Tarea 2 conduccion

Sistema Escolarizado

Conducción de hidrocarburos II Alumno: Omar Sala Martin del Campo.

Profesor: Ing. Marco Antonio Maldonado Sánchez

Fecha: 16 de Febrero del 2015

Grupo: A Tema: Calclo de !re"i#n de $ondo en n pozo cerrado Bibliografía: %ran"porte %ran"porte de &idrocarbro" por 'cto"(1))1(*araicochea 'cto"(1))1(*araicochea !etrirena+ Franci"co(&icochea Franci"co(&icochea ,ernal+ C-"ar(Ortiz opez+ O"car( 15/Feb/15

INDIC

K m . 1 4 C a r re re t e r a V i l l a h e r m o s a M a c u s p a n a , D o s M o n t e s M u n ic ic i p i o d e l Centro, Villahermosa, Tab. México ♦ Tel y Fax Fax 1!"#"$ 1!" #" $ $

*$1+

Conducci(n de hidrocarburos ))

Indicaciones!!!!!!!!!!!!!!!!!!!""!!!""P#$ %ra I&eración!!!!!!!!!!!!!!!!!!!!!"!!P#$ 'da I&eración!!!!!!!!!!!!!!!!!!!!!!!P#( $ra I&eración!!!!!!!!!!!!!!!!!!!!!!!P#) *&a I&eración!!!!!!!!!!!!!!!!!!!!!!!P#+ (&a I&eración!!!!!!!!!!!!!!!!!!!!!!!P#, )&a I&eración!!!!!!!!!!!!!!!!!!!!!!!P#%%

Calcular la presión de fondo de un po-o cerrado. dadas las siguien&es condiciones: T =140 ° F Universidad Olmeca

2


Pth=1920

lb pg

2

*$1+

|.|

h =6320 pies γg =0.85 3

Primero I&eración Paso %: /uponer una P0s:

Pws

uesta

=2500 psi

Pasó ': 1b&ener la Presión Promedio: ´= P

´ = P1

p 1 + p 2 2

1920 + 2500 =2210 psi 2

Pasó $: Calcular 234 u&ili-ando el 56&odo de /&anding !re"ione"  temperatra" !"edocritica"

Tpc =167 +316.67 ( γg ) Tpc =167 + 316.67 ( 0.85 )= 436.1695 ° R

Ppc= 702.5−50 ( γg ) Ppc= 702.5−50 ( 0.85 )=660 psi !re"ione"  temperatra" !"edoredcida"

T + 460 Tpc

Tpr =

Tpr =

140 + 460 436.1695

=1.3756

( )

Ppr =

´ P Ppc

Universidad Olmeca

*$1+


P pr =

(

Z =0.3430 +

1− 0.3430 e

2.2248

)

2210 =3.3484 660

+ 0.087681∗3.3484

0.9587

=0.693351

0.5

A =1.39∗( 1.3756 −0.92 ) −( 0.36∗1.3756 ) −0.10= 0.3430

((

B =( 0.62 −( 0.23∗1.37 ) )∗3.3484 +

0.066 1.37−0.86

)

)

2

− 0.037 ∗3.3484 +

(

0.32 10

9 ( 1.37− 1)

4

)

6

∗3.3484 = 2.2248

C =0.132−0.32∗log ( 1.37 )= 0.087681 ( (0.3106 −(0.49∗1.37 ) ) +0.1824∗1.37) D =10 =0.9587

Pasó *: Calculando la P0s:

√

´ ∗6320 0.03756∗0.85∗2210 2

2

Pws= 1920 +

´ + 460 ) 0.693351 ( 140

= 2460.74 psi

Pasó (: Anali-amos si la &olerancia en&re la P0s supues&a 7 la calculada es&8 den&ro del margen:

|2460.74 −2500|=39.25 > 0 . 0 1 !or lo e procedemo" a repetir  tilizar la !3" calclada

/egunda I&eración Universidad Olmeca

*$1+


Paso %: /uponer una P0s:

Pwsanterior =2460.74 psi Pasó ': 1b&ener la Presión Promedio:

´= P

p 1 + p 2 2

5 ´ = P1

1920 + 2460.70 =2190.37 psi 2


Tpc =167 +316.67 ( γg ) Tpc =167 + 316.67 ( 0.85 )= 436.1695 ° R

Ppc= 702.5−50 ( γg ) Ppc= 702.5−50 ( 0.85 )=660 psi !re"ione"  temperatra" !"edoredcida"

