Question

How to derive entropy transport equation from heat equation?

Answer 1

Suppose I have heat equation:

ρ(∂t+(u⋅∇))T=−∇⋅R,ρ(∂t+(u⋅∇))T=−∇⋅R,

where RR - some vector and TT - temperature.

How to get the equation for entropy SS in form of

ρT(∂t+(u⋅∇))S=−∇⋅R(2)(2)ρT(∂t+(u⋅∇))S=−∇⋅R

from (1)(1)?

It seems that I need to use some thermodynamical relations like

dT=(∂T∂S)VdS+(∂T∂V)SdV,dT=(∂T∂S)VdS+(∂T∂V)SdV,

and then to transform derivatives (∂T∂S)V=TcV(∂T∂S)V=TcV, (∂T∂V)S=−TcV(∂p∂T)V(∂T∂V)S=−TcV(∂p∂T)V, but I don't know how to get an explicit form of (2)(2).

Edit. For getting it, I've returned to definition of temperature, T≡⟨m(v−u)22⟩=ET≡⟨m(v−u)22⟩=E. Then by using relations

dE=Tds−pdV=TdS+pρ2dρ,∂tρ+∇(ρu)=0dE=Tds−pdV=TdS+pρ2dρ,∂tρ+∇(ρu)=0


I've got an equation for entropy.

Answer 2

like

dT=(∂T∂S)VdS+(∂T∂V)SdV,dT=(∂T∂S)VdS+(∂T∂V)SdV,

and then to transform derivatives (∂T∂S)V=TcV(∂T∂S)V=TcV, (∂T∂V)S=−TcV(∂p∂T)V(∂T∂V)S=−TcV(∂p∂T)V, but I don't know how to get an explicit form of (2)(2).

Edit. For getting it, I've returned to definition of temperature, T≡⟨m(v−u)22⟩=ET≡⟨m(v−u)22⟩=E. Then by using relations

dE=Tds−pdV=TdS+pρ2dρ,∂tρ+∇(ρu)=0dE=Tds−pdV=TdS+pρ2dρ,∂tρ+∇(ρu)=0