ANALYSIS  OF  GAS  AND  OIL  FIELDS    

AUTHOR:  MORETTI  CHRISTIAN,  christian.moretti@mail.polimi.it  

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

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200

300

400

500

600

700

0 10000 20000 30000 40000 50000

GAS  FIELD

y  =  0,0074x  +  289,26R²  =  0,99959

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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