Question

A wellbore has true vertical depth (TVD) of 10000 ft. The pore pressure of formation fluid in the permeable stratum at bottom of wellbore is 6500 psig. Which is/are the acceptable average static mud density value(s) (in lbm/gal) to prevent flow of formation fluid into wellbore?
[Given: 1 lbm/gal mud = 0.052 psi/ft]

• A. 13.8
• B. 11.3
• C. 11.8
• D. 13.2

Hint:

Minimum mud density equals pore pressure divided by 0.052 times TVD; only values at or above that minimum are safe.

Answer 1

The Correct Option is A, D

Step 1: Set up the balance condition needed to prevent a kick:
To stop formation fluid from flowing into the wellbore, the hydrostatic pressure exerted by the mud column at the bottom of the hole must be at least equal to the pore pressure of the formation fluid. If the mud hydrostatic pressure falls below the pore pressure, the well becomes underbalanced and formation fluid can enter the wellbore, so the mud density must be chosen to satisfy this minimum requirement.
Step 2: Write the hydrostatic pressure formula and find the minimum mud density:
The hydrostatic pressure of a mud column is given by \( P_{hyd} = 0.052 \times \rho_{mud} \times TVD \), where \(\rho_{mud}\) is in lbm/gal, TVD is in feet, and \(P_{hyd}\) is in psi. Setting this equal to the pore pressure to find the minimum acceptable mud density gives \[ \rho_{mud,min} = \frac{P_{pore}}{0.052 \times TVD} = \frac{6500}{0.052 \times 10000} = \frac{6500}{520} = 12.5 \ \text{lbm/gal} \] Any average static mud density value equal to or greater than 12.5 lbm/gal will keep the well balanced or slightly overbalanced, which is acceptable and safe against a kick, while any value below 12.5 lbm/gal leaves the well underbalanced and unsafe.
Step 3: Check option A, 13.8 lbm/gal:
The hydrostatic pressure produced is \(0.052 \times 13.8 \times 10000 = 7176\) psi, which is greater than the pore pressure of 6500 psi. Since 13.8 is above the minimum required value of 12.5 lbm/gal, the well stays overbalanced and safe, so option A is acceptable.
Step 4: Check option B, 11.3 lbm/gal:
The hydrostatic pressure produced is \(0.052 \times 11.3 \times 10000 = 5876\) psi, which is less than the pore pressure of 6500 psi. Since 11.3 is below the minimum required value of 12.5 lbm/gal, the well would be underbalanced and formation fluid could flow into the wellbore, so option B is not acceptable.
Step 5: Check option C, 11.8 lbm/gal:
The hydrostatic pressure produced is \(0.052 \times 11.8 \times 10000 = 6136\) psi, which is still less than the pore pressure of 6500 psi. Since 11.8 is below the minimum required value of 12.5 lbm/gal, this density also leaves the well underbalanced, so option C is not acceptable.
Step 6: Check option D, 13.2 lbm/gal:
The hydrostatic pressure produced is \(0.052 \times 13.2 \times 10000 = 6864\) psi, which is greater than the pore pressure of 6500 psi. Since 13.2 is above the minimum required value of 12.5 lbm/gal, the well stays overbalanced and safe, so option D is acceptable.
Final Answer:
\[ \boxed{13.8 \ \text{and} \ 13.2 \ \text{lbm/gal, options A and D}} \]