Section 2 - Calculations

2005, Halliburton 2 1 Stimulation I

Section 2

Calculations

Table of Contents Introduction ............................................................................................................................................... 2-3

Objectives .............................................................................................................................................. 2-3 Unit A: Definitions .................................................................................................................................... 2-4

Unit A Quiz............................................................................................................................................ 2-6 Unit B: Capacity, Rate, and Hydrostatic Pressure ..................................................................................... 2-7

Rectangular Volume .............................................................................................................................. 2-7 Cylindrical Volume................................................................................................................................ 2-8 Capacity ................................................................................................................................................. 2-8 Annular Capacity ................................................................................................................................... 2-9 Hydrostatic Pressure ............................................................................................................................ 2-10 Fill-Up.................................................................................................................................................. 2-10 Rate ...................................................................................................................................................... 2-10 Unit B Quiz .......................................................................................................................................... 2-12

Unit C: Fluid Flow................................................................................................................................... 2-13 Newtonian vs. Non-Newtonian Fluids................................................................................................. 2-13 Fluid Density........................................................................................................................................ 2-14 Fluid Flow Patterns .............................................................................................................................. 2-14 Friction Pressure .................................................................................................................................. 2-15 Unit C Quiz .......................................................................................................................................... 2-16

Unit D: Job Design Calculations ............................................................................................................. 2-17 Working with Equations ...................................................................................................................... 2-17 Bottomhole Treating Pressure.............................................................................................................. 2-18 Friction Loss in Pipe ............................................................................................................................ 2-18 Slurry Density and Volume.................................................................................................................. 2-19 Wellhead Pressure................................................................................................................................ 2-21 Hydraulic Horsepower ......................................................................................................................... 2-21 Pump Rate............................................................................................................................................ 2-22 Unit D Quiz.......................................................................................................................................... 2-23

Self-Check Test: Calculations ................................................................................................................. 2-25 Answers to Unit Quizzes ......................................................................................................................... 2-27

Self-Check Test Answer Key............................................................................................................... 2-32

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Use for Section notes

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Introduction

Stimulation work today ranges from very small, one transport acid jobs to large frac jobs where more than 1 million gallons of fluid are pumped. Since the best job for a given set of conditions needs to be run, the design of these jobs is critical. Although it may seem that small and large jobs have little in common, this is not the case. Every stimulation job is affected by some of the same factors such as fluid properties, flow rates, and well configurations. These factors are the basis for job calculations, which are essential to stimulation work. Job design relies on the values that these calculations give. This section is designed to help you understand the how and why of the calculations necessary for stimulation work.

Objectives After completing this section, you will be able to

Calculate the capacity of tubing

Calculate the capacity of an annular volume

Calculate tank volumes

Calculate wellhead, friction, hydrostatic and bottom hole treating pressures

Calculate hydraulic horsepower requirements

Calculate slurry density and volumes

Calculate the size of additive pump needed for a given additive concentration.

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Unit A: Definitions

There are a variety of terms used in calculations for stimulation work. These terms need to be clearly defined and understood before a job design can be attempted. This unit defines many of these terms and can be used as a reference when necessary. Absolute Permeability -Absolute Permeability is the Darcys law permeability. Absolute Volume Factor - Absolute Volume factors typically refer to units of gallons per pound (liters per kilogram). This is the absolute volume that a solid will take up in water. One pound of Ottawa sand will take up 0.0452 gallons of space in a liquid environment. One kilogram of Ottawa sand will take up 0.3774 liters of space in a liquid environment. For example, in pouring one pound of sand into a one gallon jar of water, 0.0452 gallons of water will be displaced from the jar. Barrel Oil field barrel is 42 gallons. BHTP - The Bottom Hole Treating Pressure, or BHTP, is the amount of pressure required at the perforations to cause fracture extension. Many times this value is reported as the frac gradient. The gradient is calculated by dividing the BHTP by the depth to the center of the perforations. bbl/min - This term refers to the pump rate or Barrels Per Minute (use bpm instead of bbl/min). bpm - This term refers to the pump rate or Barrels Per Minute. Closure Pressure - Closure Pressure is the amount fluid pressure required to reopen an existing fracture. This pressure is equal to, and counteracts, the stress in the rock perpendicular to the fracture plane. This stress is the minimum principal in-situ stress and is often called the closure stress. Clean Volume - Clean Volume refers to the volume of the treating fluid without taking into account proppant.

Darcys Law - For linear flow as in through a sand plug in casing.

LPkA

where:

K = PermeabilityA = AreaP = Delta Pressure = ViscosityL = Length

Density - The Density of a body is its mass per unit volume. Water density is 8.33 lb per gallon at 70F. Dirty Volume - Dirty Volume is the "clean volume plus the volume of the proppant. Effective Permeability - Effective Permeability is the permeability to one fluid in a multi-fluid system and is a function of the fluid saturation. Flash Point - Flash Point refers to the lowest temperature at which vapors above a volatile combustible substance ignite in air when exposed to spark or flame. Frac Gradient - (Hydrostatic pressure at perforation mid point + ISIP) divided by depth of perforation mid point. Hydrostatic Pressure - Hydrostatic Pressure reflects the pressure exerted by a vertical column of fluid. This pressure is calculated from the true vertical height and density of the fluid. Hydrostatic pressure is not area sensitive. ISIP ISIP (PISIP) is the instantaneous shut-in pressure. It can be determined during a pump-in test. The pumps are brought on line at a rate that will cause the formation to fracture ("break down"). Fluid is pumped into the formation for a short time then pumping is stopped. ISIP reflect the amount of pressure recorded immediately after shutting the pumps down. ISIP values can be hard to determine if the bottom hole slurry

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rate is not zero and/or water hammer is introduced. Graphical methods are used to determine an ISIP when water hammer is present by extrapolating back along a straight line section to the intersection of the first rise of the first oscillation of the water hammer. HHP - Hydraulic Horsepower is a unit of measurement for the amount of work that is or can be done by hydraulic equipment. HHP can be calculated by (pressure rate)/40.8 Mgal - The M is the Roman numeral for one-thousand. Therefore, this refers to Thousands of Gallons. Used in concentration statements. Net Pressure - Net Pressure is defined as the difference in ISIP pressure and closure pressure. Permeability - Permeability is a function of the geometry, configuration, and scalar dimensions of the voids or pores and is not as such a physical property derived from a dynamic system.

Ph - This symbol is used for hydrostatic pressure, the pressure exerted at the bottom of a fluid column. (Note that the P in this and the following symbols refers to pressure.) Pw - The Wellhead Pressure is the gauge measured treating pressure at the surface.

