GP 202 Reservoir Geomechanics Spring 2014, OpenEdX Homework 4 – Estimating Estimating Limits on S hmax hmax SOLUTIONS Please direct any questions to the forum on the OpenEdX Courseware page
Question 1: Which Stress States Are Possible?
Which stress states are possible in each of the following cases at d epth of 12000 ft? Case 1: μ = 0.6, S v = 12000 psi, S hmin hmin = 8000 psi (d) Normal and/or strike-slip faulting. In this case, S v > S hmin hmin; thus, the minimum principal stress cannot be the vertical stress, so only reverse faulting is not possible. Case 2: μ = 0.6, S v = 12000 psi, S hmin hmin = 9500 psi (d) Normal and/or strike-slip faulting. In this case, S v > S hmin hmin; thus, the minimum principal stress cannot be the vertical stress, so only reverse faulting is not possible. Case 3: μ = 0.6, S v = 12000 psi, S hmin hmin = 10000 psi (d) Normal and/or strike-slip faulting. In this case, S v > S hmin hmin; thus, the minimum principal stress cannot be the vertical stress, so only reverse faulting is not possible. Case 4: μ = 0.6, S v = 12000 psi, S hmin hmin = 12000 psi (e) Any faulting regime is possible. In this case, S v = S hmin hmin ; thus, depending on the value of S hmax hmax, any of the three faulting regimes are possible because if S hmax hmax > S v = S hmin hmin, both reverse and strike-slip faulting are possible, and if S hmax hmax = S v = S hmin hmin, any of the three faulting regimes could take place due to perturbations of any of the components of the stress around an equal value.
Question 2: Constraints on
S hmax
Case 1: μ = 0.6, S v = 12000 psi, S hmin hmin = 8000 psi Since normal and/or strike-slip faulting are possible in this case, we will refer to equations 4.45 and 4.46 on Lecture 6 pg. 29 to determine the constraints on S hmax hmax based on frictional faulting theory. In addition, ad dition, since we know the value of Shmin and S hmax hmax cannot be less S hmin hmin , we know that the lower bound for S hmax hmax must be S hmin hmin = 8000 psi. To calculate the upper bound on S hmax hmax, we must use the relationship for strike-slip faulting equation (equation 4.46).
(S hmax – P p) / (S hmin – P p) = 3.1
(Equation 4.46)
S hmax = 3.1(S hmin – P p) + P p = 13712 psi In the case of strike-slip faulting, S hmax must be greater than S v = 12000 psi, but cannot be greater than 3.1(S hmin - P p) + P p, thus 8000 psi ≤ S hmax ≤ 13712 psi .
Case 2: μ = 0.6, S v = 12000 psi, S hmin = 9500 psi Since normal and/or strike-slip faulting are possible in this case, we will refer to equations 4.45 and 4.46 on Lecture 6 pg. 29 to determine the constraints on S hmax based on frictional faulting theory. In addition, since we know the value of Shmin and S hmax cannot be less S hmin, we know that the lower bound for S hmax must be S hmin = 9500 psi. To calculate the upper bound on S hmax, we must use the relationship for strike-slip faulting equation (equation 4.46). (S hmax – P p) / (S hmin – P p) = 3.1
(Equation 4.46)
S hmax = 3.1(S hmin – P p) + P p = 18362 psi In the case of strike-slip faulting, S hmax must be greater than S v = 12000 psi, but cannot be greater than 3.1(S hmin - P p) + P p, thus 9500 psi ≤ S hmax ≤ 18362 psi .
Case 3: μ = 0.6, S v = 12000 psi, S hmin = 10000 psi Since normal and/or strike-slip faulting are possible in this case, we will refer to equations 4.45 and 4.46 on Lecture 6 pg. 29 to determine the constraints on S hmax based on frictional faulting theory. In addition, since we know the value of Shmin and S hmax cannot be less S hmin, we know that the lower bound for S hmax must be S hmin = 10000 psi. To calculate the upper bound on S hmax, we must use the relationship for strike-slip faulting equation (equation 4.46). (S hmax – P p) / (S hmin – P p) = 3.1
(Equation 4.46)
S hmax = 3.1(S hmin – P p) + P p = 19912 psi In the case of strike-slip faulting, S hmax must be greater than S v = 12000 psi, but cannot be greater than 3.1(S hmin - P p) + P p, thus 10000 psi ≤ S hmax ≤ 19912 psi .
Case 4: μ = 0.6, S v = 12000 psi, S hmin = 12000 psi
Since all faulting regimes are possible in this case, we will refer to all equations on Lecture 6 pg. 29 to determine the constraints on S hmax based on frictional faulting theory. In addition, since we know the value of Shmin and S hmax cannot be less S hmin, we know that the lower bound for S hmax must be S hmin = 12000 psi. In this case, since reverse faulting is possible and is the most compressive stress regime, we must use the relationship for reverse faulting to calculate the upper bound of S hmax. (S hmax – P p) / (S v – P p) = 3.1
(Equation 4.47)
S hmax = 3.1(S v – P p) + P p = 26112 psi In the case of reverse faulting, S hmax must be greater than S v = S hmin = 12000 psi, but cannot be greater than 3.1(S v - P p) + P p, thus 12000 psi ≤ S hmax ≤ 26112 psi .
