ANALYSIS OF GAS AND OIL FIELDS
AUTHOR: MORETTI CHRISTIAN, [email protected]
1.i We can define as ‘gas field’: a field where there are not changes of phase, in the reservoir itself, when pressure drops and so the fluid expands. At surface (where I have lower temperature and pressure) some liquid may be produced and the cause is the condensation of the gas steam when the thermal variables changes (lower temperature). Between the gas fields we can so distinguish two categories:
• Dry gas fields where there is not any production of liquid • Wet gas fields where the liquid is produced only at surface and not in the reservoir.
Regarding gas condensate fields, contrariwise in these fields there is the production of oil in the reservoir itself as pressure drops (the pressure at which there is the change of phase is called dew pressure). This type of fields is not an oil field even if there is the production of liquid, because in this case when we reach the transition pressure there is a process of transformation from a single phase (gas) to a double phases where the denser phase is oil and the less dense gas. When we have the combined production of oil and gas, the oil volume shrinks because of the exsolution of the gas dissolved in the solution with oil at reservoir condition. The solution gas is this part of the produced gas (the gas produced is in fact both solution gas from the oil and free gas, which was gas also at reservoir condition). This concept bring to the classification of two types of oils: saturated and undersaturated oil. A saturated oil is defined as one that cannot hold any more gas and so this is below the bubble point and the volume of oil decreases as pressure drops. Instead an undersaturated oil can dissolve more gas if it is available and the volume of oil increases as the pressure drops, until the bubble point is reached and the oil becomes saturated. The free water level (Zw) is the point at which the water and oil have the same pressure and so it is the meeting point in the graph (depth-‐pressure) between the pressure line of the gas and the pressure line of the water. If we define (P2,Z2) and (P3,Z3) as two points respectively of oil (with density Do) line and of water (with density Dw) line we can calculate the free water level as:
𝑍𝑤 =𝑃% − 𝑃' + 𝑍'𝐷*𝑔 − 𝑍%𝐷,𝑔
𝐷*𝑔 − 𝐷,𝑔
It should be considered that: the free water level is not the point of contact between oil and water but only the point with the same pressure because the physical point of contact can be measured only experimentally (e.g. by log measurements). 1.ii I have a gas field and it might be produced by a natural water drive. So, we consider the impact of the acquifer influx on the behaviour of the reservoir. Compared to the case without water influx, aquifer support decreases the reservoir volume of gas present in the field by We (the volume of water encroachement measured at reservoir conditions so the strength of the acquifer) and the surface volume by We/Bg.
𝐺 − 𝐺. =𝑆01 ∅𝑉𝐵0
−𝑊7𝐵0
The resulting material balance equation is:
𝐺. = 𝐺 1 −𝐵01𝐵0
+𝑊7𝐵0
The equation shows that the water influx increases production for a given pressure drop, and this is an advantage, but there is also the disadvantage that the residual gas leads to a lower final recovery. In conclusion: the gas is produced faster but the cumulative is generally lower. The pot aquifer model assumes that the water influx is due to the expansion of the rock and the water that saturates the pore space. Now we can introduce the compressibility c (of water and rock) and passing for this definition:
𝑊7 = 𝑊𝑐∆𝑃 We can obtain a new formulation of the equation and it can be rearranged to get a straight line in this way:
𝐺.
1 −𝐵01𝐵0
= 𝐺 +𝑊𝑐∆𝑃
𝐵0 − 𝐵01
Where the axis of the graph are ;<
=>?@A?@
and ∆BC@>C@A
and the slope 𝑊𝑐 (we do not necessarily need to
know W and c separately) and the intercept G. The original gas in place G and Wc have to be estimated, starting from the assigned data. In the analysis the connate water and the rock compressibility are considered negligible. Using the last equation the numerical result is:
Gp (10^6 scf) p (Mpa) Bg (rb/scf) Y (MPa scf/rb) X (10^6 scf)
0 41 0,000203 60 40 0,000224 640,00 47619,04
118 39 0,000255 578,65 38461,53 170 38 0,000305 508,33 29411,76 208 37 0,000362 473,55 25157,23
Now I can plot these results:
Figure 1 Results from the material balance
Starting from these points, it is possible to do a linear interpolation and the resulting right line is 𝑦 = 0,0074𝑥 + 289,26. If I insert the value x=0 I can obtain y=G=289,26 MMscf and within an acceptable range 260,33-318,18, whereas the slope 𝑊𝑐 turns to be 0,0074 rb/Pa and within an acceptable range 0,0067-0,0081 rb/Pa.
