general hydraulics 01

Chapter

1
Introduction to Hydraulics
This chapter covers many basic hydraulic principles that you will need to under-stand before you use the General Hydraulics Manual. After a brief description ofhydraulics concepts, the following topics will be covered in this chapter alongwith the calculations for each one:

• Area • Differential pressure• Force • Pipe capacity• Pressure • Pipe fill-up• Effective area • Buoyancy• Hydrostatic pressure • Open-ended pipe hydraulics

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

Essential Hydraulics ConceptsHydraulics are the principles governing the power generated by the movementand force of liquid.

Hydraulics concepts are primarily an application of Pascal’s Law: if a fluid has aconstant density and the fluid is at rest, all points at the same depth below theliquid’s surface are under equal pressure. For example, studying a cross-sec-tional slice of a cylinder of water demonstrates that all points on the cross-sectional surface are under equal pressure. Familiarity with fluid pressures isimportant for understanding how oilfield tools work.

Force is another important concept in hydraulics. To calculate force, multiplypressure 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 calcula-tion is area, including cross-sectional and effective areas.

October 1996 Introduction to Hydraulics 1-1

Halliburton Note

This file has been created for Halliburton use only. All information contained in this publication is confidential and proprietary property of Halliburton Energy Services, a division of Halliburton Company.

Calculating Area

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

Calculating the Area of a CircleThe area of a circle equals pi (π) times the radius times the radius, as stated inEquation 1.2:

Area of a circle = πππππ 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 = πr2 .......................................................... (1.2)

= 3.141592654 x 1.25 in. x 1.25 in.= 4.9087 in. 2 ≈≈≈≈≈ 4.909 in. 2

NOTE In this manual, all areas are rounded to three decimal places.

Most downhole tools are measured by diameter rather than radius. Use Equation1.3 to calculate the area of a circle using the circle's diameter rather than itsradius. Since pi divided by 4 equals 0.7854, and radius times radius multipliedby 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.

Figure 1.1

2 1/2-in. Diameter

1-2 General Hydraulics Manual October 1996

Figure 1.2

4 3/8-in. Diameter

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

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 ex-pressed in square feet (ft2) or square centimeters (cm2), respectively. All ex-amples 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 answersto the “Solutions to Problems” on Page 1-61.


1-4

NO

Problem 1


Figure 1.3

Problem 2


Figure 1.4

Calculating Cross-Sectional AreaSometimes 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 calcu-lating 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 casingis 10 3/4 in.

When you know the OD and weight of a pipe, its ID can be found in severalsections in the Halliburton Cementing Tables manual including “Displacement,”“Dimensions and Strengths,” “Capacity,” and “Volume and Height.”

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

Work Space

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

Work Space

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

1 1/2-in. Diameter

2 7/8-in. Diameter

General Hydraulics Manual October 1996

Problem 3

What is the ID area of 10 3/4-in., 55.5-lb/ft casing? The ID of this casing is 9.760in. (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 pieceof 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 circlesand 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. x6.366 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

Problem 4

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

Work Space


Figure 1.5


Estimating Tensile Strength with the Cross-Sectional Area

Being able to calculate cross-sectional area is sometimes useful when an oilfieldhandbook is not available. For example, if you know the cross-sectional area, youcan quickly estimate the tensile strength of drillpipe, tubing, or casing withEquation 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 thegrade designation indicates the yield strength of the pipe in thousands ofpounds-per-square inch (psi) of cross-sectional area. For example, new N-80tubing or casing has a yield strength of 80,000 psi. Example 4 illustrates thecalculation for tensile strength.

Example 4: How to calculate tensile strength

What is the approximate tensile strength of the pipe body of2.875-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 Use these calculations carefully since no safety factor is included. Be aware thatthe threaded sections on nonupset tubing, casing, etc., have a thinner wall thanthe pipe body. Use the thinner section for calculating tensile strength since it willbe the weakest point.


Calculating Force and PressureThis section defines force and pressure and gives examples and problems forcalculating force in hydraulic cylinders.

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

Definition of PressureA liquid or gas exerts a force against any surface it contacts; the force per unitarea is defined as pressure. Pressure is usually expressed in pounds per squareinch (psi). Pascal’s Law states that pressure acts equally in all directions. In otherwords, pressure in a hydraulic cylinder is acting (or exerting a force) equally oneach 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 hydrauliccylinder or a hydraulic jack. The amount of force created by a hydraulic cylinderis 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 beaccurate. In this text, pressure is expressed in pounds per square inch (psi), andarea is expressed in square inches (in.2).

Figure 1.6 on Page 1-8 shows a hydraulic cylinder mounted on a solid wall. Thepiston rod acts against a set of scales that measures force.


1-8 Gener

Calculating Force in a Hydraulic Cylinder

Example 5: How to calculate force in a hydraulic cylinder

What force is created in thecylinder in Figure 1.6?

Solution

Piston OD area = 0.7854 x 3.0 in. x3.0 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

Problem 5

What force is exerted by the cylinder in Figure 1.7?

Work Space


Figure 1.6

Figure 1.7

al Hydraulics Manual October 1996

October 1996

Calculating Pressure in a Hydraulic CylinderPressure multiplied by area equals force; therefore, force divided by area equalspressure.

When force is applied to the rod on a hydraulic cylinder, pressure is created. Pullapplied to a hydraulic jar creates pressure in the hydraulic chamber of the jar. Tocalculate 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 the2.0-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

Introduction to Hydraulics 1-9

Problem 6

If the force in Figure 1.9 is 1,000 lb and the diameter of the piston is 3.0 in., whatis the pressure in the cylinder?

