Drilling Hydraulics 4n4t1u
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Chapter 7
H YDRAULICS
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 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 bit cleaning (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 have been developed 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" by some in drilling. Commonly, bit hydraulic horsepower is optimized, or rather maximized, in order to improve bit cleaning. Hydraulic horsepower can be computed by the following equation: Hp =
PQ 1,714
Equation 7-1
Bit hydraulic horsepower, Hp bit , then is given by Equation 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 used to maximize Hp bit : 1. Write an equation relating Hp bit to the available power in the system. 2. Differentiate the equation with respect to independent variables and set the first differential equal to zero. 3. Solve the equation developed in Step 2 to see if a maximum or minimum has resulted.
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7-1
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 PQ P Q = c + bit 1,714 1,714 1,714
Equation 7-6
Rearranging and canceling the 1,714: Pbit Q = Ps Q − Pc Q
Equation 7-7
Substituting Equation 7-4 into Equation 7-7: Pbit Q = Ps Q − KQ s +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:
7-2
Ps − (s + 1)Pc = 0
Equation 7-11
Ps = (s + 1)Pc
Equation 7-12
⎛ 1 ⎞ Pc = ⎜ ⎟Ps ⎝ s + 1⎠
Equation 7-13
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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 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 alternative to maximizing Hp bit . The procedure 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 =
ρ mVn2
Equation 7-17
1,120
Substituting Equation 7-3 and 7-4 into Equation 7-17 yields:
ρ mVn2
= Ps − KQ s
1,120
Equation 7-18
Rearranging Equation 7-18:
⎡ ⎢ P − KQ s Vn = ⎢ s ρm ⎢ ⎢⎣ 1,120
(
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⎤ ⎥ ⎥ ⎥ ⎥⎦
)
1 2
Equation 7-19
7-3
Drilling Practices Chapter 7
Substituting Equation 7-16 into Equation 7-19: ⎡ ρ Q ⎢ P − KQ s IF = m ⎢ s ρm 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=
ρm 3,333
Step 2. Differentiating and setting equal to zero:
[
d (IF ) d APs Q 2 − AKQ s + 2 = dQ dQ
0=
[
1 APs Q 2 − AKQ s +2 2
]
1/ 2
] [2AP Q − (s + 2)AKQ ] −1/ 2
s +1
s
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 ⎞ Pc = ⎜ ⎟Ps ⎝s + 2⎠
Equation 7-22
Using the common value s = 1.86 then maximum impact force occurs when: Pc = 0.52Ps
Equation 7-23
Pbit = 0.48Ps
Equation 7-24
Similar to bit horsepower, the maximum impact force possible is seen to be dependent upon the maximum standpipe pressure used. Figure 7-1 shows a plot of horsepower and impact force versus flow rate. At some flow rate, the horsepower and impact force are a maximum. The
7-4
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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 always be at a higher flow rate than the maximum horsepower. 1400
1200 IMPACT FORCE
1000
Hp or IF
800
600
400 HORSEPOWER 200
0 0
100
200
300
400
500
600
700
Flow Rate, Q Figure 7-1. Plot of
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 losses are calculated at various depths and nozzle sizes are determined for various depth ranges.
Example 7-3 Given:
Hole size is 8½" Interval to be drilled is from 9,000 ft. to 16,000 feet. Drill pipe is 4½", 16.60 ppf, with XH connections The inside diameter is 3.826"
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7-5
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 150 spm Table 7-1. Liner Ratings and Output Volumes for an F-800 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 at 9,000, 12,000 and 15,000 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 pressure losses in the surface connections will be ignored. Calculate the pressure losses in the drill pipe. The length of drill pipe at a total depth of 9,000 feet will be 8,400 feet. (Total depth less the length of the drill collars.) Assume a flow rate of 200 gpm.
Pdp =
Pdp =
7-6
(
)
7.68 × 10 −5 ρ m 0.81Q 1.81PV 0.19 l D 4.83
(
)
7.68 × 10 −5 (16 )
(200 )1.81 (25 )0.19 (8,400 ) = 252 psi (3.826 )4.83 0.81
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Drilling Practices Hydraulics
Calculate the pressure losses in the drill collars:
( )
7.68 × 10−5 ρm
Pdc =
0.81 1.81
PV 0.19l
Q
D 4.83
(
)
7.68 × 10 −5 (16 )
Pdc =
(200 )1.81 (25 )0.19 (600 ) = 233 psi (2.25 )4.83 0.81
Calculate the pressure losses in the drill collar annulus. The rheology constants ‘n’ and ‘k’ must be calculated first. ⎛ 2PV + Yp ⎞ ⎟⎟ n = 3.32 log⎜⎜ ⎝ PV + Yp ⎠
⎡ (2)(25) + 12 ⎤ n = 3.32 log⎢ ⎥ = 0.7743 ⎣ 25 + 12 ⎦ k=
k =
PV + Yp 511n
25 + 12 5110.7743
= 0.3567
Calculate the annular velocity around the drill collars.
