University of Tripoli Faculty of Engineering Petroleum Engineering
November 22, 2022 | Author: Anonymous | Category: N/A
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University of Tripoli Faculty of Engineering Petroleum Engineering
Applied Reservoir Engineering PE453
Name: ESRA
NASSER
Student number: 022160140
Due Date: 30/1/2020 HW2
1
Q1.A 50 cu f tank contains gas at 50 psi and 50°F. It is connected to anoth another er tank that
contains gas at 25 psi and 50°F. When the valve between the two is opened, the pressure equalizes at 35 psi at 50°F. What is the volume o the second tank ? Soluton From ideal gas law
PV =nRT PV n= RT ntotal=¿ n +n ¿ 2
1
PV P 1 V 1 1 P 2 V 2 2 = + RT RT RT P V total= P1 V 1 + P2 V 2 35(50+V 2)=50x50+25V 2 By using solve:
V 2=75 f t 3 Q2.A high-pressure cell has a volume o 0.330 cu f and contains gas at 2500 psia and
130°F, at which condions its deviaon actor is 0.75. When 43.6 SCF measured at 14.7 psia and 60°F were bled rom the cell through a wet test meter, the pressure dropped to 1000 psia, the temperature remaining at 130°F. What is the gas deviaon actor at 1000 psia and 130°F? Soluton From real gas law:
PV = znRT A inital sae: sae: ni =
∗2500 0.75∗10.73∗( 130 + 460 ) 0.33
= 0.174 mol For sandard conditon z = 1
Psc V 2 =Zn2R T sc 14.7
n p = 2
43.6
∗ 1∗10.73∗( 60 + 460 )
=0.115mol nr =¿ n + n i
p
¿
=0.174-0.115 =0.059mol So The deviaton facor is equal : z r
=
PV n r RT
=
1000∗0.33 0.059∗10.73∗( 130 + 460 )
¿ 0.8835 mol.
Q3.A volumetric gas eld has an inial pressure o 4200 psia, a porosity o 17.2%, and
connate water o 23%. The gas volume actor at 4200 psia is 292 SCF/cu f and at 750 psia is 54 SCF/cu f. a) Calculate the inial in-place gas in standard cubic eet on a unit basis. b) Calculate the inial gas reserve in standard cubic eet on a unit basis, assuming an
abandonment pressure o 750 psia. c) Calculate the inial reserve o a 640-acre unit whose average net producve
ormaon thickness is 34 f, assuming an abandonment pressure o 750 psia. d) Calculate the recovery actor based on an abandonment pressure o 750 psia.
Soluton he inital in-place gas:
a) G=
3
43560∗⌀∗( 1− Swc ) βg
¿
43560
∗0.172∗( 1− 0.23 ) 1
( )
292
cuft scf
scf ac ft
¿ 1.68 MM
b)
(
Reserve=Gi −Ga =43560∗ ∅∗( 1− Swc )∗
1 1 − βi βa
) (
¿ 43560∗0.172∗( 1−0.23 )∗
1 1 − 0.003425 0.01852
)
scf acft
¿ 1.368 MM c)
initial init ial Reserve Reserve=1.37∗640∗34
¿ 29.8112 MMscf
d)
Rf =
reserve ∗100 Gi
¿
1368
∗100 =81.23%
1684
Q4)The ollowing pressure and cumulave producon data are available or a natural
gas reservoir:
4
Reservoir pressure ,psia
Gas deviation factor,z
2080 1885 1620 1205
0.759 0.767 0.787 0.828
cumulative production,MMMscf 0 6.873 14.002 23.687
888 645 a. Esmate the inial gas in place.
0.866 0.900
31.009 36.207
b. Esmate the Gp at an abandonment pressure o 500 psia. Assume za = 1.00.
