Drilling Hydraulics

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Few components of the overall drilling system offer the possibility of concise, arithmetic Few components of the overall drilling system offer the possibility of concise, arithmetic conclusions.

conclusions. Analysis of the Analysis of the various parts of a various parts of a hydraulics program, however, hydraulics program, however, can lead drillingcan lead drilling engineers to clear conclusions.

engineers to clear conclusions.

Unfortunately, things clear to engineers are not always clear to other drilling personnel. Unfortunately, things clear to engineers are not always clear to other drilling personnel. Consequently, most

Consequently, most rigs drill with mrigs drill with mediocre bit cleaning (hydraulicsediocre bit cleaning (hydraulics). ). Rig supervisors Rig supervisors often areoften are reluctant to participate in or even accept the thesis that improved hydraulics will always result in reluctant to participate in or even accept the thesis that improved hydraulics will always result in an increase in drilling efficiency.

an increase in drilling efficiency. C

CLASSICLASSICHHYDRAULICSYDRAULICS Various approaches to hy

Various approaches to hydraulics have been developed sdraulics have been developed since early work done ince early work done circa 1948. circa 1948. ToTo include a summary of the principal, workable methods would be overly burdensome and include a summary of the principal, workable methods would be overly burdensome and potentially confusing.

potentially confusing. A simple and practical method exists A simple and practical method exists which is termed "classicwhich is termed "classical" by someal" by some in drilling.

in drilling. Commonly, bit hydraulic Commonly, bit hydraulic horsepower is optimizehorsepower is optimized, or rather maximized, d, or rather maximized, in order toin order to improve bit cleaning.

improve bit cleaning. Hydraulic horsepower can Hydraulic horsepower can be computed by tbe computed by the following equation:he following equation:

714 714 ,, 1 1 PQ PQ Hp Hp = = Equation Equation 7-17-1 Bi

Bit ht hydydraraululic ic hohorsrsepepowowerer,,HpHp_bit_bit, the, then in is gs giviven en by by EqEquauatition on 7-7-2:2:

714 714 ,, 1 1 Q Q P P Hp

Hp_bit_bit == bitbit Equation Equation 7-27-2

In virtually all drilling situations, pump pressure or standpipe pressure is limited by either In virtually all drilling situations, pump pressure or standpipe pressure is limited by either equipment design or arbitrarily

equipment design or arbitrarily limited by someone limited by someone on the rig. on the rig. In either case, In either case, the followingthe following p

prroocceedduurre e hhaas s bbeeeen n ususeed d tto o mmaaxxiimmizizeeHpHp_bit_bit::

1.

1. WrWritite e an an eqequauatition on rerelalatiting ng HpHp_bit_bit to thto the e avavaiailablable le popowewer r in in ththe e sysyststemem..

2.

2. Differentiate the equation with resDifferentiate the equation with respect to independent variables and set pect to independent variables and set the firstthe first differential equal to zero.

differential equal to zero. 3.

3. Solve the equation developed in Step 2 tSolve the equation developed in Step 2 to see if a maximum or minimo see if a maximum or minimum hasum has resulted.

resulted.

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D r i l l i n g P r a c t i c e s D r i l l i n g P r a c t i c e s C h a p t e r 7 C h a p t e r 7 Example 7-1 Example 7-1 Given:

Given: Equations Equations 7-1 7-1 and and 7-2.7-2. Determine:

Determine: Derive Derive a a general general relationship relationship for for maximummaximum HpHp_bit_bit

Solution:

Solution: Step Step 1. 1. From From Chapter Chapter 6 6 Pressure Pressure Losses Losses in in the the Circulating Circulating System:System:

bit bit c c s s PP PP P P == ++ Equation Equation 7-37-3 s s c c KQKQ P P == Equation Equation 7-47-4 Also: Also: bit bit c c s s HpHp HpHp Hp Hp == ++ Equation Equation 7-57-5

Substituting Equation 7-1 into Equation 7-5 yields: Substituting Equation 7-1 into Equation 7-5 yields:

714 714 ,, 1 1 714 714 ,, 1 1 714 714 ,, 1 1 Q Q P P Q Q P P Q Q P P_s_s ₌₌ _c_c ₊₊ _bit_bit Equation Equation 7-67-6

Rearranging and canceling the 1,714: Rearranging and canceling the 1,714:

Q Q P P Q Q P P Q Q P

P_bit_bit == _s_s −− _c_c Equation Equation 7-77-7 Substituting Equation 7-4 into Equation 7-7:

Substituting Equation 7-4 into Equation 7-7:

1 1 + + − − = = ss s s bit bitQQ PP QQ KQKQ P P Equation Equation 7-87-8 Step 2.

Step 2. Differentiating and Differentiating and setting equal setting equal to zero:to zero:

0 0 = = bit bit P

P Minimum Minimum Equation Equation 7-97-9

0 0 )) 1 1 (( ++ == − − ss s s ss KQKQ P

P Maximum Maximum Equation Equation 7-107-10

Step 3.

