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Example 43 - Perfect Gas Modeling with Polynomial EOS

Example 43 - Perfect Gas Modeling with Polynomial EOS

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Example 43 - Perfect Gas Modeling with Polynomial EOS

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ex43_perfect_gas_model

Summary


Polynomial EOS is often used by RADIOSS to compute hydrodynamic pressure. It is cubic in compression and linear in expansion.

ex43_polynomial_eq

where,

ex43_eq1   (1)   and ex_13_equation2     (2)

Material law 6 (/MAT/HYDRO) uses this equation to compute hydrostatic pressure. It is possible to consider absolute values or relative variation (Table 1). This example shows how to build material control cards for each of the following cases:

Case

Mathematical model

Pressure

Energy

1

absolute

absolute

2

relative

absolute

3

relative

relative

4

absolute

relative

Table 1: Modeling formulation for perfect gas with /MAT/HYDRO

A simple test of compression/expansion is made to compare these formulation outputs with theoretical results.

 

Title

Perfect Gas Modeling with
Polynomial EOS

ex43-1

Number

43.1

Brief Description

Polynomial EOS is used to model perfect gas. Pressure or energy can be absolute values or relative. Material law 6 (/MAT/HYDRO) is used to build material cards for each of these cases.

Keywords

Perfect gas
Polynomial EOS
Absolute/Relative formulations
Pressure shift

RADIOSS Options

Hydrodynamic fluid material (/MAT/LAW6 (HYDRO))
Imposed displacement (/IMPDISP)
Boundary conditions (/ALE/BCS)

Compare to / Validation method

Input File

Model 1: <install_directory>/demos/hwsolvers/radioss/43_perfect_gas_polynomial_eos/01-Pabsolute_Eabsolute/*

Model 2: <install_directory>/demos/hwsolvers/radioss/43_perfect_gas_polynomial_eos/02-Prelative_Eabsolute/*

Model 3: <install_directory>/demos/hwsolvers/radioss/43_perfect_gas_polynomial_eos/03-Prelative_Erelative/*

Model 4: <install_directory>/demos/hwsolvers/radioss/43_perfect_gas_polynomial_eos/04-Pabsolute_Erelative/*

Technical / Theoretical Level

Beginner

Overview


Aim of the Problem

The purpose of this example is to plot numerical pressure, internal energy, and sound speed for a perfect gas material law. Comparison to theoretical results is made.

Physical Problem Description

This test consists with an elementary volume of perfect gas undergoing spherical expansion and compression.

ex43_cube

Initial conditions are listed below:

P0 = 1e5 Pa

V0 = 1000 m3

density0 = 1.204 kg/m3

symbol_u0 = 0

The fluid will be assumed to be a perfect gas. Volume is changed in the three directions to consider a pure compression (-1 < symbol_u <  0) followed by an expansion of matter (0 < symbol_u). See Figure 1.

This test will be modeled with a single ALE element (8 node brick) and polynomial EOS.

Evolutions of pressure, internal energy and sound speed will be compared between numerical output and theoretical results.

ex43_elementary_volume_change

Fig 1: Elementary volume change. Length is modified with /IMPDISP card; its influences on V and symbol_u are plotted.

Analysis, Assumptions and Modeling Description


RADIOSS Options Used

Nodes on each of the faces are moved with imposed displacement (/IMPDISP).

Boundary nodes are defined as Lagrangian with the /ALE/BCS card.

Element pressure, density and internal energy density are saved in the Time History file.

 

Polynomial EOS

Polynomial EOS is used in material law 6 (/MAT/HYDRO) to compute hydrodynamic pressure. It is cubic in compression and linear in expansion.

ex43_polynomial_eq

Where, P is the hydrodynamic pressure.

ex43_eq1       (1)

and

ex_13_equation2     (2)

 

ex43_coefficient are called hydrodynamic coefficients and they are input flags. Hypothesis on the material behavior allows determining of these coefficients:

General case corresponds to Mie-Guneisen EOS (see Appendix C of the Theory Manual)
Incompressible gas
Linear elastic material
Perfect gas

This example is focused only on Perfect Gas modeling.

 

Theoretical Results

The purpose of this section is to plot pressure, internal energy, and sound speed in function of the single parameter V or symbol_u.