24.5Q
v =
(D
v =
(24.5)(200 )
2 h
(8.5
− D p2
2
)
− 6.5 2
) = 163 fpm
Calculate pressure losses in the drill collar annulus. n
⎞ ⎛ 2n + 1 ⎞⎤ kl ⎟⎜ ⎥ ⎟ ⎝ 3n ⎟⎠⎥ 300 (Dh − D p ) ⎠ ⎦
Pdca
⎡⎛ 2.4v = ⎢⎜ ⎜ ⎣⎢⎝ Dh − D p
Pdca
⎧⎡ (2.4)(163 ) ⎤ ⎡ (2)(0.7743 ) + 1⎤ ⎫ = ⎨⎢ ⎥⎬ ⎥⎢ ⎩⎣ 8.5 − 6.5 ⎦ ⎣ (3)(0.7743 ) ⎦ ⎭
0.7743
⎡ (0.3567 )(600 ) ⎤ ⎢ 300(8.5 − 6.5 ) ⎥ = 20 psi ⎣ ⎦
Calculate the pressure losses in the drill pipe annulus. v =
( 24.5)(200 )
(8.5
2
− 4.5 2
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) = 94fpm 7-7
Drilling Practices Chapter 7
⎧⎡ ( 2.4)(94 ) ⎤ ⎡ ( 2)(0.7743 ) + 1⎤ ⎫ Pdpa = ⎨⎢ ⎥⎬ ⎥⎢ ⎩⎣ 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, a flow rate of 500 gpm was selected. Any reasonable flow rate will suffice. Table 7-2 shows the calculated results at 9,000 feet. Table 7-2. Pressure Losses at 9,000 ft
Q
PC =
Pdp +
Pdc +
Pdca +
Pdpa
200
560
252
233
20
55
500
2,734
1,321
1,226
57
130
The same 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. Pressure Losses at 12,000 ft
Q
PC =
Pdp +
Pdc +
Pdca +
Pdpa
200
669
342
233
20
74
500
3,252
1,793
1,226
57
176
Table 7-4. Pressure Losses at 15,000 ft
Q
PC =
Pdp +
Pdc +
Pdca +
Pdpa
200
777
431
233
20
93
500
3,771
2,265
1,226
57
223
The point where Hp bit is maximum is defined by Equation 7-14 and Equation 7-15 when the maximum surface pressure is used. In Case 1,
7-8
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Drilling Practices Hydraulics
the 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 rate versus pressure losses in the circulating system. The pressure losses in the circulating system include all pressure losses 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 where the Pc line for horsepower and impact force crosses the Pc line for each depth is the desired flow rate. The desired flow rates can also be determined mathematically. The slope of the Pc line can be determined from the information in Table 7-2.
s=
s=
Log (P2 ) − Log (P1 ) Log (Q2 ) − Log (Q1 )
Equation 7-25
Log (2,734 ) − Log (560 ) = 1.7304 Log (500 ) − Log (200 )
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7-9
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
Circulating Pressure Loss, psi
270
9,000' 325
Pc Max IF = 1300 1000
Pc Max Hp = 875
214
259
234
Q Max = 363 100 100
1000 Flow Rate, Q
Figure 7-2. Plot of Flow Rate versus Circulating Pressure Loss for Case 1
7-10
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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. Results based on Horsepower Method for Case 1
DEPTH
Pc
Pbit
Q
NOZ's
Hp bit /in2
IF/in2
9,000
875
1,625
259
10,10,11
4.32
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. Results based on Impact Force Method for Case 1
DEPTH
Pc
Pbit
Q
NOZ's
Hp bit
IF
9,000
1,300
1,200
325
12,12,13
4.02
13.76
12,000
1,300
1,200
294
12,12,12
3.63
12.44
15,000
1,300
1,200
270
12,11,11
3.33
11.40
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 ρm Q 2 An2
For the horsepower method at 9,000 feet:
1,625 =
(
)
9.14 10 −5 (16)(259 ) 2 An2
An = 0.2457in2 Nozzle diameters are available in 32nd's of an inch and most bits are run with three nozzles. Equation 7-26 can be used to calculate the average nozzle diameters with any number of nozzles.
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7-11
Drilling Practices Chapter 7
Sn =
1,304 An Nn
Equation 7-26
The average nozzle diameter would then be: Sn =
(1,304 )(0.2457 ) = 10.33 3
The required nozzles would be 10,10,11. Case 2: The same calculations are made for Case 2 as Case 1 except that 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: Pc = (0.35 )(3,590 ) = 1,256 psi Pbit = (0.65 )(3,590 ) = 2,334 psi
For the impact force method: Pc = (0.52 )(3,590 ) = 1,867 psi Pbit = (0.48 )(3,590 ) = 1,723 psi
Figure 7-3 is a graphical plot of the data for Case 2. Table 7-7 and Table 7-8 show the results of the analysis. Table 7-7. Results for Case 2 based on HorsePower Method
7-12
DEPTH
Pc
Pbit
Q
NOZ's
Hp bit /in2
IF/in2
9,000
1,256
2,334
319
11,11,10
7.65
18.82
12,000
1,256
2,334
288
10,10,10
6.91
17.01
15,000
1,256
2,334
264
10,10,9
6.34
15.60
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Drilling Practices Hydraulics
The jet sizes are calculated the same way as in Case 1. Note in Figure 7-3 that the circulation rate at 9,000 and 12,000 feet exceed the maximum flow rate with 5 inch liners. Therefore, Table 7-8 shows the maximum flow rate of 343 gpm at these depths. Table 7-8. Results for Case 2 based on Impact Force Method
DEPTH
Pc
Pbit
Q
NOZ's
Hp bit /in2
IF/in2
9,000
1,378
2,212
343
11,11,11
7.79
19.72
12,000
1,637
1,953
343
11,11,12
6.87
18.52
15,000
1,867
1,723
333
11,12,12
5.89
16.71
10000
15,000' 362 12,000' 333
401
Circulating Pressure Loss, psi
9,000'
Pc Max IF = 1867 Pc Max Hp = 1256 1000
264
319
288
Q Max = 343 100 100
1000 Flow Rate, Q
Figure 7-3. Plot of Flow Rate versus Circulating Pressure Loss for Case 2
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7-13
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 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 design limits of the rig's pumps. If the contractor pays for a stated capacity and the operator contracts for a stated capacity, how can either reasonably justify using less? , 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 behavior at various depths of interest. The depths used can be arbitrarily selected or selected based on anticipated trips from the bit program. 3. Mark lines representing the desired system pressure loss, Pc , which is dependent upon the hydraulic design criterion (maximized Hp bit or maximized IF ). 4. From the intercepts of desired Pc and Pc 's predicted, determine flow rates, Q, to be used at various depths. 5. The difference between the selected maximum pressure, Ps , and the desired circulating pressure, Pc , is the bit nozzle pressure loss, Pbit , to be used at a specific depth and circulating rate. Select nozzles that most nearly provide the Pbit needed at the Q selected for each bit run. Hydraulic slide rules or charts provided by bit manufacturers have simplified hydraulics planning. Approximately one hour of engineering time would be required to plan hydraulics for a typical 15,000 foot well with the use of charts or hydraulic slide rules. Hand-held calculators and computers can reduce the planning time to approximately 15 minutes for a typical 15,000 ft. well. Further, the use of well written programs with such calculators places the power for good planning in the hands of rig personnel. The industry no longer must tolerate mediocre hydraulics in drilling operations.