Work Space

Answer _______________(See Page 1-61 for thesolution.)

Calculating Effective Area

Definition of Effective AreaAs shown in Figure 1.10, Page 1-11, pressure applied to the piston’s back sidecannot act over the entire cross-sectional area of the piston because of the rodposition. 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 areaand the rod OD area. This difference is the effective area that the pressure worksagainst in a hydraulic cylinder.

Figure 1.9


October 1996

Example 7: How to calculate the effective area

What is the effective area of the pressure on the cylinder in Figure1.11?

Solution

Piston OD area = 0.7854 x 3.0in. 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

Problem 7

Applying 1,000 psi to the cylinder will exert what pull (force) on the scales inFigure 1.11?

Work Space


Figure 1.10

Figure 1.11


1-12 Gen

Problem 8

Figure 1.12 shows different pressuresacting on each side of the piston.

What force is registered on thescales? Is this force up or down?

Hint: Work this problem as if itwere two separate problems. Oneforce is upward and the other force isdownward. Subtract the smallerforce from the larger force. Theremaining force on the scales will bein the same direction as the largerforce.

Work Space


Problem 9

Assume both pressures in Figure1.13 are 1,000 psi and that thecylinder is the one from Problem 8.What is the force, and in whichdirection does it act?

Work Space

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

Figure 1.13

Figure 1.12

eral Hydraulics Manual October 1996

October 1996

Calculating Differential Areas and DifferentialPressureThe Solution and Alternate Solution (Page 1-62) of Problem 9 (Page 1-12) showthat you can eliminate calculation time by canceling opposing forces. Situationswith equal pressures acting on different areas, as in Problem 9, are normallycalled differential areas. If equal pressures act on different areas, you can cancelthe opposing forces in the equation. Example 8 demonstrates differentialpressure 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 forceact in an upward or downward direction?

SolutionWorking with top side ofpiston:Piston area = 0.7854 x 4 in. x4 in. = 12.566 in.²Rod area = 0.7854 x 1.5 in. x1.5 in. = 1.767 in.²Effective area= Piston area - Rod area= 12.566 in.² - 1.767 in.²= 10.799 in.²Pressure = 1,500 psi

Force = 1,500 psi x 10.799in.² = 16,198.5 lbâWorking with bottom side ofpiston:Effective area = 10.799 in.²(calculated above)Pressure = 2,000 psiForce = 2,000 psi x 10.799 in.² = 21,598 lbáNet force = 21,598 lbá - 16,198.5 lbâ = 5,399.5 lbá

Alternate SolutionPressure on bottom side = 2,000 psiPressure on top side = 1,500 psi

Differential pressure = Pressure on bottom side - Pressure on top side= 500 psiEffective area = 10.799 in.² (from Solution)Net force = 500 psi x 10.799 in.² = 5,399.5 lbá

Figure 1.14


1-14 Ge

Example 8 demonstrates how you can save calculation time by cancelingpressures or forces. Equal amounts of opposing pressures can be canceled if the areasare equal.

Since the pressures in Figure 1.14, Page 1-13 are acting on equal areas, 1,500 psiof 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 isthe differential pressure.

Problem 10

What force is the conventional hydraulic cylinder in Figure 1.15 exerting on thescale?

Work Space


Figure 1.15

neral Hydraulics Manual October 1996

Problem 11

While the cylinder illustrated in Figure 1.15, Page 1-14 is being used, the rodbreaks. Since the piston and rod on this cylinder are one solid piece, both arereplaced. The new piston is not machined correctly. Instead of having a diameterof 3 3/4 in., the new piston end has a 3 5 /8-in. diameter. The new rod end has a1 5/8-in. diameter, as specified. What effect, if any, do the new piston and rodhave on the system?

Work Space


Calculating Hydrostatic PressureThis section introduces the concepts necessary to convert mud weight tohydrostatic pressure.

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

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

Converting Mud Weight to Hydrostatic PressureTo convert mud weight to hydrostatic pressure, multiply mud weight by aconstant, 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)


Example 9: How to calculate hydrostatic pressure from mud weight

What is the hydrostatic pressure of a column of 9.6-lb/gal mud at6,450 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


Calculating Hydrostatic Pressure from FluidGradientsMultiplying 0.05195 by the mud weight (lb/gal) yields a fluid gradient—hydrostatic pressure in psi per foot of depth (psi/ft). These fluid gradientssimplify hydrostatic pressure calculations.

Fluid gradients for a range of fluid weights have been calculated and can befound in Table 4.1, Page 4-2 (lb/gal) and Table 4.2, Page 4-3 (lb/ft3) of thismanual (or at the back of the Halliburton Cementing Tables in the “HydrostaticPressure and Fluid Weight Conversion Tables”). Chapter 4 of this manualcontains this table and many others for reference.

NOTE When using oilfield handbooks to look up fluid gradients, check the unit ofmeasurement because many handbooks express fluid gradients in psi per hun-dred feet.

To use Table 4.1, find the mud weight in the left column of the table, then readthe fluid gradient in the right column. For example, the fluid gradient for12.4-lb/gal mud is 0.6442 psi/ft of depth. To change the fluid gradient tohydrostatic pressure, multiply by the depth as in Equation 1.9.

Hydrostatic pressure = Fluid gradient x Depth .................................................. (1.9)


October 1996


Example 10: How to calculate hydrostatic pressure using fluid gradi-ents

Using Equation 1.9, what is the hydrostatic pressure at 12,650 ft in awell 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


Problem 14

A drillstem test has just been run on the well in Problem 13. As equipment isremoved from the hole, the hole is not kept full. The fluid level in the well drops1,500 ft. What is the hydrostatic pressure at the 6,000-ft depth now?

