SIMPLIFIED EQUATIONS ESTIMATE FRICTION LOADS ON DOWNHOLE TUBULARS

William Dangerfield
Consultant
Hafrsfjord, Norway

A simplified calculation method of estimating friction loads on downhole tubular strings can help in planning wells, in designing casing and tubing strings, and in specifying drillstring and hoisting requirements.

These equations can be solved using an ordinary pocket calculator and do not require a computer.

Frictional loads are imposed on downhole tubulars as a result of contact between the tubular string and the hole wall or previously run casing string. The frictional resistance will add to the total axial tension on the string when it is being pulled upward (tripping out of the hole, lifting pipe out of the slips, working a stuck string, reciprocating a string, etc.). As wells become deeper, more highly deviated, with faster build-up rates and more complex hole geometry, the frictional loads will increase.

The calculation method is based on some simplifying assumptions regarding hole geometry. It is assumed the borehole is made up of vertical sections, straight tangential sections, and circular build-up and dropping sections. It is further assumed that there are no directional changes.

Obviously, in a real well, the "vertical" section will not be perfectly straight and vertical, the build-up section will not be constant and without doglegs, and the tangential section will not be a perfectly straight line at constant angle. It is not possible to determine beforehand the extent of such uncertainties. This calculation method provides a good starting point for estimating frictional resistance in the planning stages, however. After making the calculations for a particular well plan, a percentage can be added to account for hole crookedness, dogleg, etc.

VERTICAL SECTION

For this article, it is assumed that the vertical sections are straight, vertical holes. There would be no frictional resistance in a vertical hole because there would be no contact force between the tubular and the wall of the hole. The only forces on the tubular in a vertical section would be the tension imposed from below the vertical section, the weight of the tubular and the buoyancy imposed by the drilling fluid in the vertical section, and the resulting tension at the top of the section.

This assumption is considered valid because unless the "-vertical" section is unusually crooked, helical, or doglegged, the frictional forces will be minimal compared to those in the non-vertical sections.

TANGENTIAL SECTION

It is assumed that the tangential section is a straight line with constant inclination angle. Fig. 1 shows a tubular section in a tangential hole with inclination angle 0. A free body diagram of a section of length L is shown in Fig. 2.

The external forces acting on the tubular section include gravitational and buoyancy forces, the tensile loads at each end, the normal force resulting from contact with the wall of the hole, and the friction force,

For equilibrium, the horizontal and vertical forces must be in balance. Summing the horizontal force components gives Equations 1 and 2, and summing the vertical components gives Equations 3 and 4. Solving Equation 2 for n yields Equation 5. Substituting Equation 5 into Equation 4 and rearranging terms gives the solution for the tension in the tangential section (Equation 6).

ANGLE BUILD-UP SECTION

It is assumed that the angle build-up section is perfectly circular with a constant radius of curvature determined by the given angle build-up rate.

Analysis of the friction on the tubular in the angle build-up section becomes more complicated because the contact force between the tubular and the wall of the hole (normal force) will be affected by the frictional force on the tubular, which is itself dependent on the contact force.

The forces are analyzed with a free body diagram of a differential element of the tubular in an angle build-up section. The forces on the free body include tension, buoyed weight, contact or normal force, and friction force. For equilibrium, the net force on the free body must be zero. The sum of the horizontal components of the forces and the sum of the vertical components of the forces must both be equal to zero.

Figs. 3 and 4 show the normal force distribution over a typical build section. In Fig. 3, the tubular string has been run to the end of the build section. The only external force with horizontal components is the normal force distribution. For equilibrium, there must be normal forces (wall contact) on both the upper and lower sides of the tubular string. The horizontal components of the normal force distribution on the upper side must be equal and opposite to those on the lower side of the string. There is a point at which the wall contact, and therefore the normal force distribution, changes from the upper to the lower side. This point occurs at the critical inclination angle, Oc.

Fig. 4 shows a situation in which there is an axial (tensile) load on the string from below the end of the buildup section. This load might, for example, be from the portion of the tubular in the tangential section of the hole. Again, for equilibrium, the horizontal components of the external forces must balance. If the tension load on the string at the end of the build-up section is great enough, then the string will be in contact with the upper side of the hole over the entire build-up section.

Fig. 5 shows the free-body diagram for a differential element of tubular when the string is pulled against the high side of the hole (Case 1). By resolving the various forces on the element into horizontal and vertical components, a differential equation can be developed, which can be solved to provide an expression for tension as a function of the inclination angle.

Summing the horizontal force components gives Equation 7, which, after rearranging and considering the differential limits, leads to Equation 8. In a similar manner, summing the vertical components gives Equations 9 and 10. Equation 19 determines the tension when the tubular string is pulled against the upper side of the hole (Case 1). (Equations 7-18 in the accompanying box show the derivation of Equation 19.) At the kick-off point, 0 = 0, and Equation 19 can be simplified into Equation 19a.

