The performance of saturated reservoir with a 437 scf/stb gas-oil ratio is given by the following equation,
Qo, sc
4
533 53 3.3 0.133 13 3 pwf 6.67 10
p
2
wf
The formation is located at 2987 ft (TVD) and the well is completed packerless. pack erless. The completion consists of 4.5” 4.5” (OD) x 3.958” 3.958” (ID) tubing which is installed in a 6.33” ID 6.33” ID casing. The pump is set at 2705 ft. The pump is an ESP whose stage performance is given by the following equation 2
3
4
q q q q q p 0.25 3 4.40 3 6.78 5 3.43 2 1 0.20 2 pmax q q q q q max max max max max
5
pmax 24.4 psi qmax
10523.3 bpd
BEP= 0.6
This equation is only valid at 3,600 rpm. The pump has 72 stages and it operating at 1,500 rpm The wellhead operating conditions are 90 psi and 110 deg F, while the downhole temperature is 140 deg F. Estimate the oil flow rate for this well using the following information: Fluid properties, Gas Specific Gravity = 0.65, average temperature = 177 deg F Z= 0.931, Bg = 0.0346 ft3/scf, gas density=1.544 lb/ft^3, gas viscosity = 0.012cP Rs= 35.6 scf/stb, Bo=1.04 bbl/stb, oil density = 60.6 lb/ft^3, oil viscosity = 194 cP, Surface tension = 34 dyne/cm Pressure Gradient Casing GOR=437 scf/stb Qosc (stbpd)
Two-Phase Gradient -dp/dL (psi/ft)
25
0.1320
250
0.1321
500
0.1322
750
0.1323
Pressure Gradient Tubing Two-Phase Gradient -dp/dL (psi/ft)
Qosc (stbpd)
GORt (scf/stb)
25
437
0.1320
250
437
0.1324
500
437
0.1333
750
437
0.1345
25
300
0.1704
250
300
0.1707
500
300
0.1713
750
300
0.1723
5
50
0.3885
25
50
0.3885
100
50
0.3886
250
50
0.3887
500
50
0.3889
750
50
0.3893
Solution: The first step is to determine the reservoir pressure for this well. The reservoir pressure can be found by making oil flow rate zero and solving the resulting quadratic equation, 0
4
53 3.3 0.133 pwf 6.67 10
p
2
wf
The reservoir pressure is around 800 psi. We split the well flowing pressure in different values, Pwf (psig) 784 400 0
We use the IPR equation for estimating the oil flow rates, Qosc (stbpd)
Pwf (psig)
19 373.4 533.3
784 400 0
Since this is a packerless completion, the natural separation efficiency is obtained as
L g v 0.11 5 2 L A p ,h
0.5
0.08 5
18 3.3465 5
v sl 6.49 810 E sep
s
IDcas 2 ODtub 2 6.332 4.52
ft
v v sl v
BoQo, sc A p,h
18 3.3465
4
6.25 710 Qo, sc
0.08 5 4
0.08 5 6.25 7 10 Qo, sc
Thus, natural separation efficiency is obtained as, Qosc Pwf (psig) E sep (stbpd)
2
0.10 8 ft
19 373.4 533.3
784 400 0
0.88 0.267 0.203
The tubing GOR is then calculated as, GORt 1 E sep GOR f E sep Rs Thus, Qosc (stbpd)
Pwf (psig)
E sep
GORt
19 373.4 533.3
784 400 0
0.90 0.266 0.203
84.8 329.8 355.5
Since the pump is located above the perforation, we need to estimate the pressure drop in 6.33” casing as dp pip pwf ,1 L L pump dL 2 ,C While the two-phase pressure gradient in the tubing is required to estimate the discharge pressure, dp pdp pwh L pump dL 2 ,T These pressure gradient and pressure must be calculated for each flow rate using linear or Bilinear interpolation
Qosc (stbpd)
Pwf (psig)
E sep
GOR f
dp dL 2 ,C
GORt
dp dL 2 ,T
pip
pdp
19 373.4 533.3
784 400 0
0.90 0.266 0.203
437 437 437
0.1320 0.1328 0.1335
84.8 329.8 355.5
0.358 0.163 0.156
746.8 362.6 -37.62
1058.4 530.9 512
Note that the intake pressure for 533.3 stbpd is negative. It implies that this flow rate cannot be produced. The new maximum oil flow rate can be obtained as, dp dp pip 0 pwf ,1 L L pump pwf ,1 L L pump dL 2 ,C dL 2 ,C
The required or system pressure increment is calculated as, p sys pdp pip While the volumetric flow rate for this system is calculated as,
B g
qtotal
5.61
Rs E B 1 f sep o Qo , sc
GOR
So that Qosc (stbpd)
Pwf (psig)
E sep
GORt
dp dL 2 ,C
dp dL 2 ,T
pip
pdp
p sys
qtotal
19 373.4 533.3
784 400 0
0.90 0.266 0.203
75.3 330.2 355.5
0.1320 0.1328 0.1334
0.366 0.168 0.160
746.8 362.6 -37.62
1058.4 530.9 512
311.6 168.3 549.6
24.46 1066.9 1606.9
The maximum pressure increment and maximum flow rate at 3,600 rpm is given, pmax 24.4 psi qmax
10523.3 bpd
We can predict the performance at 1,500 rpm using the affinity law q1
N 1
q2
N 2
p1
p2
N 1 N 2
q2
q1
N 2
.3 10523
2
p2 p1
3600
N 1
2
1500
N 2 N 1
2
2
4384.7 bpd
1500 24.4 3600
2
2
4.24 psi
ESP performance is by the following polynomial, 2
3
4
q q q q q p 0.25 3 4.40 3 6.78 5 3.43 2 1 0.20 2 pmax qmax qmax qmax qmax qmax Let’s evaluate this equation for each of the qtotal calculated (q=qtotal), Qosc (stbpd)
qtotal
qtotal q
max
p
p stg
p pump
pmax
19
24.46
0.0055
0.999
4.23
304.56
373.4
1066.9
0.243
0.923
3.91
281.52
533.3
1606.9
0.367
0.843
3.57
257.04
The error is calculated as,
p sys p pump
5
Qosc (stbpd)
Pwf (psig)
E sep
GORt
19 373.4 533.3
784 400 0
0.90 0.266 0.203
75.3 330.2 355.5
Qo, sc E sep
0.08 5
4
0.08 5 6.25 710 Qo, sc
B g GOR f Rs 5.61 Bo
qo 1 GVF
qn
dp dL 2 ,T
pip
pdp
p sys
p pump
0.1320 0.1328 0.1334
0.366 0.168 0.160
746.8 362.6 -37.62
1058.4 530.9 512
311.6 168.3 549.6
304.56 281.52 257.04
7.04 -113.2 350.8
39.74 stbpd
GVF
qtotal
dp dL 2 ,C
qtotal qmax@1500
0.77 4
1 E sep
B g GOR f Rs1 E sep
Bo Qo, sc 1 GVF
0.35
63.6 bpd
0.01 4
This normalized flow rate is way too small for this application. It may imply that this pump is oversized for this well