1.Pressure

Perfect gas pressure is given by:

  (3)

Then,

ex43_eq2

RADIOSS assumes the hypothesis of an isentropic process to compute the change in internal energy:

dEint = -PdV

This theory gives the following differential equation:

This has the form and the general solution is:

Pressure is also polytropic:

(4)

Here, Y is the material constant (ratio of heat capacity). For diatomic gas Y =1.4. Air is made mainly of diatomic gas, so set gamma to 1.4 for air.

2.Internal Energy

Equations (3) and (4) lead to the immediate result:

ex43_internal_energy

3.Sound Speed

Perfect gas sound speed is:

ex43_perfect_gas_sound_speed2 (5)

Equation (4) gives its expression in term of volume:

ex43_perfect_gas_sound_speed3

The theoretical results are listed in the table below. Pressure, internal energy, and sound speed are expressed both in function of V and symbol_u.

Pressure (Pa)

Internal Energy Density (J)

Sound Speed (m/s)

PREF(V)

PREF(symbol_u)

densityeREF(V)

densityeREF(symbol_u)

cREF(V)

cREF(symbol_u)

ex43_Pa_V

ex43_Pa_u

ex43_Pa_V2

ex43_Pa_u2

ex43_ms_V

ex43_ms_u

 

Corresponding plots are shown below:

ex43_perfect_gas_pressure

Fig 2: Perfect Gas Pressure

ex43_perfect_gas_internal_energy

Fig 3: Perfect Gas Internal Energy

ex43_perfect_gas_sound_speed

Fig 4: Perfect Gas Sound Speed

 

Modeling Methodology

A single ALE brick element is used. Material is confined inside the element by defining brick nodes as Lagrangian. For each face, displacement is imposed on the four nodes along the normal.

Material law 6 (/MAT/HYDRO) is used and describes the hydrodynamic viscous fluid material.

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

(10)

/MAT/LAW6/mat_ID/unit_ID or /MAT/HYDRO/mat_ID/unit_ID

mat_title

 

 

 

 

 

 

 

ratio

 

 

 

 

 

 

 

 

C0

C1

C2

C3

 

 

Pmin

Psh

 

 

 

 

 

 

C4

C5

E0

 

 

 

 

 

Pressure Shift

Material law 6 introduces flag Psh which allows shifting computed pressure in the polynomial equation of state:

ex43_pressure_shift

RADIOSS Engine shifts C0 flag and computed pressure P(symbol_u ,E) with an offset of -Psh.

 

Minimum Pressure

ex43_min_pressure

The theoretical value is Pmin = 0 Pa (absolute pressure) with a default value of -1030, to accept a negative value in relative pressure formulation.

This flag has to be manually offset with -Psh.

Simulation Results and Conclusions


Material Control Cards

Material is supposed to be a perfect gas. The following cases have been investigated:

Case 1: Both Pressure and Energy are absolute values:
Case 2: Pressure is relative and Energy is absolute:
Case 3: Both Pressure and Energy are relative:
Case 4: Pressure is absolute and Energy is relative:

 

hmtoggle_plus1Case 1: Both Pressure and Energy are absolute values
1.Equation of State

Equation of state can be written:

ex43_eos_case1

with

ex43_eos_case1A

Expanding this expression and identifying the polynomial coefficients leads to:

where,

ex43_eos_case1B

2.Corresponding Input

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

(10)

/MAT/LAW6/mat_ID/unit_ID or /MAT/HYDRO/mat_ID/unit_ID

AbsolutePRESSURE_AbsoluteENERGY

 

 

 

 

 

 

 

ratio

 

 

 

 

 

 

 

 

0

0

0

0

 

 

0

0

 

 

 

 

 

 

C4 = Y - 1

C5 = Y - 1

ex43_input_eq

 

 

 

 

3.Output Results

Time History

Measure

Initial Value

Unit

/TH/BRICK  (P)

P

P0

Pressure

/TH (IE)

Eint (= E x V0)

E0V0

Energy

/TH/BRICK (IE)

Eint / V

E0

Pressure

4.Comparison with Theoretical Result

Numerical result for perfect gas pressure is given by time history. Element time history (RADIOSS /TH/BRICK) allows displaying it. This result is compared to a theoretical one. Curves are superimposed.

ex43_numerical_pressure_model1

Fig 5: Numerical pressure, model 1:

Internal energy can be obtained through two different ways. The first one is internal energy density (Eint / V) recorded by element time history (RADIOSS /TH/BRICK). The second one is the internal energy from the global time history ex43_global_time_history because the model is composed of a single element.

ex43_numerical_internal_energy_model1

Fig 6: Numerical internal energy, model 1:

hmtoggle_plus1Case 2: Pressure is relative and Energy is absolute
1.Equation of State

Equation of state for a perfect gas is:

ex43_eos_case2

Calculating Pressure from a reference one provides relative pressure:

ex43_eos_case2A

Expanding this expression and identifying with polynomial coefficients leads to:

symbol_triP(symbol_u,E) = P(symbol_u,E) = Psh = -Psh + (C4 + C5symbol_u)E

where,

ex43_eos_case1B

2.Minimum Pressure

ex43_eos_case2B

Then, the minimum pressure must be set to a non-zero value Pmin = -P0.

3.Corresponding Input

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

(10)

/MAT/LAW6/mat_ID/unit_ID or /MAT/HYDRO/mat_ID/unit_ID

RelativePRESSURE_AbsoluteENERGY

 

 

 

 

 

 

 

ratio

 

 

 

 

 

 

 

 

0

0

0

0

 

 

-P0

P0

 

 

 

 

 

 

C4 = Y - 1

C5 = Y - 1

ex43_input_eq

 

 

 

 

4.Output Result

Time History

Measure

Initial Value

Unit

/TH/BRICK  (P)

symbol_triP

0

Pressure

/TH (IE)

Eint (= E x V0)

E0V0

Energy

/TH/BRICK (IE)

Eint / V

E0

Pressure

5.Comparison with Theoretical Result

Element time history (/TH/BRICK) is the pressure relative to Psh. The resulting curve is then shifted with Psh value and starts from 0.

ex43_numerical_pressure_model2

Fig 7: Numerical pressure, model 2:

Internal energy can be obtained through two different ways. The first one is internal energy density (Eint / V) recorded by element time history (/TH/BRICK). The second one is the internal energy from the global time history ex43_global_time_history because the model is composed of a single element.

ex43_numerical_internal_energy_model2

Fig 8: Numerical internal energy, model 2:

hmtoggle_plus1Case 3: Both Pressure and Energy are relative
1.Equation of State

Equation of state for a perfect gas is:

ex43_eos_case2

Initial internal energy can be introduced:

ex43_eos_case3

Calculating pressure from a reference one provides:

P(symbol_u,E) - P0 = symbol_triP = (Y - 1)(1 + symbol_u)(symbol_triE + E0) - P0

Where,

ex43_eos_case3A

Expanding this expression and identifying with polynomial coefficients leads to:

symbol_triP(symbol_u,symbol_triE) = P(symbol_u ,E) - Psh = C0 - Psh + C1symbol_u + (C4 + C5symbol_u)symbol_triE

where,

C0 = C1 = E0(Y - 1)

C4 = C5 = Y - 1

symbol_triE0 = 0

Psh = P0

2.Minimum Pressure

ex43_eos_case2B

The minimum pressure must be set to a non-zero value Pmin = -P0

3.Corresponding Input

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

(10)

/MAT/LAW6/mat_ID/unit_ID or /MAT/HYDRO/mat_ID/unit_ID

RelativePRESSURE_RelativeENERGY

 

 

 

 

 

 

 

ratio

 

 

 

 

 

 

 

 

E0(Y - 1)

E0(Y - 1)

0

0

 

 

-P0

P0

 

 

 

 

 

 

C4 = Y - 1

C5 = Y - 1

0

 

 

 

 

4.Output Results

Time History

Measure

Initial Value

Unit

/TH/BRICK  (P)

symbol_triP

0

Pressure

/TH  (IE)

symbol_triEint (= symbol_triE x V0)

0

Energy

/TH/BRICK (IE)

symbol_triEint /  V

0

Pressure

5.Comparison with Theoretical Result

Element time history (/TH/BRICK) is the pressure relative to Psh. The resulting curve is then shifted with Psh value and starts also from 0.

ex43_numerical_pressure_model3

Fig 9: Numerical pressure, model 3:

Internal energy can be obtained through two different ways. The first one is internal energy density (Eint / V) recorded by element time history (/TH/BRICK). The second one is the internal energy from the global time history ex43_global_time_history because the model is composed of a single element. This numerical internal energy is relative to its initial value; it is shifted with the E0V0 value from the absolute theoretical one and also starts from 0.