Pfrict - The symbol indicates delta (or incremental) change; therefore, P means the gradual change in pressure. Pfrict stands for friction loss in pipe, as measured by units of

psi. The movement of fluid past a stationary object causes this friction, which in this case is the pipe wall. Pperf - The friction caused by fluid flow through a perforation or group of perforations. This symbol stands for perforation friction. Porosity A fractional or percentage value Referring to the void spaces inside a rock or the part of the rock that is not rock. Relative Permeability - Relative permeability is the ratio of the effective permeability to the absolute permeability of the porous medium. Slurry Volume - Slurry Volume is the total volume of fluid, additives, and proppants. This reflects the total volume of fluid that is pumped also referred to as Dirty Volume. Specific Gravity - Specific Gravity is a unit-less ratio relationship between a substance and a base substance. For liquids, the base is water, so the specific gravity of water is 1.0 (8.33/8.33). For a 10 lb/gal brine the specific gravity will be 10.0/8.33=1.2. For gases, air is the base substance. Temperature Gradient - Temperature Gradient defines a linear relationship of temperature to depth. Temperature Gradient from a well at 10,000 feet at 200F and surface temperature of 68F would be (200-68) /10 = 13.21F per 1000 feet.

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Unit A Quiz

Fill in the blanks with one or more words to check your progress in Unit A. 1. The term BHTP stands for the bottomhole _____________________ _______________________.

2. The BHTP gradient is also referred to as the ______________________ gradient.

3. bbl/min refers to the pump rate in ___________________________________.

4. ISIP is the _______________________ ______________________ pressure, which can be determined during a __________________________ test. In this test, the formation is fractured.

5. Pw stands for ____________________________ pressure.

6. Pfrict is the ______________________________ loss in pipe.

7. ________________________ is defined as the part of the rock that is not rock.

8. Dirty volume is the _______________________ plus the __________________________.

9. Hydrostatic pressure is calculated from _________________________ and ____________________.

10. Net pressure is defined as the difference between ___________________ and __________________.

Now, compare your answers with the Answer Key.

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Unit B: Capacity, Rate, and Hydrostatic Pressure

Capacity calculations are important in stimulation work. They are used in calculating displacement volume as well as pit or tank volume. Hydrostatic pressure is equally important in basic stimulation design equations. At the end of this unit you should be able to

calculate open pit or unmarked tank volume

volume of pipe based on its inner diameter

rate of pumping from observing pits or tanks

displacement volume

hydrostatic pressure at a certain point in the hole.

Rectangular Volume

Looking first at rectangular objects, volume can be calculated by multiplying length, by width, by height. Figure 2.1 illustrates these dimensions.

Height

LengthWidth

Figure 2.1 The three basic dimensions.

Tank Example: The tank illustrated in Figure 2.1 is 10 feet high, 20 feet long and 16 feet wide. What is the volume, expressed in cubic feet (ft3)? What is the volume expressed in barrels (bbl)?

Solution:

(a) 3ft3200ft10ft16ft20HWLVolume

=

==

(b) Conversion factor for ft3 to bbl = 0.1781 bbl/ft3

bbl569.92=

= 3ftbbl1781.03ft3200Volume

This can also be used to calculate the volume of a rectangular open pit. Pit Example: A pit has the dimensions of 12 ft deep, 30 ft wide and 40 ft long. How many barrels will it hold? How many gallons will it hold? Solution:

bbl2,564.64=

= 3ftbbl1781.0ft40ft30ft12Volume

gal107,714.88=bblgal42bbl64.564,2

A useful way to gauge how much fluid remains in a tank or pit is to get a bbl/in. of depth or bbl/ft of depth factor. In the tank example, what is the bbl/in. factor? Solution: A uniform tank that is 10 ft high has a total volume of 569.92 bbl. Therefore,

depthofin.bbl4.7493factor rate ==

=

in120bbl92.569

deepin120ftin12ft10

If you measure the fluid level in the tank and find 66 inches of fluid, how many barrels are there?

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bbl313.456==inbbl7493.4in66Volume

In our pit example, what is the bbl/ft factor?

ftbbl213.72==

ft12bbl64.2564Factor

Cylindrical Volume

You can calculate the volume of cylindrical objects by multiplying the circular flat surface area by the height. Figure 2.2 illustrates these dimensions. For oilfield calculations, you will determine areas based on diameter (d), so the equation for the area of a circle is:

27854.0 dcA =

So, the calculation for the volume of a cylinder is: Volume = (Area) (Height) so,

Heightdd = 7854.0Volume

Diameter

Height

Radius

Figure 2.2

Cylindrical Tank Example: What is the volume of a cylindrical tank 15 feet in diameter and 20 feet high in barrels? Solution:

bbl629.459=

==

33

3

ftbbl1781.0ft3.3534

ft3.3534ft20ft15ft157854.0V

We can calculate a bbl/in. or bbl/ft factor for a vertical cylindrical tank. If the tank is horizontal (such as an acid transport) the volume factor changes for each inch. This method will not work for containers that change in area as they change in height. Horizontal cylindrical tanks should have a gauge stick or a table that shows volume remaining per in. or ft of depth. What is the bbl/in. factor for the previous cylindrical tank example? What is the bbl/ft factor? Solution:

ftbbl31.473

in.bbl2.623

==

==

=

ft20bbl459.629factor

.in240bbl459.629factor

.in240ftin12ft20

If we are pumping from a tank and we know the bbl/in. or bbl/ft factor, we can calculate the pumping rate. Use a watch to time how long it takes to pump out a certain depth of fluid (i.e., one inch, six inches, one foot, etc.). Since we have a rate in inches or feet per minute, and know our factor, we can then calculate a rate. Using the cylindrical tank example above, what is our pump rate if we are pumping from the tank at 1 ft/10 minutes? Solution:

BPM3.1473Rate ==min10

ft0.1ft

bbl473.31

Capacity

Capacity is a term frequently used when talking about volume. When referring to the oilfield, it is the volume a certain length of pipe will hold. When knowing the shape of a pipe is round, the volume can be calculated by hand. This calculation can be greatly simplified by using a handbook, such as the Halliburton Cementing Tables (the Red Book). In the Capacity Section (Section 210), youll find capacity factors for various sizes of drill pipe,

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tubing and casing. Currently, these are listed as gallons per foot, barrels per foot, and cubic feet per foot. To apply this information, locate the table for the type of pipe; drill pipe, tubing or casing. Next, locate the size and weight of a pipe in the two left columns. (For tubing, it is four columns.) Then find the volume units desired across the top. Read the conversion factor where the columns intersect. For example, to find the capacity of 4 1/2 in., 16.60 lb/ft internal upset drillpipe in gallons, locate 4 1/2 in. 16.60 lb/ft in the two left columns. Then locate gallons per foot at the top (third column from left) and read the capacity factor at the intersection. The capacity factor is 0.5972 gal/ft. Multiply the capacity factor by the length of pipe in feet to calculate the capacity of this pipe. Capacity Example: What is the capacity of 5000 feet of a 5 1/2 in., 17.0 lb/ft casing in gallons? What is the capacity in barrels? Solution:

bbl116

gal4882

==

==

ft5000ft

bbl0232.0bbl)Capacity (

ft5000ft

gal9764.0gal)Capacity (

This is the amount of fluid needed to displace all the treating fluids out of the casing or to load it.

