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PETE 625Well Control
Lesson 4Gas Solubility
2
Contents
Solubility of Hydrocarbon Gases in Oil
Solubility of Non-Hydrocarbon Gases in Oil
Solubility in Water
Solution Volume Factors
Oil Mud Recommendations
Assignments
Homework #2:
Ch 1, Problems 11-21
Homework #3:
Ch 2, Problems 1-10
Read: Chapter 2 to p.45
4
Gas Solubility
Gas will dissolve to some extent in any drilling fluid, but this can generally
be ignored with a water base fluid.
Gas dissolves readily in oil base muds.
An operator drilling with a diesel or mineral oil must understand
this!
5
Gas Solubility
The solubility of a gas/liquid mixture may be expressed as the amount
of free gas (scf/bbl) that can go into solution at a given temperature
and pressure.
In general, solubility will increase as the pressure increases, and as the temperature decreases.
6
Gas Solubility
Solubility also increases as the molecular similarity between the gas
and liquid composition increases.
The bubble point pressure is the pressure at which the first bubble of
free gas breaks out of solution with a given solution gas/liquid ratio at a
given temperature.
7
Gas Solubility
Free gas cannot coexist with the liquid at pressures in excess of the
bubble point
At pressures above the bubble point, gas solubility approaches infinity. Only liquid is present.
8
Example 1.10
Using the data from Fig. 1.17, determine the amount of free gas remaining if
8,000 scf of methane are blended with
10 bbl of diesel.
p = 3,000 psia and
T = 100 oF.
9
Example 1.10
The system gas/oil ratio,
R = 8,000 scf/10 bbl = 800 scf/bbl
From Fig. 1.17, at 3,000 psia, the gas solubility is 530 scf/bbl (at 100 oF)
Therefore, 800 - 530 = 270 scf/bbl remain free
i.e. 2,700 scf of gas remain free (10*270)
10
Solubility of Methane in diesel (Fig. 1.17)
T = 100 oF
530
Sol
ubili
ty,
scf
/bbl
Pressure, psia
11
Methane solubility in # 2 diesel (Fig. 1.18)
Is anything wrong here?
At lower temperature, the solubility is higher (p.14)!
Sol
ubili
ty,
scf
/bbl
Pressure, psia
12
Methane solubility (Figs. 1.19 & 1.20)
Methane is most soluble in Conoco LTV oil, least in Mentor 28
Mentor 28
Higher solubility at lower temperature
Conoco LTV oil
100 oF
300 oF
Mentor 28
100 oF
13
Solubility of Gases
in Mentor 28(Fig. 1.21)
Ethane
Methane
14
Solubility in various fluids
13 ppgOil base mud
18 ppgOil base mud
Mentor 28
15
Solubility of Methane in distilled water (Fig. 1.22)
10,0
00 p
sia
5,000 psia
1,000 psia
Solubility Correction
Factors (Fig. 1.23)
70 oF
250 oF
Total Dissolved Solids, %
Temperature, oF
16Solubility of Gases in Diesel at 250 oF
Sol
ubili
ty in
1,0
00 s
cf/b
bl
17
Example 1.11 A 13.0 ppg 70:30 invert emulsion oil
mud consists of (by volume) 54% diesel, 23% CaCl2, 4% emulsifier, and 19% solids.
Estimate the natural gas solubility in the mud at 150 oF and 2,000 psia
Assume the gas is 95% hydrocarbon and 5% CO2
Water salinity is 200,000 ppm TDS Gas specific gravity is 0.65
18
Solution
First determine the carbon dioxide solubility in the oil and emuslifiers
In oil,
c
bso aT
pR
a, b and c are constants listed in the next slide
19
Solution
TABLE 1.3 – EQ. 1.45 CONSTANTS
20
CO2 Solubility
bblscfR
oilinCO
so /950150*059.0
000,21
7134.0
2
c
bso aT
pR
bbl/scf241150*135.0
000,2R
EmulsifiertheinCO1
8217.0so
2
For CO2, c = 1.0
21
Hydrocarbon Solubility in Oil
Next determine the hydrocarbon solubility in the oil and emulsifiers.