7-14
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Drilling Practices Hydraulics
MAXIMIZING HYDRAULICS USING FIELD DATA Classical hydraulics will remain 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 known. Pipe diameters and roughness values vary form t-tot. 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: Log Pc = Log K + sLog Q
Equation 7-27
Historically, the value of s has been taken to be 1.86. This value may have been statistically correct over many situations when it was given to the industry decades ago; however, the use of s = 1.86 often results in errors that can be avoided if s can be determined correctly. That the value of s is easy to determine on the rig is shown in Example 7-4. Example 7-4 also demonstrates a method of improving hydraulics performance on the rig as a result of improved knowledge of the flow characteristics of a specific rig at a certain time.
Example 7-4 Given:
Rig is drilling at 8,430 feet and is preparing to trip for bit. The following information is given: Hole size = 9½" Bit is STC F3 w/ 3x18/32" nozzles Drill collars are 7¾" by 2¾" and 582 feet long Drill pipe is 4½", Grade E, 16.6 ppf, with 4½" XH connections The inside diameter of the pipe is 3.826 inches Mud Properties: ρ m = 12 . 0 ppg , PV = 20 , Yp = 18
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7-15
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. 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: 2 π ⎡⎛ S1 ⎞
2 2 ⎛ S3 ⎞ ⎤ ⎛ S2 ⎞ An = ⎢⎜ ⎟ + ⎜ ⎟ ⎥ ⎟ + ⎜⎜ 4 ⎢⎝ 32 ⎠ 32 ⎠ 32 ⎟⎠ ⎥ ⎝ ⎝ ⎣ ⎦
An =
Equation 7-28
2 2 ⎛ 18 ⎞ ⎛ 18 ⎞ ⎤ ⎢⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎥ = 0.7455 in 2 4 ⎢⎝ 32 ⎠ ⎝ 32 ⎠ ⎝ 32 ⎠ ⎥⎦ ⎣
2 π ⎡⎛ 18 ⎞
Calculate the pressure drop across the bit nozzles at the two flow rates given in Table 7-9.
Pbit =
Pbit =
Pbit =
(
)
9.14 10 −5 ρ m Q 2 An2
(
)
2
)
2
9.14 10 −5 (12)(500 )
(
(0.7455 )2
9.14 10 −5 (12)(300 )
(0.7455 )2
= 493 psi
= 178 psi
The pressure losses in the circulating system, Pc , can be calculated as shown in Table 7-10. Next, the slope of the Pc versus Q line must be determined.
7-16
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Drilling Practices Hydraulics
s=
Log (P2 ) − Log (P1 ) Log (Q2 ) − Log (Q1 )
s=
Log (2,507 ) − Log (1,167 ) = 1.4969 Log (500 ) − Log (300 ) Table 7-10. Calculation of Circulating Pressures
Q
Ps
- Pbit
= PC
500
3,000
493
2,507
300
1,345
178
1,167
The value of Pc where horsepower and impact force are a maximum must be determined. From Equation 7-13: ⎛ 1 ⎞ Pc = ⎜ ⎟Ps ⎝ s + 1⎠ 1 ⎛ ⎞ Pc = ⎜ ⎟(3,000 ) = 1,201psi ⎝ 1.4969 + 1 ⎠ Pbit = Ps − Pc Pbit = 3,000 − 1,201 = 1,799 psi
From Equation 7-22 for impact force: ⎛ 2 ⎞ Pc = ⎜ ⎟Ps ⎝s + 2⎠ 2 ⎛ ⎞ Pc = ⎜ ⎟(3,000 ) = 1,716 psi ⎝ 1.4969 + 2 ⎠ Pbit = Ps − Pc Pbit = 3,000 − 1,716 = 1,284 psi
The data in Table 7-10 and the above calculations can be used to construct Figure 7-4. This figure can be used to determine the flow rate where horsepower or impact force will be a maximum. Alternatively, the flow rates can be calculated as in Example 7-3. The nozzles sizes are also determined as in Example 7-3. Table 7-11 shows the results of maximizing the hydraulics on the drilling rig.
