new - natural gas properties
TRANSCRIPT
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CONSTANTS
R = 10.73 (psi*ft3)/(lb-mol*R
o)
Mair= 28.97 lb/lb-mol
Psc = 14.65 psia
Tsc = 60 oF 520 oR
Convert Gas Gravity in separator to Gas Gravity in reservoir (need to account for liquids):
FIRST METHOD IS VIA SEPARATOR EQUATIONS - SECOND IS VIA YIELD EQNS
1 stage separator 2 or 3 stages: dry gas sg at end of 1st, liq sg at end of last
API, cond = 50 a3 -0.0001785
y,o = 0.779614 a2 1.25
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M,o = 144.8385 a1 9.13875382
Vo [scf/stb] 717.5929 scf/stb Rs 11.4048379 scf/stb gas in sol af
ps1 55 psia
rho,g b4 sep= 0.108294 lbm/ft3 ts1 520 oR
Rg scf/stb = 3333.333 Vo [scf/stb 728.997688 scf/stb
yg,sp = 0.65
rho,l = 48.64793 lbm/ft3
y,g res = 1.417462 y,g res = 1.418556
EQUATION USING YIELD TO DETERMINE y,g res
YIELD = 300 STB/MMSCF
y,g res = 1.421
FOR COND WELLS WHERE THERE IS 2 PHASE IN RES:
above Dew Point - liq only - same as wet gas above
below dew point - dual phase
need to use two-phase z-factor:
when c7+ > 4%. Knowing sg of flowing gas
yc7+ = 11.14%
pr 1.81 ?
tr 1.55 ?
z,tp = 0.66847
Rt = 4062.331
Bt (total) = 43.93421 cf/stb
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T = 100 oF 560 oR
P = 1216 psi
yg,corr
for nonHC
MOLECULAR Pc Tc Vc
COMPOUND WEIGHT psiaoR ft
3/lb Zc yi
Methane C 1 H 4 16.04 667.8 343 0.0091 0.2884 0.786
Ethane C 2 H 6 30.07 707.8 549.8 0.0788 0.2843 0.07
Propane C 3 H 8 44.10 616.3 665.7 0.0737 0.2804 0.01
n-Butane C 4 H 10 58.12 550.7 765.3 0.0702 0.2736 0.002
iso-Butane C 4 H 10 58.12 529.1 734.7 0.0724 0.2824 0.006
n-Pentane C 5 H 12 72.15 488.6 845.4 0.0675 0.2623 0.003
iso-Pentane C 5 H 12 72.15 490.4 828.8 0.0679 0.2701 0.001
neo-Pentane C 5 H 12 72.15 464 781.11 0.0674 0.2537
n-Hexane C 6 H 14 86.18 436.9 913.4 0.0688 0.2643 0.001
n-Heptane C 7 H 16 100.20 396.8 972.5 0.0691 0.2633 0.001
n-Octane C 8 H 18 114.23 360.6 1023.9 0.069 0.2587
n-Nonane C 9 H 20 128.26 332 1070.3 0.0684 0.2536
n-Decane C 10 H 22 142.28 304 1111.8 0.0679 0.2462
Ethylene C 2 H 4 28.05 729.8 508.6 0.0737 0.2765
Propene C 3 H 6 42.08 669 656.9 0.0689 0.2752
Acetylene C2
H2
26.04 890.4 555.3 0.0695 0.2704
Carbon Dioxide C 1 O 2 44.01 1071 547.6 0.0342 0.2742 0.05
Hydrogen Sulfide H 2 S 1 34.08 1306 672.4 0.0459 0.2831 0.02
Sulfur Dioxide S 1 O 2 64.06 1145 775.5 0.0306 0.2697 0
Nitrogen N 2 28.13 493 227.3 0.0514 0.2916 0.05
Water H 2 O 1 18.02 3208 1165 0.05 0.235 0
1
yhc =
A =
B =
e =
FORMULA
PHYSICAL CONSTANTS FOR PURE COMPONENTS
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ter 1st stg
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yg Ppr Tpr Z (graph) rhog [lb/ft3] B mu,g [cp] mu,g [cp]
0.70 1.81 1.55 0.830 4.43 0.010815 0.013354 0.014715
0.63 112.4068 K 124.1448
5.464533 X 0.007225
0.70 1.307093 Y 1.367877
6.324058
0.07109
yiMWi yiPci yiTci added high
12.61 524.89 269.60
2.10 49.55 38.49
0.44 6.16 6.66
0.12 1.10 1.53
0.35 3.17 4.41
0.22 1.47 2.54
0.07 0.49 0.83
0.00 0.00 0.00
0.09 0.44 0.91
0.10 0.40 0.97
0.00 0.00 0.00
0.00 0.00 0.00
0.00 0.00 0.00
0.00 0.00 0.00
0.00 0.00 0.00
0.00 0.00 0.00
2.20 53.55 27.38
0.68 26.12 13.45
0.00 0.00 0.00
1.41 24.65 11.37
0.00 0.00 0.00
20.38 691.99 378.12
663.29 377.59 SUTTON: HIGH MW, RICH IN C7+, MINOR C02 AND N2, NO H2S. 0.57