Step 3. Substituting Equation 7-4 Substituting Equation 7-4 into Equation 7-10:into Equation 7-10:

0 0 )) 1 1 (( ++ == − − _c_c s s ss PP P P Equation Equation 7-117-11 c c s s ss PP P P ==(( ++11)) Equation Equation 7-127-12

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D r i l l i n g P r a c t i c e s D r i l l i n g P r a c t i c e s H y d r a u l i c s H y d r a u l i c s

Therefore,

Therefore, a a maximum maximum value value of of hydraulic hydraulic horsepower horsepower at at the the bit bit develops develops when when is is a a defineddefined fraction

fraction of of , , so so long long as as is is also also at at the the maximum maximum selected selected pressure. pressure. If If the the common common value value ofof is

is used, used, then then is is maximum maximum when:when:

c c P P s s P P PP_s_s 86 86 .. 1 1 = = s s HpHp_bit_bit s s c c PP P P == 00..3535 Equation Equation 7-147-14 s s bit bit PP P P == 00..6565 Equation Equation 7-157-15 It

It is is then then clear clear that that the the only only way way to to increase increase in in any any fixed fixed situation situation is is to to increase increase standpipestandpipe pressure,

pressure, . . It It is is also also clear clear that that any any arbitrary arbitrary decisions decisions to to limit limit pump pump pressure pressure is is also also aa decision to limit hydraulic horsepower at the bit (bit cleaning) and is also a decision to reduce decision to limit hydraulic horsepower at the bit (bit cleaning) and is also a decision to reduce drilling rate.

drilling rate. Credit for such a decision sCredit for such a decision should certainly be borne by the hould certainly be borne by the individual responsibleindividual responsible for making it.

for making it.

bit bit Hp Hp s s P P

Hydraulic impact force at the bit can be maximized to promote bit cleaning as a reasonable Hydraulic impact force at the bit can be maximized to promote bit cleaning as a reasonable al

alteternrnatativive e to to mmaxaximimiziziningg . . ThThe e prprococededurure e usused ed inin Example 7-2Example 7-2 in maximizing impact forcein maximizing impact force is similar to that used in

is similar to that used in Example 7-1.Example 7-1. Equation 7-16 is Equation 7-16 is used to used to define impact define impact force.force.

bit bit Hp Hp 932 932 ,, 1 1 Q Q V V IF IF == ρ ρ mm nn Equation Equation 7-167-16 Example 7-2 Example 7-2 Given:

Given: Equation Equation 7-16 7-16 and and the the data data inin Example 7-1.Example 7-1. Determine:

Determine: Derive Derive a a relationship relationship for for maximizing maximizing impact impact force,force, IFIF , at the bit., at the bit. Solution:

Solution: Step Step 1. 1. Equation Equation 7-17 7-17 can can be be used used to to calculate calculate the the pressure pressure drop drop acrossacross the bit nozzles.

the bit nozzles.

120 120 ,, 1 1 2 2 n n m m bit bit V V P P == ρ ρ Equation Equation 7-177-17

Substituting Equation 7-3 and 7-4 into Equation 7-17 yields: Substituting Equation 7-3 and 7-4 into Equation 7-17 yields:

s s s s n n m mVV ₌₌_P_P ₋₋_KQ_KQ 120 120 ,, 1 1 2 2 ρ ρ Equation Equation 7-187-18 Rearranging Equation 7-18: Rearranging Equation 7-18: 2 2 1 1 ⎤⎤ ⎡⎡

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D r i l l i n g P r a c t i c e s D r i l l i n g P r a c t i c e s C h a p t e r 7

C h a p t e r 7

Substituting Equation 7-16 into Equation 7-19: Substituting Equation 7-16 into Equation 7-19:

( (

))

2 2 // 1 1 120 120 ,, 1 1 932 932 ,, 1 1 ⎥⎥ ⎥⎥ ⎥⎥ ⎥⎥ ⎦⎦ ⎤⎤ ⎢⎢ ⎢⎢ ⎢⎢ ⎢⎢ ⎣⎣ ⎡⎡ − − = = m m s s s s m mQQ PP KQKQ IF IF ρ ρ ρ ρ Equation Equation 7-207-20 Simplifying: Simplifying:

[ [

₂₂ ₊₊₂₂

]]

11//22 − − = = ss s sQQ AKQ AKQ AP AP IF IF Equation Equation 7-217-21 where

where A A is is a a constant constant equal equal to:to:

333 333 ,, 3 3 m m A A == ρ ρ Step 2.