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7-17
Drilling Practices Chapter 7 Table 7-11. Results of Example 7-3
Q
Pc
Pbit
Ps
v
NOZ's
Hp bit /in2
Vn
IF/in2
Hp t
Before Trip
500
2,507
494
3,000
175
18,18,18
2.03
215
9.42
875
Max. Hp
306
1,201
1,799
3,000
107
10,10,11
4.53
410
10.99
536
Max. IF
388
1,716
1,284
3,000
136
12,12,13
4.11
346
11.78
679
Circulating Pressure Loss, psi
10000
388 306
Pc Max IF = 1716 Pc Max Hp = 1201 1000 100
1000 Flow Rate, Q
Figure 7-4. Plot of Flow Rate versus Circulating Pressure Loss for Example 7-4
As seen from Example 7-4, improving the hydraulics parameters on the rig results in the following improvements: 1. Bit hydraulic horsepower per square inch, Hp bit /in2, is increased by 123% (2.03 Hp to 4.53 Hp) while simultaneously reducing total pump output power by 39% (875 Hp to 536 Hp). 2. By adjusting for maximum impact force, a gain of 25% is realized (9.42 lbf /in2 to 11.78 lbf /in2) while reducing total pump power by 22% (875 Hp to 679 Hp).
7-18
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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 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 has been either incorrect or of 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 Dh Dp
=
Inside diameter of pipe or drill collar, inches
=
Diameter of hole, inches
=
Outside diameter of pipe or drill collar, inches
Hp
=
Horsepower, Hp
Hp bit
=
Horsepower at the bit, Hp
Hp c
=
Horsepower in circulating system, Hp
Hp s
=
Horsepower at the surface (pump hydraulic horsepower), Hp
Hp t
=
IF
=
Total horsepower required at the pump fluid end, Hp Impact force, lbf
K
=
Constant
k
=
Power Law constant
l
=
Length of pipe or drill collars, feet
Nn
=
Number of nozzles in the bit
n
=
Power Law constant, normally the slope of shear stress-shear rate diagram on log-log plot
P P1
= =
Pressure, psi Circulating pressure loss corresponding to flow rate Q1 , psi
P2
=
Circulating pressure loss corresponding to flow rate Q2 , psi
Pbit
=
Pc
=
Pressure drop at the bit (through jet nozzles), psi Pressure loss in the circulating system ( Ps − Pbit ), psi
Pdc
=
Pressure loss in the drill collars, psi
Pdca Pdp
=
Pressure loss in the drill collar annulus, psi
=
Pressure loss in the drill pipe, psi
Pdpa
=
Pressure loss in the drill pipe annulus, psi
Ps
=
Pressure at the surface (standpipe pressure), psi
PV
=
Plastic viscosity,
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7-19
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 log-log paper
Vn
=
Nozzle velocity, ft/sec
v Yp
= =
Average fluid velocity, fpm Yield point, lbf/100 ft2
=
Mud weight, ppg
ρm
SI UNITS The equations given in the chapter are converted to SI units below
PQ 60
Equation 7-1:
Hp =
Equation 7-2:
Hp bit =
Equation 7-16:
IF =
Pbit Q 60
Equation 7-30
ρ mVnQ
Equation 7-31
60
Equation 7-17:
Pbit =
Equation 7-26:
Sn =
Equation 7-28:
An =
7-20
Equation 7-29
ρ mVn2
Equation 7-32
1,810
1.27 An Nn
[(S ) 4
π
2
1
+ (S 2 )2 + (S3 )2
Equation 7-33
]
Equation 7-34
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Drilling Practices Hydraulics
NOMENCLATURE FOR EQUATIONS IN SI UNITS A An
=
Constant
=
Area of the nozzles, mm2
D Dh Dp
=
Inside diameter of pipe or dill collar, mm
=
Diameter of hole, mm
=
Outside diameter of pipe or drill collar, mm
Hp
=
Horsepower, kWatts
Hp bit
=
Horsepower at the bit, kWatts
Hp c
=
Horsepower mm circulating system, kWatts
Hp s
=
Horsepower at the surface (pump hydraulic horsepower), kWatts
Hp t
=
Total horsepower required at the pump fluid end, kWatts
IF
=
Impact force, Nt
K
=
Constant
k
=
Consistency index, Pa secn
l
=
Length of pipe or drill collars, meters
Nn
=
Number of nozzles mm the bit
n
=
Power Law constant, normally the slope of shear stress-shear rate diagram on log-log plot
P P1
= =
Pressure, kPa Circulating pressure loss corresponding to flow rate Q1 , kPa
P2
=
Circulating pressure loss corresponding to flow rate Q2 , kPa
Pbit
=
Pc
=
Pressure drop at the bit (through jet nozzles), kPa Pressure loss mm the circulating system ( Ps − Pbit ), kPa
Pdc
=
Pressure loss mm the drill collars, kPa
Pdca Pdp
=
Pressure loss mm the drill collar annulus, kPa
=
Pressure loss mm the drill pipe, kPa
Pdpa
=
Pressure loss mm the drill pipe annulus, kPa
Ps
=
Pressure at the surface (standpipe pressure), kPa
PV Q
=
Plastic viscosity, Pa sec
=
Q1
=
Flow rate, cubic meters per minute Flow rate corresponding to circulating pressure loss P1 , cubic meters per
=
minute Flow rate corresponding to circulating pressure loss P2 , cubic meters per
Q2
minute Qmax
=
Maximum Flow rate, cubic meters per minute
S1
=
Diameter of nozzle 1, mm
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7-21
Drilling Practices Chapter 7
S2
=
Diameter of nozzle 2, mm
S3
=
Diameter of nozzle 3, mm
Sn
=
Average nozzle size, mm
s
=
Slope of pressure versus flow rate on log-log paper
Vn
=
Nozzle velocity, m/sec
v Yp
=
Average fluid velocity, meters per minute
=
Yield point, Pa
=
Mud weight, kg per cubic meter
ρm
REFERENCE 1
Scott, K. F.; "A New Practical Approach to Rotary Drilling Hydraulics," SPE 3530, 1971.