Figure 2. BEST FIT LINE MATERIAL BALANCE
1.iii Once known G, we can evaluate the value of the recovery factor as the ratio between the cumulative gas produced (GpNOW) and the original gas in place just calculated (G):
𝑅O =𝐺.PQR
𝐺 =208289,26 = 72%
From the following expression we can obtain the average gas saturation:
𝑆0 =𝐵0PQR 𝑆01
𝐵011 − 𝑅O = 0,38
Where: 𝑆01 = 1 − 𝑆*1 = 0,76.
0
100
200
300
400
500
600
700
0 10000 20000 30000 40000 50000
GAS FIELD
y = 0,0074x + 289,26R² = 0,99959
0
100
200
300
400
500
600
700
0 10000 20000 30000 40000 50000
GAS FIELD
The water influx necessary to sweep the entire gas field to residual gas saturation 𝑆0V is Wc∆𝑃 that is equal to the change in reservoir gas volume (1-‐𝑆*W − 𝑆0V)∅𝑉 hence:
Wc∆𝑃 =𝐺𝐵01 1 − 𝑆*W − 𝑆0V
1 − 𝑆*W
𝑃O = 𝑃1 −𝐺𝐵01 1 − 𝑆*W − 𝑆0V
𝑊𝑐 1 − 𝑆*W= 36 𝑀𝑝𝑎
The maximum recovery factor can be expressed as:
𝑅O^_` = 1 −𝑆0V
𝐵0PQR𝑆01𝐵01
The calculation of this parameter requires the estimation of Bg at the final pressure Pf=36 MPa and we can obtain it from an (exponential) interpolation of the values of Bg with respect to the pressure: BgNOW=0,00040 rb/scf and so Rf max is 0,81.
1.iv Some condensate liquid oil is produced dropping the pressure, it means that at our current pressure we are below the dew point. In this case (gas condensate field) the management is complex. We have two options to avoid the most valuable fraction of the hydrocarbon (the liquid) to be left behind in the reservoir:
• Maintain pressure as in an oil field (e.g. by injecting a dry heavier gas as CO2 which forms a gas cushion at the bottom of the reservoir);
• Drop pressure further until we are back in the single phase gas region. Then the reservoir can be produced as a normal gas field.
y = 0,0801e-‐0,147xR² = 0,98643
0
0,00005
0,0001
0,00015
0,0002
0,00025
0,0003
0,00035
0,0004
36,5 37 37,5 38 38,5 39 39,5 40 40,5 41 41,5
Bg-‐P
2.i The pressure in the reservoir depends on depth and density of the material above that interesting point. This dependence is shown by this relationship:
𝜕𝑃𝜕𝑧 = 𝜌𝑔
Gas and oil pressure are typically higher than water pressure in the surrounding aquifer because the density of this water is higher than gas and water densities.
