Deber de Pozos Tipo J , S y H

ESCUELA SUPERIOR POLITECNICA DEL LITORAL

FACULTAD DE INGENIERIA EN CIENCIAS DE LA TIERRA

PERFORACIÓNDIRECCIONAL

CÁLCULOS DE POZOS TIPO “J” “S” y “H”

RAMOS MEREJILDO HÉCTOR

ING. XAVIER VARGAS

PARALELO 1

Calculo para diseño Pozo Tipo JEjercicio #1

Información del pozo

Coordenadas de superficie9987272.047 N349831.46 E

Coordenada Objetivo9987547 N349961 E

KOP= 500’

BUR=2.5°/100’

Profundidad total al objetivo= 9000’ (TVD)

1. Calculo de dirección y desplazamiento horizontal

N= + 274.95

E= + 129.54

DH=2√ [ 274.95 ]2+ [129.54 ]2= 303.94 mts * 3.28 = 996.92’

Θ = arctang (E-W/N-S) = arctang (129.54/274.95)= 25.23°

Direccion = N 25° E = 25° AZM

2. Calculo de radio de curvature (R)

R= 180 x 1003.14 x BUR = 5729

BUR =57292.5 = 2291.6 ‘

3. Calcular DC Si D2 > R entonces DC = D2 – R Si D2 < R entonces DC = R – D2

D2=DH=996.92’

DC=R – D2= 2291.6’ – 996.92’= 1294.68’

4. Calculo del DO

DO=V3 –V1= 9000’ – 500’= 8500’

5. Calcular angulo DOC

Angulo DOC = arctang (DC/DO)= arctang(1294.68’/8500’)= 8.66°

6. Calcular OC

OC = DOcosDOC = 8500

cos8.66 ° = 8598’

7. Calcular angulo BOC

Angulo BOC = arccos (R/OC)= arccos(2291.6’/8598’)= 74.55°

8. Calcular angulo BOD

Si R < D2 entonces BOD = BOC - DOC Si R > D2 entonces BOD = BOC + DOC

Angulo BOD = 74.55°+ 8.66° = 83.21°

9. Calcular máximo Angulo (Θ)

Θ = 90° - Ang BOD = 90° - 83.21° = 6.79°

10. Calcular EOB (TVD)

V2 = V1 + R sen Θ = 500’ +(2291.6’ sen 6.79°)= 770.94’

11. Calcular EOB (MD)

EOB = V1 + ΘBUR

x100 = 500’ + 6.792.5

x100 = 771.6’

12. Calcular EOB (DH)

D1 = R – R cosΘ = 2291.6’ - (2291.6’ x cos 6.79°) = 16.07’

13. Calcular profundidad total al objetivo (MD)

PROF (MD) = V1 + ( ΘBUR

x100) + BC

OB= R

BC = √ (OC2−OB2 ) = √ (85982−2291.62 ) = 8287’

PROF (MD) = 500’ + 271.6’ + 8287’= 9058.6’

Ejercicio #2

Given informationTarget location S1638.3 ft , W 1147.15ftKOP (V1) 1981.00 ftTarget’s TVD (V3) 8500.00 ftBuild Radius (R) 2291.83 ft

Unknown Profile VariablesVariables Respuestas

Polar coordinate S 35° WBUR 2.5°/100 ftMAX Hold Angle (θ) 18°EOB (TVD) (V2) 2689.21 ftEOB (MD) 2701.00 ftEOB displacement (D1) 112.17 ftTotal MD to Target 8810.80 ft

CalculationsFormula respuesta

Polar coordinateDirection <B=arctg(b/c) <B= 35°Distance: a= 2√b2+c2 a = 2000 ft

BURBUR= 5729.58/R BUR= 2.5°/100 ft

MAX Hold Angle (θ)DC=R – D2 DC= 291.83 ftDO=V3 –V1 DO= 6519.00 ft<DOC=arctg (DC/DO) <DOC= 2.56°OC = DO/ cos <DOC OC= 6525.51 ft<BOC= arcos (R/OC) <BOC= 69.44°<BOD=<BOC+<DOC <BOD= 72°Θ= AOD – BOD Θ = 18°

EOB TVD (V2)V2 = V1 + R sen Θ V2= 2689.21 ft

EOB MD

EOB = V1 + ΘBUR

x100 EOB MD =2701.00 ft

EOB displacement (D1)D1 = R – R cosΘ D1 = 112.17 ft

Total MD to TargetBC = √ (OC2−OB2 ) BC= 6109.80 ft

TOTAL(MD)= V1 + ( ΘBUR

x100) + BC

TOTAL(MD) = 8810.80 ft

Ejercicio #3

Given the following profile variables, choose the correct answer below after calculating the maximum hold angle for a build and hold directional well.

