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# Drilling Engineering

## 1.

Drilling EngineeringDrilling Engineering – PE 311

Hydraulics of Drilling Fluids

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## 2.

Drilling EngineeringCirculating System

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## 3.

Drilling EngineeringHydrostatic Pressure in Liquid Column

P

For incompressible fluids, the specific weight of the

liquid in field unit is given by

p 0.052 D p0

P + dP

If P0 = 0 then

p 0.052 D

The fluid density

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p

0.052 D

## 4.

Drilling EngineeringHydrostatic Pressure in Liquid Column

Example: Calculate the static mud density required to prevent flow from a

permeable stratum at 12,200ft if the pore pressure of the formation fluid is

8500psig.

Solution:

p

8500

13.4lbm / gal

0.052 D 0.052 12,200

The mud density must be at least 13.4 lbm/gal

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## 5.

Drilling EngineeringHydrostatic Pressure in Gas Column

dp 0.052 dD

EOS of gas:

p

p0

P0

m pM

pM

V ZRT 80.3 Z T

dp

0.052 p M

dD

80.3 Z T

dp

M

p 1544 Z T

p p0 e

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m

RT

M

pV Z n R T Z

D

D0

P0 + dP

dD

M ( D D0 )

1544 Z T

## 6.

Drilling EngineeringHydrostatic Pressure in Gas Column

A well contains tubing filled with methane gas (MW = 16) to a vertical depth of

10000ft. The annular space is filled with a 9.0 lbm/gal brine. Assuming ideal gas

behavior, compute the amount by which the exterior pressure on the tubing

exceeds the interior tubing pressure at 10,000ft if the surface tubing pressure is

1000 psia and the mean gas temperature is 140F. If the collapse resistance of the

tubing is 8330 psi, will the tubing collapse due to the high external pressure?

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## 7.

Drilling EngineeringHydrostatic Pressure in Gas Column

The pressure in the annulus (external pressure) at D = 10,000 ft is

P2 = 0.052 * 9.0 * 10,000 + 14.7 = 4,695 psia

The pressure in the tubing (internal pressure) at D = 10,000ft

p p0 e

M ( D D0 )

1544 Z T

1000 e

16*10000

1544*( 460 140 )

1188 psia

Pressure difference = p2 – p = 4695 – 1188 = 3507 < 8330 psia

The tubing will withstand the high external pressure

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## 8.

Drilling EngineeringHydrostatic Pressure in Complex Fluid Column

pi 1 pi p pi 0.052 i 1 Di 1 Di

p1 p0 0.052 1 D1 D0

p2 p1 0.052 2 D2 D1

p3 p2 0.052 3 D3 D2

pn pn 1 0.052 n Dn Dn 1

n

p n p 0 0.052 i ( Di Di 1 )

i 1

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## 9.

Drilling EngineeringHydrostatic Pressure in Complex Fluid Column

=?

p a p 0 0.052 (10.5 7,000 8.5 300 12.7 1700

16.7 1,000 9.0 10,000) 0 1266 1266 psig

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## 10.

Drilling EngineeringEquivalent Circulating Density (ECD)

The effective density exerted by a circulating fluid against the formation that takes

into account the pressure drop in the annulus above the point being considered.

The ECD is calculated as:

– mud density, ppg

P – Sum of the hydrostatic pressure and the frictional pressure drop in the annulus

between the depth D and surface, Psig

D – the true vertical depth, ft

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## 11.

Drilling EngineeringEquivalent Circulating Density (ECD)

Example: A 9.5-PPG drilling fluid is circulated through the drill pipe and the annulus.

The frictional pressure losses gradient in the annulus is 0.15. Calculate the

equivalent circulating density in PPG.

Solution:

= 9.5 + P/0.052 = 9.5 + 0.15 / 0.052 = 12.4 PPG

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## 12.

Drilling EngineeringBuoyancy

We W Pb PT A

We o gVo l gHA

PT

Fb

H

We o gVo l gVo

We W Wbo

gV

We o gVo 1 l o

o gV

l

We Wo 1

o

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W

Pb

+

We , W, Wbo – effective weight, weight of the object

in air, and buoyant force.

l and o - densities of liquid and the object

## 13.

Drilling EngineeringBuoyancy

10,000 ft of 19.5-lbm/ft drillpipe and 600 ft of 147 lbm/ft drill collars are suspended

off bottom in a 15-lbm/gal mud. Calculate the effective hook load that must be

supported by the derrick. Density of steel is 65.5 lbm/gal

Solution:

W = 19.5 * 10000 + 147 * 600 = 283200 lbm

We = W(1 - f/ s) = 283200 * (1 - 15/65.5) = 218300 lbm

(density of steel = 65.5 lbm/gal = 490lbm/cu ft)

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## 14.

