Drilling Hydraulics

C h ap ap t e r 7

HY D R A U L I C S

INTRODUCTION The application of hydraulics in rotary drilling is simultaneously simple in concept and difficult to achieve on the rig. Simplicity results from the purity of the mathematics mathematics involved in the study of hydraulics. Few components of the overall drilling system offer the possibility of concise, arithmetic conclusions. Analysis of the various parts of a hydraulics program, however, can lead drilling engineers to clear conclusions. Unfortunately, things clear to engineers are not always clear to other drilling personnel. Consequently, most rigs drill with mediocre mediocre bit cleaning (hydraulics). (hydraulics). Rig supervisors often are reluctant to participate in or even accept the thesis that improved hydraulics will always result in an increase in drilling efficiency. CLASSIC HYDRAULICS Various approaches to hydraulics hydraulics have been developed since since early work done circa 1948. To include a summary of the principal, workable methods would be overly burdensome and potentially confusing. A simple and practical method exists which is termed "classical" "classical" by some in drilling. Commonly, bit hydraulic horsepower is optimized, optimized, or rather maximized, in order to improve bit cleaning. Hydraulic horsepower can be computed by the the following equation: Hp  =

PQ 1,714

Equation 7-1

Bit Bit hyd hydra raul ulic ic hors horsep epow ower er,, Hp bit , the then n is is giv given en by Equa Equati tion on 7-2: 7-2: Hpbit =

Pbit Q

1,714

Equation 7-2

In virtually all drilling situations, pump pressure or standpipe pressure is limited by either equipment design or arbitrarily limited by someone on the rig. In either case, the following procedure has been use used to maximize ize Hp bit : 1. Writ Write e an equa equati tion on rela relati ting ng Hp bit to the the avai availab lable le powe powerr in the the syst system em.. 2. Differentiate the equation with respect respect to independent variables and set the first differential equal to zero. 3. Solve the equation developed in Step 2 to to see if a maximum or minimum minimum has resulted.

Copyright © 2003 OGCI/PetroSkills. All rights reserved.

7-1

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Drilling Practices Chapter 7

Example 7-1 Given:

Equations 7-1 and 7-2.

Determine:

Derive a general relationship for maximum Hp bit

Solution:

Step 1. From Chapter 6 Pressure Losses in the Circulating System: Ps = Pc + Pbit

Equation 7-3

Pc = KQ s

Equation 7-4

Hp s = Hp c + Hp bit

Equation 7-5

 Also:

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

1,714

=

Pc Q

1,714

+

Pbit Q

Equation 7-6

1,714

Rearranging and canceling the 1,714: Pbit Q = Ps Q − Pc Q

Equation 7-7

Substituting Equation 7-4 into Equation 7-7: s Pbit Q = Ps Q − KQ +1

Equation 7-8

Step 2. Differentiating and setting equal to zero: Pbit = 0 Ps − (s + 1)KQ s = 0

Minimum

Equation 7-9

Maximum

Equation 7-10

Step 3. Substituting Equation 7-4 into Equation 7-10: Ps − (s + 1)Pc = 0

Equation 7-11

Ps = (s + 1)Pc

Equation 7-12

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Drilling Practices Hydraulics

Therefore, a maximum value of hydraulic horsepower at the bit develops when Pc is a defined fraction of Ps , so long as Ps is also at the maximum selected pressure. If the common value of s = 1.86 is used, then Hp bit is maximum when: Pc = 0.35Ps

Equation 7-14

Pbit = 0.65Ps

Equation 7-15

It is then clear that the only way to increase Hp bit in any fixed situation is to increase standpipe pressure, Ps . It is also clear that any arbitrary decisions to limit pump pressure is also a decision to limit hydraulic horsepower at the bit (bit cleaning) and is also a decision to reduce drilling rate. Credit for such a decision should should certainly be borne by the individual responsible for making it. Hydraulic impact force at the bit can be maximized to promote bit cleaning as a reasonable alte altern rnat ativ ive e to maxim aximiz izin ing g Hp bit . The The proc proced edur ure e used used in Example 7-2 in maximizing impact force is similar to that used in Example 7-1. Equation 7-16 is used to define impact force. IF =

 ρ mVn Q

Equation 7-16

1,932

Example 7-2 Given:

Equation 7-16 and the data in Example 7-1.

Determine:

Derive a relationship for maximizing impact force, IF , at the bit.