Annular Capacity

Annular Capacity is the volume contained between the outside of the drill pipe or tubing and the open hole or inside of the casing (Figure 2.3).

Figure 2.3 The annulus of a cased hole.

To calculate annular capacities, you need to know the size and weight of the outside tubular as well as the size and weight of the inside tubing or casing. If you know this information, you can refer back to the Red Book, Section 221, to calculate factors involving volume and height between tubing, tubing and casing, casings, or drill pipe and casing. Annular Capacity Example: We have a 2-3/8 in, 4.7 lb/ft tubing inside of 7 in., 26 lb/ft casing. There is a packer set at 7500 ft. What is the number of barrels of water needed to completely fill the annulus? Solution: To calculate the capacity factor, open the Red Book to Section 221, Vol. & Hgt. Between: Tbgs., Tbg & Csg., Csgs, D.P. & Csg. Find the table with the heading: Inside Tubing O.D. 2.375" ONE STRING Look for 7, 26.00 row From the Barrels Per Lin Ft column, the factor is 0.0328 bbl/ft.

bbl246==ft

bbl0328.0ft7500Volume

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Hydrostatic Pressure

Hydrostatic Pressure is the force exerted by the weight of a column of fluid and expressed in pounds per square inch (psi). The size or shape of the hole or container makes no difference. The true vertical height of the fluid column and the density (lb/gal) of the fluid are the only factors involved in hydrostatic pressure. Hydrostatic pressure can be calculated at any depth in a hole or container. The best method for this calculation is to use the Hydrostatic Pressure and Fluid Weight Conversion Tables in Section 230 of the Red Book. The extreme left column of the table gives the fluid densities in lb/gal. For each fluid density, the table lists its weight per cubic foot (lb/ft3) and kilogram per liter (kg/L), its specific gravity and the pressure in lb/sq in. for one ft of depth (psi/ft). To determine the density of a fluid without the Red Book you can multiply the fluids weight in lb/gal by 0.05195 to get an approximate hydrostatic pressure of the fluid. Hydrostatic Pressure Example: The fluid weight of 12.0 lb/gal times 0.05195 equals 0.6234 psi/ft. Solution: The Red Book value is 0.6234 psi/ft. Example: The density of fresh water is 8.33 lb/gal at 68F. This exerts a pressure of 0.433 psi/ft (See below). With perforations at 6000 ft, what is the hydrostatic pressure at that location?

Figure 2.4

Solution:

psi2598

P

=

=ft

psi433.0ft6000h

Fill-Up

The Fill-Up of pipe is defined as the length of pipe a specified volume will fill. Fill-up factors are listed in Section 210 (Capacity) of the Red Book. Fill-Up Example: How many feet of 2-7/8 in., External Upset (EUE), 6.5 lb/ft tubing will 25 barrels of acid fill? Solution: Fill-up Factor = 172.76 ft/bbl (from Red Book)

ft4319== bbl25bblft76.172Fill

Rate

You need the ability to calculate additive rates in order to pick the right size of pump for a job. Additive concentrations for job designs are given as gallons per thousand gallons (gal/Mgal). From this information, and the clean rate, you can calculate the gallons per minute the additive pump must deliver. Also, you need the ability to calculate the amount of time fluid takes to go from surface to perforations or the travel time for a fluid. This is typically called "pipe time" or time to perforations. To calculate the pipe time in minutes, begin with the capacity of the tubulars being used, and then divide by the pump rate. Additive Rate Example: The crosslinker has to be injected at 4 gallons per thousand gallons (4 gal/Mgal) while pumping at a "clean rate of 25 bbl/min. What is the pump rate in gal/min for the additive pump?

7 in.- 29 lb/ft Casing

6,000 ft Perf Location

8.33 lb/gal

6,100 ft Total Depth

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Solution: First convert clean rate from bbl/min to gal/min:

mingal4.2=

=

==

Mgalgal1000

Mgalgal4

mingal1050

RateAdditive

mingal1050

bblgal42

minbbl25RateClean

To shorten the above process, take the two steps and make them one step by taking the constants:

gal10001

bblgal42

combine them into:

bblgal1000gal42

to get:

bbl042.0

Now, reworking the previous example:

mingal4.2=

bbl042.0gal4

minbbl25

Pipe Time Example: We have a "slurry rate" of 25 bbl/min, pumping through 6000 ft of 3 in, 9.3 lb/ft, N-80 tubing. What is the travel time through the tubing? Solution: From the Red Books Capacity section, we have 114.99 Linear feet per barrel for the 3 tubing. So:

min.2.09==

==

minbbl25

bbl2.52TimePipe

bbl2.52ft

bbl00870.0ft6000CapacityPipe

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Unit B Quiz

Solve the following problems to check your progress in Unit B: 1. If we have a rectangular tank that is 132 in. wide, 21 ft long and 6 ft deep, what is the volume of the

tank in barrels? In gallons?

2. We are pulling fluid from a pit that is 50 ft long, 30 ft wide and 15 ft deep, what is the volume of the pit in barrels? In gallons?

3. What is the bbl/ft of depth factor for question 1? For question 2?

4. How many barrels of water is in a cylindrical tank that is 20 ft high with a diameter of 6 ft?

5. If you are pumping out the cylindrical tank in question 4 at 1 ft/minute, what is the pump rate in bbl/min?

6. You are on a job reflecting the following data:

2 7/8 in, 6.5 lb/ft external upset N-80 tubing

5 in, 15.50 lb/ft J-55 casing

A packer is on the end of the tubing and set at 8000 ft

Perforations are at 8213 ft

Treatment fluid is 9 lb/gal, 30 lb/Mgal WG-19

ClaySta XP added at 4 gal/Mgal

ScaleChek added at 1 gal/Mgal

Surface clean pump rate of 18 bbl/min

Calculate:

a. Displacement to perforations in barrels

b. Pipe time to perforations

c. Amount of fresh water (8.33 lb/gal) needed to fill annulus in barrels

d. d. Hydrostatic pressure at perforations.

e. Additive pump rate needed for the ClaySta XP? For the ScaleChek?

Now, look up the answers in the Answer Key.

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Unit C: Fluid Flow

Successful stimulation treatments are dependent on the characteristics of the stimulation fluid. Understanding these characteristics will lead to better job design and performance. Flow behavior of a fluid is affected by

the rheological properties of the fluid (viscosity and shear)

the dimensions of the tubular goods

the rate of flow through the pipe In this unit, you will learn about these topics:

Newtonian and Non-Newtonian fluids

Fluid density

Fluid flow patterns

Friction pressure

Newtonian vs. Non-Newtonian Fluids

Fluids such as water, acid, and most crude oils that contain no additives are classified as Newtonian (or true) fluids. To understand the definition of a Newtonian fluid, you must understand the definitions of two other terms, viscosity and shear. The viscosity of a fluid is the physical property that characterizes the flow resistance of simple (Newtonian) fluids. Viscosity is responsible for the frictional drag (or viscous force) which one part of the fluid exerts on an adjacent part if the two parts are in relative motion. Viscosity is a measure of a fluid's resistance to the deformation rate. Said another way, viscosity is the measure of a fluid's resistance to flow. Viscosity is generally written with the Greek symbol mu () and reported in units of centipoise (cp). The higher the viscosity, the higher the fluid's resistance is to flow.

