Benchmark solution
=====================

Remember that the water mass input is written as: :math:`{m}_{w}=\varphi \mathrm{.}{\rho }_{w}\mathrm{.}(1+{\varepsilon }_{v})`, which can be derived in the following form: :math:`{\mathrm{dm}}_{w}=d\varphi {\rho }_{w}(1+{\varepsilon }_{v})+d{\rho }_{w}\varphi (1+{\varepsilon }_{v})+{\rho }_{w}\varphi d{\varepsilon }_{v}` with :math:`\varphi` the Eulerian porosity.

If we assume small trips, we will therefore have:

.. math::
 :label: eq-1

    {\ mathrm {dm}} _ {w} =d\ varphi {\ rho} _ {w} +d {\ rho} _ {w}\ varphi + {\ rho} _ {w} _ {w}\ varphi d {\ varepsilon} _ {v}

The variation in porosity is written according to the relationship:

.. math::
 :label: eq-1

    d\ varphi\ text {=} (b-\ varphi) (d {\ varepsilon} _ {V}\ text {-} 3 {\ alpha} _ {0}\ text {dT}\ text {dT}\ text {+}\ text {+}}\ frac {{\ text {dp}}} _ {s}})

with :math:`{\alpha }_{0}` the linear dilation of the skeleton (assimilable to the porous medium). It is recalled that the Biot coefficient :math:`b` and the compressibility module of solid grains :math:`{K}_{s}` and are linked to the "drained" compressibility module of the porous medium :math:`{K}_{0}`, such that:

:math:`b\text{=}1\text{-}\frac{{K}_{0}}{{K}_{s}}`

Moreover, the variation in the density of water is written as:

.. math::
 :label: eq-1

    \ frac {d {\ rho} _ {w}} {{\ rho}} {{\ rho}} {{w}}\ text {=}\ frac {{\ text {dp}} _ {w}} {{k}}} {{K} _ {w}} {{K} _ {w}} {K} _ {w}} {K} _ {w}} {K} _ {w} _ {w}} {K} _ {w} _ {w} _ {w} _ {k} _ {w} _ {w} _ {w} _ {w} _ {w} _ {w}}

with the water compressibility module :math:`{K}_{w}` and its expansion module :math:`{\alpha }_{w}`.

Finally, if the law of behavior is elastic, we recall that the deformation is linked to the effective stress such that:

.. math::
 :label: eq-1

    d {\ epsilon} _ {V} =\ frac {d\ sigma\ text {'}} {{K} _ {0}} +3 {\ alpha} _ {0} _ {0}\ text {dT}

Moreover, the formulation under total constraint tells us that:

:math:`d\sigma \text{'}\mathrm{=}d\sigma +b{\text{dp}}_{w}`, considering here that the environment is in constant confinement, we therefore have:

:math:`d\sigma \text{'}\mathrm{=}b{\text{dp}}_{w}`, which in the end gives us

.. math::
 :label: eq-1

    d {\ varepsilon} _ {V}\ mathrm {=}\ mathrm {=}\ frac {b {\ text {dp}} _ {0}} _ {0}} +3 {\ alpha}}} +3 {\ alpha}} _ {0}\ text {dT}}

We can now inject (2), (3), (4) and (5) into equation (1) and we get that:

.. math::
 :label: eq-1

    \ frac {{\ mathit {dm}} _ {w}} {w}} {{w}} {{\ rho}} _ {w}}\ mathrm {=} (\ frac {{b} ^ {2}}} {{K}}} {{K} _ {2}}} {{K} _ {2}}} {{K} _ {2}}} {{K} _ {0}}} +\ frac {0}}} +\ frac {0}}} +\ frac {\ varphi}} +\ frac {\ varphi}} {K} _ {w}}) {\ text {dp}}} _ {w}} +\ varphi (3 {\ alpha} _ {0}\ text {-} 3 {\ alpha} _ {w})\ text {dT})\ text {dT}

Considering that the medium is not drained, we therefore have:

.. math::
 :label: eq-1

    (\ frac {{b} ^ {2}}} {{K} _ {0}}} +\ frac {(b\ mathrm {-}\ varphi)} {{K} _ {s}}} +\ frac {\ varphi}} {\ varphi}} {{K}} {{K} _ {w}}) {\ text {dp}}} _ {w}\ mathrm {=}} +\ frac {\ varphi}} {\ varphi} {{K}}} +\ frac {\ varphi}} {\ varphi} {{K}}}} _ {w}\ text {-} 3 {\ alpha} _ {0})\ text {dT}

This can be written in the form:


.. math::
 :label: eq-1

    {\ text {dp}} _ {w} =\ Lambda\ text {dT}

With :math:`\Lambda` the thermal pressurization coefficient such as:

:math:`\Lambda =\frac{\phi \left(3{\alpha }_{w}\text{-}3{\alpha }_{0}\right)}{\left(\frac{{b}^{2}}{{K}_{0}}+\frac{\left(b-\phi \right)}{{K}_{s}}+\frac{\phi }{{K}_{w}}\right)}`

Note that this coefficient reveals the thermal differential :math:`({\alpha }_{w}-{\alpha }_{0})`


Constant thermal expansion of water
------------------------

In case :math:`{\alpha }_{w}=\mathit{cte}` the digital application is immediate.

**Digital application:**

With the data defined above, we obtain:

:math:`\Lambda =\mathrm{2,25}{10}^{5}\mathit{Pa.}{K}^{-1}`

This gives for a temperature variation :math:`\Delta T=40°C`, a pressure variation of :math:`\Delta p=\mathrm{9,01}\mathit{Mpa}` (i.e. a pressure of 13.01 MPa).

.. _RefNumPara__898_670900974:


Thermal expansion of water as a function of temperature
--------------------------------------------------------

In this case, we will write

:math:`\Lambda (T)=\frac{\phi \left(3{\alpha }_{w}(T)\text{-}3{\alpha }_{0}\right)}{\left(\frac{{b}^{2}}{{K}_{0}}+\frac{\left(b-\phi \right)}{{K}_{s}}+\frac{\phi }{{K}_{w}}\right)}` and

:math:`{\text{dp}}_{w}=\Lambda (T)\text{dT}`

To be able to integrate this formula, we consider that the thermal expansion of water is considered to be a linear function of temperature such as :math:`{\alpha }_{w}=\mathit{a.}T+b`. :math:`\Lambda (T)` is then also a linear function that is integrated between the initial temperature T0 and the final temperature T1, which gives:

:math:`\Delta p=\frac{A}{2}\mathrm{.}({\mathit{T2}}^{2}-{\mathit{T1}}^{2})+B(\mathit{T2}-\mathit{T1})` with

:math:`A=\frac{3\phi a}{\left(\frac{{b}^{2}}{{K}_{0}}+\frac{\left(b-\phi \right)}{{K}_{s}}+\frac{\phi }{{K}_{w}}\right)}`

and

:math:`B=\frac{\phi \left(\mathrm{3b}\text{-}3{\alpha }_{0}\right)}{\left(\frac{{b}^{2}}{{K}_{0}}+\frac{\left(b-\phi \right)}{{K}_{s}}+\frac{\phi }{{K}_{w}}\right)}`

**Digital application:**

We take it here

:math:`a=\mathrm{2,63}{10}^{-6}` and :math:`b=-7{10}^{-4}` which correspond to :math:`{\alpha }_{w}(293)=\mathrm{6,67}\mathrm{.}{10}^{-5}{K}^{-1}` and :math:`{\alpha }_{w}(333)=\mathrm{1,72}\mathrm{.}{10}^{-4}{K}^{-1}` and which are realistic values for water.

All calculations done, for a temperature variation between 20° C. and 60° C., we obtain a pressure variation of :math:`\Delta p=\mathrm{10,9}\mathit{MPa}` and therefore a final pressure of :math:`\mathrm{14,9}\mathit{MPa}`.