2.ii We have to compare the expected pressure of water for the depth at which it is measured with the measured value in order to see if the reservoir is over pressure, under pressured or normally pressured. The expected value can be calculated with this expression:
𝑝R7`.7Wd7e = 𝑃_d^ + 𝜌*𝑔𝑧*
It turns to be 23.84 MPa. Since this value is higher than the measured 18.01 MPa, we can say the reservoir is under pressured. The reservoir so has been down thrown over geological time: more sediments have been deposited over the field since it was charged with oil and gas. 2.iii The free oil level and free water level are so calculated:
Depth
Pressure
Water Oil Gas
𝑍,(𝐹𝑂𝐿) =𝑝0 − 𝑝, + 𝜌,𝑔𝑧, − 𝜌0𝑔𝑧0
𝜌, − 𝜌0 𝑔= 2280 𝑚
𝑍*(𝐹𝑊𝐿) =𝑝, − 𝑝* + 𝜌*𝑔𝑧* − 𝜌,𝑔𝑧,
𝜌* − 𝜌, 𝑔= 2297 𝑚
Hence the depth of the oil column is Ho=Zw-Zo=2297-2280=17m. 2.iv We have now a further detail by log measurements: the depth of the oil/water contact, which is 2290 m (different from the calculated free level). The true contact lies above the point where the oil and the water pressures are the same, where capillary pressure is equal to the capillary entry pressure. If we calculate the new depth of the oil column considering this new parameter we would obtain Ho=Zw-Zo=2290-2280=10m 2.v Area 36000000 m^2 Avarage porosity 0,22 -‐ Net to gross 0,85 -‐ Oil formation volume factor 1,41 -‐
From these data, the original oil in place is:
𝑁 =𝐴 𝑁; ∅ (1 − 𝑆*)
𝐵,1= 61,56 𝑀𝑀𝑚'
3.i The Darcy’s law is used in order to solve the problem of finding the velocity profile in the pore spaces, considering the average flow. For a multiphase flow:
𝑞. = −𝐾 ∙ 𝑘V.𝜇.
∇𝑝. − 𝜌.𝑔
The subscript p in the equation refers to the phase for which the relation is written. Where: • 𝑞.
^t: is the Darcy’s velocity of phase p; it is not actually a local flow velocity, but it is the
volume of fluid flowing per unit area (including both solid and pore) per unit time; • ∇𝑝.
B_^
: is the gradient of the pressure of the fluid phase p;
• 𝜌.𝑔 B_^ : is the effect on the Darcy’s velocity due to gravity;
• 𝜌. u0^v : is the density of the fluid phase p;
• 𝜇. 𝑃𝑎 ∙ 𝑠 : is the viscosity of the fluid phase p; • 𝐾 𝑚% 𝑜𝑟 𝐷 : K is the permeability and it is a property of the porous medium. It indicates how
easily the fluids are able to flow throughout it. It is measured in [m2] but also in an another unit
of measurement which is the darcy [1𝐷 ≅ 10−12𝑚2 ] • 𝑘V. − : it is the relative permeability of phase p. It is not a property of the porous medium and
it is whithin the range 0 ≤ 𝑘V. ≤ 1 . It represents the weight with respect to the case in which there is only the phase p flowing. Thanks to this parameter it is possible to extend the Darcy’s law to multiphase flow assuming that each phase flows in its own sub-‐network of pore space, without affecting the flow in the other phases.
3.ii We consider the following equations:
𝑘V* ={|>}.% ~
}.�� 𝑘V, = 0.8 {�>}.% �
}.�� and fractional flow 𝑓* =
u�|�|
��|�|
������
Sw krw kro fw Dfw/DSw
0,2 0 0,255102041 0 0,23 2,36152E-‐06 0,211883673 2,22902E-‐05 0,000743006 0,26 3,77843E-‐05 0,173844898 0,0004345 0,007241673 0,29 0,000191283 0,140655102 0,002712507 0,030138962 0,32 0,000604548 0,111983673 0,010681744 0,089014535 0,35 0,001475948 0,0875 0,032634972 0,217566479 0,38 0,003060525 0,066873469 0,08385629 0,465868279 0,41 0,00567 0,049773469 0,185556476 0,883602265 0,44 0,00967277 0,035869388 0,350367923 1,459866348 0,47 0,015493907 0,024830612 0,555153841 2,056125338 0,5 0,02361516 0,016326531 0,743119266 2,47706422
0,53 0,034574956 0,010026531 0,873364725 2,646559774 0,56 0,048968397 0,0056 0,945912947 2,627535964 0,59 0,067447259 0,002716327 0,980260811 2,513489258 0,62 0,09072 0,001044898 0,994274058 2,367319185 0,65 0,119551749 0,000255102 0,998934227 2,219853837 0,68 0,154764315 1,63265E-‐05 0,999947256 2,083223451
Giving as input a range of values of Sw (and so also of So=1.Sw) we obtain the data above and the following plots. Near the value of Sw=0,53 there is the maximum of Dfw/DSw at which it is seen the change in concavity. Krp,the relative permeability of phase p represents the mobility of the phase as a fraction of what it would be for single-phase flow. Probably the wettability of the rock considered is water-wet because the maximum relative permeability is less than 0.2.
