PSI - Issue 6

Yurii Meshcheryakov / Procedia Structural Integrity 6 (2017) 109–114 Author name / Structural Integrity Procedia 00 (2017) 000 – 000

5

113

Thefirstdiffusioncoefficientisseentodependontherate dispersion. Thiscoefficientcharacterizesa mean changemesoparticle velocity because of action of random stress fields.If the averaging is carried out for the time of relaxation (see Eq. (16)) the first diffusion coefficient defines the change ofmeanvelocity Δ u 1 (velocity defect): 2 1 u D    (17) This means,thatsought non-equilibrium part of particle velocity is a sequence of change of the particle velocity dispersion in the velocity space: of change of the velocity 1 2 u 

2

D

1 2



. (18)

pl

n u



u



Let usconsider the second item in the right hand side of Eq.(3). Incommoncase, thenormalstress  can also be subdivided by equilibrium and non-equilibrium parts:  = р н    , (19) where р  isthestressattheHugoniotadiabate р  = 0 р C u  , (20)

Substitutionof (18) and (20) into Eq. (3) yields:

pl

2

р u

2

2 D x u

2



С

2    

2 x t 

2

2 u u

 

2 D      





t    

0 pl p C u

2

C

2

C

 

0

0 (21) Thefirstandthirditemsintherighthandsideofthisequationwhicharerelatedtoequilibriumimpulsetransportationarecancell edsinceprocessofimpulsetransportationattheHugoniot adiabatic curveis a steady. ThenEq. (21) takestheform: 2 2 x   t 0 2 x  + 2 -   - x .

2       - D u

2 D       . x

2

2 u u

2  

 

t    

2

2

С

С

 

(22)

0

0

2 x   t

2

2 x  

Thepartialderivativesin the right hand side of equation can be written as the following:   2 D t x           = 2 D D u t u x            = 2 u D D t u x         + 2 u D D t u x         ,

  2 D

x      

  

(23)

= 2

u



  

  

     + 2 u

D D x u u       

  

D D 

u D D x u u 





= 2

,

(24)

u



x

u u    2 D





2 2       x

  

u x

 

    .

  

 

2 D D x u u     2

2

D D x u u       

  

D D 



+2 u

= 2

 + 4

(25)

u



2

x

u u 



u



Taking into account Eqs (23) – (25) the Eq. (22) takes the form:   2 2 2 0 2 2 1 D D u u С u u x t                2 2 2 D D u x u u            + 4 u x   D D x u u          

+