Question

Exercise 3 An oil well in the X Field is producing at a stabilized rate of 600 STB/day at a stabilized bottom-hole flowing pressure of 1800 psi. Analysis of the pressure buildup test data indicates that the pay zone is characterized by a permeability of 120 md and a uniform thickness of 25 ft. The well drains an area of approximately 40 acres. The following additional data is available: r.= 0.25 ft B. = 1.25 bbl/STB A = 40 acres He =2.5 cp Calculate the pressure profile (distribution) and list the pressure drop across 1 ft intervals from rw to 1.25 ft, 4 to 5 ft, 19 to 20 ft, 99 to 100 ft. and 744 to 745 ft.

Answer 1

so) Given that 600 STB/day Pw 1800 psi Z R 120 Mo = 205 cp ru 0.25 ft N! 40 acres Bo 1:25 NV .bbl 1 STB we know that, Q. 0.00708x Rok *C Pe-Peo) MoxBox en re .ro marios MoeBoxGo xln/re = 0.00705 * R* (Pe-P) ra (pressure) ? Peio it lo * Bo * 20 r In 0.00 708 * R *h ra 1800 psi + (2.5) *(1.25)* (Goo) 0.00708 * (120) * (25) r en 0.25 1800 + 1875 21.24 0.25 encontre -> +60) P 1800 +(88.276) en X 0.25

ē 0.25 -1.25 interval. 0 ore Y=0.25 P = 180o + 88.276 en ent O 25 P. =: 1800 psi r = 1.25 1.25 po 1800 + 88276 en/ Bezs p=1942 psi = pressure drop = 1942. -1800 = 1142 psi 2) 4 - 5 interval rz u po 1800 + 88.276 en 0.25 pa nous psi ras PE 1800 + 88.276 en کی 5 8.25 P 2064 psi = preesure drop=19 psi رلي 19-20 internal r = 19 1800 + $8:276 en P. i (1995) 0.25 p = 2182 psi Y=20, P = 1800 +88.276 In 20 0.25 po 2186 Psi pressure drop = 14 psi 99-100 interval r=99 p = 1800 + 88.276 in (99 0-25 2328 psi

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