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

21
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Stimulation I

Calculations

Use for Section notes

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22

Stimulation I

Calculations

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.

2005, Halliburton

Objectives
After completing this section, you will be able to

23

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.

Stimulation I

Calculations

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.

Darcys Law - For linear flow as in through a

sand plug in casing.

kAP
L
where:

Absolute Permeability -Absolute Permeability

is the Darcys law permeability.

K
A
P

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.

Permeability
Area
Delta Pressure
Viscosity
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.

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.

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.

bbl/min - This term refers to the pump rate or

Barrels Per Minute (use bpm instead of
bbl/min).

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.

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.

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

Clean Volume - Clean Volume refers to the

volume of the treating fluid without taking into
account proppant.

2005, Halliburton

=
=
=
=
=

24

Stimulation I

Calculations

psi. The movement of fluid past a stationary

object causes this friction, which in this case is
the pipe wall.

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.

Pperf - The friction caused by fluid flow through

a perforation or group of perforations. This
symbol stands for perforation friction.

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

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.

Mgal - The M is the Roman numeral for onethousand. Therefore, this refers to Thousands of
Gallons. Used in concentration statements.

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.

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.

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.

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.)

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.

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

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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.

Solution:
(a)

Volume = L W H = 20 ft 16 ft 10 ft
= 3200 ft 3

(b) Conversion factor for ft3 to bbl = 0.1781

bbl/ft3
bbl
Volume = 3200 ft 3 0.1781 3
ft
= 569.92 bbl
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?

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

Solution:

bbl
ft 3
= 2,564.64 bbl

Volume = 12 ft 30 ft 40 ft 0.1781

2 ,564 .64 bbl 42

Height

gal
= 107,714.88
bbl

gal

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.

Width
Length

In the tank example, what is the bbl/in. factor?

Figure 2.1 The three basic dimensions.

Solution:
Tank Example:

A uniform tank that is 10 ft high has a total

volume of 569.92 bbl. Therefore,

The tank illustrated in Figure 2.1 is 10 feet high,

20 feet long and 16 feet wide.

10ft 12

in
= 120 in deep
ft
569.92bbl
bbl
rate factor =
= 4.7493
120in
in. of depth

What is the volume, expressed in cubic feet

(ft3)? What is the volume expressed in barrels
(bbl)?

If you measure the fluid level in the tank and

find 66 inches of fluid, how many barrels are
there?

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Calculations

Volume = 66in 4.7493

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.

bbl
= 313.456 bbl
in

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

Factor =

2564.64bbl
bbl
= 213.72
12ft
ft

Cylindrical Volume

What is the bbl/in. factor for the previous

cylindrical tank example? What is the bbl/ft
factor?

You can calculate the volume of cylindrical

objects by multiplying the circular flat surface
area by the height. Figure 2.2 illustrates these
dimensions.

Solution:

For oilfield calculations, you will determine

areas based on diameter (d), so the equation for
the area of a circle is:

in
= 240 in.
ft
629.459 bbl
bbl
= 2.623
factor =
in.
240 in.

Ac = 0.7854 d 2

factor =

20ft 12

So, the calculation for the volume of a cylinder

is:

629.459 bbl
bbl
= 31.473
ft
20ft

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.

Volume = (Area) (Height)

so,
Volume = 0.7854 d d Height

Using the cylindrical tank example above, what

is our pump rate if we are pumping from the
tank at 1 ft/10 minutes?

Diameter

Solution:

Height

Rate = 31 .473
Radius

bbl
1.0ft

= 3.1473 BPM
ft 10 min

Figure 2.2

Capacity
Cylindrical Tank Example:

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.

What is the volume of a cylindrical tank 15 feet

in diameter and 20 feet high in barrels?
Solution:
V = 0.7854 15ft 15ft 20ft = 3534.3ft 3

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,

bbl
3534.3ft 3 0.1781 3 = 629.459 bbl
ft

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Calculations

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.

Figure 2.3 The annulus of a cased hole.

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?

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.

Solution:

Annular Capacity Example:

gal
5000ft = 4882 gal
ft
bbl
Capacity (bbl) = 0.0232
5000ft = 116 bbl
ft
This is the amount of fluid needed to displace all
the treating fluids out of the casing or to load it.