7-22
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H YDRAULICS
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 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 bit cleaning (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 have been developed 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" by some in drilling. Commonly, bit hydraulic horsepower is optimized, or rather maximized, in order to improve bit cleaning. Hydraulic horsepower can be computed by the following equation: Hp =
PQ 1,714
Equation 7-1
Bit hydraulic horsepower, Hp bit , then is given by Equation 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 used to maximize Hp bit : 1. Write an equation relating Hp bit to the available power in the system. 2. Differentiate the equation with respect to independent variables and set the first differential equal to zero. 3. Solve the equation developed in Step 2 to see if a maximum or minimum has resulted.
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7-1
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 PQ P Q = c + bit 1,714 1,714 1,714
Equation 7-6
Rearranging and canceling the 1,714: Pbit Q = Ps Q − Pc Q
Equation 7-7
Substituting Equation 7-4 into Equation 7-7: Pbit Q = Ps Q − KQ s +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:
7-2
Ps − (s + 1)Pc = 0
Equation 7-11
Ps = (s + 1)Pc
Equation 7-12
⎛ 1 ⎞ Pc = ⎜ ⎟Ps ⎝ s + 1⎠
Equation 7-13
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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 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 alternative to maximizing Hp bit . The procedure 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 =
ρ mVn2
Equation 7-17
1,120
Substituting Equation 7-3 and 7-4 into Equation 7-17 yields:
ρ mVn2
= Ps − KQ s
1,120
Equation 7-18
Rearranging Equation 7-18:
⎡ ⎢ P − KQ s Vn = ⎢ s ρm ⎢ ⎢⎣ 1,120
(
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⎤ ⎥ ⎥ ⎥ ⎥⎦
)
1 2
Equation 7-19
7-3
Drilling Practices Chapter 7
Substituting Equation 7-16 into Equation 7-19: ⎡ ρ Q ⎢ P − KQ s IF = m ⎢ s ρm 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=
ρm 3,333
Step 2. Differentiating and setting equal to zero:
[
d (IF ) d APs Q 2 − AKQ s + 2 = dQ dQ
0=
[
1 APs Q 2 − AKQ s +2 2
]
1/ 2
] [2AP Q − (s + 2)AKQ ] −1/ 2
s +1
s
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 ⎞ Pc = ⎜ ⎟Ps ⎝s + 2⎠
Equation 7-22
Using the common value s = 1.86 then maximum impact force occurs when: Pc = 0.52Ps
Equation 7-23
Pbit = 0.48Ps
Equation 7-24
Similar to bit horsepower, the maximum impact force possible is seen to be dependent upon the maximum standpipe pressure used. Figure 7-1 shows a plot of horsepower and impact force versus flow rate. At some flow rate, the horsepower and impact force are a maximum. The
7-4
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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 always be at a higher flow rate than the maximum horsepower. 1400
1200 IMPACT FORCE
1000
Hp or IF
800
600
400 HORSEPOWER 200
0 0
100
200
300
400
500
600
700
Flow Rate, Q Figure 7-1. Plot of
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 losses are calculated at various depths and nozzle sizes are determined for various depth ranges.
Example 7-3 Given:
Hole size is 8½" Interval to be drilled is from 9,000 ft. to 16,000 feet. Drill pipe is 4½", 16.60 ppf, with XH connections The inside diameter is 3.826"
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7-5
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 150 spm Table 7-1. Liner Ratings and Output Volumes for an F-800 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 at 9,000, 12,000 and 15,000 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 pressure losses in the surface connections will be ignored. Calculate the pressure losses in the drill pipe. The length of drill pipe at a total depth of 9,000 feet will be 8,400 feet. (Total depth less the length of the drill collars.) Assume a flow rate of 200 gpm.
Pdp =
Pdp =
7-6
(
)
7.68 × 10 −5 ρ m 0.81Q 1.81PV 0.19 l D 4.83
(
)
7.68 × 10 −5 (16 )
(200 )1.81 (25 )0.19 (8,400 ) = 252 psi (3.826 )4.83 0.81
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Drilling Practices Hydraulics
Calculate the pressure losses in the drill collars:
( )
7.68 × 10−5 ρm
Pdc =
0.81 1.81
PV 0.19l
Q
D 4.83
(
)
7.68 × 10 −5 (16 )
Pdc =
(200 )1.81 (25 )0.19 (600 ) = 233 psi (2.25 )4.83 0.81
Calculate the pressure losses in the drill collar annulus. The rheology constants ‘n’ and ‘k’ must be calculated first. ⎛ 2PV + Yp ⎞ ⎟⎟ n = 3.32 log⎜⎜ ⎝ PV + Yp ⎠
⎡ (2)(25) + 12 ⎤ n = 3.32 log⎢ ⎥ = 0.7743 ⎣ 25 + 12 ⎦ k=
k =
PV + Yp 511n
25 + 12 5110.7743
= 0.3567
Calculate the annular velocity around the drill collars.
