Pump System Analysis and Centrifugal Pump Sizing

by J. Chaurette, engineer

PUMP SYSTEM ANALYSIS AND SIZING

BY JACQUES CHAURETTE p. eng.

5th Edition February 2003 Published by Fluide Design Inc. www.fluidedesign.com  Copyright 1994

I

TABLE OF CONTENTS Foreword Introduction Symbols Chapter 1 - An Introduction to pump systems 1.0 Hydrostatic pressure and fluid column height ................................................... 1.1 1.1 The three forms of energy ................................................................................. 1.2 1.2 The relationship between elevation, pressure and velocity in a fluid ................. 1.4 1.3 The difference between pressure and head ...................................................... 1.7 1.4 Fluid systems ................................................................................................... 1.8 1.5 The driving force of the fluid system ................................................................. 1.9 1.6 The components of Total Head ........................................................................ 1.10 1.7 Negative (relative) pressure ................................................................................ 1.13 1.8 The siphon effect ................................................................................................ 1.17 1.9 Specific gravity .................................................................................................. 1.22 Chapter 2 - The application of thermodynamics to pump systems 2.0 Energy and thermodynamic properties .............................................................. 2.1 2.1 Closed systems and internal energy .................................................................. 2.2 2.2 Closed systems, internal energy and work ......................................................... 2.3 2.3 Open systems and enthalpy .............................................................................. 2.4 2.4 Open systems, enthalpy, kinetic and potential energy ....................................... 2.5 2.5 Work done by the pump .................................................................................... 2.6 2.6 Fluid and equipment friction loss ........................................................................ 2.6 2.7 The control volume ............................................................................................ 2.7 2.8 The determination of Total Head from the energy balance................................. 2.9 2.9 System or Total Head equation for a single inlet-single outlet system ............... 2.9 Example 2.1-Calculate the Total Head for a typical pumping system ................. 2.13 2.10 Method for determining the pressure head at any location .............................. 2.18 Example 2.2 Calculate the pressure head at the inlet of the control valve ...... 2.21 2.11 System or Total Head equation for a single inlet-double outlet system ............ 2.26 2.12 General method for determining Total Head in a system with multiple inlets and outlets .................................................................................. 2.29 2.13 General method for determining Total Head in a system with multiple pumps, inlets and outlets ..................................................................... 2.31 2.14 General method for determining the pressure head anywhere in a system with multiple pumps, inlets and outlets............................................................. 2.34

II

Chapter 3 - The Components of Total Head 3.0 The Components of Total Head ......................................................................... 3.1 3.l Total Static Head (∆HTS) ...................................................................................... 3.1 3.2 Suction Static Head (∆HSS) ................................................................................. 3.2 3.3 Net Positive Suction Head Available (N.P.S.H.A.) .............................................. 3.3 Example 3.1-Calculate the Net Positive Suction Head Available ...................... 3.12 3.4 Pump intake suction submergence..................................................................... 3.13 3.5 Discharge Static Head (∆HDS) ............................................................................. 3.16 Example 3.2 – Calculate the Suction & Discharge Static Head .......................... 3.17 3.6 Velocity Head Difference (∆HV)........................................................................... 3.18 3.7 Equipment Pressure Head Difference (∆HEQ) ..................................................... 3.18 3.8 Pipe Friction Head Difference for newtonian fluids (∆HFP) .................................. 3.20 3.9 Fitting Friction Head Difference for newtonian fluids, K method and 2K method (∆HFF).................................................................................................... 3.23 3.10 Pipe Friction Head Difference for wood fiber suspensions (∆HFP) .................... 3.28 Chapter 4 - Pump Selection, Sizing, Interpretation of Performance Curves 4.0 Pump classes .................................................................................................... 4.1 4.1 Coverage chart for centrifugal pumps ................................................................ 4.2 4.2 Performance curve chart ................................................................................... 4.2 4.3 Impeller diameter selection ................................................................................ 4.5 4.4 System curve ..................................................................................................... 4.6 4.5 Operating point ................................................................................................... 4.7 4.6 Safety factor on Total Head or capacity ............................................................. 4.9 4.7 Pump operation to the right or left of Best Efficiency Point (B..E.P.) .................. 4.11 4.8 Pump shut-off head ........................................................................................... 4.14 4.9 Pump power ....................................................................................................... 4.15 4.9 Affinity laws ........................................................................................................ 4.17 Chapter 5 - Field Measurements 5.0 Real live measurements .................................................................................... 5.1 5.1 Total Head ......................................................................................................... 5.1 5.2 Net Positive Suction Head Available (N.P.S.H.A.) .............................................. 5.4 5.3 Shut-off head ..................................................................................................... 5.5 5.4 Equipment head difference................................................................................. 5.7 5.5 Flow measurement ............................................................................................ 5.8 5.6 Calculating flow based on power consumed by the motor.................................. 5.9

