/MAT/LAW6 (HYDRO or HYD_VISC)
Block Format Keyword Describes a fluid material. Pressure is computed using Equation of State provided by definition of /EOS option.
Format
(1)  (2)  (3)  (4)  (5)  (6)  (7)  (8)  (9)  (10) 

/MAT/LAW6/mat_ID/unit_ID or /MAT/HYDRO/mat_ID/unit_ID or /MAT/HYD_VISC/mat_ID/unit_ID  
mat_title  
${\rho}_{i}$  ${\rho}_{0}$  
$\nu $  P_{min} 
Definitions
Field  Contents  SI Unit Example 

mat_ID  Material identifier. (Integer, maximum 10 digits) 

unit_ID  Unit Identifier. (Integer, maximum 10 digits) 

mat_title  Material title. (Character, maximum 100 characters) 

${\rho}_{i}$  Initial density. (Real) 
$\left[\frac{\text{kg}}{{\text{m}}^{\text{3}}}\right]$ 
${\rho}_{0}$  Reference density used in E.O.S (equation of
state). Default = ${\rho}_{0}={\rho}_{i}$ (Real) 
$\left[\frac{\text{kg}}{{\text{m}}^{\text{3}}}\right]$ 
$\nu $  Kinematic viscosity. (Real) 
$\left[\frac{{\text{m}}^{\text{2}}}{\text{s}}\right]$ 
P_{min}  Pressure cutoff. Default = 1.0 x 10^{20} (Real) 
$\left[\text{Pa}\right]$ 
Example (Air)
#RADIOSS STARTER
/UNIT/1
unit for mat
kg m s
#12345678910
/MAT/HYDRO/4/1
AIR
# RHO_I RHO_0
1.22 0
# Knu Pmin
1.5E5 0
/EOS/POLYNOMIAL/4/1
AIR
# C0 C1 C2 C3
0 0 0 0
# C4 C5 E0 Psh RHO_0
0.4 0.4 253300 0 1.22
/EULER/MAT/4
# Modif. factor.
0
#12345678910
#enddata
/END
#12345678910
Comments

${S}_{\mathit{ij}}=2\rho {\nu}_{\mathit{eq}}{\dot{e}}_{\mathit{ij}}$
Where,
 ${\nu}_{\mathit{eq}}=\nu $
 No turbulence
 ${S}_{ij}$
 Deviatoric stress tensor
 ${\dot{e}}_{\mathit{ij}}$
 Deviatoric strain tensor
 Equation of state for hydrodynamic pressure has to be prescribed via the /EOS card.
 In case of a linear material with a
volumetric dilatation:
${C}_{1}=\frac{E}{3\left(12\nu \right)}$ and ${C}_{4}=\frac{{\alpha}_{\nu}{C}_{1}}{\rho {C}_{\nu}T}$
${C}_{4}={C}_{5}=\gamma 1$ and ${C}_{0}={C}_{2}={C}_{3}=0$
then:(1) $$\begin{array}{ll}p\hfill & ={C}_{1}\mu +({C}_{4}+{C}_{5}\mu )E\hfill \\ \hfill & ={C}_{1}\mu +{C}_{4}(1+\mu )E\hfill \\ \hfill & ={C}_{1}\mu +{C}_{4}(1+\mu ){\rho}_{0}e\hfill \\ \hfill & ={C}_{1}\mu +{C}_{4}(1+\mu ){\rho}_{0}{C}_{\nu}T\hfill \end{array}$$(2) $$\begin{array}{ll}p\hfill & ={C}_{1}\mu +{C}_{4}\rho {C}_{\nu}T\hfill \\ \hfill & ={C}_{1}\mu +{\alpha}_{\nu}T\hfill \end{array}$$If $p=cst=0$ , then ${C}_{1}\mu +{\alpha}_{\nu}T=0$ l=; so $\mu =\frac{{\alpha}_{\nu}T}{{C}_{1}}$
Where, $\mu $
 Dilatation coefficient
 $\mu <0$
 Dilatation
 All thermal data ( ${\rho}_{0}{C}_{p},{T}_{0},A,\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}B$ ) can be defined with keyword /HEAT/MAT.
 If using LAW6 coupled with LAW37 for
liquid phase (without gas phase), the
compatibility of the liquid EOS is:
 $\Delta {P}_{1}={C}_{1}\mu $ for /MAT/LAW37 (BIPHAS)
 $p={C}_{0}+{C}_{1}\mu +{C}_{2}{\mu}^{2}+{C}_{3}{\mu}^{3}+({C}_{4}+{C}_{5}\mu )E$ for LAW6 via a polynomial EOS defined as in the Example above.
with ${C}_{0}={C}_{1}={C}_{2}={C}_{3}={C}_{4}={C}_{5}=E=0$
then, $p={C}_{1}\mu $
 If using LAW6 coupled with LAW37 for gas
phase (without liquid phase), the compatibility of
the gas EOS is:
 $PV\gamma =const.$ for LAW37
 $p=\left(\gamma 1\right)\left(\mu +1\right)E$ for LAW6, via the /EOS/IDEALGAS equation of state.
Where, $E$ is the energy per unit volume.