Part 3 Kicks and Gas Migration By Prof. Dr. Abdel-Alim Hashem
Contents •
Density of real gases
•
Equivalent Mud Weight (EMW)
•
Wellbore pressure before and after kick
•
Gas migration rate - first order approx.
•
Gas migration rate – with temperature, mud compressibility and Z-fact Z -factor or considerations 2
Contents •
Density of real gases
•
Equivalent Mud Weight (EMW)
•
Wellbore pressure before and after kick
•
Gas migration rate - first order approx.
•
Gas migration rate – with temperature, mud compressibility and Z-fact Z -factor or considerations 2
Density of Real Gases g
n
m
nM
V pV
V
ZRT
pV g
g
(Real Gas Law)
m = mass M
ZRT V M M M a ir
M = molecular weight n = no. of moles
pM
gg = S.G. of gas
ZRT
29 29 g
g
p
ZRT
3
Density of Real Gases •
What is the density dens ity of a 0.6 gra gravity vity gas at o 10,000 psig and 200 F?
•
From Lesson 2, following figure
•
ppr = p/ppc = 10,015/671 = 14.93
•
Tpr = (200+460)/358 =
•
Z = 1.413
1.84
4
1.84
1.413
14.93 5
Density of Real Gases g
g
29 g g p ZR T
{
p = 10,000 psig o
T = 200 F gg = 0.6
29 (0.6 ) 10,015 1.413 (80 .28 ) 660 g =
2.33 ppg 6
Equivalent Mud Weight, EMW
•
The pressure, p (psig) in a wellbore, at a depth of x (ft) can always be expressed in terms of an equivalent mud density or weight.
•
EMW = p / (0.052 * x) in ppg
7
EMW po=0
0
•
EMW is the density of the mud that, in a column of height, x (ft) will generate the pressure, p (psig) at the bottom, if the pressure at top = 0 psig
x
or, at TD: •
p = 0.052 * EMW * TVD TVD
p 8
0
2,000
4,000 t f , h t p e D
6,000
8,000 EMW
p
0.052 * Depth
10,000
12,000 0.0
10.0
20.0
30.0
40.0
50.0
EMW, ppg 9
0
SICP = 500 psig 2,000
4,000 t f , h t p e D
After Kick 6,000
8,000
Before Kick 10,000
12,000 0
1,000
2,000
3,000
4,000
5,000
Annulus Pressure, psig
6,000
7,000
8,000 10
Gas Migration •
Gas generally has a much lower density than the drilling mud in the well, causing the gas to rise when the well is shut in.
•
Since the gas, cannot expand in a closed wellbore, it will maintain its pressure as it rises (ignoring temp, fluid loss to formation, compressibility of gas, mud, and formation)
•
This causes pressures everywhere in the wellbore to increase. 11
P3
P2
1
P1
2
3 12
Gas Migration Example 1: A 0.7 gravity gas bubble enters the bottom of a 9,000 ft vertical well when the drill collars are being pulled through the rotary table.
Flow is noted and the well is shut in with an initial recorded casing pressure of 50 psig. Influx height is 350 ft. Mud weight o = 9.6 ppg. Assume surface temperature of 70 F. Temp o gradient = 1.1 F/100 ft. Surface pressure = 14 psia •
Determine the final casing pressure if the gas bubble is allowed to reach the surface without expanding
•
Determine the pressure and equivalent density at total depth under these final conditions 13
Gas Properties at Bottom •
First assumption: BHP is brought to the surface
•
Pressure at the top of the bubble
•
P8,650 = 14 + 50 + 0.052 * 9.6 * (9,000-350) = 4,378 psia
•
T9,000 = 70 + (1.1/100) * 9,000 + 460 o
= 629 R
14
Gas Properties at Bottom •
ppc = 666 psia
•
Tpc = 389 deg R
•
ppr = 4,378/666 = 6.57
•
Tpr = 629/389 = 1.62
•
Z = 0.925 pseudocritical
-
pseudoreduced
15
Bottomhole Pressure
29 g p g
•
g = 29*0.7*4,378 / (0.925 * 80.28 * 629) = 1.90 ppg
•
DpKICK = 0.052 * 1.9 * 350 = 35 psi
•
BHP = 4,378 + 35 BHP = 4,413 psia
ZRT
(~surface press.? 16
Pressure at Surface •
Assume, at first, that Zf = 1.0 (at the surface)
•
Then,
pV ZnRT
BOTTOM
SURFACE
4,378V po V 0.925 * nR * 629 1.0 * nR * 70 460
4,378 0.925 * 629
po 1.0 * 70 460
so, po = 3,988 psia
(with Temp. corr.) 17
Solution with Z-factor Corr. •
At surface: – ppr =
3,988 / 666 = 6.00
– Tpr =
530 / 389 = 1.36
– Zf =
0.817
p0 = 3,258 psia 18
Solution with Z-factor •
A few more iterative steps result in – Z0 =
•
0.705 and p0 = 2,812 psia
At the surface
29 g p g
•
ZRT
f = 29*0.7*2,812 / (0.705*80.28*530) = 1.9 ppg 19
New BHP & EMW •
New BHP = 2,812 + 0.052 * 1.9 * 350 + 0.052 * 9.6 * 8,650
•
New BHP = 7,165 psia
•
EMW = (7,165 - 14)/(0.052 * 9,000)
•
EMW = 15.3 ppg
20
64 psia
2,812 psia
530 R
Mud 9.6 ppg
ΔPmud=2,626
8,650’
7,378 psia
9,000’
4,413 psia
psi
Gas 0.1 bbl/ft
1. 2. 3. 4. 5.
