Chapter 3
CONDENSATE RESERVOIRS
P-T phase diagram of a reservoir
fluid
Calculation of initial gas and oil from
separator data
• Recombine the produced gas (Gp) and oil (Np) in the correct
ratio to find the average SG of the total well fluid (γwf).
• The well fluid is presumed to be produced initially in one
phase reservoir
For a two-stage separator system:
Well fluid
Gps
Gst
Primary
separator
Stock Tank
Gp(surf)
Np
Two-stage separation systems
The average specific gravity of the well fluid is given by
Where,
R1 1  4602. 0  R3 3
 wf 
133,316 0
R1 
 R3
M wo
R1, R3 – producing gas-oil ratios from the separator (1) and stock tank (3)
γ1, γ3 – specific gravities of separator and stock tank gases
γo – specific gravity of the stock tank oil (water = 1.00)
Well fluid
γwf
R1, γ1
R3, γ3
Gps
Gst
Primary
separator
Stock Tank
Gp(surf)
γ0
Np
Two-stage separation systems
Recall,
o  o
141.5
API  131.5
The molecular weight of the stock tank oil, Mwo is given by
42.43 o
5954
M wo  o

API  8.811 1.008   o
Steps
Find γwf
Find Z
Find G
Initial gas in place
Initial oil in place
Find fg
Example 4.1 (Craft & Hawkins)
Calculate the initial oil and gas in place per acre-foot for a
gas condensate reservoir
Given:
Initial pressure
reservoir temperature
average porosity
average connate water
Daily tank oil
Oil gravity, 60oF
Daily separator gas
Separator gas gravity
Daily tank gas
Tank gas gravity
2740 psia
215oF
25%
30%
242 STB
48.0oAPI
3100 MCF
0.650
120 MCF
1.20
Example 4.1
Gp
Separator
Stock tank
Np
Reservoir
Example 4.1 (solution)
From the given production data;
R1 = 3,100,000 SCF/242 STB = 12,810 SCF/STB
R3 = 120,000 SCF/242 STB = 496 SCF/STB
Oil gravity and Molecular weight;
o 
141.5
 0.788
48.0  131.5
M wo 
Therefore;
5954
 151.9
48.0  8.811
 wf 
12810(0.650)  4602(0.788)  (496)(1.20)
 0.896
133,316(0.788)
12810 
 496
151.9
Finding pseudo-critical properties
With γg = 0.896
Tpc = 425oR
Ppc = 650 psia
Tpr = T/Tpc = 675/425 = 1.60
Ppr = P/Ppc = 2740/650 = 4.23
Finding z-factor
With
Tpr = 1.60
Ppr = 4.23
Z = 0.825
Initial reservoir fluid per acre-foot
43560 (1  S w )
G
Bgi
and
So,
[SCF/ac.ft]
Psc Z iT
Bgi 
Tsc Pi
43560 (1  S w )Tsc Pi
G
Psc Z iT
43560(0.25)(1  0.3)(60  460)(2740)
G
 1326 MCF / ac. ft
(14.7)(0.825)(215  460)
Mole fractions
R1
n
Well fluid
Gps
R3 Gst
Primary
separator
Stock Tank
Gp(surf) ng
γ0
Np no
Volume fraction equals the mole fraction in the gas state
On one barrel basis;
R3
R1

ng
379.4 379.4
fg 

R
350 o
R1
ng  no
 3 
379.4 379.4 M wo
12810 496

379.4 379.4
fg 
 0.951
12810 496 350(0.788)


379.4 379.4
151.9
Initial gas and oil in place
Initial gas in place = f g  G  0.951(1326)  1261MCF / ac. ft
fg G
1261103
Initial oil in place =

 94.8STB / ac. ft
R1  R3 12810  496
The total daily gas-condensate production;
daily_ gas 3100  120
G p 

 3386MCF / day
0.951
0.951
The total daily reservoir voidage;
 675 14.7  0.825 
V  3386 10 
  19,450cu. ft / day
520  2740 