Shear is the movement of one fluid particle past another. Shear rate is computed by the equation of Shear Rate = Velocity / Length. Units for shear rate are reciprocal seconds (sec-1). Figure 2.5 shows the ideal system of two parallel plates with a distance between them of L and with one plate moving at a velocity V.

Figure 2.5

In pipe flow, pressure drop represents shear stress and velocity of the shear rate. When using a Fann Viscometer, shear stress can be determined from the dial reading and the shear rate from the rotational speed of the sleeve. The most common rheological test performed on fracturing fluids is the shear stress/shear rate test. This data is used to construct a flow curve of which the slope is the fluid's viscosity. Higher rates of shear result from faster movement of the fluid particles. Temperature, however, has a strong effect on the viscosity of fluids. Liquid viscosity decreases with the increase of temperature. Gas viscosity increases with an increase in temperature. The definition of a Newtonian fluid, then, is that it has the same viscosity at all flow rates or shear rates. In comparison, non-Newtonian fluids do not have constant viscosity at all flow rates or shear rates.

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Most of the fluids we use in the oilfield are non-Newtonian "pseudo plastic" or shear thinning fluids. This behavior is represented graphically in the figure below.

0 100 200 300 400 500 600 700 800

10

0

20

30

40

50

60

Shear Rate

Figure 2.6

In general, the addition of chemicals such as fluid loss additives, gelling agents, friction reducers, and emulsifiers to a Newtonian fluid tends to change the fluid to a non-Newtonian type. The viscosity of a Newtonian fluid is a constant ratio of shear stress to shear rate. As for non-Newtonian fluids, because their flow curves are not linear or linear but not passing through the origin viscosity is not constant but is a function of shear rate. Apparent viscosity, or a, is often used when referring to the consistency of non-Newtonian fluids. The apparent viscosity of non-Newtonian fluids at any shear rate represents the viscosity of Newtonian fluids at the same shear stress and shear rate (Figure 2.7).

0 100 200 300 400 500 600 700 800

10

0

20

30

40

50

60

Shear Rate

Figure 2.7

Apparent Viscosity then, is a simplistic view of the consistency of a non-Newtonian fluid and

only relevant at a given shear stress or shear rate.

From the shear rate equation, Shear Rate = Velocity Length there will be a different shear rate and as a result, a different viscosity for different geometrys. So the shear rate down the tubing, casing and fracture will all have different viscosities due to the different shear rates To help minimize the confusion of reporting apparent viscosity at arbitrary shear rates, it has become standard practice to report apparent viscosity based on either 100 or 300 rpm (revolution per minute) speeds of the Model 35A Fann Viscometer. Halliburton assumes that all apparent viscosity values are at the 300 rpm with a B1 bob for linear gels and 100 rpm with a B2 bob for crosslinked gels unless otherwise stated.

Fluid Density

The density of fracturing fluids must be considered since it affects hydrostatic pressure. The density of a fluid is expressed in units of pounds per gallon (lb/gal). The proppant concentration added to fracturing fluids affects the density of the treating slurry. Therefore, this value must be known when performing calculations to find density and hydrostatic pressures.

Adding proppant to a fluid will also increase the fluids apparent viscosity and thus its friction characteristics will increase.

Fluid Flow Patterns

Two types of fluid flow patterns will be discussed here: Laminar and Turbulent. Both are depicted in Figure 2.8. Laminar flow is the smooth steady flow of a fluid. Turbulent flow is fluctuating and agitated. When a fluid is in turbulent flow, friction is at maximum. Eddies and currents are in the flow

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stream. Lower viscosity fluids change from laminar to turbulent flow at lower velocities. As the viscosity of a system goes up it will take a greater velocity to achieve turbulence. The distinction between the two flow patterns was first demonstrated by a classic experiment performed by the British physicist Osborne Reynolds. By injecting a colored dye into a stream of fluid moving at a low flow rate, Reynolds found that the jet of the dye flowed intact along with the main stream and no cross mixing occurring. When the flow rate was increased to critical velocity, the velocity at which turbulent flow starts, the thread of color disappeared and the color diffused uniformly throughout the entire cross-section.

Figure 2.8- Fluid flow types.

Friction Pressure

As a fluid is pumped through tubing or casing, a certain amount of friction is created. This is due to fluid moving past the pipe wall (shear).

Friction is affected mainly by rate, pipe diameter, pipe roughness, pipe length, viscosity and density. As the flow rate increases for a given fluid, the friction pressure increases. As a fluid moves into turbulent flow, the friction pressure also increases. As a pipes diameter increases, friction pressure decreases due to the decrease in velocity. To determine the friction pressures of a fluid, use the Halwin\StimWin program "Friction." To use this program, you will need to select the fluid you are interested in and input the tubular sizes and lengths. Then hit the "DO" button and you can view the results in graphical or text format. Figure 2.9 is the graphical output for WG-11 pumped through 10,000 feet of 3 in., 9.3 lb/ft tubing. Read pump rate across the bottom (X axis) and the corresponding pressure for a particular rate on the left hand (Y axis).

Friction Pressure

2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 91 10 100

Rate (bpm)

2

3

4

5

6

7

89

2

3

4

5

6

7

89

100

1000

10000

Fric

tion

Pres

sure

(ps

i)

WG-11, 40.0

1Pressure

1

Rate5.00

W4279.7

StimWin v4.3.020-Jul-00 14:34

Figure 2.9 StimWin output.

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Unit C Quiz

Fill in the blanks with one or more words to check your progress in Unit C. 1. A Newtonian fluid has the same ____________________________ regardless of the rate of

___________________________.

2. Density of fracturing fluids must be considered since it affects ______________________________.

3. Two fluid flow patterns of fluids are _______________________ flow and ____________________ flow.

4. Friction pressure is dependent upon _________________, ________________, _________________, __________________, and __________________.

5. Halliburton assumes that all apparent viscosity values for linear gels are at _____________ rpm with a B1 bob, unless otherwise stated.

6. The Halwin/StimWin program that is used to calculate friction is _________________________.

Now, look up the suggested answers in the Answer Key.

