General Hydraulics 01

Chapter

1 Introduction to Hydraulics This chapter covers many basic hydraulic principles that you will need to understand before you use the General Hydraulics Manual. After a brief description of hydraulics concepts, the following topics will be covered in this chapter along with the calculations for each one: • Area • Force • Pressure • Effective area • Hydrostatic pressure

• Differential pressure • Pipe capacity • Pipe fill-up • Buoyancy • Open-ended pipe hydraulics

This manual provides examples and practice problems for the calculations. Use the areas labelled “Work Space” for your personal calculations. Then compare your answers to the “Solutions to Problems” section beginning on Page 1-61.

Essential Hydraulics Concepts Hydraulics are the principles governing the power generated by the movement and force of liquid. Hydraulics concepts are primarily an application of Pascal’s Law: if a fluid has a constant density and the fluid is at rest, all points at the same depth below the liquid’s surface are under equal pressure. For example, studying a cross-sectional slice of a cylinder of water demonstrates that all points on the crosssectional surface are under equal pressure. Familiarity with fluid pressures is important for understanding how oilfield tools work. Force is another important concept in hydraulics. To calculate force, multiply pressure by area as stated in Equation 1.1: Force = Pressure x Area ........................................................................................... (1.1) Most tool hydraulics can be calculated with Pascal’s Law and the force formula. Some additional basic calculations are covered in this chapter. The first calculation is area, including cross-sectional and effective areas.

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Calculating Area Definition of Area Since downhole tools are usually round, in oilfield applications the term area generally refers to the area of a circle. A related concept, cross-sectional area, is the area of an exposed surface.

Calculating the Area of a Circle The area of a circle equals pi (π) times the radius times the radius, as stated in Equation 1.2: Area of a circle = π r2 ...................................................................................................................................................................... (1.2) The value of pi is 3.141592654. Example 1 is an application of Equation 1.2. Example 1: How to calculate the area of a circle

What is the area of the circle in Figure 1.1? Hint: The radius (r) of a circle is half its diameter.

Solution Area = πr 2 .......................................................... (1.2) = 3.141592654 x 1.25 in. x 1.25 in. = 4.9087 in.2 ≈ 4.909 in.2

NOTE

2 1 /2-in. Diameter

Figure 1.1

In this manual, all areas are rounded to three decimal places. Most downhole tools are measured by diameter rather than radius. Use Equation 1.3 to calculate the area of a circle using the circle's diameter rather than its radius. Since pi divided by 4 equals 0.7854, and radius times radius multiplied by 4 equals diameter times diameter, a simpler equation is Area of a circle = (π ÷ 4) x D2 or Area of a circle = 0.7854 x D² ................................................................................... (1.3) Example 2 is an application of Equation 1.3.

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Example 2: How to calculate the area of a rod end

What is the area of the rod end in Figure 1.2? Solution Area = 0.7854 x D² ...................... (1.3) = 0.7854 x 4.375 in. x 4.375 in. = 15.033 in.2 4 3/ 8-in. Diameter

Figure 1.2

NOTE

If diameter is expressed in inches, area will be expressed in square inches (in.2). If diameter is expressed in units such as feet or centimeters, area will be expressed in square feet (ft2) or square centimeters (cm2), respectively. All examples and problems in this manual show diameters in inches. Practice calculating the areas of two rod ends in Problems 1 and 2 on Page 1-4. Use the Work Space area to work out the problems; then compare your answers to the “Solutions to Problems” on Page 1-61.

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Problem 1 What is the area of the rod end in Figure 1.3? Work Space

1

1 /2 -in. Diameter

Figure 1.3

Answer _______________ (See Page 1-61 for the solution to Problem 1.)

Problem 2 What is the area of the rod end in Figure 1.4? Work Space

2 7/8 -in. Diameter

Figure 1.4

Answer _______________ (See Page 1-61 for the solution to Problem 2.)

Calculating Cross-Sectional Area Sometimes it is necessary to calculate cross-sectional area of pipe. For example, tubing’s cross-sectional area is equal to the area of the tubing’s outside diameter (OD) minus the area of the tubing’s inside diameter (ID) or Cross-sectional area of pipe = Pipe OD area - Pipe ID area............................. (1.4) The formula for calculating a pipe’s OD area or ID area is the same as for calculating the area of a circle: OD area (or ID area) = 0.7854 x D2 where D is the OD (or ID). The OD is the size of the pipe. For example, the OD of 10 3/ 4-in., 55.5-lb/ft casing is 10 3/4 in. When you know the OD and weight of a pipe, its ID can be found in several sections in the Halliburton Cementing Tables manual including “Displacement,” “Dimensions and Strengths,” “Capacity,” and “Volume and Height.” NOTE

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You can order a copy of the Halliburton Cementing Tables from Mastercraft.

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Problem 3 What is the ID area of 10 3/4-in., 55.5-lb/ft casing? The ID of this casing is 9.760 in. (from the Halliburton Cementing Tables). Work Space

Answer _______________ (See Page 1-61 for the solution.) An easy way to understand a tubing cross section is to look at the end of a piece of pipe. The hatched portion in Figure 1.5 is the cross section. NOTE

The laws of mathematics do not allow you to subtract the diameters of the circles and then find the area of that difference.

Example 3: How to calculate the cross-sectional area of tubing

What is the cross-sectional area of this tubing? Solution = 0.7854 x 7 in. x 7 in. = 38.485 in.² Tubing ID area = 0.7854 x 6.366 in. x 6.366 in. = 31.829 in.² Tubing cross-sectional area = OD area - ID area = 38.485 in.² - 31.829 in.² = 6.656 in.²

Figure 1.5

Problem 4 What is the cross-sectional area of 2 3/8-in., 4.7-lb/ft tubing? The ID of this tubing is 1.995 in. (from Halliburton Cementing Tables). Work Space

Answer _______________ (See Page 1-61 for the solution.)

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Estimating Tensile Strength with the Cross-Sectional Area Being able to calculate cross-sectional area is sometimes useful when an oilfield handbook is not available. For example, if you know the cross-sectional area, you can quickly estimate the tensile strength of drillpipe, tubing, or casing with Equation 1.5. Tensile strength (estimated) = Yield strength x Cross-sectional area ............ (1.5) Tensile strengths of tubing and casing are particularly easy to calculate since the grade designation indicates the yield strength of the pipe in thousands of pounds-per-square inch (psi) of cross-sectional area. For example, new N-80 tubing or casing has a yield strength of 80,000 psi. Example 4 illustrates the calculation for tensile strength. Example 4: How to calculate tensile strength

What is the approximate tensile strength of the pipe body of 2.875-in. OD, 2.441-in. ID, 6.5-lb/ft, N-80 tubing? Solution Tubing OD area = 0.7854 x 2.875 in. x 2.875 in. = 6.492 in.2 Tubing ID area = 0.7854 x 2.441 in. x 2.441 in. = 4.680 in.2 Cross-sectional area = Tubing OD area - Tubing ID area = 6.492 in.2 - 4.680 in.2 = 1.812 in.2 Yield strength = 80,000 psi Tensile strength (estimated) = Yield strength x Cross-sectional area (1.5) = 80,000 psi x 1.812 in.2 = 144,960 lb

NOTE

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Use these calculations carefully since no safety factor is included. Be aware that the threaded sections on nonupset tubing, casing, etc., have a thinner wall than the pipe body. Use the thinner section for calculating tensile strength since it will be the weakest point.

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Calculating Force and Pressure This section defines force and pressure and gives examples and problems for calculating force in hydraulic cylinders.

Definition of Force Force is defined in the dictionary as active power. This definition means that an item, such as the piston in a hydraulic cylinder, will move when a force adequate to overcome the resistance is applied. Usually, force is expressed in pounds (lb), and its upward or downward direction on the equipment is specified (lbá or lbâ).

Definition of Pressure A liquid or gas exerts a force against any surface it contacts; the force per unit area is defined as pressure. Pressure is usually expressed in pounds per square inch (psi). Pascal’s Law states that pressure acts equally in all directions. In other words, pressure in a hydraulic cylinder is acting (or exerting a force) equally on each square inch of the piston, cylinder cap, and cylinder walls. A force may be created when pressure acts across an area, such as in a hydraulic cylinder or a hydraulic jack. The amount of force created by a hydraulic cylinder is equal to the pressure multiplied by the area of the piston. Force = Pressure x Area of piston .......................................................................... (1.6)

NOTE

Pressure and area must be expressed in comparable units for calculations to be accurate. In this text, pressure is expressed in pounds per square inch (psi), and area is expressed in square inches (in.2). Figure 1.6 on Page 1-8 shows a hydraulic cylinder mounted on a solid wall. The piston rod acts against a set of scales that measures force.

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Calculating Force in a Hydraulic Cylinder Example 5: How to calculate force in a hydraulic cylinder

What force is created in the cylinder in Figure 1.6? Solution Piston OD area = 0.7854 x 3.0 in. x 3.0 in. = 7.069 in.2 Pressure = 3,000 psi Force = Pressure x Area (1.6) Force = 3,000 psi x 7.069 in.2 = 21,207 lb

Figure 1.6

Problem 5 What force is exerted by the cylinder in Figure 1.7? Work Space

Answer _______________ (See Page 1-61 for the solution.)

Figure 1.7

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Calculating Pressure in a Hydraulic Cylinder Pressure multiplied by area equals force; therefore, force divided by area equals pressure. When force is applied to the rod on a hydraulic cylinder, pressure is created. Pull applied to a hydraulic jar creates pressure in the hydraulic chamber of the jar. To calculate the amount of pressure created, divide force by area: Pressure = Force ÷ Area ........................................................................................... (1.7) Example 6 and Problem 6 are applications of Equation 1.7. Example 6: How to calculate pressure in a hydraulic cylinder

What is the pressure created when a 3,142-lb force is applied to the 2.0-in. diameter piston in Figure 1.8? Solution Piston OD area = 0.7854 x 2.0 in. x 2.0 in. = 3.142 in.2 Force = 3,142 lb Pressure = Force ÷ Area .................. (1.7) Pressure = 3,142 lb ÷ 3.142 in.2 = 1,000 psi

Figure 1.8

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Problem 6 If the force in Figure 1.9 is 1,000 lb and the diameter of the piston is 3.0 in., what is the pressure in the cylinder? Work Space

Answer _______________ (See Page 1-61 for the solution.)

Figure 1.9

Calculating Effective Area Definition of Effective Area As shown in Figure 1.10, Page 1-11, pressure applied to the piston’s back side cannot act over the entire cross-sectional area of the piston because of the rod position. The area on which the pressure acts is known as the effective area. In Figure 1.10, the effective area is equal to the difference in the piston OD area and the rod OD area. This difference is the effective area that the pressure works against in a hydraulic cylinder.

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Figure 1.10

Example 7: How to calculate the effective area

What is the effective area of the pressure on the cylinder in Figure 1.11? Solution Piston OD area = 0.7854 x 3.0 in. x 3.0 in. = 7.069 in.2 Rod OD area = 0.7854 x 1.5 in. x 1.5 in. = 1.767 in.2 Effective area = Piston OD area - Rod OD area = 5.302 in.2 Figure 1.11

Problem 7 Applying 1,000 psi to the cylinder will exert what pull (force) on the scales in Figure 1.11? Work Space

Answer _______________ (See Page 1-62 for the solution.)