24.5Q
v =
(D
v =
(24.5)(200 )
2 h
(8.5
− D p2
2
)
− 6.5 2
) = 163 fpm
Calculate pressure losses in the drill collar annulus. n
⎞ ⎛ 2n + 1 ⎞⎤ kl ⎟⎜ ⎥ ⎟ ⎝ 3n ⎟⎠⎥ 300 (Dh − D p ) ⎠ ⎦
Pdca
⎡⎛ 2.4v = ⎢⎜ ⎜ ⎣⎢⎝ Dh − D p
Pdca
⎧⎡ (2.4)(163 ) ⎤ ⎡ (2)(0.7743 ) + 1⎤ ⎫ = ⎨⎢ ⎥⎬ ⎥⎢ ⎩⎣ 8.5 − 6.5 ⎦ ⎣ (3)(0.7743 ) ⎦ ⎭
0.7743
⎡ (0.3567 )(600 ) ⎤ ⎢ 300(8.5 − 6.5 ) ⎥ = 20 psi ⎣ ⎦
Calculate the pressure losses in the drill pipe annulus. v =
( 24.5)(200 )
(8.5
2
− 4.5 2
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) = 94fpm 7-7
Drilling Practices Chapter 7
⎧⎡ ( 2.4)(94 ) ⎤ ⎡ ( 2)(0.7743 ) + 1⎤ ⎫ Pdpa = ⎨⎢ ⎥⎬ ⎥⎢ ⎩⎣ 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, a flow rate of 500 gpm was selected. Any reasonable flow rate will suffice. Table 7-2 shows the calculated results at 9,000 feet. Table 7-2. Pressure Losses at 9,000 ft
Q
PC =
Pdp +
Pdc +
Pdca +
Pdpa
200
560
252
233
20
55
500
2,734
1,321
1,226
57
130
The same 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. Pressure Losses at 12,000 ft
Q
PC =
Pdp +
Pdc +
Pdca +
Pdpa
200
669
342
233
20
74
500
3,252
1,793
1,226
57
176
Table 7-4. Pressure Losses at 15,000 ft
Q
PC =
Pdp +
Pdc +
Pdca +
Pdpa
200
777
431
233
20
93
500
3,771
2,265
1,226
57
223
The point where Hp bit is maximum is defined by Equation 7-14 and Equation 7-15 when the maximum surface pressure is used. In Case 1,
7-8
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Drilling Practices Hydraulics
the 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 rate versus pressure losses in the circulating system. The pressure losses in the circulating system include all pressure losses 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 where the Pc line for horsepower and impact force crosses the Pc line for each depth is the desired flow rate. The desired flow rates can also be determined mathematically. The slope of the Pc line can be determined from the information in Table 7-2.
s=
s=
Log (P2 ) − Log (P1 ) Log (Q2 ) − Log (Q1 )
Equation 7-25
Log (2,734 ) − Log (560 ) = 1.7304 Log (500 ) − Log (200 )
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7-9
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
Circulating Pressure Loss, psi
270
9,000' 325
Pc Max IF = 1300 1000
Pc Max Hp = 875
214
259
234
Q Max = 363 100 100
1000 Flow Rate, Q
Figure 7-2. Plot of Flow Rate versus Circulating Pressure Loss for Case 1
7-10
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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. Results based on Horsepower Method for Case 1
DEPTH
Pc
Pbit
Q
NOZ's
Hp bit /in2
IF/in2
9,000
875
1,625
259
10,10,11
4.32
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. Results based on Impact Force Method for Case 1
DEPTH
Pc
Pbit
Q
NOZ's
Hp bit
IF
9,000
1,300
1,200
325
12,12,13
4.02
13.76
12,000
1,300
1,200
294
12,12,12
3.63
12.44
15,000
1,300
1,200
270
12,11,11
3.33
11.40
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 ρm Q 2 An2
For the horsepower method at 9,000 feet:
1,625 =
(
)
9.14 10 −5 (16)(259 ) 2 An2
An = 0.2457in2 Nozzle diameters are available in 32nd's of an inch and most bits are run with three nozzles. Equation 7-26 can be used to calculate the average nozzle diameters with any number of nozzles.
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7-11
Drilling Practices Chapter 7
Sn =
1,304 An Nn
Equation 7-26
The average nozzle diameter would then be: Sn =
(1,304 )(0.2457 ) = 10.33 3
The required nozzles would be 10,10,11. Case 2: The same calculations are made for Case 2 as Case 1 except that 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: Pc = (0.35 )(3,590 ) = 1,256 psi Pbit = (0.65 )(3,590 ) = 2,334 psi
For the impact force method: Pc = (0.52 )(3,590 ) = 1,867 psi Pbit = (0.48 )(3,590 ) = 1,723 psi
Figure 7-3 is a graphical plot of the data for Case 2. Table 7-7 and Table 7-8 show the results of the analysis. Table 7-7. Results for Case 2 based on HorsePower Method
7-12
DEPTH
Pc
Pbit
Q
NOZ's
Hp bit /in2
IF/in2
9,000
1,256
2,334
319
11,11,10
7.65
18.82
12,000
1,256
2,334
288
10,10,10
6.91
17.01
15,000
1,256
2,334
264
10,10,9
6.34
15.60
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Drilling Practices Hydraulics
The jet sizes are calculated the same way as in Case 1. Note in Figure 7-3 that the circulation rate at 9,000 and 12,000 feet exceed the maximum flow rate with 5 inch liners. Therefore, Table 7-8 shows the maximum flow rate of 343 gpm at these depths. Table 7-8. Results for Case 2 based on Impact Force Method
DEPTH
Pc
Pbit
Q
NOZ's
Hp bit /in2
IF/in2
9,000
1,378
2,212
343
11,11,11
7.79
19.72
12,000
1,637
1,953
343
11,11,12
6.87
18.52
15,000
1,867
1,723
333
11,12,12
5.89
16.71
10000
15,000' 362 12,000' 333
401
Circulating Pressure Loss, psi
9,000'
Pc Max IF = 1867 Pc Max Hp = 1256 1000
264
319
288
Q Max = 343 100 100
1000 Flow Rate, Q
Figure 7-3. Plot of Flow Rate versus Circulating Pressure Loss for Case 2
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7-13
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 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 design limits of the rig's pumps. If the contractor pays for a stated capacity and the operator contracts for a stated capacity, how can either reasonably justify using less? , 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 behavior at various depths of interest. The depths used can be arbitrarily selected or selected based on anticipated trips from the bit program. 3. Mark lines representing the desired system pressure loss, Pc , which is dependent upon the hydraulic design criterion (maximized Hp bit or maximized IF ). 4. From the intercepts of desired Pc and Pc 's predicted, determine flow rates, Q, to be used at various depths. 5. The difference between the selected maximum pressure, Ps , and the desired circulating pressure, Pc , is the bit nozzle pressure loss, Pbit , to be used at a specific depth and circulating rate. Select nozzles that most nearly provide the Pbit needed at the Q selected for each bit run. Hydraulic slide rules or charts provided by bit manufacturers have simplified hydraulics planning. Approximately one hour of engineering time would be required to plan hydraulics for a typical 15,000 foot well with the use of charts or hydraulic slide rules. Hand-held calculators and computers can reduce the planning time to approximately 15 minutes for a typical 15,000 ft. well. Further, the use of well written programs with such calculators places the power for good planning in the hands of rig personnel. The industry no longer must tolerate mediocre hydraulics in drilling operations.
