Fluid-Flow - Operations Training

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OPERATIONS TRAINING PROGRAM

STUDENT TEXT

Rev. 0

FLUID FLOW




Contents:---

Table of Contents:

NOTICE: If you plan to use this material in a classroom setting, then please purchase

the exam bank and answer key from the Scribd store for $4.99 or visit

marathonjohnb at Scribd. The exam is given at the end of the course and has

specific questions for each chapter..............................................................................ii

Contents:---..........................................................................................................................iii

Chapter 1 INTRODUCTION TO FLUIDS......................................................................12

Introduction ...........................................................................................................................................12

Description of Fluids..............................................................................................................................13

Humidity.................................................................................................................................................14

Relative Humidity..................................................................................................................................14

Density () and Specific Volume ().........................................................................................................14

Density Differences for Non-Mixable (Non-Miscible) Fluids.............................................................15

Specific Gravity......................................................................................................................................18

Pressure (p) ............................................................................................................................................21

Pressure Measurements.........................................................................................................................23

Absolute, Gage, and Vacuum Pressure Relations...............................................................................30

Buoyancy.................................................................................................................................................32

Hydrostatic Pressure.............................................................................................................................34

Pascal's Law (the law of hydraulics)....................................................................................................39

Pressure Difference for Fluid Flow......................................................................................................41

Chapter 1 Summary..............................................................................................................................43

Chapter 2 Compression of Fluids......................................................................................45Compressibility......................................................................................................................................45

The Combined Gas Law........................................................................................................................45

Effects of Pressure Changes on Confined Fluids ...............................................................................47

Effects of Temperature Changes on Confined Fluids........................................................................48

Filling and Venting................................................................................................................................48

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Chapter 2 Summary..............................................................................................................................51

Chapter 3 NATURAL CIRCULATION FLOW..............................................................53

Natural Circulation................................................................................................................................53

Conditions Required For Natural Circulation....................................................................................54

Chapter 3 Summary..............................................................................................................................56

Chapter 4 VOLUMETRIC AND MASS FLOW RATE..................................................57

Volume (V)..............................................................................................................................................57

Volumetric Flow Rate ()........................................................................................................................59

Mass, Density, and Specific Volume.....................................................................................................64

Mass Flow Rate ()...................................................................................................................................66

The Steady Flow Condition...................................................................................................................68

Continuity of Flow.................................................................................................................................68

Chapter 4 Summary..............................................................................................................................75

Chapter 5 TYPES OF FLOW............................................................................................77

Laminar Flow.........................................................................................................................................77

Turbulent Flow.......................................................................................................................................77

Factors Influencing Type of Flow.........................................................................................................78

Ideal Fluid...............................................................................................................................................79

Noise Level and Flow Rate....................................................................................................................79

Chapter 5 Summary..............................................................................................................................80

Chapter 6 FORMS OF ENERGY &THE GENERAL ENERGY EQUATION......... ..81General Energy Equation......................................................................................................................81

Potential Energy (PE)............................................................................................................................83

Kinetic Energy (KE)..............................................................................................................................84

Flow Energy (FE)...................................................................................................................................84

Internal Energy (U)................................................................................................................................87

Heat, as an operator controlled input or output (Q)..........................................................................88

Work, as an operator controlled input or output (W)........................................................................89

General Energy Equation......................................................................................................................89

A Special Case of the General Energy Equation: Bernoulli's Principle...........................................92

Simplified Bernoulli's Equation............................................................................................................94

Specific Energies....................................................................................................................................96

Chapter 6 Summary:.............................................................................................................................98

ENERGY CONVERSIONS IN IDEAL FLUID SYSTEMS..........................................99

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Energy Conversions in Ideal Fluid Systems........................................................................................99

Energy Conversions for Changes in Cross-Sectional Area (Flow Area)..........................................99

Energy Conversions for Changes in Elevation..................................................................................102

Chapter 7 Summary:...........................................................................................................................105

Chapter 7 Energy Conversions in Real Fluid Systems..................................................107

Friction..................................................................................................................................................107

Fluid Friction........................................................................................................................................107

Viscosity................................................................................................................................................108

Energy Conversion by Fluid Friction in Real Fluids........................................................................108

Energy Conversion by Fluid Friction................................................................................................110

Open versus Closed Fluid Flow Systems...........................................................................................116

Energy Conversions in Closed Systems.............................................................................................116

_Head_ ...............................................................................................................................................120

Head Loss due to Friction...................................................................................................................125

Throttling..............................................................................................................................................126

Overcoming Head Losses....................................................................................................................127

Centrifugal Pump Operation..............................................................................................................128

Positive Displacement Pump Operation............................................................................................129

Using the General Energy Equation to Analyse Real Fluids...........................................................130

Specific Rules Using Arrow Analysis.................................................................................................135

The General Energy Equation and Diagnosis using Arrow Analysis.............................................137

Chapter 8 Summary:...........................................................................................................................149

Chapter 8 Fluid Flow Measurement................................................................................151

Flow Measuring Devices......................................................................................................................151

Differential Pressure Meters...............................................................................................................151

Orifice Plates........................................................................................................................................151

Flow Nozzles.........................................................................................................................................153

Venturi Tubes.......................................................................................................................................153

Other Applications of the Venturi Principle.....................................................................................154

Chapter 9 Summary:...........................................................................................................................157

Water Hammer and Pipe Whip......................................................................................159

Mechanisms of Water Hammer..........................................................................................................159

Occurrence of Water Hammer (and Steam Hammer).....................................................................159

Cavitation..............................................................................................................................................165

Cavitation in Centrifugal Pumps........................................................................................................165

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Net Positive Suction Head (NPSH).....................................................................................................167

Conditions Causing Cavitation...........................................................................................................168

Minimizing Gas Formation in Liquid Piping Systems.....................................................................170

Other Pump Problems.........................................................................................................................170Possible Results of Water Hammer....................................................................................................171

Methods of Water (and Steam) Hammer/ Pipe Jet & Pipe Whip Prevention ..............................174

Chapter 10 Summary..........................................................................................................................176

Chapter 9 Unintended Siphoning.....................................................................................177

Introduction..........................................................................................................................................177

Siphoning..............................................................................................................................................177

Chapter 11 Summary..........................................................................................................................180

List of Figures:

Figure 1-1 Example of non-miscible fluids.......................................................................15

Figure 1-2 Pressure caused by Molecules.........................................................................22

Figure 1-3 Force versus Pressure......................................................................................22

Figure 1-4 Pressure Scales.................................................................................................24

Figure 1-5 Typical Pressure Gage.....................................................................................24

Figure 1-6 Liquid Supported by Atmospheric Pressure.................................................25Figure 1-7 Buoyancy Forces on an Object........................................................................33

Figure 1-8 Relationship between Liquid Level and Pressure.........................................34

Figure 1-9 Pressure Versus Height....................................................................................35

Figure 1-10 Static Head versus Pressure..........................................................................36

Figure 1-11 Head and Pressure Illustration....................................................................38

Figure 1-12 Pressurizing a.................................................................................................40

Figure 1-13 Hydraulic System Forces...............................................................................40

Figure 1-14 A Simple Hydraulic System..........................................................................41

Figure 3-15 Air Baloon Buoyancy....................................................................................53

Figure 3-16 Heat Source / Heat Sink.................................................................................54

Figure 4-17 Volume of an Object.......................................................................................57

Figure 4-18 Volume of Pipe Section A..............................................................................58

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Figure 4-19 Volume of Pipe Section B...............................................................................58

Figure 4-20 Volumetric flow Rate Visual..........................................................................59

Figure 4-21 Volumetric Flow rate Between Two Points..................................................60

Figure 4-22 Volumetric Flow Rate Example 1..................................................................61

Figure 4-23 Volumetric Flow Rate Example 2..................................................................63

Figure 4-24 Mass Flow Rate Example...............................................................................67

Figure 4-25 Continuity of Flow..........................................................................................68

Figure 4-26 Continuity Example 2.....................................................................................71

Figure 5-27 The Two Basic Types of Fluid Flow..............................................................78

Figure 6-28 A Visual of Potential Energy..........................................................................83

Figure 6-29 Visual of Kinetic Energy................................................................................84

Figure 6-30 Flow Energy in Compressing Piston.............................................................85

Figure 6-31 Flow Energy in Fluid Flow through a Pipe..................................................85

Figure 6-32 Visual of Flow Energy...................................................................................86

Figure 6-33 Visual of Internal Energy..............................................................................87

Figure 6-34 Visual of Heat Energy.....................................................................................88

Figure 6-35 Visual of Work Energy...................................................................................89

Figure 6-36 Fluid Energies 'IN' versus 'OUT'..................................................................90

Figure 6-37 Energies Added versus Energies Removed..................................................90Figure 6-38 Visual of the General Energy Equation........................................................91

Figure 6-39 Bernoulli's Principle.......................................................................................92

Figure 6-40 Ping Pong Ball Floating in Air Stream.........................................................93

Figure 6-41 Air Passing Above and Below Airplane Wing .............................................93

Figure 6-42 Air Passing by a Thrown Baseball................................................................94

Figure 7-43 Pipe Section with a Reduction in Area........................................................101

Figure 7-44 Pipe Section With Increase in Area.............................................................101

Figure 7-45 Pipe Section with Increasing Elevation.......................................................103

Figure 7-46 Pipe Section with Decreasing Elevation......................................................103

Figure 8-47 Straight Pipe Section....................................................................................109

Figure 8-48 Pipe Section with Changes in size and Elevation.......................................109

Figure 8-49 The Pressure Drop from a 1°F Temperature Rise....................................111

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Figure 8-50 Pressure Drop and Fluid Friction...............................................................114

Figure 8-51 Energy Conversions in a Closed System.....................................................117

Figure 8-52 A Simple Closed Loop System.....................................................................119

Figure 8-53 Closed Loop Example...................................................................................119

Figure 8-54 Pressure is Proportional to Column Height...............................................120

Figure 8-55 Pressures Within a Fluid Flow System (exaggerated)...............................123

Figure 8-56 Total Static Head Examples.........................................................................125

Figure 8-57 Typical Valve.................................................................................................127

Figure 8-58 A Centrifugal Pump......................................................................................128

Figure 8-59 Pressures Within a Centrifugal Pump........................................................129

Figure 8-60 Positive Displacement Pump........................................................................130

Figure 8-61 General Energy Equation in Mental Form................................................130

Figure 9-62 A Simple Orifice Plate..................................................................................152

Figure 9-63 A Simple Flow Nozzle...................................................................................153

Figure 9-64 Simple Venturi Tube....................................................................................154

Figure 9-65 Auto Carburetor Uses Venturi Principle....................................................154

Figure 9-66 A Typical Steam Jet......................................................................................155

Figure 9-67 A Simple Eductor..........................................................................................156

Figure 10-68 Case 1 Valve Quickly Closed.....................................................................162Figure 10-69 Case 2 Valve Quickly Opened....................................................................162

Figure 10-70 Case 3: Cold Condensate in Steam Line..................................................163

Figure 10-71 Case 4: Hot Condensate in Steam Line....................................................163

Figure 10-72 Case 5: Boiling...........................................................................................164

Figure 10-73 Cavitation in a Centrifugal Pump............................................................166

Figure 10-74 Cavitation and the Collapsing Bubble.....................................................167

Figure 10-75 Pump Runout .............................................................................................168

Figure 10-76 Low Suction Pressure.................................................................................169

Figure 10-77 Pipe Rocket / Pipe Jet.................................................................................172

Figure 10-78 Pipe Whip....................................................................................................173

Figure 11-79 Example of a Siphon..................................................................................178

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List of Tables:

Table 1-1 Densities of Common Materials.......................................................................15

Table 1-2 Densities of Common Fluids.............................................................................21

Table 1-3 Common Pressure Units...................................................................................26

Table 1-4 Absolute, Gage and Vacuum Pressure............................................................30

List of Terminal Objectives:

TO 1.0Given the necessary fluid system parameters, SOLVE for unknown fluidparameter values as system conditions are varied...................................................12

TO 2.0Given the necessary fluid system parameters and using the Combined Ideal

Gas Law, DESCRIBE the compressibility or incompressibility of a fluid when a

pressure is exerted.......................................................................................................45

TO 3.0For any natural circulation fluid system, DESCRIBE the mechanism that

allows for fluid flow.....................................................................................................53

TO 4.0Using fluid system volumetric and mass flow rates, SOLVE for unknown fluid

parameters values to predict fluid system characteristics.......................................57

TO 5.0Given the necessary fluid system parameters, DETERMINE the fluid flow typeand the flow characteristics of that fluid system......................................................77

TO 6.0Given a fluid system, IDENTIFY the forms of energy using the General Energy

Equation.......................................................................................................................81

TO 7.0GIVEN an Ideal fluid system where no heat is transferred in or out, and no

work is performed on or by the fluid, EXPLAIN the energy conversions that

occur ............................................................................................................................99

TO 8.0GIVEN a Real fluid system, DESCRIBE the effects of fluid friction to predict

energy conversions....................................................................................................107

TO 9.0EXPLAIN the energy conversions that occur as fluid flows through the Venturi

tube, flow nozzle, and orifice plate flow measuring devices .................................151

TO 10.0IDENTIFY the conditions and prevention methods for both "water hammer"

and "pipe whip" in fluid systems.............................................................................159

TO 11.0IDENTIFY the conditions and prevention methods of a fluid siphon for a fluid

system.........................................................................................................................177

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References:

ARITHMETIC : Student Text, TTFGMAPA.H0102, rev. 2 / Westinghouse Savannah River Company, Aiken, SC

MATHEMATICS : Student Text, TTFGMA1A.H0104, rev. 4 / Westinghouse Savannah River Company, Aiken, SC

Bay, Denise and Horton, Robert B., Macmillan Physical Science, Teacher's Edition, Macmillan Publishing Co., New

York, (1988).

Cline, John W., Thermodynamics, Heat Transfer, and Fluid Flow, Westinghouse Savannah River Company HLW

Fundamentals Training Program, (1993).

Driskell, Les., Control Valve Selection and Sizing , Instrument Society of America, North Carolina, (1983).

Driskell, Les, Control-Valve Selection and Sizing , Independent Learning Module, Instrument Society of America,

Publishers Creative Services Inc., Research Triangle Park, North Carolina, (1983).

Durham, Franklin P., Thermodynamics, 2nd ed., Prentice-Hall, Inc., New Jersey, (1959).

Freeman, Ira M., Physics Made Simple, Revised Edition, Bantan Doubleday Dell Publishing Group, Inc., New York,

(1990).Giancoli, Douglas C., Physics, 3rd ed, Prentice Hall, New Jersey, (1991).

Glasstone, Samuel and Sesonske, Alexander, Nuclear Reactor Engineering, 3rd ed., Van Nostrand Reinhold Co.,

New York, (1981).

Heimler, Charles H. and Price, Jack S., Focus on Physical Science, Teacher's Edition, Charles E. Merrill Publishing

Co., Ohio (1984).

Hewitt, Paul G., Conceptual Physics....a new introduction to your environment , 3rd ed., Little Brown and Company,

Inc., Boston, (1977).

Holman, J. P., Thermodynamics, 4th ed., McGraw Hill, Inc., New York, (1988).

Julty, Sam, How Your Car Works, Book Division, Times Mirror Magazines, Inc., New York (1974).

Murphy, James T., Zizewitz, Paul W., and Hollon, James Max, Physics Principles & Problems, Charles E. Merrill

Publishing Co., Ohio, (1986).Serway, Raymond A. and Faughn, Jerry S., College Physics, 2nd ed., Saunders College Publishing, Philadelphia,

(1989).

U.S. Department of Energy, DOE Fundamentals Handbook, Thermodynamics, Heat Transfer, and Fluid Flow,

Vols. 1 through 3, U.S. Department of Energy, (1992).

Wiedner, Richard T. and Sells, Robert L., Elementary Classical Physics, College Physics Series, Vol. 1, Allyn and

Bacon, Inc, Boston, (1965).

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OPERATIONS TRAINING PROGRAM Student Guide: Fluid Flow Chapter 1: Introduction to Fluids

Chapter 1 INTRODUCTION TO FLUIDS

This chapter introduces various terms used to describe the characteristics of a fluid and some basic flow

characteristics of given fluids in a typical application. It also presents the relationship between

various parameters within a given fluid system under various conditions

TO 1.0 Given the necessary fluid system parameters, SOLVE for unknown

fluid parameter values as system conditions are varied

EO 1.1 DEFINE the following Fluid Flow terms to include their

typical units: specific volume, density, and specific gravity

EO 1.2 EXPLAIN what will occur when two non-mixable fluids are

placed in the same container

EO 1.3 CALCULATE a fluid’s density, specific volume, or specific

gravity when given any one of the other quantities

EO 1.4 DEFINE the Fluid Flow term “Pressure” to include units

EO 1.5 Given the necessary fluid parameters,

CALCULATE/CONVERT absolute pressure, gage pressure,

feet of head, or vacuum pressure for a fluid system

EO 1.6 EXPLAIN Archimede’s Principle and relate it to the term

“Buoyancy”

EO 1.7 DESCRIBE the relationship between the pressure in a fluid

column and the density and depth of the fluid

EO 1.8 DEFINE the Fluid Flow term “Head” to include units

EO 1.9 EXPLAIN the concept of Pascal’s law, including its

applications.

Introduction

Fluid flow is an important part of most industrial processes; especially those involving the transfer of

heat. Frequently, when it is desired to remove heat from the point at which it is generated, some type of

fluid is involved in the heat transfer process. Examples of this are the cooling water circulated through a

gasoline or diesel engine, the air flow past the windings of a motor, and the flow of water through the

core of a nuclear reactor. Fluid flow systems are also commonly used to provide lubrication.

Fluid flow in the nuclear field can be complex and is not always subject to rigorous mathematical

analysis. Unlike solids, the particles of fluids move through piping and components at different velocities

and are often subjected to different accelerations.

Page 12 of 170 Rev. 0




Even though a detailed analysis of fluid flow can be extremely difficult, the basic concepts involved influid flow problems are fairly straightforward. These basic concepts can be applied in solving fluid flow

problems through the use of simplifying assumptions and average values, where appropriate. Even though

this type of analysis would not be sufficient in the engineering design of systems, it is very useful in

understanding the operation of systems and predicting the approximate response of fluid systems tochanges in operating parameters.

The basic principles of fluid flow include three concepts or principles; the first two of which the student

has been exposed to in previous manuals. The first is the principle of momentum (leading to equations

of fluid forces) which was covered in the manual on Classical Physics. The second is the conservation of

energy (leading to the First Law of Thermodynamics) which was studied in thermodynamics (Heat

Transfer). The third is the conservation of mass (leading to the continuity equation) which will be

explained in this module.

Description of Fluids

A fluid is any substance that flows. The molecules of fluids are not rigidly attached to each other.Essentially, fluids are materials which have no repeating crystalline structure. Fluids include both liquids

and gases. Liquids are fluids which have a definite volume and take the shape of their container. Gases

also take the shape of their container; however, they will expand to completely fill the container thus theydo not have a definite volume.

Several properties of fluids are discussed in the Heat Transfer course. These include temperature,

pressure, mass, specific volume and density.

Temperature is defined as the relative measure of how hot or cold a material is. It can be used to

predict the direction that heat will be transferred.

Pressure is defined as the force per unit area. Common units for pressure are pounds force per square inch (psi).

Mass is defined as the quantity of matter contained in a body and is to be distinguished from

weight, which is measured by the pull of gravity on a body.

The specific volume of a substance is the volume per unit mass of the substance. Typical units

are ft3 /lbm.

Densit y, on the other hand, is the mass of a substance per unit volume. Typical units are lbm/ ft3

Density and specific volume are the inverse of one another. Both density and specific volume are

dependant on the temperature and somewhat on the pressure of the fluid. As the temperature of the fluid increases, the density decreases, and the specific volume increases. Since liquids are

considered incompressible, an increase in pressure will result in no change in density or specific

volume of the liquid. In actuality, liquids can be slightly compressed at high pressures, resulting

in a slight increase in density and a slight decrease in specific volume of the liquid.





Humidity

Humidity is the amount of liquid vapor suspended in a gas (or the amount of water in air). The units of

humidity are grains per cubic foot. (A grain is the weight of a wheat seed.)

Relative Humidity

Relative humidity is the percentage of liquid that a gas contains compared to being 100% saturated,

(where it can hold no additional liquid). Units are percent.

Relative humidity is a percentage measurement of humidity up to and including saturation at 100% at any

particular temperature. Since air holds more water when it is at a higher temperature, air that is saturated

and then heated will have the capacity to hold more water and will no longer be termed "saturated". The

relative humidity of air will then be less than 100% if it's temperature is increased. As a result, without

changing the amount of liquid suspended within a gas, and by only changing the temperature, the relative

humidity can vary from saturated at 100% relative humidity to something considerably less than

saturated.

A gas can not contain more than 100% of its liquid holding capacity. If the temperature of a 100%

saturated gas is decreased then its capacity to hold moisture decreases and the liquid precipitates. This is

why dew accumulates on leaves and grass when the temperature goes down in the early morning hours.

Density ( ρ ) and Specific Volume ( υ )

Density, ρ , is the amount of mass contained in one cubic foot of space; units are mass per unit Volume.

Specific volume, υ , is the amount of space occupied by one pound mass (the force of one pound

converted to mass by dividing by g c ); units are Volume per unit mass.

Specific volume is the inverse of density;

υ =1

ρ & ρ =

1

υ .

Where:

ρ = Density (Greek letter rho), lbm/ft3

or kg/m3

, etc.

m = mass, lbm or kg, etc.

V = Volume, ft3

or m3, etc.

υ = Specific volume, ft3

/lbm or m3

/kg , etc.

Both density and specific volume measure the same property: how close the molecules or atoms of a

substance are to each other.

Volume (V ) is the amount of space occupied by a three-dimensional figure. Volume is represented by

length units cubed





( ft 3 , in

3 , m

3 , etc.).

The specific volume is the amount of space occupied by a unit of mass. Specific volume is the total

volume V divided by the total mass m of an object.

υ = V

m

Where:

υ = specific volume, ft 3 /lbm

V = volume, ft 3

m = mass, lbm

A low value of density (or high value of specific volume)

means the molecules or atoms in the substance are relativelyfar apart. This is true of gases (hydrogen, oxygen) and for

vapors such as steam. Conversely, a high value of density (or

low value of specific volume) means that the molecules or

atoms are relatively close together. This is true of liquids

(such as water) and solids such as ice.

The density of a material will govern the way it behaves

when put in contact with other materials. Table 1-1 lists the

densities of some common materials. If a material that is

very dense is placed into a container containing a less-denseliquid, the material will sink. For example, if a piece of iron

is placed into a container of water, the iron will sink because

it is more dense than water.

If, however, that same piece of iron is placed in a liquid that

is more dense, such as mercury, the iron will float. Even

though iron is relatively dense, it is not as dense as the

mercury.

Density Differences for Non-Mixable (Non-Miscible) Fluids

Miscibility is the property of two substances, which

makes them "mixable". Salt and water are miscible so

when they are mixed together they make salt water and

stay mixed until separated by evaporation. But when


Densities of Some Common Materials:

MaterialDensity, g/cm

3

hydrogen9.0 x 10

-5

helium2 x 10

-4

air 1.3 x 10

-3

Styrofoam 0.1

wood 0.7alcohol 0.8

ice 0.92

water 1.0

sea water 1.03

aluminum 2.7

rock 3

iron 7

mercury 13.6

Table 1-1 Densities of Common Materials

Figure 1-1 Example of non-miscible fluids




two substances are non-miscible (not mixable), like oil and water, the water, being the highest density

liquid sinks to the bottom of the container, and the oil being less dense rises to the top. They "unmix"themselves very quickly. Oil and vinegar salad dressing is an example of this.

Why do some fluids "unmix" themselves? Vinegar is more dense than olive oil; therefore, under the

attraction of gravity the vinegar moves to the bottom of the container. The object with greater masscreates a greater pressure around itself than an object with a smaller mass. As a result, the lighter objects

get “pushed” out of the way. This pressure then forces the lighter oil molecules out of the lowest regions

and upward where the pressure is lower. A layering effect is created in a salad dressing bottle where the

denser vinegar, under the influence of earth’s gravitation, occupies the bottom of a container while the

less dense olive oil is forced to rest on top.

Some gases do not mix well with other gases. As an example, certain subterranean bunkers that

contained poisonous chlorine gas used during the Second World War is still a potential health hazard for

Europeans. Chlorine gas is a nerve agent and is heavier than air so it tends to pool in the lowest areas. It

does not deteriorate nor dissipate, so it remains active, ready to permanently destroy the nervous system

of anyone who may step into it. In this example it is good for one to know his or her density

fundamentals. In industries today, there are an abundance of chemicals in fluid form (liquid or gas) thatcan be equally as dangerous to their surrounding areas.

Like chlorine gas, phosgene gas and carbon monoxide are also heavier than air. Phosgene gas both

suffocates and creates hydrochloric acid in the lungs. It is created in many industrial processes where

foods may rot, or even where an animal decomposes near a confined space. It has the odor of new mown

hay or green corn. Phosgene gas has killed and caused pneumonia in workers who entered unventilated

confined spaces without wearing self contained breathing devices. Carbon monoxide exits from the

exhaust pipe of a vehicle and migrates downward into confined spaces where it displaces the air. It

suffocates a victim by displacing the oxygen in red blood cells.

Radioactive tritium gas is a heavy form of hydrogen gas (an isotope). Tritium is many times lighter thanair so it escapes upward when released. Gas bubbles of tritium in air act like bubbles of air rising from

the bottom of a fish tank. This light gas rises to occupy a thin layer in the highest regions of the gas

envelope that covers the earth.

A comparison of the densities of two (non-mixable) items will allow us to predict which item will float

and which will sink. Consider the following examples:





Example:

Using information from Table 1-1, Densities of Common Materials, calculate the specific volumeof mercury.

ρ =

1

υ

υ =1

ρ

υ =1

13.6g

3cm

υ = 0.0735cm3

g

Example:

An unknown material has a specific volume of

0 373

.cm

g calculate its density.

ρ = 1

υ

ρ =1

0 373

.cm

g

ρ = 27033

. g

cm

Example:

An unknown material has a specific volume of 00200

3

.ft

lbm calculate its density.

ρ = 1

υ

ρ = 1

002003

.ft

lbm

ρ = 503

lb

ft

m





Specific Gravity

Specific gravity is the ratio of the density of a fluid or solid to the density of a standard fluid. Water is the

standard of comparison for liquids and solids, and air is the standard of comparison for gases.

For a liquid the ratio ρ liquid

ρ water

is the specific gravity.

For a gas the ratio ρ

gas

ρ air

is the specific gravity.

Specific gravity units are a dimensionless number and can be reported without units. When specific

gravity is calculated to be a number greater than the number 1.0 (one) the liquid or gas is more dense than

its standard so may tend to drop to the bottom of a container. When specific gravity is calculated to be a

number smaller than the number 1.0 (one) the liquid or gas is less dense than its standard and may tend torise to the top of the container.

