143006135-Petroleum-Production-Engineering.pdf

UNIVERSITY OF KIRKUK College of Engineering - Pet. Eng. Department

PETROLEUM PRODUCTION ENGINEERING BOYUN GUO  –  WILLIAM  WILLIAM C. LYONS  –  ALI  ALI GHALAMBOR 

CHPTER -3RESERVOIR DELIVERABILITY

Solution of Problems page 3/43 By: Hawar AbdulKhaliq Hamma Gul

1

3.1

Construct IPR of a vertical well in an oil reservoir.

Consider (1) transient flow at 1 month , (2) steady  –  state  state flow , and (3) pseudo  –  steady state flow. the following data are given: Porosity , ϕ = 0.25 Effective horizontal permeability , k = 10 md Pay zone thickness , h = 50 ft Reservoir pressure , pe = 5000 psi Bubble point pressure , p b = 100 psi Fluid formation volume factor , Bo = 1.2 Fluid viscosity , µ o = 1.5 cp. Total compressibility , Ct = 0.0000125 psi -1 Drainage area , A = 640 acre (re =2.980 ft) ft) Wellbore radius , r w = 0.328 ft Skin factor , S = 5

Solution : time = 30 day = 24 x 30 = 720 hr.

1. For transient flow, *

J=

 

   ( (   )     =   (( ) J*  = 0.1515215 bbl/day/psi

2

Calculated data points are: *

qo = J  (Pe –  pwf)

Pwf (psi) 0 1000 2000 3000 4000 5000

qo (stb/day) 757.6075 606.086 454.5645 303.043 151.5215 0

6000

5000

4000

    )    i    s    p     (     f 3000    w    p

2000

1000

0 800

700

600

500

400

300

qo (stb/day)

200

"Transient IPR curve"

3

100

0

2. For steady state flow, *

J=

     =   *+  *+

*

J = 0.13938 bbl/day/psi Calculated points are:

Pwf (psi) 0 1000 2000 3000 4000 5000

qo (stb/day) 696.9 557.52 418.14 278.76 139.38 0 6000

5000

4000     )    i    s    p     (     f 3000    w    p 2000

1000

0 800

700

600

500

400

300

200

100

qo (stb/day)

"Steady state IPR curve" 4

0

3. For pseudo steady state, *

J=

      =       

J*= 0.147202 bbl/day/psi

Calculated points are: Pwf (psi) 0 1000 2000 3000 4000 5000

qo (stb/day) 736.01 588.808 441.8 294.404 147.202 0 6000

5000

4000

    )    i    s    p 3000     (     f    w    p 2000

1000

0 800

700

600

500

400

300

200

100

qo (stb/day)

"

Pseudo-steady state IPR curve" 5

0

3.2 Construct IPR of a vertical well in an

saturated oil reservoir

Using Vogel's equation .the following data are given: Porosity , ϕ = 0.20 Effective horizontal permeability , k = 80 md Pay zone thickness , h = 55 ft Reservoir pressure , pe = 4500 psi Bubble point pressure , p b = 4500 psi Fluid formation volume factor , Bo = 1.1 Fluid viscosity , µ o = 1.8 cp. Total compressibility , Ct = 0.000013 psi -1 Drainage area , A = 640 acre (re =2.980 ft) Wellbore radius , rw = 0.328 ft Skin factor , S = 2

Solution : Assume pseudo-steady state flow , *

J=

     =       1.518476 STB/day/psi

=

         = 

q max =

= 3796.2 STB/day.