Drilling EngineeringFlow Through Jet Bits

Energy balance:

p2 p1 0.052 D2 D1 8.074 *10 4 v22 v12 Pp p f

Pp is heat entering the system

Pf is heat loss due to friction

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## 15.

Drilling EngineeringFlow Through Jet Bits

Applying the energy equation for a flow through a nozzle with neglecting:

(1)effects of elevation: D2 - D1 = 0

(2)effects of uptream velocity v o = 0

(3)Heat entering the system Pp = 0 and friction loss Pf = 0

p 2 p1 0.052 D2 D1 8.074 *10 4 v 22 v12 Pp p f

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## 16.

Drilling EngineeringFlow Through Jet Bits

p 2 p1 8.074 * 10 4 v n2

vn

vn C d

pb

8.074 10 4

pb

8.074 10 4

C d correction factor

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## 17.

Drilling EngineeringFlow Through Jet Bits

Flow Through Parallel Nozzles

Assuming a constant Pb

through all the nozzles

vn Cd

vn

pb

8.074 10 4

q1 q2 q3

q

.... n

A1 A2 A3

An

Pressure drop across the bit

Δpbit

8.311 *10-5 q 2

C d2 At2

– lbm/gal ; q – gpm ; At - in2

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q

A

i

i

q

At

## 18.

Drilling EngineeringFlow Through Jet Bits

Hydraulic Impact Force

The purpose of the jet nozzles is to improve the cleaning action of the drilling fluid at the bottom of

the hole. Since the fluid is traveling at a vertical velocity v n , before reaching to the hole and

traveling at zero vertical velocity after striking the hole bottom hence all the fluid momentum is

transferred to the hole bottom.

Force is time rate of change of momentum, hence:

q vn

mv m

Fj

v

t

32.17 * 60

t

Substitute vn to the equation above gives

F j 0.01823 c d q p

Where Fj is the hydraulic impact force given in pounds.

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## 19.

Drilling EngineeringFlow Through Jet Bits

Flow Through Parallel Nozzles

Example: A 12.0 lbm/gal drilling fluid is flowing through a bit containing three 13/32 in nozzles at a

rate of 400 gal/min. Calculate the pressure drop across the bit and the impact force developed by

the bit.

Solution: Assume Cd = 0.95

2

13

At 3 0.3889in 2

4 32

Δpbit

8.311 *10 -5 q 2

C d2 At2

8.311 *10 5 *12 * 400 2

1169 psi

2

2

0.95 * 0.3889

Hydraulic impact force:

F j 0.01823 cd q p 0.01823 0.95 400 12 *1,169 820 lbf

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## 20.

Drilling EngineeringRheological Model

Flow curves of time-independent fluids

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Newtonian fluids:

Power law fluids:

K n

Bingham fluids:

y p

Herschel-Bulkley

(Yield power law fluids)

y K n

## 21.

Drilling EngineeringRheological Model

Newtonian Model

Non-Newtonian Model

Bingham-plastic model

p y ; y

0;

y y

p y ; y

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n

Power Law model:

K

Yield power law model:

y K

n

## 22.

Drilling EngineeringClassification of Drilling Fluids

Pseudoplastic

(Time-independent

shear thinning fluids)

If

the

apparent

viscosity

decreases with increasing shear

rate

Dilatant

(Time-independent

shear

thickening fluids)

If the apparent viscosity increases

with increasing shear rate

Flow curves of time-independent fluids

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## 23.

Drilling EngineeringClassification of Drilling Fluids

Thixotropic

(Time-dependent

shear

thinning fluids): If the apparent viscosity

decreases with time after the shear rate is

or time-dependent thinning

increased to a new constant value

Rheopectic

(Time-dependent

shear

thickening fluids): If the apparent viscosity

increases with time after the shear rate is

increased to a new constant value

Drilling fluids and cement slurries are

generally thixotropic

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or time-dependent thickening

## 24.

Drilling EngineeringRotational Viscometer

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## 25.

Drilling EngineeringRotational Viscometer

A rotational viscometer is used to determine type of the fluid and the rheological model of the

fluid. This can be done by varying the speed of the rotor (varying the shear rate) and reading the

dial reading (shear stress). To convert the speed to shear rate and dial reading to shear stress,

simply use these corellations:

= 1.703 x rpm, 1/s

= 1.06 x Dial Reading

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## 26.

Drilling EngineeringRotational Viscometer

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## 27.

Drilling EngineeringRotational Viscometer

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## 28.

Drilling EngineeringRotational Viscometer

The data below are obtained from a rotational viscometer. Determine type of fluid and the

rheological model of this fluid.

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## 29.

Drilling EngineeringRotational Viscometer

LogK = 0.3913 --> K = 2.49

n = 0.39

y = 5 lbf/100ft2

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