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Problem 8 Figure 1.12 shows different pressures acting on each side of the piston. What force is registered on the scales? Is this force up or down? Hint: Work this problem as if it were two separate problems. One force is upward and the other force is downward. Subtract the smaller force from the larger force. The remaining force on the scales will be in the same direction as the larger force. Work Space Figure 1.12

Answer _______________ (See Page 1-62 for the solution.)

Problem 9 Assume both pressures in Figure 1.13 are 1,000 psi and that the cylinder is the one from Problem 8. What is the force, and in which direction does it act? Work Space

Figure 1.13

Answer _______________ (See Page 1-62 for the solution and alternate solution.)

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Calculating Differential Areas and Differential Pressure The Solution and Alternate Solution (Page 1-62) of Problem 9 (Page 1-12) show that you can eliminate calculation time by canceling opposing forces. Situations with equal pressures acting on different areas, as in Problem 9, are normally called differential areas. If equal pressures act on different areas, you can cancel the opposing forces in the equation. Example 8 demonstrates differential pressure on a double-rod cylinder. Example 8: How to calculate force in a double-rod cylinder

In Figure 1.14, what force is exerted on the scale? Does the net force act in an upward or downward direction? Solution Working with top side of piston: Piston area = 0.7854 x 4 in. x 4 in. = 12.566 in.² Rod area = 0.7854 x 1.5 in. x 1.5 in. = 1.767 in.² Effective area = Piston area - Rod area = 12.566 in.² - 1.767 in.² = 10.799 in.² Pressure = 1,500 psi Force = 1,500 psi x 10.799 in.² = 16,198.5 lbâ Working with bottom side of piston: Figure 1.14 Effective area = 10.799 in.² (calculated above) Pressure = 2,000 psi Force = 2,000 psi x 10.799 in.² = 21,598 lbá Net force = 21,598 lbá - 16,198.5 lbâ = 5,399.5 lbá Alternate Solution Pressure on bottom side = 2,000 psi Pressure on top side = 1,500 psi Differential pressure = Pressure on bottom side - Pressure on top side = 500 psi Effective area = 10.799 in.² (from Solution) Net force = 500 psi x 10.799 in.² = 5,399.5 lbá

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Example 8 demonstrates how you can save calculation time by canceling pressures or forces. Equal amounts of opposing pressures can be canceled if the areas are equal. Since the pressures in Figure 1.14, Page 1-13 are acting on equal areas, 1,500 psi of the pressure on the bottom side will be balancing the 1,500 psi on the top side. This situation leaves only 500 psi effective pressure on the bottom side—this is the differential pressure.

Problem 10 What force is the conventional hydraulic cylinder in Figure 1.15 exerting on the scale? Work Space

Answer _______________ (See Page 1-63 for the solution.)

Figure 1.15

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Problem 11 While the cylinder illustrated in Figure 1.15, Page 1-14 is being used, the rod breaks. Since the piston and rod on this cylinder are one solid piece, both are replaced. The new piston is not machined correctly. Instead of having a diameter of 3 3/4 in., the new piston end has a 3 5 /8-in. diameter. The new rod end has a 1 5/8-in. diameter, as specified. What effect, if any, do the new piston and rod have on the system? Work Space

Answer _______________ (See Page 1-63 for the solution.)

Calculating Hydrostatic Pressure This section introduces the concepts necessary to convert mud weight to hydrostatic pressure.

Definition of Hydrostatic Pressure Hydrostatic pressure is the pressure created by a column of fluid. This column of fluid may be the mud in a well or the water in a lake. The taller the column or the heavier the fluid, the higher the hydrostatic pressure at the bottom of the column. Hydrostatic pressure is usually expressed in pounds per square inch (psi).

Definition of Mud Weight Mud weight, as the term implies, is the weight of a standard volume of mud. In the United States, mud weight is usually expressed in pounds per gallon (lb/gal). In a few areas, such as California and some international locations, mud weight is expressed in pounds per cubic foot (lb/ft³).

Converting Mud Weight to Hydrostatic Pressure To convert mud weight to hydrostatic pressure, multiply mud weight by a constant, 0.05195, and then multiply by depth in feet. Hydrostatic pressure = Mud weight in lb/gal x 0.05195 x Depth in ft ........... (1.8)

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Example 9: How to calculate hydrostatic pressure from mud weight

What is the hydrostatic pressure of a column of 9.6-lb/gal mud at 6,450 ft? Solution Hydrostatic pressure = Mud weight x 0.05195 x Depth ......................... (1.8) = 9.6 lb/gal x 0.05195 x 6,450 ft = 3,216.744 psi

Problem 12 What is the hydrostatic pressure of a column of 16.5-lb/gal mud at 10,000 ft? Work Space

Answer _______________ (See Page 1-63 for the solution.)

Calculating Hydrostatic Pressure from Fluid Gradients Multiplying 0.05195 by the mud weight (lb/gal) yields a fluid gradient— hydrostatic pressure in psi per foot of depth (psi/ft). These fluid gradients simplify hydrostatic pressure calculations. Fluid gradients for a range of fluid weights have been calculated and can be found in Table 4.1, Page 4-2 (lb/gal) and Table 4.2, Page 4-3 (lb/ft 3) of this manual (or at the back of the Halliburton Cementing Tables in the “Hydrostatic Pressure and Fluid Weight Conversion Tables”). Chapter 4 of this manual contains this table and many others for reference. NOTE

When using oilfield handbooks to look up fluid gradients, check the unit of measurement because many handbooks express fluid gradients in psi per hundred feet. To use Table 4.1, find the mud weight in the left column of the table, then read the fluid gradient in the right column. For example, the fluid gradient for 12.4-lb/gal mud is 0.6442 psi/ft of depth. To change the fluid gradient to hydrostatic pressure, multiply by the depth as in Equation 1.9. Hydrostatic pressure = Fluid gradient x Depth .................................................. (1.9)

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Example 10 is an application of Equation 1.9.

Example 10: How to calculate hydrostatic pressure using fluid gradients

Using Equation 1.9, what is the hydrostatic pressure at 12,650 ft in a well containing 10.6-lb/gal mud? Solution Fluid gradient for 10.6-lb/gal mud = 0.5506 psi/ft (from Table 4.1, Page 4-2) Hydrostatic pressure = 0.5506 psi/ft x 12,650 ft = 6,965.09 psi

Problem 13 What is the hydrostatic pressure at 6,000 ft in a well containing 12.5-lb/gal mud? Work Space

Answer _______________ (See Page 1-63 for the solution.)

Problem 14 A drillstem test has just been run on the well in Problem 13. As equipment is removed from the hole, the hole is not kept full. The fluid level in the well drops 1,500 ft. What is the hydrostatic pressure at the 6,000-ft depth now? Work Space

Answer _______________ (See Page 1-63 for the solution.)

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Calculating Hydrostatic Pressure with Different Fluids in a Fluid Column Often, several different fluids are present in a fluid column. To calculate total hydrostatic pressure, calculate the hydrostatic pressure for each fluid, then add these hydrostatic pressures together.

Problem 15 The 10,000-ft tubing string in Figure 1.16 contains 9,000 ft of 15.6-lb/gal cement; the remainder of the tubing contains 9.0-lb/gal water. What is the total hydrostatic pressure at the bottom of the tubing? Work Space

Figure 1.16

Answer _______________ (See Page 1-63 for the solution.)

It is possible for parallel columns of different-weight fluids to exist. This situation occurs when cement is spotted in or reversed out of tubing. Fluids of different weights in parallel columns have different hydrostatic pressures. These differences can be equalized when pressure is trapped or applied to the lighterweight fluid. To calculate the amount of pressure needed to equalize the hydrostatic pressure of two fluid columns, find the difference between the hydrostatic pressures. For more information about this type of calculation, see “Calculating Changes in Hydrostatic Pressure,” Page 1-21 and “Calculating Weight-Indicator Readings when Spotting Fluids,” Page 1-54.

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Problem 16 If the well described in Problem 15 has 9.0-lb/gal water on the outside of the tubing, how much surface pressure is needed to reverse the cement out of the tubing? Hint: The pressure needed to reverse the cement out of the tubing equals the difference in the hydrostatic pressures of fluids in the tubing and casing (without accounting for friction). Work Space

Answer _______________ (See Page 1-64 for the solution.)

If mud weight is expressed in pounds per cubic foot (lb/ft3), calculate the fluid gradient by dividing the mud weight by 144. Then, to obtain hydrostatic pressure, multiply this fluid gradient by the depth in feet. When mud weight is given in lb/ft3, use Equation 1.10 to find hydrostatic pressure. Hydrostatic pressure = Mud weight in lb/ft³ x Depth in ft ........................ (1.10) 144 Equation 1.11 is another way of calculating hydrostatic pressure: Hydrostatic pressure = Mud weight in lb/ft³ x 0.006944 x Depth in ft ......... (1.11)

Example 11: How to calculate hydrostatic pressure using Equation 1.11 Using Equation 1.11, what is the hydrostatic pressure at 6,000 ft in a well containing 72-lb/ft³ mud? Solution Hydrostatic Pressure = Mud weight in lb/ft³ x 0.006944 x Depth in ft .................................. (1.11) = 72 lb/ft³ x 0.006944 x 6,000 ft = 2,999.808 psi

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Problem 17 What is the hydrostatic pressure at 8,000 ft in a well containing 104-lb/ft³ mud? Use Equation 1.11. Work Space

Answer _______________ (See Page 1-64 for the solution.)

Use Table 4.2, Page 4-3 to eliminate part of Equation 1.11. The numbers in Table 4.2 reflect the mud weight (in lb/ft³) already multiplied by 0.006944 (or divided by 144) to give the fluid gradient in psi/ft of depth. To use Table 4.2, find the mud weight in the left-hand column, and read the fluid gradient in the right column. For example, the fluid gradient for 100-lb/ft³ mud is 0.6944 psi/ft. To translate this psi/ft to hydrostatic pressure, multiply the fluid gradient by the depth.

Example 12: How to calculate hydrostatic pressure

What is the hydrostatic pressure at 10,000 ft in a well containing 120-lb/ft³ mud? Use Table 4.2, Page 4-3. Solution Fluid gradient for 120-lb/ft3 mud = 0.8333 psi/ft (from Table 4.2, Page 4-3) Hydrostatic pressure = 0.8333 psi/ft x 10,000 ft = 8,333 psi

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Problem 18 What is the hydrostatic pressure created by a 9,000-ft column of 80-lb/ft³ fluid? Use Table 4.2, Page 4-3. Work Space

Answer _______________ (See Page 1-64 for the solution.)

Calculating Changes in Hydrostatic Pressure There are at least two ways to calculate a change in hydrostatic pressure; one method is shorter than the other one. To determine changes in hydrostatic pressures, calculate the hydrostatic pressure before and after the change, and find the difference. Long Method Change in hydrostatic pressure= Hydrostatic pressure after - Hydrostatic pressure before ........................................................................................................ (1.12) Find the fluid gradients in Table 4.1, Page 4-2 (for lb/gal fluid weights) or Table 4.2, Page 4-3, (for lb/ft3 fluid weights). Short Method First, find the difference in the fluid gradients of the two fluids; then multiply this difference by the length of fluid column changed. Equation 1.13 expresses this calculation. Change in hydrostatic pressure = (Fluid 1 gradient - Fluid 2 gradient) x Length of fluid column changed ...................................................................... (1.13) Compare the long and the short methods of calculating changes in hydrostatic pressure by studying the solutions for Example 13 on Page 1-22. Example 14, Page 1-23, explains how to calculate change in hydrostatic pressure when only part of the fluid column changes.

