4.3 WP6: Applications of Thermoelasticity

Theory thermo-Elasticity:

$$\left\lbrace \begin{array}{ll} \nabla \uuline{\sigma} + \ulin... ...ilon_{vol}}{\partial t} & \text{Diffusivity} \end{array}\right.$$ (4.1)

Theory thermo-poroelasticity:

$$\left\lbrace \begin{array}{ll} \cfrac{\partial p}{\partial t}... ...\partial t} & \text{Temperature diffusivity} \end{array}\right.$$ (4.2)

where the coefficient of volumetric thermal expansion for variation in fluid content at constant frame volume is

$$\beta_e = \alpha \beta_d + \phi (\beta_f - \beta_s)$$ (4.3)

where usually $\beta_f > \beta_s$. The drained volumetric thermal expansion coefficient is

$$\beta_d = \beta_s; \:$$ (4.4)

Drained thermoelastic effective stress coefficient ($K$: Bulk modulus)

$$\alpha_d = K \: \beta_d$$ (4.5)

and the drained specific heat at constant strain divided reference temperature is

$$m_d = \frac{c_d}{T_0}$$ (4.6)

4.3.1 Exercise 1: Thermal stress around a wellbore

4.3.2 Exercise 2: Thermal stress in a reservoir

[Exercise to be based on https://doi.org/10.3390/en14165054]

Constitutive equation:

$$\uuline{\sigma} = \uuline{C} \cdot \uuline{\varepsilon} \color{red} + 3 \alpha_L K \theta \uuline{I}; \:\: \theta = T - T_0$$ (4.7)

Assuming uniform cooling leads to no horizontal strains: $\varepsilon_{11} = \varepsilon_{22} = 0$.

Furthermore, assuming a compliant caprock, no change of overburden, and small change in pore pressure leads to constant effective vertical stress $\sigma_{33} =$cst.

Let us use the matrix notation for the normal stresses (shear stresses are zero):

$$%compliance matrix \left[ \begin{array}{c} \sigma_{11} \\ \... ... 3 \alpha_L K \theta \\ 3 \alpha_L K \theta \end{array}\right]$$ (4.8)

Solving for $\sigma_{11}$ and $\sigma_{33}$ results in:

$$\left\lbrace \begin{array}{ll} \sigma_{11} = \cfrac{\nu E}{(1... ...\nu)} \varepsilon_{33} + 3 \alpha_L K \theta \end{array}\right.$$ (4.9)

Replacing $\varepsilon_{33}$ in the first equation leads to

$$\sigma_{11} = \left( \frac{\nu}{1-\nu} \right) \sigma_{33} + \left( \frac{1-2\nu}{1-\nu} \right) 3 \alpha_L K \theta$$ (4.10)

which expressed in terms of the Young's modulus ( $K = E/[3(1-2\nu)]$) is:

$$\sigma_{11} = \left( \frac{\nu}{1-\nu} \right) \sigma_{33} + \frac{\alpha_L E}{1-\nu} \theta$$ (4.11)

Hence, a variation of horizontal stress with temperature is

$$\frac{\partial \sigma_{11}}{\partial \theta} = + \frac{\alpha_L E}{1-\nu}$$ (4.12)

Which is in the order of

$$\frac{\partial \sigma_{11}}{\partial \theta} = 0.1 \frac{\text{MPa}}{^{\circ}\text{C}}$$ (4.13)

for $\alpha_L = 10^{-5}$ $1/^{\circ}$C, $E = 10$ GPa, and $\nu = 0.2$.