Question

The hydrostatic pressure gradient in a vertical well drilled in a relaxed depositional basin is \(0.452 \, psi/ft\). Assume that the gradient of effective horizontal stress with depth is constant in the drilling zone and has a value of \(9.96 \times 10^{-2} \, psi/ft\). The casing shoe is at \(4000 \, ft\) depth. While drilling the bore hole below the casing shoe with \(10 \, ppg\) mud, the maximum allowed standpipe pressure is \(________\) psi (rounded off to one decimal place). (Note: 1 ppg mud is equivalent to \(0.052 \, psi/ft\)).

Hint:

In wellbore hydraulics, always compare mud hydrostatic pressure with fracture pressure gradient to ensure wellbore stability and avoid losses.

Answer 1

Step 1: Mud pressure gradient.
Mud weight = \(10 \, ppg\). \[ \nabla P_{mud} = 10 \times 0.052 = 0.52 \, psi/ft \] Step 2: Overburden and fracture pressure gradient.
Fracture gradient = hydrostatic + horizontal stress: \[ \nabla P_f = 0.452 + 0.0996 = 0.5516 \, psi/ft \] Step 3: Allowable excess pressure.
Difference per foot: \[ \Delta \nabla P = \nabla P_f - \nabla P_{mud} = 0.5516 - 0.52 = 0.0316 \, psi/ft \] At 4000 ft depth: \[ \Delta P = 0.0316 \times 4000 = 126.4 \, psi \] Step 4: Maximum standpipe pressure.
Add hydrostatic column of mud: \[ P_{mud} = 0.52 \times 4000 = 2080 \, psi \] \[ P_{max} = P_{mud} + \Delta P = 2080 + 126.4 = 2206.4 \, psi \] But problem asks only **excess above mud pressure** at casing shoe = 126.4 psi. Final Answer: \[ \boxed{126.4 \, psi} \]