III

Glossary Bibliography Appendix A Useful equations (metric and imperial systems) The definition of viscosity Rheological (viscous behavior) properties of fluids Appendix B The Newton-Raphson iteration technique applied to the Colebrook equation Appendix C The determination of slurry density based on the volume and weight concentration of the solid particles Appendix D The use of imperial system (FPS) units Appendix E Power factors and efficiency values for ABB electric motors

IV

Foreword One of my goals in writing this book was to make sense of the various terminology, equations and miscellaneous tips and tricks that are published in the general literature on centrifugal pump sizing and system head calculations. There is no lack of articles or books on the subject but usually only certain isolated aspects of the topic are treated. It seems that so far nobody has put all the relevant concepts and principles together. I had some difficulty starting this book. For starters, I was concerned about the quality of my writing skills, I think they have improved. Then I wondered if I had anything original to say. That was more difficult. At first, I believed that I could simply write down what I knew on the subject. I asked myself a few tough questions, and quickly discovered major gaps in my knowledge. That was a good starting point. Another source of inspiration came from conversations with colleagues on pump system problems (I was not much help), another gold mine. I found that most books on the subject just do not give the big picture, so here is the big picture.

V

Introduction The purpose of this book is to describe how pressure can be determined anywhere within a pump system. The inlet and outlet of a pump are two locations where pressure is of special interest. The difference in pressure head (the term pressure head refers to the energy associated with pressure divided by the weight of fluid displaced) between these two points is known as the Total Head. A system equation will be developed based on fundamental principles from which the Total Head of the pump can be calculated, as well as the pressure head anywhere within the system. These principles can be applied to very complex systems. Friction loss due to fluid flow in pipes is the most difficult component of Total head to calculate. The methods used to calculate friction loss for different types of fluids such as water and viscous fluids of the Newtonian type and wood fiber suspensions (or stock) will be explained. The fluids considered in this book belong to the categories of viscous and non-viscous Newtonian fluids. Wood fiber suspensions are a special type of slurry. There is an excellent treatment on this subject by G.G. Duffy in reference 2. For the reader’s benefit, a condensed version is provided. Slurries, which are an important class of fluids, are not considered. I recommend reference 7, which provides a complete treatment of the subject. However, all the principles for Total Head determination described in this book apply to slurry fluid systems. The only exception is the methods used to calculate pipe friction head. Centrifugal pumps are by far the most common type of pump used in industrial processes. This type of pump is the focus of the book. The challenge in pump sizing lies in determining the Total Head of the system, not the particular pump model, or the materials required for the application. The pump manufacturers are generally more than willing to help with specific recommendations. Information on models, materials, seals, etc., is available from pump manufacturer catalogs. Often when approaching a new subject, our lack of familiarity makes it difficult to formulate meaningful questions. Chapter 1 is a brief introduction to the components of Total Head. I hope it proves as useful to you as it did to me.

