HW3 Solution

PNG 410: Applied Reservoir Engineering Home work assignment #3 Solution Problem 1: We have a single phase gas reservoir with the following field data: Contour Contour Thickness Pressure Water number area (acre) (ft) psia saturation 1 551 5 4200 0.34 2 410 6 4185 0.25 3 290 5 4102 0.20 4 142 4 4055 0.20 5 80 5 4044 0.20 6 0 -

Porosity 0.22 0.22 0.22 0.22 0.22 -

1.1 Please calculate the reservoir volume using trapezoidal rule and pyramidal rule. What is difference you get using the two different integration rules? a. Calculation of reservoir volume using trapezoidal rule: Z Z Z Z Z Vtrapzoidal  1 ( A1  A2 )  2 ( A2  A3 )  3 ( A3  A4 )  4 ( A4  A5 )  5 ( A5  A6 ) 2 2 2 2 2 5 6 5 4 5 = (551  410)  (410  290)  (290  142)  (142  80)  (80  0) 2 2 2 2 2 = 6226.5 acre-ft Z Z Z V pyramidal  1 ( A1  A2  A1 A2 )  2 ( A2  A3  A2 A3 )  3 ( A3  A4  A3 A4 ) 3 3 3 Z Z  4 ( A4  A5  A4 A5 )  5 ( A5  A6  A5 A6 ) 3 3 = 6113.1 acre-ft 1.2 Please calculate the area-averaged and volume-average pressure of the reservoir.

pavg , area 

Ap i

i

i

A



551 4200  410  4185  ...  80  4044  4154.08 psia 551  410  ...  142  80



551 5  4200  410  6  4185  ...  80  5  4044  4157.58 psia 551 5  410  6  290  5  142  4  80  5

i

i

pavg , area 

V p i

i

i

V

i

i

1.3 Please calculate the original gas in place, using the trapezoidal rule for the reservoir volume.

5

Vgas   Vi  i  (1  S wi )  998.767 acre-ft , here Vi is the volume of reservoir portion i 1

calculated from trapzoidal rule, and i and Swi are the corresponding porosity and water saturation.

1.4 If the gas formation volume factor at the average initial pressure is 0.00453 cu ft/SCF, what is the original gas in place in units of SCF? Vgas  43,560 cu ft/acre-ft  998.767 acre-ft = 43.5 MM CF convert to SCF under standard condition using Bgi, G  Vgas / Bg  9.60 MMM SCF

Problem 2: A small gas reservoir has an initial pressure of 3200 psia and temperature 220 deg F. The pressure-production history and gas volume factors are as follows: Bg Pressure Gp Z factor p/z slope Estimated G (psia) (MMSCF) (MM (cu assuming volumetric SCF) ft/SCF) reservoir) 3200 0 0.0052622 0.875337 3655.733 2925 79 0.0057004 0.866741 3374.711 -3.55725 1027.685 2525 221 0.0065311 0.857427 2945.476 -3.02278 1116.427 2125 452 0.0077360 0.854542 2486.711 -1.98599 1483.124

2.1 Calculate the initial gas in place using production data at the end of each production intervals, assuming volumetric behavior. For each pressure, calculate the z factor using Bg pr zT Bg  0.02829 r , z =  0.0609Bg pr pr 0.02829  (220  460) use the above z formula to calculate z at each p, and p/z values. p ( ) z =- pi . For each interval, p/z vs Gp plot has slope of ziG G p G can be estimated for each interval, as seen in the table.

2.2 Is this reservoir a volumetric reservoir or water drive reservoir? Explain. 2.2 G value is not a constant, indicating water drive reservoir.

2.3 Plot p/z vs cumulative production.

p/z vs Gp 3500 3000

p/z

2500 2000 1500 1000 500 0 0

100

200

300

400

500

Gp (MM SCF)

2.4 If the initial volume of gas in place is 1018 MM SCF, and if there is no appreciable water production, what is the volume of water invaded at the end of each period? Using the following equation: G(Bg  Bgi )  We  G p Bg  BwW p Because Wp is negligible, We  G p Bg  G (Bg  Bgi ) at each interval, there is G p , Bg ,Bgi , and G values. First interval; We1  79 MM SCF  0.0057004 cu ft/SCF - 1018 MM SCF  (0.0057004-0.0052622) = 0.004244 MM SCF = 4244 SCF = 4244 SCF/(5.615 cu ft/bbl)= 755.8 STB We2  27004.97 STB We3  174237.5 STB

Problem 3: The production of a gas field is given below: p/z (psia) Gp (MMM SCF) 6553 0.393 6468 1.642 6393 3.226 6329 4.260 6246 5.504 6136 7.538 6080 8.749 3.1 draw p/z vs Gp. Is this a volumetric reservoir?

p/z vs Gp p/z = -56.889Gp + 6569.5 2 R = 0.998

6,600

p/z (psia)

6,500 6,400 6,300 6,200 6,100 6,000 0

2

4

6

8

10

Gp (MMM SCF)

This is a volumetric reservoir, because the p/z vs Gp figure is linear. 3.2 what is the initial gas in place? Initial gas in place is the Gp when p/z is zero. This give about 6569.5/56.889 = 115.5 MMM SCF. 3.3 What percentage of the initial gas in place will be recovered at a p/z of 900? At p/z = 900 psia, Gp = (900 – 6569.5) /(- 56.889) = 99.66 MMM SCF Recovery factor at that point = 99.66 / 115.5 * 100% = 86.3% Problem 4: Calculate the daily gas production including the condensate and water gas equivalents for a reservoir with the following daily production: Separator gas production = 6 MM SCF Condensate production = 100 STB Stock tank gas production = 21 M SCF Fresh water production = 10 bbl Initial reservoir pressure = 6000 psia Current reservoir pressure = 2000 psia Reservoir temperature = 225F Water vapor content of 6000 psia and 225 deg F = 0.86 bbl/MM SCF Condensate gravity = 50API

Solution:

API  50 141.5 o   0.78 131.5  API 42.43   o M wo   144.84 1.008   o GEgasCondensate  133000 

o M wo

 716.51 SCF

GE water  7390 SCF G p  G ps  Gst  GEgasCondensate  GEwater = 6000000+21000+716.51100+7390  10 = 6,166,551 SCF