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Rsw, scf/stb R,brine
9.967435084 7.689965
K 3.73583 50000 S, ppm
SIG 0.005420337
Y -2.43143E-07
X
rho,g
er MW particles
G
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Hall-Yarborough (1973) z-factor calculation
Hall, K.R. & Yarborough, L. (1973): A New Euation of State for Z-factor Calculations,
The Oil and Gas Journal, June 18, 82-92
Molecular weight oxygen, MO 15.9994 g/mole
Molecular weight sulfur, MS 32.07 g/mole
Molecular weight, carbon, MC 12.01 g/moleMolecular weight, hydrogen, MH 1.01 g/mole
Molecular weight air, MA 28.97 g/mole
Components Molecular weight Mole fraction Tci Tci Pci Pci
g/mole yi oR K psia Mpa
Methan, CH4 16.042 0.786 343 190 667.8 4.61
Ethan, C2H6 30.07 0.07 549.8 305 708 4.88
Propan, C3H8 44.10 0.01 665.7 370 616 4.25
i-Butane, C4H10 58.12 0.006 734.7 408 529 3.65
n-Butane, C4H10 58.12 0.002 765 425 551 3.80
i-Pentane C5H12 72.15 0.001 829 460 491 3.39
n-Pentane C5H12 72.15 0.003 845 469 489 3.37
Hexane C6H14 86.18 0.001 913 507 437 3.01
Heptane C7H16 100.21 0.001 972 540 397 2.74
Hydogen, H2 2.02 0 60 33 187 1.29
Nitrogen, N2 28.01 0.05 227.4 126 492 3.39
Oxygen, O2 32.00 0 277.8 154 731 5.04
Carbon dioxid, CO2 44.01 0.05 547.6 304 1071 7.38
Hydrogensulfid, H2S 34.08 0.02 672.4 373 1306 9.01
Dihydrogenoksid, H2O 18.02 0 1165 647 3199 22.06
Mole fraction1.0000
Total molecular weight gas 20.38 g/mole
Spesific gravity 0.70
Temperature 37.78oC 310.93 K
Pressure 8.384E+06 Pa
Method used (1, 2 or 3) 1
Compressebility factor, z 0.8298
Method 1 (Properties from composition, Key's rule)
Pseudo critical temperature
TPc (oil field units) 378.13 oR
TPc (SI units) 209.74 K
Pseudo critical pressure
PPc (oil field units) 691.90 psia
iiyPPPc
iiyTTPc
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PPc (SI units) 4.8E+6 Pa
Method 2 (Sutton's correlations)
Pseudo critical temperature
TPc (oil field units) 378.43 oRTPc (SI units) 209.90 K
Pseudo critical pressure
PPc (oil field units) 662.87 psia
PPc (SI units) 4.6E+6 Pa
Method 3 (Standing's correlation)
Pseudo critical temperature If we have a "
TPc (oil field units) 390.42 oR
TPc (SI units) 216.57 K
Pseudo critical pressure
PPc (oil field units) 659.00 psia
PPc (SI units) 4.5E+6 Pa
Pseudo reduced properties
TPR 1.482PPR 1.757
Hall-Yarborough
t 0.675
a 0.037
Reduced-density parameter, y0 0.01
Continue until f(y) < 1x10^(-5)
Iteration 1
f(y) -54.5E-3
Derivated of f(y), df(y) 0.945
Newton Rapson: Reduced- density parameter, y1 0.067670896
Compressebility factor, z 0.9486
Iteration 2
dy
ydf
yf
prTt 1
pc
prp
pP
pc
pr
T
TT
75=PPc
169.TPc
168pcHCT
667pcHCp
y
pZ
pr
a
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f(y) -7.0E-3
Derivated of f(y), df(y) 0.731
Newton Rapson: Reduced-density parameter y2 0.077219952
Compressebility factor, z 0.8313
Iteration 3
fy -95.9E-6
Derivated of f(y), df(y) 0.712
Newton Rapson: Reduced-desity parameter y3 0.077
Compressebility factor, z 0.8298
Iteration 4
f(y) -16.3E-9
Derivated of f(y), df(y) 0.712
Newton Rapson: Reduced-density parameter y4 0.077
Compressebility factor, z 0.8298
Iteration 5
f(y) -478.6E-18Derivated of f(y), df(y) 0.712
Newton Rapson: Reduced-density parameter y4 0.077
Compressebility factor, z 0.8298
yfy
yy ii'