Specific gravity is a measure of the relative density of a substance compared to the density of water. It isthe density of the substance divided by the density of water. The density of pure water at standard

temperature and pressure (32oF and 14.7 psi) is approximately 62.4 lbm/ft3. Knowing the specific gravity

of the substance provides a quick way of determining if the substance will float or sink when put in water.

Specific gravity is a unitless quantity since it is the ratio of two quantities having the same units.

The equation for specific gravity SG is stated as follows:

standard

substance

ρ

ρ =SG

Where:

SG = specific gravity of substance

ρ substance = density of substance, lbm/ft3

ρ standard = density of the standard. For a liquid the standard is water at 0oC and

1 atmosphere of pressure, (the average pressure at sea level). For a gas

the standard is air at 0oC and 1 atmosphere of pressure.

water 3 3lbft

1 gcm

= 1 kg = 1,000kgm

ρ = 62.4 m =3

If the specific gravity of a liquid is greater than 1, it will sink in water. If the specific gravity is less than

1, it will float in water. Also, if the specific gravity of a gas is greater than 1, it will sink in air and

occupy low regions, and if it is less than 1 it will rise in air until it reaches an equilibrium height where

the density of air is equal to its own density. Consider the following examples:





Example:

Given:

ρ water = density of water, 1.0

g/cm

3

ρ oil = density of oil, 0.083

g/cm3

Calculate the specific gravity

of oil.

water

oil SG ρ

ρ =

3

3

0.1

83.0

cm

g cm

g

SG =

0.1

83.0=SG

SG = 0.83

Since the specific gravity is lessthan 1, the oil will float.

Example

Given:

ρ water = density of water, 1.0

g/cm

3

ρ mercury = density of mercury,

13.6 g/cm3

Calculate the specific gravity

of mercury.

water

mercurySG

ρ

ρ =

3

3

0.1

6.13

cm

g cm

g

SG =

0.1

6.13=SG

SG = 13.6

Since the specific gravity is

greater than 1, the mercury will

sink.

Example

Given:

ρ water = density of

water, 62.4 lbm/ ft3

υ ice = specific volume

of ice, 0.0174ft

lb

3

m

Calculate the specific

gravity of ice.

ρ = 1

υ

ρ =1

001743

.ft

lbm

ρ = 57 43

.lb

ft

m

SG =ρice

ρwater

3

3

ft

lb62.4

ft

lb4.57

m

m

SG =

4.62

4.57=SG

SG = 0.92

Since this is less than 1 it will

float.









*Note: Gases are compared to the density of air and not to the density of water.

The standard of comparison for all liquids is water, and the standard for all gases is air. Table 1-2

compares the densities of common fluids, both liquids and gases

Densities of Common Fluids

Fluid Density lbm/ft3

Acetone 49.421

Air (at 1 atmosphere, 0°C) 0.0805

Alcohol (methyl) 50.544

Benzene 56.098

Bromine 198.869

Carbon tetrachloride 99.466

Chlorine (gas) 0.2005

Coconut oil 57.72

Sulfuric acid (fuming) 114.70

Gasoline 42.432

Glycerin 78.624

Helium (at 1 atmosphere, 0°C) 0.0112

Hydrogen (at 1 atmosphere, 0°C) 0.00561

Kerosene 50.544

Mercury 848.64Milk 64.272

Water (fresh) 62.4

Water (sea) 63.96

Table 1-2 Densities of Common Fluids

Pressure ( p)

All substances are made of molecules. A molecule is a chemically bonded group of atoms (or elements).

Molecules account for the general characteristics of all fluids, and knowing these characteristics is a

requirement of this course. The key to understanding of these general characteristics of a fluid is found in

comprehending just three things about a fluid:

1) - mass;

2) - the internal energy of a molecule, and

3) - the attraction of each molecule for every other molecule.





These key characteristics either alone or taken together explain all of the other characteristic of fluids of

interest in this course. Stated another way; these three characteristics are the major source of the other general characteristics of every fluid! Therefore, knowing the mass, the internal energy of a molecule,

and the attraction of molecule for molecule explains, viscosity, fluid friction, compression and

decompression pressures, atmospheric pressure, temperature, “head” pressure, buoyancy, hydraulic

pressure, condensation and boiling, and several other fluid characteristics.

Pressure is caused by the

collisions of the molecules of a

fluid with the walls of its

container (See Figure 1.2).

Except at absolute zero

temperature, where by definition

all internal energy is zero and all

movement is stopped, themolecules of any substance are

constantly moving. In a solid,the molecules are tightly bound

so that they only vibrate and

rotate. In a liquid or a gas, they also have freedom to translate (move around). The molecules are

continuously colliding with each other and with the walls of their container. As billions of molecules in

each cubic inch of a fluid collide billions of times each second with the walls, they exert forces that push

the walls outward. The forces resulting from these repeated collisions by these molecules add up to the

pressure exerted by a gas on itself and on its surroundings.

Pressure is "the force per unit of area” that a substance exerts on itself and on its surroundings. In a

confined fluid, pressure is always exerted equally in all directions. We use the lower case p for pressure.

pressure =force

area

p = F

A

We need to draw a careful

distinction between force and

pressure. Consider the example in

Figure 1.3. Two rectangularblocks, each with dimensions of 2

by 10 by 20 inches, and each

weighing 500 lb f , are placed on a

table. Each block exerts a

downward force of 500 lb f on the


Figure 1-2 Pressure caused by Molecules

A

B

10 in

2 in

20 in

2 in

20 in

10 in

Figure 1-3 Force versus Pressure




surface of the table, but because they apply that force over different areas, the pressures they exert on the

table are different.

Block A applies 500 lb f over an area of 200 in2. It will exert a pressure given by:

pA =500lb f

20 in • 10 in

pA = 2.5 lb f in2 = 2.5 psi

Block B applies 500 lb f over an area of 20 in2. It will exert a pressure given by:

pB =500 lb f

2 in • 10 in

pB = 25lb f

in2= 25 psi

The pressure exerted by Block B is ten times that exerted by block A even though the applied force is the

same. This is because the weight of block A is distributed over a larger area.

Pressure ≠ Force

Pressure Measurements

Pressure is a force divided by an area.

In the English system, we measure pressure in pounds per square inch. In SI, the pressure unit is the

pascal. These two units are given as follows:





11

12

= f

2 psi

lb

in

1 Pa =

1 N

m

kg

m s2 = •

There are two standard methods used to measure pressure:

• The amount of pressure above a perfect vacuum (absolute), and

• The amount of pressure above or below atmospheric pressure, (gage).

Pressures referenced to a perfect vacuum are called absolute pressures. In the United States, we use psi

for our pressure measurements. An absolute pressure then has the unit "pounds per square inch -

absolute" or psia. The letter "a" is not a unit; it is a label. It simply tells us what the reference pressure is.

Zero pressure on the absolute scale is the absence of all pressure. Outer space is nearly a perfect vacuum.Achieving a nearly perfect vacuum on earth is not difficult. The space above the mercury column in a

barometer is a nearly perfect vacuum. Atmospheric pressure is the amount of pressure exerted on the

earth's surface by the weight of the air molecules in the atmosphere. The atmospheric pressure at sea

level is 14.7 psia.

Pressure referenced to the earth's atmospheric pressure is called gage pressure, and is given as "pounds

per square inch - gage" or psig. Again, the "g" is simply a label telling us what the reference pressure is.

Figure 1.4 shows the relationships between absolute pressure measurements and gage pressure

measurements. In doing problems, psig will cancel with psia since the units are psi even though the

references are different.

From Figure 1.4 the following relationship can be determined:

pabsolute = patm + p gage

A pressure gage is an instrument which measures pressure. To clear up any confusion; a pressure gage is

a piece of hardware; a gage pressure is a pressure referenced to atmospheric pressure. Pressure gages

indicate the amount of pressure sensed

relative to a reference pressure.

Figure 1.5 shows a typical bellows type

pressure detector.

Other types of pressure detectors use

similar arrangements to measure the

difference between an unknown pressure

and the reference pressure. Pressure


TYPICAL PRESSURE GAGE

MEASURED PRESSURE

BELLOW

REFERENCE

PRESSURE

INDICATED P

IDIFFERENCE BETWEE

MEASURED PR ESSUR REFERENCEPRESSURE and

Figure 1-5 Typical Pressure Gage

Figure 1-4 Pressure Scales




gages referenced to atmospheric pressure indicate the amount of pressure above or below atmospheric

pressure. The units for these pressure gages are psig (lb f /in2

gage) for pressure above atmospheric and

psiv (lb f /in2

vacuum) for pressure or vacuum below atmospheric.

Finally, pressures and

vacuums may be expressed

in terms of the height of a

liquid column the pressure

will support. These include

inches of water, feet of

water, and inches of

mercury. Millimeters of

mercury is also a common

unit for measuring pressure.

Millimeters of mercury isalso given the name torr.

One torr equals one

millimeter of mercury.

Figure 1.6 shows thatatmospheric pressure at sea level, 14.7 psia, would support a mercury column 29.92 inches high or a

water column 34.0 feet high (408 inches).

Table 1-3 compares some of the common pressure units to atmospheric pressure. Unless otherwise stated

each of these pressure units is assumed to be an absolute pressure.

1 atm 14.7 psia

1 atm 408 in. of H 2 O

1 atm 29.92 in Hg

1 atm 760 mm Hg


Figure 1-6 Liquid Supported by Atmospheric Pressure

1.6 Liquid Supported By Atmospheric Pressure




1 atm 760 torr

1 atm 101 105. × Pa

Table 1-3 Common Pressure Units





The following examples utilize common pressure units:

a. Convert 14.0 psia to inches of water.

= 388.571 in H2O

b. Convert 14.7 in of H2O to psia:

= 0.5296 psia

c. Convert 50 in Hg to in of H2O:

= 682 in H20

d. Convert 300 torr to Pascals:

= 39,868 Pa or 3.9868 x 104 Pa


300 torr 1 atm 101 105. × Pa

760 torr 1 atm

50 in Hg 1 atm 408 in H2O

29.92 in Hg 1 atm

14.7 in H2O 1 atm 14.7 psia

408 in H2O 1atm

14.0 psia 1 atm 408 in H20

14.7 psi 1 atm




Referring to Table 1-3, estimate the following values. Then, using note paper (if needed), calculate each

value in the assigned units.

Examples: (answers on the following page)

a. Convert 27 psi:

Estimate Calculate

in H2O in H2O

in Hg in Hg

mm Hg mm Hg

torr torr

Pa Pa

b. Convert 340 torr:

Estimate Calculatein H2O in H2O

in Hg in Hg

mm Hg mm Hg

psi psi

Pa Pa

c. Convert 256 psi:

Estimate Calculate

in H2O in H2O

in Hg in Hg

mm Hg mm Hg

torr torr

Pa Pa

d. Convert 455 in H2O:

Estimate Calculate

in Hg in Hg

mm Hg mm Hg

psi psi

torr torr

Pa Pa





Answers-

a. Convert 27 psi:

Estimate Calculate

about 750 in H2O 749.3878 in H2O

about 55 in Hg 54.9551 in Hg

about 1400 mm Hg 1,395.9184 mm Hg

about 1400 torr 1,395.9184 torr

about 2x105

Pa 1.8551x105

Pa

b. Convert 340 torr:

Estimate Calculate

about 190 in H2O 182.5263 in H2O

about 14 in Hg 13.3853 in Hgabout 340 mm Hg 340 mm Hg

about 7 psi 6.5763 psi

about 0.45x105

Pa 0.4518x105

Pa

c. Convert 256 psi:

Estimate Calculate

about 7000 in H2O 7105.3061 in H2O


about 13,000 mm Hg 13,235.3741 mm Hg

about 13,000 torr 13,235.3741 torr

about 17x105

Pa 17.589x105

Pa

d. Convert 455 in H2O:

Estimate Calculate


about 900 mm Hg 847.5490 mm Hg

about 17 psi 16.3934 psi

about 900 torr 847.5490 torr

about 1.2x105 Pa 1.1263x105 Pa





Absolute, Gage, and Vacuum Pressure Relations

A key to understanding pressure measuring is to understand where these measurements originated. Gage

pressure is the normal way the body operates. Though there is atmospheric pressure on the body the body

ignores it; so also with gage pressure. Most operations can be performed successfully ignoring

atmospheric pressure so many gages read in “gage pressure.”

Converting a measurement of zero psi gage pressure is equal to a measurement of 14.7 psi absolute.Always add 14.7 psi to change gage pressure to absolute pressure. Subtract 14.7 psi to convert absolute

pressure to gage pressure.

A pump does work to put a pressure on the fluids in a piping system, but may also be required to produce

a vacuum to lift a fluid so the it may be pumped. The pressure pushing the fluid is work and the vacuum

lift is also work; both must be done by the pump. Both of these are seen as positive amounts of work

performed by the pump. Vacuum lift is considered to be a positive amount and the vacuum gage reads

positively under almost all situations.

Always convert gage pressure to vacuum pressure (and vice versa) by taking the same amount of pressureand reverse the sign (i.e. from positive to negative or negative to positive).

Equations give the relationship between absolute pressure, gage pressure, and vacuum pressure. We can

convert them using these equations. Absolute pressure (psia) is equal to the atmospheric pressure (psia)

plus the gage pressure (psig):

pabsolute = patm + p gage

Absolute pressure (psia) is equal to atmospheric pressure (psia) minus vacuum pressure (psiv):

pabsolute = patm - pvac

Vacuum pressure is normally used for pressures below one atmosphere where it is a positive reading.

Vacuum pressure starts with zero at one atmosphere and reaches it’s maximum value of 14.7 psiv at a

perfect vacuum where absolute pressure is zero. The relationship between absolute pressure, gage

pressure and vacuum pressure is illustrated in Table 1-4.

Absolute Pressure Gage Pressure Vacuum Pressure

19.7 psia 5 psig Not Normally Used

14.7 psia 0 psig 0 psiv

0 psia -14.7 psig 14.7 psiv

Table 1-4 Absolute, Gage and Vacuum Pressure





Example:

The pressure indicated is 25 psig. Calculate the absolute pressure.

pabsolute = patm + pgage

pabsolute = 14.7 psia + 25 psig pabsolute = 39.7 psia

Example:

The pressure indicated is 25 psia. Calculate the gage pressure.


pabsolute – patm = patm – patm + pgage

pgage = pabsolute – patm

pgage = 25 psia – 14.7 psia pgage = 10.3 psig

Example:

The pressure indicated is 12.5 psiv. Calculate the absolute pressure.


pabsolute = 14.7 psia - 12.5 psig

pabsolute = 2.2 psia

Example:

The pressure indicated is 10 psia. Calculate the vacuum pressure.


pabsolute + pvac = patm - pvac + pvac

pabsolute - pabsolute + pvac = patm - pabsolute

pvac = patm - pabsolute

pvac = 14.7 psia - 10 psia pvac = 4.7 psiv





Example:

The pressure indicated is 75 psig. Convert the pressure to torr.

Notice that torr is an absolute pressure. In order to do this conversion we must first convert psig to psia.

Once the pressure is converted to psia it is in an absolute scale and can be converted to torr


pabsolute = 14.7 psia + 75 psig

pabsolute = 89.7 psia

= 4,638 torr

Example:

The pressure indicated is 75 in of H2O. Calculate the vacuum pressure.


pabsolute + pvac = patm - pvac + pvac

pabsolute - pabsolute + pvac = patm - pabsolute

pvac = patm – pabsolute

=2.702 psia

pvac = 14.7 psia - 2.702 psia

pvac = 11.998 psiv

Buoyancy

Buoyancy is defined as the upward force on an immersed object. We have all observed the buoyant

effects of a liquid, but buoyancy also exists for gases. When we go swimming, our bodies are held up by

the water. Wood, ice, and cork float on water. When we lift a rock from a stream bed, it suddenly seems

heavier on emerging from the water because it has been buoyed up while being submerged in a fluid.

Boats rely on this buoyant force to stay afloat. A balloon filled with a light gas rises in air, a heavier gas.

This buoyant force occurs because there is a pressure inside of a fluid that exerts a force on any bodytouching that fluid. This fact causes the upward force on the bottom of a submerged object to be greater

than the downward force on its top surface. The amount of this buoyant effect was first computed and

stated by the Greek philosopher Archimedes.

Archimedes found, when he got into his bath tub which was filled with water to the rim, that water was

displaced out of the tub and onto the floor. He calculated that the volume of water displaced was equal to


89.7 psia 1 atm 760 torr

14.7 psi 1 atm

75 in H2O 1 atm 14.7 psia

408 in H2O 1 atm




the volume of his body. From these findings he determined that “any body completely or partially

submerged in a fluid is buoyed up (pushed up) by a force equal to the weight of the amount (volume) of fluid displaced by the body.” This is known as Archimedes’ Principle.

If a body weighs more than the liquid it displaces, it sinks but will appear to lose an amount of weight

equal to that of the displaced liquid, as our rock. If the body weighs less than that of the displaced liquid,the body will rise to the surface eventually floating at such a depth that will displace a volume of liquid

whose weight will just equal its own weight. A floating body displaces its own weight of the fluid in

which it floats.

If a diver’s body and diving equipment equals the weight of the water that is displaced the diver hangs

suspended at any location or depth like a fish. If a diver and equipment weigh less than the water

displaced the diver has a hard time submerging and bobs about on the surface like a boat. With a total

weight of more than the weight of the water displaced, a diver sinks to the bottom.

See Figure 1.7. A cubic foot of water weighs 62.4 pounds, and if a one pound ball displaces a cubic footof water a person using the ball to aid them in floating is also buoyed up. The more liquid the ball

displaces the greater the amount of force exerted on the swimmer. When the ball is totally submerged itwill exert an upward force of 62.4 pounds and with its own weight of one pound of force downward

canceling one of those pounds will exert a force of 61.4 pounds upward. This is true of any immersedobject that displaces a cubic foot of liquid.

Every boat is specifically designed to displace the amount of fluid weight that is

equal to or greater than its own weight.


Figure 1-7 Buoyancy Forces on an Object




Hydrostatic Pressure

Anyone who dives under the surface of the water notices that the pressure on their eardrums at a depth of

even a few feet is quite noticeable and increases with depth. Measurements have shown that the

hydrostatic (“hydro” = water, “static” = non-moving; therefore: hydrostatic = “non-moving water” )

pressure within a liquid increases directly with the depth of the liquid. The pressure at any depth is thesame in all directions. A liquid molecule does not move when the pressure is the same, but stays in one

spot until a difference in pressure causes it to move

away from the higher pressure

region toward the lower

pressure region.

Figure 1.8 illustrates the

relationship between liquid

level and pressure. If holes

are placed in the tank, the

liquid in the tank will leak out.The lower in the tank the hole

is placed, the greater the

velocity of the liquid as it

leaks from the tank due to the

increased pressure. The holes are known as “delivery points.” The greater the depth, the greater the

pressure and the greater the speed and volume of the liquid flowing out of the tank.

P total = P atmospheric + Pelevation

P total = P atmospheric + ρgz

The total pressure measured at the bottom of the tank is due to the pressure of the atmosphere (14.7 psi at

sea level) plus the pressure due to the height of molecules stacked one above another in the tank. This

last pressure is due to both the density ( ρ ) of the liquid, gravitational pull of the earth on every molecule

(g), and the height of the stack of molecules in the tank (z). (“h” represents several other properties so we

use “z” to represent height.)

Oftentimes it is necessary to obtain the liquid level in a vessel by the equivalent pressure that is measured.

A diptube, (a hollow tube usually constructed of metal), is inserted down the entire measurable depth of the vessel. The diptube is only a fraction of the diameter of the vessel in which the pressure is measured.

The actual diameter of the dibtube or of the vessel itself is not needed. The height of the liquid in the

vessel, and any pressure that is exerted on top of the liquid is the only required parameter to obtain the

vessel level. The following example helps explain why the tube sizing (or tube diameter) is not important

and does not effect the measurement.


Figure 1-8 Relationship between Liquid Level and Pressure




Example:

Recall that pressure is the measure of an applied force of a given area.

A

g V

A

mg

Area

Weight

Area

Forceessure

ρ ====Pr

where: h AV ∗=

The volume is equal to the cross-sectional area (A) times the height (h) of liquid.

Substituting this into the above equation yields:

hg A

g h A

A

g V essure ρ

ρ ρ =

×==

)(Pr

The above equation tells us that the pressure exerted by a column of water is directly proportional to the

height of the column and the density of the water and is independent of the cross-sectional area of the

column.

The pressure thirty feet below the surface of a one inch diameter standpipe is the same as the

pressure thirty feet below the surface of a large lake!

Notice the vertical column of water shown on the left side in Figure 1.9 that extends from the bottom of

the tank. This column rises almost to the 30 foot height. This illustrates that a pressure that is equivalent

to 30 feet of water can shoot a column of water 30 feet high. When dealing with flowing systems,

engineers frequently refer to pressure in terms of feet of head (static head). Head is the pressure that

will support a height (usually in feet) of a column of liquid.


Figure 1-9 Pressure Versus Height




Static head is the measure of the vertical distance from the water surface to the delivery point and thisdistance has a direct relationship to the pressure caused by the weight of water extending above that point.

The calculations of the pressure at 10 ft, 20 ft,and 30 ft due to the water in the tank shown in

Figure 1.9 follow a straight line when placed on

a graph like figure 1.10.

• At 10 feet the pressure is 4.33

psi (pounds force per square inch)


psi.


psi

Head is a measure of pressure.

Also, the pressure in a container of fluid is directly proportional to the density of the fluid as well as to the

depth. Pressure increases with depth (or static head), so at a specific depth two fluids of different

densities will have different pressures. Therefore, the higher the density of a fluid the higher the pressureat a depth due to the weight of the fluid above.

If the depth is increased (while other factors do not change) the pressure increases.

If the depth is decreased (while other factors do not change) the pressure decreases.

If the density is increased (while other factors do not change) the pressure increases.

If the density is decreased (while other factors do not change) the pressure decreases.

Example:

At 30 feet of depth a pressure of 13 psi (gage pressure) is measured. Therefore the

relationship between 13 psi and 30 feet will allow us to predict the relationship between

any other pressure and the depth that created it.

THUS-

In other words the ratio between 30 feet of head and the 13 psi it creates in a tank is

exactly the same as the ratio between the number of feet of head that can created 14.7

psi.


Figure 1-10 Static Head versus Pressure




This is an algebra problem that can be solved by multiplying both sides of the equation

by the same number in the attempt to eliminate everything on the right side of theequation except for χ

Solving for χ:

Canceling like-numbers and like-units (dimensions) in numerators and denominators

have:

So at 14.7 psi (one atmosphere of pressure) there is an equivalent depth of liquid of 33.92

feet (34 feet effectively) in a tank were we to create the same amount of pressure by

putting pure water into a tank.

This is referred to as 34 feet of head, recognizing that the pressure (head) is created by

the depth (or height) of 34 feet of pure water in a liquid column.

30ft

13psi=

33.92ft

14.7psi





Example:

Using Figure 1-11, Fill in the table below: ( answers are on the following page)

# depth Estimated

psig

Calculated

psig

Estimated

psia

Calculated

psia

1 3 ft 1.3 psig 16 psia

2 10 ft

3

4 14 ft

5

6 16 ft

7 7 ft


Figure 1-11 Head and Pressure Illustration




Student Guide: Fluid Flow Chapter 1: Introduction to Fluids

Answers to head and pressure illustration:

# depth Estimated

psig

Calculated

psig

Estimated

psia

Calculated

psia

1 3 ft about 1.5 psig 1.3 psig about 16.5 psia 16 psia

2 10 ft about 4.5 psig 4.33 psig about 19 psia 19.03 psia

3 14 ft about 6 psig 6.06 psig about 21 psia 20.76 psia



6 16 ft about 7 psig 6.93 psig about 21.7 psia 21.63 psia

7 7 ft about 3 psig 3.03 psig about 17.7 psia 17.73 psia

Estimate accuracy comes with experience. An Estimate that is within 10% of the calculated value is good

and gives a mental check on the calculation.

Pascal's Law (the law of hydraulics)

Because liquids are essentially incompressible, when we pressurize a liquid, its density does not change.

In Figure 1.12, we apply a force F to the piston on a confined liquid. This results in an increase of

pressure throughout the liquid. This pressure increase is the same everywhere in the liquid. Pascal's

principle says:

The pressure applied to a fluid confined in a container is transmitted undiminished throughout the

fluid and acts in all directions.






Conditions and Results:

• a pressure is applied to a fluid

• the fluid is enclosed

• the pressure is transmitted without loss

• the pressure is measured the same in every direction

• the fluid and the walls of the vessel receive the same measure of pressure

This principle applies to all liquids and to many gas-filled systems as well. It is the fundamental principle

of hydraulics, and is important for applications in hydraulics, such as hydraulic valve operators or

hydraulic jacks. This principle is also important in maintaining a static pressure on a totally contained

system. Quite often, an external pressure is applied to a system to maintain the entire system at a

minimum pressure. (examples: a pressurizer that helps to stop cavitation by increasing the pressure in an

enclosed loop, or a storage tank with a bag (balloon) that maintains the pressure on a water system when

the pump is temporarily off.)

A force applied to even a very small surface area is transmitted at the speed of sound (speed of pressure in

a fluid) throughout the fluid.

Any equal area anywhere within the fluid will “feel” the same amount of force over its surface.

Referring to Figure 1-13, this provides an opportunity to use a “mechanical advantage”. Like a lever, a

small amount of force applied at one location can produce an enormous amount of force at another location.


Figure 1-12 Pressurizing a

Confined Liquid

Figure 1-13 Hydraulic System Forces





Example:

The hydraulic system in Figure 1.14 consists of a small piston at A, a large piston at B, and a

fluid-filled reservoir connecting the two cylinders.

The area of piston A is 1 in2, and the area of piston B is 1 ft

2. If we apply a force of 40 lb f to the

top of piston A, how much force can we generate at piston B?

Pascal's principle says that the

pressure at A p A is equal to the

pressure at B p B. Then,

A B

A

A

B

B

BA B

A

area area

area

area

p = p

F =

F

F = F •

F B =40lb f

1 in2

1 ft2 144 in2

1 ft2

F B = 5,760 lb f

So the product of F x d is the same for both pistons. In order to move 5,760 lb f up a short distance, we

have to exert 40 lb f down a long distance. This is how hydraulic systems work.

Pressure Difference for Fluid Flow

Fluids flow, if a difference in pressure exists, and a natural flow takes place from a high pressure area to a

low pressure area, (just as heat flows if a different temperature exists flowing from the higher temperature

area to the lower temperature area).

Two common methods of creating high pressure to cause fluids to flow are by using gravity (feet of head

in tanks of liquid) and pumps. Both methods will create high pressures and instead of causing the liquids

to compress they cause them to flow.

The term “head” is a reference to an amount of “pressure” and head is measured in “feet.”