3.iii The slope of the tangent in the coordinates of the shock gives the dimensionless velocity of the shock.
0
0,05
0,1
0,15
0,2
0,25
0,3
0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8
kr
Sw
krw
kro
0
0,1
0,2
0,3
0,4
0,5
0,6
0,7
0,8
0,9
1
1,1
1,2
0,1 0,15 0,2 0,25 0,3 0,35 0,4 0,45 0,5 0,55 0,6 0,65 0,7 0,75
fw
Sw
Now we can see the different regions of the Buckeley-Leverett solution with the different regions of rarefaction and shock.
00,10,20,30,40,50,60,70,80,91
1,1
0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8
fw
Sw
fw-‐Sw
0
0,1
0,2
0,3
0,4
0,5
0,6
0,7
0,8
0 1 2 3 4 5 6
Sw
vD dimensionless velocity
Buckley -‐ Leverett solution
4.i Another fluid (gas/water) is injected into the reservoir through injection wells in order to: 1. maintain the reservoir pressure (above bubble point) and keep a high driving force to keep the oil flowing. 2.water (or gas) displaces the oil from the pore space of the rock, leading to high recovery. The water is denser than gas and oil and so it is put at the bottom of the reservoir (in the aquifer). Oil and water are immiscible (that is do not mix) and so there is a finite surface tension at the interface between the fluids. The reason why gas is injected at the top of oil column (in the gas cap) is that as reservoir drive mechanism gas cap can allow higher recoveries with respect to the case in which gas is trapped in solution in unsaturated oil. Furthermore, the gas at the top of the reservoir expands pushing oil downwards so that with the wise well completions near the base of the oil column, also the problem of excessive gas production in the oil field can be mitigated. 4.ii We assume that there is no active aquifer and we can ignore the compressibility for the formation. Hence the material balance equation is reduced to:
𝐹 = 𝑁 𝐸, + 𝑚𝐸0 That is:
𝑁. 𝐵, + 𝑅B − 𝑅t 𝐵0 = 𝑁 𝐵, − 𝐵,1 + 𝑅t1 − 𝑅t 𝐵0 + 𝐵,1𝑁𝑚𝐵0𝐵01
− 1
The approach is to reduce it to an equation of a straight line:
𝐹𝐸,
= 𝑁 + 𝑁𝑚𝐸0𝐸,
We take the production data and the fluid properties as function of the pressure and we compute F, E0 and Eg for each pressure value:
Then we plot F/E0 on the y axis and EG/E0 on the x axis.
The original oil in place N is represented by the intercept whereas the slope of the straight line is Nm. From this obtained slope than m that is the relative size of the gas cap can be easily found. We get N=102,16 MMstb and mN=5,93 MMstb. Therefore m=mN/N=0,058=5,8% and so the 5,8% of the total oil volume is the gas cap. We want now to relatively quantify how much recovery is due to gas expansion and how much to oil expansion. N [MMstb] m NE0/F mNEg/F
102,16 5,8% 34,5% 65,5% The contribution of the gas cap is much higher. Note that P��,�A�
�+ ^P�@,�A�
�≠ 1 , this is due to the fact that N and m are approximated esteems.