VI

Symbols Variable nomenclature A CW CV D F f g

E E ∆En H ∆HP ∆HDS ∆HEQ ∆HF ∆HSS ∆HTS ∆Hv ∆KE L m M ∆PE p P Re SG

area solids concentration ratio by weight in a slurry solids concentration ratio by volume in a slurry pipe diameter force pipe friction factor acceleration due to gravity: 32.17 ft/s2 energy specific energy enthalpy variation of the system head Total Head discharge static head equipment head difference friction head difference suction static head total static head velocity head difference kinetic energy variation of the system length of pipe mass mass flow rate potential energy variation of the system pressure

T

power Reynolds number specific gravity; ratio of the fluid density to the density of water at standard conditions temperature

Q q ∆U V v W z

heat loss volumetric flow rate internal energy variation of the system volume velocity work vertical position

Imperial system (FPS units) in2 (inch square) non-dimensional

Metric system (SI units) mm2 (mm square)

non-dimensional in (inch) lbf (pound force) non-dimensional ft/s2 (feet/second squared) Btu (British Thermal Unit) Btu/lbm Btu ft (feet) ft ft ft ft ft ft ft Btu ft lbm (pound mass) tn/h Btu psi (pound per square inch) hp (horsepower) non-dimensional non-dimensional °F (degrees Fahrenheit) Btu ft3/s Btu ft3 ft/s Btu ft

mm (millimeter) N (Newton) m/s2 (meter/second squared) kJ (kiloJoule) kJ/kg kJ m (meter) m m m m m m m kJ m kg (kilogram) t/h kJ kPa (kiloPascal) W (watt)

°C (degrees Celsius) kJ m3/s kJ m3 m/s kJ m

VII

Variable nomenclature

∆ ε ν

Greek terms delta: the difference between two terms epsilon: pipe roughness nu: kinematic viscosity

η µ ρ γ

eta: efficiency mu: dynamic viscosity rho: density gamma: specific weight

Imperial system (FPS units)

Metric system (SI units)

ft SSU (Saybolt Universal Second) non-dimensional

m cSt (centiStoke)

3

lbm/ft lbf/ft3

cP (centiPoise) kg/m3 N/m3

& , Q& , W& ) indicates the rate of change of the Note: A dot above the symbol (i.e. m variable. A term with multiple subscripts such as ∆HEQ1-2 means the total or sum of all equipment head between points 1 and 2. *FPS: Foot-pound-second system (Imperial) used in the U.S. and anglophone countries. **SI: Système internationale, the metric system.

AN INTRODUCTION TO PUMP SYSTEMS 1.0

HYDROSTATIC PRESSURE AND FLUID COLUMN HEIGHT Our starting point will be pressure and how it is developed within a pumping system. It is easy to build pressure within a solid, a fluid however requires containment walls. A fluid can only be pressurized if it is in a container. Sometimes the container is very big like an ocean. This does not mean that the container has to be closed. Even if the container has an outlet, it is still possible to build pressure. An experiment with a common syringe will demonstrate this fact. With the syringe full of water, it is quite easy to generate significant pressure within the fluid (this is evidenced by the amount of force applied to the plunger) even while fluid squirts from the tip. Hydrostatic pressure is the pressure associated with a motionless body of water. Pressure within the body of water varies and is directly proportional to the vertical position with respect to the free surface. Divers are well aware of this fact since each foot of vertical descent increases pressure on the eardrums.

Figure 1-1 Pressure vs. hydrostatic head. Fluid weight is the cause of hydrostatic pressure. In Figure 1-1, a thin slice of fluid is isolated so that the forces surrounding it can be visualized. If we make the slice very thin, the pressure at the top and bottom of the slice will be the same. The slice is compressed top and bottom by force vectors opposing each other. The fluid in the slice also exerts pressure in the horizontal direction against the pipe walls. These forces are balanced by stress within the pipe wall. The pressure at the bottom of the slice will be equal to the weight of fluid above it divided by the area.

1•2

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

The weight of a fluid column of height (z) is:

F =ρ g V =γ V =γ z A since V = z A The pressure (p) is equal to the fluid weight (F) divided by the cross-sectional area (A) at the point where the pressure is calculated :

γ z A p= F = =γ z A A

[1-1]

where F : force due to fluid weight V : volume g : acceleration due to gravity (32.17 ft/s2) ρ : fluid density in pound mass per unit volume γ : fluid density in pound force per unit volume

Figure 1-2 Fluid container shapes vs. pressure head. Note that the relationship between pressure (p) and fluid column height (z) is independent of the total volume in the container. The pressure generated by the weight of water at the deep end of a pool (i.e. 10 feet down) is the same as ten feet below the surface of a lake.