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?

Capacity Example:

Capacity (gal) = 0.9764

Solution:
To calculate the capacity factor, open the Red
Book to Section 221, Vol. & Hgt. Between:
Tbgs., Tbg & Csg., Csgs, D.P. & Csg.

Annular Capacity

Find the table with the heading: Inside Tubing

O.D. 2.375"

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).

ONE STRING
Look for 7, 26.00 row
From the Barrels Per Lin Ft column, the factor
is 0.0328 bbl/ft.
Volume = 7500ft 0.0328

2005, Halliburton

29

bbl
= 246 bbl
ft

Stimulation I

Calculations

Solution:

Hydrostatic Pressure

Ph = 6000ft 0.433

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.

= 2598 psi

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.

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).

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)

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.

Fill = 172.76

ft
25 bbl = 4319 ft
bbl

Rate

Hydrostatic Pressure Example:

The fluid weight of 12.0 lb/gal times 0.05195
equals 0.6234 psi/ft.

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.

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?

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.

7 in.- 29 lb/ft
8.33 lb/gal

psi
ft

Casing

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?

6,000 ft
Perf Location
6,100 ft
Total Depth

Figure 2.4

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Calculations

Solution:

Now, reworking the previous example:

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

25 bbl
0.042
gal
4 gal
= 4.2
min
bbl
min

bbl
gal
gal
42
= 1050
min
bbl
min
gal
gal
4
1050
min
Mgal
gal
Additive Rate =
= 4.2
gal
min
1000
Mgal

Clean Rate = 25

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:

To shorten the above process, take the two steps

and make them one step by taking the constants:
42 gal
1

bbl
1000 gal

bbl
= 52.2 bbl
ft
52.2 bbl
Pipe Time =
= 2.09 min.
bbl
25
min

Pipe Capacity = 6000ft 0.00870

combine them into:

42 gal
1000 gal bbl

to get:
0.042
bbl

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Stimulation I

Calculations

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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Stimulation I

Calculations

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.

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 (sec1

). 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.

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
Figure 2.5

Newtonian vs. Non-Newtonian

Fluids

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.

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 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.

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).

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.

The higher the viscosity, the higher the fluid's

resistance is to flow.

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Calculations

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

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

60
50
40
30
20
10
0

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.

100 200 300 400 500 600 700 800

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.

Fluid Density

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).

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.

60
50
40

Fluid Flow Patterns

30
20

Two types of fluid flow patterns will be

discussed here: Laminar and Turbulent. Both are
depicted in Figure 2.8.

10
0

100 200 300 400 500 600 700 800

Shear Rate

Figure 2.7

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

Apparent Viscosity then, is a simplistic view of

the consistency of a non-Newtonian fluid and

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Stimulation I

Calculations

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.

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.

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.

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.

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 2.9 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
WG-11, 40.0

100009

Pressure

8
7

Rate

W4

5.00 279.7

Friction Pressure (psi)

Figure 2.8- Fluid flow types.

10009
8
7
6
5
4

100
1

10

Rate (bpm)

Friction Pressure

StimWin v4.3.0
20-Jul-00 14:34

Figure 2.9 StimWin output.

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).

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Calculations

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.

2005, Halliburton

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Unit D: Job Design Calculations

In this unit you will learn how to calculate:

Friction Loss in Pipe ( Pfrict)

Slurry Density () and Volume

Wellhead Pressure (WHTP)

Hydraulic Horsepower (HHP)

Pump Rate (Q)

WHTP

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

BHTP

When Halliburton prepares to mobilize

equipment for a stimulation treatment, two
major job variables must be determined. These
are:

P -Hydrostatic

Bottomhole Treating Pressure (BHTP)

P - Friction

Figure 2.10 -

What is the estimated Wellhead Treating

Pressure? (WHTP)

As stated in the definitions:

What is the proposed pumping rate?

BHTP: The pressure inside the formation.

Calculating these two variables helps us

determine the Hydraulic Horsepower, blending
and proppant delivery equipment to spot on
location.

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.

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.

Therefore, WHTP is influenced by BHTP, Ph,

and Pfrict. Always remember the following:

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:

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

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English) that says two things are equal or evenly

balanced.

