**2. Measurement of rocks conductivity**

The method of determining the thermal conductivity of rocks and constituent fluids is to measure the amount of heat that passes through a system consisting of two discs of known rocks (quartz) and which includes the rock disc under analysis (**Figure 1**).

The system is placed in a thermostatic bath, measuring the temperature difference between the three discs.

After reaching the thermal equilibrium, the values Q1, Q2, and Q3 are almost equal and the heat transfer relation becomes [13]:

$$\mathbf{2Q\_2} \cong \mathbf{Q\_1} + \mathbf{Q\_3} \tag{2}$$

(3)

As explained in relation 1, we obtain the value of thermal conductivity:

$$k\_{\tau} = \frac{k\_q \frac{\Delta\_1}{x\_1} S\_1 + k\_q \frac{\Delta\_2}{x\_2} S\_3}{2 \frac{\Delta\_3}{x\_2} S\_2}$$

$$\begin{array}{|c|c|}\hline \\ \text{T=const.}\\ \hline \\ \text{Disc cooper} \\ \text{Disc quartz} \\ \text{Q\_1 S\_1.}\text{z} \\ \hline \\ \text{Disc cooper} \\ \hline \\ \text{Disc coecker} \\ \text{Q\_2 S\_2.}\text{z} \\ \hline \\ \text{Disc cooper} \\ \hline \\ \text{Disc quartz} \\ \text{Q\_3 S\_2.}\text{z} \\ \hline \\ \text{Disc cooper} \\ \text{T=const.} \end{array}$$

**Figure 1.** *Thermal conductivity measurement system (split bar method) [1, 9, 13].*

where *kr* is the thermal conductivity of the rock sample with cross section *S2* and thickness *z2*, *kq* is the thermal conductivity of quartz discs with cross section *S1, S3* and thickness z1, z3.

The thermal conductivity of crystalline quartz can be determined by the relation [10, 12–14]:

$$k\_q = \frac{1}{60\,\text{J} + 0.242t} \tag{4}$$

**Figure 2** shows a device for determining the thermal conductivity by the plate method, with a single rock sample.

The mathematical model for determining the thermal conductivity of rocks in the exploitation areas of oil and gas deposits starts from the equation:

$$k\_r = k\_q \left(\frac{\Delta t\_1 + \Delta t\_3}{2\Delta t\_2}\right) \left(\frac{z\_2}{z}\,\frac{s}{s\_2}\right) \tag{5}$$

where *kr* is the thermal conductivity of the rock sample with cross section *S2* and thickness *z2*, and *kq* is the thermal conductivity of quartz discs with cross section *S1, S3* and thickness z1, z3.

For ease of determinations, identical quartz dikes were made, and then:

$$\mathcal{S}\_1 = \mathcal{S}\_3 = \mathcal{S} \tag{6}$$

$$\mathbf{z}\_1 = \mathbf{z}\_3 = \mathbf{z} \tag{7}$$

Logarithming Eq. (5) we get:

$$\log k\_{\tau} = \log k\_{q} + \log \left(\Delta t\_{1} + \Delta t\_{3}\right) - \log 2 - \log \Delta t\_{2} + \log z\_{2} - \log z - \log s\_{2} \tag{8}$$

To reduce the thermal resistance to the contact between the crucible and the analyzed rock, a thin layer of vaseline is used.

Another method for determining the thermal conductivity starts from the knowledge of the mineralogical composition and porosity.

Thus, knowing that rocks from oil and gas deposits are characterized by a liquid phase (the pore space saturation fluid) and a solid phase (the mineral skeleton), the thermal conductivity of the analyzed rock depends on the thermal conductivity of the two constituent phases.

If the two phases are oriented parallel to each other (a maximum thermal conductivity value is obtained), then:

$$k\_{\text{max}} = pk\_f + (1 - p)k\_m \tag{9}$$

where *k max* is the maximum conductivity of the analyzed rock, *p* represents the porosity of the rock, *kf* tests the conductivity of the fluid phase, *km* represents the conductivity of the matrix (mineral skeleton).

If the two phases are oriented in series (which gives a minimum value of the thermal conductivity), we can write the equation of the conductivity of the deposit swarm, as:

$$\frac{1}{k\_{\min}} = \frac{p}{k\_f} + \frac{1-p}{k\_m} \tag{10}$$

For an average thermal conductivity value, the following relationship is used:

$$k = k\_f^p \bullet k\_m^{1-p} \tag{11}$$

Logarithmizing Eq. (11), we can obtain a linear relationship between the effective conductivity of the porous medium and the effective conductivity of the constituent fluid, namely:

$$
\log k = p \, \log k\_f + (1 - p) \log k\_m \tag{12}
$$

For rocks with low porosity and complex mineralogical composition, their conductivity can be obtained based on the relations:

$$\frac{1}{k\_{\min}} = \frac{V\_1}{k\_1} + \frac{V\_2}{k\_2} + \dots + \frac{V\_n}{k\_n} \tag{13}$$

$$k\_{\text{max}} = V\_1 k\_1 + V\_2 k\_1 + \dots + V\_n k\_n \tag{14}$$

where *V*1,*V*<sup>2</sup> … ,*Vn*, are the volumes of the mineral fractions, 1, 2 , … , n, and *k*<sup>1</sup> ,*k*2, … , *kn*, are the thermal conductivity of minerals.