1.1

THE THREE FORMS OF ENERGY There are three forms of energy that are related and always occur together in a fluid system. They are potential, kinetic and pressure energies. This section briefly provides the definition for each to give the reader some familiarity with the terms. If we divide energy by the weight of the fluid element we obtain specific energy or head.

1•3

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

POTENTIAL SPECIFIC ENERGY

Potential specific energy = z

Figure 1-3 Potential specific energy provided by the difference in elevation of fluid particles.

KINETIC SPECIFIC ENERGY

Kinetic

specific

energy

=

2

v 2 g

Figure 1-4 Kinetic specific energy provided by moving fluid particles.

PRESSURE SPECIFIC ENERGY

Pressure specific energy =

p

γ

Figure 1-5 Pressure specific energy provided by the weight of a fluid column.

1•4

1.2

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

THE RELATIONSHIP BETWEEN ELEVATION, PRESSURE AND VELOCITY IN A FLUID

There is a relationship between the energies associated with elevation, pressure and velocity of fluid particles. The energy terms are: the elevation energy (z), the pressure energy (p/γ), and the velocity energy (v2/g). The sum of these 3 types of energy must be constant, since energy cannot be lost. Stated in another way: the energy at point 1 must be equal to the energy at point 2 (see Figure 1-6). The sum of the three forms of energy must be constant:

z +

p

2

+ v = E = CONSTANT γ 2g

Or if we wish to describe the relationship between the energy levels of fluid particles in different locations of the system such as point 1 and 2 in Figure 1-6 then: 2 2 p p v v z 1 + γ 1 + 2 g1 = z 2 + γ 2 + 2 g2

Relationship between pressure, elevation and velocity

A variation in one or two of these terms implies a variation in the third. The total energy at point 1 in a fluid system must be equal to the total energy at point 2 (see Figure 1-6). For example, if we were to increase the velocity at point 1 by reducing the section, keeping all the other terms the same, the pressure p1 will decrease.

Figure 1-6 The relationship between pressure, elevation and velocity.

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD Relationship between pressure and elevation

There are many areas in a system where the velocity is constant. In that case, it is only pressure and elevation that are related. In particular, if the velocity is zero as in a static system, we have the relationship between pressure and fluid column height previously mentioned.

Figure 1-7 The relationship between pressure and elevation. The following equation gives the relationship between elevation and pressure when the velocity is constant.

z1 +

p1

γ

= z2 +

p2

γ

Figure 1-8 shows a real system with a pressure gauge at the low part of the system (near the pump) and one in the upper part (near the discharge tank). The pressure p1 will be greater than p2 due to the elevation difference.

Figure 1-8 Pressure variation due to elevation in a real system.

1•5

1•6

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

Relationship between pressure and velocity If the elevation is constant, then there is a relationship between pressure and velocity. It is this relationship that helps us calculate the flow rate in a venturi tube. In Figure 1-9, the pressure p2 will be lower than p1 due to the increase in velocity at point 2.

Figure 1-9 The relationship between pressure and velocity. The following equation gives the relationship between pressure and velocity when the elevation is constant. 2 p2 v2 2 + v1 = + γ γ 2g 2g

p1

A venturi tube is used to measure flow rate. The flow rate (q) is proportional to the difference in pressure at points 1 and 2 ( see Figure 1-9).

q =K

p2 − p1

γ

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

1•7

The 3 forms of energy (elevation, pressure and velocity) are always present in a fluid system. Using a simple container (see Figure 1-10), these types of energy can be clearly demonstrated. 1. Potential energy. The fluid particles at elevation z vs. those at the bottom have potential energy. We know this type of energy is present because we must have spent energy moving the fluid particles up to that level. 2. Pressure energy. The weight of the fluid column produces pressure p at the bottom of the tank. The pressure energy is transformed to kinetic energy when we open the valve at the bottom of the tank. 3. Kinetic energy. If fluid is allowed to exit the container at the bottom of the tank, it will exit the tank with a velocity v. Pressure energy has been converted to kinetic energy.

