Petroleum Engineering 405 Drilling Engineering
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Transcript Petroleum Engineering 405 Drilling Engineering
PETE 411
Well Drilling
Lesson 26
* Well Control *
* Variable Geometry *
1
Well Control-Variable Geometry
Initial Shut-In Conditions
Driller’s Method
Kick at Casing Seat
Kick at Surface
Wait and Weight Method
Kick at Casing Seat
Kick at Surface
2
Read:
Applied Drilling Engineering, Ch. 4
HW #14: Driller’s Method - due 11-11-02
HW #15: W&W Method - due 11-15-02
Quiz B
Thursday, Nov. 14, 7 - 9 p.m. Rm. 101
Closed Book
1 Equation sheet allowed, 8 1/2”x 11” (both sides)
3
Example Problem
1. Determine the pressure at the casing seat
at 4,000’ when using the old mud
(Driller’s) versus using the kill mud
(Wait and Weight) to circulate a gas
kick out of the hole.
2. Determine the casing pressure at the
surface when the top of the gas
bubble has just reached the surface,
for the same two mud weights used
above.
4
Example Problem
Well depth
Hole size
Drill pipe
= 10,000’
= 10.5”
= 4.5”, 16.60 #/ft
Drill Collars
= 8” * 3.5” * 500 ft
Surface casing = 4,000’, 13-3/8”, 68 #/ft
Mud Weight
= 10 ppg
Fracture gradient @ 4,000’ = 0.7 psi/ft
5
Example Problem
Shut in Annulus Pressure,
SICP = 400 psi
Shut in Drillpipe Pressure, SIDPP = 200 psi
Pit level increase = 20 barrels (kick size)
Drop the Z terms but consider Temperature
T at surface = 70 deg. F
Temperature gradient = 1.2 deg.F/100 ft
6
Initial (Closed-In) Conditions:
SIDPP 200 psi
SICP
400 psi
Pit Level Increase
20 bbl
Initial Mud Weight 10 # /gal
Initial mud gradient = 0.520 psi/ft
(0.052 * 10 = 0.520)
7
Initial (Closed-In) Conditions:
Bottom Hole Pressure,
P10,000
psi
200 (10,000 ft ) * (0.520
) 5,400 psi
ft
Annular Vol/ft outside Drill Collars,
vDC_Ann
gal bbl
2
2
2
(10.5 8 ) in (12 in)
2
4
231in 42 gal
VDC_Ann 0.04493 bbl/ft
( 20.28 ft/bbl )
8
vdp.csg = 0.13006 bbl/ft
4,000’
vdp,hole = 0.08743 bbl/ft
9,500’
vdc,hole = 0.04493 bbl/ft
10,000’
9
Height of Kick Fluid,
20 bbl
h10,000
445 ft
0.04493 bbl/ft
Hydrostatics in the Annulus,
5,400 400 0.520 * 9,555 Pkick_10,00 0
Hydrostatic Pressure across Kick Fluid,
Pkick_10,000 5,400 400 0.520 * 9,555
Pkick_10,000 31. 4 psi
10
Driller’s Method - kick at bottom
BHP P0 Δ PK0 Δ PMA
- - - (2)
Weight of Kick Fluid Pressure * area
lb
2
2
2
31.4 2 * (10.5 8 ) in
in 4
W 1,141 lb
F = P * A = W
11
SICP = 400 psi
4,000’
SIDPP = 200 psi
9,555’
9,500’
445’
10,000’
31.4
psi
PB = P 10,000 = 5,400 psi
12
Driller’s Method - kick at csg. seat
What is the pressure at 4,000 ft when the
top of the kick fluid first reaches that point?
