### Heat transfer model

The heat transfer phenomena between the hot rock and the water circulating in a deep borehole heat exchanger takes place by conduction and convection. In a previous study (Alimonti and Soldo 2016) the model was developed using an analytical solution of the Fourier equation and it was implemented in a C computation code.

In the following paragraphs, the equations of the model are reported.

#### Rock temperature

Using a ground surface temperature *T*
_{o} of 25 °C, the rock temperature at the depth *z* has been evaluated using the following relation:

$$T_{\text{w}} \left( z \right) = T_{\text{o}} + {\text{GT}} \cdot z$$

(1)

where GT is the geothermal gradient of the site.

#### Heat transfer in the downward pipe

In the model, the heat transfer from the rock is due mainly to the conduction; no convection takes place. Then the heat moves still by conduction from the reservoir to the external casing of the WBHX, which is separated from the rock wall by a layer of cement. The convection takes place in the heat transfer between the casing and the water in the borehole heat exchanger.

The heat acquired by the water in the downward pipe is directly proportional to the length of the pipe (Δ*z*), the external radius of the borehole (*r*
_{w}), the total heat exchange coefficient (*k*
_{t}), the difference between the rock temperature at depth *z* (*T*
_{w}) the temperature of the fluid in the outer pipe (*T*
_{f,down}). The thermal power can be calculated with the following relation:

$$\dot{Q}_{\text{down}} = 2\pi r_{\text{w}} k_{\text{t}} \left( {T_{\text{w}} \left( z \right) - T_{\text{f,down}} } \right)\Delta z$$

(2)

The total heat transfer coefficient is the reciprocal of the total thermal resistance, which can be expressed as:

$$R_{\text{t}} = R_{\text{a}} + R_{\text{c}} + R_{\text{s}}$$

(3)

The three terms in Eq. (3) are the thermal resistance due to the convection into the annular space of the WBHX (*R*
_{a}), the thermal resistance due to the conduction through the casings (*R*
_{c}); the thermal resistance due to the conduction in the rock (*R*
_{s}).

To evaluate the conductive thermal resistance in the rock the thermal conductivity of the rock *λ*
_{s}, the thermal diffusivity of the rock *a*
_{s}, the external radius of the well *r*
_{w}, and the elapsed time since the start *t*′ must be known:

$$R_{\text{s}} = \frac{1}{{2\lambda_{\text{s}} }}\ln \frac{{2\sqrt {a_{\text{s}} t^{{\prime }} } }}{{r_{\text{w}} }}$$

(4)

This equation arises from the analytical solution of heat transfer equation given in Carslaw and Jaeger (1959). The term \(2\sqrt {a_{\text{s}} t^{{\prime }} }\) represents the travelling distance of the temperature front. At distance >\(2\sqrt {a_{\text{s}} t^{\prime } }\) in the rock the temperature is undisturbed and equal to *T*
_{w}.

The conductive thermal resistance of the rock (Fig. 2) increases very rapidly in the first year of operation. After the second year of work, *R*
_{s} is 1.1 m^{2}K/W and increases up to 1.2 m^{2}K/W after 10 years. This behavior is due to the exponential growth of the interested volume of rock by the heat transfer.

To evaluate the thermal resistance *R*
_{a} the radius of the external casing *r*
_{c} and the convective heat transfer coefficient *h* must be known:

$$R_{\text{a}} = \frac{1}{{2 \cdot r_{\text{c}} \cdot h}}$$

(5)

The convective heat transfer coefficient is calculated using the definition of the Nusselt number and a form of Dittus-Boelter equation, having assumed turbulent flow inside tubes (Davis and Michaelides 2009; Bennett and Myers 1982)

$$h = \frac{{0.023 \cdot \lambda_{\text{f}} \cdot Re^{0.8} \cdot Pr^{0.4} }}{{2 \cdot r_{\text{c}} }}$$

(6)

The pipes are steel, which has a high thermal conductivity, so in the model the thermal resistance of the casings has been neglected compared to the rock resistance.

The effect of the cementing ring was studied by evaluating the mutual thermal resistance (to heat conduction) of the concrete and of the rock. The results show that, despite the thermal conductivity of the concrete being less than that of the rock, given the small thickness of the grouting layer with respect to the extension of the rock, the presence of cement is negligible. The heat resistance of the cement is 0.0233 W/m^{2}K instead for the rock is equal to 0.11 W/m^{2}K. Therefore, the diameters of the casing and of the well are assumed to be similar.

