Page 206 - Physical Principles of Sedimentary Basin Analysis
P. 206

188                             Heat flow

                 (Notice that the symmetry of the stress tensor was used between the second and the third
                 equality.) We therefore have

                                d
                                  (W + Q) =    (b · v +∇ · (σv) −∇ · q + S) dV     (6.314)
                                dt           V
                 where σ ij v j is written as the matrix–vector product σv. Combining the left-hand
                 side (6.312) and the right-hand side (6.314) of the first law of thermodynamics (6.306)
                 gives the temperature equation

                               De               1 Dv  2
                                  +∇ · q − S =−        +∇ · (σv) + b · v
                               dt               2   dt

                                                     Dv i  ∂σ ij         ∂v i
                                            =−v i        −     − b i  + σ ij
                                                     dt    ∂x j          ∂x j
                                                 ∂v i
                                            = σ ij  .                              (6.315)
                                                 ∂x j
                 Newton’s second law is identified in the second step, and the contribution from work done
                 on the system is the term σ ij ∂v i /∂x j .
                   We are seeking an equation for temperature. The next step is therefore to replace the
                 internal energy by an expression in temperature, and to do that some thermodynamics is

                 needed. We begin by splitting the stress σ ij into deviatoric stress σ and pressure p:
                                                                      ij

                                        σ ij = σ + σ m δ ij = σ − p δ ij           (6.316)
                                              ij
                                                          ij
                                                                                1
                 where the pressure is defined as the negative of the mean stress σ m =  σ ii . (See
                                                                                3

                 Section 3.10 for deviatoric stress.) The right-hand side (6.315) becomes σ ∂v i /∂x j when
                                                                            ij
                 −p ∂v i /∂x i is moved to the left-hand side. On the left-hand side we now look at the change
                 in the internal energy and work during a time interval dt
                                                               1

                                        de + pd  ii =   de + pd                    (6.317)

                                                 =  (de + pdν)                     (6.318)
                                                 =   Tds                           (6.319)
                 where d  ii = (∂v i /∂x i ) dt is the strain increment during the time step, dν is the change in
                 specific volume and ds is the change in specific entropy. We will look more closely at the
                 steps above, and by bringing to mind equation (3.13) we see that the strain increment d  ii
                 is the same as a relative volume change dV/V . Specific volume is defined as volume per
                 unit mass, and it is therefore the same as inverse density, ν = 1/ , which gives
                                                             1
                                     d  ii  1 dV     d
                                         =       =−     = d     = dν.              (6.320)
                                             V         2
                 The relation dV/V =−d /  follows from mass conservation in the volume V since
                 m =  V is constant. The pressure p was needed in order to introduce the change in
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