6

Calculated points by Vogel's equation:

q = q max

   [     ]

Pwf (psi) 0 1000 2000 3000 4000 4500

qo (stb/day) 3796.2 3477.507 2858.867 1940.28 721.7467 0

5000 4500 4000 3500 3000 2500 2000 1500 1000 500 0 4000

3500

3000

2500

2000

1500

1000

qo (stb) "

IPR curve"

7

500

0

    )    i    s    p     (     f    w    p

3.3

Construct IPR of a vertical well in an unsaturated oil reservoir

Using generalized Vogel's equation .the following data are given: Porosity , ϕ = 0.25 Effective horizontal permeability , k = 100 md Pay zone thickness , h = 55 ft Reservoir pressure , pe = 5000 psi Bubble point pressure , p b = 3000 psi Fluid formation volume factor , Bo = 1.2 Fluid viscosity , µ o = 1.8 cp. Total compressibility , Ct = 0.000013 psi -1 Drainage area , A = 640 acre (re =2.980 ft) Wellbore radius , rw = 0.328 ft Skin factor , S = 5.5

Solution : Assume pseudo-steady state flow ,

J*=

     =      

*

J = 1.300687 STB/day/psi qb = J * (Pe –  Pb) = (1.300687) (5000 –  3000) = 2601.4 STB/day.

   qv =  = (1.300687) (3000) / 1.8 = 2167.8 STB/day.

8

Calculated points by:

         q = J  (Pe  –  Pb) +  [     ] *

Pwf (psi) 0 500 1000 1500 2000 2500 3000 5000

qo (stb/day) 4769.2 4648.767 4431.987 4118.86 3709.387 3203.567 2601.4 0

6000

5000    i 4000     )    s    p     (     f    w 3000    p

2000

1000

0 6000

5000

4000

3000

2000

qo (stb)

9

1000

0

3.4

Construct IPR of a vertical well in an unsaturated oil reservoir

Using generalized Vogel's equation .the following data are given: Reservoir pressure , pe = 5500 psi Bubble point pressure , pb = 3500 psi Tested following bottom-hole pressure in well A , pwf 1 = 4000 psi Tested production rate from well A , q1 = 400 stb / day Tested following bottom-hole pressure in well B, pwf 1 = 2000 psi Tested production rate from well B , q1 = 1000 stb/day

Solution : Well A : Pwf1 > pb *

J  =

 

= 400 / (5500 - 4000) = 0.2667 stb/day/psi. *

qb = J  (Pe  –  Pb) = 0.2667 (5500  –  3500) = 533.4 stb/day

    qv =  = (0.2667) (3500) / 1.8 = 518.6 stb/day

Calculated points by:

         q = J  (Pe  –  Pb) +  [     ] *

10

Pwf (psi) 0 500 1000 1500 2000 2500 3000 3500 5000

qo (stb/day) 1052 1028.716 988.498 931.3461 857.2604 766.2408 658.2873 533.4 0

6000

5000

4000

    )    i    s    p 3000     (     f    w    p 2000

1000

0 1200

1000

800

600

400

qo (stb/day)

11

200

0

Well B : Pwf1 < pb

   []  = ]       [       

*

J  =

= 0.3111 stb/day/psi *

qb = J  (Pe  –  Pb) = 0.3111 (5500  –  3500) = 622.2 stb/day

  

qv =

= (0.3111) (3500) / 1.8 = 604.92 stb/day Calculated points by:

         q = J  (Pe  –  Pb) +  [     ] *

o

qo

   = 622.2 + .92  [  ] Pwf (psi) 0 500 1000 1500 2000 2500 3000 3500 5000

qo (stb/day) 1227.12 1199.96 1153.048 1086.384 999.9664 893.7967 767.8746 622.2 0

12

6000 5000 4000

    )    i    s    p     (     f 2000    w    p 3000

1000 0 1400

1200

1000

800

600

qo (stb/day)

13

400

200

0

3.5 Construct IPR of a well in saturated oil reservoir using both and Fetkovich's equation . The following data are given: Reservoir pressure , pe = 3500 psi Bubble point pressure , p b = 3500 psi Tested following bottom-hole pressure , pwf 1 = 2500 psi Tested production rate at pwf 1 , q1 = 600 stb / day Tested following bottom-hole pressure , pwf 2 = 1500 psi Tested production rate at pwf 2, q2 = 900 stb/day

Solution:

                =     

Vogel's equation: qomax.=

qomax.

