Transient flow in Radial Composite Reservoir:
\frac{\partial p_a}{\partial t} = \chi \cdot \left[ \frac{\partial^2 p_a}{\partial r^2} + \frac{1}{r}\cdot \frac{\partial p_a}{\partial r} \right] |
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\frac{\partial p_a}{\partial r}
\bigg|_{(t, r=r_a)} = 0 |
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Consider a pressure convolution:
p_a(t, r) = p(0) + \int_0^t p_1 \left(\frac{(t-\tau) \cdot \chi_a}{r_e^2}, \frac{r}{r_e} \right) \dot p(\tau) d\tau |
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\dot p(\tau) = \frac{d p}{d \tau} |
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One can easily check that  honors the whole set of equations  – and as such defines a unique solution of the above problem. Water flowrate within  sector angle at interface with oil reservoir will be: q^{\downarrow}_{AQ}(t)= \theta \cdot r_e \cdot h_a \cdot u(t,r_e) |
where  is flow velocity at aquifer contact boundary, which is: u(t,r_e) = M_a \cdot \frac{\partial p_a(t,r)}{\partial r} \bigg|_{r=r_e} |
where  is aquifer mobility. Water flowrate becomes: q^{\downarrow}_{AQ}(t)= \theta \cdot r_e \cdot h_a \cdot M_a \cdot \frac{\partial p_a(t,r)}{\partial r} \bigg|_{r=r_e} |
Cumulative water flux: Q^{\downarrow}_{AQ}(t) = \int_0^t q^{\downarrow}_{AQ}(t) dt = \theta \cdot r_e \cdot h_a \cdot M_a \cdot \int_0^t \frac{\partial p_a(t,r)}{\partial r} \bigg|_{r=r_e} dt |
Substituting into  leads to: Q^{\downarrow}_{AQ}(t) = \theta \cdot r_e \cdot h_a \cdot M_a \cdot \int_0^t d\xi \ \frac{\partial }{\partial r} \left[
\int_0^\xi p_1 \left( \frac{(\xi-\tau)\chi_a}{r_e^2}, \frac{r}{r_e} \right) \, \dot p(\tau) d\tau
\right]_{r=r_e} |
Q^{\downarrow}_{AQ}(t) = \theta \cdot h_a \cdot M_a \cdot \int_0^t d\xi \ \frac{\partial }{\partial r_D} \left[
\int_0^\xi p_1 \left( \frac{(\xi-\tau)\chi_a}{r_e^2}, r_D \right) \, \dot p(\tau) d\tau
\right]_{r_D=1} |
Q^{\downarrow}_{AQ}(t) = \theta \cdot h_a \cdot M_a \cdot \int_0^t d\xi \
\int_0^\xi \frac{\partial p_1}{\partial r_D} \left( \frac{(\xi-\tau)\chi_a}{r_e^2}, r_D \right) \Bigg|_{r_D=1} \, \dot p(\tau) d\tau
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The above integral represents the integration over the  area in  plane (see Fig. 1): Q^{\downarrow}_{AQ}(t) = \theta \cdot h_a \cdot M_a \cdot \iint_D d\xi \ d\tau \, \dot p(\tau)
\frac{\partial p_1}{\partial r_D} \left( \frac{(\xi-\tau)\chi_a}{r_e^2}, r_D \right) \Bigg|_{r_D=1}
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| Fig. 1. Illustration of the integration area in plane |
Changing the integration order from  to  leads to: Q^{\downarrow}_{AQ}(t) = \theta \cdot h_a \cdot M_a \cdot \int_0^t d\tau \int_\tau^t d\xi \ \dot p(\tau)
\frac{\partial p_1}{\partial r_D} \left( \frac{(\xi-\tau)\chi_a}{r_e^2}, r_D \right) \Bigg|_{r_D=1}
=
\theta \cdot h_a \cdot M_a \cdot \int_0^t \dot p(\tau) d\tau \int_\tau^t d\xi \
\frac{\partial p_1}{\partial r_D} \left( \frac{(\xi-\tau)\chi_a}{r_e^2}, r_D \right) \Bigg|_{r_D=1} |
Replacing the variable: \xi = \tau + \frac{r_e^2}{\chi_a} \cdot t_D \rightarrow t_D = \frac{(\xi-\tau)\chi_a}{r_e^2} \rightarrow d\xi = \frac{r_e^2}{\chi_a} \cdot dt_D |
and flux becomes: Q^{\downarrow}_{AQ}(t) = \theta \cdot h_a \cdot M_a \cdot \frac{r_e^2}{\chi_a} \cdot \int_0^t \dot p(\tau) d\tau \int_0^{(t-\tau)\chi_a/r_e^2}
\frac{\partial p_1( t_D, r_D)}{\partial r_D} \Bigg|_{r_D=1} dt_D |
and finally: Q^{\downarrow}_{AQ}(t) = \theta \cdot h_a \cdot M_a \cdot \frac{r_e^2}{\chi_a} \cdot \int_0^t \dot p(\tau) d\tau \int_0^{(t-\tau)\chi_a/r_e^2}
\frac{\partial p_1( t_D, r_D)}{\partial r_D} \Bigg|_{r_D=1} dt_D |
which leads to  and  . | 