@@ -13,8 +13,8 @@ project = "/Parabolic/LiquidFlow/BuildupTest/buildup_test.prj" {{< data-link >}} Problem description {#problem-description .unnumbered .unnumbered} =================== ## Problem description The pressure buildup test is performed by shutting in a producing well at time $t=t_p$, after which a smooth rise of the well head pressure can @@ -26,8 +26,7 @@ the model, a time dependent nodal source term was set up to represent the shut-in operation. The simulated pressure profile is then verified against the analytical solution. Model Setup {#model-setup .unnumbered .unnumbered} =========== ## Model Setup This benchmark represents a scenario in which the well had been producing geothermal brine for $118\ \mathrm{h}$ at a rate of @@ -70,8 +69,7 @@ which corresponds to the infinite shut-in time $(\Delta t)$. This leads to an extrapolated pressure $p_0$ of $67.5~\mathrm{kPa}$, which is the undisturbed reservoir pressure . Input files {#input-files .unnumbered .unnumbered} =========== ## Input files The benchmark project is defined in the input file buildup_test.prj. It defines the process to be solved as "LiquidFlow" and the primary variable is hence pressure. @@ -85,8 +83,7 @@ conditions, and source term can be found in line_1000_axi.gml file. The mesh is specified in line_1000_axi.vtu, which is stored in the VTK format and can be directly visualized in Paraview. Analytical solution {#analytical-solution .unnumbered .unnumbered} =================== ## Analytical solution The pressure buildup test is comparable to a pumping recovery test as the extraction rate is first kept constant at $Q$, and then becomes zero @@ -101,8 +98,8 @@ $$\Delta p=\rho g \frac{-Q}{4\pi T}W\left(\frac{r^2S}{4Tt}\right)$$ and for $t>t_p$, $$\Delta p=\rho g \frac{-Q}{4\pi T}W\left(\frac{r^2S}{4Tt}\right)+\rho g \frac{Q}{4\pi T}W\left(\frac{r^2S}{4T(t-t_p)}\right)$$ Results and evaluation {#results-and-evaluation .unnumbered .unnumbered} ====================== ## Results and evaluation The pressure evolution is simulated throughout the domain and the result is compared with the analytical solution at $r=10.287\ \mathrm{m}$. In @@ -121,12 +118,11 @@ Figure 2: OGS 6 result compared with analytical solution Figure 3: Absolute and relative error References {#references .unnumbered .unnumbered} ======== ## References  RN Horne. Characterization, evaluation, and interpretation of well data. In: R DiPippo, editor,Geothermal Power Generation, chapter 6, pages 141–163.Elsevier, 2016. Appendix {#appendix .unnumbered .unnumbered} ======== ## Appendix \centering | $\Delta t$ (h) | $\Delta p$ (bar) | $\Delta t$ (h) | $\Delta p$ (bar) |