Work Space



1-18 Gen

Calculating Hydrostatic Pressure with DifferentFluids in a Fluid ColumnOften, several different fluids are present in a fluid column. To calculate totalhydrostatic pressure, calculate the hydrostatic pressure for each fluid, then addthese hydrostatic pressures together.

Problem 15

The 10,000-ft tubing string inFigure 1.16 contains 9,000 ft of15.6-lb/gal cement; theremainder of the tubingcontains 9.0-lb/gal water.What is the total hydrostaticpressure at the bottom of thetubing?

Work Space


It is possible for parallel columns of different-weight fluids to exist. Thissituation occurs when cement is spotted in or reversed out of tubing. Fluids ofdifferent weights in parallel columns have different hydrostatic pressures. Thesedifferences can be equalized when pressure is trapped or applied to the lighter-weight fluid.

To calculate the amount of pressure needed to equalize the hydrostatic pressureof two fluid columns, find the difference between the hydrostatic pressures. Formore information about this type of calculation, see “Calculating Changes inHydrostatic Pressure,” Page 1-21 and “Calculating Weight-Indicator Readingswhen Spotting Fluids,” Page 1-54.

Figure 1.16


October 1996

Problem 16

If the well described in Problem 15 has 9.0-lb/gal water on the outside of thetubing, how much surface pressure is needed to reverse the cement out of thetubing?

Hint: The pressure needed to reverse the cement out of the tubing equals thedifference in the hydrostatic pressures of fluids in the tubing and casing(without accounting for friction).

Work Space


If mud weight is expressed in pounds per cubic foot (lb/ft3), calculate the fluidgradient by dividing the mud weight by 144. Then, to obtain hydrostaticpressure, multiply this fluid gradient by the depth in feet.

When mud weight is given in lb/ft3, use Equation 1.10 to find hydrostaticpressure.

Hydrostatic pressure = x Depth in ft ........................ (1.10)

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 usingEquation 1.11

Using Equation 1.11, what is the hydrostatic pressure at 6,000 ft in awell 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

Mud weight in lb/ft³ 144


1-20 Gen

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


Use Table 4.2, Page 4-3 to eliminate part of Equation 1.11. The numbers in Table4.2 reflect the mud weight (in lb/ft³) already multiplied by 0.006944 (or dividedby 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 thefluid gradient in the right column. For example, the fluid gradient for 100-lb/ft³mud is 0.6944 psi/ft. To translate this psi/ft to hydrostatic pressure, multiplythe fluid gradient by the depth.

Example 12: How to calculate hydrostatic pressure

What is the hydrostatic pressure at 10,000 ft in a well containing120-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


October 1996

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


Calculating Changes in Hydrostatic PressureThere are at least two ways to calculate a change in hydrostatic pressure; onemethod is shorter than the other one.

To determine changes in hydrostatic pressures, calculate the hydrostaticpressure before and after the change, and find the difference.

Long Method

Change in hydrostatic pressure= Hydrostatic pressure after - Hydrostaticpressure before ........................................................................................................ (1.12)

Find the fluid gradients in Table 4.1, Page 4-2 (for lb/gal fluid weights) orTable 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 multiplythis difference by the length of fluid column changed. Equation 1.13 expressesthis 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 hydrostaticpressure 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 onlypart of the fluid column changes.


1-22 Gene

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

If the 8.33-lb/gal water in a 10,000-ft well is replaced with 10.0-lb/galsalt 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 - Hydrostaticpressure 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

ral Hydraulics Manual October 1996

October 1996

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

With 10,000 ft of tubing in the hole, enough 15.6-lb/gal cement isspotted to the bottom to fill 5,000 ft of tubing. The original fluid in thehole is 10.0-lb/gal brine water. What is the change in hydrostaticpressure?

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/galfluid is replaced with a 16.0-lb/gal fluid?

Work Space



1-24 Gene

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 thetubing has been replaced with 42° API oil weighing 6.8 lb/gal?

Work Space


Hydrostatic Pressure in Directionally Drilled HolesThis part of Chapter 1 explains how to calculate hydrostatic pressure indirectionally drilled holes.

In directionally drilled holes, actual vertical depth may vary greatly from drilleddepth. In a directional hole, only the vertical depth is used to calculatehydrostatic pressure—regardless of the path taken to reach this depth, the truevertical height of the fluid column is equal to the vertical depth. Example 15shows that the vertical depth measurement provides accurate results.

Example 15: How to calculate hydrostatic pressure in a directionallydrilled hole

Figure 1.17 illustrates an angled 30-ft joint of tubing in a lake. The bottom ofthe 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 freshwater = 0.433 psi/ft(from Table 4.1, Page4-2)

Hydrostatic pressure= 0.433 psi/ft x 20 ft= 8.66 psi

Figure 1.17


October 1996

Figure 1.17, Page 1-24 shows that the hydrostatic pressure at the bottom of thetubing is equal to the hydrostatic pressure in 20 ft of water. In Example 15, if the30-ft length of the tubing had been used for the calculation instead of the 20-ftactual vertical depth, the hydrostatic pressure would have been 12.99 psi—50%too high.

In the oilfield, only vertical depth and mud weight influence hydrostaticpressure. Hole volume and diameter do not influence hydrostatic pressure, asdemonstrated by Figure 1.18.

