Certainly! Let's break down the problem **step-by-step**: --- ## **Given** The Fetkovich IPR for a fully two-phase reservoir: \[ \frac{q}{q_{\text{max}}} = 1 - \left(\frac{p_{wf}}{p_r}\right)^2 \] Where: - \( q \): well rate - \( q_{\text{max}} \): maximum well rate (at \( p_{wf} = \)) - \( p_{wf} \): bottomhole flowing pressure - \( p_r \): reservoir pressure --- ## **(a) Modify the Fetkovich IPR for the entire pressure range \([p_b, p_r]\)** ### **Background** - **For \( p_{wf} \geq p_b \)** (single-phase, above bubble point): Flow is linear: \[ q = q_{max,lin} \left( \frac{p_r - p_{wf}}{p_r - p_b} \right) \] - **For \( p_{wf} < p_b \)** (two-phase, below bubble point): Use Fetkovich equation, but shift it to start from \( p_b \). ### **Step 1: Linear portion (above bubble point)** \[ q = q_{b} \left( \frac{p_r - p_{wf}}{p_r - p_b} \right) \quad \text{for } p_{wf} \geq p_b \] Where \( q_b \) is the well rate at \( p_{wf} = p_b \). --- ### **Step 2: Modified Fetkovich (below bubble point)** We want the equation to start at \( p_b \) (not \( \)), so: \[ \frac{q}{q_{max}} = 1 - \left( \frac{p_{wf}}{p_r} \right)^2 \] But for \( p_{wf} < p_b \), we need to **normalize** the pressure drop: Let’s shift the origin to \( p_b \), so the equation becomes: \[ \frac{q - q_b}{q_{max} - q_b} = 1 - \left( \frac{p_{wf} - }{p_b - } \right)^2 \] But \( p_b \) is not zero. Let’s generalize: Let \( q = q_b \) at \( p_{wf} = p_b \) and \( q = q_{max} \) at \( p_{wf} = \): So, \[ \frac{q - q_b}{q_{max} - q_b} = 1 - \left( \frac{p_{wf}}{p_b} \right)^2 \] Or, \[ q = q_b + (q_{max} - q_b) \left[ 1 - \left( \frac{p_{wf}}{p_b} \right)^2 \right] \] for \( \leq p_{wf} < p_b \) --- ## **(b) Match slopes at the bubble point** At \( p_{wf} = p_b \), the **slope** of the two equations must match. ### **Step 1: Compute slope of linear IPR at \( p_{wf} = p_b \):** \[ q = q_b \left( \frac{p_r - p_{wf}}{p_r - p_b} \right) \] \[ \left. \frac{dq}{dp_{wf}} \right|_{p_{wf}=p_b} = -\frac{q_b}{p_r - p_b} \] --- ### **Step 2: Compute slope of modified Fetkovich at \( p_{wf} = p_b \):** \[ q = q_b + (q_{max} - q_b) \left[ 1 - \left( \frac{p_{wf}}{p_b} \right)^2 \right] \] \[ \frac{dq}{dp_{wf}} = (q_{max} - q_b) \left[ -2 \frac{p_{wf}}{p_b^2} \right] \] At \( p_{wf} = p_b \): \[ \left. \frac{dq}{dp_{wf}} \right|_{p_{wf}=p_b} = (q_{max} - q_b) \left[ -2 \frac{p_b}{p_b^2} \right] = -\frac{2(q_{max} - q_b)}{p_b} \] --- ### **Step 3: Set slopes equal and solve for \( q_{max} \):** \[ -\frac{q_b}{p_r - p_b} = -\frac{2(q_{max} - q_b)}{p_b} \] \[ \frac{q_b}{p_r - p_b} = \frac{2(q_{max} - q_b)}{p_b} \] \[ q_b p_b = 2(q_{max} - q_b)(p_r - p_b) \] \[ q_b p_b = 2 q_{max} (p_r - p_b) - 2 q_b (p_r - p_b) \] \[ q_b p_b + 2 q_b (p_r - p_b) = 2 q_{max} (p_r - p_b) \] \[ q_b [p_b + 2(p_r - p_b)] = 2 q_{max} (p_r - p_b) \] \[ q_b [2 p_r - p_b] = 2 q_{max} (p_r - p_b) \] \[ q_{max} = \frac{q_b [2 p_r - p_b]}{2 (p_r - p_b)} \] --- ## **Final Answers** ### **(a) The modified Fetkovich IPR is:** - **For \( p_b \leq p_{wf} \leq p_r \):** \[ q = q_b \left( \frac{p_r - p_{wf}}{p_r - p_b} \right) \] - **For \( \leq p_{wf} < p_b \):** \[ q = q_b + (q_{max} - q_b) \left[ 1 - \left( \frac{p_{wf}}{p_b} \right)^2 \right] \] --- ### **(b) The expression for \( q_{max} \) is:** \[ \boxed{ q_{max} = \frac{q_b [2 p_r - p_b]}{2 (p_r - p_b)} } \] Where \( q_b \) is the well rate at \( p_{wf} = p_b \). --- ### **Summary Table** | Pressure Range | Equation | |----------------------------|--------------------------------------------------------------------------------------------| | \( p_b \leq p_{wf} \leq p_r \) | \( q = q_b \left( \frac{p_r - p_{wf}}{p_r - p_b} \right) \) | | \( \leq p_{wf} < p_b \) | \( q = q_b + (q_{max} - q_b) \left[ 1 - \left( \frac{p_{wf}}{p_b} \right)^2 \right] \) | | \( q_{max} \) | \( q_{max} = \frac{q_b [2 p_r - p_b]}{2 (p_r - p_b)} \) | --- If you need a worked numerical example or further clarification, let me know!