The constant c must first be calculated.
coil = 0.3576 + 1.168 g
+ (0.0027 - 0.00492 g)T
- (4.51*10-6 - 8.198*10-6 g)T2
22
Hydrocarbon Solubility in Oil
coil = 0.3576 + 1.168
+ (0.0027 - 0.00492)150
- (4.51*10-6 - 8.198*10-6)1502
coil = 1.0605
bblscfRso /408150*922.1
000,20605.1
2552.0
c
bso aT
pR
23
Hydrocarbon Solubility in Emulsifier
cemul = 0.4 + 1.65 g - 1.01g2
= 0.4 + 1.65 * 0.65 - 1.01 * 0.652 = 1.0458
Thus,
bblscfRse /252150*162.4
000,20458.1
1770.0
c
bso aT
pR
24
Solution
Solubility of CO2 in oil = 950 scf/bbl
Solubility of CO2 in emulsifiers = 241 scf/bbl
Solubility of HC in oil = 408 scf/bbl
Solubility of HC in emulsifiers = 252 scf/bbl
25
Solution
Mixture solubility in the oil
= 0.95 * 408 + 0.05 * 950 = 392 scf/bbl
Mixture solubility in the emulsifier
= 0.95 * 252 + 0.05 * 241 = 251 scf/bbl
From Fig. 1.22, at 150 oF and 2,000 psia,
HC solubility in fresh water = 12 scf/bbl
95% Hydrocarbons 5% CO2
26
12
Temperature, oF
Solubility of Methane in distilled water (Fig. 1.22)
2,000 psi
150 oF
27
Solubility Correction Factor for Salinity(Fig. 1.23)
150 oF
200,000 ppm
28
Solution
From Fig 1.23 the salinity correction factor is 0.4
Solubility of HC in salt water
= 12 * 0.4 = 5 scf/bbl
29
The solubility
of CO2
in fresh water
is 145 scf/bbl
(Fig. 1.25)
2,000 psi
150 oF
145
30
Salinity correction factor
is 0.45 so solubility of CO2
in salt water
= 145 * 0.45
= 65 scf/bbl
Fig. 1.26
31
Solution
Mixture solubility in the salt water,
= 0.95 * 5.0 + 0.05 * 65 = 8 scf/bbl
Finally, mixture solubility in whole mud
= 0.54 * 395 + 0.23 * 8 + 0.04 * 251
= 213 + 1.8 + 10
= 224 scf/bbl
oil water emulsifier
32
Example 1.12
Mud: 94% fresh water + 6% solids
Gas: 0.92 mole fraction of Methane
0.06 mole fraction of CO2
0.02 mole fraction of H2S
Estimate the natural gas solubility in the mud at 180 oF and 5,200 psia
33
Solution
The only component capable of dissolving any gas is the fresh water.
From Fig. 1.22, the solubility of methane in fresh water = 21 scf/bbl
From Fig. 1.25, the solubility of CO2 in fresh water = 182 scf/bbl
34
Solution The H2S partial
pressure = 0.02 * 5,200 = 104 psia
From Fig 1.27, the partial solubility of H2S is about 36 scf/bbl
~36
Solubility = 0.92*21+0.06*182+36 = 66 scf/bblMethane CO2 H2S
Solubility in whole mud = 0.94 * 66 = 62 scf/bbl
35
Gas in solution
Some free gas
Domino effect
36
Drilled gas
Rock removal rate
hr/ftRin144
ftind
4 2
22
b
hr
ft
3.183
Rd 32b
R ft
db in
37
Drilled gas
Entry rate of drilled gas
bb
bg2
b
TZ65.14*hrmin/60
520*0.1pS
3.183
Rd
bb
bgbgsc TZ
pSRdq
9.309
2
min
scf
38
Drilled gas
If circ. Rate = qm bbl/min,
then the ratio of gas to mud
bbl
scf
TZq
pSRdr
bbm
bgm 9.309
2
surfacetheat
39
Example 1.13
Gas sand thickness = 50 ft
Bit diameter = 12.25”
Drilling rate, R = 250 ft/hr
Depth = 6,000’
BHP = 3,000 psia
BHT = 140 oF
Mud Density = 10.5, ppg
Sand porosity = 25%
Gas Saturation = 80%
Circulation Rate = 8 bbl/min
40
Solution
What is the drilled gas concentration?