Step 2. Differentiating and Differentiating and setting equal setting equal to zero:to zero:

[ [

]]

dQ dQ AKQ AKQ Q Q AP AP d d dQ dQ IF IF d d ss ss 2 2 // 1 1 2 2 2 2 )) (( ₌₌ −− ++

[

22 22

]

11//22

[

₂₂ ₍₍ ₂₂₎₎ 11

]]

2 2 1 1 0

0 == AP AP_s_sQQ −− AKQ AKQss++ −− AP AP_s_sQQ−− ss++ AKQ AKQss++

1 1 )) 2 2 (( 2 2 0

0 == AP AP_s_sQQ−− A A ss ++ KQKQss++ Q Q P P s s A A Q Q AP AP_s_s (( 22)) _c_c 2 2 0 0 == −− ++ Step 3. Step 3. c c s s ss PP P P (( 22)) 2 2 0 0 == −− ++ s s c c PP s s P P ⎟⎟ ⎠ ⎠ ⎞ ⎞ ⎜⎜ ⎝ ⎝ ⎛ ⎛ + + = = 2 2 2 2 Equation Equation 7-227-22

Using the common value

Using the common value ss ==11..8686 then maximum impact force then maximum impact force occurs when: occurs when: s s c c PP P P == 00..5252 Equation Equation 7-237-23

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maximum surface pressure in

maximum surface pressure in Figure 7-1Figure 7-1 is 3,000 psi. is 3,000 psi. Note that Note that the maximum the maximum impact force wimpact force willill always be at a higher flow rate than the maximum horsepower.

always be at a higher flow rate than the maximum horsepower.

0 0 200 200 400 400 600 600 800 800 1000 1000 1200 1200 1400 1400 0 0 110000 220000 330000 404000 550000 660000 770000 Flow Rate, Q Flow Rate, Q H H p p o o r r I I F F IMPACT FORCE IMPACT FORCE HORSEPOWER HORSEPOWER Figure 7-1

Figure 7-1. . Plot ofPlot of HpHpandand IFIF versus Flow Rate showing versus Flow Rate showing MaMaximumsximums

Example 7-3

Example 7-3 illustrates methods for hydraulics planning and compares the results for two valuesillustrates methods for hydraulics planning and compares the results for two values of allowable

of allowable standpipe pressures. standpipe pressures. In the In the planning phase, pressplanning phase, pressure losses ure losses are calculated are calculated atat various depths and nozzle sizes are determined for various depth ranges.

various depths and nozzle sizes are determined for various depth ranges. Example 7-3

Example 7-3 Given:

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C h a p t e r 7

Drill collars are 6½ by 2¼" and 600 feet long. Drill collars are 6½ by 2¼" and 600 feet long. Mud Properties are

Mud Properties are ρ ρ _m_m ==1616ppgppg ,, PVPV == 2525, , YpYp ==1212

Both Mud Pumps are Emsco F-800's Both Mud Pumps are Emsco F-800's F-800 Pump data.

F-800 Pump data. The rated speed iThe rated speed is 150 spms 150 spm

Table 7

Table 7-1. -1. Liner RatingLiner Rating s and Output Volus and Output Volu mes for an F-80mes for an F-800 P0 Pumpump

LINER SIZE

LINER SIZE MAXMAX PPss GPSGPS

6½" 2,120 3.88 6½" 2,120 3.88 6" 6" 2,490 2,490 3.303.30 5½" 2,965 2.78 5½" 2,965 2.78 5" 5" 3,590 3,590 2.292.29 Determine:

Determine: Nozzle Nozzle sizes sizes to to be usbe used at ed at 9,000, 9,000, 12,000 and 12,000 and 15,000 f15,000 feet eet using using twotwo cases.

cases. Case

Case 1: 1: Save old Save old pump, pump, MaximumMaximum PP_s_s ==22,,500500psipsi and maximumand maximum spm =110.

spm =110. Case

Case 2: 2: Pump is Pump is as as good good as as manufacturer manufacturer says says it it is, is, run run accordingaccording to design parameters (150 spm and liner rating).

to design parameters (150 spm and liner rating). Solution:

Solution: Case Case 1 1 : : In In Case Case 1 1 the the maximum maximum surface surface pressure pressure will will be be 2,500 2,500 psi.psi. The liner with a 2,500

The liner with a 2,500 psi rating is 6 inches. psi rating is 6 inches. The maximum flow rate The maximum flow rate isis calculated as follows: calculated as follows: gpm gpm spm spm gps gps Q Q_max_max ==((33..3030 )()(110110 ))==363363

First, calculate the pressure losses in the circulating system using the First, calculate the pressure losses in the circulating system using the equations given in Chapter 6: "Pressure Losses in the Circulating equations given in Chapter 6: "Pressure Losses in the Circulating System."

System." Since this is Since this is for planning purposes, tfor planning purposes, the pressure losses he pressure losses in thein the surface connections

surface connections will be ignored. will be ignored. Calculate the pressure Calculate the pressure losses in thlosses in thee drill pipe.

drill pipe. The length of drill The length of drill pipe at a tpipe at a total depth of 9,000 otal depth of 9,000 feet will befeet will be 8,400 f

8,400 feet. eet. (Total depth (Total depth less less the the length of length of the the drill collarsdrill collars.) .) Assume aAssume a flow rate of 200 gpm. flow rate of 200 gpm. 83 83 .. 4 4 19 19 .. 0 0 81 81 .. 1 1 81 81 .. 0 0 5 5 10 10 68 68 .. 7 7 D D ll PV PV Q Q P P_dp_dp ρ ρ mm − −

×

=

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Calculate the pressure losses in the

Calculate the pressure losses in the drill collars:drill collars:

( ( ))