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Example 13: How to calculate change in hydrostatic pressure when the entire fluid column changes

If the 8.33-lb/gal water in a 10,000-ft well is replaced with 10.0-lb/gal salt water, what is the change in hydrostatic pressure? Solution (Long Method) Fluid gradient for 8.33-lb/gal water = 0.433 psi/ft (from Table 4.1, Page 4-2) Hydrostatic pressure before changing fluids = 0.433 psi/ft x 10,000 ft = 4,330 psi Fluid gradient for 10.0-lb/gal salt water = 0.5195 psi/ft (from Table 4.1, Page 4-2) Hydrostatic pressure after changing fluids = 0.5195 psi/ft x 10,000 ft = 5,195 psi Change in hydrostatic pressure= Hydrostatic pressure after - Hydrostatic pressure before ................................................................................. (1.12) = 5,195 psi - 4,330 psi = 865 psi Alternate Solution (Short Method) Fluid gradient for 10.0-lb/gal salt water = 0.5195 psi/ft (from Table 4.1, Page 4-2) Fluid gradient for 8.33-lb/gal water = 0.4330 psi/ft (from Table 4.1, Page 4-2) Length of fluid column changed = 10,000 ft Change in hydrostatic pressure = (Fluid gradient - Fluid gradient) x Length of fluid column changed ........................................................ (1.13) = (0.5195 psi/ft - 0.4330 psi/ft) x 10,000 ft = 0.0865 psi/ft x 10,000 ft = 865 psi

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Example 14: How to calculate hydrostatic pressure change when part of the fluid column changes

With 10,000 ft of tubing in the hole, enough 15.6-lb/gal cement is spotted to the bottom to fill 5,000 ft of tubing. The original fluid in the hole is 10.0-lb/gal brine water. What is the change in hydrostatic pressure? Solution Fluid gradient for 15.6-lb/gal cement = 0.8104 psi/ft (from Table 4.1, Page 4-2) Fluid gradient for 10.0-lb/gal brine = 0.5195 psi/ft (from Table 4.1, Page 4-2) Length of column changed = 5,000 ft Change in hydrostatic pressure = (Fluid gradient - Fluid gradient) x Length of fluid column changed ........................................................ (1.13) = (0.8104 psi/ft - 0.5195 psi/ft) x 5,000 ft = 0.2909 psi/ft x 5,000 ft = 1,454.5 psi

Problem 19 What is the change in hydrostatic pressure if a 10,000-ft column of 8.33-lb/gal fluid is replaced with a 16.0-lb/gal fluid? Work Space

Answer _______________ (See Page 1-64 for the solution.)

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Problem 20 During a flow test with a packer set at 5,000 ft with 10.0-lb/gal brine in the hole, what is the change in hydrostatic pressure when all the 10.0-lb/gal brine in the tubing has been replaced with 42° API oil weighing 6.8 lb/gal? Work Space

Answer _______________ (See Page 1-64 for the solution.)

Hydrostatic Pressure in Directionally Drilled Holes This part of Chapter 1 explains how to calculate hydrostatic pressure in directionally drilled holes. In directionally drilled holes, actual vertical depth may vary greatly from drilled depth. In a directional hole, only the vertical depth is used to calculate hydrostatic pressure—regardless of the path taken to reach this depth, the true vertical height of the fluid column is equal to the vertical depth. Example 15 shows that the vertical depth measurement provides accurate results. Example 15: How to calculate hydrostatic pressure in a directionally drilled hole Figure 1.17 illustrates an angled 30-ft joint of tubing in a lake. The bottom of the tubing is 20 ft below the lake’s surface. What is the hydrostatic pressure at the bottom of the tubing joint? Solution Fluid gradient of fresh water = 0.433 psi/ft (from Table 4.1, Page 4-2) Hydrostatic pressure = 0.433 psi/ft x 20 ft = 8.66 psi

Figure 1.17

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Figure 1.17, Page 1-24 shows that the hydrostatic pressure at the bottom of the tubing is equal to the hydrostatic pressure in 20 ft of water. In Example 15, if the 30-ft length of the tubing had been used for the calculation instead of the 20-ft actual vertical depth, the hydrostatic pressure would have been 12.99 psi—50% too high. In the oilfield, only vertical depth and mud weight influence hydrostatic pressure. Hole volume and diameter do not influence hydrostatic pressure, as demonstrated by Figure 1.18. Figure 1.18 shows four pieces of pipe with lengths of 1,000 ft and diameters of 1 ft. Pipes A, B, and C contain fresh water, and pipe D contains mud. Since all four pipes contain equal volumes of fluid, any difference in hydrostatic pressure is not influenced by volume. First, compare the hydrostatic pressures of pipes B and C, and note the importance of calculating hydrostatic pressure based on true vertical depth rather than total depth. Although the two pipes are the same length, the true vertical depth of pipe C is half that of pipe B. Therefore, the hydrostatic pressure of pipe C is half that of pipe B. Notice that pipe A has the lowest hydrostatic pressure because of its shallow vertical depth. Also compare pipes B and D, and note how mud weight greatly influences hydrostatic pressure.

Figure 1.18

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Example 16: How to calculate hydrostatic pressure in a directionally drilled well Figure 1.19 is a schematic of a directionally drilled well that was drilled straight down for 1,500 ft, kicked out (off vertical) for 10,000 ft, and then straightened up for the last 2,000 ft. The actual vertical depth of the well is 8,000 ft. If the fluid in the hole is 15.0-lb/gal mud, what is the hydrostatic pressure at the total depth (TD)? Solution Fluid gradient for 15.0-lb/gal mud = 0.7792 psi/ft (from Table 4.1, Page 4-2) Vertical depth = 8,000 ft Hydrostatic pressure at TD = 0.7792 psi/ft x 8,000 ft = 6,233.6 psi

If the calculation for the solution for Example 16 had been based on total depth (TD) instead of vertical depth, the hydrostatic pressure would be calculated as 0.7792 psi/ft x 13,500 ft = 10,519.2 psi The calculation based on TD results in an incorrect and significantly higher pressure than the calculation based on actual vertical depth.

Figure 1.19

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Problem 21 What is the hydrostatic pressure at TD for the well illustrated in Figure 1.20? Work Space

Answer _______________ (See Page 1-64 for the solution.)

Figure 1.20

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To find the total hydrostatic pressure of a slanted pipe at TD, follow these steps: 1. Determine a factor for the slanted portion of the hole using Equation 1.14. Slant factor = True vertical height of slant ÷ Length of slant .............. (1.14) 2. Calculate the length of the cement in the slanted pipe using Equation 1.15. Length of cement column in slant = Total cement length - Length of straight hole with cement .............. (1.15) 3. Determine the height of the cement in slant using Equation 1.16. Height of cement in slant = Slant factor x Length of cement column in slant................................. (1.16) 4. Calculate the hydrostatic pressure of the cement. Hydrostatic pressure of cement = Total cement height x Cement fluid gradient 5. Calculate the hydrostatic pressure of the brine. Hydrostatic pressure of brine = Height of brine column x Brine fluid gradient 6. Find the total hydrostatic pressure. Total hydrostatic pressure = Hydrostatic pressure of brine + Hydrostatic pressure of cement

Example 17: How to calculate hydrostatic pressure in a directionally drilled well (slant factor considered) Figure 1.21, Page 1-29 illustrates a 10,000-ft TD hole. Since the hole is directionally drilled, the actual vertical depth is only 8,000 ft. The bottom 4,500 ft of tubing is spotted with 15.6-lb/gal cement. What is the hydrostatic pressure at TD? Solution When the geometry of the hole is not available, estimate the hydrostatic pressure by making a schematic as shown in Figure 1.21. Then proceed with Steps 1 through 6 above. 1. Use Equation 1.14 for slant factor: 4,000 ft ÷ 6,000 ft = 0.6667 2. Use Equation 1.15 for length of cement column in slant: 4,500 ft - 3,000 ft = 1,500 ft The true vertical height of the cement in the slant is proportional to the length of cement in the slant by the same factor. To find the vertical height of the cement in the slant, use Equation 1.16. Solution for Example 17 continued on Page 1-29

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Example 17 Solution—continued 3. Height of cement in slant = 0.6667 x 1,500 ft = 1,000 ft 4. Total height of cement = Height of cement in slant + Length of straight hole containing cement = 1,000 ft + 3,000 ft = 4,000 ft Fluid gradient for 15.6-lb/gal cement = 0.8104 psi/ft (from Table 4.1, Page 4-2) Hydrostatic pressure of cement = 0.8104 psi/ft x 4,000 ft = 3,241.6 psi 5. Height of brine column = True vertical depth - Total cement height = 8,000 ft - 4,000 ft = 4,000 ft Fluid gradient for 10.0-lb/gal brine = 0.5195 psi/ft (from Table 4.1, Page 4-2) Hydrostatic pressure of brine = 0.5195 psi/ft x 4,000 ft = 2,078 psi 6. Total hydrostatic pressure = 3,241.6 psi + 2,078 psi = 5,319.6 psi

Figure 1.21

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Problem 22 Figure 1.22 illustrates the conditions for a 9,000-ft TD hole. The vertical depth is 7,000 ft. Enough 9.0-lb/gal fluid is spotted to fill the bottom 6,000 ft of tubing. What is the hydrostatic pressure inside the tubing at TD? Work Space

Answer _______________ (See Page 1-65 for the solution.)

Figure 1.22

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Problem 23 Figure 1.23 shows a well with a 10,000-ft vertical depth. Drilled depth (TD) is 15,000 ft. If enough 15.6-lb/gal cement to fill 5,000 ft of tubing is spotted to the bottom, what is the hydrostatic pressure at TD if the cement is displaced with 18.0-lb/gal mud? Work Space

Answer _______________ (See Page 1-65 for the solution.)

Figure 1.23

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Measurements for Directionally Drilled Holes Following is a list of terms that refer to various measurements for directionally drilled holes. Be familiar with these terms, and correct all data for calculations accordingly. • VSS (vertical subsea)—VSS refers to the vertical depth of the well measured from sea level. Geologists generally use this concept to identify and correlate formations. • VSF (vertical depth from surface flange)—Many oil companies currently measure vertical depth from the surface flange since the flange will be a permanent part of the completion. • MSF (measured depth from surface flange)—MSF is the total drilled depth measured from the surface flange. • MKB (measured depth from kelly bushing)—Some companies use the kelly bushing as a reference point. These companies use MKB to abbreviate total depth measured from the bushing. • VKB (vertical depth from kelly bushing)—VKB refers to vertical depth measured from the kelly bushing.

Calculating Pipe Capacity Definition of Pipe Capacity Pipe capacity is the volume of fluid or cement required to fill a specified length of pipe. Capacity, therefore, depends upon the pipe’s ID and length in each situation. Since pipe ID is determined by pipe weight, capacity is also determined by pipe weight. The Halliburton Cementing Tables list capacities of most commonly used pipes in several units of measure including bbl/ft, gal/ft, and ft³/ft. Since the most common measure of capacity and displacement volumes is barrels, most examples in this manual use barrels (bbl).