Tpr =

T + 460 Tpc

140 + 460

Tpr =

436.1695

=1.3756

( )

Ppr =

Ppr =

(

Z =0.3430 + A =1.39∗( 1.3756 −0.92 )

1− 0.3430 e

0.5

−

2.1879

´ P Ppc

2190.37 =3.3187 660

)

+ 0.087681∗3.3187

0.9587

=0.693639

( 0.36∗1.3756 ) −0.10= 0.3430

Universidad Olmeca

*$1+


B =( 0.62 −( 0.23∗1.37 ) )∗3.3187 +

((

0.066 1.37 −0.86

)

)

2

(

−0.037 ∗3.3187 +

0.32 10

9 ( 1.37 −1 )

)

6

∗3.3187 =2.1879

C =0.132−0.32∗log ( 1.37 )= 0.087681

D=10

( (0.3106−(0.49∗1.37 )) +0.1824∗1.37)

=0.9587

6


√

´ ∗ 6320 0.03756∗0.85∗2190.37 2

2

Pws= 1920 +

´ + 460 ) 0.693639 ( 140

=2452.02 psi


|2452.02−2460.74|= 8.71> 0.01 !or lo e procedemo" a repetir  tilizar la !3" calclada

Tercer I&eración Paso %: /uponer una P0s:

Pwsanterior =2452.02 psi Pasó ': 1b&ener la Presión Promedio:

´= P

p 1 + p 2 2 ´ = P1

1920 + 2452.02 =2186.01 psi 2


Tpc =167 +316.67 ( γg ) Tpc =167 + 316.67 ( 0.85 )= 436.1695 ° R

Ppc= 702.5−50 ( γg ) Ppc= 702.5−50 ( 0.85 )=660 psi

Universidad Olmeca

*$1+


!re"ione"  temperatra" !"edoredcida"

Tpr =

Tpr =

T + 460 Tpc

140 + 460 =1.3756 436.1695

7

( )

Ppr =

P pr =

(

1− 0.3430

Z =0.3430 + A =1.39∗( 1.3756 −0.92 )

e

0.5

−

2.1798

´ P Ppc

2186.01 =3.3121 660

)

0.9587

+ 0.087681∗3.3121

=0.69371

( 0.36∗1.3756 ) −0.10= 0.3430

B =( 0.62 −( 0.23∗1.37 ) )∗3.3121 +

((

0.066 1.37 −0.86

)

)

2

(

−0.037 ∗3.3121 +

0.32 10

9( 1.37− 1)

)

6

∗3.3121 =2.1798

C =0.132−0.32∗log ( 1.37 )= 0.087681

D=10

( (0.3106−(0.49∗1.37 )) +0.1824∗1.37)

=0.9587


√

´ ∗6320 0.03756∗0.85∗2186.01 2

2

Pws= 1920 +

´ + 460 ) 0.69371 ( 140

=2450.08 psi


|2452.02−2450.08|=1.93 > 0 . 0 1 !or lo e procedemo" a repetir  tilizar la !3" calclada

Universidad Olmeca


*$1+

Cuar&a I&eración Paso %: /uponer una P0s:

Pwsanterior =2450.08 8

Pasó ': 1b&ener la Presión Promedio:

´= P

p 1 + p 2 2 ´ = P1

1920 + 2450.082 =2185.04 psi 2


Tpc =167 + 316.67 ( γg )

Tpc =167 + 316.67 ( 0.85 )= 436.1695 ° R Ppc= 702.5− 50 ( γg )

Ppc= 702.5−50 ( 0.85 )=660 psi


Tpr =

T + 460 Tpc

140 + 460

Tpr =

436.1695

=1.3756

( )

Ppr =

Ppr =

(

Z =0.3430 +

1− 0.3430 e

2.1780

)

´ P Ppc

2185.04 =3.31066 660

+ 0.087681∗3.31066

0.9587

=0.69372

Universidad Olmeca

*$1+


A =1.39∗( 1.3756 −0.92 )

0.5

−

( 0.36∗1.3756 ) −0.10= 0.3430

B =( 0.62 −( 0.23∗1.37 ) )∗3.31066 +

((

0.066 1.37 −0.86

)

