Bài giảng Well drilling engineering - Chapter 5: Surge and Swab Pressures (Part 6) - Đỗ Quang Khánh

1 Well Drilling Engineering  Surge and Swab Pressures Dr. DO QUANG KHANH 2  15 - Surge and Swab Pressures  Surge and Swab Pressures - Closed Pipe - Fully Open Pipe - Pipe with Bit Example General Case (complex geometry, etc.) Example 3 READ:   APPLIED DRILLING ENGINEERING Chapter 4 HW # ADE # 4.52, 4.53, 4.54 Surge Pressure due to Pipe Movement When a string of pipe is being lowered into the wellbore, drilling fluid is being displaced and forced out of the wellbore. The pressure required to force the displaced fluid out of the wellbore is called the surge pressure . Surge Pressure due to Pipe Movement An excessively high surge pressure can result in breakdown of a formation. When pipe is being withdrawn a similar reduction is pressure is experienced. This is called a swab pressure , and may be high enough to suck fluids into the wellbore, resulting in a kick. 6 7 Surge Pressure - Closed PipeNewtonian The velocity profile developed for the slot approximation is valid for the flow conditions in the annulus; but the boundary conditions are different, because the pipe is moving: V = 0 V = -V p 8 When y = 0, v = - v p , When y = h, v = 0, Substituting for c 1 and c 2 : At Drillpipe Wall At Hole Wall 9 Velocity profile in the slot h 0 W 10 Changing from SLOT to ANNULAR notation A = Wh = Substitute in: 11 Or, in field units or, in field units: Frictional Pressure Gradient Same as for pure slot flow if v p = o (K c = 0.5) Results in: 12 How do we evaluate v ? For closed pipe, flow rate in annulus = pipe displacement rate: d 1 d 2 v p 13 Open Pipe Pulling out of Hole 14 Surge Pressure - Open Pipe Pressure at top and bottom is the same inside and outside the pipe. i.e., From Equations (4.88) and (4.90d): 15 i.e., Surge Pressure - Open Pipe Valid for laminar flow, constant geometry, Newtonian 16 Example Calculate the surge pressures that result when 4,000 ft of 10 3/4 inch OD (10 inch ID) casing is lowered inside a 12 inch hole at 1 ft/s if the hole is filled with 9.0 lbm/gal brine with a viscosity of 2.0 cp. Assume laminar flow. 1. Closed pipe 2. Open ended 17 Derivation of Equation (4.94) From Equation (4.92): 18 Derivation of Eq. (4.94) cont’d From Equation (4.93): Substituting for v i : 19 So, i.e., 20 Surge Pressure - General Case The slot approximation discussed earlier is not appropriate if the pipe ID or OD varies , if the fluid is non-Newtonian , or if the flow is turbulent . In the general case - an iterative solution technique may be used. 21 Fig. 4.42Simplified hydraulic representation of the lower part of a drillstring 22 General Solution Method 1. Start at the bottom of the drillstring and determine the rate of fluid displacement. 2. Assume a split of this flow stream with a fraction, f a , going to the annulus, and (1-f a ) going through the inside of the pipe. 23 3. Calculate the resulting total frictional pressure loss in the annulus, using the established pressure loss calculation procedures. 4. Calculate the total frictional pressure loss inside the drill string. General Solution Method 24 5. Compare the results from 3 and 4, and if they are unequal, repeat the above steps with a different split between q a and q p . i.e., repeat with different values of f a , until the two pressure loss values agree within a small margin. The average of these two values is the surge pressure. General Solution Method 25 NOTE : The flow rate along the annulus need not be constant, it varies whenever the cross- sectional area varies. The same holds for the drill string. An appropriate average fluid velocity must be determined for each section. This velocity is further modified to arrive at an effective mean velocity. 26 Fig. 4.42Simplified hydraulic representation of the lower part of a drillstring 27 Burkhardt Has suggested using an effective mean annular velocity given by: Where is the average annular velocity based on q a K c is a constant called the mud clinging constant ; it depends on the annular geometry. (Not related to Power-law K!) 28 The value of K c lies between 0.4 and 0.5 for most typical flow conditions, and is often taken to be 0.45 . Establishing the onset of turbulence under these conditions is not easy.The usual procedure is to calculate surge or swab pressures for both the laminar and the turbulent flow patterns and then to use the larger value. 29 K c K c 30 For very small values of a , K = 0.45 is not a good approximation K c Fig. 4.41 - Mud clinging constant, K c, for computing surge-and-swab pressure. K c 31 Table 4.8. Summary of Swab Pressure Calculation for Example 4.35 Variable f a =(q a /q t ) 1 0.5 0.75 0.70 0.692 (q p ) 1 , cu ft/s 0.422 0.211 0.251 0.260 (q p ) 2 , cu ft/s 0.265 0.054 0.093 0.103 (q p ) 3 , cu ft/s 0.111 -0.101 -0.061 -0.052 32 Table 4.8 Summary of Swab Pressure Calculation Inside Pipe Variable f a =(q a /q t ) 1 0.5 0.75 0.70 0.692 D p BIT , psi 442 115 160 171 D p DC , psi 104 33 44 46 D p DP , psi 449 273 293 297 Total D p i , psi 995 421 497 514 33 Table 4.8 Summary of Swab Pressure Calculation in Annulus Variable f a =(q a /q t ) 1 0.5 0.75 0.70 0.692 0.422 0.633 0.594 0.585 0.012 0.223 0.183 0.174 104 139 128 126 335 405 392 389 Total D p a , psi 439 544 520 515 Total D p i , psi 995 421 497 514 34 Table 4.8 Summary of Swab Pressure Calculation for Example 4.35 f a : 0.5 0.75 0.70 0.692 35 v p 36 VELOCITY SURGE PRESSURE ACCELERATION 37 Inertial EffectsExample 4.36 Compute the surge pressure due to inertial effects caused by downward 0.5 ft/s 2 acceleration of 10,000 ft of 10.75” csg. with a closed end through a 12.25 borehole containing 10 lbm/gal. Ref. ADE, pp. 171-172 38 From Equation (4.99) Inertial Effects - Example 4.36 39 END of Lesson