Figure 1.18 shows four pieces of pipe with lengths of 1,000 ft and diameters of1 ft. Pipes A, B, and C contain fresh water, and pipe D contains mud. Since allfour pipes contain equal volumes of fluid, any difference in hydrostatic pressureis not influenced by volume.

First, compare the hydrostatic pressures of pipes B and C, and note theimportance of calculating hydrostatic pressure based on true vertical depthrather than total depth. Although the two pipes are the same length, the truevertical depth of pipe C is half that of pipe B. Therefore, the hydrostatic pressureof pipe C is half that of pipe B.

Notice that pipe A has the lowest hydrostatic pressure because of its shallowvertical depth.

Also compare pipes B and D, and note how mud weight greatly influenceshydrostatic pressure.

Figure 1.18


1-26 Gener

Example 16: How to calculate hydrostatic pressure in a directionallydrilled well

Figure 1.19 is a schematic of a directionally drilled well that was drilledstraight down for 1,500 ft, kicked out (off vertical) for 10,000 ft, and thenstraightened up for the last 2,000 ft. The actual vertical depth of the well is8,000 ft. If the fluid in the hole is 15.0-lb/gal mud, what is the hydro-static 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 higherpressure than the calculation based on actual vertical depth.

Figure 1.19


October 1996

Problem 21

What is the hydrostatic pressure at TD for the well illustrated in Figure 1.20?

Work Space


Figure 1.20


1-28 Gene

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 fluidgradient

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 + Hydrostaticpressure of cement

Example 17: How to calculate hydrostatic pressure in a directionallydrilled well (slant factor considered)

Figure 1.21, Page 1-29 illustrates a 10,000-ft TD hole. Since the hole isdirectionally drilled, the actual vertical depth is only 8,000 ft. The bottom4,500 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 hydrostaticpressure by making a schematic as shown in Figure 1.21. Then proceedwith 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 thelength of cement in the slant by the same factor. To find the vertical heightof the cement in the slant, use Equation 1.16.

Solution for Example 17 continued on Page 1-29


October 1996

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


1-30 Gener

Problem 22

Figure 1.22 illustrates the conditions for a 9,000-ft TD hole. The vertical depth is7,000 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


Figure 1.22


October 1996

Problem 23

Figure 1.23 shows a well with a 10,000-ft vertical depth. Drilled depth (TD) is15,000 ft. If enough 15.6-lb/gal cement to fill 5,000 ft of tubing is spotted to thebottom, what is the hydrostatic pressure at TD if the cement is displaced with18.0-lb/gal mud?

Work Space


Figure 1.23


Measurements for Directionally Drilled Holes

Following is a list of terms that refer to various measurements for directionallydrilled holes. Be familiar with these terms, and correct all data for calculationsaccordingly.

• VSS (vertical subsea)—VSS refers to the vertical depth of the well measuredfrom sea level. Geologists generally use this concept to identify and correlateformations.

• VSF (vertical depth from surface flange)—Many oil companies currentlymeasure vertical depth from the surface flange since the flange will be apermanent part of the completion.

• MSF (measured depth from surface flange)—MSF is the total drilled depthmeasured from the surface flange.

• MKB (measured depth from kelly bushing)—Some companies use the kellybushing as a reference point. These companies use MKB to abbreviate totaldepth measured from the bushing.

• VKB (vertical depth from kelly bushing)—VKB refers to vertical depthmeasured from the kelly bushing.

Calculating Pipe Capacity

Definition of Pipe CapacityPipe capacity is the volume of fluid or cement required to fill a specified length ofpipe. Capacity, therefore, depends upon the pipe’s ID and length in each situa-tion. Since pipe ID is determined by pipe weight, capacity is also determined bypipe weight.

The Halliburton Cementing Tables list capacities of most commonly used pipes inseveral units of measure including bbl/ft, gal/ft, and ft³/ft. Since the mostcommon measure of capacity and displacement volumes is barrels, most ex-amples in this manual use barrels (bbl).

Using Capacity Factors to Calculate Pipe CapacityA capacity factor allows you to calculate easily the amount of fluid required tofill 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.


October 1996

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 inbarrels?

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 FactorsHere are the equations used to calculate pipe capacity factors. The capacityfactors 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 doesnot list a factor for the specific type of pipe on the job.

πππππ ID² (gal)

77 (ft)

πππππ ID² (bbl)

3,234 (ft)

πππππ ID² (ft³)

576 (ft)

whereπ = 3.141592654ID = 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? Thecapacity factor for this drillpipe is 0.00658 bbl/ft (from the Halliburton CementingTables).

Work Space


Capacity factor of pipe in gal/ft = ............................................... (1.18)

Capacity factor of pipe in bbl/ft = ............................................... (1.19)

Capacity factor of pipe in ft³/ft = ............................................... (1.20)


Problem 25

What is the capacity in barrels of 600 ft of 6.0-in. OD, 2 1/2-in. ID drill collars? Thecapacity factor for these drill collars is 0.0061 bbl/ft (from the HalliburtonCementing Tables).

Work Space


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

Calculating Pipe Fill-Up

Definition of Pipe Fill-UpFill-up is the length of pipe that a given volume of fluid will fill. Fill-up factorssimplify fill-up calculations and may be found in the “Capacity” section of theHalliburton 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 FactorsTo calculate fill-up, multiply the fill-up factor (in ft/bbl) by volume (in bbl) toobtain the length of pipe filled (in ft). If volumes are measured in gallons orcubic feet, multiply the appropriate fill-up factor by the volume to yield thelength 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).