4,413 psia 4,378 3,988 (T) 3,258 (Z) 2,812 (Z)
6.
2,024 (mud)
7,004 psia 7,179 psia 21
Compression of Mud in Annulus vA = 0.1 bbl/ft) •
DV = compressibility * volume * Dp
•
= -6 * 10-6 (1/psi) * 0.1(9,000-350)*2,626
•
DV = -13.63 bbls
•
Initial kick volume = 0.1 * 350 = 35 bbls
•
New kick volume = 35 + 13.63 = 48.63 bbl 22
Compression of Mud in Annulus • From Boyle’s Law, pV = const –
p2 * 48.63 = 2,812 * 35
–
p2 = 2,024 psia
p8650
poA
poB
poC
Consider:
V,p,Z const. p,Z change mud comp.
2nd iteration ? ……………. 3rd or, Is there a better way ? 23
Gas Migration Rate •
A well is shut in after taking a 30 bbl kick. The SIDPP appears to stabilize at 1,000 psig. One hour later the pressure is 2,000 psig. Hole Cap =
0.1 bbl/ft
MW
=
14 ppg
TVD
=
10,000 ft
24
Gas Migration Rate
•
How fast is the kick migrating?
•
What assumptions do we need to make to analyze this question?
25
P3
P2
1 Hr
1
P1
2
3 26
First Attempt •
If the kick rises x ft. in 1 hr and the pressure in the kick = constant, then the pressure increases everywhere,
•
Dp = 0.052 * 14 * x
•
x = (2,000 - 1,000) / (0.052 * 14)
•
x = 1,374 ft
•
Rise velocity = 1,374 ft/hr 27
Gas Migration Rate •
Field rule of thumb ~ 1,000 ft/hr
•
Laboratory studies ~ 2,000 – 6,000 ft/hr
•
Who is right?
•
Field results?
•
Is the previous calculation correct? 28
Second Attempt •
Consider mud compressibility
•
Hole capacity = 0.1 bbl/ft * 10,000 ft = 1,000 bbl of mud
•
Volume change due to compressibility and increase in pressure of 1,000 psi,
•
DV = 6*10-6 (1/psi) * 1,000 psi * 1,000 bbl = 6 bbl 29
Second Attempt •
i.e. gas could expand by 6 bbl, to 36 bbl
•
Initial kick pressure =1,000 + 0.052 * 14 * 10,000 (approx.) = 8,280 psig = 8,295 psia 30
Second Attempt •
A 20% expansion would reduce the pressure in the kick to ~ 0.8*8,295 = 6,636 psia = 6,621 psig
•
So, the kick must have migrated more than 1,374 ft!
31
Second Attempt •
How far did it migrate in 1 hour?
•
The pressure reduction in kick fluid = 8,260 - 6,621=1,659 psi
•
The kick must therefore have risen an additional x2 ft, given by: 1,659 = 0.052 * 14 * x 2 x2 = 2,279 ft 32
Second Attempt •
2nd estimate = 1,374 + 2,279 = 3,653 ft/hr
•
What if the kick size is only 12 bbl?
•
What about balooning of the wellbore?
•
What about fluid loss to permeable formations? T? Z?... 33
Bore Hole Ballooning
34
Example •
Kick occurs. After shut-in, initial csg. Press = 500 psig. 30 minutes later, p = 800 psig
•
What is the slip velocity if the kick volume remains constant?
•
MW = 10.0 ppg
35
Simple Solution
p 2 p1 psi v slip psi g t 2 t1 hr ft v slip
Ignoring temperature, compressibility and other effects.
800 500 0.052 10.0 0.5
v slip 1,154 ft / hr
What factors affect gas slip velocity, or migration rate? 36
Gas slip velocity •
The bubble size, and the size of the gas void fraction, will influence bubble slip velocity.
• The “void fraction” is defined as the ratio (or
percentage) of the gas cross-sectional area to the total flow area.
37
Gas Slip Velocity Bubbles
with a void fraction > 25% assume a bullet nose shape and migrate upwards along the high side of the wellbore concurrent with liquid backflow, on the opposite side of the wellbore
38
Gas slip velocity •
Large bubbles rise faster than small bubbles • Other factors: – Density differences – Hole geometry – Mud viscosity – Circulation rate – Hole inclination o • One lab study showed max. rate at 45 . 39