)

2

−0.037 ∗3.31066 +

(

0.32 9 ( 1.37−1 )

10

)

6

∗3.31066 =2.1780

C =0.132−0.32∗log ( 1.37 )= 0.087681 9

D=10

( (0.3106−(0.49∗1.37 )) +0.1824∗1.37)

=0.9587


√

´ ∗6320 0.03756∗0.85∗2185.04 2

2

Pws= 1920 +

´ + 460 ) 0.69372 ( 140

=2449.65 psi


|2449.65− 2450.08|=0.42 > 0.01 !or lo e procedemo" a repetir  tilizar la !3" calclada

9uin&a I&eración Paso %: /uponer una P0s:

Pwsanterior =2449.65 Pasó ': 1b&ener la Presión Promedio:

´= P

p 1 + p 2 2 ´ = P1

1920 + 2449.65 =2184.65 psi 2


Tpc =167 + 316.67 ( γg ) Tpc =167 + 316.67 ( 0.85 )= 436.1695 ° R

Ppc= 702.5−50 ( γg )

Universidad Olmeca

*$1+


Ppc= 702.5−50 ( 0.85 )=660 psi


Tpr =

Tpr =

T + 460 Tpc

140 + 460 436.1695

10

=1.3756

( )

Ppr =

P pr =

(

Z =0.3430 + A =1.39∗( 1.3756 −0.92 )

2184.65 =3.31 $ 660

1− 0.3430

0.5

−

e

2.1776

´ P Ppc

)

0.9587

+ 0.087681∗3.3103

=0.6937

( 0.36∗1.3756 ) −0.10= 0.3430

B =( 0.62 −( 0.23∗1.37 ) )∗3.3103 +

((

)

)

(

)

0.066 0.32 −0.037 ∗3.31032 + ∗3.31036 =2.1776 9( 1.37−1) 1.37 −0.86 10

C =0.132−0.32∗log ( 1.37 )= 0.087681 D =10

( (0.3106− (0.49∗1.37 )) +0.1824∗1.37)

=0.9587


√

´ 0.03756∗0.85∗2184.65 ∗6320 2

2

Pws= 1920 +

´ + 460 ) 0.6937 ( 140

=2449.55 psi


|2449.65− 2449.55|=0.09 > 0.01 !or lo e procedemo" a repetir  tilizar la !3" calclada

Universidad Olmeca


*$1+

/e&a I&eración

11

Paso %: /uponer una P0s:

Pwsanterior =2449. 55 Pasó ': 1b&ener la Presión Promedio:

´= P

p 1 + p 2 2 ´ 1= P

1920 + 2449.55 2

=2184.775 ps i


Tpc = 167 + 316.67 ( γg )

Tpc =167 + 316.67 ( 0.85 )= 436.1695 ° R Ppc= 702.5−50 ( γg )

Ppc= 702.5−50 ( 0.85 )=660 psi


Tpr =

Tpr =

T + 460 Tpc

140 + 460 436.1695

=1.3756

( )

Ppr =

´ P Ppc

Universidad Olmeca

*$1+


P pr =

(

Z =0.3430 +

A =1.39∗( 1.3756 −0.92 )

2184.775 =3.3102 660

1− 0.3430

0.5

−

e

2.1775

)

+ 0.087681∗3.31020.9587=0.6937

( 0.36∗1.3756 ) −0.10= 0.3430

B =( 0.62 −( 0.23∗1.37 ) )∗3.3102 +

((

0.066 1.37 −0.86

)

)

12

2

(

−0.037 ∗3.3102 +

0.32 10

9( 1.37− 1)

)

6

∗3.3102 =2.1775

C =0.132−0.32∗log ( 1.37 )= 0.087681 D =10

( (0.3106 − (0.49∗1.37 )) +0.1824∗1.37)

=0.9587


√

´ ∗6320 0.03756∗0.85∗2184.775 2

2

Pws= 1920 +

´ + 460 ) 0.6937 ( 140

= 2449.53 psi


|2449.53− 2449.55|=0.01 > 0.01 a tolerancia "e cmple por lo e ne"tra pre"i#n de $ondo en el pozo cerrado e" de 244).5p"i

Universidad Olmeca

Tarea 2 conduccion

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