Figure 1-14 A Simple Hydraulic System





Gravity and Pumps both provide “Head” or Pressure






Chapter 1 Summary

Specific volume - amount of space occupied by unit of mass. υ = V

mor υ =

1

ρ

Density - amount of mass in a unit of volume. ρ = m

V or ρ =

1

υ

Specific gravity - the ratio of the density of a fluid to the density of a standard fluid

SG =ρliquid

ρwater

and SG =ρgas

ρair

Two non-mixable (non-miscible) fluids will separate when placed in the same container. The fluid with

the highest density will sink to the bottom.

Pressure is a force acting over an area.

p = F

A

1 atm = 14.7 psia

1 atm = 408 in. of H O

1 atm = 29.92 in Hg

1 atm = 760 mm Hg

1 atm = 760 torr

1 atm = Pa

pabs = patm + pgage

pabs = patm - pvac

From these findings he determined that “any body completely or partially submerged in a fluid is buoyed

up (pushed up) by a force equal to the weight of the amount (volume) of fluid displaced by the body.”

This is known as Archimedes’ Principle.

Buoyancy is the upward force on an immersed object.

The pressure of a liquid is directly proportional to the depth of the liquid. The area of a container of

liquid has no effect on the pressure; the depth and density of the liquid determines the pressure at the

bottom of the container.

Head is a measure of pressure in units of feet since it defines the depth at which a pressure is measured.

When converting pressure units change first to number of atmospheres.

Pascal's law states in effect that a pressure applied to a contained fluid is transmitted without decreasing

throughout the container. It is experienced both in the fluid as well as upon the walls of the container no

matter in what direction the measurement is taken.






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OPERATIONS TRAINING PROGRAM Student Guide: Fluid Flow Chapter 2: Compression of Fluids

Chapter 2 Compression of Fluids

Compressibility is the measure of the change in volume a substance undergoes when a pressure is exerted

on the substance. This chapter covers the fundamental concepts of the Combined Ideal Gas Law

and the compressibility and incompressibility of fluids.

TO 2.0 Given the necessary fluid system parameters and using the

Combined Ideal Gas Law, DESCRIBE the compressibility or

incompressibility of a fluid when a pressure is exerted

EO 2.1 STATE the Combined Ideal Gas Law

EO 2.2 SOLVE for fluid parameter using the Combined Ideal Gas Law

EO 2.3 DESCRIBE when a fluid may be considered to be

incompressible

EO 2.4 DESCRIBE the effects of a pressure or temperature change on

a confined fluid

EO 2.5 EXPLAIN how to prevent over pressurization accidents caused

by gas or liquid confinement.

Compressibility

Compressibility is the measure of the change in volume a substance undergoes when a pressure is exerted

on the substance. Liquids are generally considered to be incompressible. For instance, a pressure of

16,400 psig will cause a given volume of water to decrease by only 5% from its volume at atmospheric

pressure. Gases on the other hand, are very compressible. The volume of a gas can be readily changed

by exerting an external pressure on the gas

The Combined Gas Law

The combined gas law relates to the properties of a (non-existent) so-called ideal or perfect gas. An ideal

gas has the same properties at every point throughout its mass and is not influenced by chemical or

external forces. To be mathematically correct, the combined gas law can only be used on gases of low

densities that do not undergo a change to solids or to liquids. In addition, absolute temperatures and

pressures must always be used during these calculations or the calculations will be incorrect. (See

example on following page)

The combined gas law, as its name implies, is a combination of two laws of nature observed by the

physicists, Charles and Boyle. Charles determined that at constant low pressures, the volume of a gas, V ,is directly proportional to the absolute temperature, T , of the gas. At any time the ratio of volume to

absolute temperature remains the same,




Student Guide: Fluid Flow Chapter 2: Compression of Fluids

V 1

T 1=

V 2

T 2,

whereV

1 andT

1 are the initial volume and absolute temperature, andV

2 andT

2 are the final volumeand absolute temperature.

Boyle determined that at low pressures the product of volume and pressure of constant temperature gas is

always the same.

P1V 1 = P2V 2

These two laws are combined to be stated as:

For a given mass of any gas, the product of the absolute pressure and volume occupied by the gas,divided by its absolute temperature, is a constant.

Combined Gas Law: a “constant”.

Where:

P1

= initial absolute pressure,

V1 = initial volume

T1

= initial absolute temperature, in absolute terms ˚R or

K (degrees Rankine, or Kelvin )

P2

= final absolute pressure

V2

= final volume

T2 = final absolute temperature, ˚R or K

This means that the ratio of absolute pressure times volume when compared to absolute temperature

always yields a number that does not change. The ratio does not change with any change in any variable.

Example:

(using the English System in degrees Rankine)

140˚F air is charged into a tank until its pressure is 160 psia (“psia” is the designation for absolute

pressure). Over the course of the day the air temperature cools so the gas in the tank slows down,

has less energy and therefore exerts less pressure on the walls of the tank as time goes on. Whatwill be the temperature of the air, in degrees Fahrenheit, when the pressure in the tank reaches

150 psia? Assume no air is added or removed from the tank during this time.





Solution:

The ratio of the absolute Pressure times Volume divided by the absolute Temperature is an

unchanging constant number for a specific gas that says that the ratio at any time will be equal to

the ratio at any other time for the same gas.

The combined gas law is:P1V 1

T 1=

P2V 2

T 2= Constant,

and we can choose to eliminate the “= Constant” part of the expression.

To make things even simpler, the tank volume is constant, so V1 and V2 are equal. This allows us

to see that V1 and V2 are constant and may be eliminated from the equation, allowing us to write

the equation as:P1

T 1=

P2

T 2

T2 will be the final temperature and T1 the initial temperature. We then decide to solve the

equation for final temperature T2: T 2 = T 1P2

P1

.

Converting the initial temperature to absolute: (add 460˚F to any Fahrenheit temperature to make

it Rankine- absolute)

T1 = 460˚ + 140˚F = 600˚R

Substitute and solve for final temperature:

T 2 =

(600 ° R)(150 psi)

160 psi =562.5

° R

Converting back to °F : T 2 = 562.5° R − 460° F = 102.5° F

Effects of Pressure Changes on Confined Fluids

Fluids may be classified as compressible or incompressible. Gases are highly compressible. A fluid is

considered incompressible when it is a liquid, because its volume and density remain essentially constantwith changes in pressure. It is this incompressibility that allows a hydraulic system to operate.

Although liquids are generally considered to be incompressible, in reality, liquids really do compress, but

they compress so little that they are still considered to be “incompressible”. A very large pressure must

be applied to see a significant change in volume and density. For example, a pressure of 16,400 psi will

cause a volume of water to decrease by only 5 percent from its volume at atmospheric pressure. The fact

that liquids don’t compress can have significance in our operations. For example, if we enclosed a tank

and fill it with water without venting it we can exceed pressure limits possibly to the point of rupturing

the tank, valves, or vents. Worst Case Expected: Pressure Up - Hydraulic bursting / Pressure Down –

crumple the tank, or (if the fluid placed in the container is hot) upon release of pressure an explosion.





Gases, on the other hand, are very compressible. If a fluid increases its density significantly when

pressure is applied, then the fluid is considered compressible. This occurs with fluids of very low

densities, such as gases. The volume of a gas can be readily changed by applying an external pressure on

the gas. Gases may be compressed until they become liquids but at that point further attempts to reduce

volume may have catastrophic results. Worst Case Expected: Pressure Up – Tank becomes a rocket, or

explosion / Pressure Down – Crumple the tank (implosion).

Gasses are COMPRESSIBLE.

Liquids are INCOMPRESSIBLE.

If we put pressure on them and they can’t flow away they may burst the confining vessel!!

Effects of Temperature Changes on Confined Fluids

An increase in temperature will tend to decrease the density of many fluids as the molecules become more

active and “bump into each other more often” driving the molecules further away from one another. If

the fluid is confined within a container of fixed volume, the effect of a temperature change will depend on

the compressible nature of the fluid.

If the fluid in a closed container is a “compressible” gas, it will respond to a temperature change in the

same manner predicted by the ideal gas law. A 5 percent increase in absolute temperature will cause a

corresponding 5 percent increase in absolute pressure; a 5 percent decrease in absolute temperature will

cause a corresponding 5 percent decrease in absolute pressure.

Worst Case Expected: Temperature Up – Tank becomes a rocket, or explosion / Temperature

Down – Crumple the tank (implosion).

If the fluid is an “incompressible” liquid in a closed container, changes in temperature cause a much more

dramatic effect. If a container is filled with a liquid as the temperature increases the liquid attempts to

expand and change into a gas (a decrease in density). Since the liquid is confined within the container,

the walls of the container are “bumped into more often” so pressure increases. This results in a

tremendous increase in pressure for a relatively minor increase in temperature. This has a greater

potential of causing an explosion.

Worst Case Expected: Temperature Up – Tank becomes a rocket, or explosion, or burst tank /Temperature Down – Freeze, and if water: burst tank since water expands upon freezing)

Filling and Venting

Proper filling and venting techniques can prevent serious problems caused by gas trapped in closed

systems which may degrade system performance. Air in pumps can cause gas binding. Air in heat

exchangers reduces the heat transfer capability because air effectively creates an insulation barrier. Tank

explosions may occur with over pressurization. A steam incident death at Hanford could have been

prevented if the operator had been able to use proper steam line draining/venting measures.





Gas and liquid filled tanks and lines are pressure tested by filling with cold liquid only, since an hydraulic

over pressurization will simply burst the container. If a hot liquid or a gas were used the container might

explode and the expanding gas released might propel shrapnel on the expanding wave front at a rate faster

than the speed of sound.

When to be Concerned

Filling a “closed” liquid filled system; previously drained:

Over-pressurization accidents may be eliminated by opening high point vents to allow gas to be

forced from the system. When a steady stream of liquid issues from the vents, the high points are

then closed. After filling a system, the pump may be jogged to circulate the liquid for a few

seconds to attempt to move gas in the lines to the high points where the vents are again opened to

release any gas that may have been trapped. (See appendix; Crowder experiences)

Draining a “closed” liquid filled system:

It is also important to open vent valves to allow gas to enter when draining a system. If ventvalves are not opened, all the liquid may not be able to drain from the system. Later during

maintenance activities the trapped and perhaps heated or contaminated liquid may be released and

cause harm to people and equipment.

Heating a “closed” liquid filled system:

If a confined liquid is heated it will attempt to expand and pressure will build very rapidly. If

pressure exceeds the limits of the weakest portion of the containment system it will rupture or

explode violently. Safety or relief valves must be maintained in proper operating condition to

protect the system.

Cooling and confining an “open” liquid-filled system:

An open system containing a heated liquid which is then confined and cooled may also exceed

design limits and by depressurization cause the system to implode. Also draining a 10,000 gallon

tank with the vents closed can quickly turned it into a 3,000 gallon tank.. KEEP Vents OPEN

when draining and cooling!

Warming up a gas filled system:

Example: Steam line- All lines must be warmed up slowly and have no liquid present.

Splitting of metal or explosion can result from these.

Accident Prevention:

Proper filling, venting, draining, and temperature control.

Indicators of gas disturbance in a closed liquid filled system:

Some major indicators of gas in a liquid filled system could be:• Fluctuating pump discharge pressures





• Fluctuating flow rates

• Fluctuating motor currents

• Increased noise levels

• Excessive equipment vibration

• Higher than normal heat exchanger temperatures (air blanket causing decease in

heat transfer)• Bubbles present in sight flow indicators

• Increasing levels in surge tanks (as liquid is displaced by gas)





Chapter 2 Summary

Combined Ideal Gas Law: For a given mass of any gas, the product of the absolute pressure and volume

occupied by the gas, divided by its absolute temperature, is a constant

To use the combined Ideal gas law, absolute temperatures and absolute pressures must be used during the

calculations.

P1V 1

T 1=

P2V 2

T 2= Constant

Combined Ideal Gas Law

Liquids are considered incompressible; gases are compressible.

Effects of a pressure, temperature, or volume change on a confined fluid:

• A pressure or temperature increase of a confined gas (compressible fluid) will increase internal

pressure, and rupture or explosion may occur but are not likely.

• A pressure or temperature decrease of a confined gas (compressible fluid) will decrease internal

pressure, and rupture or implosion may occur but are not likely.

• A pressure or temperature increase of a confined liquid (incompressible fluid) will increase internal

pressure dramatically, and will likely cause rupture or explosion.

• A pressure or temperature decrease of a confined liquid (incompressible fluid) will decrease internal pressure dramatically, and will likely cause rupture or implosion.

Proper filling, venting, draining and temperature control can prevent pressurization accidents.





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OPERATIONS TRAINING PROGRAM Student Guide: Fluid Flow Chapter 3: Natural Circulation Flow

Chapter 3 NATURAL CIRCULATION FLOW

It is possible to design some fluid systems in a manner that does not require the presence of pumps to

provide circulation. This chapter describes the mechanism for natural circulation flow of a fluid

TO 3.0 For any natural circulation fluid system, DESCRIBE the

mechanism that allows for fluid flow

EO 3.1 LIST the conditions necessary for natural circulation to occur

EO 3.2 EXPLAIN how fluid flows in natural circulation fluid systems

Natural Circulation

The head (or pressure) required to compensate for the head losses is created by density gradients and

elevation changes. Flow that occurs under these circumstances is called natural circulation. Natural

circulation is circulation of a fluid without the use of mechanical devices, such as pumps. The driving

force for natural circulation flow is the difference in density between two bodies or areas of fluid.

Thermal driving head is the force that causes natural circulation to take place. It is caused by the

difference in density between two bodies or areas of fluid. In the figure below this force causing the

balloon to rise is a result of a difference in density between the hot air inside the balloon and the cooler air

surrounding it. This is an example of buoyancy.

Heat added to the air in the balloon makes it less dense or lighter than the surrounding air. Since the air in

the balloon is less dense, gravity has less effect on it. The result is that the gas in the balloon weighs less

than the surrounding air. Gravity tends to pull cooler air down into the space occupied by the balloon.

W ar m A ir

Co ld Air Co ld Air

Figure 3-15 Air Baloon Buoyancy




Student Guide: Fluid Flow Chapter 3: Natural Circulation Flow

This downward movement of cooler air may force the balloon out of the space it occupied and the balloon

may rise if the weight of the material of the balloon is light enough. Like a bubble in water the gas

bubble (balloon) tends also to rise in air.

Conditions Required For Natural Circulation

Natural circulation will only occur if the correct conditions exist. Even after natural circulation begins,

removal of any one of these conditions will stop natural circulation. The conditions for natural circulation

are:

• Temperature difference (Heat sink=where the heat goes; Heat source=where the heat comes from)

• Height difference

• Fluids are in contact with each other

There must be two bodies of fluid at different temperatures. This could also be one body of fluid with

areas of different temperatures. The difference in temperature is necessary to cause a density difference

in the fluid. The density difference is the driving force for natural circulation flow.

The difference in temperature must be maintained for the natural circulation to continue. Addition of heat

by a heat source must exist at the high temperature area. Continuous removal of heat by a heat sink must

exist at the low temperature area. Otherwise, the temperatures would equalize and no further circulation

would occur.

The warm area must be at a lower elevation than the cool area. As seen by the examples of the balloon

and the closed loop, a warmer fluid is less dense and will tend to rise and a cooler fluid is more dense and

will tend to sink. To take advantage of the natural movement of warm and cool fluids, the heat source

and heat sink must be at the proper elevations.

Hot air rises

Cooler air

Elevation

Figure 3-16 Heat Source / Heat Sink





The hot and cold areas must be in contact so that flow between the areas is possible. If the flow path is

obstructed or blocked, then natural circulation cannot occur. A vapor bubble may be caught at one point

causing a vapor lock (where a gas bubble is created at a hot spot and stops the liquid from which it was

created from flowing), or a valve may be closed.





Chapter 3 Summary

Natural circulation flow is the circulation of a fluid without the use of mechanical devices (i.e. pumps.

etc.).

The driving force for natural circulation is the difference in the densities between two bodies or layers of

fluid.

For natural circulation to occur the system must have all the following conditions:

Temperature difference (causing density difference)Height difference (heat source located below the heat sink - cooling)

Fluids are in contact (flow path exists between the warm fluid and cold fluid)

Natural circulation will occur with the conditions state above because:

Hotter fluids tend to become less dense and rise

Colder fluids tend to become more dense and sink



OPERATIONS TRAINING PROGRAM Student Guide: Fluid Flow Chapter 4: Volumetric and Mass Flow Rate

Chapter 4 VOLUMETRIC AND MASS FLOW RATE

Understanding the quantities measured by the volumetric flow rate and mass flow rate is crucial to

understanding other fluid flow topics. This chapter calculates the volumetric and mass flow rates

of fluids to determine various fluid parameters.

TO 4.0 Using fluid system volumetric and mass flow rates, SOLVE for

unknown fluid parameters values to predict fluid system

characteristics

EO 4.1 DEFINE the fluid flow terms ‘Mass Flow Rate’ and

‘Volumetric Flow Rate’ to include their typical units

EO 4.2 Given the necessary fluid parameters, DETERMINE the mass

flow rate or the volumetric flow rate of a given fluid system

EO 4.3 DEFINE the fluid flow term ‘Steady-State’

EO 4.4 EXPLAIN the Law of Conservation of Mass and Energy as

pertains to a fluid flow system

EO 4.5 Given a steady-flow system, APPLY the Continuity Equation to

determine fluid system parameters

Volume (V)

Volume is the amount of space occupied by a

fluid or an object. Mathematically, volume is

the length of an object times its cross-sectional

area. See Figure 4.1.

arealengthV ×=

Fluids are usually in pipes or cylindrical tanks.

We will focus on cylinders. The area A of a

cylinder is:

A = r = D

2=

D

4

2

2 2

π π π

•

VOLUME

Space Occupied By A Three-Dimensional Object

UNITS = LENGTH CUBED = In , ft , m

r

h

h

l

w

V rectangle = l w h

V circle = π r h

Figure 4-17 Volume of an Object




The volume of a cylinder is:

V = l r = l D

2=

l D

4

2

2 2

• •

π π

π .

Example:

Consider the following sections of pipes:

In both of the above pipe sections, A of fig 4-2 and B of fig 4-3, the volumes of each pipe section

are equal. Recall the mathematical expression for the volume of the given shape, a right cylinder,

is:

Volume = cross-sectional area x length

Whether the expression is 6 x 8 OR 8 x 6, the product is still 48.

Common units of volume are gallons, liters, cubic inches, and cubic feet.

Area =

6 in2

Length

= 8 in

Volume = 48 in

Figure 4-18 Volume of Pipe

Section A Length= 6 in

rea= 8 in2

Volume = 48 in

Figure 4-19 Volume of PipeSection B




Volumetric Flow Rate (V •

)

Volumetric flow rate (V •

) is the volume of fluid that flows past a fixed point per unit of time. Since

volumetric flow rate is a measure of volume with respect to time, its units must also be with respect totime.

Examples of units for volumetric flow rate include

gallons per minute (gpm), cubic feet per second

(ft3/sec), cubic inches per second (in3/sec), and liters per hour (L/hr).

The volumetric flow rate of a fluid through a pipe can

be determined if the cross-sectional area of the pipe

and the velocity of the fluid are known.

Volumetric flow rate V • is given by:

V =V

t =

lA

t

•

For a round pipe (which is a cylinder),

V =l 2r

t =

l 2 D

4t

• π π

.

This equation for V •

can be rearranged to look like a velocity times an area:

V =l r

t =

l D

4t

V =l

t r

2 2

2

•

•

•

π π

π

We know that π r 2

is the cross-sectional area A of the pipe (assume circular pipe area).

We recognize thatt

l as the dimensions of a velocity or speed.

In fact,t

l is the Flow velocity, expressed as vflow.

We can rewrite the volumetric flow rate as:

Figure 4-20 Volumetric flow Rate Visual




V = v A•

•flow flow

Where:

V •

= volumetric flow rate, m3/s or ft3/s

vflow = flow velocity, m/s or ft/s

Aflow = flow area: the cross-sectional area of the pipe, m2

or ft2

We find the volumetric flow rate V •

from the pipe size and the fluid velocity.

Example:

Figure 4.5 shows water with a flow velocity vflow of 2 feet/sec., moving from point X through

point Y.

To find volumetric flow rate:

The inside area of the pipe is 3 ft2. Water flowing through the pipe will fill the pipe, occupying

the entire cross-sectional area of 3 ft2; therefore, the area at point X is 3 ft2.

The velocity of the water is 2 ft/sec. This means that in one second, water will travel 2 feet down

the pipe.

If water is allowed to flow past point X for one second, a length of water 2 feet long will have

passed point X. The volume of the water is equal to the length ( 2 ft ) times the area ( 3 ft2 ),

which equals 6 ft3.

Volume of a cylinder = (Length of cylinder) × (cross-sectional Area of cylinder):

V = L × A.

Figure 4-21 Volumetric Flow rate Between Two Points




V = 2 feet in length x 3 square foot cross-sectional area equals 6 cubic feet of volume:

2 ft × 3 ft2 = 6 ft3 (six cubic feet)

If the fluid is allowed to flow freely at 2 ft/sec, 6 ft 3 will flow past point X each second.

Therefore the volumetric flow rate would be 6 ft3 per second.

Example:

The inside diameter of the pipe is 3 feet and the velocity of the fluid is 5 ft/sec. What is the

volumetric flow rate of the fluid? Refer to Figure 4.6.

We use the equation for volumetric flow rate:

V •

= vflow Aflow

The area of the pipe is not given, so we first find the cross-sectional area of the pipe. The area of

a circle A = π r 2. The radius is equal to one-half the diameter.

A = r = D

2

2

2

π π

D = 3 ftVFLOW =5 fts

Figure 4-22 Volumetric Flow Rate Example 1




Substituting values:

A =3

2

A =9

4

A = 7.0686

2

2

2

π

π

ft

ft

ft

•

Substitute the values for area and fluid velocity into the volumetric flow rate formula:

V •

= 7.0686 ft2•5 ft/s

Solve for volumetric flow rate:

V •

= 35.343 ft3/s = 35 ft

3/s

Anytime the cross-sectional area of a pipe and the velocity of the fluid are known, the

volumetric flow rate can be determined.

Remember:

Volumetric flow rate, V •

=Volume

Time=

Area × Length

Time= Area × Velocit

Note: All volumetric flow rates and mass flow rates hereafter will have a dot above the letter to

separate them from the normal symbols for volume and mass, (i.e. V • , m

• )

Also recall that: Velocity,ν

=

Length

Time




So, mathematically:

(Area × Velocity)

Volumetric Flow Rate

Where: V •

= volumetric flow rate, ft3/sec

A = cross-sectional flow area, ft2

= fluid velocity, ft/sec

Example:

The inside diameter of the pipe is 1 foot and the velocity of the fluid passing through the pipe is 5

ft/sec. What is the volumetric flow rate of the fluid?

Use the equation for volumetric flow rate: V •

= (Area × Velocity)

The area of the pipe is not given, so first find the cross-sectional area of the pipe. The area of a

circle is equal to π (we will use 3.14) times the square of the radius or one half of the diameter.

V •

= A × ν (Area × Velocity)

V •

= π D

2

2

× ν

Substitute values: V •

= 3.14 ×1ft

2

2

× 5ft

sec= 3.14×

1ft( )2

22 × 5

ft

sec

V •

= 3.14 ×12 ft 2

22 × 5

ft

sec= 3.14 ×

1 ft 2

4× 5

ft

sec

The volumetric flow rate is: V • = 3.93

ft 3

sec

This is the volume of fluid that flows past a point in the pipe every second.

5 ft/sec 1 foot

diameter

Figure 4-23 Volumetric Flow Rate Example 2




Summation:

Volumetric flow rate, V •

=Volume

Time; can take four forms:

V •

=V

t

V •

= Aν

V •

= 3.14d

2

2

ν

V •

= 3.14r 2

ν

Where:

is volumetric flow rate in cubic feet per second

d is inner diameter of a pipe in feet

r is inner radius of a pipe in feet

is fluid velocity in feet per second

V is volume in cubic feet

t is time in seconds

A is area in square feet

Mass, Density, and Specific Volume

Mass is the amount of matter in a substance. The density of a material relates mass to the volume it

occupies.

The amount of mass in a volume is determined by multiplying the density of the material times the

volume it occupies.






Mass Flow Rate (m•

)

Mass flow rate (m•

) is the amount of mass that flows past a fixed point per unit time. Since mass can be

defined in terms of volume and density, mass flow rate can be defined in terms of volumetric flow rateand density. The mass flow rate is the density of the fluid times the volumetric flow rate.

Where:

m•

= mass flow rate (lbm/sec) (units are “pounds mass per second”)

ρ = density (lbm/ft3) (units are “pounds mass per cubic feet”)

V •

= volumetric flow rate (ft3/sec) (units are “cubic feet per second”)

( Normal units for mass flow rate include lbm/hr and lbm/sec.)

Since volumetric flow rate can be defined in terms of area and velocity,

,

values of area and velocity may be substituted for volumetric flow rate and the following equation is also

true for mass flow rate:

or

Mass Flow Rate

Where:

m•

= mass flow rate (lbm/sec) (units are “pounds mass per second”)

ρ = density (lbm/ft3) (units are “pounds mass per cubic feet”)

A = cross-sectional flow area (ft2) (units are “square feet”)

ν = fluid velocity (ft/sec) (units are “feet per second”)




Example:

If the volumetric flow rate of the fluid in a pipeline is 4.0 ft3/sec, and the density of the fluid is

62.4 lbm/ft3, what is the mass flow rate of the system?

Use the equation which relates mass flow rate and volumetric flow rate:

m•

= ρ V •

Substituting values of density and volumetric flow rate:

m•

= ρV•

= 62.4lbm

ft3 × 4.0

ft3

sec

m•

= 249.6lbm

sec

Example:

Refer to Figure 4.8. If the velocity of the fluid is 5.0 ft/s, the pipe diameter is 3.0 ft, and the

density of the fluid is 62.4 lbm/ft3, what is the mass flow rate of the system?

Using the equation relating mass flow rate and volumetric flow rate:

Substituting values of density and volumetric flow rate:

D = 3 ft

p = 62.4

lbm

ft3

VFLOW =5 f t

s

Figure 4-24 Mass Flow Rate Example




m•

= 2,205.4 lbm/s

The Steady Flow Condition

When the flow velocity vflow at every point is constant with time, we say that the fluid motion is at a

steady-state. At any given point in a steady flow system, the velocity of each passing fluid particle is

always the same. At a different point in the system, the fluid velocity may be different from the velocity

at the first point. But the velocity at both points is constant over time.

Continuity of Flow

Continuity of flow states that the same mass flow exists everywhere in a pipe when the pipe is full andthere are no leaks into or out of the pipe (i.e. closed system). If a pump is pushing 100 lbm/hr of fluid

into a full pipe, the flow from the outlet at the other end of the pipe must be 100 lbm/hr.

