Question

Calculate the productivity index (PI) for a well producing under semi-steady state flow condition with following reservoir parameters. qo300 STB/day (constant) S -5 Bo.06 bbl/STB Ho-2.5 cp k =60 md h107 ft w 0.29 ft pe5000 psi at 1000 ft pwf3000 psi

Answer 1

Given data q_0 = 300 STB/day (constant) S = 5 B_0 = 1.06 bbl/STB mu_0 = 2.5 cp k = 60 md h = 107 ft r_w = 0.29 ft p_e = 5000 psi at 1000 ft p_wf = 3000 psi Q_0 = 0.00708 k_0 h (p_e - p_wf)/mu_0 B(lu(r_e/r_w) - 0.75 + 5) 300 = 0.00708 times 60 times 107 (5000 -3000)/2.5 times 1.06 (ln(r_e/0.29) - 0.75 + 5) r_e = 1.89 times 10^47 ft Productivity index for semi steady state flow J = 0.00708 kh/mu_0 B_0 [lu (r_e/r_w) - 0.75 + 5] = 0.00708 times 60 times 07/2.5 times 1.06 [lu (1.89 times 10^47/0.29) - 0.75 + 5] J = 0.15 bbl/(psi - D)