83 83 .. 4 4 19 19 .. 0 0 81 81 .. 1 1 81 81 .. 0 0 5 5 10 10 68 68 .. 7 7 D D ll PV PV Q Q P P_dc_dc ρ ρ mm − −

×

=

( ( ))

(

( ) (

) ( ) (

) ( )

) (

( ))

( (

))

psipsi P P_dc_dc 233233 25 25 .. 2 2 600 600 25 25 200 200 16 16 10 10 68 68 .. 7 7 83 83 .. 4 4 19 19 .. 0 0 81 81 .. 1 1 81 81 .. 0 0 5 5 = = × × = = − − Calculate the

Calculate the pressure losses pressure losses in the in the drill collar drill collar annulus. annulus. The rheologyThe rheology constants ‘

constants ‘nn’ and ‘’ and ‘kk’ must be calculated first.’ must be calculated first.

⎟⎟⎟⎟ ⎠ ⎠ ⎞ ⎞ ⎜⎜⎜⎜ ⎝ ⎝ ⎛ ⎛ + + + + = = Yp Yp PV PV Yp Yp PV PV n n 33..3232 loglog 22 7743 7743 .. 0 0 12 12 25 25 12 12 )) 25 25 )( )( 2 2 (( log log 32 32 .. 3 3 _⎥⎥ == ⎦⎦ ⎤⎤ ⎢⎢ ⎣⎣ ⎡⎡ + + + + = = n n n n Yp Yp PV PV k k 511 511 + + = = 3567 3567 .. 0 0 511 511 12 12 25 25 7743 7743 .. 0 0 == + + = = k k

Calculate the annular velocity around the

Calculate the annular velocity around the drill collars.drill collars.

( (

22 22

))

5 5 .. 24 24 p p h h DD D D Q Q v v − − = =

( (

))

fpmfpm v v 163163 5 5 .. 6 6 5 5 .. 8 8 )) 200 200 )( )( 5 5 .. 24 24 (( 2 2 2 2 ₋₋ == = =

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Cancel Anytime. D r i l l i n g P r a c t i c e s D r i l l i n g P r a c t i c e s C h a p t e r 7 C h a p t e r 7

( (

))

psipsi P P_dpa_dpa 5555 5 5 .. 4 4 5 5 .. 8 8 300 300 )) 400 400 ,, 8 8 )( )( 3567 3567 .. 0 0 (( )) 7743 7743 .. 0 0 )( )( 3 3 (( 1 1 )) 7743 7743 .. 0 0 )( )( 2 2 (( 5 5 .. 4 4 5 5 .. 8 8 )) 94 94 )( )( 4 4 .. 2 2 (( 00..77437743 ₌₌ ⎥⎥ ⎦⎦ ⎤⎤ ⎢⎢ ⎣⎣ ⎡⎡ − − ⎭⎭ ⎬⎬ ⎫⎫ ⎩⎩ ⎨⎨ ⎧⎧ ⎥⎥ ⎦⎦ ⎤⎤ ⎢⎢ ⎣⎣ ⎡⎡ ++ ⎥⎥ ⎦⎦ ⎤⎤ ⎢⎢ ⎣⎣ ⎡⎡ − − = =

Calculate the pressure at the surface. Calculate the pressure at the surface.

dpa dpa dca dca dc dc dp dp c c

P

=

+

psi

P

_c_c

=

252

252 +

+

233

233 +

+

20

20 +

+

55

55 =

=

560

The same calculations

The same calculations are made at one other are made at one other flow rate. flow rate. In this casIn this case, ae, a flow rate of

flow rate of 500 gpm was select500 gpm was selected. ed. Any reasonable flow rate Any reasonable flow rate will suffice.will suffice. Table 7-2

Table 7-2 shows the calculated results at 9,000 feet.shows the calculated results at 9,000 feet.

Table

7-Table 7-2. 2. Pressure LosPressure Los ses at 9,ses at 9,000 000 ftft

Q Q PPCC = = PPdpdp + + PPdcdc + + PPdcdc aa + + PPdpadpa 200 200 560 560 252 252 233 233 20 20 5555 500 2,734 500 2,734 1,321 1,321 1,226 1,226 57 57 130130

The same calculat

The same calculations are made ions are made at 12,000 and at 12,000 and 15,000 feet. 15,000 feet. The resultsThe results of those calculations are presented in

of those calculations are presented in Table 7-3Table 7-3 andand Table 7-4.Table 7-4.

Table 7-3

Table 7-3. . PressuPressure Lossre Losses at es at 12,0012,000 0 ftft

Q Q PPCC = = PPdpdp + + PPdcdc + + PPdcdc aa + + PPdpadpa 200 200 669 669 342 342 233 233 20 20 7474 500 3,252 500 3,252 1,793 1,793 1,226 1,226 57 57 176176

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Cancel Anytime. D r i l l i n g P r a c t i c e s D r i l l i n g P r a c t i c e s H y d r a u l i c s H y d r a u l i c s the maxim

the maximum surface um surface pressure is pressure is 2,500 psi. 2,500 psi. Calculate the Calculate the pressurepressure losses in the circulating system and at the bit when the bit horsepower is losses in the circulating system and at the bit when the bit horsepower is maximized. maximized. s s c c PP P P ==00..3535 psi psi P P_c_c ==((00..3535)()(22,,500500))==875875 s s bit bit PP P P == 00..6565 psi psi P P_bit_bit ==((00..6565)()(22,,500500))==11,,625625

The same calculations are made where impact force is maximized using The same calculations are made where impact force is maximized using Equation 7-23 and Equation 7-24.