“What goes in must come out as fast as it went in”, or said another way, “you can not create or destroymatter by pumping it through a pipe”.

The same mass flow exists throughout the pipe system, regardless of changes in the

configuration of the pipe.

This statement supports the law of conservation of mass and energy which states: neither mass nor

energy can be created nor destroyed, only altered in form.

Continuity of flow is commonly represented by the continuity equation. The continuity equation states

that the mass flow rate at one point is equal to the mass flow rate at any other point. Matter doesn’t just

appear nor disappear for no reason so the volume passing one point equals the volume passing any other

point.

mass flow rate @

point 1 = 100 lbm/hrmass flow rate @

point 2 = 100 lbm/hr

mass flow rate is the same

throughout system

Figure 4-25 Continuity of Flow




This is only true if the system is closed (i.e. does not have any leaks in or out).

Mass Flow Rate does not equal Velocity!!

Continuity Equation:

Where:

m1

•= mass flow past a point 1 (lbm/hr) (pounds mass per hour)

m2

•= mass flow past a point 2 (lbm/hr) (pounds mass per hour)

Substituting the mass flow rate equation, , on both sides produces the continuity equation:

Continuity Equation

Where:

ρ 1 = density at point 1 ρ 2 = density at point 2 (lbm/ft3)

A1 = area of pipe at point 1 A2 = area of pipe at point 2 (ft2)v1 = average fluid velocity at point 1 v2 = average fluid velocity at point 2 (ft/sec)

The continuity equation can be applied to the flow of any fluid, liquid, or vapor.




In applying it to the flow of liquids, it can be assumed that the density does not change significantly even

though there are changes in the cross-sectional area of flow.

As stated earlier liquids are considered incompressible.

If

then not only does mass flow rate in = mass flow rate out,

but also volumetric flow rate in = volumetric flow rate out.

The continuity of flow equation can be simplified:

In a fully filled enclosed system, changes in pipe area will cause changes in the fluid velocity to ensure

that system mass flow rate remains constant.

Example:

Refer to Figure 4.10 on the next page

The average velocity of water flowing in a 6-inch pipe is 25 feet per second. If the mass

flow rate is constant, what would be the velocity of the water if the size of the pipe

decreases to 4 inches?

(The cross-sectional area of the 6-inch pipe is 0.20 square feet, and the cross-sectional area

of the 4-inch pipe is 0.09 square feet.)




Using the simplified continuity equation:

Rearranging the equation to solve for velocity at point 2:

Substituting values for area and velocity at points 1 and 2:

Calculating/ Simplifying:

Comparing the velocities for the 6-inch pipe (25 feet per second) and the 4-inch pipe (55.56 feet per

second), a significant increase in fluid velocity occurs as the pipe size decreases.

This effect can be seen by the increase in velocity of water from a garden hose if the flow area is

reduced by placing a finger over the end of the hose.

If mass flow rate is constant and area decreases, flow velocity must increase.

If mass flow rate is constant and area increases, flow velocity must decrease.

Another way to see this change without actually doing the mathematics is called arrow analysis:

Refer to Figure 4.10 again. Using the simplified continuity equation:

Where:

A1 represents the cross-sectional area at point 1;

Figure 4-26 Continuity Example 2




ν 1 represents the fluid velocity at point 1;

- and -

A2 represents the cross-sectional area at point 2;

ν 2 represents the fluid velocity at point 2.

A2 is less than A1. In order to keep the equation equal, the velocity on the right side (ν 2) must be

greater than ν 1. In other words, to keep both sides equal the velocity on side 2 must increase to

compensate for the decrease in cross-sectional area. This can be shown with arrows in the

equation,

,

where the arrows indicate the decrease in area and the increase in velocity.

So without actually doing the mathematics a prediction can be made that the velocity will increase whenthe area decreases.

Calculations of Velocity, Area and Volumetric Flow Rate in a cylindrical pipe.

Use the equations and (when needed)




Fill in the empty spaces in the table: (answers on following page)

V⋅

Volumetric

Flow Rate

A

Area

v

Velocity

r

Radius

1.5 ft2

15ft

sec

10ft

sec0.5 ft

0.5 ft2

2.5ft

sec

18 ftsec

.25 ft

0.25 ft2

12ft

sec

30ft

sec2.2 ft




Student Guide: Fluid Flow Chapter 4: Volumetric and Mass Flow Rate

Answers:

V⋅

VolumetricFlow Rate

A

Area

v

Velocity

r

Radius

22.5ft3

sec

1.5 ft2

15ft

sec

7.85ft3

sec

0.79 ft2

10ft

sec0.5 ft

1.25ft3

sec

0.5 ft2

2.5ft

sec

3.53ft3

sec

0.2 ft2

18ft

sec.25 ft

3.0 ft3

sec0.25 ft2 12 ft

sec

455.9ft3

sec

15.2 ft2

30ft

sec2.2 ft





Chapter 4 Summary

• Volumetric flow rate (

V

•):

The volume of fluid that flows past a fixed point per unit time.

Units are gallons per minute (gpm), cubic feet per second (ft3/sec), and liters per hour.

Equals the cross-sectional area times fluid velocity or V •

= Aν (Area x Velocity).

• Mass flow rate ( ):

The amount of mass that flows past a fixed point per unit time.

Units include pounds mass per hour (lbm/hr) and pounds mass per second (lbm/sec).

Equal to density times volumetric flow rate or m• = ρ V • .

Also equal to density times area times the velocity or m•

= ρ Aν .

• The Principle of Conservation of Mass and Energy states that neither mass nor energy can be created or

destroyed, only altered in form.

• Continuity of flow exists when the same mass flow rate exists everywhere in a pipe.

m1

•= m2

•

ρ 1 A1v1 = ρ 2 A2v2

Therefore when mass flow rate is constant,

-if pipe size decreases, fluid velocity must increase;

-if pipe size increases, fluid velocity must decrease.

• The velocity of a fluid is given by ν =V•

A, or ν =

V•

πr2

(where A = πr2 ).





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OPERATIONS TRAINING PROGRAM Student Guide: Fluid Flow Chapter 5: Types of Flow

Chapter 5 TYPES OF FLOW

This chapter describes the two basic types of fluid flow: laminar and turbulent. The factors that

determine the type of flow is also discussed.

TO 5.0 Given the necessary fluid system parameters, DETERMINE the

fluid flow type and the flow characteristics of that fluid system

EO 5.1 DEFINE the two basic types of fluid flow: laminar, turbulent

EO 5.2 EXPLAIN how fluid flow parameters influence the type of fluid

flow

EO 5.3 DEFINE the Fluid Flow term “viscosity”

EO 5.4 EXPLAIN how the viscosity of a liquid varies with temperature

Laminar Flow

Laminar flow is the flow of fluid in parallel layers (refer to Figure 5.1 on next page). The layers of fluid

move smoothly over adjacent layers. There is little or no mixing between them. The fluid particles move

in definite paths or streams. This type of fluid flow is relatively silent, with low noise levels.

Turbulent Flow

Turbulent flow is characterized by random movement of the fluid particles (refer to Figure 5.1 on next

page). The particles travel in irregular paths with no distinguishable pattern this type of fluid flow has

high noise levels as compared to laminar flow.




Student Guide: Fluid Flow Chapter 5: Types of Flow

Factors Influencing Type of Flow

The type of flow experienced by a fluid depends on several factors, the velocity of the fluid, the diameter

of the pipe, the viscosity of the fluid, the roughness of the pipe, and the size of the pipe.

As fluid velocity and roughness increases, the movement of the particles becomes more random and

turbulent flow is more likely.

A larger diameter pipe allows more movement of particles making it more likely that turbulent flow willoccur.

Viscosity is a measure of a fluid's resistance to flow . A thick oil has a high viscosity, water has a low

viscosity. A more viscous fluid is more likely to experience laminar flow while a “thinner” less viscous

fluid is more likely to experience turbulent flow. As temperature increases, the “thickness” of a liquid

will decrease, causing viscosity to decrease, which will increase the possibility of turbulent flow.

Fluid friction increases with an increase in turbulence. Increased friction means that more energy must be

added to the system to cause the desired flow.

The amount of mixing within the fluid depends on the type of flow. Turbulent flow causes mixing of the

fluid. Mixing of the fluid enhances heat transfer and is an important consideration for convectionheat transfer systems. This can be a benefit in facilitation of heat transfer in heat exchangers, boilers,

and cooling towers. Turbulent flow may also enhance heat loss when you don’t want it so turbulent flow

within piping may indicate a need to periodically check for proper piping insulation.

Laminar Flow (Layers)

Turbulent Flow (Random)

Silence

Noise!

High Viscosity - Low Velocity - Smooth Walled Piping - Small Dia. Pipe

Low Viscosity - High Velocity - Rough Walled Piping - Large Dia. Pipe

Laminar Flow (Layers)

Turbulent Flow (Random)

Silence

Noise!

Figure 5-27 The Two Basic Types of Fluid Flow





Ideal Fluid

An ideal fluid is one that is incompressible and has no viscosity. An Ideal fluid is imaginary and does not

actually exist, but sometimes it is useful to consider what would happen to an ideal fluid in a particular

fluid flow problem in order to simplify the problem. Engineers do this all the time because it is almost

true in real fluids, so their results are still quite accurate though not perfect. It helps to simplify complexmathematics.

Noise Level and Flow Rate

Increasing the velocity of a fluid causes an increase in turbulence. An increase in turbulence will cause an

increase in noise level and vibration as the fluid flows through system piping and components. This

relationship between noise level and flow rate can be used by an operator to gauge changes in the flow

rate of a system.

Noise level and vibration can be valuable indicators of flow conditions





Chapter 5 Summary

Laminar flow is flow in parallel layers, while turbulent flow is random movement of fluid particles.

Also:

An increase in turbulence causes an increase in fluid friction.

Turbulent flow is better than laminar flow for heat transfer.

Noise and vibration are indicators of turbulence.

As fluid velocity and roughness increases, the movement of the particles becomes more random andturbulent flow is more likely.

A larger diameter pipe allows more movement of particles making it more likely that turbulent flow will

occur.

Fluid friction increases with an increase in turbulence. Increased friction means that more energy must be

added to the system to cause the desired flow.

An ideal fluid is one that is incompressible and has no viscosity.

Viscosity is a measure of a fluid's resistance to flow.

• As temperature increases a fluid will become thinner , so the measure of it’s resistance to flow,

viscosity, will decrease.• As temperature decreases viscosity will increase.



OPERATIONS TRAINING PROGRAM Student Guide: Fluid Flow Chapter 6: Forms of Energy & The General Energy

Equation

Chapter 6 FORMS OF ENERGY &THE GENERAL

ENERGY EQUATION

This chapter discusses the forms of energy in fluid flow systems. It also discuss the General Energy

Equation and the Law a Conservation of Mass and Energy in fluid flow systems.

TO 6.0 Given a fluid system, IDENTIFY the forms of energy using the

General Energy Equation

EO 6.1 DEFINE the following terms to include their units: potential

energy, kinetic energy, flow energy, internal energy, heat

energy, work energy, specific energy

EO 6.2 DESCRIBE the relationship between the General Energy

Equation and the Law of Conservation of Mass and Energy


The general energy equation is an attempt to sum up all of the energies that a fluid possesses. However,

the general energy equation does not take into consideration nuclear energies, nor chemical energy so is

not a perfect reflection of all energy in nature. It is therefore an incomplete mathematical expression of

the law of nature called the law of conservation of mass and energy.

The general energy equation is, however, a very useful tool when talking about conventional engineering

energy concerns. It means that for a steady flow system, the total energy (conventional engineering

energy only) at one point must equal the total energy at any other point in the system. As a result, even

though the amount of any particular type of energy may go up or down due to a change in other types of

energy, the total amount of energy will not have changed.

The general energy equation only considers the change or the difference in certain energies, and includes

some energies added from outside and some energies removed from the system. This may sound

complex, yet if written it may be easier to understand.

Energies at a beginning point 1, added to all energies that come from outside the system equate to:

Energies at an ending point 2, added to all energies that are removed and leave the system-

Substituting equivalencies for the energy at point 1 and the energy added:




Student Guide: Fluid Flow Chapter 6: Forms of Energy & The General Energy Equation

E point1 is PE1 + KE1 + FE1 + U1 and

Eadded is Qin + W in

Substituting equivalencies for the energy at point 2 and the energy removed:

E point2 is PE2 + KE2 + FE2 + U2 and

E removed is Qout + Wout

All of these energies added together make up the general energy equation.


Where:

PE = potential energy (ft.-lbf)

KE = kinetic energy (ft.-lbf)

FE = flow energy (ft.-lbf).

U = internal energy (Btu)

Q = heat (Btu).

W = work (ft.-lbf).

Before proceeding further, let’s review the different forms of energy in the general energy equation. Youhave seen potential and kinetic energy in earlier fundamentals courses.





Potential Energy (PE)

Potential energy (PE) is the energy an object (in this case a mass of fluid) possesses due to its position

above a reference level. The energy possessed by a fluid due to the work of the force of gravity upon it is

a large part of the potential energy of a fluid.

PE =mgz

g c

when using the English System,

(PE = mgz when using the International System)

Where:

PE = potential energy (ft-lbf)

m = mass (lbm)

z = elevation above (or below) reference level (ft)

g = acceleration of gravity 32.17ft / sec2( )

gc = gravitational constant 32.17ft − lbm / lbf − sec2( )

The drawing below is to represent a volume of fluid in a tank that has a height “z” that creates a pressure

at the bottom of the tank. It is to be a visual representation of potential energy.

Whenever a liquid is elevated above some reference point, its

potential energy is increased. When a liquid is lowered, its

potential energy is decreased.

Specific potential energy is the amount of potential energy possessed by a working fluid per pound-mass

of the fluid. It is denoted by the small letters pe.

Since g and gc are numerically equal, the specific potential energy is numerically equal to the height

above a reference level expressed in terms of ft-lbf/lbm.

pe =PE

m=

mgz

mg c

=gz

g c

pe = specific potential energy (ft-lbf/lbm)

Figure 6-28 A Visual of

Potential

Energy

Z







Attempting to compress a liquid that is

virtually incompressible does not cause it to

compress but causes it to move out of theway. This is much like trying to compress a

water melon seed between the fingers causes

it to slip from between the fingers and fly

away. In addition as Volume 1 (Figure 6.4)

moves through the pipe it must push Volume

2 ahead of it. In this example Volume 1 is

doing work on Volume 2 in raising its

elevation and in moving it from one location

to another. The pressure driving the flow is

what pushes the volume or fluid. The work

is being done by flow energy. The flow

energy is dependent on the pressure of the

fluid and volume of the fluid

moved.

Flow energy is shown in the following equation:

d

l u i dl u i d

o r v

l u i da n d

a t i

l u i da n d

a t i

F L O

l u i dp i s t ow o r k l

F I

F O

Figure 6-30 Flow Energy in Compressing Piston

F

V lV l

P iP i

- V o lp o i- F l up r- W

Figure 6-31 Flow Energy in Fluid Flow through a Pipe





PV FE =

Where:

FE = flow energy ft − lbf ( ) P = pressure lbf / ft 2( ) V = volume ft

3( )

The more fluid that is moved and the further the distance, the more flow energy that is needed to move it.

Flow energy is entirely different from kinetic energy because flow energy is not dependent on velocity. It

is only dependent upon pressure and volume.

Flow energy is provided by pumps or compressors - Flow energy pushes fluids around and compresses

them. Flow energy (FE) is the energy possessed by a fluid due to the pressure that is stored in a volume

of the fluid.

The drawing below is to represent a volume of fluid under pressure and is to be a visual representation of

flow energy.

Example: Determine the flow energy (FE) of a system where water with a pressure of 2 lbf / ft 2

passes

through a 4-foot section of pipe with an internal area of 0.5 feet.

Solution: FE = PV= P x A x L (Pressure times Area times Length)

FE = 2lbf

ft 2 x

0.5 ft 2

1 x

4 ft

1= 4 ft − lbf

Figure 6-32 Visual of Flow

Energy





Specific Flow Energy is the flow energy a working fluid possesses per pound-mass of the fluid. It is

denoted by the small letters fe. It is equal to the total flow energy divided by the total mass.

FE

m= fe =

PV

m= Pυ

Here we also have defined “υ ”, specific volume, whereV

m= υ , and where pressure is designated by the

letter “p”.

Internal Energy (U)

Potential energy and kinetic energy are visible forms of energy. These can be seen in terms of the

position and the velocity of an object. In addition to these visible forms of energy, a substance possesses

other forms of energy. For example, suppose we have a bucket of water sitting on the ground. The

bucket of water has no potential energy, because it is sitting on the ground. The bucket of water has no

kinetic energy because the bucket is not moving relative to its environment. However, the individualmolecules of water have energy due to their random motion with the bucket. The random motion of the

water molecules is a result of the internal energy (U). This is a microscopic form of energy that can not be measured directly.

Internal energy is the energy of the molecules of a substance due to their rotation, vibration, translational

motion, and intermolecular attractions. These forms of energy can not be observed directly, but the sumof these forms of energy can be indirectly measured. These microscopic forms of energy are grouped

together and are called internal energy (U). Temperature is an indirect measure of the internal energy of

an object. Therefore, we can measure internal energy indirectly with a thermometer.

The following drawing shows random molecular energies, with their intermolecular attractions and is to

be a visual representation of internal energy (U).

The random movement of atoms is of interest to us, and is increased and decreased by adding or

removing sensible heat energy to change the “temperature” of a fluid. Sensible heat energy is heat you

can measure with a thermometer. Also the “strength of attraction” of the atoms for each other is of

interest to us, and this is increased or decreased as latent heat energy is added or removed to change the

“state” of a fluid. Latent heat is heat that is hidden from measurement by a thermometer. (A more

complete explanation for these energies is given in the heat transfer course.)

Figure 6-33 Visual of Internal

Energy





Specific Internal Energy is the internal energy a working fluid possesses per pound-mass of the fluid. It

is denoted by the small letter u. It is equal to the total internal energy divided by the total mass.

Heat, as an operator controlled input or output (Q)

We are only concerned with the change in heat measurement that we read on a thermometer due to our

actions. This change in temperature is defined as “heat” and is written with the capital letter “Q” and has

units of Btu’s, (British thermal units) when using the English system. Heat is the energy transferred dueonly to a difference in temperature. If heat is transferred into a system the systems energy goes up, and

the energy is stored inside. If heat is transferred out of a system the systems energy goes down. (A more

through definition is given in the Heat Transfer course to include “latent” heat).

The drawing below is a visual representation of heat energy.

Specific Heat Energy is the measure of the difference in two temperature measurements of thermal

energy a working fluid possesses per pound-mass of the fluid. It is denoted by the small letter q. It is

equal to the heat energy divided by the total mass.

Q

Figure 6-34 Visual of Heat Energy





Work, as an operator controlled input or output (W)

Work is mechanical energy transferred due to a “force acting over a distance”. It is a measure of the work

that is done by the liquid acting on the container (like a turbine blade being driven by steam) or the work

that the container does on the liquid (like a piston compressing a gas into a smaller volume). It is writtenwith the capital letter “W”, and is measured in Btu’s (in the English system).

The drawing is of a force moving an object a distance. It is a visual representation of work energy.

Specific Work Energy is the work energy a working fluid possesses per pound-mass of the fluid. It isdenoted by the small letter w. It is equal to the work energy divided by the mass.


Again looking at the general energy equation, there are six energies on each side of the general energy

equation all added together.

PE 1+ KE

1+ FE

1+U

1+ Q

in+ W

in= PE

2+ KE

2+ FE

2+U

2+ Q

out + W

out

Where: Symbol Name Where? Most Common UnitsPE1 = potential energy at point 1 (ft.-lbf)

KE1 = kinetic energy at point 1 (ft.-lbf)

FE1 = flow energy at point 1 (ft.-lbf)

U1 = internal energy at point 1 (Btu)

Qin = heat energy in (Btu)

W in = work energy in (ft.-lbf)

and:PE2 = potential energy at point 2 (ft.-lbf)

KE2 = kinetic energy at point 2 (ft.-lbf)

FE2 = flow energy at point 2 (ft.-lbf)U 2 = internal energy at point 2 (Btu)

Qout = heat energy out (Btu)

Wout = work energy out (ft.-lbf)

Referring to the figure below, Point 1 is the beginning point for measurement of energies (PE, KE, FE

and U) contained in a given fluid flowing through a pipe section. These energies are on the “in” side

Figure 6-35 Visual of Work Energy





where energy is introduced into the system. Point 2 is the final or ending point for measurement of

energies (PE, KE, FE and U) contained in the fluid. These energies are on the “out” side where energy is

taken out of the system. When the difference between these measurement is determined, the general

energy equation begins to take shape. We will have defined “the difference in” or “change in” potential

energy, kinetic energy, flow energy, and internal energy that are contained in the fluid itself.

There are other energies from outside of the fluid that feed those energies within the fluid from outside,

Qin and Win. (Refer to Fig 6-11). Also there are other energies that take energy away from the energies

within the fluid, Qout and Wout. An example of Win may be a system pump. An example of Qout could bewhere some heat is transferred from the process fluid to another fluid, such as the cooling water in a heat

exchange system.

Put them together and you have an engineer’s mental picture of the general energy equation!

PE1

KE1

FE1

U1

PE2

KE2

FE2

U2

Point 1

Point 2IN

OUT

Figure 6-36 Fluid Energies 'IN' versus 'OUT'

Q i Q o

W i

W o

Flow

Figure 6-37 Energies Added versus Energies Removed





All the energy that goes

into the system equals all

the energy that comes

out of the system. (Thisis the conservation of

energy part!

You lose no energy- at

least theoretically

speaking.)

The General Energy Equation has units that can look like this:

PE 1+ KE

1+ FE

1+U

1+ Q

in+ W

in= PE

2+ KE

2+ FE

2+U

2+ Q

out + W

out

(ft-lbf) + (ft-lbf) + (ft-lbf) + (ft-lbf) + (ft-lbf) + (ft-lbf) = (ft-lbf) + (ft-lbf) + (ft-lbf) + (ft-lbf) + (ft-lbf) + (ft-lbf)

Anything that has units of “ft-lbf” is “energy.”

or they can have another name for the very same thing called “British thermal units”

PE 1+ KE

1+ FE

1+U

1+ Q

in+ W

in= PE

2+ KE

2+ FE

2+U

2+ Q

out + W

out

Btu’s + Btu’s + Btu’s + Btu’s + Btu’s + Btu’s = Btu’s + Btu’s + Btu’s + Btu’s + Btu’s + Btu’s

Anything that has units of “British thermal units” is “energy”

Q i Q o

W i W o

UF

KP

(System)

Figure 6-38 Visual of the General Energy Equation





A Special Case of the General Energy Equation: Bernoulli's Principle

Have you ever wondered why a car's convertible top bulges upward at high speeds or why smoke goes up

a chimney? These are examples of a principle discovered by Daniel Bernoulli (1700-1782). Bernoulli's principle says that where the velocity of a fluid is high, the pressure is low, and where the velocity is low,

the pressure is high. (Bernoulli was a Swiss mathematician who was the first to study this phenomena in

1738. He also developed an equation that expressed this principle quantitatively. ) By way of illustration,

in a case where water is flowing continuously through the pipe at Figure 6.13. If the pressure at points 1and 2 were measured, you would find that the pressure is lower at point 2, where the velocity is higher,

than it is at point 1, where the velocity is lower.

At first glance, this might seem strange. You might expect that the higher velocity at point 2 would imply

a greater pressure. But this cannot be the case. Because if the pressure at point 2 were higher than at

point 1, this higher pressure would slow the fluid down, where in fact it has speeded up in going from

point 1 to point 2. (Just as a wide river's velocity increases as it goes through a narrow gorge). So the

pressure at point 2 must be less than at point 1, which will allow the fluid to accelerate.

Bernoulli's principle explains many common phenomena, for example the convertible top in the previous

question. The air velocity over the convertible top is higher than inside the car. The pressure is therefore

lower above the top than below, and it bulges upward. Smoke goes up a chimney partly because hot air

rises, but Bernoulli's principle also helps. Because wind blows across the top of a chimney, the pressure

is less there than inside the house. So, air and smoke are pushed up the chimney. Even on a still night,

there is enough air flow to assist upward flow of smoke. (Bernoulli's principle applies to both liquids and

gases.) There are many examples of this principle that can be observed in everyday life.

Point 1

Point 2

Flow

Higher PressureLower Flow Velocity

Lower PressureHigher Flow Velocity

Figure 6-39 Bernoulli's Principle





Example:

A ping pong ball can be made to float above a blowing jet of air, see Figure 6.13. (Remember: Where

the fluid velocity is higher, the pressure is lower.) If the ball starts to leave the jet of air, the higher

pressure outside the jet pushes the ball back in. (Giancolli: 252)

Example:

In a properly designed airplane wing, the air stream separates at the front of the wing and rejoins

smoothly at the rear as in Figure 6.14. The air that flows over the upper surface must travel further, so its

average velocity is greater than the velocity of the air below. (Fluid velocity is higher so the pressure is

lower.) As a result, the air pressure above the wing is less than the air pressure below the wing. Thiscauses a net upward force on the wing called "lift." The forces on the lower side of the wing may account

for over 80% of the entire lift force on the wing. (Freeman: 78; Serway & Faughn: 234).

Example:

A pitcher throwing a curve ball and a golfer hitting a slice are both examples of Bernoulli's principle. A

curve ball results when a baseball is thrown with proper spin. The ball in Figure 6.16 is moving to the

right and is spinning clockwise. From the point of view of the baseball, the air is streaming by it toward

the left. However, because the ball is moving it is compressing air in front, and because the ball is

Low Pressure

High Pressure

Ping Pong

all

Figure 6-40 Ping Pong Ball Floating in Air

Stream

Airplane Wing1

12

2

GoUS AirForce!!!

Figure 6-41 Air Passing Above and Below Airplane Wing





spinning, some air is "dragged" along with the ball because the surface is rough. This causes the air

above the ball at B to be slowed down and bunched up (compressed) while the air below the ball is helped

along by the direction of spin (decompressed). Because the speed of the air is greater at A than at B, it

follows from Bernoulli's principle. The pressure at A is less than at B because there are more molecules at

B to exert pressure on the ball. The pressure difference produces a net force on the ball, causing it to

curve downward faster than under the force of gravity alone.

If the ball were to spin in the opposite direction the ball would create a compressed air front before it and

also a compressed air pocket below it and a decompressed air pocket above it. It would tend to rise incomparison to the trajectory of a normal ball.

So, Bernoulli's Principle, "where the velocity of a fluid is high, the pressure is low; and where the velocity

is low, the pressure is high" helps to explain why some everyday things happen. Lets look at the

scientific basis of Bernoulli's principle.