1
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Convertion
oC -> K 273.15
atm -> bar 1.01325
cp -> Pas 1.00E-03
bar -> Pa 1.00E+05
m m -> m 1.00E-03
oR -> K 0.56
psig -> bar 0.07
oF -> oR 460
Instructions
1. Insert mole fractions.
2. Insert temperature [C] and pressure [Pa].
3. Choose which pseudo method you will use.
4. Read your z-factor beneath.
(Insert values in cells where the font color is red)
Methods for finding pseudo critical pressure and
pseudo critical temperature
1. Properties from composition, Key's rule (Preferred choice).
Newton-Rapson iteration.
2. Sutton's correlations
3. Standing's correlations
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ry" gas, SG < 0,75 If we have a "wet" gas, SG 0,75
tttyttt
y
yyyy 2324
432
2,2427,9082,218,216,952,1952,291
4441
pr ytttytttyyyyy
ap 32232
3
432
4,422,2427,9058,476,976,1410
2
g3.6-131,0-6.8 g
2
gg 74,0-349.5+2
25,1225 gHCgHC
25,370,5 gHCgHC
25,71330187 gHCgHCpcHCT
21,117,51706 gHCgHCpcHCp
212,106125,0 tet
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tyt 82,218,134,2
t82,218,
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Drainage Area Ad = ac
Net Pay h = ft
Porosity
Initial W.Sat Swi =
Temp T = 123oF
Specific Gravity yg = 0.64Initial Press. pi = 1422 psia
Water Compress cw=
Date Pressure z p/z Gp Bg F/Eg
psia MMSCF cf/scf MMscfD
1 1661 0.8287 2004 166 0.008232 #DIV/0!
2 1406 0.8456 1663 293 0.009923 1719.085
3 1135 0.8687 1307 406 0.012628 1166.195
4 847 0.8841 958 490 0.017222 938.6614
5 0 0 #DIV/0! 0 #DIV/0! #DIV/0!
6 0 0 #DIV/0! 0 #DIV/0! #DIV/0!
7 0 0 #DIV/0! 0 #DIV/0! #DIV/0!
8 0 0 #DIV/0! 0 #DIV/0! #DIV/0!
9 0 0 #DIV/0! 0 #DIV/0! #DIV/0!
10 0 0 #DIV/0! 0 #DIV/0! #DIV/0!
11 0 0 #DIV/0! 0 #DIV/0! #DIV/0!
12 0 0 #DIV/0! 0 #DIV/0! #DIV/0!
13 0 0 #DIV/0! 0 #DIV/0! #DIV/0!
14 0 0 #DIV/0! 0 #DIV/0! #DIV/0!
15 0 0 #DIV/0! 0 #DIV/0! #DIV/0!16 0 0 #DIV/0! 0 #DIV/0! #DIV/0!
17 #DIV/0! #DIV/0! #DIV/0!
18 #DIV/0! #DIV/0! #DIV/0! Second plo
19 #DIV/0! #DIV/0! #DIV/0! First plot -
VOLUMETRIC: 0 MMSCF
P/Z PLOT: 13000 MMSCF
F/EG PLOT: 13000 MMSCF
GENERAL INPUT
PRODUCTION CALCULATED0
500
1000
1500
2000
2500
p/z,psia
2
4
6
8
10
12
14
16
18
20
F/Eg,
MMscf
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leads to less error for VOLUMETRIC DEPLETIONS
asks any forms of depletion, the F/Eg plot gives more insight
0 100 200 300 400 500 600 700 800 900 1000
Gp, MMscf
0
0
0
0
0
0
0
0
0
0
0
0 10000 20000 30000 40000 50000
Gp, MMscf
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Res. Press pr = 3500 psia Gas Viscosity
Res. Boundary re = 1589 ft
Gross hg = 24 ft Velocity Coeff
Net hn = 24 ft Non-D Coeff.
