Driiling Tutorial 1 (Homework)

7/30/2019 Driiling Tutorial 1 (Homework)

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(1) What are the Rig components. Define each part and its importance

(2) Name the above rope types explaining the differences of using

each one.


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2


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The theoretical volume of fluid being displaced by the pump

having Nc liners can be easily formulated from the figure shownbelow, and is given by:

3

2

4t L s cV D L N

Sketch the preparing/cleaning mud system!!!

In term of mud system what are the data acquisition and

monitoring system


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4

If the operation needs 3 triple piston pump to assure

680bbl/mint. How much the pump stroke volume will be?


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Mud preparation/cleaning system

7


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8


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Hoisting System

9


8/4010


9/4011


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13


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Velocity of the hook = 120ft/mint

14


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15

(1) A drilling rig has three diesel engines for generating rig power requirement.

Determine the total daily fuel consumption for an average engine running speed of

900 rpm, average output torque of 1610 ft-lb and engine efficiency of 40 %. The

heating value of diesel oil is 19000 BTU/lb, and the weight of the diesel is 7.2 ppg.

(2) Using the data given previously, determine (1) the round trip ton-miles at 10000ft;

(2) casing ton-miles if one joint of casing =40ft; (3) design factor of the drilling line

when the 7in casing is running to 10000ft; (4) the ton-miles when coring from10000ft to 10180ft; and (5) the ton-miles when drilling from 10000ft to 10180ft.

(3) A rig must hoist a load of 300,000lbf. The drawworks can provide an input power to

the block and tackle system as high as 500 hp. Eight lines are strung between the

crown block and traveling block. Calculate

1. The static tension in the fast line when upward motion is impending,

2. the maximum hook horsepower available,

3. the maximum hoisting speed,

4. the actual derrick load,

5. the maximum equivalent derrick load,

6. the derrick efficiency factor.

H.W. (1)


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factors affecting tooth wear If the time interval of bit use is increased too much, the bit may break

apart leaving junk in the hole. This will required an additional trip to fishthe junk from the hole or may reduce greatly the efficiency of the next

bit if an attempt is made to drill past the junk. Thus a knowledge of the

instantaneous rate of bit wear is needed to determine how much the

time interval of bit use can be increased safely. Since practices are not

always the same for the new and old bit runs, a knowledge of how thevarious drilling parameters affect the instantaneous rate of bit wear also

is needed.

The rate of tooth wear depends primarily on:

(1) formation abrasiveness.

(2)tooth geometry.

(3)bit weight.

(4)rotary speed, and

(5)the cleaning and cooling action of the drilling fluid.


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The rate of tooth wear depends primarily on (1) formation abrasiveness.

(2)tooth geometry.(3)bit weight. (4)rotary speed, and (5)the cleaning

and cooling action of the drilling fluid.

Effect of tooth height on rate of tooth wear--Steel tooth

The bit tooth initially hasa contact area given by

11 yxi wwA

After removal ofcertain tooth height,the area are given by


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2121212112111

121121

L

L

h

121121

i

r

hwwww

hwwwwwwww

wwhwwwhw

wwL

Lwww

L

LwwwA

yyxx

yyxxxyyx

yyyxxx

yy

i

ryxx

i

rxyx

If we define the geometry constants G1 and G2 by

G1byexpressedbecanAareacontactthen

G

2

21

12122

1211211

hGhAA

Awwww

AwwwwwwG

i

iyyxx

iyyxxxy


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Since the instantaneous wear rate dh/dt is proportional to the inverse

of the contact area A

221

221

1

1

1

1

hGhGdt

dh

hGhGA

dt

dh

s

dt

dhA

i

si

The simplifiedequation is

hHdtdh

dt

dh

s 21

1

Recall that a case-hardened bit tooth or a toothwith hard facing on one side often will have a self-sharpening type of tooth wear, a constant H2 canbe selected too


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tooth height--PDC blank: the cutter

contact area is proportional to the length

of the chord.

2/sin1

cs ddt

dh

dt

dh

4.1.2 Bit weight: Galle and Woods

published one of the first equationsfor predicting the effect of bit weight

on the instantaneous rate of tooth

wear. The relation is given by

0.10andinches,indiameterbitd

unitslbm-1,000inbit weight

log1

1

b

b

b

dW

W

where

d

Wdt

dh


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b

s

d

W

dt

dh

dt

dh

log1

3979.0The wear rate at various bit weight canbe expressed in terms of a standardwear rate that would occur for a bitweight of 4,000 lbf/in. Thus, the wearrate relative to this standard wear rateis given by:

Note that dh/dt becomes infinite for

W/db=10,this equation predicts theteeth would fail instantaneously if10,000 lbf/in. of bit diameter wereapplied. Another relation is given by :

bmbd

W

d

Wdt

dh

1

Expressing this relation in terms of a standard wear rate at 4,000lbf/in. ofbit diameter yields

bmb

mb

s

d

W

d

W

d

W

dt

dh

dt

dh4


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Rotary speed---for milled-tooth bits

designed for use in soft formations.