The total heat exchange coefficient can be determined as:

$$\frac{1}{{k_{\text{t}} }} = \frac{{D_{\text{c}} }}{{2 \cdot \lambda_{\text{s}} }} \cdot \ln \frac{{4\sqrt {a_{\text{s}} t^{\prime } } }}{{D_{\text{c}} }} + \frac{1}{h}$$

(7)

#### Heat transfer in the upward pipe

The heated water enters in the internal pipe and flows upward exchanging the heat with the wall of the composite pipe.

The thermal power is proportional to the radius of the inner tube (*r*
_{i}), the overall heat transfer coefficient (*k*
_{o}), the temperature of the water in the inner pipe (*T*
_{f,up}), the temperature of the fluid (*T*
_{f,down}) in the outer pipe, the length of the pipe (Δ*z*):

$$\dot{Q}_{\text{up}} = 2\pi r_{\text{i}} k_{\text{o}} \left( {T_{\text{f,up}} - T_{\text{f,down}} } \right)\Delta z$$

(8)

Using the theory of the heat exchange in the multi-layer cylindrical wall, the total heat exchange coefficient *k*
_{o} can be calculated with the relation:

$$\frac{1}{{k_{\text{o}} }} = \frac{{r_{\text{i}} }}{{r_{\text{i}} + t}} \cdot \frac{1}{{h_{\text{o}} }} + r_{\text{i}} \mathop \sum \limits_{i = 1}^{n} \ln \left( {\frac{{r_{{{\text{j}} + 1}} }}{{r_{\text{j}} }}} \right) \cdot \frac{1}{{\lambda_{\text{j}} }} + \frac{1}{{h_{\text{i}} }}$$

(9)

The first element in Eq. (9) is due to the convective heat transfer to the outer wall: *r*
_{i} is the radius of the inner tube, *t* is the thickness of the composite pipe, *h*
_{o} is the coefficient of convective heat transfer to the outer wall. The second term of Eq. (9) is related to the conductive heat transfer through the composite pipe: *λ*
_{j}
*r*
_{j} are the thermal conductivity and the radius of the material (air and steel). The third element of Eq. (9) is due to the convection to the inner wall.

#### The thermo-siphon effect

The WBHX has been optimized to produce the maximum thermal power using the minimum electrical energy to pump the fluid in the downward pipe. This condition is achieved through the spontaneous circulation due to the thermo-siphon effect: when the fluid is heated, it goes back on top naturally through the inner tube. The pressure enhancement is due to the variation of the density that is lesser in the downward pipe than in the upward pipe.

The following relations have been used to evaluate the pressure losses:

$$\Delta P = \rho g\Delta z - \Delta P_{f} \;\;{\text{downward}}$$

(10)

$$\Delta P = - \rho g\Delta z - \Delta P_{f} \;\;{\text{upward}}$$

(11)

$$\Delta P_{f} = f\frac{\Delta z}{D}\rho \left( T \right)\frac{{v^{2} }}{2}$$

(12)

where Δ*P*
_{
f
} are the friction losses, *f* is the friction factor calculated with the explicit correlation of Churchill (1977), Δ*z* and *D* are respectively the length and the diameter of the pipe, *ρ* and *v* are, respectively the density and the velocity of the fluid.

Considering that the pipes are very large in length, the hypothesis of none local pressure losses has been used.

### Energy conversion systems

Two different systems have been selected to convert the thermal power into electrical one: an organic Rankine cycle power plant and a Stirling motor.

#### ORC plant model

Figure 3 shows the schematic of the ORC plant. The black lines indicate the pattern of the water, the green lines indicate the pattern of the working fluid of the ORC plant.

The hot water exiting from the deep borehole, flows through an heat exchanger and transfers the heat to the ORC’s working fluid. Then the water passes in the preheater and it is re-injected into the well by means of the pump P.

The ORC’s working fluid is heated up to the boiling point in the preheater PH and then it attains the condition of saturated vapor in the evaporator E. The saturated vapor is sent to the turbine T where the expansion takes place: the thermal energy is converted in kinetic and then in electrical in the generator G. Exiting from the turbine the working fluid is condensed (C) and then it is pumped (CP) to the preheater.

Figure 4 shows the thermodynamic cycle of the ORC plant.