Simplified Bernoulli's Equation

Bernoulli's equation results in a simplification of the general energy equation. It applies to a steady flow

system containing an ideal fluid (is incompressible and has no viscosity or friction) and in which no

work is done on or by the fluid, no heat is transferred to or from the fluid, and no change occurs in the

internal energy (no temperature change) of the fluid. As we will see later these are serious limitations toreal world applications but they are necessary assumptions when we are not concerned about absolute

perfection.

Under these conditions, the General Energy Equation can be simplified to become the simplified

Bernoulli's equation:

Original General Energy Equation:

B

A

Force on

ball

Spin

Path of

ball

Air dragged

around

by spin

Figure 6-42 Air Passing by a Thrown Baseball





PE 1+ KE

1+ FE

1+U

1+ Q

in+ W

in= PE

2+ KE

2+ FE

2+U

2+ Q

out + W

out

Simplification Process:

PE 1+ KE

1+ FE

1+U

1+ Q

in+ W

in= PE

2+ KE

2+ FE

2+U

2+ Q

out + W

out

0 0

Note: U does not change from point one to point two, therefore it is the same and may be

canceled out. This only occurs because an ideal fluid has no viscosity (or friction), therefore “no

change” occurs in the internal energy of the fluid.

PE 1+ KE

1+ FE

1+U

1+ Q

in+ W

in= PE

2+ KE

2+ FE

2+U

2+ Q

out + W

out

0 0 0 0

independent of one another

Also Q (both in and out) becomes zero for an ideal fluid. No heat is transferred to or from the

fluid. (Operators control real fluids by operating heaters to add heat or by operating heat

exchangers to take heat out. Either action may take place with or without the other. An operator

can add heat without taking it out and vise versa.)

PE 1+ KE

1+ FE

1+U

1+ Q

in+ W

in= PE

2+ KE

2+ FE

2+U

2+ Q

out + W

out

0 0 0 0 0 0

independent of one anotherindependent of one another

Also W (both in and out) becomes zero for an ideal fluid. No work is transferred to or from the

fluid. (Operators control real fluids by operating compressors and pumps to add work or by

operating turbines and paddle wheels to take work out. An operator can add work without taking

it out and vise versa.)





Therefore: for an ideal fluid

PE 1+ KE

1+ FE

1= PE

2+ KE

2+ FE

2

(since ∆ U=0, ∆ Q=0, ∆ W=0) (“∆ “ means “change in”)(In other words, since there is no change in U it is given a zero value for an ideal fluid)

Substituting equivalent expressions for potential, kinetic, and flow energies, the equation becomes:

mgz1

gc

+mv1

2

2gc

+ P1V1 =mgz2

gc

+mv2

2gc

2

+ P2V 2

Which means that the total energy in a system at point one equals the total energy at any point two. Or

stated another way, the total energy in such a system is constant. (i.e. No loss of energy! A perfect

system! That is why it is called an ideal fluid!) Of course an ideal fluid does not exist.

mgz

gc

+mv2

2gc

+ PV = Constant

Simplified Bernoulli's Equation

Where: m = mass (lbm) g = acceleration due to gravity 32.17ft / sec2( )

v = velocity z = height above or below a reference

g c = gravitational constant, 32.17ft − lbm / lbf − sec2( )

P= pressure lbf / ft 2

( ) V = volume ft 3

( )

Specific Energies

For flowing systems, it is useful to express forms of energy in terms of energy per pound mass. Rather

than talk about the total energy in the system we can then talk about the energy in a single mass unit of

the fluid. Dividing an energy by mass turns it into a “specific” energy. Each term of Bernoulli's

equation can be expressed as a specific energy by dividing the term by the total mass of the system.

Energy per pound mass(lbm) is shown by small letters:

pe1 + ke1 + fe1 = pe2 + ke2 + fe2





Specific flow energy (fe) is the flow energy per unit of mass. It is equal to the total flow energy divided

by the total mass. Since specific volume (υ ) is equal to total volume divided by mass, specific flow

energy is equal to pressure times the specific volume.

fe =FE

m=

PV

m= Pυ

Where:

fe = specific flow energy ft − lbf

lbm

FE = flow energy ft − lbf ( )

V = volume ft 3( ) m = mass (lbm)

P = pressure lb f / ft2( ) υ = specific volume ft

3 / lbm( )

Substituting equivalent expressions after dividing by the mass of the system gives the specific energy

form of the equation.

gz1

gc

+v1

2

2gc

+ P1υ1 =gz2

gc

+v2

2

2gc

+ P2 υ2

Simplified Bernoulli's Equation-Specific Energy form

Each term in the above Equation represents a form of energy possessed by a moving fluid (potential,

kinetic, and flow related energies). The equation physically represents a balance of the potential, kinetic,

and flow energies so that if one form of energy increases, one or more of the others will decrease to

compensate, and vice versa.





Chapter 6 Summary:

Potential energy is the energy a fluid possesses due to its height (position) relative to other bodies.

Kinetic energy is the energy a fluid possesses due to its velocity.

Flow energy is the energy a fluid possesses due to its pressure and volume.

Internal energy is the energy of the molecules of a fluid due to rotation, vibration, translational motion

and intermolecular attractions.

Heat and Work are both outside inputs performed by an operator for the purpose of increasing or

decreasing the four fluid energies listed above; their use will be discussed in greater depth in the heattransfer course.

Rather than talk about the total energy in the system we can then talk about the energy in a single mass

unit of the fluid. Dividing an energy by mass turns it into a “specific” energy.

The Law of Conservation of Mass and Energy states; “Energy can neither be created nor destroyed, only

altered in form.” This law includes all energies in the universe.

The General Energy Equation includes only those energies pertinent to conventional engineering practice,

and is an incomplete mathematical expression of the law of conservation of mass and energy:PE

1+ KE

1+ FE

1+U

1+ Q

in+ W

in= PE

2+ KE

2+ FE

2+U

2+ Q

out + W

out

Dividing mass out of the General Energy Equation gives the specific form of the General Energy

Equation:

pe1 + k e1 + f e1 + u1 + qin + win = pe2 + k e2 + f e2 +u2 + qout + wout

All energy has units of Btu’s or Force times Distance.



OPERATIONS TRAINING PROGRAM Student Guide: Fluid Flow Chapter 7: Energy Conversions in Ideal Fluid Systems

ENERGY CONVERSIONS IN IDEAL FLUID SYSTEMS

This chapter describes the various energy conversions that can occur in Ideal Fluid Systems. An Ideal

Fluid System is one where no heat or work is transferred into or out of the fluid. Although not in

real world applications, the Ideal Fluid concept is frequently used to understand and predict

system behavior.

TO 7.0 GIVEN an Ideal fluid system where no heat is transferred in or out,

and no work is performed on or by the fluid, EXPLAIN the energy

conversions that occur

EO 7.1 Given an Ideal fluid system, DETERMINE the energy

conversions that occur using arrow analysis

Energy Conversions in Ideal Fluid Systems

As discussed earlier, energy may neither be created nor destroyed. However, the potential, kinetic, andflow energies in a fluid system may be converted from one form to another depending on the changes that

occur to the elevation or flow area (pipe size) of the piping system. Bernoulli's equation helps explain

how these energy conversions take place and how the energy balance is affected. (DOE Vol. III, p. 23)

Energy Conversions for Changes in Cross-Sectional Area (Flow Area)

The energy conversions that occur when pipe flow area (pipe size) changes can be determined using the

simplified form of Bernoulli's equation discussed earlier.

gz1

gc

+v1

2

2gc

+ P1υ1 =gz2

gc

+v2

2

2gc

+ P2 υ2

Remember, the above equation assumes an ideal fluid and no heat transferred in or out (perfectly

insulating), and that no work is done on or by the system (no compression, no decompression), andtherefore, no change occurs in the internal energy of the system (no loses due to turbulence, no internal

friction).

Since we are concerned here only with a change in pipe size, we do not consider a change in pipe

elevation. With no change in pipe elevation, there is no change in potential energy (nor specific potential

energy).

Since potential energy (gz

gc

) is the same at all points, it can be canceled out of the equation.




Student Guide: Fluid Flow Chapter 7: Energy Conversions in Ideal Fluid Systems

So the above equation can be simplified to:

v1

2

2gc

+ P1υ1 =v2

2

2gc

+ P2υ2

From the previous discussion of Bernoulli's principle, we know that as pipe size gets smaller or larger,

flow velocity increases or decreases, respectively. (Remember how velocity increases as a river goes

through a narrow gorge and slows down as the river widens again.)

As the velocity of a fluid increases, the kinetic energy of the fluid increases. The increase in kinetic

energy must come from some place. The only other energy that can change in this example is flow

energy (Pυ ). Therefore, as kinetic energy increases, flow energy must decrease to offset the increase in

kinetic energy. This is in accordance with the conservation of energy principle (energy can not be created

nor destroyed). This relationship is shown by the arrows in the equation below.

Notes regarding use of arrow analysis:

An easy way to get used to using arrow analysis is to remember that the analysis always looks back from point 2 to

where it was earlier at point 1, and it is a comparison of the difference that takes place.

Only three comparisons can be made, either “increase”, “decrease”, or “no change”.

• When the arrow points up, ↑ , it means an increase when looking back from pnt 2 to pnt 1.

• When the arrow points down, ↓ , it means a decrease when looking back from pnt 2 to pnt 1.

• When the arrow points sideways, → or ↔ , it means no change when looking back from pnt 2 to pnt 1.

It shows that as one form of energy goes up, the other must go down in order for the equation to remain in

balance.

v1

2gc

2

+ P1υ1 =v2

2gc

2

↑ +P 2υ2 ↓

This means that, in this instance where only the pipe size changes, a change in kinetic energy is exactly

offset by a corresponding change in flow energy. Or stated mathematically:

feke ∆=∆

(Where the "∆ " stands for the Greek "d", delta, and means "difference" or “change in”.)

***(The following examples assume a continuously flowing ideal fluid.)***





Example:

Explain the energy conversions that occur between points 1 and 2 below.

Flow energy is converted to kinetic energy.

Explanation:

1. Since pipe elevation does not change, potential energy remains the same and is not a

consideration. Only kinetic energy and flow energy are involved.2. Area 1 is greater than area 2.

3. Since area decreases, velocity must increase to maintain continuity of flow.

4. If velocity increases, then kinetic energy increases. The energy increase comes from a

conversion of flow energy.

Example:

Explain the energy conversions that occur in an ideal fluid between points 1 and 2 below.

Kinetic energy is converted to flow energy.

Explanation:

1 2

Figure 7-43 Pipe Section with a Reduction in Area

1 2

Figure 7-44 Pipe Section With Increase in Area





1. Since pipe elevation does not change, potential energy remains the same and is not a

consideration. Only kinetic energy and flow energy are involved.

2. Area 1 is smaller than area 2.

3. Since area increases, velocity must decrease to maintain continuity of flow.

4. If velocity decreases, kinetic energy decreases. The energy decrease is a conversion of kinetic energy to flow energy.

Energy Conversions for Changes in Elevation

The energy conversions that occur for a change in fluid elevation can also be determined using the

simplified form of Bernoulli's equation.

gz1

gc

+v1

2

2gc

+ P1υ1 =gz2

gc

+v2

2

2gc

+ P2 υ2

Since we are concerned here only with a change in elevation (the pipe goes up or down ), we do not

consider a change in pipe size. With no change in pipe size, there is no change in velocity or kinetic

energy. Since the kinetic energy term (v2

2gc

) is the same at all points, it can be canceled out of the

equation. So the above equation can be simplified to:

gz1

gc

+ P1υ1 =gz2

g c

+ P2υ2

As the elevation of a fluid increases, the potential energy of the fluid increases. Since energy can neither

be created nor destroyed, the increase must come from someplace. The only other energy subject to

change in this instance is flow energy. Therefore, if the potential energy increases, flow energy must

decrease to offset the change in potential energy. This is in accordance with the conservation of energy

principle. This is illustrated below by the arrows. They show how as one form of energy goes up the

other must go down to keep the equation in balance.gz1

gc

+ P1υ1 =gz2

gc

↑ +P2υ 2 ↓

This means that in this instance, where only the pipe elevation changes, a change in potential energy is

exactly offset by a change in flow energy. Or stated mathematically:

∆ pe = ∆ fe





Example:

Explain the energy conversion that occurs as an ideal fluid flows from point 1 to point 2.

Flow energy is converted to potential energy.

Explanation:

1. Since there is no pipe area (size) change, there is no change in fluid velocity so kinetic

energy remains the same and is not a consideration. Only potential energy and flow

energy are involved.2. Elevation 1 is lower than elevation 2.

3. Because elevation increases, potential energy must increase.

4. If potential energy increases, flow energy must decrease. In the equation below the

arrows illustrate the relative increase in potential energy and the corresponding decrease

in flow energy necessary to keep the equation in balance.

gz1

gc+ P1υ1 = ↑ gz2

gc+ ↓P 2υ2

Example:

Explain the energy conversion that occurs as an ideal fluid flows from point 1 to point 2 below.

Potential energy is converted to flow energy.

1

2

Z1

< Z2

Z1

Z2

Figure 7-45 Pipe Section with Increasing Elevation

1

2

Z < Z

Z1

Z2

Figure 7-46 Pipe Section with Decreasing Elevation





Explanation:

1. Since there is no pipe area (size) change, there is no change in fluid velocity so kinetic

energy remains the same and is not a consideration. Only potential energy and flow

energy are involved.2. Elevation 1 is higher than elevation 2.

3. If elevation decreases, potential energy must decrease.

4. If potential energy decreases, flow energy must increase. In the equation below the

arrows illustrate the relative increase in flow energy and the corresponding decrease in

potential energy necessary to keep the equation in balance in this example.

gz1

gc

+ P1υ1 = ↓ gz2

gc

+ ↑P 2υ2





Chapter 7 Summary:

Potential energy is the energy a fluid possesses due to its height (position) relative to another body.

If potential energy increases flow energy decreases, and if potential energy decreases flow energyincreases. (If pe ↑ then fe ↓ ; If pe ↓ then fe ↑ )

Potential energy increases when height increases, and decreases when height decreases.

Kinetic energy is the energy a fluid possesses due to its velocity.

If kinetic energy increases flow energy decreases, and if kinetic energy decreases flow energy increases.

(If ke ↑ then fe ↓ ; If ke ↓ then fe ↑ )

Bernoulli's principle says : Where the velocity of a fluid is high, the pressure is low; and where the

velocity is low, the pressure is high.

Velocity changes with pipe size changes. (A reducer causes an increase in velocity and a reduction in

pressure; and vice versa.)

Flow energy is the energy a fluid possesses due to its pressure and volume.

Flow energy takes the hit from potential energy, kinetic energy and internal energy. Whatever each of

these energies do flow energy compensates and does the opposite.

Advance information for comparison and ease of review-


and intermolecular attractions.Internal energy always increases and flow energy always compensates by decreasing. (u ↑ and fe ↓

100% of the time) Internal energy exists in a real fluid where there is always viscosity and fluid

friction due to intermolecular attractions.





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OPERATIONS TRAINING PROGRAM Student Guide: Fluid Flow Chapter 8: Energy Conversions in Real Fluid Systems

Chapter 7 Energy Conversions in Real Fluid Systems

This chapter describes fluid friction as well as the use of arrow analysis to determine the various energy

conversions that occur in real fluid systems where head loss is always present.

TO 8.0 GIVEN a Real fluid system, DESCRIBE the effects of fluid friction

to predict energy conversions

EO 8.1 DEFINE the fluid flow term “fluid friction”

EO 8.2 IDENTIFY the factors that effect fluid friction in Real fluid

systems

EO 8.3 DEFINE the fluid flow term “head loss”

EO 8.4 DESCRIBE the effects of fluid friction on a flowing fluid in a

piping system

EO 8.5 DEFINE the fluid flow term “throttling”

EO 8.6 DESCRIBE the effects of throttling on the fluid flow in a

piping system

EO 8.7 USE arrow analysis and the general energy equation to predict

energy conversions in a real fluid system

Friction

Friction is the force which opposes movement. Friction forces are created whenever two objects are

touching and there is movement in opposition to one another. If we slide a box along the floor, some of

the mechanical energy necessary to move the box is converted into internal energy. We measure this asan increase in the temperature of the box and the floor. When you rub your hands together, the friction

between your hands will cause an increase in the temperature you can feel.

Fluid Friction

Fluid friction is the force which opposes the movement of a fluid. A good example of fluid friction is theresistance we feel when stirring paint with a flat stick. Fluid friction converts flow energy into internal

energy.

Fluid friction occurs between a fluid and the walls of a pipe, and between each of the molecules of a fluid,

since each molecule hinders the movement of every other molecule it touches.




Viscosity

Viscosity was discussed in the chapter on Types of Flow. It is mentioned again here because viscosity is

often identified with fluid friction. Viscosity was defined earlier as “a measure of a fluid's resistance to

flow". Viscosity is tied to the internal friction of a fluid which makes it resist flowing past a solid surface

or other layers of fluid. If there is no viscosity there is no fluid friction and vice versa.

Factors Affecting Fluid Friction

The following factors will cause an increase in fluid friction:

• Increase in fluid velocity

• Increase in roughness of pipe

• Higher viscosity fluids

• Smaller diameter pipe

• Increase in turbulent flow

• Increase in pipe length

• Increased numbers of valves, fittings, and bends

Energy Conversion by Fluid Friction in Real Fluids

In a previous chapter we assumed an ideal fluid. For an ideal fluid, there is no viscosity and no fluid

friction and therefore, no change in internal energy of the system. But for a real fluid, the effects of

fluid friction must be considered.

This is Bernoulli's equation written to include internal energy "u", assuming no work is done on or by the

system, and no heat enters into or moves out of the system (∆ Q=0, ∆ W=0)

pe1+ ke

1+ fe

1+u

1= pe

2+ ke

2+ fe

2+ u

2

Bernoulli's equation can be applied to a straight horizontal pipe, with no change in elevation and no

change in pipe size (Figure 8.1). These assumptions eliminate the kinetic and potential energy terms from

the equation so that it becomes:

fe1+ u

1= fe

2+ u

2




Fluid friction resists the fluid flow and causes flow energy to be converted to internal energy as the fluid

flows through the piping system. Internal energy is the energy associated with the motion of the

molecules of a substance. Temperature is a measure of the molecular motion of molecules. An increase

in internal energy will increase the temperature of the fluid.

Fluid friction causes internal energy to increase, and flow energy to decrease. The change in internal

energy in this example is equal to the change in flow energy or:

∆u = ∆ fe

The amount of flow energy converted to internal energy by fluid friction is called head loss. (Also

equivalent to a pressure loss since fe = Pυ ).

Bernoulli's equation can also help show the effect of friction on flow in a pipe that has changes in both

elevation and pipe size.

Example:

Explain what happens to the energy levels and

temperature and pressure of the system shown

in figure 8.2. as water flows from point 1 to

point 2?

Solution:

Begin by evaluating the energy conversionsthat occur. In this case, the pipe is not

horizontal and the pipe size changes. The three

factors to consider are changes to elevation,

pipe size, and the presence of fluid friction.

PointPoint

Figure 8-47 Straight Pipe Section

Figure 8-48 Pipe Section with Changes in

size and Elevation




Potential Energy: Goes up. Because elevation increases from point 1 to point 2, the potential

energy of the fluid increases. Flow energy was converted to potential energy. This means

pressure decreases due to the decrease in fe.

Kinetic Energy: Goes up. Because the pipe flow area decreases from point 1 to point 2, thefluid velocity increases, which increases kinetic energy. This increase came from flow energy.

Pressure decreases again due to the decrease in fe.

Internal Energy: Goes up. Because fluid friction is present, flow energy is converted to internal

energy. This means pressure decreases a third way, and temperature increases.

-As a Result-

Flow Energy: Goes down. The changes in elevation, pipe size, and the presence of friction all

contribute to drops in the flow energy (pressure decrease) at point 2. This is characteristic of an

open flow (non recirculating) system. A pump must supply enough pressure to overcome the

pressure decreases caused by fluid friction and changes in pipe elevation and pipe size.

In the equation below, the arrows show the net effect of the energy conversions that took place in

this open/non-recirculating system.

pe1 + ke1 + fe1 +u1 = pe2 ↑ +ke2 ↑ + fe2 ↓↓ ↓+u2 ↑

Substituting equivalent terms for specific potential, specific kinetic, and specific flow energies:

gz1

gc

+v1

2

2gc

+ P1υ1 + u1 =gz2

gc

↑ +v2

2

2gc

↑ +P2υ 2 ↓ ↓↓ +u2 ↑

Energy Conversion by Fluid Friction

So far, in our discussions on energy conversions for changes in area and elevation, we assumed a

frictionless fluid. For an ideal fluid, there is no fluid friction and no change in internal energy of the

system. For a real fluid, the effects of fluid friction are real.

We will start with the delta form of the general energy equation.

∆ pe + ∆ ke + ∆ u + ∆ (Pυ) + wk,net = qnet

Fluid friction produces a pressure drop and an increase in internal energy. We can measure the pressure

drop. We can take pressure readings at both ends of a long pipe run, for example. We can also take pressure readings on both sides of a valve or other component.

We would like to know whether we can detect fluid friction as a change in fluid temperature. Recall that

Joule's constant J is the relationship between heat and mechanical work. We would like to answer the




following question: how much of a pressure drop is required to produce a 1ο F increase in temperature?The answer to this question is found using the mechanical equivalent of heat and some dimensional

analysis.

Example:

The pipe shown in Figure 8.4 has a constant

flow rate. There is no change in pipe size or

elevation.. What is the pressure drop from

point 1 to point 2?

First we write down the delta form of the

general energy equation:

∆ pe + ∆ ke + ∆ u + ∆ (Pυ ) + wk,net = qnet

Then we simplify the general energy equation

by analysis:

wk,net = 0

qnet = 0

∆ pe = 0

∆ ke = 0

This leaves us with

∆ u + ∆ (Pυ ) = 0

The change in internal energy ∆ u is the change associated with the change in fluid temperature from

70ο F to 71ο F.

In a separate Heat Transfer course, regarding sensible and latent heat, we will find that the heat added to a

system is given by:

q = c∆ T

where:

q = heat per unit mass in g

Jor lb

Btu

m

c = specific heat capacity inCg

Jor

Flb

Btu

°•°•m

∆ T = change in temperature in ο F or ο C

For water at room temperature, c = 1.0Flb

Btu

°•m. So for a 1ο F increase in temperature,

Pipe

1 2

T 1 = 70oF

p 1 =D 1 = 10 in

Z 1 = 50 ft

T 2 = 71oF

p 2 =D 2 = 10 in

Z 2 = 50 ft

Figure 8-49 The Pressure Drop from a 1°F

Temperature Rise




q = c∆T =1.0Btu

lbm⋅°F

1°F= 1

Btu

lbm

This energy--<which is transferred as heat at the molecular level>--is converted from flow energy

(pressure) to internal energy (temperature). So we are looking for a pressure drop equal to 1 Btu/lbm.

We do this by dimensional analysis:

Btu 1 Btu 778 ft•lb f 62.4 lbm 1 ft2

lbm lbm 1 Btu ft3 144 in2 1 =

Btu

lbm

1 = 337 psi

Although it's possible to prove that the density belongs in this calculation, dimensional analysis gave us

the correct answer. We just used unit conversion factors which took us in the right direction.

This says that a 1ο F rise in fluid temperature corresponds to a 337 psi change in pressure. Note that this

result is only valid for water at room temperature. For other fluids and at other temperatures, this is not

the right answer.

In other words, if the pressure drops 337 psi through a pipe, a fitting, or an obstruction, the friction will

produce a 1ο F rise in temperature. For this reason, we cannot measure the effects of fluid friction by

observing a temperature rise. We can only see the pressure drop. The internal energy increase is real. It's

just not measurable with a thermometer.

A decrease in pressure produces a decrease in flow energy ( p ↓ •υ↔

= ↓ fe ).

That decrease in the flow energy must be exactly matched by an increase in internal energy.

(↓ ↑∆ ∆ fe + u = 0 ).

The amount of flow energy converted to internal energy by fluid friction is called head loss. (Head loss

will be discussed later). Fluid friction converts flow energy to internal energy as fluid flows through a

piping system. Internal energy is the energy associated with the motion of the molecules of a substance.

Temperature is a measure of that molecular motion. An increase in internal energy is seen as an increase

in the temperature of the fluid.

Energy Conversions in a Real Fluid System with No Change in Flow Area or Elevation:




If the area is constant, the velocity can not change, in accordance with the continuity equation. Since

specific kinetic energy consists of the velocity squared divided by 2, if the velocity is constant the kinetic

energy must also be constant

Specific potential energy consists of gravity times height (z). Gravity does not change. Since there is no

change in elevation (z stays constant), there is no change in specific potential energy.

In a real fluid system that has viscosity there is always friction. Friction ALWAYS causes an increase in

the specific internal energy.

Since specific internal energy has increased and energy can neither be created nor destroyed the energymust be transformed into another form. Since the specific kinetic energy and the specific potential energy

did not change, the only energy form that can change is the specific flow energy. Since specific internal

energy increased, the specific flow energy must decrease. Specific flow energy consists of specific

volume times pressure. For an incompressible fluid the density and specific volume remain constant. If the specific volume remains constant and the specific flow energy decreases, the pressure must also

decrease.

Remember, there are several factors will cause an increase in fluid friction in real fluids:

• Increase in fluid velocity

• Increase in roughness of pipe

• Higher viscosity fluids

• Smaller diameter pipe

• Presence of turbulent flow instead of laminar flow

• Increase in turbulence

• Presence of valves, fittings, and bends

• Length of pipe




Example:

The system shown in Figure 8.5 has a pressure of

70 psig at point 1 and a pressure of 45 psig at point

2. What is the flow energy loss due to fluid

friction?

As always, we first write the delta form of the

general energy equation:


Analyzing the system, we find:

qnet = 0

wk,net = 0

∆ ke = 0

The general energy equation then reduces to:

∆ pe + ∆ u + ∆ (Pυ ) = 0

We identify ∆ u as the flow energy loss due to fluid friction. So we solve for ∆ u:

∆ u = -∆ pe - ∆ (Pυ)

Next, we get numerical values for υ ∆ P and ∆ pe.

∆ PSince: P2 - P1 = 45 psig - 70 psig

∆ P= -25 psig =2

25

in

lbf

υ

υ =1

ρ=

1

62.4 lbm

ft 3

=0.01603ft3

lbm

therefore,

∆ (Pυ)

υ ∆ P=

υ ∆ P = -57.708m

f

lb

lbft •

D 1 = 6 in

V 1 = 500 gpm

Z 1 = 80 ftp 1

= 70 psig

T 1 = 70 F

1

2

D 2 = 6 in

V 2 = 500 gpm

Z 2 = 125 ft

p 2 = 45 psig

T 2 = 70 F •

•

Pipe

Figure 8-50 Pressure Drop and Fluid Friction

0.01603 ft3 -25 lb f

144 in2

lbmin

21 ft

2




Now solving for ∆ pe:

∆ pe = g •( z 2 - z 1) = ( ) ft ft

s

ft 80125

17.322

−•

∆ pe =

∆ pe = 45m

f

lb

lbft •

Since

∆ u = -∆ pe - ∆ (Pυ)

Then,

∆ u = −(−57.708m

f

lb

lbft •) −45

m

f

lb

lbft •

∆ u = 12.708m

f

lb

lbft •

That is,

u2 - u1 = 12.708m

f

lb

lbft •

The internal energy at point 2 is greater than the internal energy at point 1 by 12.708m

f lb

lbft • .