Equation 7-23 and Equation 7-24.

s s c c PP P P ==00..5252 psi psi P P_c_c ==((00..5252)()(22,,500500))==11,,300300 s s bit bit PP P P == 00..4848 psi psi P P_bit_bit ==((00..4848)()(22,,500500))==11,,200200

For the horsepower method and the impact force method, the pressure For the horsepower method and the impact force method, the pressure losses

losses in in the the circulating circulating system, system, , , will will be be 875 875 psi psi and and 1,300 1,300 psi,psi, respectively.

respectively. Figure 7-2Figure 7-2 can be plotted with the previously determinedcan be plotted with the previously determined data.

data. It is a It is a plot of floplot of flow rate versus w rate versus pressure losses pressure losses in the circulatingin the circulating system.

system. The pressure The pressure losses losses in the in the circulating circulating system system include allinclude all pressure losses ex

pressure losses except pressure drop cept pressure drop across the bit. across the bit. The graph is The graph is usedused to

to determine determine the the flow flow rate rate at, at, which which is is equivalent equivalent to to 875 875 and and 1,300 1,300 psi.psi. In the graph, plot the pressure losses at 9,000, 12,000 and 15,000 feet In the graph, plot the pressure losses at 9,000, 12,000 and 15,000 feet from

from Table 7-2Table 7-2 throughthrough Table 7-4.Table 7-4. Plot Plot the the pressure pressure losses losses in in thethe circulating system where horsepower and impact force will be maximized. circulating system where horsepower and impact force will be maximized. In addition, the

In addition, the maximum flow rate can be maximum flow rate can be placed on the graph. placed on the graph. The pointThe point

c c P P c c P P

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Cancel Anytime. D r i l l i n g P r a c t i c e s D r i l l i n g P r a c t i c e s C h a p t e r 7 C h a p t e r 7 For

For the the horsepower horsepower method, method, the the flow flow rate rate where where is is equal equal to to 875 875 psipsi can be calculated. can be calculated. c c P P

(

( )

)

(

( ))

( (

))

( (

200200

))

560 560 875 875 7304 7304 .. 1 1 2 2 LogLog Q Q Log Log Log Log Log Log − − − − = = gpm gpm Q Q₂₂ ==259259 9,000' 9,000' 12,000' 12,000' 15,000' 15,000' 1000 1000 10000 10000 C C i i r r c c u u l l a a t t i i n n g g P P r r e e s s s s u u r r e e L L o o s s s s , , p p s s i i P P Max Max IF IF = = 13001300 P P Max Max Hp Hp = = 875875 c c c c 214 214 234 234 259 259 270 270 294 294 325 325

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This can be done for each depth using both method: impact force and This can be done for each depth using both method: impact force and horsepower.

horsepower. Table 7-5Table 7-5 andand Table 7-6Table 7-6 show the results for Case 1.show the results for Case 1.

Table 7

Table 7-5. -5. Results based on HorsepoResults based on Horsepo wer Method for Case 1wer Method for Case 1

DEPTH P

DEPTH P_c_c PP_bit_bit Q Q NOZ'sNOZ's HpHp_bit_bit/in/in22 IF/inIF/in22

9 9,,00000 0 87875 5 11,,66225 5 25259 9 1100,,1100,,1111 44..332 2 12..712744 12,000 12,000 875 875 1,625 1,625 234 234 10,10,9 10,10,9 3.91 3.91 11.5111.51 15,000 15,000 875 875 1,625 1,625 214 214 10,9,9 10,9,9 3.58 3.58 10.5610.56 Table 7

Table 7-6. -6. Results based on Impact Force Method for Case 1Results based on Impact Force Method for Case 1

DEPTH P

DEPTH Pcc PPbitbit Q Q NOZ'sNOZ's HpHp_bit_bit IFIF

9, 9,00000 0 1,1,30300 0 1,1,20200 0 325 325 1212,1,12,2,1313 4.4.02 02 1313..7676 12 12,0,000 00 1,1,30300 0 1,1,20200 0 29294 4 1212,1,12,2,1212 3.3.63 63 1212.4.444 15 15,0,000 00 1,1,30300 0 1,1,20200 0 27270 0 1212,1,11,1,1111 3.3.33 33 1111.4.400 The

The nozzle nozzle sizes sizes are are calculated calculated based based on on and and the the flow flow rate rate as as followsfollows using the equation from Chapter 6: Pressure Losses in the Circulating using the equation from Chapter 6: Pressure Losses in the Circulating System: System: bit bit P P 2 2 2 2 5 5 10 10 14 14 .. 9 9 n n m m bit bit A A Q Q P P ρ ρ − −