BHTP = Pisip + Ph
PISIP = 1800 psi (given)

For example, the equation BHTP = PISIP + Ph

says that bottomhole treating pressure is equal to
instantaneous shut-in pressure (PISIP) plus
hydrostatic pressure. (Ph)

Hydrostatic pressure for 8.33 lb/gal

fresh water = 0.4330 psi/ft (from Red Book)

psi
7050ft
ft
= 3052.65 psi
BHTP = 1800psi + 3052.65psi
Ph = 0.4330

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.

= 4852.65psi

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:

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.

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.

Example:

Bottomhole Treating Pressure

What is the friction pressure in the tubing under

these conditions?

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.

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.

Example:

a. Solution:
a. In StimWin choose Fresh Water

What is the BHTP under the following

conditions?

b. Set the rate from 1 to 10 bbl/min,

Tubing is 2 3/8 in., 4.7 lb/ft, EUE, J-55 to 7000

ft.

d. Use Internal n and K,

c. Set Increment to 1
e. Go to the Wellbore tab by clicking the
right or left arrow on the toolbar.

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.

f.

PISIP = 1800 psi

Navigation icons

g. Fill in the tubing and casing information

Solution:

h. Hit F5 key or click the DO icon

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i.
j.

DO icon

So, the Bulk Density (or Specific Gravity) is

measured as if the proppant were a solid and not
made up of individual particles.

Click the Text Output Icon

k.
l.

specific gravity is measured in grams per cc

(cubic centimeter).

Text icon

Example:

h. The program arrives at the value of

10318.6 psi at 10 bbl/min.

What is the slurry density (lb/gal) and slurry

volume (gal) of fresh water with 2 lb/gal Ottawa
proppant added?

Slurry Density and Volume

Solution:

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.

Set up a table as shown:

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

Absolute
Volume
(gallons)

Fresh
Water

8.33

----

Sand

0.0452

0.0912

TOTALS

10.33 lb

1.0912 gal

Divide total pounds by total gallons to calculate

slurry density.

Table 2.1 Absolute Volume Factors

Bulk
Density
(lb/ft3)

Absolute
Volume
Factor
(gal/lb)

Materials

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.

PROPPANT TYPE

Materials
(pounds)

Slurry Density =

10.33 lb
lb
lb
=
= 9.4666
gal 1.0912 gal
gal

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:
gal

Dirty Vol = 2000 gal + 4000 lb sand 0.0452

lb

= 2000 gal + 180.8 gal

= 2180.8 gal

(1 ft is equal to one sack of proppant)

The absolute volume of proppant is calculated

from the specific gravity of the proppant. The

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maximum allowable value. Usually, the job is

shut down at that point.

Example:

What is the slurry density and "dirty" volume?

Example:

Fracturing fluid is Diesel #2 with a density

of 7.33 lb/gal.

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?

Sand concentration is 10 lb/gal.

Stage size is 10,000 gallons "clean" volume.

Solution:
Materials

Diesel #2
Sand
TOTALS

Materials
(pounds)

Absolute
Volume
Factor
(gal/lb)

Absolute
Volume
(gallons)

7.33

----

10

0.0452

0.452

17.33 lb

Slurry Density =

Casing Capacity = 3000ft 0.6699

= 2009.7 gal

Therefore, we have 2009.7 gallons of slurry in

the casing. In order to calculate the sand volume
we need to use the equation.

1.452 gal

lb
17.33lb
lb
=
= 11.935
gal 1.452gal
gal

Volume Factor = 1 + (Prop Conc Abs Vol)

lb
gal
0.0452
lb
gal
= 1 + 0.1808 = 1.1808
= 1 + (4

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

10,000 gal + 10
10,000 gal 0.0452
lb
gal
= 14,520 gal " dirty" volume

To calculate the clean volume, rearrange the

following equation:
Slurry Volume = Clean Vol Volume Factor
Slurry Volume
Volume Factor
2009.7 gal
Clean Vol =
1.808
= 1701.9817 gal 2% KCL water
Clean Vol =

Instead of using a table you can use the

following equations for Slurry Density, Slurry
Volume, and Volume Factor:
Slurry =

BaseFluid (lb/gal) + Prop Conc(lb/gal)