Knowing the mass flow rate \(\dot{m}_{\text{b}}\) and the outlet (*T*
_{a}) and inlet (*T*
_{c}) temperature of the WBHX, the mass flow rate of the working fluid \(\dot{m}_{\text{wf}}\) can be calculated using the following equation:

$$\dot{m}_{\text{wf}} = \dot{m}_{\text{b}} \frac{{c_{\text{p}} \left( {T_{\text{a}} - T_{\text{c}} } \right)}}{{h_{1} - h_{4} }}$$

(13)

where *h*
_{1} is the enthalpy at the outlet of the evaporator, *h*
_{4} is the enthalpy at the inlet of the preheater.

Indicating with *h*
_{2} the enthalpy at the inlet of the condenser C, the electrical power available to the turbine T can been evaluated using the following equation:

$$\dot{W}_{t} = \dot{m}_{\text{wf}} \left( {h_{1} - h_{2} } \right)$$

(14)

The WBHX model explained in the previous paragraph evaluates the temperature of the water at the wellhead, which is the also inlet temperature at the heat exchange unit. The inlet temperature of water in the WBHX the exit temperature *T*
_{c} of the heat carrier fluid from the heat exchanger is fixed at 40 °C. The mass flow rate of the working fluid in the ORC plant is calculated according the temperature profile in the WBHX and fixing pinch point temperature around 5 °C (Fig. 5).

The ratio of the net electrical power produced from the cycle \(\dot{W}_{\text{net}}\) and the heat transfer rate in the heat exchanger unit *Q*
_{ht} has been used to evaluate the thermal efficiency:

$$\eta_{\text{th}} = \frac{{\dot{W}_{\text{net}} }}{{Q_{\text{ht}} }} = 1 - \frac{{\left( {h_{2} - h_{3} } \right)}}{{\left( {h_{1} - h_{4} } \right)}}$$

(15)

#### Stirling motor model

According to Kolin et al. (2000) when compared to the classic Clausius–Rankine cycle, mostly used in the present geothermal plants, Stirling cycle offers many theoretical and practical advantages. From thermodynamic point of view, Stirling cycle is equivalent to the optimal Carnot cycle, having the highest possible efficiency. The thermodynamic Stirling motor model follows the indications of Lloyd (2009).

In Fig. 6 the pressure–volume diagram for the Stirling cycle is shown. The real cycle has a lower efficiency compared to the ideal cycle. The assumptions of an ideal Stirling cycle are the use of a perfect gas as a working fluid, absence of flow resistance, perfect regeneration, no conduction heat losses, isothermal expansion and compression, non-sinusoidal piston motion, absence of mechanical friction, dead space assumed to be zero.

The amount of the net work per cycle can be evaluated as the sum of the work done during the gas compression stage (*W*
_{c}) and the work done by the gas during the expansion stage (*W*
_{e}):

$$W_{\text{net}} = W_{\text{c}} + W_{\text{e}} = nR_{\text{gas}} \left( {T_{\text{h}} - T_{\text{c}} } \right)\ln \left( {\frac{{V_{\hbox{max} } }}{{V_{\hbox{min} } }}} \right)$$

(16)

where *n* is the mole number of gas, *R*
_{gas} is the universal gas constant, equal to 8.314472 J/kg mol, *T*
_{h} is the hot source temperature, *T*
_{c} is the cold sink temperature, *V*
_{max} is the maximum volume, *V*
_{min} is the minimum volume.

Because in the ideal cycle the losses are absent, the produced work is equal at the supplied heat. Substituting inside Eq. (17) the value of the net work and of the supplied heat, the efficiency of the ideal Stirling cycle can be calculated as:

$$\eta = \frac{{T_{\text{h}} - T_{\text{c}} }}{{T_{\text{h}} }}$$

(17)

The ideal power is the product between the net work per cycle and the number of revolutions per minute:

$${\text{EP}} = {\text{rpm}} \cdot W_{\text{net}}$$

(18)

The reduction in power compared to the ideal cycle with no dead space can be evaluated with the empirical formula of the Schmidt factor:

$$F_{\text{s}} = 0.74 - 0.68 \delta$$

(19)

where the dead space ratio *δ* is the ratio between the total dead space volume *V*
_{d} and the total volume of the gas swept by the displacer *V*
_{sw}:

$$\delta = \frac{{V_{\text{d}} }}{{V_{\text{sw}} }}$$

(20)

The real power can be calculated with the following relation:

$${\text{EP}}_{\text{real}} = {\text{EP}} \cdot F_{\text{s}}$$

(21)