October 1996

Example 19: How to calculate fill-up

How many feet of 2 7/8-in., 10.4-lb/ft internal upset drillpipe will 60 bblof cement fill?

Solution

Fill-up factor for 2 7/8-in., 10.4-lb/ft drillpipe = 222.49 ft/bbl (from theHalliburton 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 factorfor this tubing is 258.65 ft/bbl (from the Halliburton Cementing Tables).

Work Space



1-36 Gener

Problem 27

On a squeeze job, the bottomhole pressure is limited to 9,000 psi. Conditions areas 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) followingthe water with 50 bbl of cement, (3) placing 10 bbl of water, and (4) displacingthe 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


October 1996

Problem 27—continued

D. What is the maximum pump pressure when the water ahead of the cementreaches 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 cementis in the tubing?

Work Space

Answer _______________

G. What is the maximum pump pressure when the water behind the cementreaches the tool?

Work Space

Answer _______________

H. What is the maximum pump pressure when the mud behind the cementreaches 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.)


Calculating Buoyancy

Definition of BuoyancyThe dictionary defines buoyancy as the power of a fluid to exert an upward forceon a body placed in it. Fluid tries to float anything placed in it. In oilfield appli-cations, buoyancy is the force that causes even very heavy items, such asdrillpipe 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 displacedwhen pipe is run in the hole. Normally, weights for tubing and drillpipe aregiven in pounds per foot (lb/ft), as weighed on dry land (or air).

This section of Chapter 1 provides examples and problems for calculatingweights of both steel and aluminum pipe in liquid. Later, two methods ofcalculating the weight of pipe in fluid are described: the buoyancy-factormethod and the area/hydrostatic-pressure method.

Calculating Weight in Fluid for Steel PipeWhen mud weight is expressed in pounds per gallon (lb/gal), the weight of open-ended 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 inpounds 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


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 in12.3-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 mudwas calculated to be 3.8167 lb/ft. For the 1,000 ft of this tubing, the weightindicator 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 asTable 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 ofopen-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 Equation 1.23 applies only to steel pipe. Mud weight must be expressed in poundsper cubic foot.


Calculating Weight in Fluid for Aluminum PipeCalculate the weight of open-ended aluminum pipe suspended in fluid by usingEquation 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 inpounds per gallon.

Buoyancy FactorsTable 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)] fromEquation 1.22, Page 1-38. To use Table 4.3, find the mud weight in the leftcolumn, and read the buoyancy factor in the right column.

Table 4.4, Page 4-5 lists buoyancy factors when mud weight is expressed inlb/ft.³ Buoyancy factors for aluminum drillpipe are not provided because theyare 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)


October 1996

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 stepas 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


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


1-42 Gene

Buoyancy-Factor MethodOften in the oilfield, the hole is not completely full of fluid. Pipe weight can becalculated with the buoyancy factor for only the portion of pipe that is suspendedin fluid. The remainder of the pipe's weight in air must be accounted for. Aftercalculating the weights of the portion of pipe in fluid and the portion of pipe inair, add the two results together for the total pipe weight in the hole.

Example 23: How to calculate pipe weight in fluid with the buoyancy-factor method

Figure 1.24 shows a well filled to 4,000 ft with 8.33-lb/gal water. Then10,000 ft of 2 3/8-in., 4.7-lb/ft tubing is run in the well. What is the weight-indicator 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

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 MethodAnother way to calculate the weight-indicator reading for pipe in fluid is thearea/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 thelower 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 thebuoyant force from the pipe weight in air yields the weight-indicator reading orstring weight.

Figure 1.24


October 1996

To find the string weight using the area/hydrostatic-pressure method, followthese 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 1and 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 3/8-in., 4.7-lb/ft tubing is suspendedin 12.3-lb/gal mud. What is theweight-indicator reading?

Solution

Fluid gradient for12.3-lb/gal mud= 0.6390 psi/ft (fromTable 4.1, Page 4-2)

Hydrostatic pressure= 0.6390 psi/ft x 1,000 ft= 639 psi

Tubing OD area= 4.430 in.² (fromTable 4.5, Page 4-6)

Tubing ID area = 3.126 in.² (fromTable 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 lbáPipe 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

Figure 1.25


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

Calculating Weight-Indicator Readings for Mixed-Pipe StringsA mixed-pipe string is made up of two or more sizes or weights of pipe. Whenfluid is in the hole, a mixed string’s weight can be calculated with either of thetwo 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 onbottom. Figure 1.27, Page 1-45, shows where hydrostatic pressures are acting onthe string in Figure 1.26.

Looking only at the 3 1/2-in. drillpipe in Figure 1.27, the hydrostatic pressure at1,500 ft is pushing down across the pipe wall thickness. Since the areas at the topand bottom of the 3 1/2-in. pipe are equal, use differential pressures to calculateforce.

Examining the joint between the two sizes of pipe in Figure 1.27 reveals that theonly downward force not accounted for is the area between the 3 1/2-in. pipe ODand the 4 1/2-in. pipe ID. Since the hydrostatic pressures inside and outside are

Figure 1.26


October 1996

Figure 1.27

equal, the downward force from the 3 1/2-in. pipe OD to the 4 1/2-in. pipe ID iscanceled by an equal upward force. The only remaining force is the upwardforce 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, followthese steps:

1. Multiply the fluid’s buoyancy factor by the pipe weight in air (lb/ft) toobtain the pipe’s weight in liquid.

2. For each size of pipe, multiply the pipe’s weight in liquid by the length ofthat 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.