Volume Factor

Now to calculate the sand volume:

where:

Wsand = 1701.9817 gal 4

Slurry = Slurry Density

BaseFluid = Base Fluid Density

lb
= 6807.927 lb
gal

Since there are 95.9 lb of Ottawa sand in one

sack: (Table 2.1):

Prop Conc = Proppant Concentration

Volume Factor =

Vs =

lb
gal
Abs Vol factor
1 + Prop Conc

lb
gal

6807.9268 lb
= 71 sacks of sand
lb
95.9
sk

To calculate the hydrostatic pressure, we need to

use a different equation:

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

2005, Halliburton

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Slurry Density =

Base fluid density + sand concentration

volume factor

The volume factor (1.1808) has already been

calculated.
lb
lb
+4
gal
gal
Slurry Density =
1.1808
lb
12.43
gal
=
1.1808
lb
= 10.5268
gal
8.43

lb
0.05195
gal
psi
= 0.5469
ft
psi
Ph = 3000 ft 0.5469
ft
= 1640.6 psi
or

Ph Gradient = 10.5268

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:
PISIP = 1775 psi (given)
Pfrict = 2966.1 psi (from the Friction Program)
Pw = 1775psi + 2966.1psi + 0psi

Ph = 3000 ft 0.5455 (RedBook)

= 4741.1psi

= 1636.5 psi

Hydraulic Horsepower

Wellhead Pressure

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.

The equation for calculating pressure at the

wellhead is
Pw = BHTP - Ph + Pfrict + Pperf or
= PISIP + Pfrict + Pperf (since PISIP = BHTP - Ph)

bbl
Pw (psi ) Rate

min
HHP =
40.8
or

Where:
BHTP = Bottomhole Treating Pressure
Ph = Hydrostatic Pressure

gal
Pw (psi) Rate

min
HHP =
1713.6

Pfrict = Fluid friction from Surface to the top

perforation
Pperf = Fluid friction across all perforations

The value 1713.6 is 40.8 42 gal/bbl

PISIP = Instantaneous Shut In Pressure

Example:

Example:

What is the HHP under these conditions?

Pressure at the wellhead is 3000 psi

Injection rate is 30 bbl/min.

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

7700 ft.

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Solution:

lb
h- 8.33
gal

bbl
Pw (psi ) Rate

min
HHP =
40.8
bbl
3000 psi 30
min
=
40.8
= 2205.88 HHP

P
h-10

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

ISIP-10

psi
9100 ft = 4727.45 psi
ft

lb
gal

= 1800 psi 787.15 psi = 1012.85 psi

Pperf = 150psi (given)

Pw = 1012.85 psi + 6300 psi + 150 psi
= 7462.85 psi
bbl
7462.85 psi 30
Pw Rate
min
=
HHP =
40.8
40.8
= 5487.390 HHP

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:
HHP 40.8 = Pw Rate

Now divide both sides by Pw. This gives you rate

in bbl/min.

Solution:
Pfrict = 70

= 0.5195

The change is an increase in Ph

What is the Pfrict, Pw, and HHP under these

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

psi
9100 ft = 3940.3 psi
ft

Ph = 4727.45 psi 3940.3 psi = 787.15 psi

Example:

lb
gal

= 0.433

psi
9000ft = 6300psi
100ft

HHP 40.8
bbl
= Rate

Pw
min

Pw = PISIP + Pfrict + Pperf

Example:

Pisip with fresh water = 1800 psi (given).

What is the maximum pump rate in bbl/min that

can be delivered at maximum psi under these
conditions?

Fracturing fluid is 10 lb/gal.

Solution to the problem requires PISIP be
calculated with 10 lb/gal fluid.

Treating fluid is 15% HC1 acid, 8.962 lb/gal

1000 HHP is available at the location.

Maximum wellhead pressure is 5700 psi.

Solution:
bbl
bbl 1000 HHP 40.8
Rate
= 7.1579
=
5700 psi
min
min

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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?