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Unit D: Job Design Calculations

In this unit you will learn how to calculate:

Bottomhole Treating Pressure (BHTP) Friction Loss in Pipe ( Pfrict) Slurry Density () and Volume Wellhead Pressure (WHTP) Hydraulic Horsepower (HHP) Pump Rate (Q) When Halliburton prepares to mobilize equipment for a stimulation treatment, two major job variables must be determined. These are:

What is the estimated Wellhead Treating Pressure? (WHTP)

What is the proposed pumping rate? Calculating these two variables helps us determine the Hydraulic Horsepower, blending and proppant delivery equipment to spot on location. To make these calculations, it is advisable to always draw a wellbore sketch. This helps you to visualize fluid movement through the wellbore and the resulting forces which must be overcome to properly place the stimulation treatment.

Where does WHTP come from? Simply stated, WHTP is the surface pressure required to pump into the formation. Looking at the basic wellbore diagram helps to define the problem:

Figure 2.10 -

As stated in the definitions: BHTP: The pressure inside the formation. Hydrostatic Pressure, Ph : The fluid columns pressure (as a function of the fluid density). Friction Pressure, Pfrict : Pressure due to fluid movement in the pipe. The faster we pump, the higher the velocity and the higher the Pfrict. Therefore, WHTP is influenced by BHTP, Ph, and Pfrict. Always remember the following:

The higher the BHTP, the higher the WHTP.

The higher the pump rate, the higher the fluid velocity which causes higher Pfrict and results in higher WHTP.

The higher the fluid column density, the higher the Ph and the lower the WHTP.

Working with Equations

Before beginning the actual calculations in this unit, two basic principles about equations must be understood. First, an equation is a mathematical statement (simple expression in

BHTP

P - Fr

ictio

n

P -H

ydro

stat

ic

Maximum friction pressure occurs at the top of the well. Maximum hydrostatic pressure occurs at the bottom of the well.

WHTP

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English) that says two things are equal or evenly balanced. For example, the equation BHTP = PISIP + Ph says that bottomhole treating pressure is equal to instantaneous shut-in pressure (PISIP) plus hydrostatic pressure. (Ph) Keep in mind that you can rewrite an equation and not affect its value. You can perform the same operation (that is, add, subtract, multiply, or divide by the same number or symbol) on both sides of an equation. In another example, assume you know the value of BHTP and the Ph. You need to calculate the value of PISIP. You can rewrite the equation for BHTP (presented above) by subtracting Ph from both sides: BHTP - Ph = PISIP + Ph Ph

On the right side of the equation, Ph minus Ph cancels out, so you are left with BHTP - Ph = PISIP. You can now solve for PISIP by subtracting Ph from BHTP.

Bottomhole Treating Pressure

To calculate bottomhole treating pressure (BHTP), you will also need to know fluid density and the depth of the perforations. Knowing the fluids density, you can then use the Hydrostatic Pressure and Fluid Weight conversion tables from the Red Book to find the psi/ft pressure gradient. Hydrostatic pressure (Ph) can be calculated by multiplying the psi/ft value and the depth of the perforations. Example: What is the BHTP under the following conditions? Tubing is 2 3/8 in., 4.7 lb/ft, EUE, J-55 to 7000 ft. Casing is 5 1/2 in., 20 lb/ft, J-55 to 7100 Perforations are at 7050 ft. Well fluid is 8.33 lb/gal fresh water. PISIP = 1800 psi Solution:

BHTP = Pisip + Ph

PISIP = 1800 psi (given) Hydrostatic pressure for 8.33 lb/gal fresh water = 0.4330 psi/ft (from Red Book)

4852.65psi3052.65psi1800psiBHTP

psi3052.65

7050ftft

psi0.4330Ph

=

+=

=

=

Friction Loss in Pipe

To calculate the friction loss for a treating tubular, you will use the StimWin program Friction. Keep in mind that the fluids in Friction do not have breakers in them, the fluids on location may be off by some percentage. Also be aware that the roughness for the tubular has not been taken into account. Example: What is the friction pressure in the tubing under these conditions? Tubing is 2 3/8 in. OD, 1.995 in. ID, 4.7 lb/ft, EUE, J-55 with a packer at 8500 ft. Casing is 5 1/2 in., 4.892 ID, 17 lb/ft, J-55, LTC Perforations are at 8560 ft. Treating fluid is fresh water at 8.33 lb/gal. Pump rate is 10 bbl/min.

a. Solution: a. In StimWin choose Fresh Water b. Set the rate from 1 to 10 bbl/min, c. Set Increment to 1 d. Use Internal n and K, e. Go to the Wellbore tab by clicking the

right or left arrow on the toolbar.

f. Navigation icons g. Fill in the tubing and casing information h. Hit F5 key or click the DO icon

Calculations


i. DO icon j. Click the Text Output Icon

k. Text icon l. h. The program arrives at the value of

10318.6 psi at 10 bbl/min.

Slurry Density and Volume

Slurry density is an extremely important factor in stimulation. It is used during the calculations of BHTP and friction pressure while running sand-laden fluid. On a fracturing job, proppant is added to the gel on a lb/gal basis. For example, one pound of dry sand will be added to one gallon of fluid. Because sand adds density and volume, the resulting slurry density and volume will change. The absolute volume factors in Table 2.1 will be used to help calculate slurry density and volume in the following example problems.

Table 2.1 Absolute Volume Factors

PROPPANT TYPE Bulk

Density (lb/ft3)

Specific Gravity (g/cc)

Absolute Volume (gal/lb)

20/40 Ottawa 95.9 2.65 0.0452

20/40 AcFRAC BLACK 102 2.55 0.0470

20/40 AcFRAC BLACK 100 2.57 0.0466

20/40 SUPER HS 95.5 2.55 0.0470

20/40 ECONO- PROP 96 2.70 0.0444

20/40 CARBO- LITE 97 2.71 0.0442

16/20 CARBO- LITE 97 2.71 0.0442

20/40 CARBO- PROP 117 3.27 0.0366

16/30 INTER- PROP 120 3.32 0.03671

20/40 INTER- PROP 120 3.13 0.0383

12/18 CARBO HSP 2000 128 3.56 0.3366

16/30 CARBO HSP 2000 128 3.56 0.3366

20/40 CARBO HSP 2000 128 3.56 0.3366

30/60 CARBO HSP 2000 128 3.56 0.3366

(1 ft3 is equal to one sack of proppant) The absolute volume of proppant is calculated from the specific gravity of the proppant. The

specific gravity is measured in grams per cc (cubic centimeter). So, the Bulk Density (or Specific Gravity) is measured as if the proppant were a solid and not made up of individual particles.