We can convert this to Btu or an equivalent pressure drop.

m f f

f Btu/lb0164.0

lbft778

Btu1

lb

lbft708.12=

•

•

T∆= cq

Rearrange to solve for T∆ :c

q=∆T

F0164.0Btu1.0

Flb

lb

Btu0.0164T °=

°•=∆

m

m

u2-u1=2

2

2

3in/lb507.5

in144

ft1

ft

lb4.62

lb

lbft12.708 f

m

m

f =

•= 5.507 psi

32.17 ft 45 ftlb f •s

2

s2 32.17 lbm•ft




The internal energy change due to friction is 0.0164 F° (not a measurable temperature

rise), however, it is 5.507 psi (a detectable pressure drop!).

Energy Conversion In a Real Fluid System with an Increase in Elevation

If the area is constant, the velocity can not change, in accordance with the continuity equation. Since

specific kinetic energy consists of the velocity squared divided by 2, if the velocity is constant the kineticenergy must also be constant.

Specific potential energy consists of gravity times height (z). Gravity does not change. Thus if elevation

increases (z increases), the specific potential energy must increase.

In a real fluid (or system) there is always friction. Friction ALWAYS causes an increase in the specific

internal energy.

Since specific potential energy and the specific internal energy have both increased and energy canneither be created nor destroyed the energy must be transformed into another form. Since the specific

kinetic energy did not change, the only energy form that can change is the specific flow energy. Since

specific potential energy and specific internal energy have both increased, the specific flow energy must

decrease. Specific flow energy consists of specific volume times pressure. For an incompressible fluid

the density and specific volume remain constant. If the specific volume remains constant and the specific

flow energy decreases, the pressure must also decrease.

Open versus Closed Fluid Flow Systems

An open fluid system is, as the name implies, a fluid system where the fluid source and fluid destination

are separate. One example of an open system could be a typical drinking water system where the clean pressurized water comes from one location but never returns to that same destination. A “once through”

design is another way of viewing an open system.

A closed fluid system is a fluid system where the fluid source and fluid destination are the same. In other

words, the fluid re-circulates in some “loop” and is returned to the pressure source (pump) for to be re-

distributed into the system. One example of a closed system could be the radiator coolant system in an

automobile where the water is drawn from the radiator bottom into the engine block by the water pump.

The pump returns the heated fluid through the opened thermostat into the radiator top, where the cycle is

repeated.

Energy Conversions in Closed Systems

In a closed or recirculating flow system, changes in elevation and pipe size do not cause significant net

losses in system pressure. Fluid friction causes the most significant losses in system pressure in a

closed/recirculating flow system. Where viscosity, and therefore fluid friction (e.g. ideal fluid), do not

exist the following takes place:

Example:




Explain the energy conversions and pressure and temperature changes for the closed/recirculating

system shown in Figure 8.3, as water flows from point 1 to point 5.

Solution:

Begin by considering the energy conversions that occur and noting which conversions causeincreases in pressure (flow energy) and which cause decreases in pressure (flow energy).

Point 1 to Point 2 - Flow energy (pressure) is converted to potential energy. Pressure

decreases.

EXPLANATION: As the fluid is elevated from point 1 to point 2, its potential energy is

increased. This increase in potential energy has to come from somewhere. (Per the Lawof Conservation of Energy - you don't get something for nothing). The increased

potential energy comes from flow energy being converted to potential energy. Flow

energy (pressure times specific volume, Pυ ) has decreased. Therefore, pressure has

decreased. The arrows in the equation below show the relative changes in energy from

point 1 to point 2.

gz1

gc

+v1

2

2gc

+ P1υ1 + u1 =gz2 ↑

gc

+v2

2

2gc

+ ↓P2υ2 + u2

Point 2 to Point 3 - Flow energy (pressure) is converted to kinetic energy. Pressure

decreases.

EXPLANATION: The pipe size decreases at point 3, which causes an increase in

velocity. When velocity increases, kinetic energy increases. The increase in kinetic

energy has to come from somewhere (again - you don't get something for nothing). Itcomes from flow energy ( Pυ ) causing a pressure decrease.

Figure 8-51 Energy Conversions in a Closed System




gz1

gc

+v1

2

2gc

+ P1υ1 + u1 =gz2

gc

+ ↑v2

2

2gc

+ ↓P2υ2 + u2

Point 3 to Point 4 - Kinetic energy (velocity) is converted to flow energy. Pressureincreases.

EXPLANATION: The pipe flow area increases as the fluid leaves point 3, which causes

velocity to decrease. When velocity decreases, kinetic energy decreases. The lost kinetic

energy is converted back to flow energy ( Pυ) and pressure increases an amount equal to

the decrease from point 2 to point 3.

gz1

gc

+v1

2

2gc

+ P1υ1 + u1 =gz2

gc

+↓v2

2

2gc

+ ↑P2 υ2 + u2

Point 4 to Point 5 - Potential (height) energy is converted to flow energy. Pressure

increases.

EXPLANATION: As the fluid elevation drops from point 4 to point 5, its potential

energy is decreased. The decrease in potential energy is equivalent to the increase in

potential energy as the fluid moved from point 1 to point 2. The potential energy lost isconverted back to flow energy ( Pυ ). The pressure therefore increases by an amount

equal to the pressure decrease from point 1 to point 2.

gz1

gc

+v1

2

2gc

+ P1υ1 + u1 =gz2 ↓

gc

+v2

2

2gc

+ ↑P2υ2 + u2

Net Result: No losses in Pressure (e.g. ideal fluid) If we did not consider the effects of fluid friction, the above analysis would show that the pressure at point 1 and point 5

would be the same. The decrease in pressure when the pipe size decreases is exactly

offset by the increased in pressure when the pipe size increases. The decrease in pressure

when elevation increases is exactly offset when the fluid returns to the same elevation it

started from.

Fluid friction is present in all real fluids including all gases and liquids. Fluid friction results in the

conversion of flow energy to internal energy, as evidenced by a decrease in pressure and an increase in

temperature.

In the equation below the arrows show the net result of the energy conversion in a real fluid that took place in a closed/recirculating system.

gz1

gc

+v1

2

2gc

+ P1υ1 + u1 =gz2

gc

+v2

2

2gc

+ ↓P 2υ2 + ↑u2

Flow Energy goes down and Internal Energy goes up by the same amount.




Unlike the pressure drop caused by the changes in pipe size and elevation, fluid friction results in an

unrecovered pressure drop. This pressure drop is referred to as head loss. This drop in pressure must

be replaced by work on the fluid, usually provided by a pump, or the fluid will stop flowing.

Fluid Friction results in an un-recovered pressure drop

Let's look next at another closed loop. Many fluid systems are closed loops. A typical closed loop has a

pump to supply pressure, some load

(like a heat exchanger), andconnecting piping. The system in

Figure 8.6 is a simple closed loop.

Water leaves the pump at point 1, flows through the loop, and returns to the pump at point 2. Out in the

loop, the system experiences changes in elevation

(z), pipe size(A), and direction. It probablytransfers heat. If it is a power system (like a

hydraulic system) it may also perform work.

We want to find out what the effect of fluid friction

is on this real system.

Example:

Figure 8.7 shows the fluid properties of a system. Determine the fluid friction energy drop in this

system. The density of water at 104ο F is 61.9 lbm/ft3.

We start with the delta form of the general energy equation:


The figure shows that the system has heat in and heat out. However, since T 1 = T 2, we

may conclude that qin = qout, and that qnet = 0. Flowing from point 1 to point 2 there is

no work, so wk,net = 0. The pipe diameters are the same, so v1 = v2, and ∆ ke = 0. The

elevation changes from point 1 to point 2. However, a 2-ft change in height produces

a negligible change in the pressure, compared to the rest of the problem. (If we were

W k, out

Q in Q out

2 1

Figure 8-52 A Simple Closed Loop System

Figure 8-53 Closed Loop Example




in doubt on this point, we could include pe in the calculation.) Therefore, the general

energy equation reduces to:

∆ u + υ ∆ P = 0

Again, we identify ∆ u as the fluid friction term. Friction converts flow energy into internalenergy.

∆ u = -υ ∆ P

∆ u = -0.01613 ft3 -65 lb f 144 in2

lbm in2 1 ft2

∆ u = 151.2 m

f

lb

lbft •

The fact that ∆ u is positive tells us that internal energy increases as the water flows from point 1 to point

2. At point 2, the pressure has decreased, and the internal energy has increased by exactly the same

amount. The pump raises the fluid pressure, and friction reduces it. The reason we need a pump in a

closed loop is to overcome the effects of fluid friction.

Unlike the pressure drop caused

by the changes in area and

elevation, fluid friction results in

an unrecovered pressure drop.

This pressure drop is referred toas head loss. This drop in

pressure must be compensated by

work on the fluid by the pump.

_Head_

Engineers and pump

manufacturers often use the term

head to mean pressure.

Head is "pressure expressed as an equivalent water column height."

Head is related to pressure in psi by the height of an equivalent water column. Recall the equation we

developed in the chapter on fluids, and refer to Figure 8.8. The relationship between pressure and column

height is given by:

Patm

10 ft

30 ft

20 ft

30 ft

Figure 8-54 Pressure is Proportional to Column Height




P = ρ gz

Pressure is the product of density ρ , gravity g , and height z . Head and pressure describe the same fluid

property.




"Head" is another term for fluid pressure. Each of the energies possessed by a fluid can also be

expressed in terms of "head”. The term "head" is used to refer to pressure. It is a reference to the height,

typically in feet, of a column of water that a given pressure will support. If atmospheric pressure at sea

level (14.7 psia) is expressed as head, it would be about 33.92 feet. Head is measured in feet because of

its direct relationship to the pressure at the bottom of a column of fluid. Pumps move fluid from one

point to another and from one level to another. When pumps were first built, it was easier to measure the pressures as differences in height between components. The differences in height can be used to measure

pressure, because a column of water has a pressure associated with it.

Elevation head - represents the potential energy of a fluid due to its elevation above or below a

reference level. (It is the energy a fluid possesses due to its height).

Velocity head - represents the kinetic energy of the fluid. (It is the energy a fluid possesses due

to its velocity). Velocity head is the height in feet that a flowing fluid would rise in a column if

all its kinetic energy were totally converted to potential energy.

Pressure head - represents the flow energy of the fluid. (It is the energy a fluid possesses due to

its pressure). Pressure head is the height in feet a column of water would have if all its flowenergy were totally converted into potential energy.

When all the energies of a system are turned into heights as though all energies were potential energies,

we have a convenient way of discussing them in terms of the pressure at the bottom of a tank of water.

We may then talk about this total head pressure in terms of a number of feet of height.

How it is done mathematically:

Elevation Head pe = pegz

gc

=gz

gc

solving for z; z = z in feet

Pressure Head fe = pe Pυ =

gz

gcsolving for z; z=

Pυgc

g in feet

Velocity Head ke = pev2

2gc

=gz

gc

solving for z; z=v2

2gin feet

When we wish to determine the delivery pressure of a fluid we talk about several specific pressures which

we express as head, in units of feet:

Static head is the potential energy of the fluid, expressed in feet (this is also an elevation head,

but of the tank only, or of a pump).

Velocity head is the kinetic energy of the fluid, expressed in feet.

Pressure head is the flow energy of the fluid, expressed in feet.

Total head is the sum of all elevation heads above the delivery point (e.g. level of liquid within a

tank “static head of tank” + the height of the floor level upon which the tank sits “elevation head

of floor” or the hill upon which it sits “elevation head of hill”)




Head loss is loss of pressure. Head loss shows up in the general energy equation as an increase in

internal energy. It is the amount of pressure loss due to friction as a fluid moves through the

system. It is caused by valves, pipe, elbows, orifices, and other flow restrictions. Pressure

measured at the pump discharge of a given system is higher than pressure downstream because of

the head loss that occurs along the length of the pipe.

Figure 8-9 describes the effects fluid flow at some constant velocity to the system pressures. The

following can be derived referencing the above illustration:

• When the valve is fully shut the Gage A, B, and C pressures are equal to the static head

of the tank plus the elevation head of the tank (height that the tank is above the pressure gage)

• When the valve is opened to allow some fluid flow Gage A, B, and C pressures Decrease

due to head losses. Note that Head losses are the combination of velocity head and friction

head. Velocity head and friction head both take pressure away from pressure head.

However, velocity head is the same value at all three gage measurements, while friction head

increases down the length of the pipe.

• This measurement or calculation can be taken at any point in the piping system. Noticethat the amount of velocity head does not increase as the fluid flows through the pipe once

flow has been initiated. Also notice that the amount of friction head does indeed increase as

the fluid continues to flow greater and greater distances through the piping system.

The figure 8-9 is exaggerated to better define the different heads involved when a tank causes a fluid to

flow. A pump can replace the tank in the drawing and the very same heads will be present. Both tanks

and pumps (including air compressors) can create pressure head. However, moving any mass of fluid to

give it a velocity comes at the cost of some of that pressure head in the form of velocity head . Also, since

flow

Static head

y

y

y

z

x xx

zz

Velocity Head =

Friction Head =

Pressure Head =Elevation

head

Gage

C

Gage

B

Gage

A

alve

Total

head

Figure 8-55 Pressures Within a Fluid Flow System (exaggerated)




every real fluid has a resistance to flow called fluid friction there is an additional unrecoverable loss of

pressure head which goes into the internal energy of the fluid. This is eventually dissipated into the

surroundings and lost as heat. The energy lost due to friction head is not recoverable and does not return

as an eventual increase in pressure head.

Loss due to friction head is easily noticed in a closed loop system where a fluid returns to a pump. The pump must continue to make up the head losses in the system due to friction head. If this were not so a

fluid could be started moving in a closed loop system and it would never stop.

The term “head loss” is usually used as a synonym for friction head loss, so when you hear of losses of

head or losses of head pressure in a system immediately think of fluid friction, internal energy, andfriction head.

Total head (is a sum) =

Static head of the tank + Elevation head of the tank to measure point such as a tank platform

height.

(i.e. the sum of all elevation heads)

Delivery head (is a difference) =

Total head – Friction head – Velocity head

(i.e. the pressure left after velocity and friction have been paid their price)

When a fluid comes from a tank or is delivered via a pump there is a static head that is created and sinceall real fluids have fluid friction there is also a small amount of friction head. The longer the piping run

and the more the fluid is agitated by the molecules of the fluid bumping into fittings and bends etc. the

greater the losses of pressure head (due to friction head).

When fluid flows for short distances head losses are small and the head losses are not significant. That is

the reason many engineers like to think of water as having virtually no fluid friction. As a result theengineering profession created the concept of an imaginary ideal fluid where fluid friction does not exist.

Please remember that we often are not just concerned with pumping water short distances but may pump

very viscose fluids with significant losses due to fluid friction. Even water pumped for large distances

and passing through many fittings and bends will have significant friction losses. An example could be

fire fighters count the number of hoses from the pump to the nozzle and turn up the pump 15 psi for each

100 feet of 2 1/2 inch hose to make up for the loss of head. The high velocity of the water is the culpritthat causes this action to become necessary. High viscosity slurries may also require pumps to carry a

heavier load as in radioactive waste, (for example), where fluids are purposely concentrated to a higher

concentration of heavy metals.

Referring to Figure 8.10, the illustration shows that total head is the sum of the elevation head produced

by the tank (or pump) itself and the actual elevation of the tank or pump above the delivery point. The

delivery head is this total head after subtracting the losses due to friction head and velocity head. For




water pumped short distances friction head is very small or insignificant. For large distances, high

velocity fluids, and high viscosity fluids friction head must be considered.

Refer to figure 8-10. When a pump is lifting fluid from a sump or a tank that is lower than the pump

itself, and is delivering fluid over the top of a tank it is working more than if the fluid were being taken

from a tank that is higher than the pump where the fluid is being forced into the pump, and pumping to

the side of a tank. Where the pump is delivering to a tank through the side of the tank it is pushing onlyagainst the actual head pressure (static head) of the fluid already in the tank. In the first case the pump

works harder. The pump does less work in the second case shown.

By studying the total static head in the two drawings you will be able to see pump A doing more work

than pump B because it is lifting fluid and also delivering fluid higher than the level needed. Pump B

benefits from the higher elevation of the supply tank as well as from the smaller elevation to be delivered.

Head Loss due to Friction

Friction head loss is unavoidable in real fluids. Head loss occurs because of the friction between the

fluid and the walls of the pipe; or between adjacent particles as they move relative to one another. Friction

also occurs in turbulence caused whenever the flow is redirected or affected in any way by such

components as piping entrances and exits, pumps, valves, flow reducers, and fittings.

Work must be done on a fluid to overcome fluid friction. This is typically done by a pump, and is

referred to as “work done on a fluid”.

(Actually helpsthe pump)

Static SuctionLift (head)

TotalStaticHead

Pump A

DischargeHead

(Actually countersthe pump work)

Static Suction

Lift (head)

DischargeHead

TotalStaticHead

Pump BFluid is

‘pushed’ intothe pump

Fluid is‘pulled’ intothe pump

Figure 8-56 Total Static Head Examples




Head losses due to friction occur throughout a piping system but they are greatest as the fluid flows

through small openings in valves, fittings, and any piping with rough inner surfaces.

Studies of the flow of liquids show that head losses due to friction:

Increase as:

• The length of pipe increases - a longer pipe has more surface area.• The fluid velocity increases.

• The pipe diameter decreases - a pipe with a smaller diameter has more surface area per

unit of cross-sectional area.

• The roughness increase in piping inner surfaces.

• Controlling Flow

Throttling

Valves are used to control the flow rate and pressure of fluid systems. Throttling is the term used to

describe the control of fluid flow by manipulating a valve. It is commonly used to mean partially closinga valve to restrict flow. Throttling is the brakes of a fluid system and just like the brakes on a car, heat is

given off.

Throttling is the brakes of a fluid system.

Without fluid friction throttling could not take place.




A valve causes a reduction in flow area followed by area increasing to the previous size. (See Figure

8.11.) As fluid flows from upstream of the valve, fluid enters the reduced area of the valve. A sudden

increase in velocity and decrease in pressure occurs as flow energy is converted to kinetic energy.

On the downstream side of the valve, the flow area increases abruptly. Velocity decreases and pressure

increases as kinetic energy is converted back to flow energy.

If we truly had ideal fluids in our piping systems and therefore the conversions between flow energy and

potential and kinetic energy were the only processes occurring, there would be no pressure drop across a

valve. Valves would not work. We could not limit flow and could not stop a fluid from flowing. If wecould create a truly ideal fluid we would have to throw away all piping systems and valves.

Due to the abrupt increase in area downstream of the valve, flow becomes highly turbulent. The

increased turbulence causes an increase in fluid friction. The increase in fluid friction causes a

conversion of flow energy to internal energy and a sustained pressure drop occurs. Manipulation of

valves is a common method of controlling the flow rate and pressure of fluid systems by putting on the

brakes. Chattering of the sprayer unit for a kitchen sink shows that the flow stops before the valve is

fully closed.

Overcoming Head Losses

Pumps are machinery used to overcome head losses. These devices are the opposite of pressure reducers

and could easily be called “pressure increasers”. The centrifugal pump is even built the opposite of a

pressure reducer with an expanding volute to increase pressure as the fluid passes through it. Althoughcentrifugal pump theory is covered in detail in another fundamentals course, the following section

discusses centrifugal pump operation.

c t h oFigure 8-57 Typical Valve




Centrifugal Pump Operation

A centrifugal pump converts mechanical energy to flow energy using the centrifugal force created by a

rotating shaft. Figure 8.12 shows the major parts of a centrifugal pump.

Here's how a centrifugal pump works to increase the head (or pressure) of the fluid:

1. A motor driven shaft is connected to and rotates the impeller.

2. The fluid being pumped enters through the suction eye at the center of the impeller hub.

3. The impeller vanes catch the fluid and spin it in the direction of the impeller and shaft rotation.The vanes of the impeller are shaped so that they throw the fluid out of the impeller – (they do

not scoop or catch the fluid.)

4. As the fluid gains velocity (kinetic energy), it pushes outward against the casing walls. This

action is due to centrifugal force.5. As the fluid moves outward and around within the pump casing, it makes room for more fluid at

the center of the impeller, creating a suction which draws more fluid into the suction eye of the

impeller.

6. The water discharged at the tip of the impeller into the pump volute is moving rapidly (it contains

a large amount of kinetic energy). In the volute, essentially a gradually-widening chamber, the

fluid spreads out to fill the chamber and thus slows down. The increasing cross-sectional area of

the volute causes fluid velocity to decrease. As the fluid slows, the kinetic energy is converted

into flow energy (pressure energy) according to Bernoulli’s principle.7. This conversion results in an increase in fluid pressure at the pump discharge.

The overall result of this process is that the fluid pressure at the pump discharge is much higher than it

was at the pump suction. The pressure changes, which occur as the fluid passes through the centrifugal

pump, are illustrated in Figure 8.13. (The kinetic energy of the fluid entering is virtually equal to the

kinetic energy of the fluid leaving where the inlet and discharge piping are the same size.) The

Mechanical Sciences course will discuss pumps in more detail.

Volute

Casing

Impeller

Vanes

Eye

Suction

Motor-DrivenShaft

Figure 8-58 A Centrifugal Pump




Normally, a

centrifugal

pump produces a

relatively low

pressure

increase in the

fluid. This pressure

increase can

be anywhere

from several

dozen to

several

hundred psidacross a

centrifugal

pump with a

single stage

impeller. The

term PSID (Pounds Force Per Square Inch Differential) is equivalent to DP. In this context, it is the

pressure difference between the suction and discharge of a pump. PSID can also be used to describe a pressure drop across a system component (strainers, filters, heat exchangers, valves, demineralizers, etc.).

When a centrifugal pump is operating at a constant speed, an increase in the system back pressure on the

flowing stream causes a reduction in the magnitude of volumetric flow rate that the centrifugal pump can

maintain.

Positive Displacement Pump Operation

Positive displacement pumps add energy to fluid by a pushing action that applies a direct pressure to the

volume of fluid being pumped. The positive displacement pump moves fluid by pushing it out of the pump due to the mechanical action of the pump. There are two basic types of positive displacement

pumps: reciprocating and rotary, and there are many different versions of each.

The reciprocating positive displacement pump is shown in Figure 8.14. For each cycle, the piston pushes

exactly one cylinder volume of fluid into the discharge line. The valve arrangement of this type of pump

is such that the suction valve is drawn off its seat when the piston moves out of the cylinder drawing

liquid into the cylinder. On the discharge stroke the suction valve shuts and fluid is forced out of the

pump through the discharge valve. The positive displacement pump converts the mechanical energy of

the pump directly to flow energy as the piston applies a pressure to the volume of water in the cylinder.

Pressure Curve

Suction Eye Impeller Volute Piping

Turbulence and gradual

head loss due to fluid friction

Pressure within the System

Figure 8-59 Pressures Within a Centrifugal Pump




The rotary-type positive displacement pump uses gears, vanes or screw-type rotors that rotate to move

fluids. The flow from a rotary pump is fairly steady compared to the pulsating flow of a reciprocating

pump. They are used primarily in oil systems and chemical systems.

Using the General Energy Equation to Analyse Real Fluids

Remember that the General Energy Equation is strictly an attempt to express the “difference” between

measurements of eight kinds of energies. With this information and using arrow analysis a prediction can be made about the “energy changes” within the system between the two points.

The specific energies pictured in the figure below are a mental image of the energy contained in a system

(or the General Energy Equation in mental

form). Those energies inside of the square are

pictured as being contained inside the fluid.

Those energies outside of the square are

pictured as entering from outside or leaving the

fluid.

Because nature will not allow us to destroy nor

create mass or energy by simply pumping a

fluid through a pipe the total energy in a steady

flowing system at an initial point 1 must also be

the total energy at another final point 2. We

can equate everything initial on one side of an

equation to everything final on the other side of

an equation and can create a mathematical expression for the energy in this mental picture of a system.

S uS t

sS t

s s

S uV S uV

O

O

C

C

Figure 8-60 Positive Displacement Pump

Figure 8-61 General Energy Equation in

Mental Form




We thus have created the specific form of the General Energy Equation with the initial side on the left and

the final side on the right of the equation sign:

Why use the specific form? Because in the specific form, mass is divided out of every energy term, so it

allows us to speak generally about what is happening within a system without being tied to any particular

fluid (because of its “mass,” or its “mass flow rate”). This means that the equation we have defined

applies to all fluids and is truly general.

The next several pages contain examples of the energy conversions within fluid systems under a variety

of circumstances:

Training Note for examples which follow:

Canceling unnecessary energy terms helps to simplify diagnosis.

n o t u

r b i n

e s , o

r p a d d l e

w h e

e l s

pe1+ ke

1+ fe

1+ u

1+ q

in+ w

in= pe

2+ ke

2+ fe

2+ u

2+ q

out + w

out

n o

p u m p s o r c

o m p r e s s

o r s

n o h

e a t e x

c h a

n g e r s ( c o

o l e

r s )

n o

p i p

e e l e

v a t i o

n c

h a n g

e s

n o p i p

e s i z

e c h

a n g

e s

n o p r e s s u r e c

h a n g

e s

i f a n i d

e a l f l u i

d

n o h e a t e x

c h a n g e r

s , o r s t e

a m

g e n

e r a t o

r s

0 0 0 0 0 0 00 0 000

These energy terms may bezeroed-out when a specificcircumstance occurs

The energy terms that are tied together by solid lines are paired. This means that they are

equal to each other and cancel so may be eliminated from the equation. Those not tied

together are not paired, or not equal to one another so must be eliminated individually

from the General Energy Equation. This is a mathematical way of treating energies that

in reality are not zero except under very special circumstances. For instance, internal

energy u is hardly every zero, but where there is no change in internal energy the values

on the left and the right side of the equation are identical and cancel each other out, so are




effectively zeroed out. This logic also holds for pe, ke, and fe. If there is no change in

energy from point 1 to point 2 the energy values are the same and cancel.

As a mathematical convention we will therefore indicate that energy values for

paired energy terms in the General Energy Equation will be set to zero when there

is no change.

Example:

In the drawing below, which of the 8 (eight) types of energy are not paired?

qin qout

win wout

u fe

ke pe

Answer:

The inputs and outputs of q and w are not paired. If any of these energies become zero this does

not effect the value or existence of the other. Each one may exist without being effected by the

existence of any other. In other words, q (both in or out) can exist without the other. Also w(both in or out) can exist without the other term.

