Page 49 - Physical Principles of Sedimentary Basin Analysis
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2.12 Fourier’s law and heat conductivity         31

                                           k 1 h 1 + k 2 h 2
                                      k   =                                    (2.98)
                                            h 1 + h 2
                                               h 1 + h 2
                                     k ⊥ =                                     (2.99)
                                           (h 1 /k 1 ) + (h 2 /k 2 )
            respectively. The anisotropy is expressed by the ratio of the permeabilities R = k   /k ⊥ .
            (a) Show that R can be written as
                                      (h 1 k 1 + h 2 k 2 )(h 1 k 2 + h 2 k 1 )
                                  R =                        .                (2.100)
                                             (h 1 + h 2 ) 2
            (b) Use the result from (a) to rewrite R as
                                        (1 + R h R k )(1 + R h /R k )
                                    R =                                       (2.101)
                                              (1 + R h ) 2
            where R h = h 2 /h 1 and R k = k 2 /k 1 . The anisotropy ratio R depends only on the thickness
            ratio h 2 /h 1 and permeability ratio k 2 /k 1 as we would expect. The anisotropy should be
            independent of both length scale and permeability scale of the formation.
            (c) Verify that R = 1for R k = 1 regardless of R h .



                              2.12 Fourier’s law and heat conductivity

            The heat flux q through a solid material is given by Fourier’s law
                                                  ∂T
                                           q =−λ                              (2.102)
                                                  ∂x
            as the product of the heat conductivity λ of the material and the temperature gradient in the
                                                                               2
            material, ∂T/∂x. The heat flux is measured as energy (J) through a cross-section (m ) per
            time (s), and it is therefore given in units W m −2 . The thermal gradient has units C/m,
                                                                              ◦
            which implies that units for the thermal conductivity is W m −1  K −1 .
              The heat conductivities of the sediments and rocks show a large variability because of
            pore fluids and different mineral compositions of the solid skeleton (see Table 2.2). There
            are also cases of heat conductivity measurements on the same (or nearly the same) rocks
            by different laboratories that are quite different. For example Midttømme et al. (1998)
            measured lower heat conductivities than Bloomer (1981) for claystones and mudstones
            from the UK, and Norden and Förster (2006) measured higher heat conductivities than in
            previous studies for sandstones in the Northeast German Basin.
              The heat conductivity of minerals are much better constrained than those for rocks, since
            they have a specific crystal structure and a chemical formula. Clauser and Huenges (1995)
            have collected the heat conductivities for a large number of minerals. The heat conduc-
            tivity of water is considerably lower than for most minerals, which makes the porosity an
            important controlling factor for the average heat conductivity of fluid saturated rocks and
            sediments. The average bulk heat conductivity is often related to the heat conductivities
            of the solid and the fluid using the porosity as a normalizing weight. A commonly used
            average is the geometric mean
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