Example: What is the slurry density (lb/gal) and slurry volume (gal) of fresh water with 2 lb/gal Ottawa proppant added? Solution: Set up a table as shown:

Materials Materials (pounds) Absolute Volume Factor (gal/lb)

Absolute Volume

(gallons)

Fresh Water 8.33 ---- 1

Sand 2 0.0452 0.0912

TOTALS 10.33 lb 1.0912 gal

Divide total pounds by total gallons to calculate slurry density.

gallb9.4666

gal1.0912lb10.33

gallbDensitySlurry ===

The total of the absolute volume column (in gals) is also referred to as "dirty" volume. If you were to run 2,000 gallons of water with 2 lb/gal Ottawa sand, then "clean" volume is 2,000 gallons. The "dirty" volume is the "clean" volume plus the sand volume (in gallons). Total pounds of sand would be 2000 gal 2 lb/gal = 4000 lb. Sand volume (in gallons) is the total pounds of sand times the absolute volume factor for sand. In this case the sand volume is 4000 lb 0.0452 gal/lb. To calculate "dirty" volume:

gal2180.8gal180.8gal2000

lbgal0.0452sandlb4000gal2000VolDirty

=

+=

+=

Calculations


Example: What is the slurry density and "dirty" volume?

Fracturing fluid is Diesel #2 with a density of 7.33 lb/gal.

Sand concentration is 10 lb/gal. Stage size is 10,000 gallons "clean" volume. Solution:

Materials Materials (pounds) Absolute Volume Factor (gal/lb)

Absolute Volume

(gallons)

Diesel #2 7.33 ---- 1

Sand 10 0.0452 0.452

TOTALS 17.33 lb 1.452 gal

gallb11.935===

1.452gal17.33lb

gallbDensitySlurry

"Dirty" Volume = "Clean" volume + (sand concentration clean volume absolute volume factor)

volume dirty""gal14,520=

+lbgal0452.0gal000,10

gallb10gal000,10

Instead of using a table you can use the following equations for Slurry Density, Slurry Volume, and Volume Factor:

FactorVolumel)Conc(lb/gaProp(lb/gal) BaseFluidSlurry

+=

where: Slurry = Slurry Density BaseFluid = Base Fluid Density Prop Conc = Proppant Concentration

+

=

lbgalfactorVolAbs

gallbConcProp1

FactorVolume

One place where an understanding of slurry density and volume is necessary is when a well "screens out". A screen out occurs when fluid and proppant can no longer be pumped into the formation and causes the pressure to reach its

maximum allowable value. Usually, the job is shut down at that point. Example: We are pumping 2% KC1 water (8.43 lb/gal) with 4 lb/gal 20/40 Ottawa sand. The casing is 4-1/2 inch, 10.5 lb/ft. Perforations are at 3,000 ft. As soon as the 4 lb/gal stage gets to the perfs, the well screens out. How many sacks of sand are left in the casing? What is the hydrostatic pressure at the perforations?

gal2009.7=

=ft

gal6699.0ft3000Capacity Casing

Therefore, we have 2009.7 gallons of slurry in the casing. In order to calculate the sand volume we need to use the equation.

1.18080.18081lbgal0.0452

gallb(41

Vol)AbsConc(Prop1Factor Volume

=+=

+=

+=

To calculate the clean volume, rearrange the following equation:

waterKCL2%gal1701.98171.808

gal2009.7VolClean

Factor VolumeVolumeSlurry VolClean

Factor VolumeVolClean VolumeSlurry

=

=

=

=

Now to calculate the sand volume:

lb6807.927==gallb4gal1701.9817Wsand

Since there are 95.9 lb of Ottawa sand in one sack: (Table 2.1):

sandofsacks71==

sklb9.95

lb9268.6807Vs

To calculate the hydrostatic pressure, we need to use a different equation:

Calculations


factorvolumeionconcentratsanddensityfluidBaseDensitySlurry +=

The volume factor (1.1808) has already been calculated.

gallb10.5268

1.1808gallb12.43

1.1808gallb4

gallb8.43

DensitySlurry

=

=

+

=

psi1636.5(RedBook)0.5455ft3000P

or

psi1640.6ft

psi0.5469ft3000P

ftpsi0.5469

0.05195gallb10.5268Gradient P

h

h

h

=

=

=

=

=

=

Wellhead Pressure

The equation for calculating pressure at the wellhead is Pw = BHTP - Ph + Pfrict + Pperf or = PISIP + Pfrict + Pperf (since PISIP = BHTP - Ph) Where: BHTP = Bottomhole Treating Pressure Ph = Hydrostatic Pressure Pfrict = Fluid friction from Surface to the top perforation Pperf = Fluid friction across all perforations PISIP = Instantaneous Shut In Pressure Example:

Tubing is 2 7/8 in., 6.5 lb/ft, EUE, J-55 to 7700 ft.

Casing is 7 in., 20 lb/ft, J-55 to 7900 ft. Packer is at 7700 ft.

Flow rate is 20 bbl/min.

Perforations are two shots per foot, 0.40 in., at 7750 ft to 7775 ft (50 shots).

Treating fluid is fresh water mixed with WG-18, at 30 lb/1000gal. From the StimWin Frict Program, we should get a total pipe friction value of 2966.1 psi to the top perf.

Assume that perforation friction is zero.

Instantaneous shut-in pressure with fresh water is 1775 psi.

Calculate pressure at the wellhead (Pw) by using this formula: Pw = PISIP + Pf rict+ Pperf

Solution:

4741.1psi0psi2966.1psi1775psiP

Program)Friction the(from psi2966.1P(given) psi1775P

w

frict

ISIP

=

++=

=

=

Hydraulic Horsepower

Two equations may be used to determine hydraulic pressure (HHP). The unit in which the flow rate is given in (bbl/min or gal/min) should determine the equation used.

( )

1713.6mingalRate(psi)P

HHP

or

40.8minbblRatepsiP

HP

w

w

=

=H

The value 1713.6 is 40.8 42 gal/bbl Example: What is the HHP under these conditions?

Pressure at the wellhead is 3000 psi

Injection rate is 30 bbl/min.

Calculations


Solution:

( )

HHP2205.88=

=

=

8.40minbbl30psi3000

8.40minbblRatepsiP

HHPw

Example: What is the Pfrict, Pw, and HHP under these conditions?

Tubing is 2 3/8 in., 4.7 lb/ft, EUE, N-80

Packer at 9000 ft

Casing is 5 1/2 in., 17 lb/ft, N-80 to 9500 ft Perforations are at 9100 ft

Well fluid is fresh water. PISIP with fresh water = 1800 psi

Frac using 10 lb/gal salt water with WG-6 mixed at 40 lb/1000 gal and CW-1 mixed at 10 lb/1000 gal

Injection rate is 5 bbl/min. Pfrict gradient for the tubing is 70 psi/100 ft

Assume Pfrict in the casing is zero

Assume Pperf to be 150 psi Solution:

6300psi9000ft100ft

psi70Pfrict ==

Pw = PISIP + Pfrict + Pperf

Pisip with fresh water = 1800 psi (given). Fracturing fluid is 10 lb/gal. Solution to the problem requires PISIP be calculated with 10 lb/gal fluid.

psi7462.85=++=

=

==

==

==

==

psi150psi6300psi85.1012P)psi (given150P

psi85.1012psi15.787psi1800P

rease in P is an incThe change

psi15.787psi3.3940psi45.4727P

psi45.4727ft9100ft

psi5195.0P

psi3.3940ft9100ft

psi433.0P

w

perf

gallb

10ISIP-

h

h

gallb10h-

gallb33.8h-

HHP5487.390=

=

=

8.40minbbl30psi85.7462

8.40RateP

HHP w

Pump Rate

By rewriting the base equation for HHP, you can obtain an equation for calculating bbl/min. Multiply both sides of the equation by 40.8:

RateP40.8HHP w =

Now divide both sides by Pw. This gives you rate in bbl/min.