Fig. 6 shows the free-body diagram for a differential element of a tubular when the string is in contact with the low side of the hole (Case 2). An analysis similar to that for Case 1 yields an expression for the tension in the tubular when it is in contact with the low side (Equations 20-26 show the derivation). Equation 27 determines the tension when the tubular string is in contact with the low side of the hole (Case 2).

The critical angle, Oc, is the inclination angle at which the normal force is zero (Fig. 3). The normal force when the tubular is in contact with the low side is given by Equation 24.

The normal force distribution, n, will be zero when the critical angle, Oc, is found which satisfies Equation 29. The term on the left side of Equation 29 is a constant, depending on the tension in the tubular at the end of the angle build-up (Tb), the inclination angle at the end of the build-up (4)), the friction factor (0, the tubular weight per unit length including buoyancy effect (w), and the radius of curvature (R).

A direct solution is not possible; therefore, reiterative methods must be used. There may be no solution for Oc in the range of inclination angle over the build-up section in question (Fig. 4). To determine if a solution exists, the value of the inclination angle at the end of the build-up (0) is substituted for Oc in the terms on the right side of Equation 29, and the result simplified. For a solution to exist, the inequality in Equation 30 must be true.

ANGLE DROPPING SECTION

Figs. 7 and 8 show the forces in an angle dropping section. An analysis of the free-body diagram (Fig. 8) yields an expression for tension in an angle dropping section (Equation 31). (The detailed derivation is not presented, but it follows the method for the angle buildup section.)

In the special case in which the inclination drops to vertical, the angle 0 = 0, and Equation 31 simplifies into Equation 31a.

EXAMPLES

The accompanying examples demonstrate the use of the equations in determining the friction loads. The solution method divides the borehole into sections, based on geometry changes (vertical, tangential, build-up, or dropping sections), tubular weight changes, or friction factor changes (cased/open hole).

Starting from the bottom, the tensile load added at each section change is calculated using the appropriate equation. The tension determined at each section change becomes an input value for the next section calculation, continuing in this manner all the way to surface.

RUNNING CASING IN A 60 WELL

A string of 18-5/8-in., 87.5-lb/ft casing will be run in a hole drilled vertically to 500 ft, then kicked off with an angle built at 3/100 ft to a maximum inclination angle of 60. For casing and equipment design, the engineer must estimate the tensile load from frictional resistance. The following data are given:

Mud density, 9 ppg
Buoyancy factor, 0.864
Friction factor, 0.35 (cased and open hole)
Maximum inclination angle, 60 (1.047 radian)
Angle build-up rate, 3/100 ft
Radius of curvature, 180/(PI-0.03) = 1,910 ft
Casing weight in air, 87.5 lb/ft
Casing weight in mud, 75.6 lb/ft.

The casing will be run to the end of the angle buildup section. It is assumed that there is no tensile load imposed at the shoe. The inequality in Equation 30 is true, and Equation 29 is used to solve for the critical angle Oc. Substituting the given data into Equation 34 yields.

(sinOc - 0.35cosOc)e-0.350c = 0.1421

By trial-and-error, Oc = 28.47 (0.4969 radian). The section of casing from 28.47 inclination to the end of the build-up section contacts the low side of the hole. The frictional resistance in this section is calculated using Equation 27. Substituting 0.4969 radian for 0 and substituting the other given data into Equation 27 gives Toc = 68,837 lb at the critical inclination angle of 28.47.

The section of casing from the kick-off point to 28.47 inclination contacts the upper side of the hole. The frictional resistance in this section is calculated using Equation 19a. In this case, the angle (b is equal to Oc (0.4969 radian), and is equal to the tension at Oc, or 68,837 lb. Substituting these values and the other given data into Equation 19a gives TKOP = 150,085 lb at the kick-off point.

The remaining section is the 500 ft vertical section. The added tension in this section is simply the length multiplied by the buoyed weight:

Tsurf = Tkop + wL

TSL,RF = 150,085 + (75.6)500 = 187,885 lb at surface

The calculated tensions are valid for the simplified borehole geometry and assumed friction factor. Because of the inevitable presence of hole crookedness or dogleg, the engineer might want to add a percentage factor (20%, for example) to the calculated values to account for uncertainties.

S-PROFILE WELL

A development well will have an S-shaped profile to penetrate two offset targets. The well will be kicked off at 500 ft, with the angle built at 3/100 ft to a maximum inclination of 60, and then drilled tangentially to 9,500 ft measured depth (MD). At this point intermediate casing will be set, and drilling will continue with the angle dropping at a rate of 3/100 ft to 10 inclination. The well path will continue tangentially another 500 ft to a total depth of 11,667 ft MD. A string of 9-5/8-in., 53.5-lb/ft casing will be run to total depth.

The engineer wishes to determine the tensile loads on the casing while reciprocating at total depth, to check the casing design, and to determine hoisting equipment requirements. The following data are given:

Mud density, 12 ppg
Buoyancy factor, 0.819
Friction factor, 0.35 (open hole) and 0.30 (cased hole)
Angle build-up rate, 3/100 ft
Maximum inclination angle, 60 (= 1.047 radian)
Angle dropping rate, 30/100 ft
Final angle, 10 (0.1745 radian)
Radius of curvature, 180/(PI-0.03) = 1,910 ft (both sections)
Casing weight in air, 53.5 lb/ft
Casing weight in mud, 43.82 lb/ft.