Solution:

Step 1. Equation 7-17 can be used to calculate the pressure drop across the bit nozzles. Pbit =

2 ρ mVn

Equation 7-17

1,120

Substituting Equation 7-3 and 7-4 into Equation 7-17 yields: 2  ρ mV n

1,120

= Ps − KQ s

Equation 7-18

Rearranging Equation 7-18:

⎡

⎤

1 2

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Drilling Practices Chapter 7

Substituting Equation 7-16 into Equation 7-19:

⎡ s  ρ mQ ⎢ Ps − KQ ⎢ IF =  ρ m 1,932 ⎢ ⎢⎣ 1,120

(

⎤ ⎥ ⎥ ⎥ ⎥⎦

1/ 2

)

Equation 7-20

Simplifying:

[

IF =  APs Q 2 − AKQ s +2

]

1/ 2

Equation 7-21

where  A is a constant equal to:  A =

ρ m

3,333

Step 2. Differentiating and setting equal to zero:

[

2 s +2 d (IF ) d  APs Q − AKQ = dQ dQ

0=

[

1  APs Q 2 − AKQ s + 2 2

]

1/ 2

]− [2 AP Q − (s + 2) AKQ + ] 1/ 2

s

s 1

0 = 2 APs Q − A(s + 2)KQ s +1 0 = 2 APs Q − A(s + 2)Pc Q

Step 3. 0 = 2Ps − (s + 2)Pc

⎛  2  ⎞ ⎟Ps ⎝ s + 2 ⎠

Pc = ⎜

Equation 7-22

Using the common value s = 1.86   then maximum impact force occurs when: Pc = 0.52Ps

Equation 7-23

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Drilling Practices Hydraulics

maximum surface pressure in Figure 7-1 is 3,000 psi. Note that the maximum impact force will will always be at a higher flow rate than the maximum horsepower. 1400

1200 IMPACT FORCE

1000

800    F    I   r   o   p    H

600

400 HORSEPOWER

200

0 0

100

200

300

400

500

600

700

Flow Rate, Q

Figure 7-1. 7-1. Plot of

Maximums Hp and IF versus Flow Rate showing Maximums

Example 7-3 illustrates methods for hydraulics planning and compares the results for two values of allowable standpipe pressures. In the planning phase, pressure pressure losses are calculated at various depths and nozzle sizes are determined for various depth ranges.

Example 7-3 Given:

Hole size is 8½"

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Drilling Practices Chapter 7

Drill collars are 6½ by 2¼" and 600 feet long. Mud Properties are  ρ m = 16 ppg , PV = 25 , Yp = 12 Both Mud Pumps are Emsco F-800's F-800 Pump data. The rated speed is is 150 spm Table 7-1. 7-1. Liner Rating s and Output Volu mes for an F-800 F-800 Pump Pump

Determine:

Solution:

LINER SIZE

MAX Ps

GPS

6½"

2,120

3.88

6"

2,490

3.30

5½"

2,965

2.78

5"

3,590

2.29

Nozzle sizes to be used used at 9,000, 12,000 and 15,000 feet feet using two cases. Case 1:

Save old pump, Maximum Ps = 2,500 psi and maximum spm =110.

Case 2:

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

Case 1 : In Case 1 the maximum surface pressure will be 2,500 psi. The liner with a 2,500 psi rating is 6 inches. The maximum flow rate is calculated as follows: Qmax = (3.30 gps )(110 spm ) = 363 gpm

First, calculate the pressure losses in the circulating system using the equations given in Chapter 6: "Pressure Losses in the Circulating System." Since this is for planning purposes, the the pressure losses in the surface connections will be ignored. Calculate the pressure losses in the the drill pipe. The length of drill pipe at a total total depth of 9,000 feet will be 8,400 feet. feet. (Total depth less the length of the drill collars.) collars.) Assume a flow rate of 200 gpm.

Pdp =

7.68 × 10 −5

0.81 1.81  ρ m Q PV 0.19 l 4.83

D

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Drilling Practices Hydraulics

Calculate the pressure losses in the drill collars:

Pdc =

Pdc =

(

7.68 × 10−5

)

0.81 1.81  ρ m Q PV 0.19l 4.83

D

(

)

7.68 × 10 −5 (16 )

0.81

(200 )1.81 (25)0.19 (600) 233 psi = (2.25)4.83

Calculate the pressure losses in the drill collar annulus. The rheology constants ‘n’ and ‘k’ must be calculated first.