=

minbbl

RateP

40.8HHPw

Example: What is the maximum pump rate in bbl/min that can be delivered at maximum psi under these conditions?

Treating fluid is 15% HC1 acid, 8.962 lb/gal 1000 HHP is available at the location.

Maximum wellhead pressure is 5700 psi. Solution:

minbbl7.1579==

psi57008.40HHP1000

minbblRate

Calculations


Unit D Quiz

Solve the following problems to check your progress in Unit D. 1. What is the BHTP under these conditions?

Perforations are at 8000 ft. Well fluid is 9.3 lb/gal salt water. Pisip = 1200 psi.

2. What is the BHTP gradient under these conditions? Perforations are at 9050 ft. Well fluid is 9.7 lb/gal salt water. Pisip = 1975 psi.

3. What is the Pisip with sand-laden fluid? (Assuming we might have an unexpected shutdown.) Perforations are at 7450 ft. BHTP gradient is 0.65 psi/ft Fracturing fluid is 2% KC1 water mixed with WG-11 at 60 lb/1000 gal, WAC-11 at 20 lb/Mgal and 20/40 Ottawa sand at 5.5 lb/gal. Density of 2% KC1 water is 8.42 lb/gal.

4. Tubing is 2 7/8 in., 6.5 lb/ft, EUE, N-80 with packer at 9000 ft. Casing is 7 in., 23 lb/ft, J-55 to 9200 ft. Perforations at 9050 ft Well fluid is 10 lb/gal salt water. PISIP with 10 lb/gal fluid is 2000 psi. Fracture using 10% salt water at 8.93 lb/gal mixed with WG-17 at 40 lb/1000 gal Proppant is 20/40 Econoprop Injection rate is 20 bbl/min. Pfrict gradient is 38.3 psi/100 ft. (Disregard Pfrict in casing and Pperf)

a) What is the displacement to the perfs in barrels?

b) How many barrels of fresh water are needed to fill the annulus?

c) What is the tubing friction pressure?

d) What is the wellhead pressure?

e) What is the required HHP?

Calculations


f) If you are on the 5 lb/gal proppant stage and the well screens out with the well full of slurry, what is the hydrostatic pressure at the perfs?

g) How much proppant is left in the well (sacks)?


Calculations


Self-Check Test: Calculations

Fill the blanks with the best answer to the following items. (NOTE: You will need a Red Book for reference during this self-check test.) 1. The flow pattern of fluid where fluid velocity and friction are high, and the fluid moves primarily as

one unit is called what? ___________________ ______________________.

2. bbl/min stands for _______________ _____ ______________.

3. Perforations are at 8,000 ft. The well fluid is 2% KC1 water which is 8.42 lb/gal. PISIP = 2,575 psi. Calculate BHTP: _______________ psi/ft 8,000 ft = _______________ psi BHTP = 2,575 psi + _______________ psi = ______________ psi

4. Perforations are at 11,000 ft. BHTP gradient is 0.82 psi/ft. BHTP = _______________ psi

5. Perforations are at 9,060 ft. BHTP gradient is 0.72 psi/ft. Fracturing fluid is 2% KC1 water mixed with WG-19 at 40 lb/Mgal Proppant is 20/40 Econoprop at 3 lb/gal Calculate PISIP with sand-laden fluids.

Materials Materials Absolute Absolute (Pounds) Volume Factor (Gal/Lb) Volume (Gallons) 2% KC1 _______________________ _________________________ ____________________

Proppant _______________________ _________________________ ____________________

TOTAL _______________________ lb _________________________gal/lb ____________________ gal

Slurry Density = __________________lb/gal

BHTP = _______________________ ft _________________________psi/ft = ____________________ psi

Ph = _______________________ ft _________________________psi/ft = ____________________ psi

PISIP = _______________________ psi - _________________________psi = ____________________ psi

6. Casing is 5 1/2 in., 20 lb/ft, J-55 to 6300 ft. Perforations are at 6300 ft. Treating fluid is salt water mixed with WG-17 at 40 lb/1000 gal. Injection rate is 40 bbl/min. Pfrict gradient is 7.68 psi/100 ft. Pfrict = _________________ psi

Calculations


7. Tubing is 2 7/8 in., 6.5 lb/ft, EUE, J-55 to 6600 ft Perforations are at 6750 ft. Well fluid is 2% KC1 water, 8.43 lb/gal Pw = 6000 psi. Injection rate is 12 bbl/min. Assume Pperf = 0. What is the hydraulic horsepower required? = _______________ HHP 8. Casing is 4 1/2 in., 11.6 lb/ft, N-80 to 11,000 ft. Tubing is 2 3/8 in, 4.7 lb/ft. Perforations are at 10,875 ft. Packer is at 10,500 ft. Well fluid is 2% KC1 water (8.42 lb/gal). Pisip with well fluid is 2900 psi. Fracturing fluid is 25 lb Delta fluid(using WG-22) in 2% KC1 water. Crosslinker is being added at 2 gal/Mgal. Sandwedge is being added at 4 gal per sack Injection rate is 10 bbl/min. You are pumping out of a rectangular tank 20 ft long, 10 ft wide, and 8 ft deep. Pperf is 200 psi. Pfrict gradient is 41.96 psi/100 ft in the tubing. Pfrict in casing can be assumed to be negligible. Calculate:

a. Displacement to perfs in bbl? b. Water needed to fill annulus in bbl? c. Tank volume in bbl? d. How fast the tank level is dropping (in./min)? e. Pfrict = ________________ psi/ _______________ ft ______________ ft = _____________ psi f. Pw = __________ psi + __________ psi + ___________ psi = _________ psi g. HHP = _____________ h. You are going to pump 15,000 gals with 8 lb/gal Interprop 20/40. If the well screens out as soon as

the 8 lb/gal stage gets to the perfs, how many sacks of Interprop are left in the well? i. What is the hydrostatic pressure at the perfs in question h? j. What is the top of proppant in wellbore? k. What is the pump time to the top perf? l. What rate will be required of the liquid additive pump running crosslinker?