Figure 1-10 The 3 forms of energy in a fluid system, potential, kinetic and pressure. 1.3

THE DIFFERENCE BETWEEN PRESSURE AND HEAD In a moving fluid, the velocity of the particles must be considered. The principle of conservation of energy states that the energy of a fluid particle as it travels through a system must be constant. This can he expressed as:

m gz + m g

p

γ

+

1 m v 2 = E = C O N STAN T 2

[1-2]

where (E) is the total energy of fluid particles with mass (m) and velocity (v). The total energy consists of the potential energy (mg z), the pressure energy (mg p/γ), and the kinetic energy (m v2/2g). By dividing all the terms in the above equation by (mg), we obtain equation [1-3] which is known as Bernoulli’s equation. E now becomes the specific energy of the fluid particle or the energy per unit weight E . All the terms on the left-hand

1•8

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

side of Bernoulli’s equation are known as head. Bernoulli’s equation expresses the relationship between the elevation head (z), the pressure head (p/γ), and the velocity head (v2/2g). Bernoulli’s equation:

z +

p

γ

+

v2 = E = C O N STAN T 2g

[1-3]

Bernoulli’s equation will be expanded later to include the pump Total Head and the friction losses. It is important at this point to make a clear distinction between pressure and head. Head is a generic term for a type of specific energy (i.e. elevation, pressure or velocity head). When we need to calculate pressure at a specific point in a system, we will refer to it as pressure head. Pressure head can be converted to pressure by equation [1-1]. Pressure can be measured anywhere in the system quite easily and can provide valuable information. However, since it is not an energy term and cannot be used for calculations involving head, especially Total head. The pressure measurement must be converted to pressure head (see equation [1-5]) to be useful in these calculations. 1.4

FLUID SYSTEMS The pump is the heart of a fluid system. It is impossible (or at the very least impractical) to move fluid from one place to another without the energy provided by a pump. Figure 1-11 shows a simple but typical system. All the fluid in the suction tank will eventually be transferred to the discharge tank. The purpose of the system is to displace fluid and sometimes to alter it by filtering, heating or other with the appropriate equipment. The box with the term EQ symbolizes equipment such as: control valves, filters, etc. Any device (or equipment) introduced into Figure 1-11 A typical pumping system. the line will have the effect of reducing the pressure head in the line. This requires more head, or energy, from the pump.

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

1•9

How is a system designed? A.

The flow rate is determined based on the process and production requirements.

B.

The location and size of the suction and discharge tanks is established.

C.

The location, capacity and size of the equipment to be installed in the line is determined.

D.

The pump location is fixed.

E.

The line sizes are determined and the auxiliary equipment such as manual valves are sized and located.

F.

The Total Head of the pump is determined as well as the size, model, type, and power requirement.

Where does a system begin and end?

We can imagine a boundary (also called control volume) which envelops and determines the extent of a system. A complete system contains fluid that is continuous from inlet to outlet. There can be no gaps or empty spaces between parts of the fluid. In Figure 1-11, the system inlet is at point 1 and the outlet is at point 2. Point 1 is located on the liquid surface of the suction tank. Normally there would be a pipe inserted into the suction tank providing fluid to maintain the elevation of point 1. This fill pipe is not considered part of the system. The outlet of the system, or point 2, is located on the liquid surface of the discharge tank. Again, there is normally a pipe, which controls the level of point 2. This discharge pipe is not part of the system. The reasons for this will be carefully explained when we review control volumes in section 2.7. 1.5

THE DRIVING FORCE OF THE FLUID SYSTEM The pump supplies the energy to move the fluid through a system at a certain flow rate. The energy is transferred to the fluid by a rotating a disk with curved vanes called an impeller (see Figure 1-12). This movement drives the liquid in a circular path and imparts centrifugal force to it. The pump pushes and pressurizes the fluid up against the casing (or volute). This ability of the pump to pressurize fluids at a certain flow rate provides us with the means to design systems that will meet the process goals (for example, flow and pressure at the desired locations). The Total Head is the difference in head between the inlet and outlet of the pump that produces the flow rate required of the system.