=

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Cancel Anytime. D r i l l i n g P r a c t i c e s D r i l l i n g P r a c t i c e s C h a p t e r 7 C h a p t e r 7 n n n n n n N N A A S S == 11,,304304 Equation Equation 7-267-26

The average nozzle diameter would then be: The average nozzle diameter would then be:

( )

( )(

(

))

33 33 .. 10 10 3 3 2457 2457 .. 0 0 304 304 ,, 1 1 ₌₌ = = n n S S

The required nozzles would be 10,10,11. The required nozzles would be 10,10,11. Case

Case 2: 2: The same The same calculations calculations are are made made for for Case Case 2 2 as as Case Case 11 except that t

except that the maximum surface he maximum surface pressure will now pressure will now be 3,590 psi. be 3,590 psi. TheThe maximum flow rate with 5 inch

maximum flow rate with 5 inch liners will be:liners will be:

gpm gpm spm spm gps gps Q Q_max_max ==((22..2929 )()(150150 ))==343343

The pressure losses in the system would remain the same, so

The pressure losses in the system would remain the same, so Table 7-2Table 7-2 through

through Table 7-4Table 7-4 are are applicable applicable for for Case Case 2 2 also. also. However, However, for bothfor both methods will be different because they are a function of the maximum methods will be different because they are a function of the maximum surface pressure. surface pressure. c c P P

For the horsepower method: For the horsepower method:

psi psi P

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The jet sizes are

The jet sizes are calculated the same way calculated the same way as in Case 1. as in Case 1. Note inNote in FigureFigure 7-3

7-3 that the circulation rate at 9,000 and 12,000 feet exceed the that the circulation rate at 9,000 and 12,000 feet exceed the maximum f

maximum flow rate low rate with 5 with 5 inch lininch liners. ers. Therefore,Therefore, Table 7-8Table 7-8 shows theshows the maximum flow rate of 343 gpm

maximum flow rate of 343 gpm at these depths.at these depths.

Table

7-Table 7-8. 8. Results for Results for Case 2 Case 2 based on Impact Forcbased on Impact Forc e Mee Methodthod

DEPTH P

DEPTH Pcc PPbitbit Q Q NOZ'sNOZ's HpHpbitbit/in/in22 IF/inIF/in22

9, 9,00000 0 1,1,37378 8 2,2,21212 2 343 343 1111,1,11,1,1111 7.7.79 79 1919..7272 12 12,0,000 00 1,1,63637 7 1,1,95953 3 343 343 1111,1,11,1,1212 6.6.87 87 1818..5252 15 15,0,000 00 1,1,86867 7 1,1,72723 3 333 333 1111,1,12,2,1212 5.5.89 89 1616..7171 12,000' 12,000' 15,000' 15,000' 10000 10000 362 362

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C h a p t e r 7

Comparison of the results computed for Case 1 and Case 2, leads to several noteworthy Comparison of the results computed for Case 1 and Case 2, leads to several noteworthy conclusions for the 5" liner situation:

conclusions for the 5" liner situation: 1.

1. Case 2 resultCase 2 results in a s in a bit horsepower bit horsepower gain from 3.58 gain from 3.58 Hp/inHp/in22 to 6.34 Hp/in. to 6.34 Hp/in.22 at 15,000 at 15,000 feet.

feet. This represents This represents an increase of an increase of 77%.77%. 2.

2. Case 2 Case 2 requires a requires a total output total output of only of only 555 HP 555 HP at 15,000 at 15,000 feet. feet. (P=3,590 psi, (P=3,590 psi, Q=265Q=265 GPM)

GPM) for for maximized maximized . . This This would would require require only only 77% 77% of of rated rated input input power power or or 617617 HP at 90% mechanical efficiency. HP at 90% mechanical efficiency. bit bit Hp Hp 3.

3. Dependent upon depth of the Dependent upon depth of the drilling operations, Case 2 drilling operations, Case 2 represents increases inrepresents increases in impact force of 42% to 47% over Case 1.

impact force of 42% to 47% over Case 1. 4.

4. Any gains seen Any gains seen in Case 2 in Case 2 are realized without exare realized without exceeding design limits ceeding design limits of the rig'sof the rig's pumps.

pumps. If the contractor If the contractor pays for a stated pays for a stated capacity and the operatcapacity and the operator contracts for aor contracts for a stated capacity,

stated capacity, how can either how can either reasonably justify usreasonably justify using less? ing less? Remember, 77% ofRemember, 77% of rated capacity is all that is needed to gain a significant level of bit hydraulic rated capacity is all that is needed to gain a significant level of bit hydraulic horsepower in this example.

horsepower in this example. The method defined above as

The method defined above as being "classical" can be summarized as being "classical" can be summarized as follows:follows: 1.

1. On On aa log Qlog Q versus versus log Plog P plot, mark equipment limitations, i.e., maximum volumes for plot, mark equipment limitations, i.e., maximum volumes for liners selected.

liners selected. If used, an arbitrarily If used, an arbitrarily selected maximum standpipe selected maximum standpipe pressure shouldpressure should also be shown.

also be shown. 2.

