Page 53 - Physical Principles of Sedimentary Basin Analysis

2.12 Fourier’s law and heat conductivity 35
Table 2.5. Density, heat conductivity and specific heat capacity for common minerals in
sediments. The data are taken from Goto and Matsubayashi (2008).

Density Thermal conductivity Specific heat Thermal diffusivity
Mineral kg/m3 W m −1 K −1 Jkg −1 K −1 10 −6 2 −1
m s
Quartz 2648 7.69 741 3.92
Albite 2620 2.20 776 1.08
Anorthite 2760 1.68 745 0.82
Orthoclase 2570 2.32 707 1.28
Muscovite 2831 2.32 796 1.03
Illite 2660 1.85 808 0.86
Smectite 2608 1.88 795 0.91
Chlorite 2800 5.15 818 2.25
Calcite 2710 3.59 820 1.62
Seawater 1024 0.59 3993 0.15
Mud (grain) 2731 3.40 758 1.64

where V tot = V i is the total volume of the solid. The average density is therefore calcu-
lated with respect to the volume fraction V i /V tot of each mineral. The density of common
minerals are ∼2600 kg m −3 according to Table 2.5, which is a fair estimate for the average
density of a sediment matrix. The average specific heat capacity is calculated similarly, but
with respect to the mass fraction m i /m tot of each mineral in the bulk solid

c i m i m i
= (2.108)
c av = c i
m i m tot

where m tot = m i is the total mass. Table 2.5 shows that a typical average specific heat
capacity is 800 J kg −1 K −1 .
We notice from Table 2.5 that water has a heat conductivity that is less than half the heat
conductivity of minerals, and that it has a specific heat capacity that is about five times
higher than for minerals. Water has less than half the density of the minerals and the heat
capacity per unit volume for water is therefore approximately double that for minerals.
The porosity is therefore an important parameter for a sediment’s ability to conduct and
store heat.
Exercise 2.22 The temperature at the basin surface and at the depth 2000 m is mea-
◦
◦
suredtobe0 C and 90 C, respectively. The heat flux through the sedimentary column
is q = 0.035 W m −2 . What is the average heat conductivity of the interval?
Exercise 2.23 How much does the average heat conductivity given by function (2.103)
change when the porosity is changed by φ?
Exercise 2.24 (a) Show that a linear average heat conductivity function

λ(φ) = λ s · (1 − c φ) (2.109)