Using Capacity Factors to Calculate Pipe Capacity A capacity factor allows you to calculate easily the amount of fluid required to fill a pipe string. Use Equation 1.17 to calculate capacity. Capacity = Capacity factor x Pipe length ........................................................... (1.17) Capacity factors can be found in the Halliburton Cementing Tables. Example 18 shows a calculation for pipe capacity. 1-32

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Example 18: How to calculate tubing capacity

What is the capacity of 6,000 ft of 3 1/2 -in., 9.3-lb/ft tubing, expressed in barrels? Solution Capacity factor for 3 1/2-in., 9.3-lb/ft tubing = 0.00870 bbl/ft (from the Halliburton Cementing Tables) Capacity = Capacity factor x Pipe length............................................. (1.17) = 0.00870 bbl/ft x 6,000 ft = 52.20 bbl

Calculating Capacity Factors Here are the equations used to calculate pipe capacity factors. The capacity factors listed in the Halliburton Cementing Tables are based on these equations. You may need them to calculate factors if the book is not available or if it does not list a factor for the specific type of pipe on the job. Capacity factor of pipe in gal/ft = π ID² (gal) ............................................... (1.18) 77 (ft) Capacity factor of pipe in bbl/ft = π ID² (bbl) ............................................... (1.19) 3,234 (ft) Capacity factor of pipe in ft³/ft =

π ID² (ft³)

............................................... (1.20)

576 (ft) where π = 3.141592654 ID = internal diameter of the pipe

Problem 24 What is the capacity (bbl) of 10,000 ft of 3 1/2-in., 15.50-lb/ft drillpipe? The capacity factor for this drillpipe is 0.00658 bbl/ft (from the Halliburton Cementing Tables). Work Space

Answer _______________ (See Page 1-65 for the solution.)

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Problem 25 What is the capacity in barrels of 600 ft of 6.0-in. OD, 2 1/2-in. ID drill collars? The capacity factor for these drill collars is 0.0061 bbl/ft (from the Halliburton Cementing Tables). Work Space

Answer _______________ (See Page 1-65 for the solution.) NOTE

To find capacity factors for drill collars in the field, find the capacity factor for an open hole or tubing with the same ID in the Halliburton Cementing Tables.

Calculating Pipe Fill-Up Definition of Pipe Fill-Up Fill-up is the length of pipe that a given volume of fluid will fill. Fill-up factors simplify fill-up calculations and may be found in the “Capacity” section of the Halliburton Cementing Tables. These fill-up factors are expressed in various units: ft/gal, ft/ft,3 and ft/bbl.

Calculating Pipe Fill-Up using Fill-Up Factors To calculate fill-up, multiply the fill-up factor (in ft/bbl) by volume (in bbl) to obtain the length of pipe filled (in ft). If volumes are measured in gallons or cubic feet, multiply the appropriate fill-up factor by the volume to yield the length of pipe filled (Equation 1.21). Length of pipe fill-up = Fill-up factor x Volume ............................................. (1.21) Fill-up factors can be found in the Halliburton Cementing Tables in the “Capacity” section (in the lin. ft/bbl column). Example 19 is an application of Equation 1.21.

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Example 19: How to calculate fill-up

How many feet of 2 7/8-in., 10.4-lb/ft internal upset drillpipe will 60 bbl of cement fill? Solution Fill-up factor for 2 7/8-in., 10.4-lb/ft drillpipe = 222.49 ft/bbl (from the Halliburton Cementing Tables) Length of pipe fill-up = Fill-up factor x Volume .................................... (1.21) Fill-up = 222.49 ft/bbl x 60 bbl = 13,349.4 ft

Problem 26 How many feet of 2 3/ 8-in., 4.7-lb/ft tubing will 50 bbl of oil fill? The fill-up factor for this tubing is 258.65 ft/bbl (from the Halliburton Cementing Tables). Work Space

Answer _______________ (See Page 1-66 for the solution.)

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Problem 27 On a squeeze job, the bottomhole pressure is limited to 9,000 psi. Conditions are as follows:

• • • • •

Packer depth = 10,000 ft Mud weight = 14.0 lb/gal Cement weight = 16.0 lb/gal Spacer fluid (water) = 8.33 lb/gal Tubing = 2 3/8 in., 4.7 lb/ft (from the Halliburton Cementing Tables, capacity factor = 0.00387 bbl/ft; fill-up factor = 258.65 ft/bbl)

The job involves (1) placing 10 bbl of water ahead of the cement, (2) following the water with 50 bbl of cement, (3) placing 10 bbl of water, and (4) displacing the cement with mud. A. What is the capacity of the tubing? Work Space

Answer _______________

B. What is the maximum pump pressure with a full column of mud? Work Space

Answer _______________

C. What is the maximum pump pressure when cement mixing begins? Work Space

Answer _______________ Problem 27 continued on Page 1-37 1-36

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Problem 27—continued D. What is the maximum pump pressure when the water ahead of the cement reaches the tool? Work Space

Answer _______________ E. What is the maximum pump pressure with a full column of cement? Work Space

Answer _______________ F. What is the maximum pump pressure when all the water behind the cement is in the tubing? Work Space

Answer _______________ G. What is the maximum pump pressure when the water behind the cement reaches the tool? Work Space

Answer _______________ H. What is the maximum pump pressure when the mud behind the cement reaches the tool (with a full column of mud)? Work Space

Answer _______________ (See Pages 1-66 and 1-67 for the solutions to Problem 27, Parts A through H.)

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Calculating Buoyancy Definition of Buoyancy The dictionary defines buoyancy as the power of a fluid to exert an upward force on a body placed in it. Fluid tries to float anything placed in it. In oilfield applications, buoyancy is the force that causes even very heavy items, such as drillpipe and drill collars, to weigh less in fluid than in air. The buoyant, or upward force on pipe is equal to the weight of fluid displaced when pipe is run in the hole. Normally, weights for tubing and drillpipe are given in pounds per foot (lb/ft), as weighed on dry land (or air). This section of Chapter 1 provides examples and problems for calculating weights of both steel and aluminum pipe in liquid. Later, two methods of calculating the weight of pipe in fluid are described: the buoyancy-factor method and the area/hydrostatic-pressure method.

Calculating Weight in Fluid for Steel Pipe When mud weight is expressed in pounds per gallon (lb/gal), the weight of openended steel pipe suspended in fluid can be calculated with Equation 1.22. WL = (WA) x [1 - (0.01528 x MW)] ....................................................................... (1.22) where: WL = weight of pipe suspended in liquid (lb/ft) WA = weight of pipe in air (lb/ft) MW = mud weight (lb/gal) NOTE

Equation 1.22 applies only to steel pipe. The mud weight must be expressed in pounds per gallon.

Examples 20 and 21 show applications of Equation 1.22. Example 20: How to calculate weight of steel pipe in fluid

What does 2 3/ 8-in., 4.70-lb/ft tubing weigh in 12.3-lb/gal mud? Solution WL = (WA) x [1 - (0.01528 x MW)] ...................................................... (1.22) = 4.70 x [1 - (0.01528 x 12.3)] = 4.70 x (1 - 0.1879) = 4.70 x 0.8121 = 3.8167 lb/ft

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Example 21: How to calculate the reading of the total-weight indicator

1,000 ft of 2 3/8-in., 4.70-lb/ft tubing is suspended in 12.3-lb/gal mud. What is the reading of the total-weight indicator? Solution In Example 20, the weight of 4.7-lb/ft pipe suspended in 12.3-lb/gal mud was calculated to be 3.8167 lb/ft. For the 1,000 ft of this tubing, the weight indicator would show 3.8167 lb/ft x 1,000 ft = 3,816.7 lbâ

NOTE

To simplify the above calculations, find the buoyancy factor in a chart such as Table 4.3, Page 4-4. This method will be explained on Page 1-42.

Problem 28 A. What does 1,700 ft of 3 1/2-in., 15.50-lb/ft drillpipe weigh in 14.7-lb/gal mud? Work Space

Answer

_______________

B. What does the same drillpipe weigh in air? Work Space

Answer _______________ (Solutions for Problem 28 are on Page 1-67). When mud weight is expressed in pounds per cubic foot (lb/ft3), the weight of open-ended steel pipe suspended in fluid can be calculated with Equation 1.23. WL = WA [1 - (0.002045) x (MC)] .......................................................................... (1.23) where: WL = weight of pipe suspended in liquid (lb/ft) WA = weight of pipe in air (lb/ft) MC = mud weight (lb/ft³) NOTE

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Equation 1.23 applies only to steel pipe. Mud weight must be expressed in pounds per cubic foot.

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Calculating Weight in Fluid for Aluminum Pipe Calculate the weight of open-ended aluminum pipe suspended in fluid by using Equation 1.24 when mud weight is expressed in lb/gal. WL = WA [1 - (0.044) x (MW)] .............................................................................. (1.24) where: WL = weight of pipe suspended in liquid (lb/ft) WA = weight of pipe in air (lb/ft) MW = mud weight (lb/gal) NOTE

Equation 1.24 applies only to aluminum pipe. Mud weight must be expressed in pounds per gallon.

Buoyancy Factors Table 4.3, Page 4-4 lists the buoyancy factors for various weights of fluids. These buoyancy factors are based on calculations for [1 - (0.01528) x (MW)] from Equation 1.22, Page 1-38. To use Table 4.3, find the mud weight in the left column, and read the buoyancy factor in the right column. Table 4.4, Page 4-5 lists buoyancy factors when mud weight is expressed in lb/ft.³ Buoyancy factors for aluminum drillpipe are not provided because they are not used often. You can simplify calculations for steel pipe weight in fluid with Equation 1.25. Pipe weight in liquid = Buoyancy factor x Pipe weight in air ...................... (1.25)

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Example 22: Calculating pipe weight in liquid with Equation 1.25

What is the weight-indicator reading when 1,500 ft of 3 1 /2-in., 15.50-lb/ft drillpipe is suspended in 10.2-lb/gal mud? Solution Buoyancy factor for 10.2-lb/gal mud = 0.8441 (from Table 4.3, Page 4-4) Pipe weight in 10.2-lb/gal mud = 0.8441 x 15.50 lb/ft = 13.08355 lb/ft Pipe weight in liquid = Buoyancy factor x Pipe weight in air................ (1.25) = 13.08355 lb/ft x 1,500 ft = 19,625.325 lbâ Alternate Solution If you use a calculator, you may prefer to make the calculation in one step as follows. Indicator reading = Buoyancy factor x Pipe lb/ft x Pipe ft = 0.8441 x 15.50 lb/ft x 1,500 ft = 19,625.325 lbâ

Problem 29 What does 7,000 ft of 4 1/2-in., 16.60-lb/ft drillpipe weigh in 19.0-lb/gal mud? Work Space

Answer _______________ (See Page 1-67 for the solution.)

The following examples and problems demonstrate how weight-indicator readings with fluid in the pipe can be calculated with two methods. These two interchangeable methods are the buoyancy-factor method and the area/hydrostatic-pressure method.

