Home > Documents > 1 PETE 411 Well Drilling Lesson 35 Wellbore Surveying Methods.

# 1 PETE 411 Well Drilling Lesson 35 Wellbore Surveying Methods.

Date post: 23-Dec-2015
Category:
Author: jacob-morton
View: 280 times
Embed Size (px)
of 44 /44
1 PETE 411 Well Drilling Lesson 35 Wellbore Surveying Methods
Transcript 1

PETE 411

Well Drilling

Lesson 35

Wellbore Surveying Methods 2

Wellbore Surveying Methods

Average Angle Balanced Tangential Minimum Curvature Radius of Curvature Tangential

Other Topics Kicking off from Vertical Controlling Hole Angle 3

Applied Drilling Engineering, Ch.8 (~ first 20 pages)

Projects:

Due Monday, December 9, 5 p.m.

( See comments on previous years’ design projects ) 4

Homework Problem #18

Balanced Cement Plug

Due Friday, December 6 5

I, A, MD 6

Example - Wellbore Survey Calculations

The table below gives data from a directional survey.

Survey Point Measured Depth Inclination Azimuth along the wellbore Angle Angle

ft I, deg A, deg

A 3,000 0 20 B 3,200 6 6 C 3,600 14 20 D 4,000 24 80

Based on known coordinates for point C we’ll calculate the coordinates of point D using the above information. 7

Example - Wellbore Survey Calculations

Point C has coordinates:

x = 1,000 (ft) positive towards the east

y = 1,000 (ft) positive towards the north

z = 3,500 (ft) TVD, positive downwards

z

E (x)

N (y)C

Dz

N

D

C

yx 8

Example - Wellbore Survey Calculations

I. Calculate the x, y, and z coordinates of points D using:

(i) The Average Angle method

(ii) The Balanced Tangential method

(iii) The Minimum Curvature method

(iv) The Radius of Curvature method

(v) The Tangential method 9

The Average Angle Method

Find the coordinates of point D using

the Average Angle Method

At point C, x = 1,000 ft

y = 1,000 ft

z = 3,500 ft

80 A 24I

20 A 14I

DD

CC

ft 400MD D, to C from depth Measured 10

The Average Angle Method

80 A 24I

20 A 14I

ft 400MD D, to C from depth Measured

DD

CC

z

E (x)

N (y)

C

D

zN

D

C

yx 11

The Average Angle Method 12

The Average Angle Method

This method utilizes the average of I1 and I2 as an inclination, the average of A1 and A2 as a direction, and assumes the entire survey interval (MD) to be tangent to the average angle.

From: API Bulletin D20. Dec. 31, 1985

2

III 21AVG

AVGAVG AsinIsinMDEast

AVGIcosMDVert

2

AAA 21

AVG

AVGAVG AcosIsinMDNorth 13

192

2414

2

III DCAVG

The Average Angle Method

502

8020

2

AAA DC

AVG

AVEAVG AsinIsinMDEast

50sinsin19400x

ft76.99x 14

The Average Angle Method

AVGIcos400Vert cos19400z

AVGAVG AcosIsinMDNorth

ft 71.83y

50cossin19400y

ft21.378z 15

The Average Angle Method

At Point D,

x = 1,000 + 99.76 = 1,099.76 ft

y = 1,000 + 83.71 = 1,083.71 ft

z = 3,500 + 378.21 = 3,878.21 ft 16

The Balanced Tangential Method

This method treats half the measured distance (MD/2) as being tangent to I1 and A1 and the remainder of the measured distance (MD/2) as being tangent to I2 and A2.

From: API Bulletin D20. Dec. 31, 1985

2211 AsinIsinAsinIsin2

MDEast

2211 AcosIsinAcosIsin2

MDNorth

12 IcosIcos2

MDVert 17

The Balanced Tangential Method

DDCC AsinIsinAsinIsin2

MDEast

oooo 80sin24sin20sin14sin2

400

ft66.96x 18

The Balanced Tangential Method

DDCC AcosIsinAcosIsin2

MDNorth

oooo 80cos24sin20cos14sin2

400

ft59.59y 19

The Balanced Tangential Method

CD IcosIcos2

MDVert

oo 14cos24cos2

400

ft77.376z 20

The Balanced Tangential Method

At Point D,

x = 1,000 + 96.66 = 1,096.66 ft

y = 1,000 + 59.59 = 1,059.59 ft

z = 3,500 + 376.77 = 3,876.77 ft 21

Minimum Curvature Method 22

Minimum Curvature Method

This method smooths the two straight-line segments of the Balanced Tangential Method using the Ratio Factor RF.

(DL= and must be in radians)2tan

2RF

RFAcosIsinAcosIsin2

MDNorth 2211

RFAsinIsinAsinIsin2

MDEast 2211

RFIcosIcos2

MDVert 21 23

Minimum Curvature Method

)AAcos(1IsinIsinIIcoscos CDDCCD

)2080cos(124sin14sin1424cos o00ooo

cos = 0.9356

The Dogleg Angle, , is given by: 24

Minimum Curvature Method

The Ratio Factor,

2tan

2RF

2

67.20tan

3608.0

2RF

o

0110.1RF 25

Minimum Curvature Method

RFAsinIsinAsinIsin2

MDEast DDCC

0110.180sin24sin20sin14sin2

400 oooo

ft72.97x

ft72.97011.1*66.96 26

Minimum Curvature Method

RFAcosIsinAcosIsin2

MDNorth DDCC

ft25.60y

ft25.60011.1*59.59

0110.180cos24sin20cos14sin2

400 oooo 27

Minimum Curvature Method

RFIcosIcos2

MDVert CD

0110.114cos24cos2

400 oo

ft91.380z

ft91.3800110.1*77.376 28

Minimum Curvature Method

At Point D,

x = 1,000 + 97.72 = 1,097.72 ft

y = 1,000 + 60.25 = 1,060.25 ft

z = 3,500 + 380.91 = 3,880.91 ft 29

2

CDCD

DCDC 180

AAII

AcosAcosIcosIcosMDEast

2oooo 180

20801424

80cos20cos24cos14cos400

ft 14.59 x 30

2

CDCD

CDDC 180

)AA()II(

)AsinA(sin)IcosI(cosMDNorth

2180

)2080)(1424(

)20sin80)(sin24cos400(cos14

ft 79.83 y 31

180

II

)IsinI(sinMDVert

CD

CD

ft 73.773 z

180

1424

)14sin24(sin400 oo 32

At Point D,

x = 1,000 + 95.14 = 1,095.14 ft

y = 1,000 + 79.83 = 1,079.83 ft

z = 3,500 + 377.73 = 3,877.73 ft 33

The Tangential Method

ft 400MD D, to C from depth Measured

80 A 24I

20 A 14I

DD

CC

80sinsin24400

DD AsinIsinMDEast

ft 22.160x 34

The Tangential Method

DIcosMDVert 24cos400

ft 42.365z

DD AcosIsinMDNorth

ft 25.28y

oo 80cos24sin400 35

The Tangential Method

ft 3,865.42365.423,500z

ft 1,028.2528.251,000 y

ft 1,160.22160.221,000x

D,Point At 36

Summary of Results (to the nearest ft)

x y z

Average Angle 1,100 1,084 3,878

Balanced Tangential 1,097 1,060 3,877

Minimum Curvature 1,098 1,060 3,881

Radius of Curvature 1,095 1,080 3,878

Tangential Method 1,160 1,028 3,865 37 38 39

Building Hole Angle 40

Holding Hole Angle 41 42

CLOSURE

(HORIZONTAL) DEPARTURE 43 44

Tool Face Angle

Recommended