NORTE DE SANTANDER
2.2 DESCRIPCIÓN DE LA RED VIAL
2.2.2 Red Vial Secundaria
during which gas is circulated out the choke line, free gas will remain trapped below the closed preventer. If the closed preventer is an annular preventer, it is possible for this volume of gas to be quite significant. In order to prevent a rapid unloading of the riser due to trapped gas when the annular preventer is opened or the introduction of a secondary kick due to light density drilling fluid in the riser, close the uppermost rams below the choke line and close the diverter. Open the preventer above the trapped gas and allow this gas to rise toward the surface. Displace the riser with kill fluid and reopen the rams. It may be necessary in extreme cases to close the bottom rams to isolate the hole and fill the riser by circulating through the kill line. This problem becomes more severe with increased water depth and/or preventer size.
2.7 - WORKSHOP 2
SCORE
1. There are a variety of mechanisms that can cause abnormal formation fluid pressures. List 4 of the principle causes below.
Answer (a) ————————
(b) ———————— (c) ————————
(d) ———————— 2
2. Shown below is a pressure versus volume plot of a leak off test.
The leak off was carried out with a 10.6 ppg mud. The casing shoe is at 4000ft TVD.
a. What is the maximum pressure that the exposed formations below the shoe can support?
ANSWER... 2 b. What is the “Fracture Gradient?”
ANSWER... 2
c. What is the maximum mud weight?
ANSWER... 2 d. If drilling was resumed and the mud weight was increased to
12.6 ppg. Calculate M.A.A.S.P ANSWER... 2 1200 1000 800 600 400 200 0 PRESSURE (PSI) VOLUME
SCORE
3. M.A.A.S.P. The maximum allowable annular surface pressure should be re-calculated..
a. At the start of each shift
b. As soon as possible after a drilling break c. When approaching a suspected transition zone
d. When the mud weight has been increased in the system e. If a kick has occurred and the well is shut-in
ANSWER... 2 4. The calculated M.A.A.S.P. value is relevant..
a. When the influx is in the open-hole section b. As the influx approaches the surface
ANSWER... 2
SCORE
5. Given the following data:
Depth 10000ft TVD Bit size 8 1/2" Shoe depth 8500ft TVD Mud weight 12.6 ppg Collars - 600ft. capacity = 0.0077 bbl/ft Metal displacement = 0.03 bbl/ft Drill-pipe 5" capacity = 0.0178 bbl/ft Metal displacement = 0.0075 bbl/ft
Casing/pipe annular capacity = 0.0476 bbl/ft
Casing capacity = 0.0729 bbl/ft
One stand of drill-pipe = 94ft.
Assuming the 12.6 ppg mud gives an over-balance of 200 psi.
a. If 10 stands of pipe are removed “dry” without filling the hole, what would be the resultant reduction in bottom-hole pressure?
ANSWER... 3 b. If 5 stands of pipe had been pulled “wet” without filling the
hole, the resultant reduction in bottom-hole pressure would be. ANSWER... 3 c. If prior to tripping a 20 barrel slug of 14.6 ppg mud was
displaced to prevent a wet trip, what would be the expected volume return due to the U-tubing of the heavy mud?
6. Prior to tripping out of the hole a trip tank and pump are lined up to keep the hole full as the pipe is removed. The trip tank contains 30 barrels of mud. After pulling 10 stands of pipe the level in the trip tank is 27 barrels. (Use data given in Question 6). Would the safest option be..
a. To continue tripping but flow-check when bits at shoe. b. Stop and shut the well in. If no pressures seen open
the well up and continue tripping.
c. Flow-check. If no flow, go back to bottom and circulate. d. Flow-check. If no flow, continue tripping
ANSWER... 2
SCORE
7. A well can be induced to flow by swabbing. Swabbing is the
reduction of bottom hole pressure due to the effects of pulling pipe. List below 3 conditions that can cause swabbing.
Answer (a) ———————
(b) ———————
(c) ——————— 2
8. A drill string consist of 5" 20 lb/ft drill-pipe and 8 1/2" drill-collars. The spare kelly cock has 4 1/2" I. F. thread connections. What crossover sub is required for the collars?
ANSWER... 2 9. A fixed rig is set in 300ft of sea water. The marine conductor has
been set X ft below the sea-bed. The flow line is 65ft above the mean sea-level. The strength of the sub-sea formations is 0.68 psi/ft. Sea-water gradient is 0.445 psi/ft. It is proposed to drill with 9.2 ppg mud. What is the minimum depth that the conductor has to be set below sea-bed to prevent losses?
ANSWER... 8 10. An over-balance or trip margin is added to the mud. When
tripping this will prevent a loss of B.H.P. due to the swabbing effect of pulling the pipe.
ANSWER. TRUE/FALSE 2 11. Assume casing is set at 4800ft TVD/MD and that gas sands were
encountered at 5000ft and at 8500ft. If the formation pressure gradient at 5000ft is 0.47 psi/ft and at 8500ft it is 0.476 psi/ft. What mud weight is required to give an over-balance or trip margin of 200 psi?
WORKSHOP 2 - Answers
1. (a) Under compacted shales
(b) Thick gas sands (c) Faults
(d) Diaprism salt domes
(e) Shape of reservoir structure 2. Surface pressure = 1100 psi
a. (CSG TVD x MUD WT x .052) + Surface pressure = (4000 x 10.6 x .052) + 1100 = 3305 psi
b. Frac g = Max press ÷ CSG TVD = 3305 ÷ 4000 = 0.826 psi/ft c. Max Mud Wt = Frac g ÷ .052
= .826 ÷ .052 = 1 5.88ppg
d. MAASP = (Max mud wt - Drlg mud wt) x .052 x CSG TVD
= (15.88 - 12.6) x .052 x 4000 = 682 psi
3. d. 4. a.
5. a. Mud g x Met Disp
————————— CSG Cap - Met Disp
= .655 x .0075 = .0049 = .0751 psi/ft
————— ———
.0729 – .0075 .0654
.0751 x 940 = 71 psi
b. Mud g x (Met Disp + pipe cap) ——————————————— Ann Cap = .655 (.0075 + .0178) = .3525 psi/ft ————————— .047 470ft x .3525psi/ft = 166 psi
c. Dry pipe vol = Slug vol x (slug wt) ————— - 1 (mud wt) = 14.6 20 X ———– – 1 12.6 = 3.17 bbls 6. c.
7. (a) Pulling speed.
(b) Mud Properties, viscosity - Gel strength. (c) Profile of hole (Wellbore geometry). 8. 4 1/2" if box - 6 5/8" reg pin.
9. (Hyd mud to sea bed) - (Hyd sea water) —————————————————— (Frac g - Mud g) = (365 x 9.2 x .052) - (300 x .445) ————————————— (.68 - .478) = 41.1 —— .202 203 ft. 10. False.
11. Mud Wt to give 200 psi overbalance = 5000 x .47 psi/ft = 2350 + 200
= 2550
∴ 2550 ÷ 5000 ÷ .052 = 9.8 ppg
If the 200 psi is to overbalance formation pressure at 8500ft mud wt would be 9.6 ppg. This would overbalance the sands at 5000ft by 148 psi.