Permeability k = 0.13625 md
WB Radius rw = 0.34 ft
SGU gas yg = 0.7Skin S = 0 Linear flow ends
Res. Temp T = 190 oF 650 P.Radial Begins
Porosity phi = 0.11 PSS time start
Compressibilit ct = 0.00024 /psi DIM FRAC COND =
Water Sat. Sw = 0.35 Eff WB Radius
Fract Cond kf*w = 582 mD-ft
Half Length xf = 369 ft
25
P2 VALID UP TO 3000 PSI
pwf p2
psia qsc McfD
3500 0
3000 73
2000 185
1000 252
500 269
0 275
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ug = 0.020578 cp
z = 1.00
Beta = 2.5528E+11 1/ft
D 0.0003548
ug, avg= 0.01872 D/MSCF Find u at 1750 psi
agp = 44626.0538 2177120.07
bgp = 2.06440509 100.3209782tL,end = 10 days
t,pr beg= 257 days
tpss = 419 days in transient flow
cfd = 11.5760424
rw'/xf = 0.44 from figure 4.24 based on cfd
rw' = 162.36 ft
agt = 312110
p2-nonD m(p)
qsc McfD qsc McfD qsc McfD
0 7.82E+08 0 0 0
73 5.61E+08 102 101 709.5342
183 3.10E+08 217 215 1511.796
249 7.70E+07 324 319 2258.874
266 1.87E+07 351 345 2445.585
271 0.00E+00 359 353 2505.504
0
500
1000
1500
2000
2500
3000
3500
4000
0 100 200 300
pwf,psia
Flow Rate, MSCFD
P2 method
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A B C
1212678 1915.7 0.878
1212678 1915.7 0.878
5118.7 355.9 1.4871
5118.7 355.9 1.4871
111 0 1.9365
108.03 0 1.9635
400
p2 Darcy
p2 non-Darcy
m(p) Darcy
m(p) non-Darcy
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CALCULATING BHP FROM WHP GIVEN RATE: Dry Gas:
Single Phase water or Undersaturated oil: No Viscosity/Density changes with T,P
WHP = 2534 PSI Qg, MMcf/
Depth 8500 ft Depth
Visc of water uw = 1 cp BHP =
Water Rate qw 1680 bpd 1.17 bpm ID =
roughness e = 0.00005 ft 70 bph e =Angle (+90 up) theta = 90 degrees T =
Inside Diameter ID = 3.826 in SGU =
Density pg = 62.5 lbm/ft3 Theta =
1500
Superficial Velocity 8500 Assumptio
v = 1.37 ft/sec
Re = 40548.49 TURBULENT IF > 2100 TURBULENT P avg =
f = 0.022316 z (from INP
0.127134 psf/ft friction first term ug (from IN
neglect kinetic second term
62.5 psf/ft gravity third term Nre =
dp/dl= 0.434911 psi/ft TOTAL f =
3696.741 Delta P from 1 to 2 if (-) then WHP < BHP C2 =
BHP = 6231 psi 7.504422
3689.236 c1/c2 =
WHP =
IN SINGLE PHASE, BOTH FORCES ACT DOWNWARD WHEN PRODUCING AT
INCREASING RATES, SO VLP CURVE ALWAYS INCREASES.
0 784010000 7920
20000 8064
50000 9115
90000 11811
95000 122510
5000
10000
15000
0 20000 40000 60000 80000 100000
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HP from BHP while flowing:
d 10 Input Yellow boxes and Green assumption
5000 ft Find z and ug
2500 psi Use WHP at bottom as new assumption
2.91 inches Type in Green Box.
0.00005 ft Find new z and ug110 F repeat til convergence.
0.7
90
= 1500 psi WHP
2000
UT) = 0.754 Type Pavg, T, and SGU into I&R sheet
PUT) = 0.0174 Type Pavg, T, and SGU into I&R sheet
2765493.542
0.0142
3.05389E-05
843753.6892
2094
30 40 50 60 70
1500 0.29 0.72 1 1.32
8500 1.67 4.12 5.69 7.5
1.96 0 4.84 6.69 8.82
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BOTTOMHOLE PRESSURE FROM SURFACE
Pwh = 1050 psig p2
Depth = 9500 ft
yg = 0.72
T_bar = 100 oF
Theta = 90 degrees (going up) find p1 (BHP)
Assume
p1 (bhp) = 1383 psig Guess
pbar = 1216.5 psig
z = 0.83 Pick
C2 = 2.9E-05
p1 = 1384 Make this new guess, find new z, til convergence
1383.589 alternative equation
This is how gas lift injection at bottom is calculated
0.15 psi/ft
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BOTTOMHOLE PRESSURE FROM SURFACE
Pwh = 1050 psig p2
Depth = 9500 ft
yg = 0.72
T_bar = 100 oF
Theta = 90 degrees (going up) find p1 (BHP)
Assume
p1 (bhp) = 1383 psig Guess