1-46 Gene

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

What is the weight-indicator reading for the mixed string in Figure1.26 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 mixedstring 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



October 1996

Problem 31

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

Work Space


Example 26: How to calculate the effect of a shoe on the weight-indicator reading

How does adding a shoe on the bottom of the tubing in Figure 1.28affect the weight-indicator reading?

Solution

Figure 1.29 (Page1-48) shows thehydraulic forcesacting on the shoe.Notice the downwardforce fromhydrostatic pressureacting across thearea from the shoeOD to the tubing OD.Note the equalupward force on theshoe bottom. Theseforces cancel eachother, leaving onlythe upward force toact across the areafrom the tubing ODto the tubing ID. InExample 24 onPage 1-43, this upward force (hydrostatic pressure acting from tubing ODto tubing ID) was defined as buoyancy; therefore, the shoe will have noeffect on the weight-indicator reading.

Figure 1.28


1-48 Genera

Figure 1.29

Problem 32

Does the 2 1/8-in. OD, 1.0-in ID stinger in Figure 1.30 affect the weight-indicatorreading?


Figure 1.30

l Hydraulics Manual October 1996

October 1996

Other Factors Influencing Weight Calculations

All weight-indicator calculations presented have assumed the same fluid wasinside and outside the tubing. However, when tubing is run in dry (no fluidinside), other forces must be considered. Hydrostatic pressure changes when drytubing, such as bull-plugged tubing, a string of testing tools, or a retrievablepacker with a closed tubing valve, is run in.

Figure 1.31 illustrates bull-plugged tubing. Notice that with the bottom of thetubing bull-plugged, hydrostatic pressure works across the entire tubing ODarea. Example 27 shows the calculation of the weight-indicator reading.

Example 27: How to calculate the indicator reading in bull-pluggedtubing

Figure 1.31 shows 1,000 ft of 2 3/8-in., 4.7-lb/ft bull-plugged tubing run in8.33-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 for8.33 lb/gal fluid= 0.433 psi/ft (fromTable 4.1, Page 4-2)

Hydrostatic pressure at1,000 ft = 0.433 psi/ft x1,000 ft = 433 psi

Upward force on tubing OD= 4.430 in.² x 433 psi =1,918.19 lbá

Indicator reading = 4,700 lbâ - 1,918.19 lbá = 2,781.81 lbâââââ

Figure 1.31


Calculating Hydraulics of Open-Ended PipeYou can apply the principles of area, pressure, and buoyancy to downholeconditions when pumping through open-ended tubing.

Calculating Surface PressuresWhen pipe rams are closed around tubing, the casing becomes a large hydrauliccylinder, and the tubing acts as a piston. Applying pump pressure to the systemcan move the piston (tubing) upward. To actually move the piston, the weight ofthe piston itself (pipe weight minus the buoyant forces) must be overcome. Inother words, to pump tubing out of the hole, pump pressure must create anupward force equal to or greater than the weight-indicator reading with thetubing in fluid.

For these calculations, assume that there is no friction between the pipe andrams. Friction varies widely depending on factors such as the types of rams usedand the condition of the pipe.

Since pressure acts equally in all directions, any surface pressure acts at thebottom of the tubing, across the area from tubing OD to tubing ID. Pressure alsoacts at the top of the tubing across the tubing ID. The effective area is equal tothe tubing OD. Figure 1.32 illustrates this force. The upward force caused by thesurface pressure that acts on open-ended pipe is measured on the weightindicator 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


October 1996

drillpipe and tubing sizes. To use Table 4.5, locate the pipe OD in the left col-umn. Next, find pipe weight in the second column, and move horizontallyacross the columns. The third column is the OD area in square inches. Thefourth column gives ID in inches, and the fifth column is ID area in squareinches. The two right columns repeat the weight and ODs.

Example 28 shows the calculation of an indicator reading during pumpingoperations.

Example 28: How to calculate weight-indicator readings for open-ended tubing during pumping

Figure 1.33 shows 1,000 ft of 2 3/8-in., 4.7-lb/ft of open-ended tubinghanging in fresh water.

If 500-psi surface pressure is applied, what does the weight indicatorread?

Solution

Buoyancy factor for freshwater = 0.8727 (fromTable 4.3, Page 4-4)

Pipe weight in liquid= 0.8727 x 4.7 lb/ft= 4.10169 lb/ft

Indicator reading beforepumping = 1,000 ft x4.10169 lb/ft= 4,101.69 lbâ

Tubing OD area = 4.430 in.2

(from Table 4.5, Page 4-6)

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

Figure 1.33


1-52 Gene

Figure 1.34

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-indicatorreading before the pump isstarted?

Work Space

Answer ______________

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 PressuresThe maximum allowable surface pressure (without pumping the pipe out of thehole) can be calculated as the pressure required to create a force equal to the pipeweight in liquid (weight accounting for buoyancy). This pressure is equal tofloating pipe weight divided by tubing OD area (Equation 1.26).

Maximum allowable surface pressure = ................ (1.26)


Pipe weight in liquidTubing OD area


October 1996

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-endedtubing hanging in fresh water. Use Equation 1.26 to find the maximumpump pressure that can be applied without pumping the tubing out ofthe hole.

Solution

Buoyancy factor for 8.33-lb/galfluid = 0.8727 (from Table 4.3,Page 4-4)Pipe weight in liquid = 4.7 lb/ftx 0.8727 = 4.10169 lb/ft

Available weight = 4.10169 lb/ftx 1,000 ft = 4101.69 lbâ

Tubing OD area = 4.430 in.2 (fromTable 4.5, Page 4-6)

Maximum pump pressure= 4,101.69 lb ÷ 4.430 in.2

= 925.890 psi

Problem 34

Figure 1.36 shows 1,000 ft of 4 1/2-in., 16.60-lb/ft of open-ended drillpipe hangingin fresh water. What is the maximum pump pressure that can be applied withoutpumping the drillpipe out of the hole?

