PNGE 333 Homework 6

Phillip Colasessano 2/28/2015

HOMEWORK  6  

1.       The   following   information   is   available   from   an   oil   reservoir.   There   were   two  producing  wells  in  the  reservoir  during  the  first  year.  Each  well  was  produced  at  a  constant  rate  of  300  STB/D.      

t  Days  

P  Pisa  

Bo  RB/STB  

Bg    RB/MCF  

RS  SCF/STB  

Rp  SCF/STB  

Wp  Bbls  

0   2500  (Pi  =Ps)   1.50   1.05   720   -­‐   -­‐  365   2450   1.45   1.15   670   720   0    

Determine   the   initial   oil   in   place   assuming   that   this   reservoir   is   a   solution-­‐gas  drive  reservoir.  You  may  ignore  the  rock  and  the  formation  water  expansion.      

2.   Repeat  problem  2  assuming  the  reservoir  is  surrounded  by  an  aquifer.  The  aquifer  properties  are  given  below:  

 

A  (Reservoir)   1200   acres  φ   0.1   fraction  h   25   ft.  

ct   7.00E-­‐06   1/psi  μ,  water   1   cp  

k   75   md  α   360   degrees  

 3.   The   reservoir   in   Problem   2  was   produced   for   another   year.   During   the   second  

year,  there  were  five  procuring  wells  in  the  reservoir.  Each  well  was  produced  at  a   constant   rate   of   300   STB/D.   The   following   information  were   obtained   at   the  end  of  the  second  year:    

t  Days  

P  Pisa  

Bo  RB/STB  

Bg    RB/MCF  

RS  SCF/STB  

Rp  SCF/STB  

Wp  Bbls  

730   2370   1.43   1.30   620   1125   0    Which   of   the   two   assumptions   (solution-­‐gas   drive   vs.   water   drive)   is   more  accurate?  

 4.   Calculate  the  initial  oil,  the  solution  gas,  and  the  free  gas  in  place  assuming  the  

reservoir  in  problem  3  has  also  an  initial  gas  cap.    

a.  Solution  Method:  

1. Find the Cumulative Oil Produced (Np), then the Bt , and use these values along with the given data to calculate the initial oil in place (N). 𝑁! = (𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛  𝑟𝑎𝑡𝑒 ∗ 𝑑𝑎𝑦𝑠  𝑒𝑙𝑎𝑝𝑠𝑒𝑑 ∗ 𝑛𝑢𝑚𝑏𝑒𝑟  𝑜𝑓  𝑤𝑒𝑙𝑙𝑠)

𝐵! = 𝐵! + (𝑅!" − 𝑅!)𝐵!

𝑁 =𝑁![𝐵! + 𝑅! − 𝑅!" 𝐵!]

(𝐵! − 𝐵!")

2. Find r2w , B , tD , pD , p’D , We , Gp , then use these values to find N

𝑟!! =𝐴 ∗ 43,560

𝛂360 π

𝐵 = 1.12ϕℎ𝑐!𝑟!!α360

𝑡! =0.00633𝑘𝑡ϕµμ𝑐𝑟!!

𝑝! =370.529𝑡!

!! + 137.582𝑡! + 5.6949𝑡!

!!

328.834+ 265.488𝑡!!! + 45.217𝑡! + 𝑡!

!!

𝑝!! =716.441+ 46.798𝑡!

!! + 270.038𝑡! + 71.0098𝑡!

!!

1296.86𝑡!!! + 1204.73𝑡! + 618.618𝑡!

!! + 538.072𝑡!! + 142.41𝑡!

!!

𝑊! =𝑊!!!! + 𝑡!! − 𝑡!!!!𝐵 𝑝! − 𝑝! −𝑊!!!!𝑝!

𝑝! − 𝑡!!!!𝑝!!

𝐺! = 𝑅! ∗ 𝑁!

𝑁 =𝑁!𝐵! + 𝐵! 𝐺! − 𝑁!𝑅! − (𝑊! −𝑊!)

𝐵! − 𝐵!" + (𝑅!" − 𝑅!)𝐵!

3. Repeat steps from 1. and 2. To compare but with 5 wells and new given data.

𝑁! = 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛  𝑟𝑎𝑡𝑒 ∗ 𝑑𝑎𝑦𝑠  𝑒𝑙𝑎𝑝𝑠𝑒𝑑 ∗ 𝑛𝑢𝑚𝑏𝑒𝑟  𝑜𝑓  𝑤𝑒𝑙𝑙𝑠

Solution-gas Drive

𝐵! = 𝐵! + (𝑅!" − 𝑅!)𝐵!