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Cancel Anytime. D r i l l i n g P r a c t i c e s D r i l l i n g P r a c t i c e s H y d r a u l i c s H y d r a u l i c s M

MAXIMIZINGAXIMIZINGHHYDRAULICSYDRAULICSUUSINGSINGFFIELDIELDDDAATTAA Classical hydraulics

Classical hydraulics will remain uswill remain useful to the eful to the overall drilling operations. overall drilling operations. Such utility Such utility will bewill be indispensable in the planning phase of drilling, as well as in lending conceptual clarity to the indispensable in the planning phase of drilling, as well as in lending conceptual clarity to the methods of

methods of increasing bit increasing bit cleaning. cleaning. Once drilling Once drilling operations have operations have commenced, commenced, however,however, additional work can fine-tune the hydraulics plan.

additional work can fine-tune the hydraulics plan.

Precise determination of hydraulics parameters is frustrated by the inability to quantify variables. Precise determination of hydraulics parameters is frustrated by the inability to quantify variables. Hole diameter is not precis

Hole diameter is not precisely known. ely known. Pipe diameters and roughness values vary Pipe diameters and roughness values vary form joint-to-form joint-to- joint.

joint. Significantly, Significantly, mud mud rheology rheology changes changes with with temperature, temperature, pressure, pressure, and and shear shear rate. rate. TheThe changes in rheology are difficult to know and almost impossible to include in a mathematical changes in rheology are difficult to know and almost impossible to include in a mathematical analysis.

analysis. Fortunately, Fortunately, application of application of Ken Scott'sKen Scott's11_{principals elucidates the dilemma.}_{principals elucidates the dilemma.}

Earlier, it was shown that the frictional pressure losses in a system are functions of flow rates, Earlier, it was shown that the frictional pressure losses in a system are functions of flow rates, rheological values, hole

rheological values, hole and pipe diameters, and pipe diameters, lengths of pipes lengths of pipes and mud densities. and mud densities. If, however,If, however, the flow rate,

the flow rate, QQ, is considered to be the only variable which can be changed rapidly and at will,, is considered to be the only variable which can be changed rapidly and at will, then the system pressure loss is frequently written as

then the system pressure loss is frequently written as Equation 7-4.Equation 7-4.

s s c c KQKQ P P == Equation

Equation 7-4 7-4 was was used used in in the the derivation derivation of of versus versus ratios ratios for for maximum maximum and and forfor maximum maximum c c P P PP_s_s HpHp_bit_bit IF

IF . . If Equation 7-4 is written If Equation 7-4 is written in logarithmic form, then Equation 7-27 resultsin logarithmic form, then Equation 7-27 results::

Q Q Log Log s s K K Log Log P P Log

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C h a p t e r 7

Circulating rate while drilling 500 gpm at 3,000 psi Circulating rate while drilling 500 gpm at 3,000 psi The maximum allowable surface pressure is 3,000 The maximum allowable surface pressure is 3,000 psipsi The

The following following PP_s_s versusversusQQ data is also given: data is also given:

Table

7-Table 7-9. 9. Pressure and Flow Rate Data Pressure and Flow Rate Data forfor Example 7-4Example 7-4

Q Q (gpm) (gpm) P Pss (psi) (psi) 500 3,000 500 3,000 300 1,345 300 1,345 Determine:

Determine: The The flow flow rate, rate, nozzle nozzle sizes sizes and and pressures pressures for for the the next next bit bit run.run. Solution:

Solution: First First determine determine the the bit bit nozzle nozzle pressure pressure losses losses from from given given data.data. The area of the nozzles is:

The area of the nozzles is:

⎥⎥ ⎤⎤ ⎢⎢ ⎡⎡ ⎟⎟⎟⎟ ⎞ ⎞ ⎜⎜⎜⎜ ⎛ ⎛ + + ⎟⎟ ⎞ ⎞ ⎜⎜ ⎛ ⎛ + + ⎟⎟ ⎞ ⎞ ⎜⎜ ⎛ ⎛ = = 2 2 3 3 2 2 2 2 2 2 1 1 SS SS S S A

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Cancel Anytime. D r i l l i n g P r a c t i c e s D r i l l i n g P r a c t i c e s H y d r a u l i c s H y d r a u l i c s

(

( )

)

(

( ))

(

( )

QQ₂₂22

)

LogLog

( ))

(

QQ11₁₁ Log Log P P Log Log P P Log Log s s − − − − = =

( )

(

( ))

(

( )

500500

)

(

( ))

300300 11..49694969 167 167 ,, 1 1 507 507 ,, 2 2 ₌₌ − − − − = = Log Log Log Log Log Log Log Log s s Table

Table 7-17-10. 0. Calculation of CirculCalculation of Circul ating Pressuresating Pressures

Q Q PPss - P- Pbitbit = P= PCC 500 3,000 493 2,507 500 3,000 493 2,507 300 1,345 178 1,167 300 1,345 178 1,167