V4,000 V10,000
0.08743 * h4,000
h4,000
P10,000
*
P4,000
T4,000
*
T
10,000
5,400 70 48 460
*
20 *
P
650
4,000
1,098,444
P4,000
13
Driller’s
Method
Top of
Kick at
Casing
Seat
4,000’
9,500’
10,000’
P = 5,400 psi
14
Driller’s Method - kick at csg. seat
Again,
BHP P4,000 Δ PK_4,000 Δ PMA
PK_4,000
weight
1,141 lbs
π
area
10.5 2 4.5 2 in
4
P K_4,000 = 16.1 psi
15
Driller’s Method - kick at csg. seat
BHP P4,000 Δ PK_4,000 Δ PMA
5,400 P4,000 16 0.52 * (6,000 h4,000 )
5,384 P4,000 3,120 0.52 * h4,000
2,264 P4,000
1,098,444
(0.52) *
P
4,000
16
Driller’s Method - kick at csg. seat
This results in the quadratic Eqn:
P
2
4,000
2,264 P4,000 571,191 0
With the solutions:
P4,000
2,264 2,264 4 * 571,191
2
2
P4,000 2,493 psi 0.6233 psi/ft
0.5
0.7
17
Driller’s Method - Top of Kick at Casing Seat
P0,ann = ?
P4,000 = 2,493 psi
4,000’
h4,000 = 441 ft
P = 16 psi
9,500’
10,000’
BHP = 5,400 psi
1,098,444
h4,000
P4,000
18
Driller’s
Method
Top of
Kick at
Surface
We need two
equations…
P0 = ?
4,000’
9,500’
10,000’
BHP = 5,400 psi
19
Driller’s Method - kick at surface
When the bubble rises, it expands. The
volume of the bubble at the surface is given by:
P10,000
V0 V10,000
P0
T0
T
10,000
(Z = const.)
5400 70 460
(0.13006) h0 20
P0 70 120 460
677,084
h0
Po
- - - (1)
20
Driller’s Method - kick at surface
Δ PK,0
weight
1,141 lb
π
area
2
2
2
12.415 4.5 in
4
Δ PK0 10.85 11 psi
21
Driller’s Method - kick at surface
From Eq. (2),
BHP P0 Δ PK0 Δ PMA
5,400 P0 11 0.52 * (10,000 h0 )
But, from Eq. (1),
677,084
h0
P0
22
Driller’s Method - kick at surface
BHP P0 Δ PK0 Δ PMA
So,
677,084
5,400 P0 11 0.52 10,000
P0
5,400 5,200 11 P0 P0 352,084
2
P0 189 P0 352,084 0
2
Quadratic equation . . .
23
Driller’s Method - kick at surface
189 1892 ( 4)(352,084)
P0
2
1
2
P0 695.34 psi 695 psi
677,084
h0
973.74
695.34
h0 974 ft
4,000’
9,500’
10,000’
24
Driller’s Method - kick at surface
P4,000 P0 0.52 * (4,000 974) Δ PKO
695 1,574 11
2,280 psi
( = 0.57 psi/ft )
Alternativ ely,
4,000’
P4,000 P10,000 (0.52) * (10,000 4,000)
5,400 - 3,120
2,280 psi
9,500’
10,000’
25
Driller’s Method. Top of Kick at Surface
P0,ann = 695 psi
h0 = 974 ft
PK,0 = 11 psi
4,000’
P 4,000 = 2,280 psi
9,500’
10,000’
P10,000 = ?
26
Wait and Weight Method
- kick at casing seat
Calculate the pressure at 4,000 ft when the
top of the bubble reaches this point.
Volume of bubble at 4,000 ft
V4,000 V10,000
P10,000
P
4,000
T4,000
T
10,000
27
Wait and
Weight
Method
Top of
Kick at
Casing
Seat
Old Mud
Kill Mud
4,000’
9,500’
10,000’
BHP = 5,400 psi
28
W&W - Pressure at top of kick at 4,000 ft
V4,000 V10,000
P10,000
P
4,000
T4,000
T
10,000
5,400
(0.08743) h4,000 20
P
4,000
1,098,444
h4,000
P4,000
578
650
- - - (6)
Also,
BHP P4,000 ΔPK_4,000 ΔPM ΔPM1
- - - (7)
29
Wait and Weight Method
- kick at surface
4,000’
Capacity inside drill string
= DP_cap. + DC_cap.