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

2005, Halliburton

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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
(Pounds)

Absolute
Volume Factor (Gal/Lb)

Absolute
Volume (Gallons)

2% KC1 _______________________

_________________________

____________________

Proppant _______________________

_________________________

____________________

TOTAL _______________________ lb

_________________________gal/lb

____________________ gal

BHTP = _______________________ ft

_________________________psi/ft =

____________________ psi

Ph

_______________________ ft

_________________________psi/ft =

____________________ psi

_______________________ psi -

_________________________psi =

____________________ psi

Slurry Density = __________________lb/gal

PISIP =

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

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

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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.

132 in.
21 ft 6 ft = 1386ft 3
in
12
ft
bbl
1386ft 3 0.1781 3
ft
= 246.847 bbl
246.8466 bbl 42

gal
bbl

= 10,367.56 gal

2.

50 ft 30 ft 15 ft = 22,500 ft 3
bbl
22,500 ft 3 0.01781 3
ft
= 4007.25 bbl
4007.25 bbl 42

gal
bbl

= 168,304.5 gal
246.8466 bbl
bbl
= 41.14
6ft
ft
4007.25 bbl
bbl
Q2=
= 267.15
15 ft
ft
Q1=

3.

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A c = 6ft 6ft 0.7854 = 28.2744ft 2

28.2744ft 2 20ft = 565.488ft 3
4.

565.488ft 3 0.1781

bbl
ft 3

= 100.713 bbl
100.71341 bbl
bbl
= 5.0357
20ft
ft
ft
bbl
1
5.0357
= 5.0357 BPM
min
ft

5.

Tubing = 8000ft .00579

bbl
ft

= 46.322 bbl
6. a.

Casing = (8213 8000)ft .0238

bbl
ft

= 5.0694 bbl
5.0694 bbl + 46.322 bbl
= 51.389 bbl
b. T

51.389 bbl
= 2.8550 min
18BPM

c. V = 8000ft 0.0158

d.

bbl
= 126.4 bbl
ft

lb
gal
0.05195 2
gal
in ft
= 3839.99psi

Ph = 8213ft 9

18bbl 42gal
4gal

min
bbl 1000gal
gal
= 3.024
ClaySta XP
min
or
Rate 1 =

gal
min
e.
gal
1gal
bbl
Rate 2 = 18
42

min
bbl 1000gal
gal
= 0.756
ScaleChek
min
or
gal
18 .042 1 = 0.756
min
18 .0 42 4 = 3.024

2005, Halliburton

2 28

Stimulation I

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.

psi
8000ft
ft
= 3864.8 psi

Ph = 0.4831

BHTP = 1200psi + 3864.8psi

= 5064.8psi
psi
9050ft
ft
= 4560.295psi

Ph = 0.5039

2.

BHTP = 1975psi + 4560.295psi

= 6535.295psi
6535.295psi
9050ft
psi
= 0.722
ft

Frac Grad. =

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

13.92 lb
lb
= 11.1485
1.2486 gal
gal
psi
BHTP = 7450 ft 0.65
ft
= 4842.5 psi
psi
3. Ph Grad = 0.5766
(RedBook)
ft
psi
Ph = 7450 ft 0.5766
ft
= 4295.67 psi
PISIP = 4842.5 psi 4295.67 psi
= 546.83 psi
Density =

2005, Halliburton

2 29

Stimulation I

Calculations

Tubing = 9000ft 0.00579

bbl
ft

= 52.11 bbl
4. a. Casing = 50ft 0.0393

bbl
ft

= 1.965 bbl
Volume = 52.11bbl + 1.965bbl
= 54.075 bbl

b.

Vann = 9000ft 0.0313

bbl
ft

= 281.7 bbl

psi
100ft 9000ft
Pfrict =
ft
100
100ft
= 3447 psi
38.3

c.

Pw = PISIP + Pfrict + Pperf

lb = 2000 psi

ISIP 10

gal

lb
h 10
gal

h 8.9

d.

lb
gal

= 0.5195

psi
9050 ft
ft

= 4701.475 psi
psi
= 0.4623
9050 ft
ft

= 4183.815 psi
Ph = 4701.475 4183.815
= 517.66 psi
lb = 2000 psi + 517.66 psi

ISIP 8.9

gal

= 2517.66 psi
Pw = 2517.66 psi + 2447 psi
= 5964.66 psi

Pw Q
40.8
5964.66psi 20BPM
=
40.8
= 2923.853 HH P
HHP =

e.