⎛ 2PV + Yp  ⎞ ⎟⎟ ⎝  PV + Yp  ⎠

n = 3.32 log⎜⎜

⎡ (2)(25) + 12 ⎤ ⎥ = 0.7743 ⎣ 25 + 12 ⎦

n = 3.32 log⎢

k=

k =

PV + Yp

511n 25 + 12 5110.7743

= 0.3567

Calculate the annular velocity around the drill collars. v =

24.5Q

(D 2 − D 2 ) h

v =

p

( 24.5)(200 )

(8.5 2 − 6.5 2 )

= 163 fpm

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Drilling Practices Chapter 7

Pdpa

⎧⎡ (2.4)(94) ⎤ ⎡ (2)(0.7743 ) + 1⎤ ⎫ = ⎨⎢ ⎥⎬ ⎥⎢ ⎩⎣ 8.5 − 4.5 ⎦ ⎣ (3)(0.7743 ) ⎦ ⎭

0.7743

⎡ (0.3567 )(8,400 ) ⎤ ⎢ 300(8.5 − 4.5 ) ⎥ = 55 psi ⎣ ⎦

Calculate the pressure at the surface.

Pc = Pdp + Pdc + Pdca + Pdpa

Pc = 252 + 233 + 20 + 55 = 560 psi The same calculations are made at one other flow rate. In this case, case, a flow rate of 500 gpm was selected. selected. Any reasonable flow rate will suffice. Table 7-2 shows the calculated results at 9,000 feet. Table 7-2. 7-2. Pressure Los ses at 9,000 9,000 ft

Q

PC =

Pdp  +

Pdc  +

Pdc a +

Pdpa

200

560

252

233

20

55

500

2,734

1,321

1,226

57

130

The same calculations calculations are made at 12,000 and 15,000 feet. The results of those calculations are presented in Table 7-3 and Table 7-4. Table 7-3. 7-3. Pressu re Losses Loss es at 12,000 12,000 ft

Q

PC =

Pdp  +

Pdc  +

Pdc a +

Pdpa

200

669

342

233

20

74

500

3,252

1,793

1,226

57

176

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Drilling Practices Hydraulics

the maximum maximum surface pressure is 2,500 psi. Calculate the pressure losses in the circulating system and at the bit when the bit horsepower is maximized. Pc = 0.35Ps Pc = (0.35 )( 2,500 ) = 875 psi Pbit = 0.65Ps Pbit = (0.65 )( 2,500 ) = 1,625 psi

The same calculations are made where impact force is maximized using Equation 7-23 and Equation 7-24. Pc = 0.52Ps Pc = (0.52)( 2,500 ) = 1,300 psi Pbit = 0.48Ps Pbit = (0.48 )( 2,500 ) = 1,200 psi

For the horsepower method and the impact force method, the pressure losses in the circulating system, Pc , will be 875 psi and 1,300 psi, respectively. Figure 7-2 can be plotted with the previously determined data. It is a plot of flow flow rate versus pressure losses in the circulating system. The pressure losses in the circulating system include all pressure losses except except pressure drop across the bit. The graph is used to determine the flow rate at, which Pc is equivalent to 875 and 1,300 psi. In the graph, plot the pressure losses at 9,000, 12,000 and 15,000 feet from Table 7-2 through Table 7-4. Plot the pressure losses in the circulating system where horsepower and impact force will be maximized. In addition, the maximum flow rate can be placed on the graph. The point

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Drilling Practices Chapter 7

For the horsepower method, the flow rate where Pc is equal to 875 psi can be calculated.

1.7304 =

Log (875) − Log (560 ) Log (Q2 − Log (200 ))

Q2 = 259 gpm

10000

15,000' 12,000' 294 270    i   s   p  ,   s   s   o    L   e   r   u   s   s   e   r    P   g   n    i    t   a    l   u   c   r    i    C

Pc Max IF = 1300

1000

Pc Max Hp = 875

214 234

259

325

9,000'

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Drilling Practices Hydraulics

This can be done for each depth using both method: impact force and horsepower. Table 7-5 and Table 7-6 show the results for Case 1. Table 7-5. 7-5. Results based on Horsepo wer Method for Case 1

