P1) A saturated reservoir with an average pressure of 2000 psi, GOR of 200 scf/stb and productivity index of 6.5 stbpd/psi is going to be produced using rod pump. The middle of perforation is 6000 ft and the tubing is 2-7/8” x 6.5 lbf/ft (2.441”). The wellhead operating conditions are 50 psi and 80 deg F while the reservoir temperature has been measured in 200 deg F. The downhole pump is a 2-1/4” x 12” with -1 fit, which will operate with a maximum stroke length of 240”. The pump is set at the middle of the perforation in a casing is 5-1/2” x 15.5 lbf/ft (4.950”). Determine the oil flow rate if the pump will operate at 8 SPM. The following average fluid properties can be used for solving the problem: Z=0.937, Bg=0.03 ft3/scf,g=1.653 lb/ft3,g=0.013 cP,Rs=94.26 scf/stb,Bo=1.07 bbl/stb,o=50.248 lb/ft3,o=2.16 cP,=26.627 dyne/cm

Solution:

The reservoir performance is obtained as,

Qo ,max=Jpr

1 . 8=7200

stbpd

Qo , sc=Qo ,max−0. 2Qo ,max

prpwf−

0 .8Qo,max

( pr )2( pwf )2=7200−0 . 72 pwf−1. 44×10−3 ( pwf )2

Now, we choose three wellflowing pressure,

Pwf (psig)19801000

0

We calculate the flow rates,

Pwf (psig) Qo,sc (stbpd)

1980 1291000 5040

0 7200

Since this is a packerless completion, the natural separation efficiency is obtained as

v∞=0 . 115( σ ( ρL−ρg )ρL

2 )0.5

=0 .082fts

Ap ,h=( IDcas )2−(ODtub )2

183 .3465=

( 4 . 950 )2−(2. 875 )2

183. 3465=0 .089 ft2

vsl=6 . 498×10−5BoQo , sc

Ap ,h

=6 . 498×10−5 1 .070 . 089

Qo , sc=7 .812×10−4Qo , sc

E sep=v∞

vsl+v∞= 0 .082

0 .082+7 . 812×10−4Qo , sc

Pwf (psig) Qo,sc (stbpd)

Esep

1980 129 0.4491000 5040 0.02

0 7200 0.014

The tubing GOR is then calculated as,GOR t=(1−Esep)GOR f +EsepRs

Pwf (psig) Qo,sc (stbpd)

Esep GORt

1980 129 0.449 152.51000 5040 0.02 197.885

0 7200 0.014 198.5

The two-phase pressure gradient is then calculated and pwf2 is estimated,

pwf ,2=pwh+( dpdL )2φ

(ΔL )

The two phase pressure gradient can be calculated from tables. Thus,

Pwf (psig) Qo,sc (stbpd)

Esep GORt ( dpdL )2φ

1980 129 0.449 152.5 0.28331000 5040 0.02 197.885 0.3232

0 7200 0.014 198.5 0.4023

Since the GOR is so close to 200 for the last two flow rate, we are going to approximate the GOR to 200.

Qosc (stbpd)

GORt (scf/stb)

Two-Phase Gradient -dp/dL

(psi/ft)25 200 0.2323

250 200 0.2327500 200 0.2338750 200 0.2355

1000 200 0.23762000 200 0.25033000 200 0.26894000 200 0.29295000 200 0.32196000 200 0.35567000 200 0.39388000 200 0.4363

25 100 0.3395250 100 0.3398500 100 0.3406750 100 0.3417

1000 100 0.34312000 100 0.35163000 100 0.36404000 100 0.38005000 100 0.39946000 100 0.42197000 100 0.44748000 100 0.4758

The discharge pressure is then estimated as,

pwf ,1 (psig)

Qo,sc (stbpd)

Esep GORt ( dpdL )2φ

pwf ,2

1980 129 0.449 152.5 0.2833 1749.81000 5040 0.02 197.885 0.3232 1989.2

0 7200 0.014 198.5 0.4023 2463.8

The required or system pressure increment is calculated as,Δp=pwf ,2−pwf ,1

pwf ,1 (psig)

Qo,sc (stbpd)

Esep GORt ( dpdL )2φ

pwf ,2 Δp

1980 129 0.449 152.5 0.2833 1749.8 -230.21000 5040 0.02 197.885 0.3232 1989.2 989.2

