALE/MAT.

 

Smooth Particle Hydrodynamics Modeling (SPH)

The 12798 particles are distributed though a hexagonal compact net. No mesh is used.

rad_ex_fig_13-6

Fig 6: Smooth Particle Hydrodynamics modeling with hexagonal compact net.

The nominal value h0 is the distance between each particle and its closest neighbor. According to the assigned property of the part, the mass of the particles should be calculated. The mass is related to the density and the size of the net, in accordance with the following equation:

rad_ex_fig_13-6B

Where:

Particle mass of low pressure part: mp  = 1.25265x10-5 g
Particle mass of high pressure part: mp = 3.13166x10-4 g

Particle mass is specified in the SPH property set.

The scale factor of the time step is set to 0.3 in order to ensure cell stability computation.

Boundary conditions are used to introduce SPH symmetry conditions (/SPHBCS). This option is specific to the SPH modeling and consists of creating ghost particles, symmetrical to the real particles with respect to the symmetry plane.

rad_ex_fig_13-7

Fig 7: SPH symmetry planes definition.

Each symmetry condition is defined according to the plane passing through the frame origin attached to the plane and is normal in relation to the local direction of this frame.

Selected nodes and SPH symmetry condition frame along (-x) axis:

clip0853

Six symmetry planes are used:

x and (-x) symmetry conditions: SLIDE without rebound (Ilev =0)
y and (-y) symmetry conditions: SLIDE without rebound (Ilev =0)
z and (-z) symmetry conditions: TIED with elastic rebound (Ilev =1)

For the SLIDE-type condition, the material is perfectly sliding along the plane

The particles must lie on the symmetry planes at t = 0.

rad_ex_fig_13-8

Fig 8: Local direction of frame

Particles should move into the positive semi-space defined as:

sphbcs_OP

Where, O is the origin of the frame, P is a point of the plane, and clip0855 is the local direction of the frame.

Simulation Results and Conclusions


Comparison of the Finite Element Results with the Analytical Solution

Simulation results along the tube axis at 0.4 ms are shown in the following diagrams.

 

Pressure

ex_13_pressureA

 

Density

ex_13_densityA

 

Velocity

ex_13_velocityA

 

Lagrangian formulation: Scale factor = 0.1
Eulerian formulation: Scale factor = 0.5

Scale Factor Effect on Eulerian Results

Case 1: Scale factor = 0.5

Case 2: Scale factor = 0.9

 

Pressure

ex_13_pressureB

 

Density

ex_13_densityB

 

Comparison of SPH Results and Analytical Solution

Simulation results along the tube axis at 0.4 ms. Scale factor: 0.3 and 0.67.

 

Pressure – Hexagonal Net and SPH Symmetry Conditions

ex_13_pressureC

 

Density Hexagonal Net and SPH Symmetry Conditions

ex_13_densityC

 

Velocity Hexagonal Net and SPH Symmetry Conditions

ex_13_velocityC

 

Indications on computation for each formulation are given in the following table (the scale factor is set to 0.5):

 

 

Finite Element approach

SPH approach

Formulation

Lagrangian

Eulerian

SPH

Normalized CPU

1.08

1

1809

Number of cycles (normalized) up to 0.4 ms

1.42

1

3.46

(DTsca=0.5)

Pressure Distribution along Tube at 0.4 ms

ex_13_eulerian_pressure

ex_13_pressure-lang

rad_ex_fig_13-9

Fig 9: Pressure wave produced in the shock-tube at t = 4 ms for different approaches and animations regarding pressure, density and velocity

Reference

[1] J. D. Anderson Jr., Modern Compressible Flow with Historical Perspective, McGraw Hill Professional Publishing, 2nd ed., Oct. 1989.