7-14
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Drilling Practices Hydraulics
MAXIMIZING HYDRAULICS USING FIELD DATA Classical hydraulics will remain 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 known. Pipe diameters and roughness values vary form t-tot. 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: Log Pc = Log K + sLog Q
Equation 7-27
Historically, the value of s has been taken to be 1.86. This value may have been statistically correct over many situations when it was given to the industry decades ago; however, the use of s = 1.86 often results in errors that can be avoided if s can be determined correctly. That the value of s is easy to determine on the rig is shown in Example 7-4. Example 7-4 also demonstrates a method of improving hydraulics performance on the rig as a result of improved knowledge of the flow characteristics of a specific rig at a certain time.
Example 7-4 Given:
Rig is drilling at 8,430 feet and is preparing to trip for bit. The following information is given: Hole size = 9½" Bit is STC F3 w/ 3x18/32" nozzles Drill collars are 7¾" by 2¾" and 582 feet long Drill pipe is 4½", Grade E, 16.6 ppf, with 4½" XH connections The inside diameter of the pipe is 3.826 inches Mud Properties: ρ m = 12 . 0 ppg , PV = 20 , Yp = 18
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7-15
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. 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: 2 π ⎡⎛ S1 ⎞
2 2 ⎛ S3 ⎞ ⎤ ⎛ S2 ⎞ An = ⎢⎜ ⎟ + ⎜ ⎟ ⎥ ⎟ + ⎜⎜ 4 ⎢⎝ 32 ⎠ 32 ⎠ 32 ⎟⎠ ⎥ ⎝ ⎝ ⎣ ⎦
An =
Equation 7-28
2 2 ⎛ 18 ⎞ ⎛ 18 ⎞ ⎤ ⎢⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎥ = 0.7455 in 2 4 ⎢⎝ 32 ⎠ ⎝ 32 ⎠ ⎝ 32 ⎠ ⎥⎦ ⎣
2 π ⎡⎛ 18 ⎞
Calculate the pressure drop across the bit nozzles at the two flow rates given in Table 7-9.
Pbit =
Pbit =
Pbit =
(
)
9.14 10 −5 ρ m Q 2 An2
(
)
2
)
2
9.14 10 −5 (12)(500 )
(
(0.7455 )2
9.14 10 −5 (12)(300 )
(0.7455 )2
= 493 psi
= 178 psi
The pressure losses in the circulating system, Pc , can be calculated as shown in Table 7-10. Next, the slope of the Pc versus Q line must be determined.
7-16
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Drilling Practices Hydraulics
s=
Log (P2 ) − Log (P1 ) Log (Q2 ) − Log (Q1 )
s=
Log (2,507 ) − Log (1,167 ) = 1.4969 Log (500 ) − Log (300 ) Table 7-10. Calculation of Circulating Pressures
Q
Ps
- Pbit
= PC
500
3,000
493
2,507
300
1,345
178
1,167
The value of Pc where horsepower and impact force are a maximum must be determined. From Equation 7-13: ⎛ 1 ⎞ Pc = ⎜ ⎟Ps ⎝ s + 1⎠ 1 ⎛ ⎞ Pc = ⎜ ⎟(3,000 ) = 1,201psi ⎝ 1.4969 + 1 ⎠ Pbit = Ps − Pc Pbit = 3,000 − 1,201 = 1,799 psi
From Equation 7-22 for impact force: ⎛ 2 ⎞ Pc = ⎜ ⎟Ps ⎝s + 2⎠ 2 ⎛ ⎞ Pc = ⎜ ⎟(3,000 ) = 1,716 psi ⎝ 1.4969 + 2 ⎠ Pbit = Ps − Pc Pbit = 3,000 − 1,716 = 1,284 psi
The data in Table 7-10 and the above calculations can be used to construct Figure 7-4. This figure can be used to determine the flow rate where horsepower or impact force will be a maximum. Alternatively, the flow rates can be calculated as in Example 7-3. The nozzles sizes are also determined as in Example 7-3. Table 7-11 shows the results of maximizing the hydraulics on the drilling rig.
Copyright © 2003 OGCI/PetroSkills. All rights reserved.