4.iv Now the recovery factor is:
𝑅𝑓 = P�P= �,%=
=}%,=�= 4%, ∆B
B= BA>B�
BA= 25 %
This value is a good value considering the ∆B
B related to our current recovery factor.
We are still producing from the field and so this is not the final recovery factor which for this type of reservoir is typically 20-70%.
y = 5,93x + 102,16R² = 0,99982
0
50
100
150
200
250
300
350
400
450
500
0 10 20 30 40 50 60 70
F/Eo
(MMstb)
Eg/Eo
Values of the final Rf near 70% are only possible for very slow production allowing the gas to displace oil down to very low final saturation. 5.i We consider a two phase flow with these characteristics: immiscible, incompressible, isothermal. This flow is through a simplified porous medium that is considered homogeneous, horizontal and one-dimensional. The material balance for these two phases leads to the equation:
𝜙𝜕𝐶𝑤𝜕𝑡 +
𝜕𝑞𝑤𝜕𝑥 = 0, 𝐶𝑛 + 𝐶𝑤 = 1
where Cw is the wetting phase saturation, Cn the non-wetting phase saturation and 𝜙 the constant porosity. The flux of the wetting phase qw can be described by the multiphase extension of Darcy’s law in one dimension:
𝑞* = −𝐾 ∙ 𝑘*𝜇*
𝜕𝑃𝑤𝜕𝑥
The same considerations are valid for the non-wetting phase:
𝑞�* = −𝐾 ∙ 𝑘�*𝜇�*
𝜕𝑃𝑛𝑤𝜕𝑥
Considering finite range of time ∆t and space ∆x :
• the input mass of tracer flowing is: 𝐴 ∆𝑡 𝑞𝑤 𝑥 𝐶 𝑥 . • the mass of tracer flowing out is: 𝐴 ∆𝑡 𝑞𝑤(𝑥 + ∆𝑥) 𝐶(𝑥 + ∆𝑥). • At time t the mass between the two fluxes: 𝐴 ∆𝑥 Ф 𝐶(𝑡) 𝑆𝑤 + 𝐴 ∆𝑥 𝑚𝐶(𝑡). • At time t+∆t the mass between the two fluxes: 𝐴 ∆𝑥 Ф 𝐶(𝑡 + ∆𝑡) 𝑆𝑤 + 𝐴 ∆𝑥 𝑚𝐶(𝑡 + ∆𝑡)
The balance becomes: A ∆t qw (C(x) – C(x + ∆x)) = Sw A ∆x Ф (C(t + ∆t) – C(t)) + A ∆x (mC(t + ∆t) – mC(t))
This balance can be also written, considering qw=qt, Sw around 1and ^
¡= e^
eW W d
as:
¢£ ¤ ¥ – ¤ ¥�∆¥
∆¥=
Ф ¤ £�∆£ –¤ £ �¦¤ £�∆£ –¦¤ £
∆£− qt §¤
§¥ = Ф §¤
§£+ §¦(¤)
§£
So the final conservation equation is:
𝑞𝑡 𝜕𝐶/𝜕𝑥 + 𝜕𝐶/𝜕𝑡 (Ф + 𝑑𝑚/𝑑𝐶) = 0 5.ii Considering that v=x/t:
∂C∂t =
dCdv∂v∂t =
dCdV
−vt
∂C∂x =
dCdv∂V∂t =
dCdV
1t
®dd e¡e¯− ¯
d e¡e¯ (Ф + e^
e¡) = 0
We can now obtain the researched expression of velocity:
𝑣 = 𝑞𝑡
Ф + 𝑑𝑚𝑑𝐶
5.iii Starting from these data: 𝑞d =
==}~ ^t
∅ = 0,25𝑚 𝐶 = '∗W
=�},%∗W
𝐶 = 1 u0^v
e^e¡= ('(=�},%¡) – },% '¡)
(= �},%¡)²
The resulting velocity of the solute is: 𝑣 = ®d
Ф�(v(³´µ,²¶) – µ,² v¶)(³ ´µ,²¶)²
= 42,37 𝜇𝑚/𝑠