0 7200 0.014 198.5 0.4023 2463.8 2463.8

The requirement of total flow rate can be calculated as,

q total=[ Bg

5 .61 (GOR f−Rs) (1−Esep )+Bo ]Qo , sc

pwf ,1 (psig)

Qo,sc (stbpd)

Esep GORt ( dpdL )2φ

pwf ,2 Δp q total

1980 129 0.449 152.5 0.2833 1749.8 -230.2 178.21000 5040 0.02 197.885 0.3232 1989.2 989.2 8185.7

0 7200 0.014 198.5 0.4023 2463.8 2463.8 11718.3

The theoretical displacement for the pump based on the equation,

q th=0 .1166 (d p )2LST (N )=0 .1166 (2 .25 )2 (240 ) ( 8 )=1133.4stbpd

The slippage can be estimated as,

S=870d p (w )3 (Δp )

(μo ) (Lp )=870

2 .25 ( 11000 )

3

2 .16 (12 )(Δp )=7 . 552×10−8 (Δp )

pwf ,1 (psig)

Qo,sc (stbpd)

Esep GORt ( dpdL )2φ

pwf ,2 Δp q totalS q pump

1980 129 0.449 152.5 0.2833 1749.8 -230.2 178.2 0 1133.41000 5040 0.02 197.885 0.3232 1989.2 989.2 8185.7 0.00007 1133.4

0 7200 0.014 198.5 0.4023 2463.8 2463.8 11718.3 0.00019 1133.4

The error is then

pwf ,1 (psig)

Qo,sc (stbpd)

Esep GORt ( dpdL )2φ

pwf ,2 Δp q total q pumpε

1980 129 0.449 152.5 0.2833 1749.8 -230.2 178.2 1133.4 -955.21000 5040 0.02 197.885 0.3232 1989.2 989.2 8185.7 1133.4 7052.3

0 7200 0.014 198.5 0.4023 2463.8 2463.8 11718.3 1133.4 10584.9

Linear Interpolation

Qo , sc=129+5040−1297052 .3−(−955 . 2 )

(0− (−955 .2 ) )=714 . 8stbpd

Let’s estimate GVF,

GVF=Bg (GOR f−Rs ) (1−E sep)

5 .61 Bo+Bg (GORf −Rs ) (1−E sep )

E sep=v∞

vsl+v∞= 0 .082

0 .082+7 .812×10−4Qo , sc

=0 .128

GVF=0. 315≈31. 5 %

P2) Let’s repeat the previous problem, but We are going to consider a PCP with a displacement of 290 bpd/100 rpm and maximum pressure increment of 2950ft H2O.The pump will work at 350 rpm.

Solution:

The entire problem is the same until we reach the pump part. Since the displacement is given,

q th=D (N )=290100

(350 )=1015stbpd

By definition, the DPmax happens when the flow rate has drifted a 80% from the theoretical displcement

S=[ q@Δp=0−q@Δpmax

( Δpmax )2 ] (Δp )2=[ q@ Δp=0−0 . 8q@ Δp=0

( Δpmax )2 ] (Δp )2 =[ 0 .2q@Δp=0

( Δpmax)2 ] ( Δp)2

Now the maximum pressure increment is given in ft of h2O,

Δpmax=0. 433 (Hmax )=0. 433 (2950 )=1277 . 4 psi

Thus,

S=[ 0. 2q@Δp=0

(Δpmax )2 ] (Δp )2=[ 0. 2 (1015 )(1277 .4 )2 ] (Δp )2=1 .244×10−4 (Δp )2

The pump performance is given as,

q pump=1015−1. 244×10−4 (Δp )2

Going back to our problem

pwf ,1 (psig)

Qo,sc (stbpd)

Esep GORt ( dpdL )2φ

pwf ,2 Δp q total q pumpε

1980 129 0.449 152.5 0.2833 1749.8 -230.2 178.2 1015 -836.81000 5040 0.02 197.885 0.3232 1989.2 989.2 8185.7 893.3 7292.4

0 7200 0.014 198.5 0.4023 2463.8 2463.8 11718.3 259.9 11458.4

Qo , sc=634 . 5 stbpd

GVF=Bg (GOR f−Rs ) (1−E sep)

5 .61 Bo+Bg (GORf −Rs ) (1−E sep )

E sep=v∞

vsl+v∞= 0 .082

0 .082+7 . 812×10−4Qo , sc

=0 . 142

GVF=0. 312≈31. 2 %