3
Finding pseudo-critical properties
Method 1: γg
Method 2: composition
Ppc   yi Pci
Tpc   yiTci
Calculation of initial gas and oil from
composition
• The composition of the total well fluid is calculated from the
analyses of the produced gas and liquid by recombining them
in the ratio in which they are produced
• When the composition of the stock tank liquid is known, a
unit of this liquid must be combined with the proper amount
of gas(es) from the separator(s) and the stock tank
• When the composition of gas and liquid in the first or high
pressure separator are known, the shrinkage the separator
liquid undergoes in passing to the stock tank must be
measured or calculated in order to know the proper
proportion in which the separator gas and liquid must be
combined.
Ex. 4-2 (Finding the initial gas and oil in place from the
composition of gas and liquid from the primary separator)
Given:
Reservoir pressure
reservoir temperature
hydrocarbon porosity
Std. con.
Separator gas
Stock tank oil
Mol. Wt. C7+ in separator liquid
Sp. Gr. C7+ separator liquid
Sp. Gr. Separator liquid
at 880 psig and 60oF
Separator liquid volume factor
Molar volume at 15.025 psia and 60oF
Composition (as given in Table 4.3)
4350 psia
217oF
17.4%
15.025 psia, 60oF
842,600 SCF/day
31.1 STB/day
185.0
0.8343
0.7675
1.235 bbl at
880 psia/STB
371.2 cu.ft/mole.
yi
xi
xiMWi
Mole/barrel of separator liquid;
nL 
 o  w 0.7675  350lb / bbl

 2.107 moles / bbl
MWo
127.48lb / lbmole
Given composition of separator
fluids
842,6000 SCF/day
ng , (2)= yi
Well fluid, zi
ng + nl
Primary
separator
nl, (3)= xi
Gst
Gp(surf)
Stock Tank
γo = 0.7675
Bosep = 1.235 sep bbl/STB
Separator liquid rate = 31.1 x 1.235 = 38.41 sep bbl/day
842,600 SCF / day
 21937 SCF / bbl
38.41sep.bbl / day
21937 SCF / bbl
In term of mole (ng) =
 59.1mol / sep.bbl
371.2 SCF / mole
Separator GOR =
Np
31.1 STB/day
zi
yi
xi
xiMWi
Column (12) and (14) are from Table 1.1 (C&H)
zi
yi
xi
xiMWi
Critical properties for C7+ is from Fig 4.3 (C&H)
Critical properties for
C7+
MW of C7+ = 185 lb/mole
Sp. Gr.C7+ = 0.8343
From the chart;
Tc of C7+ = 1227oR
Pc of C7+ = 300 psia
Finding z-factor
Tpr 
T
(217  460)

 1.79
Tpc
379.23
Ppr 
P
4350

 6.51
Ppc 668.23
From the chart
Z = 0.963
Initial in place
As in Ex. 4-1;
G
43560 (1  S w )Tsc Pi
Psc Z iT
G
43560 HCTsc Pi
Psc Z iT
G
43560(0.174)(60  460)(4350)
15.025(0.963)(217  460)
G  1.75  106 SCF / ac. ft
In term of moles;
1.75  10 6 SCF / ac. ft
G
 4715moles / ac. ft
371.2 SCF / mole
Or using gas law;
PV  nZR' T
PV
4350  (43560  0.174)
n

 4713moles / ac. ft
ZR' T 0.963(10.73)(217  460)
Initial gas and oil in place
fg 
ng
n g  nL

59.10
 0.966
59.10  2.107
Initial gas in place:
f g  G  0.966  (1.75 106 SCF / ac. ft )  1690MCF / ac. ft
Initial oil in place = (1  0.966)  4715  61.6STB / ac. ft
2.107 1.235
The total daily gas-condensate production;
842,600
G p 
 872,257 SCF / day
0.966
The daily reservoir voidage;
V  872,527 
(217  460) 15.025

 0.963  3778cu. ft / day
(60  460)
4350
Alternative method to find mole/bbl for separator liquid
If the SG of the separator
liquid is not available, Table
1.1 can be used to determine
the mole/bbl for separator
liquid.
Need to convert from Gal to
bbl
1 bbl = 42 Gal
yi
xi
xiMWi
1/0.46706 = 2.141
mole/bbl sep