𝑁 =𝑁![𝐵! + 𝑅! − 𝑅!" 𝐵!]

(𝐵! − 𝐵!")

Water Drive

𝑡! =0.00633𝑘𝑡ϕµμ𝑐𝑟!!

𝑝! =370.529𝑡!

!! + 137.582𝑡! + 5.6949𝑡!

!!

328.834+ 265.488𝑡!!! + 45.217𝑡! + 𝑡!

!!

𝑝!! =716.441+ 46.798𝑡!

!! + 270.038𝑡! + 71.0098𝑡!

!!

1296.86𝑡!!! + 1204.73𝑡! + 618.618𝑡!

!! + 538.072𝑡!! + 142.41𝑡!

!!

𝑊! =𝑊!!!! + 𝑡!! − 𝑡!!!!𝐵 𝑝! − 𝑝! −𝑊!!!!𝑝!

𝑝! − 𝑡!!!!𝑝!!

𝑁 =𝑁! 𝐵! + 𝑅! − 𝑅!" 𝐵! − (𝑊! −𝑊!)

(𝐵! − 𝐵!")

4. Simultaneously determine the initial oil and gas in place using a plot. The equation of the line is the answer. 𝐹 = 𝐵! − 𝑅!𝐵! 𝑁! + 𝐵!𝐺! +𝑊!

𝐸! = 𝐵! − 𝐵!"

𝐸! = (𝐵! − 𝐵!")

𝐺! = 𝑁 ∗ 𝑅!"

b.  Results:    

1. 𝑁! = 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛  𝑟𝑎𝑡𝑒 ∗ 𝑑𝑎𝑦𝑠  𝑒𝑙𝑎𝑝𝑠𝑒𝑑 ∗ 𝑛𝑢𝑚𝑏𝑒𝑟  𝑜𝑓  𝑤𝑒𝑙𝑙𝑠 = 300 !"#!

∗

365  𝐷𝑎𝑦𝑠 ∗ 2  𝑤𝑒𝑙𝑙𝑠 = 219000  𝑆𝑇𝐵

𝐵! = 𝐵! + 𝑅!" − 𝑅! 𝐵! = 1.45+ 720− 670 ∗ 1.15 ∗ 10!! = 1.5075𝑅𝐵𝑆𝑇𝐵

𝑁 =𝑁! 𝐵! + 𝑅! − 𝑅!" 𝐵!

𝐵! − 𝐵!"

=219000  𝑆𝑇𝐵 1.5075+ 720− 720 ∗ 1.15 ∗ 10!!

1.5075− 1.50

= 44,019,000  𝑆𝑇𝐵

2.

𝑟!! =𝐴 ∗ 43,560

𝛂360 π

=1200 ∗ 43,560

(360360)π= 16638694.37  𝑓𝑡!

𝐵 = 1.12ϕℎ𝑐!𝑟!!α360

= 1.12 ∗ 0.1 ∗ 25 ∗ 7.00 ∗ 10!! ∗ 16638694.37 ∗360360

= 326.12  𝑏𝑏𝑙𝑠𝑝𝑠𝑖

𝑡! =0.00633𝑘𝑡ϕµμ𝑐𝑟!!

=0.00633 ∗ 75 ∗ 365

0.1 ∗ 1 ∗ (7.00 ∗ 10!!) ∗ 16638694.37 = 14.88

𝑝! =370.529𝑡!

!! + 137.582𝑡! + 5.6949𝑡!

!!

328.834+ 265.488𝑡!!! + 45.217𝑡! + 𝑡!

!!= 1.83

𝑝!! =716.441+ 46.798𝑡!

!! + 270.038𝑡! + 71.0098𝑡!

!!

1296.86𝑡!!! + 1204.73𝑡! + 618.618𝑡!

!! + 538.072𝑡!! + 142.41𝑡!

!!

= 0.03

𝑊! =𝑊!!!! + 𝑡!! − 𝑡!!!!𝐵 𝑝! − 𝑝! −𝑊!!!!𝑝!

𝑝! − 𝑡!!!!𝑝!!

= 0+ 14.88− 0326.12 2500− 2450 − 0 ∗ 0.03

1.83− 0 ∗ 0.03

= 132,586.49  𝐵𝑏𝑙𝑠

𝐺! = 𝑅! ∗ 𝑁! = 157680000

𝑁 =𝑁!𝐵! + 𝐵! 𝐺! − 𝑁!𝑅! − (𝑊! −𝑊!)