Calculations


Answers to Unit Quizzes Items from Unit A Quiz 1. treating pressure 2. Frac 3. Barrels Per Minute 4. instantaneous shut-in/pump-in 5. Wellhead 6. Friction 7. Porosity 8. Clean Volume/Volume of Proppant 9. True Vertical Height and Density of Fluid 10. ISIP Pressure and Closure Pressure

Items from Unit B Quiz

1.

gal10,367.56

bbl246.847

=

=

=

bblgal42bbl246.8466

ftbbl0.17811386ft

1386ftft6ft21

ftin12

in.132

33

3

2.

gal168,304.5

bbl4007.25

=

=

=

bblgal42bbl4007.25

ftbbl0.01781ft22,500

ft22,500ft15ft30ft50

33

3

3.

ftbbl267.15

ftbbl41.14

==

==

ft15bbl4007.252Q

6ftbbl246.84661Q

Calculations


4.

bbl100.713=

=

==

33

32

2c

ftbbl0.1781565.488ft

565.488ft20ft28.2744ft28.2744ft0.78546ft6ftA

5. BPM5.0357=

=

ftbbl5.0357

minft1

ftbbl5.0357

20ftbbl100.71341

6. a. ( )

bbl51.389=+

=

=

=

=

bbl46.322bbl5.0694bbl5.0694

ftbbl

.0238ft80008213Casing

bbl46.322ft

bbl.005798000ftTubing

b. min2.8550==18BPM

bbl51.389pT

c. bbl126.4==ft

bbl0.01588000ftV

d. 3839.99psi=

=ftin

gal0.05195gallb98213ftP 2h

e.

mingal0.756

ScaleChekmingal0.756

mingal3.024

XPClayStamingal3.024

=

=

=

=

=

=

1.04218

or

1000gal1gal

bblgal42

minbbl182 Rate

442 .018

or

1000gal4gal

bbl42gal

min18bbl1 Rate

Calculations


Items from Unit C Quiz 1. viscosity/shear (flow) 2. Hydrostatic Pressure

3. Laminar/Turbulent

4. rate/pipe diameter/pipe roughness/pipe length/viscosity/density

5. 300

6. friction

Items from Unit D Quiz

1.psi3864.8

8000ftft

psi0.4831Ph

=

=

5064.8psi=+= 3864.8psi1200psiBHTP

2.

ftpsi0.722=

=

=

+=

=

=

9050fti6535.295psGrad. Frac

i6535.295psi4560.295ps1975psiBHTP

i4560.295ps

9050ftft

psi0.5039Ph

2% KCl : 8.42 lb : abs. vol. 1

Sand : 5.5 lb : abs. vol. factor 0.0452 : abs. vol. 0.2486

TOTALS: Weight = 13.92 lb

Volume = 1.2486 gal

3.

psi546.83=

=

=

=

=

=

=

==

psipsipsi

ftpsift

psift

psiftgallb

gal

67.42955.4842P67.4295

5766.07450P

(RedBook)ft

psi0.5766 GradP

5.4842

65.07450BHTP

1485.112486.1

lb13.92Density

ISIP

h

h

Calculations


4. a.

bbl54.075=

+=

=

=

=

=

1.965bbl52.11bblVolumebbl1.965

ftbbl0.039350ftCasing

bbl52.11ft

bbl0.005799000ftTubing

b. bbl281.7=

=ft

bbl0.03139000ftVann

c.

psi3447=

= 9000ft

100ftft100

100ftpsi38.3

Pfrict

d.

psi5964.66=

+==

+==

=

=

=

=

=

=

++=

psipsiPpsi

psipsipsi

psi

ftftpsi

psi

ftftpsi

psi

w 244766.251766.2517

66.5172000P66.517

815.4183475.4701P815.4183

90504623.0P

475.4701

90505195.0P

2000PPPPP

gallb8.9ISIP

h

gallb8.9h

gallb10h

gallb10ISIP

perffrictISIPw

e.

PHH2923.853=

=

=

40.820BPM5964.66psi

40.8QwPHHP

Calculations


f.

gallb11.399

1.222gallb5lb8.93Density

1.2220.2221

lbgal0.0444

gallb51factor Vol

=

+=

=

+=

+=

( )psi5359.41=

= RedBook0.59229050ftPh

g.

sks 96.8==

=

=

=

=

sklb96

9292.76lbV

lb9292.76gallb51858.55galW

gal1858.551.222

bblgal4254.075bbl

VolClean

sand

sand

Calculations


Self-Check Test Answer Key

1. laminar flow 2. barrels per minute

3. psi 6066.23491.2psi2575psiBHTP

psi3491.28000ft(RedBook) ft

psi 0.4364Ph

=+=

==

4. psi9020==ft

psi0.8211000ftBHTP

5. Material Material (lb) Abs. Vol. Factor (gal/lb) Abs. Vol. (gal) 2% KC1 8.42 1 Proppant 3 0.0444__ 0.1332 TOTALS 11.4 1.1332

psipsipsiP

psi(RedBook)ft

psiftP

psift

psiftBHTP

gallb

1.1332gal11.42lbitySlurryDens

ISIP

h

1769.424753.786523.2

4753.780.52479060

6523.20.729060

10.0777

==

==

==

==

6. psi6300ft100ft

100ftpsi7.68

Pfrict 483.84==

7. HHP40.8

minbbl126000psi

HHP 1764.706=

=

8 a.

bbl5.8125bbl40.6350bblV

bbl5.8125ft

bbl0.0155375ftV

bbl40.6350ft

bbl0.0038710,500ftV

total

casing

tubing

46.4475=+=

==

==

Calculations


b. bblft

bbl0.010110,500ftVann 106.05==

c. bbl

ftbbl0.17811600ftV(bbl)

1600ft8ft10ft20ft)V(ft3

33

284.96==

==

d.

minin

minbbl2.96833

minbbl10

mininRate

inbbl2.96833

96in284.96bblFactorTank

inbblTankFactor

minbblPumpRate

mininRate

3.369==

==

=

e. psi10,500ft100ft

psi41.96Pfrict 4405.8==

f. psipsipsipsiPw 7505.82004405.82900 =++=

g. HHP 40.8

minbbl107505.8psi

HHP 1839.657=

=

h.

sacks

sklb120

b11,946.08lV

lb11,946.08gallb8l1493.260gaW

gal1493.2601.3064

bblgal4246.4475bbl

eCleanVolum

1.3064lbgal0.0383

gallb81Factor Volume

InterProp

InterProp

99.551==

==

=

=

=

+=

Calculations


i.

psi0.05195gallb12.56910875ftP

gallb12.569

1.3064gallb8

gallb8.42

DensitySlurry

h 7100.935==

=

+

=

j.

ft3079.625ft-10,500ftProppant of Top

3079.625ft(RedBook)ftft46.06766.851ftFill

66.851ft32.7ft99.551ftV

32.7ft(RedBook)ft

ft0.0872375ftV

33

333tubinginProp

33

casing

7420.375==

==

==

==

k min

minbbl10

bbl 46.4475Time Pipe 4.64475==

l. mingal

bbl0.042

Mgal2gal

minbbl10Rate-LA 0.84==

Documents

Section 2 - Calculations