DEPTH

Pc

Pbit

Q

NOZ's

Hp bit /in 2

IF/in2

9,000

875 875

1,625

259 259

10,10,11

4.32

12. 12.74

12,000

875

1,625

234

10,10,9

3.91

11.51

15,000

875

1,625

214

10,9,9

3.58

10.56

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

DEPTH

Pc

Pbit

Q

NOZ's

Hp bit

IF

9,00 9,000 0

1,30 1,300 0

1,20 1,200 0

325 325

12,1 12,12, 2,13 13

4.02 4.02

13. 13.76

12,0 12,000 00

1,30 1,300 0

1,20 1,200 0

294 294

12,1 12,12, 2,12 12

3.63 3.63

12.4 12.44 4

15,0 15,000 00

1,30 1,300 0

1,20 1,200 0

270 270

12,1 12,11, 1,11 11

3.33 3.33

11.4 11.40 0

The nozzle sizes are calculated based on Pbit and the flow rate as follows using the equation from Chapter 6: Pressure Losses in the Circulating System:

Pbit =

9.14 10 −5  An2

 ρ m Q

2

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Drilling Practices Chapter 7

Sn =

1,304 An

Equation 7-26

Nn

The average nozzle diameter would then be: Sn =

(1,304 )()(0.2457 ) 3

= 10.33

The required nozzles would be 10,10,11. Case 2: The same calculations are made for Case 2 as Case 1 except that the the maximum surface pressure will now be 3,590 psi. The maximum flow rate with 5 inch liners will be: Qmax = (2.29 gps )(150 spm ) = 343 gpm

The pressure losses in the system would remain the same, so Table 7-2 through Table 7-4 are applicable for Case 2 also. However, Pc for both methods will be different because they are a function of the maximum surface pressure. For the horsepower method: P = (0 35 )(3 590 ) = 1 256 psi

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Drilling Practices Hydraulics

The jet sizes are calculated the same way as in Case 1. Note in Figure 7-3  7-3  that the circulation rate at 9,000 and 12,000 feet exceed the maximum flow flow rate with 5 inch liners. liners. Therefore, Table 7-8 shows the maximum flow rate of 343 gpm at these depths. Table 7-8. 7-8. Results for Case 2 based on Impact Forc e Method Method

DEPTH

Pc

Pbit

Q

NOZ's

Hp bit /in 2

IF/in2

9,00 9,000 0

1,37 1,378 8

2,21 2,212 2

343 343

11,1 11,11, 1,11 11

7.79 7.79

19. 19.72

12,0 12,000 00

1,63 1,637 7

1,95 1,953 3

343 343

11,1 11,11, 1,12 12

6.87 6.87

18. 18.52

15,0 15,000 00

1,86 1,867 7

1,72 1,723 3

333 333

11,1 11,12, 2,12 12

5.89 5.89

16. 16.71

10000

15,000' 362

12,000'

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Drilling Practices Chapter 7

Comparison of the results computed for Case 1 and Case 2, leads to several noteworthy conclusions for the 5" liner situation: 1. Case 2 results results in a bit horsepower gain from 3.58 Hp/in2  to 6.34 Hp/in. 2  at 15,000 feet. This represents an increase of 77%. 2. Case 2 requires a total output of only 555 HP at 15,000 feet. (P=3,590 psi, Q=265 GPM) for maximized Hp bit . This would require only 77% of rated input power or 617 HP at 90% mechanical efficiency. 3. Dependent upon depth of the drilling operations, Case 2 represents increases in impact force of 42% to 47% over Case 1. 4. Any gains seen in Case 2 are realized without exceeding exceeding design limits of the rig's pumps. If the contractor pays for a stated capacity and the operator operator contracts for a stated capacity, how can either reasonably justify using using less? Remember, 77% of rated capacity is all that is needed to gain a significant level of bit hydraulic horsepower in this example. The method defined above as being "classical" can be summarized as follows: 1. On a log Q versus log P plot, mark equipment limitations, i.e., maximum volumes for liners selected. If used, an arbitrarily selected maximum standpipe pressure should also be shown. 2. Predict pressure/volume pressure/volume behavior at various depths of interest. The depths used can

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Drilling Practices Hydraulics