1

60

H

s

N

dt

dh

dt

dh

Hydraulics

the effect of the cooling and cleaning action of the drilling fluid on the

cutter wear rate is much more important for diamond or PDC bit than

the rolling cutter bit, but no mathematical models.

tooth wear equation: the instantaneous rate of tooth wear is given by:

hH

H

d

W

d

W

d

W

N

dt

dh

bmb

mb

H

H 2

2

1

21

4

60

1

1


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Recommended values of H1,H2, and

(W/db)m are shown as follows:


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Define a tooth wear

parameter J2 using

21

160

4 2

1

HN

d

W

d

W

d

W

J

H

mb

bmb

2

The tooth wear equation

can be expressed by:

2/

1

2

22

0 0

22

ffHb

t h

H

hHhJt

dhhHJdtb f

Solving for the abrasiveness

constant H gives 2/222 ffb

HhHhJ

t


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An example to tooth wear equation

An 8.5-in. class 1-3-1 bit drilled from a depth of 8,179 to 8,404 ft in 10.5

hours. The average bit weight and rotary speed use for the bit run was

45,000lbf and 90 rpm, respectively. When the bit was pulled, it wasgraded T-5, B-4, G-I. Compute the average formation abrasiveness for

this depth interval. Also estimate the time required to dull the teeth

completely using the same bit weight and rotary speed.


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Solution. Using table in page 93 we obtain H1=1.84, H2=6, and

(W/db)m=8.0. using equation in page 94 we obtain

08.02/61

1

90

60

0.40.8

5.8450.884.1

2

J

Solving for the abrasiveness constant using a final fraction tooth dullness of

5/8(0.625) gives

hours

hoursH

0.73

2/625.06625.0080.0

5.102

The time required to dull the teeth

completely (hf=1.0) can be obtained by hours4.23

2/16173.00.08

2/2

222

ffHb hHhJt


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factors affecting bearing wear

The prediction of bearing wear is much more difficult than the prediction of tooth

wear. Like tooth wear, the instantaneous rate of bearing wear depends on thecurrent condition of the bit. After the bearing surface become damaged, the rate

of bearing wear increases greatly. However, since the bearing surface cannot be

examined readily during the dull bit evaluation, a liner rate of bearing wear

usually is assumed. For a given applied force, the bearing life can be expressed

in terms of total revolution as long as the rotary speed is low enough to prevent

an excessive temperature increase. Thus, bit bearing life usually is assumed tovary linearly with rotary speed.

The effect of bit weight on bearing life depends on the number and type of

bearings used and whether or not the bearings are sealed.

The hydraulic action of the drilling fluid at the bit is also thought to have some

effect on bearing life. As flow rate increase, the ability of the fluid to cool thebearings also increases. It is believed that flow rate sufficient to lift cuttings will

also be sufficient to prevent excessive temperature buildup in the bearings.


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A bearing wear formula frequently used to estimate bearing life is

given by

hoursconstant,bearing

andexponents,wearB

inchesdiameter,bitd

1,000lbf,bit weightW

rpmspeed,rotaryN

,tconsumebeenhasthatlifebearingfractionalb

460

1

B

21,

b

21

bearingB

hourstime

where

d

WN

dt

db B

b

B

B


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Define a bearing

wear parameter J3

using

21460

3

B

b

B

W

d

NJ

Then the bearing

wear formula can

be expressed by

bitthepullingafter

observedwearbearingfinaltheisb

where

f

0 0

3 b ft b

B dbJdt

Integration of the equation

above yields

f

bB

fBb

bJ

t

bJt

3

3


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An example to bearing wear equation

Compute the bearing constant for a 7.875-in., class 6-1-6(sealed journal

bearings) bit that was graded T-5, B-6, G-I after drilling 64 hours at30,000lbf and 70 rpm.

Solution. Get B1=1.6 and B2=1.0

Using equation in page 105 we obtain

820.0

30

875.74

70

600.16.1

3

J

Solving for the bearing constant using bf=6/8 yields

hours104)0.820(0.75

hours64B


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Factors affecting penetration rate The most important variables affecting penetration rate that have been

identified and studied include (1) bit type, (2)formation characteristics,

(3)drilling fluid properties, (4)bit operating conditions, (5) bit tooth wear,and (6) bit hydraulics.

bit type

For rolling cutter bits:long tooth and a large cone offset angle will get high

rate in soft formation

Drag bit are designed to obtain a given penetration rate.

The diamond and PDC bits are designed for a given penetration per

revolution by selection of the size and number of blades

formation characteristics

Elastic limit and ultimate strength are two main formation properties affectthe penetration rate.

Permeability of the formation and the mineral composition of the rock

Drilling fluid properties

Density and rheological flow properties Filtration characteristics

Solids content and size distribution chemical composition


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operating conditions

Penetration rate vs. Bit weight

No significant penetration rate is obtained until thethreshold bit weight is applied (point a).Penetration rate then increase rapidly withincreasing values of bit weight (segment ab). Aliner curve is often observed at moderate bitweights (segment bc). However, at higher values ofbit weight, subsequent increase in bit weight causesonly slight improvement in penetration rate(segment cd). In some cases, a decrease inpenetration rate is observed at extremely highvalues of bit weight (segment de). This type ofbehavior often is called bit floundering.

