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 × 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).

Hint:

The standpipe pressure includes the pressure from the mud column as well as the effective horizontal stress at depth. Ensure to calculate both components for an accurate result.

Answer 1

The maximum standpipe pressure can be calculated by adding the hydrostatic pressure due to the mud column and the additional pressure due to horizontal effective stress.
- The hydrostatic pressure is given by:
\[ P_{{hydrostatic}} = 0.052 \times {Mud Weight} \times {Depth} \] Substituting the values:
\[ P_{{hydrostatic}} = 0.052 \times 10 \times 4000 = 2080 \, {psi} \] - The horizontal effective stress is:
\[ P_{{stress}} = 9.96 \times 10^{-2} \times 4000 = 398.4 \, {psi} \] Thus, the total maximum standpipe pressure is:
\[ P_{{max}} = P_{{hydrostatic}} + P_{{stress}} = 2080 + 398.4 = 2478.4 \, {psi} \] Therefore, the maximum allowed standpipe pressure is approximately between 122.0 to 130.0 psi (rounded off).