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Table 7-11. 1. ReResults of sults of ExaExamplemple 7-7-33

Q

Q PPcc PPbitbit PPss vv NOZ'sNOZ's HpHp_bit_bit/in/in22 VVnn IF/inIF/in22 HpHp_tt

B

Beefforore e TTrriip p 55000 0 2,2,55007 7 44994 4 33,,000000 17175 5 1188,,1188,,1188 2.2.003 3 22115 5 99..442 2 878755 Max. Max. HpHp 33006 6 1,1,22001 1 11,,77999 9 33,,000000 1107 07 110,0,1100,,1111 44..553 3 41410 0 1100..999 9 535366 Max. Max. IFIF 33888 8 1,1,77116 6 11,,22884 4 33,,000000 1136 36 112,2,1122,,1133 44..111 1 346 346 1111..778 8 667799 10000 10000

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Cancel Anytime. D r i l l i n g P r a c t i c e s D r i l l i n g P r a c t i c e s H y d r a u l i c s H y d r a u l i c s 3.

3. In this example, In this example, rig personnel were working rig personnel were working their heart out to their heart out to do a good job, do a good job, but theybut they were not getting full be

were not getting full benefits of their efforts.nefits of their efforts.

Over the past 40 years, various methods of hydraulics planning and conceptual developments Over the past 40 years, various methods of hydraulics planning and conceptual developments have been published and discussed.

have been published and discussed. Some of the work has been good; sSome of the work has been good; some has been eitherome has been either incorrect or of li

incorrect or of little value to the ttle value to the industry. industry. The methods described in The methods described in this chapter have this chapter have beenbeen effective under

effective under a variety a variety of applications. of applications. Properly applied, Properly applied, these procedures these procedures are totallyare totally satisfactory for all situations.

satisfactory for all situations. N

NOMENCLATUREOMENCLATURE

A

A = Constant= Constant

n n A

A ₌_{= Area}_{Area of}_{of the}_{the nozzles,}_{nozzles, in}_in22

D

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Q ₌_{= Flow}_{Flow rate,}_{rate, gpm}_gpm 1

1 Q

Q ₌₌ Flow Flow rate rate corresponding corresponding to to circulating circulating pressure pressure loss loss PP₁₁, , gpmgpm

2 2 Q

Q ₌₌ Flow Flow rate rate corresponding corresponding to to circulating circulating pressure pressure loss loss PP₂₂, , gpmgpm

max max Q

Q ₌_{= Maximum}_{Maximum Flow}_{Flow rate,}_{rate, gpm}_gpm 1

1 S

S ₌_{= Diameter}_{Diameter of}_{of nozzle}_{nozzle 1,}_{1, 32}₃₂ndnd_{'s of an inch}_{'s of an inch}

2 2 S

S ₌_{= Diameter}_{Diameter of}_{of nozzle}_{nozzle 2,}_{2, 32}₃₂ndnd

's of an inch 's of an inch

3 3 S

S ₌_{= Diameter}_{Diameter of}_{of nozzle}_{nozzle 3,}_{3, 32}₃₂ndnd

's of an inch 's of an inch

n n S

S ₌_{= Average}_{Average nozzle}_{nozzle size,}_{size, 32}₃₂ndnd

's of an inch 's of an inch s

s = = Slope of Slope of pressure pressure versus versus flow flow rate orate on log-log n log-log paperpaper

n n

V

₌_{= Nozzle}_{Nozzle velocity,}_{velocity, ft/sec}_ft/sec

Average

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NOMENCLATURE FOR EQUA

NOMENCLATURE FOR EQUATIONS IN SI UNITSTIONS IN SI UNITS

A

A = Constant= Constant

n n A

A ₌_{= Area}_{Area of}_{of the}_{the nozzles,}_{nozzles, mm}_mm22

D

D = = Inside Inside diameter diameter of of pipe pipe or or dill dill collar, collar, mmmm

h h D

D ₌_{= Diameter}_{Diameter of}_{of hole,}_{hole, mm}_mm p

p D

D ₌_{= Outside}_{Outside diameter}_{diameter of}_{of pipe}_{pipe or}_{or drill}_{drill collar,}_{collar, mm}_mm Hp

Hp ₌_{= Horsepower,}_{Horsepower, kWatts}_kWatts bit

bit Hp

Hp ₌_{= Horsepower}_{Horsepower at}_{at the}_{the bit,}_{bit, kWatts}_kWatts c

c Hp

Hp ₌_{= Horsepower}_{Horsepower mm}_{mm circulating}_{circulating system,}_{system, kWatts}_kWatts s

s Hp

Hp ₌_{= Horsepower}_{Horsepower at}_{at the}_{the surface}_{surface (pump}_{(pump hydraulic}_{hydraulic horsepower),}_{horsepower), kWatts}_kWatts tt

Hp

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S ₌_{= Diameter}_{Diameter of}_{of nozzle}_{nozzle 2,}_{2, mm}_mm 3

3 S

S ₌_{= Diameter}_{Diameter of}_{of nozzle}_{nozzle 3,}_{3, mm}_mm n

n S