Work Space


Figure 1.35

Figure 1.36


1-54 Gener

Calculating Weight-Indicator Readings whenSpotting FluidsSometimes a heavier fluid, such as heavy mud or cement, is pumped downtubing or drillpipe while the original, lighter fluid remains between the pipe andcasing. In such a case, it is necessary to hold surface pressure on the casing sideto help prevent the heavier inside fluid from “U-tubing” or equalizing.

By holding surface pressure on the casing, total pressure at bottom (on the casingside) is made equal to the hydrostatic pressure of the heavier fluid at bottom. Apressure 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 thetubing with a heavier fluid and holding pressure on the casing. The weightindicator will now show the pipe weight to be the same as if it were suspendedwith heavier fluid both inside and outside of the drillpipe or tubing. Example 30shows the relevant calculations.

Example 30: Calculating weight-indicator readings when spottingheavy fluids

Figure 1.37 illustrates a well with 1,000 ft of 2 3/8-in., 4.7-lb/ft of open-endedtubing hanging in fresh water. After the rams are closed, a full column of15.6-lb/gal cement is spotted in the tubing, as shown in Figure 1.38. Thecasing valve is pinched down until enough pressure is trapped on thecasing to hold the cement in place in the tubing.

A. What is the weight-indicator reading beforethe cement is placed inthe tubing?

B. How much pressureshould be trapped on thecasing to hold the cementin place?

C. What is the weight-indicator reading whilethe cement is held inplace?

D. How much surfacepressure can operatorsapply to the tubing to putaway the cement withoutpumping the tubing out ofthe hole?

Example 30 continued on Page 1-55

Figure 1.37


October 1996

Example 30—continued

Solution

A. Buoyancy factor for8.33-lb/gal fluid = 0.8727(from Table 4.3, Page 4-4)

Pipe weight in 8.33-lb/galfluid = 4.7 lb/ft x 0.8727= 4.10169 lb/ft

Indicator reading =4.10169 lb/ft x 1,000 ft= 4,101.69 lbâââââ

B. Fluid gradient for15.6-lb/gal cement)= 0.8104 psi/ft (fromTable 4.2, Page 4-3)

Fluid gradient for8.33-lb/gal water =0.433 psi/ft (fromTable 4.2, Page 4-3)

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

Indicator reading = 1,000 ft x 3.57952 lb/ft = 3,579.52 lbâââââ

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

Figure 1.38


Problem 35

Figure 1.39 illustrates a well with 1,000ft of 1.315-in. OD, 1.80-lb/ft of open-ended tubing hanging in fresh water.After the rams are closed and a fullcolumn of 15.6-lb/gal cement is spot-ted, the casing valve is pinched downto hold the cement in place in thetubing.

A. What does the weightindicator show beforethe cement is pumped (whenfirst on bottom)?

Work Space

Answer _______________

B. How much pressure should be trapped on the casing to hold the cement inplace?

Work Space

Answer _______________

C. How much surface pressure can you apply to tubing containing a full col-umn of cement without pumping the tubing out of the hole?

Work Space

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

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

Figure 1.39


October 1996

Spotting a light fluid into a heavy mud system results in somewhat differentconditions than spotting the heavier fluids discussed. For example, spotting lightfluids occurs when acid or a completion fluid is pumped into a formation or acushion is pumped for a flow test. Example 31 shows calculations for spottinglight fluids.

Example 31: Calculating weight-indicator readings when spottinglight fluids

Figure 1.40 illustrates a well with 1,000 ft of 2 3/8-in., 4.7-lb/ft open-endedtubing hanging in 16.0-lb/gal mud.

A. What does the weight indicator read before pumping (when first onbottom)?

B. If tubing is spotted with fresh water until full, what is the tubingpressure if the pumps are shut down?

C. What does the indicator read with the pumps shut down and thetubing valveclosed?

D. What is themaximum pressure(total gauge) thatyou can apply todisplace the waterwithout pumpingtubing out of thehole?

Solution

A. Buoyancy factor for16.0-lb/gal mud= 0.7555 (Table 4.3,Page 4-4 )

Pipe weight in liquid= 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


Figure 1.40


1-58 Gener

Figure 1.41



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/ftChange 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 isthe hydrostatic pressure of the 16.0-lb/gal mud. Pressure at bottominside the tubing is the hydrostatic pressure of fresh water plus 398.2psi—equal to the hydrostatic pressure of 16.0-lb/gal mud. When theoriginal weight-indicator reading was calculated (in Part A), total pressure at bottom was the hydrostatic pressure of 16.0-lb/gal mud; therefore, the buoyant force on bottom was accounted for in the originalcalculation with the buoyancy factor for the 16.0-lb/gal mud. The onlyforce not accounted for is the trapped tubing pressure pushing up acrossthe 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 bytrapped tubing pressure = 3,550.85 lbâ (from Part A) - 1,244.773 lbá= 2,306.077 lbâ

D. 2,306 lb is the weightavailable to counter-act displacementpressure. 398.2 psialready on the tubingwas accounted forduring availableweight calculations;the displacementpressure calculated(based on this weight)represents the maxi-mum increase intubing pressure. So,the total maximumtubing gauge pres-sure is the trapped pressure plus the calculated pressure increase. Applied tubing pressure (pump pressure) acts across the tubing OD area asshown in Figure 1.41.