2005, Halliburton

2 30

Stimulation I

Calculations

f.

lb
gal

0.0444
Vol factor = 1 + 5
lb
gal
= 1 + 0.222
= 1.222
8.93 lb + 5 lb
lb
Density =
= 11.399
1.222gal
gal

Ph = 9050ft 0.5922(RedBook )
= 5359.41 psi

Clean Vol =

54.075bbl 42

gal
bbl

1.222
= 1858.55 gal

g.

Wsand = 1858.55gal 5

lb
gal

= 9292.76 lb
Vsand =

9292.76lb
= 96.8 sks
lb
96
sk

2005, Halliburton

2 31

Stimulation I

Calculations

Self-Check Test Answer Key

1. laminar flow
2. barrels per minute
3.

psi
(RedBook) 8000ft = 3491.2 psi
ft
BHTP = 2575psi + 3491.2psi = 6066.2 psi

4.

BHTP = 11000ft 0.82

5.

Ph = 0.4364

psi
= 9020 psi
ft

Material Material (lb)

2% KC1

Abs. Vol. Factor (gal/lb)

8.42

Proppant

TOTALS

11.4

0.0444__

Abs. Vol. (gal)

1
0.1332
1.1332

11.42lb
lb
= 10.0777
1.1332gal
gal
psi
BHTP = 9060 ft 0.72
= 6523.2 psi
ft
psi
Ph = 9060 ft 0.5247
(RedBook) = 4753.78 psi
ft
PISIP = 6523.2 psi 4753.78 psi = 1769.42 psi

SlurryDensity =

psi
100ft 6300ft = 483.84 psi
100ft

7.68

6.

7.

Pfrict =

HHP =

6000psi 12
40.8

bbl
min = 1764.706 HHP

Vtubing = 10,500ft 0.00387

8 a.

bbl
= 40.6350 bbl
ft

bbl
= 5.8125 bbl
ft
= 40.6350bbl + 5.8125bbl = 46.4475 bbl

Vcasing = 375ft 0.0155

Vtotal

2005, Halliburton

2 32

Stimulation I

Calculations

b.

Vann = 10,500ft 0.0101

bbl
= 106.05 bbl
ft

V(ft 3 ) = 20ft 10ft 8ft = 1600ft 3

c.

V(bbl) = 1600ft 3 0.1781

bbl
= 284.96 bbl
ft

d.

bbl
PumpRate

min
in

Rate
=
min TankFactor bbl

in
284.96bbl
bbl
= 2.96833
Tank Factor =
96in
in
bbl
10
in
in
min
= 3.369
Rate
=
min
min 2.96833 bbl
min

e.

Pfrict = 41.96

f.

Pw = 2900 psi + 4405.8 psi + 200 psi = 7505.8 psi

psi
10,500ft = 4405.8 psi
100ft

7505.8psi 10

bbl
min = 1839.657 HHP

g.

HHP =

h.

lb
gal
= 1.3064
Volume Factor = 1 + 8
0.0383
lb
gal
gal
46.4475bbl 42
bbl = 1493.260 gal
CleanVolume =
1.3064
lb
WInterProp = 1493.260gal 8
= 11,946.08 lb
gal
11,946.08lb
VInterProp =
= 99.551 sacks
lb
120
sk

40.8

2005, Halliburton

2 33

Stimulation I

Calculations

lb
lb
+8
gal
gal
lb
Slurry Density =
= 12.569
1.3064
gal
lb
Ph = 10875ft 12.569
0.05195 = 7100.935 psi
gal
8.42

i.

ft 3
(RedBook) = 32.7ft 3
ft
= 99.551ft 3 32.7ft 3 = 66.851ft 3

Vcasing = 375ft 0.0872

j.

VProp in tubing

ft
(RedBook) = 3079.625ft
ft 3
Top of Proppant = 10,500ft - 3079.625ft = 7420.375 ft
Fill = 66.851ft 3 46.067

l.

Pipe Time =

46.4475 bbl
= 4.64475 min
bbl
10
min

LA - Rate = 10

bbl 2gal 0.042

= 0.84
min Mgal
bbl

2005, Halliburton

gal
min

2 34

Stimulation I