Formulas and Calculations
0.7365
Ps = [ 2.4 x 207 x 2 (0.7365)+ 1 ] 8.5 — 5.0 3(0.7365) Ps = [ 2.4 x 207 x 2(0.7365) + 1 ] 8.5 — 5.0 3(0.7365
0.7365
x 0.3644 x 11,300 300(8.5 — 5.0) x 3.644 x 11,300 300(8.5 — 5.0)
Ps = (141.9 x 1.11926) 0.7365 x 3.9216 Ps = 41.78 x 3.9216 Ps = 163.8 psi 6. Pressure loss opposite drill collars: Ps = 7.7 x 10 — 5 x 12.50.8 x 4001.8 x 240.2 x 700 3 1.8 (8.5 — 6.5) x (8.5 + 6.5) Ps =37056.7 8 x 130.9 Ps = 35.4 psi Total pressure losses:
psi = 163.8 psi + 35.4 psi psi = 199.2 psi
7. Determine equivalent circulating density (ECD), ppg: ECD, ppg = 199.2 psi ÷ 0.052 ÷ 12,000 ft + 12.5 ppg ECD = 12.82 ppg 9. Fracture Gradient Determination - Surface Application Method 1: Matthews and Kelly Method F = P/D + Ki /D
where F = fracture gradient, psi/ft P= formation pore pressure, psi D = depth at σ = matrix stress at point of interest, psi point of interest, TVD, ft Ki = matrix stress coefficient, dimensionless Procedure: 1. Obtain formation pore pressure, P, from electric logs, density measurements, or from mud logging personnel.
Figure 5-1. Matrix stress coefficient chart
2. Assume 1.0 psi/ft as overburden pressure (S) and calculate as follows: S—P
5. Determine fractur e gradient, psi/ft: x
3. Determine the depth for determining Ki by: D =
D
F = P + Ki D
0.535 4. From Matrix Stress Coeffi cient chart, determine Ki:
6. Determine fractur e pressure, psi: FxD
F, psi =
7. Determine maximum mud density, ppg: ppg = F ÷ 0.052
MW,
Example: Casing setting depth = 12,000 ft
Formation pore pressure (Louisiana Gulf Coast) = 12.0 ppg 1. P = 12.0 ppg x 0.052 x 12,000 ft P = 7488 psi 2. σ = 12,000 psi — 7488 psi σ = 4512 psi 3. D = 4512 psi 0.535 D = 8434 ft 4. From chart = Ki = 0.79 psi/ft 5. F = 7488 + 0.79 x 4512 12,000 12,000
52