October 1996

Figure 1.42


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 condi-tions for a well with 3,000 ft of3 1/2-in., 13.30-lb/ft drillpipehanging open-ended in18.0-lb/gal mud.

A. What does the indicator readwhen first on bottom?

Work Space

Answer _______________Problem 36 continued on Page 1-60


1-60 Gene

Problem 36—continued

B. If the drillpipe is spotted with 9.0-lb/gal acid until full, what is the drillpipepressure if the pumps are shut down?

Work Space

Answer _______________

C. What does the indicator read when the pumps are shut down and thedrillpipe valve is closed?

Work Space

Answer _______________

D. What maximum pressure (total gauge) can be applied to displace acid with-out pumping the drillpipe out of the hole?

Work Space

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


Solutions to ProblemsPages 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 maygive 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 notused in this text.

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

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

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

Solution for Problem 4Tubing OD area = 0.7854 x 2.375 in. x 2.375 in. = 4.430 in.²Tubing 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.²Pressure = 3,000 psi

Force = Pressure x Area ............................................................................................ (1.1)= 3,000 psi x 9.621 in.² = 28,863 lbâââââ ................................................................. Answer

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


1-62 Gene

Solution for Problem 7Area = 5.302 in.² (from Example 7)

Pressure = 1,000 psiForce = Pressure x Area ............................................................................................ (1.1)= 1,000 psi x 5.302 in.² = 5,302 lbá ................................................................... Answer

Solution for Problem 8Top side of the cylinder

Piston OD area = 0.7854 x 4 in. x 4 in. = 12.566 in.²Pressure = 500 psiForce = 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.²Rod area = 0.7854 x 1 in. x 1 in. = 0.785 in.²Effective area = Piston area - Rod area= 12.566 in.² - 0.785 in.² = 11.781 in.²Pressure = 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 9Top 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 psiForce = Pressure x Area ..................................................................................... (1.1)= 1,000 psi x 11.781 in.² = 11,781 lbááááá

Total force = 12,566 lbâ- 11,781 lbá = 785 lbâ ............................................ Answer

Alternate Solution for Problem 9Rod 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


October 1996

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

Solution for Problem 11Pressure in the system acts on both sides of the piston since the piston does nottouch the cylinder. This gives an effective area equal to the rod diameter. Rodarea was calculated in Problem 10 to be 2.074 in.²Force 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 aseal has actually reversed the load on the scales.

Solution for Problem 12Hydrostatic 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 13Fluid 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 14Height of fluid column = 6,000 ft - 1,500 ft = 4,500 ftFluid 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 15Fluid 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 ftHydrostatic 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


1-64 Gene

Solution for Problem 16Fluid 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 psiReversing pressure = Total hydrostatic pressure (from Problem 15) - Casinghydrostatic pressure= 7,761.1 psi - 4,675 psi = 3,086.1 psi................................................................ Answer

Solution for Problem 17Mud weight = 104 lb/ft3

Depth = 8,000 ftHydrostatic 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 18Fluid 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 19Fluid 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 20Fluid 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 ftChange 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 21Fluid 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 ftHydrostatic pressure = 0.7273 psi/ft x 10,000 ft = 7,273 psi ....................... Answer


October 1996

Solution for Problem 221. 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 ft3. 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 ft4. 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 231. 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 ftFluid 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 psi5. 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 24Capacity of 10,000 ft of drillpipe = 10,000 ft x 0.00658 bbl/ft = 65.8 bbl ... Answer

Solution for Problem 25Capacity 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


1-66 Gene

Solution for Problem 26Fill-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 27Fluid 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 psiMaximum 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 ft2. Hydrostatic pressure of 10 bbl of water = 2,586.5 ft x 0.433 psi/ft

= 1,119.955 psi3. Length of mud column = 10,000 ft - 2,586.5 ft = 7,413.5 ft4. 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 psi6. 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 (totalpumped = 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 thetubing is3. Cement column length = 10,000 ft - 2,586.5 ft = 7,413.5 ft4. 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 psi6. Maximum pump pressure = 9,000 psi - 7,282.056 psi

= 1,717.944 psi ........................................................................................ Answer


October 1996

Solution for Problem 27—continuedE. 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 ofcement; 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 ofwater; 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 28A. Length = 1,700 ft

WA = 15.50 lb/ft

MW = 14.7 lb/galWL = (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 mudWeight 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 29Buoyancy 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/ftTotal 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


1-68 Gene

Solution for Problem 30Buoyancy 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/ftActual 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/ftActual 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= 36,007 lbâââââ ........................................................................................................ Answer

Solution for Problem 31Buoyancy 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 32Figure 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 effectivefrom tubing OD to tubing ID.Therefore, the stinger has no effect on the indicator reading. ................. Answer

Solution for Problem 33A. 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/ftIndicator 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 thepipe weight, 4,000 psi would pump the tubing out of the hole.


October 1996

Solution for Problem 34Buoyancy 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/ftIndicator reading before pumping = 14.487 lb/ft x 1,000 ft = 14,468.82 lbâ4 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 35A. 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= 1,570.86âââââ ................................................................................................... Answer

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/ftLength 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/ftIndicator 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 36A. 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/ftOriginal 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/ftDrillpipe 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


1-70 Gene

Solution for Problem 36 —continuedD. 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 psiMaximum drillpipe pressure = 1,402.8 psi (from Part B) + 2,131.453 psi= 3,534.253 psi .............................................................................................. Answer


general hydraulics 01

Documents