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DRILLING FLUID 1.

The density of 600 bbl of 12 Ibm/gal mud must be increased to 14 Ibm/gal using API barite. One gallon of water per sack of barite will be added to maintain an acceptable mud consistency. The final volume is not limited. How much barite is required? ANSWER: Initial volume V1 = 600 bbl Final density ρ2 = 14 lbm/gal Relation between V1 and V2

[ [

1+ ρwVwB − ρ2 ( 1+ ρwVwB ) V 1=V 2 1+ ρwVwB ρB ( − ρ1 1+ ρwVwB ) ρB

Initial density ρ 1 = 12 lbm/gal API of barite ρB = 35 lbm/gal API of water ρw = 8.34 lbm/gal

]

1+8.34 × 0.01 −14 ( 1+35 ×0.01 ) 600=V 2 1+8.34 × 0.01 −12 35( 1+35 ×0.01 ) 35

]

600=V 2 ( 0.8757) V 2=

600 0.8757

V 2=685.17 bbl Mass Required:

( ρB1 + ρBVwB ) × ( V 2−V 1 ) gal 35 ¿ m=( +35 ×0.01 ) ×( 685.17 −600 ) bbl× 42 1 bbl m=

lbm ×3577.14 gal gal m=92755 lbm m=25.93

2.

a.

After cementing casing in the well, it is desirable to increase the density of the 9.5 Ibm/gal mud to 14 Ibm/gal before resuming drilling operations. It also is desired to reduce the volume fraction of low specific gravity solids from 0.05 to 0.03 by dilution with water. The present mud volume is 1000 bbl, but a final mud volume of 800 bbl is considered adequate. Compute the amount of original mud that should be discarded. Answer : Initial Volume of 9.5 lbm/gal mud,

V 1=V 2

fc 2 fc 1

V 1=800 ×

( 0.03 0.05 )

V 1=480 bbl b.

Calculate the amount of water and API barite that should be added. Answer :

Amount discarded before adding API barite=1000bbl – 480 bbl Amount discarded before adding API barite=520 bbl Vw=

( ρB− ρ2 ) V 2− ( ρB− ρ1 ) V 1 (ρB−ρW )

Vw=

( 35−14 ) 800−( 35 −9.5 ) 480 (35−8.34)

Vw=171 bbl mB=( V 2−V 1−V 2 ) ρB mB=( 800 −480 −171) × 35 mB=219030 lbm 3.

It is desired to increase the density of 200 bbl of 11 Ibm/ gal mud to 11.5 Ibm/gal using API barite. The final volume is not limited. Calculate the weight of API barite required. ANSWER:

V 2=V 1

(

ρB − ρ1 ρB − ρ2

V 2=200 bbl

)

11 ( 3535−−11.5 )

V 2=204.26 bbl mB=( V 2 −V 1) × ρB mB=( 204.26 −200 ) × 42 × 35 mB=6262.2 lbm

DRILLING HYDRAULICS 1.

Pressure Loss in Laminar Flow

a.

Calculate the velocity of a fluid flowing through a 5" 19.5 lb/ft drillpipe (I.D. = 4.276") at 150 gpm. Answer :

velocity , v=

v=

Q 2.448× d 2

150 2.448 ×(4.276)2

v =3.35 ft /sec b.

Determine the pressure loss in the above situation if the fluid is a Bingham Plastic fluid with a plastic viscosity of 20 cp, a yield point of 15 lb/100 sq. ft and density is 10 ppg. Answer :

µ p V´ τy dP = + 2 dL 1500 d 225 d

15 20 ×3.35 dP + = 2 dL 1500 × 4.276 225 × 4.276 dP −3 −2 =2.44 ×10 + 1.56 × 10 dL psi dP =18 1000 ft dL c.

Calculate the pressure loss in the above situation if the fluid was a Power Law fluid with an nonNewtonian Index of 0.75 and a consistency index of 70 eq cp. ANSWER:

dP k V´ 3+1/n = ( 1+n ) × 0.0416 dL 14400 0 d

(

)

n

[ ]

1 70 × 3.35 dP 0.75 = × dL 14400 0 × 4.276( 1.75) 0.0416

0.75

3+

dP =0.0042 psi / ft ∨4.2 psi per 1000 ft dL

2.