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Buoyancy-Factor Method Often in the oilfield, the hole is not completely full of fluid. Pipe weight can be calculated with the buoyancy factor for only the portion of pipe that is suspended in fluid. The remainder of the pipe's weight in air must be accounted for. After calculating the weights of the portion of pipe in fluid and the portion of pipe in air, add the two results together for the total pipe weight in the hole. Example 23: How to calculate pipe weight in fluid with the buoyancyfactor method Figure 1.24 shows a well filled to 4,000 ft with 8.33-lb/gal water. Then 10,000 ft of 2 3/8-in., 4.7-lb/ft tubing is run in the well. What is the weightindicator reading? Solution Calculate pipe weight in water: Buoyancy factor (8.33 lb/gal) = 0.8727 (from Table 4.3, Page 4-4) Pipe weight in liquid = 0.8727 x 4.7 lb/ft = 4.10169 lb/ft Pipe length in fluid = 10,000 ft - 4,000 ft = 6,000 ft Indicator reading for pipe in liquid = 6,000 ft x 4.10169 lb/ft = 24,610.14 lbâ Figure 1.24 Calculate pipe weight in air: Length of pipe in air = 4,000 ft Indicator reading from pipe in air = 4.7 lb/ft x 4,000 ft = 18,800 lbâ Total indicator reading = 24,610.14 lbâ + 18,800 lbâ = 43,410.14 lbâ

Area/Hydrostatic-Pressure Method Another way to calculate the weight-indicator reading for pipe in fluid is the area/hydrostatic-pressure method. The basis for this method is that the upward (buoyant) force acting on the pipe is equal to the hydrostatic pressure at the lower end of the tubing that is acting on the area of the pipe-wall thickness. Pipe-wall thickness equals pipe OD area minus pipe ID area. Subtracting the buoyant force from the pipe weight in air yields the weight-indicator reading or string weight.

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To find the string weight using the area/hydrostatic-pressure method, follow these steps. 1. Calculate the hydrostatic pressure. Hydrostatic pressure = Fluid gradient x Pipe length 2. Determine the pipe’s effective area. Effective area = Pipe OD area - Pipe ID area 3. Calculate the buoyant (upward) force by multiplying the answers to Steps 1 and 2. Buoyant force = Hydrostatic pressure x Effective area 4. Determine pipe weight in air. Pipe weight in air = Pipe weight in lb/ft x Pipe length 5. Calculate total weight-indicator reading by subtracting the buoyant force (Step 3) from the pipe weight in air (Step 4). Indicator reading = Pipe weight in air lbâ- Buoyant force lbá Example 24: Calculating pipe weight in liquid using the area/ hydrostatic-pressure method. In Figure 1.25, 1,000 ft of 2 3/8in., 4.7-lb/ft tubing is suspended in 12.3-lb/gal mud. What is the weight-indicator reading? Solution Fluid gradient for 12.3-lb/gal mud = 0.6390 psi/ft (from Table 4.1, Page 4-2) Hydrostatic pressure = 0.6390 psi/ft x 1,000 ft = 639 psi Tubing OD area = 4.430 in.² (from Table 4.5, Page 4-6) Tubing ID area Figure 1.25 = 3.126 in.² (from Table 4.5, Page 4-6) Effective area = 4.430 in.² - 3.126 in.² = 1.304 in.² Buoyant force = 1.304 in.² x 639 psi = 833.256 lbá Pipe weight in air = 4.7 lb/ft x 1,000 ft = 4,700 lbâ Indicator reading = 4,700 lbâ - 833.256 lbá = 3,866.744 lbâ

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NOTE

Small variations in the calculated indicator readings result from rounding off buoyancy factor, areas, and other measurements, and from neglecting the small buoyant force on each collar. The buoyant force on each collar could be calculated, and all calculations could be carried to more decimal places; however, numbers rounded off as shown in the examples and problem solutions in this text are adequate for oilfield applications.

Calculating Weight-Indicator Readings for MixedPipe Strings A mixed-pipe string is made up of two or more sizes or weights of pipe. When fluid is in the hole, a mixed string’s weight can be calculated with either of the two methods previously described, the buoyancy-factor method or the area/ hydrostatic-pressure method. Figure 1.26 illustrates a mixed string of pipe made up of 1,500 ft of 4 1/2-in., 16.60-lb/ft drillpipe on top and 1,000 ft of 3 1/2-in., 15.50-lb/ft drillpipe on bottom. Figure 1.27, Page 1-45, shows where hydrostatic pressures are acting on the string in Figure 1.26. Looking only at the 3 1/2-in. drillpipe in Figure 1.27, the hydrostatic pressure at 1,500 ft is pushing down across the pipe wall thickness. Since the areas at the top and bottom of the 3 1/2-in. pipe are equal, use differential pressures to calculate force. Examining the joint between the two sizes of pipe in Figure 1.27 reveals that the only downward force not accounted for is the area between the 3 1/2-in. pipe OD and the 4 1/2-in. pipe ID. Since the hydrostatic pressures inside and outside are

Figure 1.26

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Figure 1.27

equal, the downward force from the 3 1/2-in. pipe OD to the 4 1/2-in. pipe ID is canceled by an equal upward force. The only remaining force is the upward force from hydrostatic pressure at 1,500 ft across the wall thickness of the 4 1/2-in. drillpipe. To calculate the actual string weight in fluid for a mixed string of pipe, follow these steps: 1. Multiply the fluid’s buoyancy factor by the pipe weight in air (lb/ft) to obtain the pipe’s weight in liquid. 2. For each size of pipe, multiply the pipe’s weight in liquid by the length of that size of pipe (ft). 3. Add the weight of each string to obtain the total weight-indicator reading. Example 25 on Page 1-46 is an application of these calculations.

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Example 25: How to calculate weight-indicator reading for a mixedpipe string

What is the weight-indicator reading for the mixed string in Figure 1.26 on Page 1-44? Solution Buoyancy factor for 10.0-lb/gal fluid = 0.8472 (from Table 4.3, Page 4-4) Weight of 3 1/2-in. pipe in fluid = 15.50 lb/ft x 0.8472 = 13.1316 lb/ft Actual weight of 1,000 ft of 3 1/ 2-in. pipe = 1,000 ft x 13.1316 lb/ft = 13,131.6 lbâ Weight of 4 1/2-in. pipe in fluid = 16.60 lb/ft x 0.8472 = 14.06352 lb/ft Actual weight of 1,500 ft of 4 1/ 2-in. pipe = 1,500 ft x 14.06352 lb/ft = 21,095.28 lbâ Indicator reading = 13,131.6 lbâ + 21,095.28 lbâ = 34,226.88 lbâ

Problem 30 If the fluid in the hole is 10 lb/gal, what is the indicator reading for a mixed string consisting of

• • •

1,000 ft of 4 1/2-in., 16.60-lb/ft drillpipe on top 1,000 ft of 3 1/2-in., 15.50-lb/ft drillpipe in the middle 1,000 ft of 2 7/8-in., 10.40-lb/ft drillpipe on bottom?

Work Space

Answer _______________ (See Page 1-68 for the solution.)

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Problem 31 If the fluid level in the well in Problem 30 is at 1,500 ft and the same mixed string of pipe is run in the hole, what is the weight-indicator reading? Work Space

Answer _______________ (See Page 1-68 for the solution.)

Example 26: How to calculate the effect of a shoe on the weightindicator reading

How does adding a shoe on the bottom of the tubing in Figure 1.28 affect the weight-indicator reading? Solution Figure 1.29 (Page 1-48) shows the hydraulic forces acting on the shoe. Notice the downward force from hydrostatic pressure acting across the area from the shoe OD to the tubing OD. Note the equal upward force on the shoe bottom. These forces cancel each other, leaving only the upward force to act across the area from the tubing OD Figure 1.28 to the tubing ID. In Example 24 on Page 1-43, this upward force (hydrostatic pressure acting from tubing OD to tubing ID) was defined as buoyancy; therefore, the shoe will have no effect on the weight-indicator reading.

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Figure 1.29

Problem 32 Does the 2 1/8-in. OD, 1.0-in ID stinger in Figure 1.30 affect the weight-indicator reading? Answer _______________ (See Page 1-68 for the solution.)

Figure 1.30

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Other Factors Influencing Weight Calculations All weight-indicator calculations presented have assumed the same fluid was inside and outside the tubing. However, when tubing is run in dry (no fluid inside), other forces must be considered. Hydrostatic pressure changes when dry tubing, such as bull-plugged tubing, a string of testing tools, or a retrievable packer with a closed tubing valve, is run in. Figure 1.31 illustrates bull-plugged tubing. Notice that with the bottom of the tubing bull-plugged, hydrostatic pressure works across the entire tubing OD area. Example 27 shows the calculation of the weight-indicator reading.

Example 27: How to calculate the indicator reading in bull-plugged tubing Figure 1.31 shows 1,000 ft of 2 3/8-in., 4.7-lb/ft bull-plugged tubing run in 8.33-lb/gal water. What does the weight indicator read? Solution Pipe weight in air = 4.7 lb/ft x 1,000 ft = 4,700 lbâ Area tubing OD = 4.430 in.² Fluid gradient for 8.33 lb/gal fluid = 0.433 psi/ft (from Table 4.1, Page 4-2) Hydrostatic pressure at 1,000 ft = 0.433 psi/ft x 1,000 ft = 433 psi Upward force on tubing OD Figure 1.31 = 4.430 in.² x 433 psi = 1,918.19 lbá â Indicator reading = 4,700 lbâ - 1,918.19 lbá = 2,781.81 lbâ

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Calculating Hydraulics of Open-Ended Pipe You can apply the principles of area, pressure, and buoyancy to downhole conditions when pumping through open-ended tubing.

Calculating Surface Pressures When pipe rams are closed around tubing, the casing becomes a large hydraulic cylinder, and the tubing acts as a piston. Applying pump pressure to the system can move the piston (tubing) upward. To actually move the piston, the weight of the piston itself (pipe weight minus the buoyant forces) must be overcome. In other words, to pump tubing out of the hole, pump pressure must create an upward force equal to or greater than the weight-indicator reading with the tubing in fluid. For these calculations, assume that there is no friction between the pipe and rams. Friction varies widely depending on factors such as the types of rams used and the condition of the pipe. Since pressure acts equally in all directions, any surface pressure acts at the bottom of the tubing, across the area from tubing OD to tubing ID. Pressure also acts at the top of the tubing across the tubing ID. The effective area is equal to the tubing OD. Figure 1.32 illustrates this force. The upward force caused by the surface pressure that acts on open-ended pipe is measured on the weight indicator because this force supports a portion of the pipe weight. Table 4.5, Page 4-6 of this manual lists the OD and ID areas of most common

Figure 1.32

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drillpipe and tubing sizes. To use Table 4.5, locate the pipe OD in the left column. Next, find pipe weight in the second column, and move horizontally across the columns. The third column is the OD area in square inches. The fourth column gives ID in inches, and the fifth column is ID area in square inches. The two right columns repeat the weight and ODs. Example 28 shows the calculation of an indicator reading during pumping operations. Example 28: How to calculate weight-indicator readings for openended tubing during pumping Figure 1.33 shows 1,000 ft of 2 3/8-in., 4.7-lb/ft of open-ended tubing hanging in fresh water.

If 500-psi surface pressure is applied, what does the weight indicator read? Solution Buoyancy factor for fresh water = 0.8727 (from Table 4.3, Page 4-4) Pipe weight in liquid = 0.8727 x 4.7 lb/ft = 4.10169 lb/ft Indicator reading before pumping = 1,000 ft x 4.10169 lb/ft = 4,101.69 lbâ Tubing OD area = 4.430 in.2 (from Table 4.5, Page 4-6) Figure 1.33 Force = 500 psi x 4.430 in.2 = 2,215 lbá Indicator reading while pumping = 4,101.69 lbâ - 2,215 lbá = 1,886.69 lbâ

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Problem 33 As shown in Figure 1.34, 4,000 ft of 2 3/8-in., 4.7-lb/ft tubing is hanging in 10.2-lb/gal mud. A.

