Lesson 15 Surge and Swab Pressures PETE 411 Well Drilling Lesson 15 Surge and Swab Pressures

Lesson 15 - Surge and Swab Pressures - Closed Pipe - Fully Open Pipe - Pipe with Bit Example General Case (complex geometry, etc.)

READ: APPLIED DRILLING ENGINEERING Chapter 4 (all) HW #8 ADE #4.46, 4.47 due 10 –14 – 02

Surge Pressure - Closed Pipe Newtonian 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 = -Vp

When y = 0, v = - vp , When y = h, v = 0, Substituting for c1 and c2: At Drillpipe Wall When y = 0, v = - vp , When y = h, v = 0, Substituting for c1 and c2:

Velocity profile in the slot W

Changing from SLOT to ANNULAR notation A = Wh = Substitute in:

Frictional Pressure Gradient Results in: Or, in field units or, in field units: Same as for pure slot flow if vp = o (Kp = 0.5)

How do we evaluate v ? vp For closed pipe, flow rate in annulus = pipe displacement rate: vp d1 d2

Open Pipe Pulling out of Hole

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):

Surge Pressure - Open Pipe i.e., Valid for laminar flow, constant geometry, Newtonian

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

1. For Closed Pipe

2. For Open Pipe,

2. For Open Pipe,

Derivation of Equation (4.94) From Equation (4.92):

Derivation of Eq. (4.94) cont’d From Equation (4.93): Substituting for vi:

So, i.e.,

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.

Fig. 4.42 Simplified hydraulic representation of the lower part of a drillstring

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, fa, going to the annulus, and (1-fa) going through the inside of the pipe.

General Solution Method 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 5. Compare the results from 3 and 4, and if they are unequal, repeat the above steps with a different split between qa and qp. i.e., repeat with different values of fa, until the two pressure loss values agree within a small margin. The average of these two values is the surge pressure.

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.

Fig. 4.42 Simplified hydraulic representation of the lower part of a drillstring

Burkhardt Has suggested using an effective mean annular velocity given by: Where is the average annular velocity based on qa Kc is a constant called the mud clinging constant; it depends on the annular geometry. (Not related to Power-law K!)

The value of Kp lies between 0. 4 and 0 The value of Kp 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.

Kc Kc

For very small values of a, K = 0.45 is not a good approximation Kc For very small values of a, K = 0.45 is not a good approximation Kc Fig. 4.41 - Mud clinging constant, Kc, for computing surge-and-swab pressure.

Table 4.8. Summary of Swab Pressure Calculation for Example 4.35 Variable fa=(qa/qt)1 0.5 0.75 0.70 0.692 (qp)1, cu ft/s 0.422 0.211 0.251 0.260 (qp)2, cu ft/s 0.265 0.054 0.093 0.103 (qp)3, cu ft/s 0.111 -0.101 -0.061 -0.052

Table 4.8 Summary of Swab Pressure Calculation Inside Pipe Variable fa=(qa/qt)1 ……… 0.5 0.75 0.70 0.692 DpBIT, psi ……… 442 115 160 171 DpDC, psi ……… 104 33 44 46 DpDP, psi ……… 449 273 293 297 Total Dpi, psi …… 995 421 497 514

Table 4.8 Summary of Swab Pressure Calculation in Annulus Variable fa=(qa/qt)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 Dpa, psi 439 544 520 515 Total Dpi, psi 995 421 497 514

Table 4.8 Summary of Swab Pressure Calculation for Example 4.35 fa: 0.5 0.75 0.70 0.692

vp

SURGE PRESSURE VELOCITY ACCELERATION

Inertial Effects Example 4.36 Compute the surge pressure due to inertial effects caused by downward 0.5 ft/s2 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

Inertial Effects - Example 4.36 From Equation (4.99)

END of Lesson 15 Surge and Swab