The well can be divided into five sections for calculating the friction loads:

500 ft tangential section at bottom (10 inclination)
1,667 ft angle dropping section (60 to 10')
7,000 ft tangential section (60 inclination)
2,000 ft angle build-up section (O to 60 inclination)
500 ft vertical section at the top.

Starting from the bottom, the tension at the top of the 500 ft tangential section is calculated using Equation 6. Substituting the given data into this equation yields T = 22,909 lb at the end of the angle dropping section.

The tension at the start of the angle dropping section is calculated using Equation 31. Substituting 22,909 lb for T, 0.1745 radian for 4), 1.047 radian for 0, and the other given data yields TD 115,994 lb at the start of the angle dropping section.

The tension at the end of the angle build-up section (start of 60 tangential section) is calculated using Equation 6. Substituting 115,994 lb for T,) and using the other given data gives T = 349,057 lb at the end of the angle build-up section.

The tension at the kick-off point is determined by first checking the inequality in Equation 30. Substituting the given data gives the following:

349,057/43.821,910

4.17

The inequality is false, so the casing is against the upper side of the hole through the entire build-up section. Equation 19a can be used to find the tension at the kickoff point. Substituting 349,057 lb for Td 1.047 radian for 4), and the other given data gives Tu = 546,208 lb at the kick-off point.

The tension at surface is then calculated:

T = Tu + wL

T = 546,208 + 43.82500 = 568,118 lb at surface

The engineer might want to add a safety margin of 20% to account for dogleg, etc., bringing the maximum load to approximately 682,000 lb (341 tons). The rig will require at least a 350-ton casing elevator and might need a higher-rated one. In addition, the engineer can combine the estimated friction forces with bending stresses and other effects to check the casing design.

HIGH-ANGLE WELL

A high-angle deviated well is planned with the kick-off point at 500 ft and an angle built at 3.5/100 ft to a maximum inclination angle of 70 followed by a long tangential section to total depth. The end of the buildup section will be at 2,500 ft MD. Casing will be set at 7,500 ft MD, which is 5,000 ft beyond the end of the buildup section. At total depth, the top of the bottom hole assembly will be 5,000 ft beyond the casing shoe, at 12,500 ft MD. The drill pipe is 5-in., 19.5-lb/ft, S-135 premium grade with a maximum yield tensile capacity of 562,000 lb. The operator wishes to determine how much tensile load can be put on the bottom hole assembly in the event of a stuck pipe situation, without exceeding 80% of the maximum capacity of the drill pipe. The following data are given:

Maximum tensile load, 450,000 lb
Mud density, 12 ppg
Buoyancy factor, 0.819
Friction factors, 0.30 (cased hole) and 0.35 (open hole)
Maximum inclination angle, 70 (1.2217 radian)
Angle build-up rate, 3.5/100 ft
Radius of curvature, 180/(-7-0.035) = 1,637 ft
Drill pipe weight in air, 19.5 lb/ft
Drill pipe weight in mud, 16 lb/ft.

In this situation, there are four sections (vertical, build-up, cased hole tangential, and open hole tangential). The calculation procedure is reversed, in that the section tensions are determined from the top working downward. Based on a maximum allowable tension at surface of 450,000 lb, the tension at the kick-off point is as follows:

Tkop = Tsurf - wL

Tkop = 450,000 - (16500) = 442,000 lb at the kick-off point

For the angle build-up section, the engineer must first determine if the critical angle (0c) exists. The inequality of Equation 30 is used (substituting given data and rearranging):

The tension (T,) at the end of the angle build-up section must be less than 24,612 lb for the critical angle (0c) to exist. In this example, it can be assumed that the long tangential section will impose a tensile load greater than 24,612 lb, and the drill pipe is against the upper side of the hole through the entire build-up section. Equation 19a is used. Equation 19a is first rearranged because in this case Tu is known, and T is to be determined. Substituting the given data yields Tb = 290,883 lb at the end of the angle build up. The next section is a 5,000-ft long tangential section in cased hole (friction factor 0.3). Substituting the given data into Equation 6 and rearranging gives T = 240,967 lb at the casing shoe.

The final section is another 5,000-ft long tangential section, but in open hole (friction factor 0.35). Equation 6 is again used to determine T = 187,294 lb at the top of the bottom hole assembly. In this example, if the driller pulls 450,000 lb on the drillstring, the maximum force exerted at the top of the bottom hole assembly' is only about 187,000 lb. The true vertical depth at the top of the bottom hole assembly is 5,458 ft.

The weight of the drill pipe in mud at this depth is 87,328 lb, and the total friction force is 175,378 lb. If a 20% margin is added to the friction force to account for doglegs, etc., the maximum force which could be exerted on the bottom hole assembly would be reduced to approximately 152,000 lb.

Issue date: 11/21/94