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### Read:Applied Drilling Engineering, Ch.4 (Drilling Hydraulics) to p. 125

### Drilling Hydraulics Applications cont’d

### Incompressible Fluids

### Incompressible fluids

### Compressible Fluids

### Compressible Fluids

### Example

### Example (i)

### Example

### Fig. 4-3. A Complex Liquid Column

### Buoyancy Force = weight of fluid displaced (Archimedes, 250 BC)

### Example

### Axial Forces in Drillstring

### Simple Example - Empty Wellbore

### Example - 15 lb/gal Mud in Wellbore

### Anywhere in the Drill Collars:Axial Tension = Wt. - Pressure Force - Bit Wt.

### Anywhere in the Drill Pipe:Axial Tension = Wts. - Pressure Forces - Bit Wt.

### Axial Tension in Drill String

### Example

### Example

### Example

### Example

### Example

### Example - Summary 4.9

Drilling Hydraulics - Hydrostatics

- Hydrostatic Pressure in Liquid Columns
- Hydrostatic Pressure in Gas Columns
- Hydrostatic Pressure in Complex Columns
- Forces on Submerged Body
- Effective (buoyed) Weight of Submerged Body
- Axial Tension in Drill String sA = FA/A

HW #4

ADE #1.18, 1.19, 1.24

Due Monday, Sept 23, 2002

Drilling Hydraulics Applications

- Calculation of subsurface hydrostatic pressures that may tend to burst or collapse well tubulars or fracture exposed formations
- Several aspects of blowout prevention
- Displacement of cement slurries and resulting stresses in the drillstring

- Bit nozzle size selection for optimum hydraulics
- Surge or swab pressures due to vertical pipe movement
- Carrying capacity of drilling fluids

Integrating,

p0

D

p

If p0 = 0 (usually the case except during well control or cementing procedures)

then,

T = temperature, R

r = density, lbm/gal

M = gas molecular wt.

m = mass of gas

…………… (1)

…………… (2)

…… (3)

But,

p= pressure of gas, psia

V= gas volume, gal

Z = gas deviation factor

n = moles of gas

R = universal gas constant = 80.3

…… (4)

from (3)

T = temperature, oR

r = density, lbm/gal

M = gas molecular wt.

m = mass of gas, lbm

p= pressure of gas, psia

V= gas volume, gal

Z = gas deviation factor

n = moles of gas

R = universal gas constant, = 80.3

Column of Methane (M = 16)

Pressure at surface = 1,000 psia Z=1, T=140 F

(i) What is pressure at 10,000 ft?

(ii) What is density at surface?

(iii) What is density at 10,000 ft?

(iv) What is psurf if p10,000 = 8,000 psia?

(i) What is pressure at 10,000 ft?

(iv) What is psurf if p10,000 = 8,000 psia?

Pa = ?

Fig. 4-4. Viewing the Well as a Manometer

Figure 4-9. Hydraulic forces acting on a foreign body

For steel,

immersed in mud,

the buoyancy factor is:

A drillstring weighs 100,000 lbs in air.

Buoyed weight = 100,000 * 0.771 = 77,100 lbs

Fb = bit weight

Drillpipe weight = 19.5 lbf/ft 10,000 ft

0 lbf

195,000 lbf

OD = 5.000 in

ID = 4.276 in

DEPTH, ft

A = 5.265 in2

AXIAL TENSION, lbf

W = 19.5 lbf/ft * 10,000 ft = 195,000 lbf

Drillpipe weight = 19.5 lbf/ft 10,000 ft

- 41,100

0

153,900

195,000 lbf

OD = 5.000 in

ID = 4.276 in

DEPTH, ft

A = 5.265 in2

AXIAL TENSION, lbf

F = P * A

= 7,800 * 5.265

= 41,100 lbf

Pressure at bottom = 0.052 * 15 * 10,000 = 7,800 psi

W = 195,000 - 41,100 = 153,900 lbf

FT

Example

A drill string consists of 10,000 ft of 19.5 #/ft drillpipe and 600 ft of 147 #/ft drill collars suspended off bottom in 15#/gal mud (Fb = bit weight = 0).

What is the axial tension in the drillstring as a function of depth?

A1

Pressure at top of collars = 0.052 (15) 10,000

= 7,800 psi

Pressure at bottom of collars = 0.052 (15) 10,600

= 8,268 psi

Cross-sectional area of pipe,

10,000’

10,600’

3

2

1. At 10,600 ft. (bottom of drill collars)

Compressive force = pA

= 357,200 lbf

[ axial tension = - 357,200 lbf ]

1

Fb = FBIT = 0

3

2

2. At 10,000 ft+ (top of collars)

FT = W2 - F2 - Fb

= 147 lbm/ft * 600 ft - 357,200

= 88,200 - 357,200

= -269,000 lbf

1

3

2

3.At 10,000 ft- (bottom of drillpipe)

FT = W1+W2+F1-F2-Fb

= 88,200 + 7800 lbf/in2 * 37.5in2 -357,200

= 88,200 + 292,500 - 357,200

= + 23,500 lbf

1

3

2

4. At Surface

FT = W1 + W2 + F1 - F2 - Fb

= 19.5 * 10,000 + 23,500

= 218,500 lbf

Also: FT = WAIR * BF = 283,200 * 0.7710

= 218,345 lbf

1

1. At 10,600 ftFT = -357,200 lbf [compression]

2. At 10,000 + ftFT = -269,000 lbf [compression]

3. At 10,000 - ftFT = +23,500 lbf [tension]

4. At SurfaceFT = +218,500 lbf [tension]

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