c. What is the recovery actor at the abandonment pressure o 500 psia? Soluton:
Reservoir pressure psi 2080 1885 1620 1205 888 645
Gas deviaon actor,z 0.759 0.767 0.787 0.828 0.866 0.9
cumulave producon,MMMsc
P/z
0 6.873 14.002 23.687 31.009 36.207
2740.448 2457.627 2058.45 1455.314 1025.404 716.6667
Object64
For mah mah soluton soluton he slope and G equal
[ ]
p z =-5.558∗10−8 m= ∆ [ G p ] ∆
P Pi = −m G p z Z i
−8
0 =2740-5.558∗10 G pmax G pmax
5
=G =49.3 MMMscf
b)
P Pi = −m G p z Z i −8
500 =2740-5.558∗10 G p
G = 40 MMMscf C) R f = reserve ∗100= 40 ∗100=82 %
Gi
49
Q5. Determine the original gas-in-place, ulmate recovery and ulmate RF at an
abandonment pressure o 500 psia or the ollowing reservoir. Gas specic gravity = 0.70(air = 1)
Reservoir temperature
= 150℉
Original Or iginal reservoir reservoir pressure = 3000 3000 psia Abandonment Abandonment reservoir reservoir pressure pressure = 500 psi Soluton:
Because gas sp.gr is a strong uncon uncon in Temperature more than pressure, pressure, so we assume that the gas sp.gr is constant. The correlaon used the calc Ppc ∧Tpc Ppc =677 + 15∗γg −12.5∗γ g 2 2 Tpc =168 + 325∗γg −12 12.. .5∗γ g
And then read the "z" rom chart :
P,psi 3000 2800 2550 2070
6
Gp, MM scf 0 580 1390 3040
Ppr 4.483 4.184 3.810 3.093
z 0.822 0.816 0.811 0.814
P/z 3647.9 3433.5 3146.2 2544.6
Object93
Gi =
−3647.9 =10055 MMscf −3.628∗10−8 500
Gp@ Gp @ 500 psi =
−3647.9 −3.628∗10−
0.9377
8
ulmate recovery ¿ 8585 MMscf
ultimate ultim ate recovery recovery factor factor =
Gp 8585 ∗100= ∗100 Gi 10055
¿ 85 % Q6. A volumetric gas reservoir produced 600 MMsc o 0.62 specic gravity gas when
the reservoir pressure declined declined rom 3600 to 2600 psi. psi. The reservoir temperature temperature is reported at 140°F. Calculate: a) Gas inially in place. b) Remaining reserves to an abandonment pressure o 500 psi. c) Ulmate gas recovery at abandonment. d) Recovery actor at abandonment. Soluton : A) Gas inially in place Ppc =677 + 15∗γg −12.5∗γ g 2 7
¿ 677 + 15∗0.62−37.5∗0.622=671.885 psi
Tpc =168 + 325∗γg −12..5∗γ g 2 ¿ 168 + 325∗0.62−12.5∗0.622=364.695 R ℉
P 3600 Ppr = = =5.358 Ppc 671.885
Tpr=
Tres ( 140 + 460 ) = =1.645 Tpc 364.695
Zi =0.8979 βgi=0.02827
Tz P
3 0.02829∗( 140 + 460 )∗ 0.8979 f t ¿ =0.004234 3600 scf 3
@ 2600 Ppr =3.869 ,Tpr =1.645 , Z @ 2600 =0.8450 , βg =0.005517
Gp@ @ 2600=( 43560∗ A∗h∗∅∗ So ) Gp 43560
Gi =
∗ A∗h∗ ∗So= ∅
[
600
βi 1− β @ 2600
1 1 − βgi βg @ 2600
=
−
1
600 0.004234
]=
f t
scf 600 MMscf
=2.580 MMMcu MMMcuft ft
0.005517
2.580∗1 2.580 = =596.67 MMMscf βi 0.004234
b)
Z@500=0.9484
f t 0.02829∗ zT 0.02829∗0.9484∗( 140 + 460 ) =0.0322 = βg @ 500 = scf 500 P 3
Gr @ 500=( 43560∗ A∗h∗∅∗So )
Gr @ 500=2.580
8
[
1 0.0322
]=
[
1
βg @ 500
]
80.12 MMMscf
recovery at 500 psi =Gi−Gr =596.67 −80.12= 516.55 MMMscf Ultimate gas recovery d) Recovery factor =
516.55 Gp ∗100 =86.57 ∗100 = 596.67 Gi
Q7. The ollowing inormaon on a water-drive gas reservoir is given:
Bulk volume = 100,000 acre-f Gas Gravity = 0.6 25% T = 140°F pi = 3500 psi
Porosity = 15%
Swi =
Reservoir pressure has declined to 3000 psi while producing 30 MMMsc o gas and no water producon. Calculate cumulave water inux. Soluton: Ppc
=677 + 15∗γg −12.5∗γ g
2
2
¿ 677+ 15∗0.62−37.5∗0. 6 = 672.5 psi
2 Tpc =168 + 325∗γg −12 12.. .5∗γ g
¿ 168 + 325∗0.62−12.5∗0. 62=358.5 R ℉ 3
f t @ 3500 Ppr =5.2 ,Tpr = 1.67 , Z @ 3500 =0.88 , βg =0.00427 scf 3
f t @ 3000 Ppr = 4.46 ,Tpr =1.67 , Z @ 3000= 0.87 , βg = 0.004922 scf G=
43560∗∅∗( 1 − S wc )∗ Ah
B gi
¿ 43560∗0.15∗100000 ( 1− 0.25 )
0.00427
¿ 115 MMMscf
Gp Bg =G ( B g− B gi) + W e + W p Bw W e =222.64 MMscf 72.68 MMscf Q8.use the empirical correlaon to calculate calculate the (RF) or libyan water drive sandstone sandstone
reservoir eld Given daa:
H(net)= 60f 9
k= 0.15darcy
Swi =1.42 bbl/stb
∅ = 0.22
μo =0.5 cp
μw = 0.4cp
Soluton: 2
K R =0.114+0.136log +0.256 Swi −1.538 ∅ −0.00035 h μ o %
=44
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