calculated data points are: Pwf (psi) 0 500 1000 1500 2000 2500 3000 3500

qo (stb/day) 602.46 575.4108 528.6894 462.2958 376.2301 270.4922 145.0822 0

14

Vogel's equation

Fetkovich's equation:

        n=         () ()         c=   calculated data points are: q= 0.0023

 

Pwf (psi) 0 500 1000 1500 2000 2500 3000 3500

qo (stb/day) 1077.052 1059.431 1006.12 915.6465 785.0386 608.4835 372.5929 0

4000 3500 3000 2500 2000 1500 1000 500 0 1200

1000

800

600

400

200

qo (stb/day)

15

0

pwf(fetkovich's model) pwf(vogel's model)

3.6

Determine the IPR for a well at the time when the average reservoir pressure

will be 1500 psig. The following data are obtained from laboratory tests of well fluid samples: Reservoir pressure Average pressure (psi) Productivity index J* (stb/day-psi) Oil viscosity (cp) Oil formation volume factor (rb/stb) Relative permeability to oil

present 2200 1.25 3.55 1.20 0.82

Solution:

                   9534 stb/day-psi     J f =         *

Vogel's equation for future IPR :

      q=  [     ]

        () ()  Vogel's equation for present IPR :

      q=  [     ]

     ( )( )      Calculated data points are:

16

future 1500 -------

3.85 1.15 0.65

Reservoir press. = 2200 psi Pwf (psi) q (stb/day) 2200 1980 1760 1540 1320 1100 880 660 440 220 0

Reservoir press. = 1500 psi Pwf (psi) q (stb/day)

0 262.7782 501.1118 715.001 904.4458 1069.446 1210.002 1326.113 1417.78 1485.002 1527.78

1500 1350 1200 1050 900 750 600 450 300 150 0

0 136.654 260.596 371.826 470.344 556.15 629.244 689.626 737.296 772.254 794.5

2500

2000

1500

pwf (present) 1000

500

0 2000

1500

1000

500

0

qo (stb/day)

*in the above table, col. 2 calculated from eq.2 & col.4 from eq.1

17

pwf (future)

3.7 Using Fetcovich's method , plot the IPR curve for a well in which '

-4

 pi is 3000 psia and J o = 4×10  stb/day-psi . Predict the IPRs of the well at well shut-in static pressures of 2500 psia, 2000 psia, 1500 psia, and 1000 psia.

Solution: The value of J'O at 2500 psia is: '

'

J O = J i '

      -4

J O = 4*10 (

  

'

The value of J O at 2000 psia is: '

-4

J O = 4*10 (

  

 and the value of J'O at 1500 psia is: '

-4

J O = 4*10 (

  

and the value of J'O at 1000 psia is: '

-4

J O = 4*10 (

  

Calculated data points are: '

qo = J O

     

18

Pe = 3000 psi Pe = 2500 psi Pe = 2000 psi  pwf (psi) q(stb/day)  pwf (psi) q(stb/day)  pwf (psi) q(stb/day) 3000 2700 2400 2100 1800 1500 1200 900 600 300 0

0 684 1296 1836 2304 2700 3024 3276 3456 3564 3600

2500 2250 2000 1750 1500 1250 1000 750 500 250 0

0 570 1080 1530 1920 2250 2520 2730 2880 2970 3000

2000 1800 1600 1400 1200 1000 800 600 400 200 0

0 456 864 1224 1536 1800 2016 2184 2304 2376 2400

Pe = 1500 psi Pe = 1000 psi  pwf (psi) q(stb/day)  pwf (psi) q(stb/day) 1500 1350 1200 1050 900 750 600 450 300 150 0

0 342 648 918 1152 1350 1512 1638 1728 1782 1800

19

1000 900 800 700 600 500 400 300 200 100 0

0 228 432 612 768 900 1008 1092 1152 1188 1200

3500

3000

2500

2000     )    i

   s    p     (     f    w 1500    p

(pe=3000 psi) (pe=2500 psi) (pe-1000 psi) (pe=2000 psi)

1000

500

0 4000

3500

3000

2500

2000

1500

1000

500

0

qo (stb/day)

"IPR curve , problem 3.7"

20

(pe=1500 psi)