Calculo de Ipr Por Standing y Fetkovich

PROBLEMA 11-A LIBRO DE BROWN TOMO 1 Las siguientes pruebas de fujo ueron tomadas en un pozo:

PRUEBA Nº Pwf (psi) 1 2 3

8 00 7 50 8 50

qo (BPD) 14 0 17 0 13 0

GRAFICAR IPR PARA EF=!"# EF=1! $ EF=1!% Pws= 1"& psi EF= '" EF= 1' EF= 1'% REOLUCION C*+,+o ./ qo02 p3 ,. p3/4 M5TODO DE TANDING 2

qo  P ' wf       P ' wf     = 1 − 0,2    − 0,8   qo max    Pws      Pws    Donde  Donde :  P ' wf   =  Pws −  EF ( Pws −  Pwf  ) PRUEBA Nº

Pwf6Pws

P7wf

1 2 3

0 .4 7 6 1  0 4 8 0 .4 4 6 4 2 8 5 7 0 .5 0 5  5 2 3 8

1 15 2 1 12 2 1 18 2

qo max =

P7wf6Pws

    

qo6qo02 EF=1

0.6857142 0.4866388 0.66785714 0.5060204 0.70357143 0.46327551

qo02PROM=

GRAFICAR IPR PARA EF=!"# EF=1! $ EF=1!%

Pwf (psi)

Pwf6Pws

1 68 0 1 60 0 1 50 0 1 40 0 1 30 0 1 20 0 1 10 0 1 00 0 0 0

1 0 . 5 2 3 8 0  5 0 .8  2 8 5 7 1 4 0 .8 3 3 3 3 3 3 3 0 .7 7 3 8 0  5 2 0 .7 1 4 2 8 5 7 1 0.654761 0.552381 0 .5 3 5 7 1 4 2 

EF=!" P7wf qo (BPD) 1680.0 1632.0 1572.0 1512.0 1452.0 132.0 1332.0 1272.0 1212.0

0 15.264122 33.721176 51.7066027 6.007577 85.650434 101.768 117.22445 132.076381

EF P7wf   1680.0 1600.0 1500.0 1400.0 1300.0 1200.0 1100.0 1000.0  0 0 .0

600 500

% 8 9 1 

0.35714286 0.276105 0.2380524 0.17857143 0.1104762 0.0552381 0

1032.0 72.0 12.0 852.0 72.0 732.0 672.0

172.36131 185.3202 17.108416 208.27424 218.826662 228.76552 238.00868

600.0 500.0 400.0 300 200 100 0

CORRECCION DEL IPR POR M5TODO DE FET FRMULA n

ql  =  Jo ' ( Pws −  Pwf   )  Donde :  Jo' =  Índice de Pr  oductividad  n = Cons tan te de turbulencia 2

2

log  y

=

2

∆ P 

mx

+

0= >= log  Jo' −

=

n

 ?o7= Pwf (psi) 1"& 1" 1 1% 18 19 11 1  &  "  % 8 9 1 

Pws9-Pwf 9 +o@(Pws9-Pwf 9" 0 262400 572400 862400 1132400 1382400 1612400 1822400 2012400 2182400 2332400 2462400 2572400 2662400 2732400 2782400 2812400 2822400

! 5.418638307 5.75766251 5.357087478 6.0538607 6.1406337251 6.207472785 6.2606437067 6.3037143087 6.33834353 6.3678030327 6.313586025 6.410338500 6.4252733046 6.4365442767 6.444415647 6.44077083 6.4506185635

qo

+o@ (qo)

0 35.0055036 75.7561853 113.166647 147.23688 177.6612 205.356714 22.40627 250.11565 267.484802 281.513726 22.20242 2.55013

! 1.54413633 1.874181 2.05371845 2.16801663 2.2503326 2.3125081 2.36060533 2.3814088 2.427211 2.44457 2.46568382 2.47647065

qo qo     o max    

qo02 EF=1 9&'"1 888'8"% 9&'"1" 8'"1& BPD

1! qo (BPD) 0 25.222068 55.2158785 83.505453 110.0018 134.72147 158.14183 17.622025 1.30672

EF=1!% P7wf qo (BPD) 1680.0 1568.0 1428.0 1288.0 1148.0 1008.0 868.0 728.0 588.0

0 35.0055036 75.7561853 113.166647 147.23688 177.6612 205.356714 22.40627 250.11565

248.471453 261.423326 272.671005 282.2144 20.053782 26.18887 300.61783

 

168.0 28.0

22.20242 2.55013

-119' -99' -89' -89' -"9'

88'1& 8%'9999 81'% 9'%9" 9&"'18

O;IC< 1

n

log q

−

log

+o@ (Pw

  j o 10

n

b 1'"&1 '%"1%8 f(2) = 1'"&" R = '&1991

8'8&%& '9"& q+  8"'%% '%%9918 111'1%% 1%8'9 18'&" 91'"98 99'"&9 9%'&%&% 9"'""& 9&'1&8 8'1 819'&%1 898'9&9" 881'9%" 88"'&%9 8%'%1&%% 8%1'"%1

+o@ (Pws9-Pwf9)

0 1.4

1.6

1.8

s9-Pwf9) s' +o@ (q)

2  8'8&%&%"9 #o$umn % Linear &#o$umn %"

2

+o@ (q)

2.2

2.4

2.6

1800

1600

1400

1200

1000

Pwf (psi@) 800

600

400

200

0 0

IPR M5TODO DE TANDING

100

200

qo (BPD)

300

'()0*6 '()1*0 '()1*4

400

IPR M5TODO DE TANDING CORREGIDO 1800

1600

1400

1200

1000

Pwf (psi@) 800

600

400

200

0 0

100

200

qo (BPD)

 POR FET:O;IC<

'()0*6 '()1*0 '()1*4

300

400

 POR FET:O;IC<

'()0*6 '()1*0 '()1*4

300

400

1800

1600

1400

1200

1000

Pwf (psi@) 800

600

1800

1600

1400

1200

1000

Pwf (psi@) 800

600

400

200

0 0

IPR M5TODO DE FET:O;IC<

100

200

qo (BPD)

300

'()1*4

400