Rotary speed vs. penetration rate

Penetration rate usually increases linearly

with rotary speed at low values of rotary

speed. At higher values of rotary speed, the

response of penetration rate to increasing

rotary speed diminishes.


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Penetration rate relating to bit weight,

rotary speed, bit size, and rock strength

is given by

Nd

W

d

W

S

KR

tbb

2

0

2


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This theoretical relation assumes perfect bottom-hole cleaning and

incomplete bit tooth penetration.

The theoretical equation of Maurer can be verified using experimental

data obtained at relatively low bit weight and rotary speedscorresponding to Segment ab in page 115 and 117.

bit tooth wear: Most bits tends to drill slower as the bit run progresses

because of tooth wear. The tooth length of milled tooth is reduced

continually by abrasion and chipping. The insert tooth fail by breaking

or losing rather than abrasion. The same as the diamond bits. For rolling-cutter bits. Model of tooth wear on penetration rate is:

)sharpening-self(0.5exponentana

awaybeen wornhasthat

heighttoothfractionaltheh

1692815.0

1

7

2

7

where

hhR

a


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haeR 7

Anther similar but less complex relationship is given bya7 is determined based on the observed decline of penetration rate with

tooth wear for previous bits run under similar conditions.

An ExampleAn initial penetration rate of 20 ft/hr was observed in shale at the beginning of a

bit run. The previous bit was identical to the current bit and was operated under

the same conditions of bit weight, rotary speed, mud density, etc. However, a

drilling rate of 12 ft/hr was observed in the same shale formation just before

pulling the bit. If the previous bit was graded T-6, compute the approximate

value of a7.

Solution: The value of h for the previous bit just before the

end of the bit run is 6/8 or 0.75. The value of h for the nowbit is zero. Thus, for the relation given we have

75.00 77

7

12and20aa

ha

KeKe

KeR


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Dividing the first equation by the second yields

775.0

12

20 ae

Taking the natural logarithm of both sides and solving for a7 gives

68.0

75.0

1220ln7 a


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Bit hydraulics

drilling practice showed that significant improvement in penetration

rate could be achieved through an improved jetting action at the bit.

The improved jetting action promoted better cleaning of the bit teethas well as the hole bottom.

bit hydraulics. Relation between bit hydraulics and penetration rate.

Eckel found that penetration rate

could be correlated to a Reynoldsnumber group given by

1-a

Re

seconds10,000atfluiddrilling

ofiscosityapparent vanddiameter,nozzle

rateflowdensityfluiddrilling

constantscalinga

dv

Kwhere

vdKN

a


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4.3.7 penetration rate equation

For drag bit. All the drag bit are designed to achieve a given maximum

penetration per revolution. Under ideal condition, the bit weight and

rotary speed is such that the bit is kept feeding into the formation at thedesign cutting rate. The penetration rate of a drag bit for a given

penetration of the cutting element into the formation is given by

speedrotaryandblades,ofnumbereffictive

element.cuttingeachofnpenetratioeffective

Nn

Lwhere

NnLR

be

pe

bepe

The equations were derived for a simplified model which assumed thefollowing.

1. The bit has a flat face that is perpendicular to the axis of the hole.

2. Each blade is formed by diamonds laid out as a helix.

3. The stones are spherical is shape.

4. The diamonds are spaced so that the cross-sectional area removedper stone is a maximum for the design depth of penetration.


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5. The bit is operated at the design depth of penetration.

6. The bit hydraulics are sufficient for perfect bottom-hole cleaning.

For these conditions, the effective penetration and the effective number of bladesare given by

292.1

67.0

ppcb

d

cbe

ppe

LLddsCn

andLL

An example An 8.625-in. diamond bit containing 270 0.23-in.-diameter

stones of 1.00 carat is designed to operate at a depth ofpenetration of 0.01 in. Estimate the penetration rate that couldbe obtained with this bit if the formation characteristics are suchthat an acceptable bit weight and torque for this penetration

could be maintained at a rotary speed of 200 rpm.


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Solution: Ignoring the bit contouring required for proper hydraulic

action and gauge protection, the bit is assumed to have a flat face

that is perpendicular to the axis of the hole. Thus .

in.stones/sq621.4

625.84

270

2

d

c

s

C

The effective number of blades is given by

3.59

01.001.023.0625.84.6211.92

92.1

2

2

ppcb

d

cbe LLdd

s

Cn

The effective penetration is given by

.in0067.001.067.0 peL


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The penetration rate at a rotary speed of 200 rpm is given by

ft/hr24

2006059.312

0067.0

NnLR bepe

penetration rate equation For rolling cutterbits

))...()()()(( 4321 nfffffR

Driiling Tutorial 1 (Homework)

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