600*86.0*8*9.309
000,3*8.0*25.0*250*25.12r
2
m
All this gas goes into solution
bblscfrm /6.17
bbm
bgm TZq
pSRdr
9.309
2
41
Volume of Drilled Gas
Bubble point is reached at 70 psia and 90 oF. What is the volume of drilled gas?
The total downhole gas volume (from drilling through the 50 ft interval),
80.0*25.0702,9
12*50*25.12
4 322
in
bbl
ft
inftinVb
bblVb 5.1
42
Find Depth of Bubblepoint From the gas law
bottombubblept ZT
pV
ZT
pV
This would happen at a depth of 100’
bpb
bp 70
55099.0
600*86.0
000,35.1V
bblVbp 68
What happens to the mud above this point?
43
Why is this a problem for well control?
Vtotal = V1+V2
Vtotal = V1+V2
Vtotal < V1+V2
44
Solution volume factor For solutions, the final volume is less
than the sum of the component volumes.
Kick sizes are determined by pit volume gain
A large gas kick that dissolves in oil mud, will not result in as much pit gain as a similarly sized kick in water based mud.
45
Solution volume factors
821 scf/bbl of methane in diesel oil at 4,075 psia would have a volume factor of 1.254 bbl/STB.
An increase in pressure to 5,070 psia will reduce the volume factor to 1.225
See Table 1.4 and Fig. 1.31
[Compressibility = (1.254-1.225)/{(1.254)*(5,070-4075)}
= 23.2*10-6 per psi ]
pVcV
46
Diesel Oil
Compressibility~ 4*10-6 per psi
T = 100 oF
47
Example 1.14
10 bbl of methane enters the wellbore. No. 2 diesel oil is used as drilling mud.
Determine the surface pit gain if 400 scf/bbl is dissolved in the diesel.
At the bottom, circ. Pressure = 5,000 psia
Circulating Temp = 200 oF
48
Solution
From Fig. 1.31, for 400 scf/bbl at 5,000 psia and 200 oF, Bog = 1.128 bbl/STB
From Fig 1.31, for gas free diesel at 5,000 psia and 200 oF, Bong = 1.012
Expansion = 1.128 - 1.012 = 0.116 bbl/STB
49
50
Solution
400 scf of gas under downhole conditions
3
3000,5 6146.5
1
1
029.1
520
660
000,5
65.14400
ft
bbl
R
R
psia
psiaftV o
o
mudofbblpergasofbbl273.0V 000,5
51
Solution i.e. Downhole solution GOR = 0.273
bbl/bbl
Thus the pit gain is 0.116 bbl for each 0.273 bbl of free gas that has been
dissolved in the diesel
For the 10 bbl gas kick,
Pit gain = 0.116 bbl*(10 bbl/0.273 bbl) = 4.2 bbl
expansion
52
Solution
From Fig. 1.31, for 600 scf/bbl at 200 oF and 5,000 psia, Rso = 1.205 bbl/STB
1.205 – 1.012 = 0.193 bbl/STB
V5,000 = 0.273*(600/400) = 0.409 bbl/bbl
So, a 10 bbl kick would result in a pit gain of 0.193*(10/0.409) = 4.7 bbl
What happens in very deep wells, at very high pressures?
53
400 scf/bblor 0.273 bbl/bblunder bottomholeconditions
4.2 bbl pit gain
10 bbl kick
54
55
56
Oil mud recommendations
Gas wells can be drilled safely with oil base drilling fluids.
Certain precautions must be taken, like drilling with a rotating head, to
direct evolved gas away from the rig floor.
57
Oil mud recommendations Set limits on quantities of drilled gas
allowed in the annulus (by limiting the penetration rates, and the
number of gas sands penetrated)
Consider keeping annular back pressure above the bubble point, if possible (consider effect on penetration rate and fracture integrity)
58
Oil mud recommendations
Mud-gas separator must be properly designed and sized for potential
well control procedures.
Remember that pit gain on the surface, for a given kick size, will be smaller than it would be for water based drilling fluids.
59
Oil mud recommendations
Pit level alarms should be set at a lower level
Educate crews on the differences between oil and water based muds,
and on how kicks behave in the two systems.
Be alert