Average reservoir pressure over the drainage volume $\begin{array}{l}V_e\end{array}$ :

(1)	$\begin{array}{l}\displaystyle p_r = \frac{1}{V_e} \iint_{A_e} p(x,y,z) dV\end{array}$

For the Steady State Radial Flow in finite reservoir the relationship between Boundary-average formation pressure $\begin{array}{l}p_e\end{array}$ and Drainarea formation pressure $\begin{array}{l}p_r\end{array}$ is going to be:

(2)	$\begin{array}{l}\displaystyle p_r = p_i - \frac{q_t}{4 \pi \sigma}\end{array}$

Derivation

(3)	$\begin{array}{l}\displaystyle V_e = \pi r_e^2 h, \quad dV = 2\pi r \, h dr\end{array}$

(4)	$\begin{array}{l}\displaystyle p_r = \frac{1}{V_e} \int p(r) dV = \frac{2}{r_e^2} \int p(r) \, r \, dr\end{array}$

For the Steady State Radial Flow in finite reservoir the reservoir pressure is going to be:

(5)	$\begin{array}{l}\displaystyle p(t,r) = p_e(t) + \frac{q_t}{2 \pi \sigma} \, \ln \frac{r}{r_e} = p_i + \frac{q_t}{2 \pi \sigma} \, \ln \frac{r}{r_e}\end{array}$

and substituting the above to (4) and integrating:

(6)	$\begin{array}{l}\displaystyle p_r = \frac{2}{r_e^2} \int \bigg[ p_i + \frac{q_t}{2\pi \sigma} \ln \frac{r}{r_e} \bigg] \, r \, dr = p_i - \frac{q_t}{4\pi \sigma}\end{array}$

For the Pseudo-Steady State Radial Flow in finite reservoir the relationship between Boundary-average formation pressure $\begin{array}{l}p_e\end{array}$ and Drainarea formation pressure $\begin{array}{l}p_r\end{array}$ is going to be:

(7)	$\begin{array}{l}\displaystyle p_r(t)= p_e(t) - 0.75 \cdot \frac{q_t}{2 \pi \sigma}\end{array}$

Derivation

(8)	$\begin{array}{l}\displaystyle V_e = \pi r_e^2 h, \quad dV = 2\pi r \, h dr\end{array}$

(9)	$\begin{array}{l}\displaystyle p_r = \frac{1}{V_e} \int p(r) dV = \frac{2}{r_e^2} \int p(r) \, r \, dr\end{array}$

For the Pseudo-Steady State Radial Flow in finite reservoir the reservoir pressure is going to be:

(10)	$\begin{array}{l}\displaystyle p(r) = p_i + \frac{q_t}{4 \pi \sigma} \, \left[ 2 \ln \frac{r}{r_e} - \frac{r^2}{r_e^2} \right]\end{array}$

and substituting the above to (4) and integrating:

(11)	$\begin{array}{l}\displaystyle p_r(t) = \frac{2}{r_e^2} \int \bigg[ p_e(t) + \frac{q_t}{4\pi \sigma} \left[ 2 \ln \frac{r}{r_e} - \frac{r^2}{r_e^2} \right] \bigg] \, r \, dr = p_i - 0.75 \cdot \frac{q_t}{2\pi \sigma}\end{array}$

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