ex43_numerical_internal_energy_model3

Fig 10: Numerical internal energy, model 3:

hmtoggle_plus1Case 4: Pressure is absolute and Energy is relative
1.Equation of State

Equation of state for a perfect gas is:

ex43_eos_case2

Initial internal energy can be introduced:

ex43_eos_case3

Which leads to:

P(symbol_u,E) = (Y -1)(1 + symbol_u)(E0 + symbol_triE)

Expanding this expression and identifying with polynomial coefficients leads to:

P(symbol_u,E) = C0 + C1symbol_u + (C4 + C5symbol_u)symbol_triE

Where,

C0 = C1 = E0 (Y - 1)

C4 = C5 = Y - 1

2.Corresponding Input

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

(10)

/MAT/LAW6/mat_ID/unit_ID or /MAT/HYDRO/mat_ID/unit_ID

AbsolutePRESSURE_RelativeENERGY

 

 

 

 

 

 

 

ratio

 

 

 

 

 

 

 

 

E0(Y - 1)

E0(Y - 1)

0

0

 

 

0

0

 

 

 

 

 

 

C4 = Y - 1

C5 = Y - 1

0

 

 

 

 

3.Output Results

Time History

Measure

Initial Value

Unit

/TH/BRICK  (P)

P

P0

Pressure

/TH  (IE)

symbol_triEint (= symbol_triE x V0)

0

Energy

/TH/BRICK (IE)

symbol_triEint /  V)

0

Pressure

4.Comparison with Theoretical Result

Element time history (/TH/BRICK) gives absolute pressure. This result is compared to a theoretical one. Curves are superimposed.

ex43_numerical_pressure_model4

Fig 11: Numerical pressure, model 4:

Internal energy can be obtained through two different ways. The first one is internal energy density (ΔEint / V) recorded by element time history (/TH/BRICK). The second one is the internal energy from the global time history ex43_global_time_history2 because the model is composed of a single element. This numerical internal energy is relative to its initial value; it is shifted with the E0V0 value from the absolute theoretical one and also starts from 0.

ex43_numerical_internal_energy_model4

Fig 12: Numerical internal energy, model 4:

Sound Speed and Time Step

Material law 6 computes sound speed through the usual expression for fluids:

ex43_soundspeed_eq1

It can be written in function of symbol_u:

ex43_soundspeed_eq2

Then,

ex43_soundspeed_eq3

The total differential of P in terms of internal energy E and symbol_u is:

ex43_soundspeed_eq4

In case of an isentropic transformation (reversible and adiabatic), the change of internal energy Eint with volume V and pressure P is given by:

dEint = -PdV

Using relation which links Eint and E leads to:

ex43_soundspeed_eq5

can be expressed in terms of volume ratio:

ex43_volume_ration_eq

its variation in function of the volume change is also:

ex43_soundspeed_eq6

Change in internal energy per unit volume E is then:

ex43_soundspeed_eq7

ex43_soundspeed_eq8

Finally, the sound speed is given by:

ex43_soundspeed

(5)

 

This expression computes the sound speed for a given equation of state P(symbol_u ,E). In the case of perfect gas, it was shown that for each type of formulation (absolute or relative), EOS can be written:

P(symbol_u,E) = C0 + C1symbol_u + (C4 + C5symbol_u)E

Equation (5) is used to compute sound speed:

ex43_compute_soundspeed

ex43_compute_soundspeed2

ex43_compute_soundspeed3

(6)

 

This calculation is then applied for each of the four cases.

Numerical Sound Speed vs. Theoretical Expression

Case

C0

C1

C4

C5

c2 from Eq (5)

Comparison with theoretical value

1

0

0

Y - 1

Y - 1

ex43_case1and2_soundspeed

c = cREF

2

0

0

Y - 1

Y - 1

ex43_case1and2_soundspeed

c = cREF

3

E0(Y - 1)

E0(Y - 1)

Y - 1

Y - 1

ex43_case3and4_soundspeed

c = cREF

4

E0(Y - 1)

E0(Y - 1)

Y - 1

Y - 1

ex43_case3and4_soundspeed

c = cREF

For each of the four formulations, the computed sound speed by RADIOSS is the same as the theoretical one. Time step and cycle number are also not affected.