pbar = 1216.5 psig
z = 0.83 Pick
C2 = 2.9E-05
p1 = 1384 Make this new guess, find new z, til convergence
1383.589 alternative equation
This is how gas lift injection at bottom is calculated
0.15 psi/ft
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GAS FLOWRATE THROUGH AN ORIFICE
yg = 0.72
Valve Size = 0.375 inch
P1, upstream, Pinj at depth = 950 psig
P2, downstream, Flowing Prod Press = 900 psig
T1, Inj gas temp at depth, TgD = 140 deg F
qgsc = 1311.23593 Equation for chart, need to correct for Temp
CgT = 1.13068277
qga = 1159.68507 Mscf/d
Rate for new choke size:
Orifice size = 0.5 inch
qgc = 2331.08609 Mscf/D
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Ground Elevation = 438 ft Patm= 14.43 psi
WH Flow Rate qsc = 2.5 MMscfD
Gas Specific Grav. yg = 0.72
Suct Pressure ps = 35 psig pabs = 49.4 psia
Suct Temp Ts = 100 oF Tabs = 560 oR
Z-factor zs = 0.72
Disch Pressure pd = 1100 psig pabs = 1114.4 psia
Z-factor zd = 0.72 zavg = 0.72
Specific Heat k = 1.25
Adiabatic Eff na = 1 (n-1)/(nk) = 0.267
Polytropic Eff np = 0.75
Mech Efficiency Em = 0.95
scan eo vs rp plot
scan type vs rate plot
THERMODYNAMIC PROPERTIES
EFFICIENCY
CALCULATED VALUESGENERAL INPUT
San Antonio, TX is 650 ft
Laredo, TX is 438 ft
GAS PROPERTIES
SUCTION SIDE
DISCHARGE SIDE
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THERMODYNAMIC COMPARISON
ISOTHERMAL
HEAD = 92,975 [(ft-lbf)/lb]
POWER REQ'D = 310 HP
ADIABATIC
HEAD = 129,025 [(ft-lbf)/lb]
POWER REQ'D = 431 HPPOLYTROPIC
HEAD = 144,941 [(ft-lbf)/lb]
POWER REQ'D = 645 HP
DESIGN PROCESS
1. NUMBER OF STAGES
Total Compression Ratio RT = 22.55
Discharge Temperature Td = 279 oF
Number of Compress Stages # STG = 3
Compression Ratio per Stage rp = 2.82
Factor F = 1.1
2. RATE
Actual cubic feet per minute ACFM = 400 cfm
Type based on rate/pressure From Figure
3. POWER - for general compressor
Quick eqution calculation BHP = 513 HP
Precise equation calculation BHP = 435 HP
SPECIAL CASE - RECIPROCATING COMPRESSOR
Engine Speed N = 1100 RPM
Stroke Length St = 3.5 in
Cylinder Diameter D = 4.3 in
Rod Diameter d = 1.25 in
Clearance C = 15 %
Overall Eff based on rp Eo = 0.84 From Figure
Piston Displacement
Single Acting Pd,s = 32 cfm
Double Acting Pd,d = 62 cfm
Select: double
Volumetric Efficiency EV = 70.9 %
Capacity Q = 44.0 cfm
Piston Speed Ps = 642 fpm Keep between 600-800
Tension Load Tl = 14098 lbf
Power Required BHP = 356 HP
-15%
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single
double
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Orifice Diameter d1 = 0.323 in.
Pipe ID d2 = 3 in.
Static Pressure pf = 200 psia
Differential hw = 100 in. of water
Specific Gravity yg = 0.6
Base Pressure p = 14.6 psiaBase Temperature T = 75
oF
z = 0.99
qsc = 1.54 MMSCFD
Fb = 341.7 TABLE
Ftb = 1.028846
Fpb = 1.008904
Fg = 1.290994
Ftf = 0.985882
b = 0.0332 TABLE Calculates Fr = 1.000235
Fpv = 1.005038
Beta = 0.107667 hw/pf = 0.5 Find Y = 1.003 TABLE
C ' = 455.2
GENERAL INPUT
CALCULATING FLOW RATE - ORIFICE METER
fwsc phCq '
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FT PSIA
0 14.70
328 14.50
500 14.40
656 14.301000 14.20
1312 14.00
1500 13.90
2000 13.70
2500 13.40
3000 13.20
3500 12.90
4000 12.70
4500 12.40
5000 12.20
5500 12.00
6000 11.80
6500 11.50
7000 11.30
7500 11.10
8000 10.90
8500 10.70
9000 10.50
10000 10.10
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y = -0.0005x + 14.665
13.60
13.80
14.00
14.20
14.40
14.60
14.80
0 500 1000 1500 2000 2500
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Series1
Linear (Series1)