Page 86 - Physical Principles of Sedimentary Basin Analysis
P. 86

68                  Linear elasticity and continuum mechanics

                       0                                5000

                                                        4000
                                                                      ζ* = ζ*(x)
                     2000
                                                        3000
                   z [m]   4000                        ζ [m]
                                                        2000

                     6000
                                                        1000

                     8000                                 0
                        0    5000   10000  15000  20000    0     5000  10000  15000  20000
                                    x [m]                              x [m]
                 Figure 3.17. Left: A 2D sedimentary basin is shown in the xz-coordinate system. Right: The same
                 basin as to the left is shown as net (porosity-free) thicknesses measured from the basement. The
                                                              ∗
                 total net thickness of the basin along the x-axis is denoted ζ . The position of a grid node in the
                 xζ-coordinate system sticks to the sedimentary grains, and is there constant regardless of deposition
                 of more sediments, compaction or eventual water depth changes.


                                        3.18 Conservation laws in 1D
                 Conservation of quantities like mass, energy or momentum are the most fundamental laws
                 of nature. We will start by showing how conservation of mass is expressed in 1D by con-
                 sidering the flow of mass through a box located from x to x + x with a cross-section area
                 A (see Figure 3.18). The mass flowing into the box (from the left side) during a time step
                  t, and the mass flowing out of the box (at the right side) during the same time step are

                                      m in = v(x, t) tA  (x, t)                    (3.154)
                                     m out = v(x +  x, t) tA  (x +  x, t)          (3.155)
                 respectively, where   is the density. The mass gain (or loss) becomes
                                                         ∂
                                      m = m in − m out ≈−   v   V  t               (3.156)
                                                        ∂x
                 where  V = A  x is the volume of the box. The increase (or decrease) of mass during
                 the time step  t is the difference between the mass m 2 in the box at time t +  t and the
                 mass m 1 in the box at time t. These masses are

                                            m 1 =  (x c , t) V                     (3.157)
                                            m 2 =  (x c , t +  t) V                (3.158)

                 respectively, where x c is the coordinate of the box center, and the difference becomes
                                                        ∂
                                         m = m 2 − m 1 ≈    V  t.                  (3.159)
                                                         ∂t
   81   82   83   84   85   86   87   88   89   90   91