Page 36 - Petroleum Production Engineering, A Computer-Assisted Approach

Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap02 Final Proof page 25 22.12.2006 7:08pm

PROPERTIES OF OIL AND NATURAL GAS 2/25
Table 2.4 Results Given by the Spreadsheet Program Brill-Beggs-Z.xls
Brill-Beggs-Z.xls
Description: This spreadsheet calculates gas compressibility factor based on the Brill and
Beggs correlation.
Instruction: (1) Select a unit system; (2) update data in the Input data section;
(3) review result in the Solution section.
U.S.
Input data Field units SI units
Pressure: 5,000 psia
Temperature: 180 8F
Gas specific gravity: 0.65 air ¼ 1
Mole fraction of N 2 : 0.1
Mole fraction of CO 2 : 0.08
Mole fraction of H 2 S: 0.02
Solution
Pseudo-critical pressure ¼ 697 psia
Pseudo-critical temperature ¼ 345 8R
Pseudo-reduced pressure ¼ 7.17
Pseudo-reduced temperature ¼ 1.95
A ¼ 0.6063
B ¼ 2.4604
C ¼ 0.0395
D ¼ 1.1162
Gas compressibility factor z ¼ 0.9960

Example Problem 2.5 A gas from oil has a specific 2.3.6 Formation Volume Factor of Gas
gravity of 0.65, estimate z-factor and gas density at Gas formation volume factor is defined as the ratio of gas
5,000 psia and 180 8F. volume at reservoir condition to the gas volume at stan-
dard condition, that is,
Solution Example Problem 2.5 is solved with the V p sc T z zT
spreadsheet program Hall-Yarborogh-z.xls. The result is B g ¼ ¼ ¼ 0:0283 , (2:62)
V sc p T sc z sc p
shown in Table 2.5.
Table 2.5 Results Given by the Spreadsheet Program Hall-Yarborogh-z.xls
Hall-Yarborogh-z.xls
Description: This spreadsheet computes gas compressibility factor with the Hall–Yarborough method.
Instruction: (1) Select a unit system; (2) update data in the Input data section;
(3) click Solution button; (4) view result.
U.S.
Input data Field units SI units
Temperature: 200 8F
Pressure: 2,000 psia
Gas-specific gravity: 0.7 air ¼ 1
Nitrogen mole fraction: 0.05
Carbon dioxide fraction: 0.05
Hydrogen sulfite fraction: 0.02
Solution
T pc ¼ 326 þ 315:7(g g 0:5) 240y N 2 83:3y CO 2 þ 133:3y H 2 S ¼ 375.641 8R
p pc ¼ 678 50(g g 0:5) 206:7y N 2 þ 440y CO 2 þ 606:7y H 2 S ¼ 691.799 psia
T pr ¼ T ¼ 1.618967
T pc
t r ¼ 1 ¼ 0.617678
T pr
p pr ¼ p ¼ 2.891013
p pc
A ¼ 0:06125t r e 1:2(1 t r ) 2 ¼ 0.031746
2
B ¼ t r (14:76 9:76t r þ 4:58t ) ¼ 6.472554
r
2
C ¼ t r (90:7 242:2t r þ 42:4t ) ¼ 26.3902
r
D ¼ 2:18 þ 2:82t r ¼ 3.921851
Y ¼ ASSUMED ¼ 0.109759
3
2
Y þ Y þ Y Y 4
D
2
f (Y) ¼ Ap pr BY þ CY ¼ 0 ¼ 4.55E-06
(1 Y) 3
Ap pr
z ¼ ¼ 0.836184
Y
2:7g g p
r g ¼ ¼ 6:849296 lb m =ft 3
zT