Let's analyze and solve the problem step-by-step: --- ## Given Data: - Mud weight used for balancing pressure: **\( \text{mud weight} \)** - Depth of the formation: **9,000 ft** - Pore pressure at that depth: **4,500 psi** --- ## Step 1: Understand what is being asked: We need to find the **necessary mud density** (in ppg) that provides the pressure to balance the formation pressure at 9,000 ft. --- ## Step 2: Recall the relation between mud weight and pressure: The pressure exerted by the mud column is given by: \[ P = \text{Mud weight (ppg)} \times .052 \times \text{Depth (ft)} \] where: - \( P \) is the pressure in psi, - .052 is a conversion factor (psi per ft per ppg), - Depth is in feet, - Mud weight is in ppg. --- ## Step 3: Rearrange the formula to find mud weight: \[ \text{Mud weight (ppg)} = \frac{P}{.052 \times \text{Depth}} \] --- ## Step 4: Substitute known values: \[ \text{Mud weight} = \frac{4,500 \text{ psi}}{.052 \times 9,000 \text{ ft}} \] Calculate denominator: \[ .052 \times 9,000 = 468 \] Now, \[ \text{Mud weight} = \frac{4,500}{468} \approx 9.62 \text{ ppg} \] --- ## **Final Answer:** \[ \boxed{ \text{Mud density} \approx \mathbf{9.62 \, ppg} } \] This is the necessary mud density to exert a pressure of 4,500 psi at 9,000 ft depth.