What is the weight-indicator reading before the pump is started? Work Space

Answer ______________ Figure 1.34

B. What is the indicator reading if the system is pumped at 4,000 psi? Work Space

Answer _______________ (See Page 1-68 for solutions to Parts A and B.)

Calculating Maximum Pump Pressures The maximum allowable surface pressure (without pumping the pipe out of the hole) can be calculated as the pressure required to create a force equal to the pipe weight in liquid (weight accounting for buoyancy). This pressure is equal to floating pipe weight divided by tubing OD area (Equation 1.26). Maximum allowable surface pressure = Pipe weight in liquid ................ (1.26) Tubing OD area Example 29 is an application of Equation 1.26.

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Example 29: How to calculate maximum pump pressure The well in Figure 1.35 has 1,000 ft of 2 3/ 8-in., 4.7-lb/ft of open-ended tubing hanging in fresh water. Use Equation 1.26 to find the maximum pump pressure that can be applied without pumping the tubing out of the hole. Solution Buoyancy factor for 8.33-lb/gal fluid = 0.8727 (from Table 4.3, Page 4-4) Pipe weight in liquid = 4.7 lb/ft x 0.8727 = 4.10169 lb/ft Available weight = 4.10169 lb/ft x 1,000 ft = 4101.69 lbâ Tubing OD area = 4.430 in.2 (from Table 4.5, Page 4-6) Maximum pump pressure = 4,101.69 lb ÷ 4.430 in.2 = 925.890 psi Figure 1.35

Problem 34 Figure 1.36 shows 1,000 ft of 4 1/2-in., 16.60-lb/ft of open-ended drillpipe hanging in fresh water. What is the maximum pump pressure that can be applied without pumping the drillpipe out of the hole? Work Space

Answer _______________ (See Page 1-69 for the solution.)

Figure 1.36

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Calculating Weight-Indicator Readings when Spotting Fluids Sometimes a heavier fluid, such as heavy mud or cement, is pumped down tubing or drillpipe while the original, lighter fluid remains between the pipe and casing. In such a case, it is necessary to hold surface pressure on the casing side to help prevent the heavier inside fluid from “U-tubing” or equalizing. By holding surface pressure on the casing, total pressure at bottom (on the casing side) is made equal to the hydrostatic pressure of the heavier fluid at bottom. A pressure equal to the hydrostatic pressure of the heavier fluid exerts an upward, buoyant force across the area from the tubing OD to tubing ID. In this case, you have changed the buoyant force on the tubing by filling the tubing with a heavier fluid and holding pressure on the casing. The weight indicator will now show the pipe weight to be the same as if it were suspended with heavier fluid both inside and outside of the drillpipe or tubing. Example 30 shows the relevant calculations. Example 30: Calculating weight-indicator readings when spotting heavy fluids Figure 1.37 illustrates a well with 1,000 ft of 2 3/8-in., 4.7-lb/ft of open-ended tubing hanging in fresh water. After the rams are closed, a full column of 15.6-lb/gal cement is spotted in the tubing, as shown in Figure 1.38. The casing valve is pinched down until enough pressure is trapped on the casing to hold the cement in place in the tubing.

A. What is the weightindicator reading before the cement is placed in the tubing? B. How much pressure should be trapped on the casing to hold the cement in place? C. What is the weightindicator reading while the cement is held in place? D. How much surface pressure can operators apply to the tubing to put away the cement without pumping the tubing out of the hole?

Figure 1.37

Example 30 continued on Page 1-55

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Example 30—continued Solution A. Buoyancy factor for 8.33-lb/gal fluid = 0.8727 (from Table 4.3, Page 4-4) Pipe weight in 8.33-lb/gal fluid = 4.7 lb/ft x 0.8727 = 4.10169 lb/ft Indicator reading = 4.10169 lb/ft x 1,000 ft â = 4,101.69 lbâ B. Fluid gradient for 15.6-lb/gal cement) = 0.8104 psi/ft (from Table 4.2, Page 4-3) Fluid gradient for 8.33-lb/gal water = 0.433 psi/ft (from Table 4.2, Page 4-3)

Figure 1.38

Change in fluid gradient = 0.8104 psi/ft - 0.433 psi/ft = 0.3774 psi/ft Length of fluid column changed to heavier fluid = 1,000 ft Pressure required on casing = 1,000 ft x 0.3774 psi = 377.4 psi C. Buoyancy factor for 15.6-lb/gal cement = 0.7616 (from Table 4.3, Page 4-4) Pipe weight in 15.6-lb/gal cement = 4.7 lb/ft x 0.7616 = 3.57952 lb/ft â Indicator reading = 1,000 ft x 3.57952 lb/ft = 3,579.52 lbâ D. Available weight = 3,579.52 lbâ (from Part C) Tubing OD area = 4.430 in.2 (from Table 4.5, Page 4-6) Maximum allowable pressure = 3,579.52 lb ÷ 4.430 in.2 = 808.018 psi

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Problem 35 Figure 1.39 illustrates a well with 1,000 ft of 1.315-in. OD, 1.80-lb/ft of openended tubing hanging in fresh water. After the rams are closed and a full column of 15.6-lb/gal cement is spotted, the casing valve is pinched down to hold the cement in place in the tubing. A. What does the weight indicator show before the cement is pumped (when first on bottom)? Work Space Figure 1.39

Answer _______________ B. How much pressure should be trapped on the casing to hold the cement in place? Work Space

Answer _______________ C. How much surface pressure can you apply to tubing containing a full column of cement without pumping the tubing out of the hole? Work Space

Answer _______________ (See Page 1-69 for the solutions to Problem 35.) NOTE

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The buoyancy-factor method is better when the pipe contains full columns of heavier fluid. The area/hydrostatic-pressure method is better for pipe containing partial columns of heavier fluid.

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Spotting a light fluid into a heavy mud system results in somewhat different conditions than spotting the heavier fluids discussed. For example, spotting light fluids occurs when acid or a completion fluid is pumped into a formation or a cushion is pumped for a flow test. Example 31 shows calculations for spotting light fluids. Example 31: Calculating weight-indicator readings when spotting light fluids Figure 1.40 illustrates a well with 1,000 ft of 2 3/8-in., 4.7-lb/ft open-ended tubing hanging in 16.0-lb/gal mud.

A. What does the weight indicator read before pumping (when first on bottom)? B. If tubing is spotted with fresh water until full, what is the tubing pressure if the pumps are shut down? C. What does the indicator read with the pumps shut down and the tubing valve closed? D. What is the maximum pressure (total gauge) that you can apply to displace the water without pumping tubing out of the hole? Solution A. Buoyancy factor for 16.0-lb/gal mud = 0.7555 (Table 4.3, Page 4-4 ) Pipe weight in liquid Figure 1.40 = 0.7555 x 4.7 lb/ft = 3.55085 lb/ft Original weight-indicator reading = 3.55085 lb/ft x 1,000 ft = 3,550.85 lbâ Solution for Example 31 continued on Page 1-58

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Example 31 Solution—continued B. Fluid gradient for 16.0-lb/gal mud = 0.8312 psi/ft (Table 4.1, Page 4-2) Fluid gradient 8.33-lb/gal water = 0.433 psi/ft (Table 4.1, Page 4-2) Change in fluid gradient = 0.8312 psi/ft - 0.433 psi/ft = 0.3982 psi/ft Change in tubing hydrostatic pressure = 0.3982 psi/ft x 1,000 ft = 398.2 psi Tubing pressure = Change in tubing hydrostatic pressure = 398.2 psi C. Figure 1.41 shows that pressure at bottom outside the tubing is the hydrostatic pressure of the 16.0-lb/gal mud. Pressure at bottom inside the tubing is the hydrostatic pressure of fresh water plus 398.2 psi—equal to the hydrostatic pressure of 16.0-lb/gal mud. When the original weight-indicator reading was calculated (in Part A), total pres sure at bottom was the hydrostatic pressure of 16.0-lb/gal mud; there fore, the buoyant force on bottom was accounted for in the original calculation with the buoyancy factor for the 16.0-lb/gal mud. The only force not accounted for is the trapped tubing pressure pushing up across the ID area of the tubing. 2 3/8-in., 4.7-lb/ft OD tubing ID area = 3.126 in. 2 (from Table 4.5, Page 4-6) Force resulting from trapped pressure = 398.2 psi x 3.126 in.2 = 1,244.773 lbá Weight-indicator reading = Original indicator reading - Force caused by trapped tubing pressure = 3,550.85 lbâ (from Part A) - 1,244.773 lbá = 2,306.077 lbâ D. 2,306 lb is the weight available to counteract displacement pressure. 398.2 psi already on the tubing was accounted for during available weight calculations; the displacement pressure calculated (based on this weight) represents the maximum increase in tubing pressure. So, Figure 1.41 the total maximum tubing gauge pressure is the trapped pressure plus the calculated pressure increase. Ap plied tubing pressure (pump pressure) acts across the tubing OD area as shown in Figure 1.41. Solution for Example 31 continued on Page 1-58

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Example 31 Solution—continued 2 3/8-in., 4.7 lb/ft tubing OD area = 4.430 in.² (from Table 4.5, Page 4-6) Maximum increase in pressure = 2,306.0768 lb (from Part C) ÷ 4.430 in.² = 520.559 psi Maximum tubing pressure = 398.2 psi (from Part A) + 520.559 psi = 908.759 psi

Problem 36 Figure 1.42 illustrates the conditions for a well with 3,000 ft of 3 1/ 2-in., 13.30-lb/ft drillpipe hanging open-ended in 18.0-lb/gal mud.

A. What does the indicator read when first on bottom? Work Space

Figure 1.42

Answer _______________ Problem 36 continued on Page 1-60

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Problem 36—continued B. If the drillpipe is spotted with 9.0-lb/gal acid until full, what is the drillpipe pressure if the pumps are shut down? Work Space

Answer _______________ C. What does the indicator read when the pumps are shut down and the drillpipe valve is closed? Work Space

Answer _______________ D. What maximum pressure (total gauge) can be applied to displace acid without pumping the drillpipe out of the hole? Work Space

Answer _______________ (Solutions to Problem 36 are on Page 1-69.) 1-60

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Solutions to Problems Pages 1-61 through 1-70 provide the solutions to the problems in Chapter 1. Most of the problems can be solved more than one way. Different methods may give slightly different answers, depending on how the numbers are rounded off.

NOTE

If you use and understand a particular method, always use it, even if it is not used in this text.