9,500’
10,000’
BHP = 5,400 psi
bbl
bbl
0.01422
* 9,500 ft 0.0119
* 500 ft
ft
ft
141 bbl
#
Quantity of 10.0
mud below the bubble.
gal
30
W&W - Pressure at top of kick
- kick at 4,000 ft
As before,
PK_4,000 16 psi
141 bbl
PM 0.052 * 10 *
0.08743 bbl/ft
PM 839 psi
SIDPP
Density of Kill Mud Old Mud Wt.
0.052 * 10,000
#
KWM 10.00 0.38 10.38
gal
31
W&W - Pressure at top of kick
- kick at 4,000 ft
141
ΔPM1 0.052 * 10.38 * 6,000 h4,000
0.08743
BHP P4,000 ΔPK_4,000 ΔPM ΔPM1
5,400 P4,000 16 839 2,368 (0.5398) h4,000
32
W&W - Pressure at top of kick
- kick at 4,000 ft
2,177 P4 000
1,098,444
(0.5398)
P
4,000
2
P4,000
2,177 P4,000 592,940 0
P4,000
2,177 2,177 4 * 592,940
2
P4,000 2,422 psi
2
0 ,5
0.61 psi/ft
33
Wait and
Weight
Method
Top of
Kick at
Surface
4,000’
9,500’
Old Mud
Kill Mud
10,000’
34
Wait and Weight Method
- kick at surface
Volume of gas bubble at surface:
P10,000
V0 V10,000
P0
T0
T
10,000
5,400 530
0.13006 * h0 20
P0 650
677,084
h0
P0
35
Wait and Weight Method
(Engineer’s Method) - kick at surface
P10,000 P0 ΔPK0 ΔPM ΔPM1
As before,
ΔPK,0 11 psi
Assume all 10 lb/gal mud is inside the
13 3/8” csg. Then the height of 10 lb mud:
141.0 bbl
hM
1,084 ft
0.13006 bbl/ft
36
Wait and Weight Method
(Engineer’s Method) - kick at surface
Hydrostatic head across the mud columns:
Old Mud:
ΔPM 0.052 * 10 * 1,084 564 psi
Kill Mud:
PM1 0.052 * 10.38 * (10,000 h0 1,084)
37
Hydrostatics in Annulus
W&W Method - kick at surface
P10,000 P0 ΔPK0 ΔPM ΔPM1
5,400 P0 11 564 0.5398 * (8,916 h0 )
12.14 P0 (0.5398) h0
38
Wait and Weight Method
kick at surface
From Eq. 4, substituting for h0
677,084
12 P0 (0.5398)
P0
P02 12P0 365,490 0
P0
12 12 2 4 * 365,490
1
2
2
P0 610.59 611 psi
39
Wait and Weight Method
- kick at surface
Height of Bubble at Surface
677,084 677,084
h0
P0
610.59
1,109 ft
40
Check Pressure at 4,000 ft
- kick at surface
P4,000 P0 ΔPK0 ΔPM ΔPM1
611 11 569
0.052 * 10.38 * (4,000 - 1,109 - 1,093)
2,161 psi
0.54 psi/ft
Looks OK
41
Wait and Weight Method
Top of Kick at Surface
Old Mud
P0,ann = 611 psi
Kill Mud
h0 = 1,109 ft
PK,0 = 11 psi
4,000’
9,500’
10,000’
POld Mud = 569 psi
P 4,000 = 2,161 psi
P10,000 = ?
42
Summary
Bubble at 10,000 ft
Driller’s
Method
P4,000
P0
Engineer’s
Method
2,480
2,480
400
400
43
Summary
Top of Bubble at 4,000 ft
Driller’s
Method
P4,000
P0
Engineer’s
Method
2,493
2,422
413
342
44
Summary
P4,000
P0
Top of Bubble at surface
Driller’s
Method
Engineer’s
Method
2,280
2,161
695
611
45
46
Why the difference?
47
Wait and Weight Method
Kick Intensity, ppg
48
Kick Intensity, ppg
Pump Strokes
49
Volume of Mud Pumped, bbl
50