MAXIMIZING HYDRAULICS USING FIELD DATA Classical hydraulics will remain useful useful to the overall drilling operations. Such utility will be indispensable in the planning phase of drilling, as well as in lending conceptual clarity to the methods of increasing bit cleaning. Once drilling operations have commenced, however, additional work can fine-tune the hydraulics plan. Precise determination of hydraulics parameters is frustrated by the inability to quantify variables. Hole diameter is not precisely precisely known. Pipe diameters and roughness values vary form joint-to joint. Significantly, mud rheology changes with temperature, pressure, and shear rate. The changes in rheology are difficult to know and almost impossible to include in a mathematical analysis. Fortunately, application of Ken Scott's1 principals elucidates the dilemma. Earlier, it was shown that the frictional pressure losses in a system are functions of flow rates, rheological values, hole and pipe diameters, lengths of pipes and mud densities. If, however, the flow rate, Q , is considered to be the only variable which can be changed rapidly and at will, then the system pressure loss is frequently written as Equation 7-4. Pc = KQ s

Equation 7-4 was used in the derivation of Pc versus Ps ratios for maximum Hp bit and for maximum IF . If Equation 7-4 is written in logarithmic form, then Equation 7-27 results: results: Log Pc = Log K + sLog Q

Equation 7-27

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Drilling Practices Chapter 7

Circulating rate while drilling 500 gpm at 3,000 psi The maximum allowable surface pressure is 3,000 psi The following Ps versus Q data is also given: Table 7-9. 7-9. Pressure and Flow Rate Data for Example 7-4

Q

Ps

(gpm)

(psi)

500

3,000

300

1,345

Determine:

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

Solution:

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

π 

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

Equation 7-28

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Drilling Practices Hydraulics

s=

s=

Log (P2 ) − Log (P1 ) Log (Q2 ) − Log (Q1 ) Log (2,507 ) − Log (1,167 ) = 1.4969 Log (500 ) − Log (300 ) Table 7-10. 7-10. Calculation of Circul ating Pressures

Q

Ps

- Pbit

= PC

500

3,000

493

2,507

300

1,345

178

1,167

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Drilling Practices Chapter 7 Table 7-11. 7-11. Results Results of Example Example 7-3 7-3

Q

Pc

Pbit

Ps

v

NOZ's

Hp bit /in 2

Vn

IF/in2

Hp t

Before ore Trip

500

2,5 2,507

494

3,000

175

18,18,18

2.0 2.03

215

9.42

875 875

Max. Hp

306

1,2 1,201

1,799

3,000

107

10,1 0,10,11

4.53

410 410

10.99

536 536

Max. IF

388

1,7 1,716

1,284

3,000

136

12,1 2,12,13

4.11

346 346

11.78

67 679

10000

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Drilling Practices Hydraulics

3. In this example, rig personnel were working their heart out to do a good job, but they were not getting full benefits be nefits of their efforts. Over the past 40 years, various methods of hydraulics planning and conceptual developments have been published and discussed. Some of the work has been good; some some has been either incorrect or of little little value to the industry. The methods described in this chapter have been effective under a variety of applications. Properly applied, these procedures are totally satisfactory for all situations. NOMENCLATURE  A  An

=

Constant

=

Area of the nozzles, in2

D

=

Inside diameter of pipe or drill collar, inches

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Drilling Practices Chapter 7

Q

=

Q1

=

Flow rate, gpm Flow rate corresponding to circulating pressure loss P1 , gpm

Q2

=

Flow rate corresponding to circulating pressure loss P2 , gpm

Qmax

=

Maximum Flow rate, gpm

S1

=

Diameter of nozzle 1, 32nd's of an inch

S2

=

Diameter of nozzle 2, 32nd's of an inch

S3

=

Diameter of nozzle 3, 32nd's of an inch

Sn

=

Average nozzle size, 32nd's of an inch

s

=

Slope of pressure versus flow rate on on log-log paper

Vn

=

Nozzle velocity, ft/sec Average fluid velocity, fpm

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Drilling Practices Hydraulics

NOMENCLATURE FOR EQUATIONS EQUA TIONS IN SI UNITS  A  An

=

Constant

=

Area of the nozzles, mm2

D Dh

=

Inside diameter of pipe or dill collar, mm

=

Diameter of hole, mm

Dp

=

Outside diameter of pipe or drill collar, mm

Hp

=

Horsepower, kWatts

Hp bit

=

Horsepower at the bit, kWatts

Hpc

=

Horsepower mm circulating system, kWatts

Hps

=

Horsepower at the surface (pump hydraulic horsepower), kWatts

Hp t

=

Total horsepower required at the pump fluid end, kWatts

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Drilling Practices Chapter 7

S2

=

Diameter of nozzle 2, mm

S3

=

Diameter of nozzle 3, mm

Sn

=

Average nozzle size, mm