___________________________________________________________________________




Example:

In the example below, which of the 8 (eight) types of energy represent the energies that are

contained within the fluid itself (and are paired)?

qin qout

win wout

u fe

ke pe

Answer:

pe, ke, fe, and u are energy differences that are contained within the fluid itself (and are paired), while q and w (both in and out) are representative of what the operator does to the fluid to either add energy or remove energy from the fluid from outside of the fluid and are not within the fluid

itself.

___________________________________________________________________________

Example:

In the example below, which of the 8 (eight) types of energy listed toggle back and forth

transferring energy between them, and which are one way transfers only?

qin qout

win wout

u fe

ke pe

Answer:

pe, ke, fe, and u easily transfer energy between them as the fluid moves through the piping, whileq and w (both in and out) are representative of what the operator does to the fluid and either add

energy or remove energy from the fluid one way.

___________________________________________________________________________




Example:

Write the specific form of the general energy equation

Answer:

pe1 + ke1 + fe1 + u1 + qin + win = pe2 + ke2 + fe2 + u2 + qout + wout

___________________________________________________________________________

Example:

Draw a representative figure the 4 types of energy that are contained within a fluid.

Answer:

u fe

ke pe

____________________________________________________________________________

Example:

What are the 4 types of energy that are operator controlled in a fluid system.

Answer:

qin qout

win wout

___________________________________________________________________________

Example:

Write how the specific form of the General Energy Equation would appear when no work is

performed on the fluid or by the fluid, and when no heat enters or exits,

Answer:




0 0 0 0

Not paired with one anotherNot paired with one another

pe1+ ke

1+ fe

1+ u

1+ q

in+ w

in= pe

2+ ke

2+ fe

2+ u

2+ q

out + w

out

pe1+ ke

1+ fe1+ u

1 = pe2+ ke

2+ fe2+ u

2

___________________________________________________________________________

Most “real fluids” act almost like the concept of an ideal fluid, but all process equipment and all piping

causes real fluids to lose a small amount of flow energy which goes into internal energy u (and thus into

unrecoverable losses which continuously remove flow energy from the system). An ideal fluid, on the

other hand, has no change in u. When there is no change in internal energy “u”, both values for u, (at

point one and at point two) are canceled. This describes a “paired” relationship where with no change in

u. The value of u at point 1 cancels with the value of u at point 2.

Specific Rules Using Arrow Analysis

Previously in this text arrow analysis has repeatedly been used without being highly defined. Somespecific rules for arrow analysis will now be introduced to allow for more in-depth use.

• When using arrow analysis with the General Energy Equation, all energies contained initially within

the fluid (paired) will be considered to be constant because they compare point 1 with point 1 andthere is no change. This means they do not change up or down. As a result an arrow will not need to

be used.

• (If an arrow is used it will be a horizontal arrow indicating no change up nor down: ↔ .)

• Those energy terms on the left side that are independent, q in and w in , must be assessed with up or

down arrows, as well as all energy terms on the right side of the General Energy Equation.




Note that the first four energy types will not need an arrow and will be considered to be constant, not

changing with movement in the piping system. All the energy types that have specific conditions listed

must be assessed by up or down arrows if they are not previously eliminated.

• An increase arrow (up arrow) will be used for any specific heat (q) or specific work (w) terms that

are not canceled out. (This represents the effect the heat or work term will have on the total energy of

the fluid, such that any heat or work in will cause energies to go up within the fluid , and any heat or work out will cause energies to go down within the fluid.)

Every arrow is assigned from the position of point 2 (two),

• pe terms can cancel each other out if there is no height change in the system.

• ke terms can cancel each other out if there is no cross-sectional area change in the system.

• fe terms always take the brunt of changes in pe and ke terms and therefore take the opposite arrow

direction. (Pressure takes the hit from height and cross-section area changes.)

(Rarely does fe ever cancel out, but if the pressures at both point one and point two are identical thiscan be canceled.)

• u terms cancel if the fluid is an ideal fluid . In an ideal fluid there is no viscosity and therefore no

fluid friction to heat up as the system changes in height, and cross-sectional area.

Note: Keep in mind that when a heat exchanger is involved, u can change . (However, usually u

is kept constant using an ideal heat exchanger system.)

n o

t u r b i n

e s , o r p a d d l e

w h

e e l s

pe1+ ke

1+ fe

1+ u

1+ q

in+ w

in= pe

2+ ke

2+ fe

2+ u

2+ q

out + w

out

n o p u m p s o r

c o m

p r e s s

o r s

n o

h e a t e x

c h a n g e r

s ( c o

o l e

r s )

n o p i p e e l e

v a t i o

n c

h a n g e s

n o

p i p

e s i z

e c h a

n g e s

n o

p r e s s u r e

c h a

n g e s

i f a n

i d e a l f l u

i d

n o h e

a t e x

c h a n

g e r s ,

o r s t e

a m g

e n e r a t o

r s

0 0 0 0 0 0 00 0 000

These energy terms may be

zeroed-out when a specificcircumstance occurs




• w (both in and out) may be canceled if it is assumed that there is no work on the ideal fluid by the

system nor by the ideal fluid on the system.

Note: Keep in mind that a pump or a compressor does do work on the fluid. If there is a pump or

compression in a system there is w in . The fluid has work done on it by pumps or

compressors.

• q (both in and out) may be canceled if it is assumed that there is no heat passing in nor out of an ideal

fluid. It is assumed that the piping has perfect insulation.

Note: Keep in mind that heat exchangers can both put heat in and take heat out of the system

(e.g., boilers and coolers).

The General Energy Equation and Diagnosis using Arrow Analysis

If we consider that there is an ideal fluid in the piping and changes in the cross-sectional area and theheight take place, the fluid will move faster or slower, and change pressure either down or up. With an

ideal fluid there is considered to be a 100% perfect transfer of energy. No loses occur because internal

energy, u, (both at point one and at point two) is the same for an ideal fluid, so cancel out. Even thoughan ideal fluid does not exist, it almost describes what does take place in a real fluid. It is a good

approximation to reality. When we must be more accurate we recognize that internal energy, u, also

exists and takes energy away from fe as energy is converted back and forth.

If there is a real fluid in the piping the internal energy, u, will always tap a small amount of flow energy

away from the system. So as the energy in a system transfers back and forth (as the piping causes the

fluid to speed up, slow down and change pressure), in a real fluid, the internal energy, u, will always

increase.




Example:

What would be the energy conversions in an “ideal system” (contains an ideal fluid) where no

heat enters nor exits the fluid, and no work is done on or by the fluid, and there is no change in

cross-sectional area (A), but there is an increase in height (z)?

pe1+ fe

1= pe

2+ fe

2

=

0 0

pe1+ ke

1+ fe

1+ u

1+ q

in+ w

inpe

2+ ke

2+ fe

2+ u

2+ q

out + w

out

00 0 0

1

2

0 0

Answer:

• pe1 + fe1 = pe2 ↑ + fe2 potential energy increases (z increase)

• pe1 + fe1 = pe2 ↑ + fe2 ↓ thus flow energy decreases (compensates) since

fe2 ↓ = (P2 υ )↓ means (P2 υ ) ↓ =P2 ↓ υ↔

pressure at point two goes down

(because υ = 1ρ , and density, ρ , must remain constant (unchanged) in an ideal fluid or an

incompressible fluid)

In summary:

z↑ pe↑ A↔

v↔

ke↔

u↔

fe↓ υ↔ P↓




Example:

What would be the energy conversions in an “ideal system” (contains an ideal fluid) where no

heat enters nor exits the fluid, and no work is done on or by the fluid, and there is no change in

cross-sectional area (A), but there is a decrease in height (z)?

+ ke 1 + q in+ w in + ke 2

1 1 2 2 pe + fe = pe + fe

+ q out + w out

0 0 00 0 0

1

2

1 1 1 2 2+ u pe + fe u = pe + fe + 2

00

Answer:

• pe1 + fe1 = pe2 ↓ + fe2 potential energy decreases (z decrease)

• pe1 + fe1 = pe2 ↓ + fe2 ↑ thus flow energy increases (compensates) since

fe2 ↑ = (P2 υ )↑ means (P2 υ )↑ =P2 ↑ pressure at point two goes up

(because υ =1

ρ , and density, ρ , must remain constant (unchanged) in an ideal fluid or an

incompressible fluid)

In summary:

z↓ pe↓ A↔

v↔

ke↔

u↔

fe↑ υ↔ P↑




Example:

What would be the energy conversions in an “ideal system”, where there is no change in height

(z), but there is a cross-sectional area (A) increase?

ke 1+ fe1 = ke 2+ fe2

=

0 0

pe1+ ke

1+ fe

1+ u

1+ q

in+ w

inpe

2+ ke

2+ fe

2+ u

2+ q

out + w

out

00 0 0

1 2

0 0

Answer:

• ke1 + fe1 = ke2 ↓ + fe2 kinetic energy decreases, (because v decreases)

since A↑ therefore v↓

• ke2 =1↔

2gc

× v2

2 ↓ and therefore ke2 ↓ because the velocity at point two

decreases

• ke1 + fe1 = ke2 ↓ + fe2 ↑ thus flow energy increases (compensates)

and since fe2 ↑ = (P2 υ )↑ means• (P2 υ )↑ =P2 ↑ υ

↔pressure at point two goes up

In summary:

z↔

pe↔

A↑ v↓ ke↓

u↔

fe↑ υ↔ P↑




Example:

What would be the energy conversions in an “ideal system”, where there is no change in height,

(z) but there is a cross-sectional area (A) decrease?

ke1+ fe 1 = ke2+ fe 2

=

0 0

pe1+ ke

1+ fe

1+ u

1+ q

in+ w

inpe

2+ ke

2+ fe

2+ u

2+ q

out + w

out

00 0 0

1 2

0 0

Answer:

• ke1 + fe1 = ke2 ↑ + fe2 kinetic energy increases, (because v increase)

since A↓ therefore v↑

• ke2 =1↔

2gc

× v2

2 ↑ and therefore ke2 ↑ because the velocity at point two increases

• ke1 + fe1 = ke2 ↑ + fe2 ↓ thus flow energy decreases (compensates) and since fe2 ↓ = (P2 υ )

↓ means (P2 υ )↓ =P2 ↓ υ↔


In summary:

z↔

pe↔

A↓ v↑ ke↑u↔

fe↓ υ↔ P↓




Example:

What would be the energy conversions in an “ideal system”, where there is an increase in height,

(z) and also a cross-sectional area (A) decrease?

ke1+ fe

1= ke

2+ fe

2

= pe1

+

ke1+ fe

1+ q

in+ w

inpe

2

+

ke2+ fe

2+ u

1+ u

2+ q

out + w

out

00 0 0

1

2

0 0

+ pe1 + pe

2

Answer:

• pe1 + ke1 + fe1 = pe2 ↑ + ke2 + fe2 increase in potential energy (z increase)

• pe1 + ke1 + fe1 = pe2 ↑ + ke2 ↑ + fe2 kinetic energy increase (v decrease)

since A↓ therefore v↑

• ke2 =1↔

2gc

× v2

2 ↑ and therefore ke2 ↑ because the velocity at point two increases

• pe1 + ke1 + fe1 = pe2 ↑ + ke2 ↑ + fe2 ↓ thus flow energy decreases (compensates)

and since fe2 ↓ = (P2 υ )↓ means (P2 υ )↓ =P2 ↓ υ↔


under the influence of two specific energy changes

In summary:

z↑ pe↑A↓ v↑ ke↑

u↔

fe↓υ

↔ P↓




Example:

What would be the energy conversions in an “ideal system”, where there is a decrease in height,

(z) and also a cross-sectional area (A) decrease?

ke1+ fe

1= ke

2+ fe

2

= pe1

+

ke1+ fe

1+ q

in+ w

inpe

2

+

ke2+ fe

2+ u

1+ u

2+ q

out + w

out

00 0 0

1

2

0 0

+ pe1 + pe

2

Answer:

• pe1

+ ke1

+ fe1

= pe2

↓ + ke + fe decrease in potential energy (z decrease)

• pe + ke + fe = pe + ke + fe kinetic energy increase (v increase)

since A therefore v • ke =

and therefore ke because the velocity at point two increases

• pe + ke + fe = pe + ke + fe ? thus flow energy both increases due to the specific

potential energy decrease and decreases due to the specific kinetic energy increase as it

compensates and since fe = (P ) means (P ) =P

• The pressure at point two goes either up or down depending upon the amount of change

of the two specific energies that affect it. (If the changes are equal they will cancel out and the pressure at point two will not change.)

In summary:

z pe

A v ke

fe? P?

This is a case where mathematical calculation is necessary to make further conclusions as to whether feincreases, decreases or remains the same.




Example:

What would be the energy conversions for a centrifugal pump which does work on an ideal fluid

(the fluid does not do work on a pump), where there is no change in height (z), and no change in

cross-sectional area (A), and no heat enters nor exits the fluid? (Assume cross-sectional areas of

entry and exit pipes to be the same, and the height change to be negligible.)

1

2

pe1+ ke

1+ fe

1+ u

1+ q

in+ w

in= pe

2+ ke

2+ fe

2+ u

2+ q

out + w

out

0 0 0 0 00 000

fe1

+ win= fe

2

Answer:

• fe

+ w = fe (Specific flow energy at point one has no change when compared to itself so

receives an arrow indicating “no change”.) The work (in) by the pump increases so when

comparing specific flow energy at point two with specific flow energy at point one there is an

increase by the amount of energy that is put in by exactly the amount of the work performed by

the pump on the fluid. (In other words, the two flow energies would be the same if there were no

work put into the fluid by the pump.)

• fe

+w

= fe

so flow energy at point two must increase, and since fe

= (P

) means (P

) =P

pressure at point two goes up.

In summary:

fe P




Example:

What would be the energy conversions for a pipe with a “real fluid” flowing through it? As a

result there is an internal energy change ( u).

Conditions: There is no change in height, (z), no change in cross-sectional area (A), no heat (q)

enters nor exits the fluid, and no work (w) is performed either on the fluid or by the fluid.

pe1+ ke 1+ fe 1+ u 1+ q in+ w in = pe2+ ke 2+ fe 2+ u 2+ q out + w out

0 0 0 00 00

1 2

0

Answer:

• fe

+u = fe + u

specific internal energy increases

• fe

+u = fe + u

that energy comes from specific flow energy (decreases)

and since fe = (P ) means (P ) =P pressure at point two goes down.

• In effect, given enough distance to travel through the pipe all the flow energy would go into

internal energy.

In summary:

u

fe P




Training Note for exercises which follow:

Any “ideal fluid” can be replaced by a “real fluid” in any exercise and the changes in flow energy

and pressure that occur in addition to what happens due to the “ideal fluid” is effectively this

horizontal pipe exercise. In other words, fe and P will always decrease because u increases in

every “real fluid” system. ___________________________________________________________________________

Example:

What would be the energy conversions that occur with a “real fluid” in a pipe where there is noheight (z) change, but there is a decrease in cross-sectional area (A)? (There is no heat (q) in nor

out, and no work (w) performed on or by the fluid where the container is a pipe. This should be

understood.)

ke1+ fe

1= ke

2+ fe

2

= pe1+ ke

1+ fe

1+ u

1+ q

in+ w

inpe

2+ ke

2+ fe

2+ u

2+ q

out + w

out

00 0 0

1 2

0 0

+ u1

+ u2

Answer:

• since A therefore v • ke =

and therefore ke because the velocity at point two increases

• ke + fe + u = ke + fe + u kinetic energy increases, (because v increases)• ke + fe + u = ke + fe + u thus flow energy decreases (compensates)

and since fe = (P ) means (P ) =P pressure at point two goes down because of a

cross-sectional area increase.

• But that is only part of the solution--

Because the fluid passes with a velocity through the pipe there is turbulence, and

turbulence within a real fluid that has fluid friction creates an increase in internal

energy.

Internal energy then takes that energy from the pressure within the system, i.e. from flow

energy, thus ke + fe + u = ke + fe + u again, flow energy compensates for an increase in

another form of energy and since fe = (P ) means




(P ) =P pressure at point two goes down also because of an internal energy

increase.

In summary:

A v ke

u

fe P

_____________________________________________________________________________

Example:

What would be the energy conversions that occur with a “real fluid” in a pipe where there is a

height (z) decrease, but no change in cross-sectional area (A)? (Where the container is a simple

pipe there is no heat (q) in nor out, and no work (w) performed on or by the fluid for all intents

and purposes. This should be understood by now.)

+ ke1

+ qin+ w

in+ ke

2

1 1 2 2 pe + fe = pe + fe

+ qout

+ wout

0 0 00 0 0

1

2

1 1 1 2 2+ u pe + fe u = pe + fe +

2

1+ uu +

2




Student Guide: Fluid Flow Chapter 8: Energy Conversions in Real Fluid Systems

Answer:

• pe + fe + u = pe + fe + u potential energy decreases (z decrease)

• pe + fe + u = pe + fe + u thus flow energy increases (compensates)

since fe = (P ) means (P ) =P pressure at point two goes up

(because = , and density, , must remain constant (unchanged) in an incompressible fluid)

• But that is not the complete answer because the fluid has internal friction, thus:

pe + fe + u = pe + fe + u internal energy increases; flow energy decreases.

• Thus pressure at point two again goes down. We can not say exactly what happens to pressure.

In summary:

z pe

u

fe? P?

This is another example where mathematical calculations would best be used to describe the final pressure amount and direction.





Chapter 8 Summary:

Fluid friction converts flow energy into internal energy.

Fluid friction causes the internal energy of a fluid to increase which is seen as an increase in system

temperature.

The following factors will cause an increase in fluid friction:

Increase in fluid velocity

Increase in roughness of pipe

Higher viscosity fluids

Smaller diameter pipeIncrease in turbulent flow

Increase in pipe length

Increased numbers of valves, fittings, and bends

The amount of flow energy converted to internal energy by fluid friction is called head loss.

If internal energy increases, temperature increases.

Valves cause a reduction in flow area when the fluid passes by a throttled disk, They control flow by

decreasing pressure at that point. Flow can be totally stopped by decreasing the pressure at the disk to

zero.

Head is another term for fluid pressure. It is the height, in feet, of a column of water that a given pressure

will support. (In other words, we turn it into potential energy.)

Friction head (head loss) = g

u

Pressure head = g

p

g

p

g

fe

ρ

ν ==

Static head is the sum of friction head + velocity head + pressure head.

Total head is the sum of all “elevation heads” (static head + elevation head).

Delivery head is the total head “minus” sum of friction head and velocity head.

Fluid friction is opposition to the movement of a fluid and converts flow energy to internal energy and

raises system temperature.

Fluid friction causes a decrease in flow energy ( fluid pressure) and an increase in internal energy which

is seen as an increase in temperature.

Chapter 8 Summary Continued





Fluid friction is the force which opposes the movement of a fluid

The flow energy converted to internal energy by fluid friction is called head loss.

Head loss is a pressure loss (an energy loss)

If flow energy decreases, pressure decreases and vice versa.

Throttling is the term used to describe the control of fluid flow and pressure by manipulating a valve. It iscommonly used to mean partially closing a valve to restrict flow.

Throttling causes a sustained pressure drop in a system due to increased turbulence and the

resulting fluid friction.

Pumps overcome head losses - they are the opposite of “pressure reducers” - they are “pressure

increasers”.

Pumps convert mechanical energy to flow energy by causing an increase in pressure.

Centrifugal pumps convert mechanical energy to flow energy using centrifugal force and an expanding

volute.

Positive displacement pumps add energy to a fluid by a pushing action that applies a direct pressure to the

volume of fluid being pumped.

Total Static Head (the work the pump does) when lifting fluid up to the pump and pumping higher than

the pump is the sum of Static Suction Lift and Discharge Head.

Static Suction Lift is read as a positive amount when read from a Vacuum Gauge. (This is the proper

way.)

Total Static Head (the work the pump does) when drawing fluid from a tank positioned higher than the

pump is the difference between Static Suction Head and Discharge Head.


and intermolecular attractions.

Internal energy always increases and flow energy always compensates by decreasing. (u ↑ and fe ↓

100% of the time) Internal energy exists in a real fluid where there is always viscosity and fluid

friction due to intermolecular attractions.





Chapter 8 Fluid Flow Measurement

This chapter discusses different types of flow measuring devices their affects to the fluid system.

TO 9.0 EXPLAIN the energy conversions that occur as fluid flows through

the Venturi tube, flow nozzle, and orifice plate flow measuring

devices

EO 9.1 DESCRIBE the basic construction of the following types of

head flow detectors: Venturi tubes, Orifice plates, Flow nozzles

Flow Measuring Devices

Many fluid systems are provided with flow-measuring devices to measure the magnitude of the

volumetric or mass flow rate. Knowledge of flow rate is required to ensure that systems are operating

acceptably.

Differential Pressure Meters

Of the many different types of flow meters, differential pressure meters are most commonly encountered

in industry. They are reliable, relatively inexpensive, and provide suitable accuracy. This type of flowmeter operates on the principle that placing a restriction in the fluid stream causes a pressure drop. (As

velocity increases through a narrow restriction, pressure decreases - Bernoulli's principle.) This

difference in pressure is measured to provide flow rate information. Some common flow measuring

devices are orifice plates, flow nozzles, and venturi tubes.

Orifice Plates

An orifice plate is simply a flat plate 1/16-inch to 1/4 inch thick, with a hole drilled in it. It is one of the

simplest of the flow path restrictions used in flow detection as well as the most economical. One of the

most common types is shown in Figure 9.1





The inlet side decreases the area of the fluid stream at the hole of the plate, causing the velocity to

increase and the pressure to decrease (flow energy is converted to kinetic energy) abruptly. The low

pressure is measured as the fluid exits the outlet side of the orifice plate since the pressure will be at itslowest. The outlet side of the plate immediately increases the area which decreases velocity and increases

pressure (kinetic energy is converted to flow energy) abruptly. The high pressure is measured upstream of

the inlet side of the orifice plate. Because the area changes for an orifice plate are so abrupt as the fluid

passes through the plate penetration, there is much turbulence caused by the orifice, and therefore some

head loss. The pressure increases to about 60% to 80% of the original input value. The pressure loss isirrecoverable; therefore, the output pressure will always be less than the input pressure. The pressure on

both sides of the orifice are measured, resulting in a differential pressure which is proportional to the flow

rate. An orifice plate is easier to install and cheaper than more complex flow measuring devices.

Besides being used to provide flow indication, orifice plates are often used as flow control devices. They

are normally installed in a straight run of smooth pipe to avoid disturbance of flow patterns caused by

fittings and valves.

Orifice plates have two distinct disadvantages; they cause a high permanent pressure drop (outlet

pressure will be 60% to 80% of inlet pressure) and they are subject to erosion, which will eventually

cause inaccuracies in the measured differential pressure.

Figure 9-62 A Simple Orifice Plate





Flow Nozzles

The flow nozzle (Figure 9-2) is commonly used for the measurement of steam flow and other high

velocity fluid flow measurements where erosion may be a problem. It is capable of measuring

approximately 60% higher flow rates than an orifice plate with the same diameter. This is due to the

streamlined contour of the throat, which is a distinct advantage for the measurement of high velocity

fluids. The flow nozzle requires less straight run piping than an orifice plate. However, the pressure drop

(head loss) is about the same for both devices.

Venturi Tubes

A venturi tube is also used to measure fluid flow. It is the most accurate flow-sensing element when

properly calibrated. A basic venturi tube is shown in Figure 9.3. It has a converging conical inlet, a

cylindrical throat, and a diverging recovery cone. It has no projections into the fluid, no sharp corners,

and no sudden changes in contour.

The inlet section decreases the area of the fluid stream, causing the velocity to increase and the pressure

to decrease (flow energy is converted to kinetic energy). The low pressure is measured in the center of

the cylindrical throat since the pressure will be at its lowest value there and neither the pressure nor the

velocity is changing. The diverging cone has a steadily increasing area which decreases velocity and

increases pressure (kinetic energy is converted to flow energy) such that the total pressure loss is only

10% to 25% due to fluid friction (head loss). The high pressure is measured upstream of the entrancecone. The major disadvantage of this type of flow detection are the high initial costs for installation and

difficulty in installation and inspection.

Figure 9-63 A Simple Flow Nozzle





A venturi is designed to minimize

turbulence and thus minimize head loss

and permanent pressure drop.

The mass flow rate through the venturi

is determined by measuring thedifferential pressure it produces. The

pressure difference is measured between

the inlet to the venturi tube and the

throat.

The square root of the pressure drop is

directly proportional to the mass flow

rate.

Pconstant)( ∆=

⋅

m

Knowing that mass flow rate is proportional to the square root of the differential pressure allows a meter

constant to be used so that a particular flow rate can be related to a particular instrument.

Other Applications of the Venturi

Principle

A Venturi is basically a tube with a narrow

constriction (the throat). It has other

applications than flow measurement.

A carburetor:

The barrel of a carburetor in a car (Figure

9.4) is an example of another common use

of the venturi principle. The flowing air

speeds up as it passes the constriction and

so the pressure is lower. Because of the

reduced pressure, gasoline under atmospheric pressure in the carburetor

reservoir is forced into the air stream and mixes with the air before entering the cylinders. (Giancoli,

252).

Steam Jets

One common method for transferring solutions between vessels in industry is to “jet” the solution with

steam jets (Figure 9-5). Steam jets require no moving parts; thus, they are reliable and need little

maintenance.

Steam, travels through piping to the steam jet nozzle located in piping near the vessel. The steam is

forced through a small opening, the nozzle portion of the jet, which increases the velocity of the steam.

Figure 9-64 Simple Venturi Tube

Figure 9-65 Auto Carburetor Uses Venturi

Principle





As the steam exits the nozzle portion, it crosses over an opening in the jet. Connected to the opening is a

dip leg leading to the process vessel. The force of the steam crossing the opening pulls a vacuum (creates

low pressure because of Bernoulli's Principle) on the process vessel dip leg and solution is drawn up

through the line from the vessel. Steam and process solution exit the jet together.

Eductors

An eductor is a type of pump or jet pump which uses the principle of reduced flow area. Figure 9.6 (next page) shows a basic eductor. During operation, the eductor has two types of fluids: a high velocity fluid,

which flows through the nozzle into the throat of the diffuser, and a fluid being pumped. The fluid being

pumped flows around the nozzle into the throat of the diffuser. As the high velocity fluid enters the

diffuser its molecules strike molecules of the low velocity fluid. The struck molecules are carried alongwith the high velocity fluid. This process is called entrainment. As the entrained molecules are carried

out of the diffuser, a low pressure area is created around the mouth of the nozzle. This low pressure area

will draw more fluid from around the nozzle into the throat of the diffuser, where it is also entrained and

continues the process.

A common example of an eductor is the chemical spray bottles that connect to the end of a garden hose.

The flow of water from the hose is directed through an eductor. The water pulls chemicals from the

bottle, mixing them with the bulk of the water flow.