Solution for Problem 1 Area = 0.7854 x D2 ...................................................................................................... (1.3) = 0.7854 x 1.5 in. x 1.5 in. = 1.767 in.² ............................................................... Answer

Solution for Problem 2 Area = 0.7854 x D2 ...................................................................................................... (1.3) = 0.7854 x 2.875 in. x 2.875 in. = 6.492 in.² ....................................................... Answer

Solution for Problem 3 Casing ID area = 0.7854 x 9.760 in. x 9.760 in. = 74.815 in.² .......................... Answer

Solution for Problem 4 Tubing OD area = 0.7854 x 2.375 in. x 2.375 in. = 4.430 in.² Tubing ID area = 0.7854 x 1.995 in. x 1.995 in. = 3.126 in.² Tubing cross-sectional area = Pipe OD area - Pipe ID area ................................ (1.4) = 4.430 in.² - 3.126 in.² = 1.304 in.² .................................................................... Answer

Solution for Problem 5 Piston area = 0.7854 x 3.5 in. x 3.5 in. = 9.621 in.² Pressure = 3,000 psi Force = Pressure x Area ............................................................................................ (1.1) â ................................................................. Answer = 3,000 psi x 9.621 in.² = 28,863 lbâ

Solution for Problem 6 Piston OD area = 0.07854 x 3 in. x 3 in. = 7.069 in.² Pressure = Force ÷ Area ............................................................................................ (1.7) = 1,000 lb ÷ 7.069 in.² = 141.463 psi .................................................................. Answer

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Solution for Problem 7 Area = 5.302 in.² (from Example 7) Pressure = 1,000 psi Force = Pressure x Area ............................................................................................ (1.1) = 1,000 psi x 5.302 in.² = 5,302 lbá ................................................................... Answer

Solution for Problem 8 Top side of the cylinder Piston OD area = 0.7854 x 4 in. x 4 in. = 12.566 in.² Pressure = 500 psi Force = Pressure x Area ..................................................................................... (1.1) = 500 psi x 12.566 in.² = 6,283 lbâ Bottom side of the cylinder Piston OD area (from above) = 12.566 in.² Rod area = 0.7854 x 1 in. x 1 in. = 0.785 in.² Effective area = Piston area - Rod area = 12.566 in.² - 0.785 in.² = 11.781 in.² Pressure = 1,000 psi Force = Pressure x Area ..................................................................................... (1.1) = 1,000 psi x 11.781 in.² = 11,781 lbá Total force = 11,781 lbá - 6,283 lbâ = 5,498 lbá .......................................... Answer

Solution for Problem 9 Top side of the cylinder Piston OD area = 12.566 in.² (from Problem 8) Pressure = 1,000 psi Force = Pressure x Area ..................................................................................... (1.1) = 1,000 psi x 12.566 in.² = 12,566 lbâ Lower side of the cylinder Effective area = 11.781 in.² (from Problem 8) Pressure = 1,000 psi Force = Pressure x Area ..................................................................................... (1.1) á = 1,000 psi x 11.781 in.² = 11,781 lbá Total force = 12,566 lbâ- 11,781 lbá = 785 lbâ ............................................ Answer

Alternate Solution for Problem 9 Rod area = 0.785 in.² (from Problem 8) Force = Pressure x Area ............................................................................................ (1.1) = 1,000 psi x 0.785 in.² = 785 lbâ ...................................................................... Answer

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Solution for Problem 10 Piston OD area = 0.7854 x 3.750 in. x 3.750 in. = 11.045 in.² Rod area = 0.7854 x 1.625 in. x 1.625 in. = 2.074 in.² Effective area = 11.045 in.² - 2.074 in.² = 8.971 in.² Pressure = 200 psi Force = Pressure x Area ............................................................................................ (1.1) = 200 psi x 8.971 in.² = 1,794.2 á ...................................................................... Answer

Solution for Problem 11 Pressure in the system acts on both sides of the piston since the piston does not touch the cylinder. This gives an effective area equal to the rod diameter. Rod area was calculated in Problem 10 to be 2.074 in.² Force on the scale is now: Force = Pressure x Area ............................................................................................ (1.1) = 200 psi x 2.074 in.² = 414.8 lbâ ...................................................................... Answer When comparing this problem with Problem 10, you can see that not having a seal has actually reversed the load on the scales.

Solution for Problem 12 Hydrostatic pressure = Mud weight x 0.05195 x Depth ...................................... (1.8) = 16.5 lb/gal x 0.05195 x 10,000 ft = 8,571.75 psi............................................ Answer

Solution for Problem 13 Fluid gradient for 12.5-lb/gal mud = 0.6493 psi/ft (from Table 4.1, Page 4-2) Hydrostatic pressure = Fluid gradient x Depth ................................................... (1.9) = 0.6493 psi/ft x 6,000 ft = 3,895.8 psi ............................................................. Answer

Solution for Problem 14 Height of fluid column = 6,000 ft - 1,500 ft = 4,500 ft Fluid gradient for 12.5-lb/gal fluid = 0.6493 psi/ft (from Table 4-2, Page 4-3) Hydrostatic pressure = Fluid gradient x Depth ................................................... (1.9) = 0.6493 psi/ft x 4,500 ft = 2,921.85 psi ........................................................... Answer

Solution for Problem 15 Fluid gradient for 9.0-lb/gal fluid = 0.4675 psi/ft (from Table 4.1, Page 4-2) Height of 9.0-lb/gal fluid = 1,000 ft Hydrostatic pressure of 9.0-lb/gal fluid = 0.4675 psi/ft x 1,000 ft = 467.5 psi Fluid gradient for 15.6-lb/gal cement = 0.8104 psi/ft (from Table 4.1, Page 4-2) Height of 15.6-lb/gal cement = 9,000 ft Hydrostatic pressure of 15.6-lb/gal cement = 0.8104 psi/ft x 9,000 ft = 7,293.6 psi Total hydrostatic = 467.5 psi + 7,293.6 psi = 7,761.1 psi ............................... Answer

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Solution for Problem 16 Fluid gradient for 9.0-lb/gal fluid = 0.4675 psi/ft (from Table 4.1, Page 4-2) Casing hydrostatic pressure = 10,000 ft x 0.4675 psi/ft = 4,675 psi Reversing pressure = Total hydrostatic pressure (from Problem 15) - Casing hydrostatic pressure = 7,761.1 psi - 4,675 psi = 3,086.1 psi................................................................ Answer

Solution for Problem 17 Mud weight = 104 lb/ft3 Depth = 8,000 ft Hydrostatic pressure = Mud weight x 0.006944 x Depth .................................. (1.11) = 104 lb/ft3 x 0.006944 x 8,000 ft = 5,777.408 psi ............................................ Answer

Solution for Problem 18 Fluid gradient for 80-lb/ft³ fluid = 0.5556 psi/ft (from Table 4.2, Page 4-3) Hydrostatic pressure = 0.5556 psi/ft x 9,000 ft = 5,000.4 psi ...................... Answer

Solution for Problem 19 Fluid gradient for 16.0-lb/gal cement = 0.8312 psi/ft (from Table 4.1, Page 4-2) Fluid gradient for 8.33-lb/gal water = 0.433 psi/ft (from Table 4.1, Page 4-2) Length of fluid column changed = 10,000 ft Change in hydrostatic pressure = (Fluid gradient - fluid gradient) x Length of fluid column changed ........... (1.13) = (0.831 psi/ft - 0.433 psi/ft) x 10,000 ft = 0.3982 psi/ft x 10,000 ft = 3,982 psi .............................................................. Answer

Solution for Problem 20 Fluid gradient for 10.0-lb/gal fluid = 0.5195 psi/ft (from Table 4.1, Page 4-2) Fluid gradient for 6.8 lb/gal oil = 0.3533 psi/ft (from Table 4.1, Page 4-2) Length of fluid column changed = 5,000 ft Change in hydrostatic pressure = (Fluid gradient - fluid gradient) x Length of fluid column changed ........... (1.13) = (0.5195 psi/ft - 0.3533 psi/ft) x 5,000 ft = 0.1662 psi/ft x 5,000 ft = 831 psi ................................................................... Answer

Solution for Problem 21 Fluid gradient for 14.0-lb/gal fluid = 0.7273 psi/ft (from Table 4.1, Page 4-2) Vertical depth = 1,000 ft + 8,000 ft + 1,000 ft = 10,000 ft Hydrostatic pressure = 0.7273 psi/ft x 10,000 ft = 7,273 psi ....................... Answer

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Solution for Problem 22 1. True vertical height of slant = 7,000 ft - 1,000 ft = 6,000 ft Slant factor = True vertical height of slant ÷ Length of slant ..................... (1.14) = 6,000 ft ÷ 8,000 ft = 0.750 2. Length of 9.0-lb/gal fluid in slant = 6,000 ft 3. Height of 9.0-lb/gal fluid = Slant factor x Length of 9.0 lb/gal in slant .. (1.16) = 0.750 ft x 6,000 ft = 4,500 ft 4. Fluid gradient for 9.0-lb/gal fluid = 0.4675 psi/ft (from Table 4.1, Page 4-2) Hydrostatic pressure of 9.0-lb/gal fluid = 0.4675 psi/ft x 4,500 ft = 2,103.75 psi Height of 15.0-lb/gal fluid = True vertical depth - Height of 9.0 lb/gal = 7,000 ft - 4,500 ft = 2,500 ft 5. Fluid gradient for 15.0-lb/gal fluid = 0.7792 psi/ft (from Table 4.1, Page 4-2) Hydrostatic pressure of 15.0-lb/gal fluid = 0.7792 psi/ft x 2,500 ft = 1,948 psi 6. Total hydrostatic pressure = 2,103.75 psi + 1,948 psi = 4,051.75 psi.... Answer

Solution for Problem 23 1. True vertical height of cement = 10,000 ft - 1,000 ft - 2,000 ft = 7,000 ft Slant factor = True vertical height of slant ÷ Length of slant .................... (1.14) = 7,000 ft ÷ 12,000 ft = 0.5833 2. Length of cement column in slant = Total cement length - Length of straight hole with cement = 5,000 ft - 2,000 ft = 3,000 ft 3. Height of cement in the slant = Slack factor x Length of cement in slant ................................................... (1.16) = 0.5833 x 3000 ft = 1,750 ft 4. Total cement height = 1,750 + 2,000 ft = 3,750 ft Fluid gradient for 15.6-lb/gal fluid = 0.8104 psi/ft (from Table 4.1, Page 4-2) Hydrostatic pressure of cement = 0.8104 psi/ft x 3,750 ft = 3,039 psi 5. Mud height = 10,000 ft - 3,750 ft = 6,250 ft Fluid gradient for 18.0-lb/gal fluid = 0.9351 psi/ft (from Table 4.1, Page 4-2) Hydrostatic pressure of mud = 0.9351 psi/ft x 6,250 ft = 5,844.375 psi 6. Total hydrostatic pressure = 3,039 psi + 5,844.375 psi = 8,883.375 psi Answer

Solution for Problem 24 Capacity of 10,000 ft of drillpipe = 10,000 ft x 0.00658 bbl/ft = 65.8 bbl ... Answer

Solution for Problem 25 Capacity of 600 ft of 6-in. OD, 2 1/2-in. ID drill collars = 600 ft x 0.0061 bbl/ft = 3.66 bbl .................................................................... Answer

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Solution for Problem 26 Fill-up of 50 bbl of oil in 2 3/8-in., 4.7-lb/ft tubing = 50 bbl x 258.65 ft/bbl = 12,932.5 ft ................................................................ Answer