Determine the hydraulic horsepower needed for the following condition using graphical method and pressure drop calculation. Compare the results between these methods. Depth = 6,000 ft (5,500 ft drill pipe, 500 ft drill collars) Drill pipe = 4.276” ID, 5” OD, internal flush Drill collars = 6 ¾ in. (ID = 2.813 in.) Mud density = 9 ppg Plastic viscosity = 25 cp Bingham yield, Yb = 18 lb/100ft2 Bit = 7 7/8-in., 3 cone, jet rock bit Annular velocity around the drill pipe = 4 ft/sec Nozzle velocity required = at least 250 ft/sec through each nozzle (this value is obtained by a commonly applied rule of thumb). Assume C = 0.95 Surface equipment type = 3 ANSWER:

Flow rate , q= ( annulus area ) × velocity q=2.45 ( d h2−d2p) ´v

(( ) )

7 2 2 −( 5 ) × 4 8 q=363 gpm q=2.45 7

a)Surface equ equipment ipment losses (Δps): Surface Equipment type 3, From Table 3.1 ,

E=5.3 × 10−5

0.8 1.8 0.2 Δ p s=E ρm q µ p −5 0.8 5 .3× 10 ¿(9) ( 363 ) (25 ) Δ p s=¿ 1.8

0.2

Δ p s=24 psi b)Pressure losses ins inside ide drill pipe:

v´ =

q 363 = =8.10 ft /sec 2 2.45 d 2.45( 4.276)2

vc=

1.08 µ p+ 1.08√ µ2p+9.3 ρm d 2 Y b ρm d

1.08(25 )+1.08√ (25 ) +9.3(9) (4.276 ) (18) vc= =5.25 ft /sec (9 )(4.276) 2

2

v´ >v c ≫ turbulent flow

N ℜ=

(use Eq.3.4)

2970 ρ v´ d ( 2970 )( 9 )(8.10 )( 4.276 ) = =37032 µp 25 N ℜ=37032, Curve II ≫ f =0.0063

From fig 7.1 ,

(( 0.0063 ) (9) ( 5500 ) ( 8.10) ) fρL v´ Δ pp = =185 psi = 25.8 d (25.8)( 4.276 ) 2

2

c)Pressure losses insi inside de drill collar:

Avg Velocity inside drill collar , ´v =

363 q = =18.72 ft /sec 2 2.45 d 2.45 ( 2.813 )2

1.08 ( 25)+ 1.08 √ ( 25 ) +9.3(9) ( 2.813) (18) =5.84 ft /sec Critical velocity , v c = (9)(2.813) 2

v´ >v c ≫ turbulent flow N ℜ=

2

(use Eq.3.4)

2970 ρ v´ d ( 2970 )( 9 )(18.72 ) ( 2.813 ) =56303 = 25 µp N ℜ=56303, Curve II ≫ f =0.0060

From fig 7.1 ,

(( 0.0060 ) (9) ( 500 ) (18.72 ) ) fρL v´ = =130 psi 25.8 d (25.8 )(2.813) 2

2

Δ pp =

d)Pressure losses thr through ough bit Δpb Three nozzles (one for each cone) will be used, hence 1/3 q will flow through each. For least 250 ft/sec through each nozzle,

√

1 q 3 d= 2.45 ´v

=

√

363 / 3 =0.44 ∈. 2.45 ×250

This nozzle allows an actual velocity of:

v´ =

121 =258 ft / sec 14 2 2.45 32

( )

v´ = at

>> Nozzle d=14/32 in is chosen:

√(

Actual nozzle diameter =

d= 3

Pressure drop across the bit, :

)

2

14 =0.758∈. 32

Δ pb =

(363 )2(9) =536 psi 4 7430 (0.95 ) 2 ( 0.758 )

e)Pressure losses aro around und drill collar:

363

v´ =

=9 ft /sec

[( ) ( ) ] 2

7 3 2.45 7 − 6 8 4

2

The hydraulically equivalent diameter of the annulus:

d a = d 1−d 2 1 3 7 d=7 −6 =1 ∈. 4 8 8

√

( )

2 1.08 (25 )+1.08 ( 25 ) + 9.3(9) 1

vc=

12 (18) 8

1 (9)(1 ) 8

=8.03 ft /sec

v´ >v c ≫ turbulent flow

N ℜ=

2970 ρ v´ d = µp

(use Eq.3.4)

( 81 )=10825

( 2970 )( 9 )(9 ) 1 25

N ℜ=10825, Curve IV ≫ f =0.000 96

From fig 7.1 ,

fρL ´v ( ( 0.0098 ) (9 ) ( 500)( 9 ) ) = Δ pac = =123 psi 25.8 d 1 (25.8)(1 ) 8 2

f)P )Pressure ressure losses ar around ound drill pipe:

2

the average velocity around drill pipe = 4 ft/sec The hydraulically equivalent diameter of the annulus:

d a = d 1−d 2 7 7 d=7 −5=2 ∈. 8 8

√

( )

2 1.08 (25 )+1.08 ( 25 ) +9.3(9) 2

vc=

72 (18) 8

7 (9)(2 ) 8

=5.82 ft / sec

v´ < v c ≫ laminar flow Δ pap =

(

(

)

(use Eq.3.6)

)

µ pv´ 25 (4 ) L 5500 = Y b+ 18+ =44 psi 5d 300 d 7 7 300(2 ) 2 5 8 8

( )

(g) The total pressure dro drop p in the system (Dpt )

Δ pt =¿ 24+185+130+536+123+44= 1042 psi (h) Horsepower output at the pump Assuming volumetric and mechanical efficiencies of the pump are 90% and 85% respectively:

HP=

363 ×1042 q × Dp =288 horsepower = 1714 ×η v ×ηm 1714 ×0.90 × 0.85

COMP COMPARI ARI ARISON SON OF PRES PRESSURE SURE DROP CALCUL CALCULA ATION V VS S GRAPHICAL METHOD : System Component Surface connections , Δps Inside drill pipe, Δpp Inside drill collar, Δpc Bit nozzles, Δpb Outside drill collar, Δpac Outside drill pipe, Δpap Total Circulation pressure, Δpt

Pressure drop (psi) 24

Graphical method (psi) 21

185 130 536 123 44 1042

135 116 539 179 56 1046

FORMA FORMATION TION PRES PRESSURE SURE

a) graph

Pore Pressure/depth 0 1000 2000

Depth (ft)

3000 4000 5000 6000 7000 8000 9000 10000 0

1

2

3

4

5

Pressure (psi) (b) The pore pressure gradients in the formations from surface are: 0 - 8000 ft: 3720 - 0/8000 - 0 = 0.465 psi/ft 0 - 8500 ft: 6800 - 0/8500 - 0 = 0.800 psi/ft 0 - 9500 ft: 6900 - 0/9500 - 0 = 0.726 psi/ft

6

7

8

(c) (c)The required mud weights are as follows: At 8000 ft: •

3720 + 200 = 3920 psi

•

3920/8000 = 0.49 psi/ft = 9.42 ppg

•

6800 + 200 = 7000 psi

•

7000/8500 = 0.82 psi/ft = 15.77 ppg

•

6900 + 200 = 7100 psi

•

7100/9500 = 0.75 psi/ft = 14.42 ppg

At 8500 ft:

At 9500 ft:

d) If the mud weight of 9.42 ppg were used to drill at 8500 ft the underbalance would be: •

6800 - (8500 x 9.42 x 0.052) = 2636 psi.

•

Hence the borehole pressure is 2636 psi less than the formation pressure.

e) If, when using 0.82 psi/ft (or 15.77 ppg) mud for the section at 8500 ft, the fluid level in the hole dropped to 500 ft the bottom hole pressure would fall by: •

500 x 0.82 = 410 psi.

•

Hence the pressure in the borehole would be 210 psi below the formation pressure.

f ) The density of the fluid in the formation between 8500 and 9500 ft is:

The fluid in the formations below 8500 ft is = gas.

Question 2 a) BIT SIZE (IN) NORMAL MUDWEIGHT (PPG) OVERBURDEN GRAD. (PPG) FORM. GRAD. (PPG)

Depth(ft) 7500 7600 7700 7800 7900 8000 8100 8200 8300 8400 8500 8600 8700

ROP(ft/hr) 125 103 77 66 45 37 40 42 41 44 34 33 32

12.25 9.5 19.2 8.9

RPM 120 120 110 110 110 110 110 110 100 100 100 100 110

WOB(000lbs 38 38 38 38 385 37 35 33 33 38 38 40 42

d EXPONENT 1.23 1.29 1.35 1.4 1.48 1.56 1.51 1.47 1.45 1.49 1.57 1.61 1.67

mud weight(ppg) 9.5 9.5 9.5 9.6 9.6 9.8 9.8 9.9 10 10.25 10.25 11 11

dc EXP0NENT 1.23 1.29 1.35 1.38 1.46 1.6 1.47 1.41 1.38 1.38 1.46 1.39 1.44

depth vs d and dc exponent d exponent

dc exponent

8800 8600 8400 8200 8000 7800 7600 7400 7200 7000 6800 1.2

1.25

1.3

1.35

1.4

1.45

1.5

1.55

1.6

1.65

1.7

b) A plot of the d and dc exponents confirms that the top of the overpressured zone is

at 8000 ft...