Figure 9-66 A Typical Steam Jet





Figure 9-67 A Simple Eductor



OPERATIONS TRAINING PROGRAM Student Guide: Fluid Flow Chapter 9: Fluid Flow Measurement

Chapter 9 Summary:

Venturi tubes, orifice plates, and flow nozzles operate on the principle that placing a restriction in a fluid

stream will cause a pressure drop. These devices measure flow rates by detecting changes in pressure.

An orifice plate is simply a flat plate 1/16-inch to 1/4 inch thick, with a hole drilled in it.

The flow nozzle is has a streamlined contour of the throat and immediately opens to the original cross-

sectional area of the pipe

The Venturi has a converging conical inlet, a cylindrical throat, and a diverging recovery cone. It has no

projections into the fluid, no sharp corners, and no sudden changes in contour.

The mass flow rate through a venturi tube is directly proportional to the square root of thedifference in pressure between the venturi tube inlet (the largest opening) and the venturi tube throat (the

constriction).




Student Guide: Fluid Flow Chapter 9: Fluid Flow Measurement

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OPERATIONS TRAINING PROGRAM Student Guide: Fluid Flow Chapter 10: Water Hammer, Steam Hammer, and Pipe

Whip

Water Hammer and Pipe Whip

The chapter identifies some of the causes and methods of minimizing water hammer (and steam hammer)

and pipe whip

TO 10.0 IDENTIFY the conditions and prevention methods for both "water

hammer" and "pipe whip" in fluid systems

EO 10.1 DESCRIBE the causes of water hammer

EO 10.2 DESCRIBE the methods used to prevent water hammer.

EO 10.3 DESCRIBE the causes of pipe whip

EO 10.4 DESCRIBE the method used to prevent pipe whip

Mechanisms of Water Hammer

Water hammer is a liquid shock wave resulting from the sudden starting or stopping of flow. It is

affected by the initial system pressure, the density of the fluid, the speed of sound in the fluid, the

elasticity of the fluid and pipe, the change in velocity of the fluid, the diameter and thickness of the pipe,

and the valve operating time.

Water hammer is a physical shock to the components in a system caused by impact of high velocity

liquids. These liquids are dense and incompressible. This is an hydraulic overpressurization.

When slugs (pockets or accumulations) of liquid are driven, or drawn by vacuum through a pipe line

these slugs of heavy fluid may move at car or truck speeds. (Slugs of liquid weighing hundreds or

thousands of pounds may slam into valves, bends, T’s, or other equipment. Liquid can slam into various

equipment breaking pipeline supports, straightening pipe bends, moving pumps out of alignment,

crushing adjacent equipment, and stressing and breaking valves, joints, and unions, while maiming and

killing workers.)

Occurrence of Water Hammer (and Steam Hammer)

Water hammer can occur when fluid flow through a pipe section is abruptly changed. This can occur

during rapid valve operations in fluids systems. This also can occur local areas of the fluid in the pipe

section undergoes a phase change. In the case of steam, different areas (pockets) of steam cool below it

saturation temperature and condenses to water. The water “slug” is pushed at high speed further down

the piping system.




Whip

The shock wave caused by water hammer can be of sufficient magnitude to cause physical damage to

piping, equipment, and personnel. Water hammer in pipes has been known to pull pipe supports from

their mounts and rupture piping,

Normally, the entire hammer process takes place in under one second!

There are various industrial processes that utilize steam. Because steam is basically water,

(H2O), in its gaseous phase, a form of water hammer should be addressed. Steam hammer issimilar to water hammer except it is for a steam system. Steam hammer is a gaseous shock wave

resulting from the sudden starting or stopping of flow. Steam hammer is not as severe as water

hammer for three reasons:

1. The compressibility of the steam dampens the shock wave

2. The speed of sound in steam is approximately one third the speed of sound in water.

3. The density of steam is approximately 1600 times less than that of water. The items

of concern that deal with steam piping are thermal shock and water slugs (i.e.,condensation in the steam system) as a result of improper warm up.

Many conditions can lead into water hammer or steam hammer in fluid systems. Listed below are

several such conditions:

(This listing is not all inclusive)

• Quickly closing a valve to stop a mass of fluid from flowing will cause pressure waves to

be transmitted through the system causing mechanical shock. During the closing of a valve,

kinetic energy of the moving fluid is converted into potential energy. Elasticity of the fluid

and pipe wall produces a wave of positive pressure (wave) back toward the fluid’s source.

When this wave reaches the source, the mass of fluid will be at rest, but under tremendous

pressure. The compressed liquid and stretched pipe walls will now start to release the liquid

in the pipe back to the source and return to the static pressure of the source. This release of

energy will form another pressure wave back to the valve. When this shockwave reaches the

valve, due to the momentum of the fluid, the pipe wall will begin to contract. The

momentum of a fluid abruptly stopped must be absorbed!.(See Figure 10-1 Case 1: Valve Quickly Closed)

• Quickly opening a valve to admit a fluid under pressure to enter a virtually empty line.(See Figure 10-2 Case 2: Valve Quickly Opened )




Whip

• Cold condensate sitting in a steam line will cause the steam to collapse and create a void

into which a mass of fluid will be drawn via vacuum to high speed movement toward bends,

T’s, valves, equipment, etc.

(See Figure 10-3 Case 3: Cold Condensate in Steam Line)

• Hot condensate sitting in a steam line will be blown by the steam into collisions with

bends, T’s, valves, equipment, etc. A confined mass of incompressible liquid moving at high

speeds will create a high hydraulic pressure when it strikes bends, T’s, valves, equipment,

like being hit with a bowling ball at high speed.

(See Figure 10-4 Case 4: Hot Condensate in Steam Line)

• Boiling of a liquid to create steam in a pipeline will result in an eventual collapse of the

steam as temperatures and pressures change. The collapse will cause a void to be created into

which a mass of fluid will be drawn via vacuum to high speed movement.

Increasing temperature, decreasing pressure, or decreasing the amount of liquid in the

piping (which effectively decreases pressure)

(See Figure 10-5 Case 5: Boiling of a Fluid in a Liquid System Line)

--The following pages illustrates each of the above five cases:--




Whip

Figure 10-68 Case 1 Valve Quickly Closed

Case 2: Conditions: Non-flowing Liquid underhead pressure from pump, tank etc.

(Fluid may or may not beflowing in this line)

Empty Line

Valve is opened quickly

Heavy liquid weighing 62.4 lbs

per cubic foot rushes forward

With no liquid or pressure tohold it back it increases velocity

At vehicle speeds a slug of liquidhits the pipe bend and heads towardother bends and equipment

W h a m !

At what amounts to both vehicle speeds and vehicle weights it is notunusual for pipes to be bent, or ripped from their mounts whenvalves are opened too quickly.

Figure 10-69 Case 2 Valve Quickly Opened




Whip

Figure 10-70 Case 3: Cold Condensate in Steam Line

Figure 10-71 Case 4: Hot Condensate in Steam Line




Whip

Case 5: Boiling - Introducing a gas into a liquid system

Water hammer can also result if high temperature systems are suddenly depressurized, or quickly

heated allowing portions of the liquid to flash to steam. This causes gas bubbles to separate the

continuous stream of liquid into individual slugs that can collide with elbows and projections in the piping. Rapid boiling may drive system flow backward or rapidly forward and may allow liquids to

move more freely within the piping. A collapse of the gas bubbles can occur causing a “steam void

collapse” water hammer.

Gas Bubbles created in liquids by decreased pressureor by increased temperature.

When a cooler liquid comes in contact the gas is collapsed and"vacuum draws" the incompressible liquid into piping components.

These gas bubbles

accumulate in high areas

Void Collapse

Cooler Liquid

Wham!

Warmer Liquid

Warm LiquidWarm Liquid

Figure 10-72 Case 5: Boiling




Whip

Cavitation

Cavitation is a special isolated form of boiling water hammer which causes enormous amounts of

erosion, and is created by myriad’s of gas bubble implosions. A gas is created within a liquid at a low- pressure area and the bubbles created are imploded at a higher pressure region. This usually is isolated

within a small area, but is very destructive.

• Cavitation is another form of induced boiling in a system, usually within a pump

Due to: Decreased Pressure, Increased Temperature, etc.

(Note: The Mechanical Sciences fundamentals course treats cavitation from the point of

view of the pump and associated equipment. This course treats cavitation from the point of

view of the fluid and operator actions on a fluid.)

Cavitation is caused by the formation of vapor bubbles in low pressure

areas and their subsequent collapse in higher pressure regions.

Cavitation occurs when the local pressure in an elbow, pipe, or pump falls below the pressure at which a

liquid will begin to boil/vaporize for the existing temperature. Usually this lowest pressure area of a

system is at the suction eye of the pump. Cavitation can cause severe damage to pumps, pipes, and other

equipment and causes a reduction in flow rate.

Cavitation is based on the principle that the lower the pressure applied to a fluid, the lower the boiling

point. For example, at sea level where the atmospheric pressure is 14.7 psi, water will boil at 212 degrees

F. However, at 5000 feet above sea level, where atmospheric pressure is approximately 13.5 psi, water

will boil at 208 degrees F

Cavitation in Centrifugal Pumps

When a liquid being pumped enters the eye of a centrifugal pump, the pressure is significantly reduced.

If the pressure drop is great enough, or if the temperature of the liquid is high enough, the pressure drop

may be sufficient to cause the liquid to boil/vaporize. When this occurs, the pump is said to be cavitating.

When it occurs, the boiling is localized and continues as the liquid and bubbles leave the suction eye. The

bubbles and the liquid travel across the impeller vanes and are subjected to increasing pressure, whichcauses the bubbles to collapse violently. (See Figure 10.6) .

Pump cavitation is recognized by excessive noise and vibration. (Sounds like pumping rocks.)




Whip

The violent implosions of the bubbles can cause substantial damage to the impeller, the casing, and other

internal pump parts. Even small amounts of cavitation can cause significant damage to a pump over an

extended period of time. In addition, cavitation can cause a significant loss of pump performance,

because the boiling action prevents fluid from being pumped to the system.

The mechanical damage that cavitation can cause is attributed to the peculiar way in which the bubbles

collapse. As a bubble progresses through the collapsing stage, its shape changes. From nearly spherical

it becomes flattened on the upstream surface, and before reaching critical size where it bursts, it assumes

a shape resembling a hamburger patty. When the bubble implodes it appears to form a cone with a tiny

central jet into which the entire energy of the bubble is concentrated. (See figure 10.7.) This is similar in

action to a “shaped charge” which creates an extremely high pressure over a very small area whenever the jet hits a solid surface. The repeated blows of these collapsing bubbles will destroy any known

engineering material. (Driskell, 140)

Figure 10-73 Cavitation in a Centrifugal Pump




Whip

To prevent cavitation , the pressure conditions at the pump suction head should be closely controlled.

In most cases the fluid must enter the pump suction eye under positive pressure to prevent cavitation. At

this point, a condition called Net Positive Suction Head becomes very important.

Net Positive Suction Head (NPSH)

The pressure at the pump suction eye must be greater than the pressure at which cavitation will occur for

the temperature of the fluid being pumped. NPSH is the minimum positive suction pressure requiredto prevent cavitation. The required NPSH is determined by testing and specified by the pump

manufacturer. Suction pressures greater than the required NPSH provide a safety margin against

cavitation.

Many factors are involved with the NPSH. Some of the more important include:

The positive pressure being applied due to the height of the fluid above the pump suction eye.

For example, storage tanks which supply the fluid to the pump are normally located high above

the pump. Another source of positive pressure being applied to the fluid would be due to factors

such as atmospheric pressure or the supply tank being pressurized with nitrogen or steam.

A negative or loss of pressure to the pump would be due to fluid friction in the piping. Most plant systems are designed so that the layout of the pump suction piping results in a minimum

loss of pressure due to short piping length.

As covered earlier, we know that for a given fluid at a certain temperature, the fluid will boil

(vaporize) sooner at a lower pressure than at a higher pressure. If the temperature of the fluid

being supplied to the pump increases beyond normal limits or if pressure drops, then cavitation is

more likely to occur.

Compressed aspressure increasesFormed in Low

Pressure

FinalImplosion

FocusedExplosion

B A N G

! !

COLLAPSING BUBBLE

Figure 10-74 Cavitation and the Collapsing Bubble




Whip

Conditions Causing Cavitation

Pump Runout.

If the discharge piping pressure is far below the pump's design pressure, the pump will drive thefluid above its maximum flow rate. This is a condition known as pump runout. Figure 10.8

shows a system experiencing pump runout.

Pump runout can occur due to misaligned system valves or pipe ruptures which greatly reduce the

pressure of the system. As the pump flow rate increases, the pressure at the suction eye of the

pump decreases. If the pressure at the suction eye goes below the required pressure (NPSH req.),

cavitation will occur.

Figure 10-75 Pump Runout




Student Guide: Fluid Flow Chapter 10: Water Hammer, Steam Hammer, and PipeWhip

Low Suction Line Pressure

The pressure existing at the pump suction is due to the height of the liquid above the pumpsuction and any applied pressure to the system (such as a pressurized expansion tank or a booster

pump). If the level of the liquid decreases or the applied pressure decreases, the pressure at the pump suction eye is reduced. If the pressure at the pump suction eye goes below the required

NPSH, cavitation will result. Figure !0.9 shows a pump cavitating as the result of low level in thesupply tank.

Pressure below the minimum required could occur also if a valve in the suction line is throttled

(partially closed) or if the pumping capacity of a pump exceeds the amount of fluid being

supplied to the pump (which "starves" the pump).

Higher Than Normal Temperature.

If the temperature of the liquid being pumped is higher than normal, the fluid could boil/vaporize

when it enters the suction eye. This could occur if the cooling water to a heat exchanger is lost or

other conditions occur that cause system fluid temperatures to rise.

Figure 10-76 Low Suction Pressure





Minimizing Gas Formation in Liquid Piping Systems

To minimize gas formation in liquid piping systems the system pressure, temperature, and volume must

be maintained such that saturation conditions are not present anywhere in the piping system.

Saturation conditions in a steam system are present when a heated liquid can flash to steam. The

saturated condition can result in steam flashing when any of the conditions occur:

• a decrease in pressure

• an increase in temperature

• a decrease in liquid amount (decreases pressure)

• or any combination of these

There are indications that can possible cavitation:

• Sounds like rocks being pumped through the system.• Fluctuating pressures, motor currents, flow rates, and

• Excessive equipment vibration

To minimize / prevent cavitation:

• Ensure system lineups are correct to prevent excessive flow rates.

• Ensure supply tanks and expansion tanks are at correct levels.

• Ensure pressures applied to expansion tanks are at correct values.

• Ensure proper suction supply and valve lineups.

• Maintain temperatures within equipment design specifications.

Other Pump Problems

In our discussion of fluid flow, it is important to discuss different conditions that can affect pump

operations. Recognizing and understanding various problems that can occur while operating pumps can

reduce troubleshooting times and even aid in the diagnosis and correction of those problems prior to

component damage and process delays.

Gas Binding

Gas binding is the buildup of gas in a centrifugal pump, resulting in a loss of pump performance.It occurs when the pump suction and/or casing accumulates enough gasses to prevent fluid from

leaving the pump decreasing flow rate, and thereby decreasing overall system performance. In

addition, because the fluid usually lubricates and cools the pump's internal parts, fluid loss canresult in serious pump damage due to overheating.





Symptoms of gas binding include:

Fluctuation and sudden drop-off of discharge pressure; sudden or gradual

silencing of normal pump noise; or a failure of discharge pressure to increase at

pump start-up.

Corrective action to prevent gas binding includes:

Venting and priming the pump prior to start-up or, if air binding is suspected,

carefully venting the pump casing. In addition, the equipment should

periodically be checked for air leaks.

Possible sources of air leaks include

Leaks on the suction side of the pump in areas such as piping joints, flanges, and

gauge lines; casing leaks; leaks into the stuffing box; and inadequate filling

and/or venting of the pump after maintenance.

Shutoff head

Shutoff head or dead head is a condition in which a centrifugal pump discharges into a pressureequal to or greater than its maximum discharge pressure. This means that there is no flow from

the pump.

This can be caused by any condition which opposes or prevents flow from the pump, such as a

shut or throttled pump discharge valve, changed system loads, or an improper valve line-up

downstream in the system.

Because of their design, centrifugal pumps can operate for short periods of time at a shutoff headcondition. Long periods of operation, however, result in the fluid temperature increasing to the

point where boiling and/or pump damage will occur. Pumps required to operate for long periods

of time in this condition usually have a recirculation line installed to maintain pump flow .

When used, recirculation lines maintain continuous fluid flow, even when a shut-off head

condition exists. However, even when a recirculating line is present, any unnecessary shutoff

head conditions should be minimized.

Possible Results of Water Hammer

Prevention Rule of Thumb for All Types of Water Hammer:

Never allow a liquid into a gas filled system!

Never allow a gas into a liquid filled system!





Pipeline Failure

Many things may cause pipeline failures other than water hammer, but water hammer may cause massive pipeline failure. Two types of pipeline failure are defined below:

Case A -Pipe Rocket or Pipe Jet

Liquid is driven under pressure at high velocity through a pipe break causing the pipe to break its

supports and jet in the opposite direction (Figure 10-10). A jet force equal and opposite to a

water cannon force is created as a mass of heavy liquid exits from the pipeline. The pipe is

initially held in place by pipe supports, but as the jet takes effect, the pipe support system may

fail and allow the pipe to rocket into and through walls. (Recall Newton’s Third Law from

physics.)

Hot condensate can heavily damage equipment by direct force and scald workers. Steam canquickly fill work areas and once the slug of liquid has been discharged the work area may be

exposed to live steam exiting from the pipeline.

Case B - Pipe Whip

Liquid is driven under pressure at high velocity through an unsecured pipeline and, like the end of

a garden hose, whips about. (Figure 10-11). When a pipeline ruptures at a weak point, like the

loose end of a garden hose suddenly filled with water, the pipeline may whip free of supports and jet about the facility crushing everything in its path. Ruptures may be caused by a water hammer,

thermal stress, or other physical shock to the piping system.

Pipeline Failure Case A: Conditions: Quantity of high temperature liquid isforced under pressure at high velocity to exit from the end of a pipeline.(Water Cannon and Pipe Jet)

Hot Condensate

SteamPipe JetForce

Water CannonForce

Figure 10-77 Pipe Rocket / Pipe Jet





Prevention is the same as prevention of water (and steam) hammer, but also includes

“properly supporting” and securing of piping systems. But what is “proper support”? Most

individuals think that welding a pipeline to the building structure is “proper support”. This is

exactly wrong!! A rigid system will be far more likely to break under impact than a system that

will allow for movement. Earthquake construction is what is needed here. A building that cansway without breaking may ride out an earthquake while a rigid structure will often break.

Seismic supports, and expansion joints that allow movement can make the difference between

containment and rupture.

Figure 10-78 Pipe Whip





Methods of Water (and Steam) Hammer/ Pipe Jet & Pipe Whip Prevention

Proper filling and venting of liquid filled systems:

All air must be removed

• Water hammer can occur if a closed loop system is improperly filled and vented

allowing air to remain in the system. Normally, the pressure exerted on water discharged

from a pump is transmitted equally throughout the piping system (disregarding head loss), but

with air introduced liquids are free to move and be driven or drawn toward equipment at highspeed. If a pump is started on a partially filled system, the water leaving the pump will

undergo little resistance to flow until it slams into a bend or component in the pipe.

Proper valve operation:

Valves must be operated slowly

• Open valves slowly by “cracking” open the seat just enough open the seal between the

seat and valve housing, wait for pressure to equalize, and then open the valve slowly.

• Use by-pass or equalizing valves as applicable to equalize pressures where water hammer

is a potential.

• Close valves slowly and deliberately, avoiding rapid closure that could send an impulsesurge (wave) of the pressurized fluid from the closed valve towards the pressure source.

Liquid removal for gas filled systems

All condensation must be removed

• Water hammer will occur in steam flow systems if water is allowed to accumulate in

the piping. This is especially a problem when steam lines are warmed too quickly or if steam

traps are clogged and fail to remove all condensation.

Pipelines are to be filled with either gas or liquid

(never both) -- or are empty.





Piping systems are to be supported and strongly secured to building framework using shock absorbing

pipe hanger systems.

• Use of expanding piping, expansion joints and seismic hangers may make the difference between total disaster and an expensive clean up.





Chapter 10 Summary

Water hammer is a physical shock to mechanical components of a system.

Water hammer may be created by the following:

• Quickly closing a valve

• Quickly opening a valve

• Cold condensate sitting in a steam line

• Hot condensate sitting in a steam line

• Boiling of a liquid to create steam in a pipeline

via:

• Increasing Temperature

• Decreasing Pressure

• Decreasing Amount of Liquid in container (which effectively decreases pressure - hot liquid flashes

to steam)

Water hammer may be prevented by:

• Removing all gas from liquid systems - Venting

• Removing all liquid from gas systems - Draining

• Operating valves slowly, allowing time for pressure to equalize• Using bypass valves and equalizing valves

• Properly aligning steam traps

• Ensuring correct pressure & temperature are maintained so that boiling point is not achieved (At

saturation temperatures and pressures condensing to liquid or flashing to steam takes place.)

• Operating boilers within limits

• Maintaining quality water chemistry program and adhering to controls

Pipe whip is caused by an broken pipeline, or any opening in a pipe containing a fluid under pressure.

This opening decreases the pressure at one point in the system sufficient to cause the pipe to move theopposite direction from the opening, like a full balloon being released from the hand. (Pipe whip is one

possible result of a water hammer.)

Prevention or decreasing damage due to pipe jet & pipe whip may be effected by securing piping systems

to building framework by using (movable) shock absorbing pipe hangers and expansion joints at

vulnerable positions. Rigid systems are to be avoided.





Chapter 9 Unintended Siphoning

This chapter discusses unintended siphoning and system cross-connections in fluid systems

TO 11.0 IDENTIFY the conditions and prevention methods of a fluid siphon

for a fluid system

EO 11.1 STATE the operating cause of a siphon and the required

conditions for its existence

EO 11.2 DESCRIBE how to prevent siphoning in a fluid system

Introduction

“Unintended” siphoning and cross connections are always surprises and are therefore always unwelcome.

However, during the course of following good operations practices it should be noted that siphons may

intentionally be created, or systems may be created to be permanently cross connected. In some instances

the expense of creating non-cross connected systems is too high to be feasible. An example of such a

system is the ventilation systems of nuclear facilities. Maintaining a one-way flow is a method of

controlling a ventilation system. Recognition of the potential dangers of a permanently cross-connected

system can help to keep incidents from occurring. (Note: Mechanical Sciences will treat one way flow

methods in greater depth.)

Siphoning

A dictionary definition for a siphon describes it as "a bent tube through which a fluid can be drawn over

the edge of one container into a lower container by means of air pressure.” The tube must be filled by

suction, (a lower pressure area than that of the liquid), before flow will start. Siphon action takes place

because of a pressure difference between one end of a tube or pipe and the other. Another simple reason

often given for siphon action is that a liquid always seeks the lowest level.

Siphon action can be compared to the sliding of a chain draped over a smooth rod. (See Figure 11-1). If

the chain is draped so that its midpoint is over the center of the rod and both ends hang at equal length it

does not slide. But if one end hangs lower than the other end, the longer and heavier end falls and pulls

the shorter end upward and over the rod.

The pressure environment to which each tube end is exposed pushes upward on both ends of the tube

(pipe) causing the liquid column to become a cohesive unit. When the pressure environments at each

tube end differs from one another, one becomes the high pressure “push” with regards to the other and the

liquid in the tube is pushed out. If / when the liquid in the tube is voided, a vacuum (suction) occurs and

pressure environment would immediately push more liquid up the tube to fill in the void; thus flow

occurs.

The path may be irregular, and the “tube” can be a run of piping or hose, but almost any air tight pathway

between containers has the potential of becoming an unintended siphon.





The operating cause of a syphon is:

A difference in pressure between the two ends of the tube or pipe.

Conditions for a siphon:

A siphon needs more than a piping connection between reservoirs to function. All the following

conditions must exist at the same time for a siphon to occur:

• Height of liquid level is higher in sending container than receiving container

• Tube or pipe must be airtight. (no vent)

• Tube or pipe must be open throughout its length (a leak may be detected if a

siphon occurs)

• Receiving reservoir must have space available or be open to the atmosphere (i.e. be vented). Empty piping can also act as a receiving reservoir.

With all of these conditions in place a siphon may be started unintentionally by simply opening a

valve or turning off a pump. Containers will continue to fill or empty and liquid levels will rise or

fall seemingly without explanation.

14.7 psi

1 psi

2 psi

14.7 psi(13.7 psi total)

(12.7 psi total)Lower Pressure Area

L i q u id P r e s su r e

At m o sp h e r i c P r e s s u r e

Fi ure 11-79 Exam le of a Si hon





How to prevent a siphon from being created:

To stop unintended siphoning the following are some general guidelines:

• An effective air gap allows liquid flow in only one direction and prevents siphoning.• Maintain the valves in a piping run so they do not leak.

• Open and close vents at the proper time.

When there is a physical pathway between two containers (vessels) of fluids, and the conditions allow

the fluid to flow in a reverse direction, a backflow of the fluid can result. The types of backflow that can

occur are:

Backsiphonage - occurs when the pressure in supply piping is decreased below the pressure in

receiving piping.

Backpressure - occurs when the process system pressure is increased above the pressure in the

supply system.

Backflow can occur through existing piping or through temporary hoses.

Here are some examples:

• Corrosion of a tube wall in a heat exchanger.

• Leaking valves

• Temporary hose connection. Temporary measures often become permanent and

can bypass safety devices.

• Process liquid in a container in contact with delivery nozzles, faucets etc.

• Opening or closing a valve at the wrong time.

Two types of devices to prevent siphons are as follows:

Approved air gap separation. (Six inches or more of air space above the highest possible fluid

level.) This is recommended in extremely hazardous situations and is considered the best way to

protect against creation of unintended cross connections. Faucets should be placed high enough

that any accumulated liquid in a sink will not reach up to the faucet to create a cross connection.

This air gap keeps process water from flowing backwards and from coming in contact with freshwater supplies.

Mechanical devices:

Reduced Pressure Principle Backflow Device (RPBD)

Double Check Valve Assembly (DCVA)Atmospheric Vacuum Breaker (AVB)

Pressure Vacuum Breaker (PVB)




Student Guide: Fluid Flow Chapter 11: Unintended Siphoning

Chapter 11 Summary

Siphoning takes place because water seeks a lower potential energy level.

When the conditions are set up which will allow it to flow downhill it will move uphill to get there.

The operating cause of a siphon is: Difference in pressure

The conditions are:

Height of liquid level is higher in sending container than receiving container

Tube or pipe must be air tight

Tube or pipe must be open throughout its length

Receiving container must have space available or be open to the atmosphere

Siphons can not occur if fluid to fluid contact is not possible.

An air gap of six (6) inches or more air space above the highest possible fluid level is one method of

controlling siphons.

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