Target Displacement = 2500 ft Target TVD = 10500 ft Build Up Rate = 3° per 100 ft Kickoff Point = 6925 ft

Select the correct answer.

a) 120.91°b) 136°34’ c) 41.18° d) 26.72°

CALCULOS

R= 180 x 1003.14 x BUR

= 5729BUR

=5729

3 = 1909.67 ft

Calculo de DC

Si D2 > R entonces DC = D2 – R

Si D2 < R entonces DC = R – D2

D2=DH=2500 ft

DC=D2 – R = 2500’ – 1909.67’= 590.33 ft

DO=V3 –V1= 10500’ – 6925’= 3575 ft

Angulo DOC = arctang (DC/DO)= arctang(590.33’/3575’)= 9.38°

OC =DO

cosDOC =

3575cos9.38 °

= 3623.45 ft

Angulo BOC = arccos (R/OC)= arccos(1909.67’/3623.45’)= 58.19 °

Angulo BOD

Si R < D2 entonces BOD = BOC - DOC

Si R > D2 entonces BOD = BOC + DOC

Angulo BOD = 58.19°- 9.38°= 48.81°

Θ = 90° - Ang BOD = 90° - 48.81° = 41.19°

Ejercicio #4

Given the following profile variables, choose the correct answer below after calculating the end of build TVD for a build and hold directional well:

Target Displacement = 2,500 ft Target TVD = 10,500 ft Build Up Rate = 3o per 100 ft Kickoff Point = 6,925 ft Maximum Hold Angle = 41.18o


a) 8,297.67 ft b) 6,783.4 ftc) 472.41 ftd) 8,182.5 ft

CALCULOS

V2 = V1 + R sen Θ = 6925’ +(1909.67’ sen 41.18°) = 8182.38 ft

Ejercicio #5

Given the following profile variables, choose the correct answer below after calculating the end of build (EOB) displacement for a build and hold directional well:

Target TVD = 10500 ft Build Up Rate = 3° per 100 ft Kickoff Point = 6925 ft Maximum hold angle ( ) = 41.18°


a) 1378.84 ftb) 376.22 ftc) 472.41 ftd) 5276.84 ft

CALCULOS

EOB (DH) = R – R cosΘ = 1909.67’ - (1909.67’ x cos 41.18°) = 472.47 ft

Ejercicio # 6

Given the following profile variables, choose the correct answer below after calculating the end of build (EOB) measured depth (MD) of a build and hold directional well:

Target Displacement = 2,500 ft Target Direction = S28°E Target TVD = 10,500 ft Build Up Rate = 3° per 100 ft Kickoff Point = 6,925 ft Maximum Hold Angle = 41.18°


a) 8,297.67 ftb) 11,376.84 ftc) 8,182.5 ftd) 472.41 ft

CALCULOS

EOB (MD) = V1 + ΘBUR

x100 = 6925’ + 41.18

3x 100 = 6925‘+ 1372’ = 8297.67 ft

Calculo para diseño Pozo Tipo s

Ejercicio #7

Given the following profile variables, choose the correct answer below after calculating the maximum hold angle for an "S"-type directional well:

Target Displacement = 3,439.84 ft Target TVD = 11,000 ft Build Up Rate = 2.5° per 100 ft Drop Off Rate = 1.5° per 100 ft Kickoff Point = 1,500 ft


a) 136°34’

b) 26.72° c) 120.91°d) 22.55°

CALCULOS

R1= 5729BUR1

= 2291.6 ft

R2=5729DOR

= 3819.3 ft

FE=R1 - (D3 – R2)= 2291.6 - (3439,84 – 3819.3)= 2671.1 ft

EO=V4 – V1= 9500 ft

Angulo FOE = arctag (FE/EO)= 15.7º

OF= √ (FE2+EO2 ) = 9868.5 ft

FG = R1 + R2 = 6110.9 ft

Angulo FOG = arcsen (FG/OF)= 38.26º

Θ = Max Angulo = Angulo FOG - Angulo FOE= 38.26º - 15.7º = 22.56 º

Ejercicio #8

Given the following profile variables, choose the correct answer below after calculating the end of build’s measured depth for an "S"-type directional well:

Target Displacement = 3,439.84 ft Target TVD = 11,000 ft Build Up Rate = 2.5° per 100 ft Drop Off Rate = 1.5° per 100 ft Kickoff Point = 1,500 ft Maximum Hold Angle = 22.55°


a) 3,439.84 ftb) 175.22 ftc) 2,378.89 ftd) 2,402 ft

CALCULOS

EOB (MD) = V1 + ΘBUR

x100 = 1500 + 22.55

2.5x100 = 1500 + 902 = 2402 ft

Ejercicio #9

Given the following profile variables, choose the correct answer below after calculating the start of drop true vertical depth for an "S"-type directional well:

Length of tangent section = 7748.36 ft Build Up Rate = 2.5° per 100 ft Drop Off Rate = 1.5° per 100 ft Kickoff Point = 1,500 ft Maximum Hold Angle = 22.55°


a) 9535.18 ftb) 7378.89 ftc) 3439.84 ftd) 175.22 ft

CALCULOS

SOD(TVD) = V2 + (sección tg x cos θ) = 2379.17 + (7748.36 cos 22.55) = 9534.76 ft

Ejercicio #10

Given the following profile variables, choose the correct answer below after calculating the end of build (EOB) true vertical depth of an "S"-type directional well:

Target Displacement = 3,439.84 ft Target TVD = 11,000 ft Build Up Rate = 2.5° per 100 ft Kickoff Point = 1,500 ft Maximum Hold Angle = 22.55°


a) 3653.69 ftb) 175.22 ftc) 1404 ftd) 2378.89 ft

CALCULOS

V2 EOB (TVD) = V1 + R1 senθ = 1500 + (2291.6 sen 22.56) = 1500 + 879.17 = 2379.17 ft

Calculo para diseño Pozo Tipo h

Ejercicio #11

Given the following profile variables, choose the correct answer below after calculating the maximum hold angle for a horizontal well:

Target Displacement = 2100 ft Target Direction = 35° AZM Target TVD = 3300 ft 1st Build Up Rate = 6° per 100 ft 2nd Build Up Rate = 8° per 100 ft Kickoff Point = 1000 ft


a) 26.72° b) 70.91°c) 2.66°34’ d) 42.88°

CALCULOS

R1= 5729BUR1

= 954.83 ft

R2=5729BUR 2

= 716.13 ft

EG= V4 – V1 – R2 = 3300-1000-716.13= 1783.87 ft

EO = DH – R1 = 2100 – 954.83= 1154.17 ft

Angulo GOE = arctang (EG/EO)= 57.18°

OG = √ (EG2+EO2 )= 2124.69ft

OF= R1 – R2= 238.7ft

Angulo GOF = arctang (OF/OG)= 83.54

Max Angulo = 180 - Angulo GOF - Angulo GOE = 180° - 83.54° - 57.18° = 39.3°

Ejercicio #12

Given the following profile variables, choose the correct answer below after calculating the first end of build’s displacement for a horizontal well:

Target Displacement = 2100 ft Target TVD = 3300 ft 1st Build Up Rate = 6° per 100 ft Kickoff Point = 1000 ft MAX Inclination Angle ( ) = 42.88°


a) 289.11 ft b) 300 ft c) 763.8 ft d) 255.21 ft

CALCULOS

EOB (DH) = R1 – R1 cos θ = R1 (1 – cos θ)= 954.83 (1-0.733)= 254.94 ft

Ejercicio #13

Given the following profile variables, choose the correct answer below after calculating the start of second build measured depth of a horizontal well:

Target Displacement = 2100 ft Target TVD = 3300 ft 1st Build Up Rate = 6° per 100 ft Tangent Section = 1939.74 ft Kickoff Point = 1000 ft Maximum Hold Angle = 44.88°


a) 3869.11 ft b) 4000 ft c) 3654.46 ftd) 2763.8 ft

CALCULOS

MD = V1 + Θ

BUR1x100 + BC= 1000 +

42.886

x 100 + 1939.74 = 3653.9 ft

Ejercicio #14

Given the following profile variables, choose the correct answer below after calculating the target’s total measured depth of a horizontal well:

Target Displacement = 2100 ft Target TVD = 3300 ft 1st Build Up Rate = 6° per 100 ft 2nd Build Up Rate = 8° per 100 ft Tangent Section = 1939.74 ft Kickoff Point = 1000 ft Maximum Hold Angle = 44.88°


a) 4000 ftb) 4243.43 ftc) 3869.11 ftd) 2763.8 ft

CALCULOS

Prof total en MD = V1 + Θ

BUR1x100 + BC +

90−ΘBUR 2

x100

Prof total en MD = 1000 + 42.88

6x 100 + 1939.74

90−42.886

x 100= 4243.46 ft

Ejercicio #15


Localización del objetivo = 1800 pies @ 135° AZM

Profundidad del objetivo (TVD) = 3800’

KOP = 2000’

BUR1 = 5.73°

BUR2= 9.54°

14. Calculo de radios de curvatura (R1 y R2)

R1= 5729BUR1

= 999.83 ft

R2=5729BUR 2

= 599.80 ft

15. Calcular máximo Angulo de inclinación (θ)

EG= V4 – V1 – R2 = 3800 – 2000 – 599.80 = 1200.1ft

EO = DH – R1 = 1800 – 999.83= 800.18 ft

Angulo GOE = arctang (EG/EO) = 56.31°

OG = √ (EG2+EO2 )= 1442.4ft

OF= R1 – R2= 399.93 ft

Angulo GOF = arctang (OF/OG) = 73.90°

Max Angulo = 180 - Angulo GOF - Angulo GOE = 180° - 56.31° - 73.90°= 49.79°

16. Calcular EOB1 en TVD/MD/DH

EOB1 (TVD) = V1 + R1 sen θ = 2763.31 ft

EOB1 (MD) = V1 + Θ

BUR1x100 = 2868.84 ft

EOB1 (DH) = R1 – R1 cos θ = 354.38 ft

17. Calcula EOH en en TVD/MD/DH

FG = √ (OG2−OF2 ) = 1385.85 ft

EOH (MD) = V1 + Θ

BUR1x100 + BC= 4238.71 ft

BC’ = BC cos θ = 893.35 ft

EOH (TVD) = V2 + BC = 3656.9 ft

EC’ =√ (BC2−BC '2 ) = 1056.78 ft

EOH (DH)= D1 + EC’ = 1411.34 ft

18. Calcular EOB2 MD

EOB2 MD = V1 + Θ

BUR1x100 + BC +

90−ΘBUR 2

x100

EOB2 MD = 4675.85 ft

19. Calcular profundidad total TOE (MD)

TOE (MD) = Total MD + Longuitud de navegación

TOE (MD) = 4675.74 + 500 = 5173.74 ft

20. Calcular profundidad total al TOE

Profund total = D3 + longitud de navegación

Profund total = 1800 + 500 = 2300 ft

Ejercicio #16


Localización del objetivo = 1800 pies @ 135° AZM

Profundidad del objetivo (TVD) = 4000’

KOP = 1800’

BUR1 = 1°

BUR2= 10°

1. Calculo de radios de curvatura (R1 y R2)

R1= 5729BUR1

= 5729 ft

R2=5729BUR 2

= 572.9 ft

2. Calcular máximo Angulo de inclinación (θ)

EG= V4 – V1 – R2 = 4000 – 1800 – 572.90 = 1627.1ft

EO = DH – R1 = 1800 – 5729 = -3929 ft

NOTA: este ejercicio no se puede realizar ya que la línea EO sale negativo porque el radio de curvatura 1 (R1) es mucho mayor que el desplazamiento horizontal

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Deber de Pozos Tipo J , S y H