𝐵! − 𝐵!" + 𝑅!" − 𝑅! 𝐵!

=219000 1.45 + 1.15 ∗ 10!! ∗ 157680000− 219000 ∗ 670 − 132,586− 0

1.45− 1.50 + 1.15 ∗ 10!! 720− 670

= 26,340,866.67  𝑆𝑇𝐵

3.

𝑁! = 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛  𝑟𝑎𝑡𝑒 ∗ 𝑑𝑎𝑦𝑠  𝑒𝑙𝑎𝑝𝑠𝑒𝑑 ∗ 𝑛𝑢𝑚𝑏𝑒𝑟  𝑜𝑓  𝑤𝑒𝑙𝑙𝑠

= 300 ∗ 365 ∗ 5  𝑤𝑒𝑙𝑙𝑠 + 219000 = 766500

Solution-gas Drive

𝐵! = 𝐵! + 𝑅!" − 𝑅! 𝐵! = 1.43+ 720− 620 1.3 ∗ 10!! = 1.56𝑅𝐵𝑆𝑇𝐵

𝑁 =𝑁![𝐵! + 𝑅! − 𝑅!" 𝐵!]

(𝐵! − 𝐵!")=(766500) 1.56+ (1.3 ∗ 10!!)(1125− 720)

(1.56− 1.50)

= 26655037.5  𝑆𝑇𝐵

Percent Error: !"#$%&"".!"!!""##$%&.!!"#$%&"".!"

∗ 100 = −1.20%

Water Drive

𝑡! =0.00633𝑘𝑡ϕµμ𝑐𝑟!!

= 29.76

𝑝! =370.529𝑡!

!! + 137.582𝑡! + 5.6949𝑡!

!!

328.834+ 265.488𝑡!!! + 45.217𝑡! + 𝑡!

!!= 2.143

𝑝!! =716.441+ 46.798𝑡!

!! + 270.038𝑡! + 71.0098𝑡!

!!

1296.86𝑡!!! + 1204.73𝑡! + 618.618𝑡!

!! + 538.072𝑡!! + 142.41𝑡!

!!

𝑊! =𝑊!!!! + 𝑡!! − 𝑡!!!!𝐵 𝑝! − 𝑝! −𝑊!!!!𝑝!

𝑝! − 𝑡!!!!𝑝!!= 588,750.84  𝐵𝑏𝑙𝑠

𝑁 =𝑁! 𝐵! + 𝑅! − 𝑅!" 𝐵! − (𝑊! −𝑊!)

(𝐵! − 𝐵!")

=(766500 1.56+ (1125− 720)(1.3 ∗ 10!! − 588,750.84  𝐵𝑏𝑙𝑠

(1.56− 1.50)

= 16,842,523.5  𝑆𝑇𝐵

Percent Error: !"#$%&"".!"!!"#$%&%'.!!"#$%&"".!"

= 36.06%

Solution Gas Drive is more accurate 1.20%<36.06%

4.

𝐹 = 𝐵! − 𝑅!𝐵! 𝑁! + 𝐵!𝐺! +𝑊!

= (1.43− 620 ∗ 1.3 ∗ 10!! ∗ 766500

+ 1.3 ∗ 10!! 157680000 = 1142358.75  𝑆𝑇𝐵

𝐸! = 𝐵! − 𝐵!" = 1.56− 1.50 = 0.06  𝑅𝐵𝑆𝑇𝐵

𝐸! = 𝐵! − 𝐵!" = 2.5 ∗ 10!!𝑅𝐵𝑆𝐶𝐹  

y = 1,894,250,454.55x + 18,762,327.27

R² = 1.00

N=1,894,250,454.55 MMSTB

G=18,762,327.27 BCF

𝐺! =𝑁 ∗ 𝑅!"10! = 13508875.63  𝐵𝐶𝐹

y  =  1,894,250,454.55x  +  18,762,327.27  R²  =  1.00  

0.00E+00  5.00E+06  1.00E+07  1.50E+07  2.00E+07  2.50E+07  3.00E+07  3.50E+07  4.00E+07  4.50E+07  5.00E+07  

0   0.002   0.004   0.006   0.008   0.01   0.012   0.014   0.016  

y  =  F/Eo  

X  =  Eg/Eo