Solution for Problem 27 Fluid gradient for 8.33-lb/gal water = 0.433 psi/ft (from Table 4.1, Page 4-2) Fluid gradient for 14.0-lb/gal mud = 0.7273 psi/ft (from Table 4.1, Page 4-2) Fluid gradient for 16.0-lb/gal cement = 0.8312 psi/ft (from Table 4.1, Page 4-2) A. Tubing capacity = 10,000 ft x 0.00387 bbl/ft = 38.7 bbl ......................... Answer B. Maximum pump pressure with a full column of mud (total pumped = 0 bbl) Mud hydrostatic pressure = 10,000 ft x 0.7273 psi/ft = 7,273 psi Maximum pump pressure = 9,000 psi - 7,273 psi = 1,727 psi................ Answer C. Maximum pump pressure when starting to mix cement (total pumped = 10 bbl of water) 1. Fill-up of 10 bbl of water = 10 bbl x 258.65 ft/bbl = 2,586.5 ft 2. Hydrostatic pressure of 10 bbl of water = 2,586.5 ft x 0.433 psi/ft = 1,119.955 psi 3. Length of mud column = 10,000 ft - 2,586.5 ft = 7,413.5 ft 4. Hydrostatic pressure of mud = 7,413.5 ft x 0.7273 psi/ft = 5,391.839 psi 5. Total hydrostatic pressure = 1,119.955 psi + 5,391.839 psi = 6,511.794 psi 6. Maximum pump pressure = 9,000 psi - 6,511.794 psi = 2,488.206 psi ........................................................................................ Answer D. Maximum pump pressure when water ahead of cement reaches tool (total pumped = 38.7 bbl [10 bbl of water and 28.7 bbl of cement]) 1. Fill-up of 10 bbl of water = 2,586.5 ft (from Part C) 2. Hydrostatic pressure of 10 bbl of water = 1,119.955 psi (from Part C) The cement will more than fill the tubing. The amount of cement in the tubing is 3. Cement column length = 10,000 ft - 2,586.5 ft = 7,413.5 ft 4. Cement hydrostatic pressure = 7,413.5 ft x 0.8312 psi/ft = 6,162.101 psi 5. Total hydrostatic pressure = 1,119.955 psi + 6,162.101 psi = 7,282.0556 psi 6. Maximum pump pressure = 9,000 psi - 7,282.056 psi = 1,717.944 psi ........................................................................................ Answer

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Solution for Problem 27—continued E. Maximum pump pressure with full cement column (Total pumped = 48.7 bbl to 60 bbl; 10 bbl of water and 38.7 to 50 bbl of cement; 38.7 bbl of cement in the tubing) 1. Hydrostatic pressure of 10,000 ft of cement = 10,000 ft x 0.8312 psi/ft = 8,312 psi 2. Maximum pump pressure = 9,000 psi - 8,312 psi = 688 psi ............ Answer F. Maximum pump pressure when all water behind cement is in tubing (Total pumped = 70 bbl; 10 bbl of water, 50 bbl of cement, and 10 bbl of water; 28.7 bbl of cement and 10 bbl of water in the tubing) Maximum pump pressure = 1,717.944 psi ............................................... Answer (same as D-1 through D-6) G. Maximum pump pressure when water behind cement reaches the tool (Total pumped = 98.7 bbl; 10 bbl of water, 50 bbl of cement, 10 bbl of water, and 28.7 bbl of mud; 10 bbl of water and 28.7 bbl of mud in the tubing) Maximum pump pressure = 2,488.206 psi ............................................... Answer (same as C-1 through C-6) H. Maximum pump pressure when mud behind cement reaches the tool (Total pumped = 108.7 bbl; 10 bbl of water, 50 bbl of cement, 10 bbl of water, and 38.7 bbl of mud; 38.7 bbl of mud in the tubing) Maximum pump pressure = 1,727 psi ...................................................... Answer (same as B-1 and B-2)

Solution for Problem 28 A. Length = 1,700 ft WA = 15.50 lb/ft MW = 14.7 lb/gal WL = (WA) x [1 - (0.01528) x (MW)] .............................................................. (1.22) = 15.50 x [1 - (0.01528) x (14.7)] = 15.50 x (1 - 0.2246) = 15.50 x 0.775 = 12.018 lb/ft in 14.7-lb/gal mud Weight in 14.7-lb/gal mud = WL x Length = 12.018 lb/ft x 1,700 ft = 20,430.6 lbâ ..................................................... Answer B. Total weight in air = WA x Length = 15.50 lb/ft x 1,700 ft = 26,350 lbâ .................................................................................................. Answer

Solution for Problem 29 Buoyancy factor for 19.0-lb/gal mud = 0.7097 (from Table 4.3, Page 4-4) Pipe weight in 19.0 lb/gal mud = Buoyancy factor x Pipe weight in air ........ (1.25) = 0.7097 x 16.60 lb/ft= 11.78102 lb/ft Total pipe weight in fluid = Length x Pipe weight in fluid per foot = 7,000 ft x 11.78102 lb/ft = 82,467.14 lbâ ..................................................... Answer

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Solution for Problem 30 Buoyancy factor for 10.0-lb/gal fluid = 0.8472 (from Table 4.3, Page 4-4) Weight of 2 7/8-in. pipe in liquid = 0.8472 x 10.4 lb/ft = 8.811 lb/ft Actual weight of 1,000 ft of 2 7/8-in. pipe = 1,000 ft x 8.811 lb/ft = 8,811 lbâ Weight of 3 1/2-in. pipe in liquid = 0.8472 x 15.50 lb/ft = 13.132 lb/ft Actual weight of 1,000 ft of 3 1/2-in. pipe = 1,000 ft x 13.132 lb/ft = 13,132 lbâ Weight of 4 1/2-in. pipe in liquid = 16.60 lb/ft x 0.8472 = 14.064 lb/ft Actual weight of 1,000 ft of 4 1/2-in. pipe = 1,000 ft x 14.064 lb/ft = 14,064 lbâ Indicator reading = 8,811 lbâ + 13,132 lbâ + 14,064 lbâ â ........................................................................................................ Answer = 36,007 lbâ

Solution for Problem 31 Buoyancy factor for 10.0 lb/gal fluid = 0.8472 (from Table 4.3, Page 4-4) Weight of 2 7/8-in. pipe in liquid = 0.8472 x 10.40 lb/ft = 8.811 lb/ft Actual weight of 1,000 ft of 2 7/8-in. pipe = 1,000 ft x 8.811 lb/ft = 8,811 lbâ Weight of 3 1/2-in. pipe in liquid = 15.50 lb/ft x 0.847 = 13.132 lb/ft Actual weight of 3 1/2-in. pipe in liquid= 13.132 lb/ft x 500 ft = 6,566 lbâ Actual weight of 500 ft of 3 1/2-in. pipe in air = 15.50 lb/ft x 500 ft = 7,750 lbâ Actual weight of 1,000 ft of 4 1/2-in. pipe in air = 16.60 lb/ft x 1,000 ft = 16,600 lbâ Indicator Reading = 8,811 lbâ + 6,566 lbâ + 7,750 lbâ + 16,600 lbâ = 39,727 lbâ ........................................................................................................ Answer

Solution for Problem 32 Figure 1.30, Page 1-48 shows all the forces resulting from hydrostatic pressure. The dashed forces on the stinger cancel out, leaving only the forces effective from tubing OD to tubing ID. Therefore, the stinger has no effect on the indicator reading. ................. Answer

Solution for Problem 33 A. Buoyancy factor for 10.2 lb/gal mud = 0.8441 (from Table 4.3, Page 4-4) Pipe weight in liquid = 0.8441 x 4.7 lb/ft = 3.96727 lb/ft Indicator reading = 3.96727 lb/ft x 4,000 ft = 15,869.08 lbâ ................. Answer B. Tubing OD area = 4.430 in.2 (from Table 4.5, Page 4-6) Force from pump pressure = 4,000 psi x 4.430 in.² = 17,720 lbá ......... Answer Since the upward force resulting from pump pressure is greater than the pipe weight, 4,000 psi would pump the tubing out of the hole.

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Solution for Problem 34 Buoyancy factor for 8.33-lb/gal water = 0.8727 (from Table 4.3, Page 4-4) Pipe weight in fluid per foot = 16.60 lb/ft x 0.8727 = 14.487 lb/ft Indicator reading before pumping = 14.487 lb/ft x 1,000 ft = 14,468.82 lbâ 4 ½-in. drillpipe OD area = 15.904 in.² (from Table 4.6, Page 4-8) Maximum pump pressure = Pipe weight in fluid ÷ Tubing OD area ............. (1.26) = 14,486.82 lb ÷ 15.904 in.² = 910.892 psi ......................................................... Answer

Solution for Problem 35 A. Buoyancy factor for 8.33-lb/gal water = 0.8727 (from Table 4.3, Page 4-4) Pipe weight in 8.33-lb/gal water = 0.8727 x 1.80 lb/ft = 1.571 lb/ft Indicator reading when first on bottom = 1.57086 lb/ft x 1,000 ft â ................................................................................................... Answer = 1,570.86â B. Fluid gradient for 15.6-lb/gal cement = 0.8104 psi/ft (from Table 4.1, Page 4-2) Fluid gradient for 8.33-lb/gal water = 0.433 psi/ft (from Table 4.1, Page 4-2) Change in gradient = 0.8104 psi/ft - 0.433 psi/ft = 0.3774 psi/ft Length of column changed from 8.33-lb/gal to 15.6-lb/gal fluid = 1,000 ft Pressure required on casing = 1,000 ft x 0.3774 psi/ft = 377.4 psi ....... Answer C. Buoyancy factor for 15.6-lb/gal cement = 0.7616 (from Table 4.3, Page 4-4) Pipe weight in 15.6-lb/gal cement = 0.7616 x 1.80 lb/ft = 1.37088 lb/ft Indicator reading when first on bottom = 1.37088 lb/ft x 1,000 ft = 1370.88 lbâ 1.315-in. tubing OD area = 1.358 in.² (from Table 4.5, Page 4-6) Maximum allowable pressure = 1,370.88 lb ÷ 1.358 in.² = 1,009.485 psi .............................................................................................. Answer

Solution for Problem 36 A. Buoyancy factor for 18.0-lb/gal mud = 0.7249 (from Table 4.3, Page 4-4) Pipe weight in liquid = 0.7249 x 13.30 lb/ft = 9.641 lb/ft Original weight-indicator reading = 9.64117 lb/ft x 3,000 ft = 28,923.51 lbâ ............................................................................................. Answer B. Fluid gradient for 18.0-lb/gal mud = 0.9351 psi/ft (from Table 4.1, Page 4-2) Fluid gradient for 9.0 lb/gal acid = 0.4675 psi/ft (from Table 4.1, Page 4-2) Change in fluid gradient = 0.9351 psi/ft - 0.4675 psi/ft = 0.4676 psi/ft Drillpipe pressure = 0.4676 psi/ft x 3,000 ft = 1,402.8 psi ..................... Answer C. 3 ½-in. OD drillpipe ID area = 6.0 in.² (from Table 4.6, Page 4-8) Force caused by trapped drillpipe pressure = 1,402.8 psi (from Part B) x 6.0 in.² = 8,416.8 lbá Weight-indicator reading = 28,923.51 lbâ (from Part A) - 8,416.8 lbá = 20,506.71 lbâ ............................................................................................. Answer Solution for Problem 36 continued on Page 1-70 October 1996

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Solution for Problem 36 —continued D. 3 1/2-in. drillpipe OD area = 9.621 in.² (from Table 4.6, Page 4-8) Increase in drillpipe pressure = 20,506.71 lb (from Part C) ÷ 9.621 in.² = 2,131.453 psi Maximum drillpipe pressure = 1,402.8 psi (from Part B) + 2,131.453 psi = 3,534.253 psi .............................................................................................. Answer

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General Hydraulics Manual

October 1996