7-17
Drilling Practices Chapter 7 Table 7-11. Results of Example 7-3
Q
Pc
Pbit
Ps
v
NOZ's
Hp bit /in2
Vn
IF/in2
Hp t
Before Trip
500
2,507
494
3,000
175
18,18,18
2.03
215
9.42
875
Max. Hp
306
1,201
1,799
3,000
107
10,10,11
4.53
410
10.99
536
Max. IF
388
1,716
1,284
3,000
136
12,12,13
4.11
346
11.78
679
Circulating Pressure Loss, psi
10000
388 306
Pc Max IF = 1716 Pc Max Hp = 1201 1000 100
1000 Flow Rate, Q
Figure 7-4. Plot of Flow Rate versus Circulating Pressure Loss for Example 7-4
As seen from Example 7-4, improving the hydraulics parameters on the rig results in the following improvements: 1. Bit hydraulic horsepower per square inch, Hp bit /in2, is increased by 123% (2.03 Hp to 4.53 Hp) while simultaneously reducing total pump output power by 39% (875 Hp to 536 Hp). 2. By adjusting for maximum impact force, a gain of 25% is realized (9.42 lbf /in2 to 11.78 lbf /in2) while reducing total pump power by 22% (875 Hp to 679 Hp).
7-18
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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 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 has been either incorrect or of 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 Dh Dp
=
Inside diameter of pipe or drill collar, inches
=
Diameter of hole, inches
=
Outside diameter of pipe or drill collar, inches
Hp
=
Horsepower, Hp
Hp bit
=
Horsepower at the bit, Hp
Hp c
=
Horsepower in circulating system, Hp
Hp s
=
Horsepower at the surface (pump hydraulic horsepower), Hp
Hp t
=
IF
=
Total horsepower required at the pump fluid end, Hp Impact force, lbf
K
=
Constant
k
=
Power Law constant
l
=
Length of pipe or drill collars, feet
Nn
=
Number of nozzles in the bit
n
=
Power Law constant, normally the slope of shear stress-shear rate diagram on log-log plot
P P1
= =
Pressure, psi Circulating pressure loss corresponding to flow rate Q1 , psi
P2
=
Circulating pressure loss corresponding to flow rate Q2 , psi
Pbit
=
Pc
=
Pressure drop at the bit (through jet nozzles), psi Pressure loss in the circulating system ( Ps − Pbit ), psi
Pdc
=
Pressure loss in the drill collars, psi
Pdca Pdp
=
Pressure loss in the drill collar annulus, psi
=
Pressure loss in the drill pipe, psi
Pdpa
=
Pressure loss in the drill pipe annulus, psi
Ps
=
Pressure at the surface (standpipe pressure), psi
PV
=
Plastic viscosity,
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7-19
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 log-log paper
Vn
=
Nozzle velocity, ft/sec
v Yp
= =
Average fluid velocity, fpm Yield point, lbf/100 ft2
=
Mud weight, ppg
ρm
SI UNITS The equations given in the chapter are converted to SI units below
PQ 60
Equation 7-1:
Hp =
Equation 7-2:
Hp bit =
Equation 7-16:
IF =
Pbit Q 60
Equation 7-30
ρ mVnQ
Equation 7-31
60
Equation 7-17:
Pbit =
Equation 7-26:
Sn =
Equation 7-28:
An =
7-20
Equation 7-29
ρ mVn2
Equation 7-32
1,810
1.27 An Nn
[(S ) 4
π
2
1
+ (S 2 )2 + (S3 )2
Equation 7-33
]
Equation 7-34
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Drilling Practices Hydraulics
NOMENCLATURE FOR EQUATIONS IN SI UNITS A An
=
Constant
=
Area of the nozzles, mm2
D Dh Dp
=
Inside diameter of pipe or dill collar, mm
=
Diameter of hole, mm
=
Outside diameter of pipe or drill collar, mm
Hp
=
Horsepower, kWatts
Hp bit
=
Horsepower at the bit, kWatts
Hp c
=
Horsepower mm circulating system, kWatts
Hp s
=
Horsepower at the surface (pump hydraulic horsepower), kWatts
Hp t
=
Total horsepower required at the pump fluid end, kWatts
IF
=
Impact force, Nt
K
=
Constant
k
=
Consistency index, Pa secn
l
=
Length of pipe or drill collars, meters
Nn
=
Number of nozzles mm the bit
n
=
Power Law constant, normally the slope of shear stress-shear rate diagram on log-log plot
P P1
= =
Pressure, kPa Circulating pressure loss corresponding to flow rate Q1 , kPa
P2
=
Circulating pressure loss corresponding to flow rate Q2 , kPa
Pbit
=
Pc
=
Pressure drop at the bit (through jet nozzles), kPa Pressure loss mm the circulating system ( Ps − Pbit ), kPa
Pdc
=
Pressure loss mm the drill collars, kPa
Pdca Pdp
=
Pressure loss mm the drill collar annulus, kPa
=
Pressure loss mm the drill pipe, kPa
Pdpa
=
Pressure loss mm the drill pipe annulus, kPa
Ps
=
Pressure at the surface (standpipe pressure), kPa
PV Q
=
Plastic viscosity, Pa sec
=
Q1
=
Flow rate, cubic meters per minute Flow rate corresponding to circulating pressure loss P1 , cubic meters per
=
minute Flow rate corresponding to circulating pressure loss P2 , cubic meters per
Q2
minute Qmax
=
Maximum Flow rate, cubic meters per minute
S1
=
Diameter of nozzle 1, mm
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7-21
Drilling Practices Chapter 7
S2
=
Diameter of nozzle 2, mm
S3
=
Diameter of nozzle 3, mm
Sn
=
Average nozzle size, mm
s
=
Slope of pressure versus flow rate on log-log paper
Vn
=
Nozzle velocity, m/sec
v Yp
=
Average fluid velocity, meters per minute
=
Yield point, Pa
=
Mud weight, kg per cubic meter
ρm
REFERENCE 1
Scott, K. F.; "A New Practical Approach to Rotary Drilling Hydraulics," SPE 3530, 1971.
7-22
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