1 • 10

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

Figure 1-12 Principal components of a centrifugal pump. The Total Head of the pump provides the energy necessary to overcome the friction loss due to the movement of the fluid through pipes and equipment. It also provides the energy to compensate for the difference in height, velocity and pressure between the inlet and the outlet. 1.6

THE COMPONENTS OF TOTAL HEAD Friction

Fluids in movement generate friction. There is a difference in pressure (pF1 - pF2 ) required to move a fluid element A (see Figure 1-13) towards the outlet. This difference in pressure is known as the friction pressure loss (∆p) (see Figure 1-13). When this term is converted to head, it is then known as friction head loss. Reference tables and charts for friction head loss are widely available (see reference 1 and 8). Viscosity is an important factor and higher viscosity fluids generate higher friction. More about viscosity in chapter 3.

Figure 1-13 The difference in pressure due to fluid friction.

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

1 • 11

Equipment

Any equipment in the line will create a reduction in pressure head. A filter is a common example of a device producing a pressure drop (see Figure 1-14). Other examples are control valves, heat exchangers, etc. Equipment introduced into an existing system will reduce the flow rate unless the pump is modified to provide more energy (for example by installing a bigger impeller).

Figure 1-14 The effect of equipment in a system. Velocity

The kinetic energy of the fluid increases when it leaves the system at a higher velocity than when it enters and this requires extra energy. The energy required for the velocity increase is typically small and is often neglected. However, certain systems are specifically designed to produce high output velocity. This is done by using nozzles (see Figure 1-15 C) and therefore require a substantial amount of energy that the pump must supply. Figure 1-15 illustrates three systems with progressively higher output velocities.

Figure 1-15 Various systems with varying output velocities. CASE A. DISCHARGE PIPE END SUBMERGED. The output velocity v2 is small and the velocity head negligible. CASE B. DISCHARGE PIPE END NOT SUBMERGED. The output velocity v2 is small and the velocity head is non-negligible. CASE C. NOZZLES ARE INSTALLED ON THE DISCHARGE PIPE END. The output velocity v2 is high and the velocity head is high.

1 • 12

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD Elevation

It takes energy to pump fluids from a lower level to a higher one. There is often a significant difference in elevation between the inlet of a system (point 1), and the outlet (point 2, see Figure 1-16). Typically, the elevation difference within a system is the largest contributor to the Total Head of the pump.

Figure 1-16 The difference in elevation between suction and discharge tank. Pressurized Tanks

The discharge and suction tank may be positively or negatively pressurized (with respect to the local atmospheric pressure). If the discharge tank is positively pressurized then the pump must provide additional energy to overcome this additional pressure. When the discharge tank is negatively pressurized, there is less resistance to the fluid entering the discharge tank and this reduces the energy required of the pump. The effect on the pump is exactly opposite when the suction tank is positively or negatively pressurized. Figure 1-17 Pressurized suction and discharge tanks.

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

1 • 13

In many applications the tanks are not pressurized and the level of pressure in these tanks is zero or the same as the local atmospheric pressure. However, there is a misuse of terminology that tends to muddy the waters occasionally. Have you ever heard someone say: “Boy, that water is really coming out, there must be allot of pressure at the end of that hose”. In reality, there is no pressure at the outlet of the hose since the fluid comes out into the atmosphere. What feels like pressure, is the mass of Figure 1-18 Ouch. water particles hitting at high velocity. The kinetic energy is converted to pressure energy, which produces a force on the hand. The pressure is zero but the water jet has a significant amount of kinetic energy. 1.7

NEGATIVE (RELATIVE) PRESSURE Figure 1-19 illustrates how easy it is to produce negative relative pressure. The fluid is stationary and the elevation of point 1 is identical to that of points 3, 6 and 9. The pressure at point 2 is higher than the pressure at point 1 due to the depth of point 2 and the pressure caused by the fluid at that depth. The pressure at point 2 is positive and decreases to zero as we reach the level of point 3 (same level as point 1) inside the tube. From 3 to 4, the pressure decreases and becomes negative. The pressure at point 5 is the same as at point 4 since we are on the same level. Figure 1-19 The pressure distribution in a static Pressure then increases system. from negative to zero at point 6 (same level as point 1). From point 6 to 7 there is a further increase, the pressure remains constant from 7 to 8. The pressure decreases to zero from point 8 to 9 since point 9 is at the same level as point 1.

1 • 14

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

We create relative negative pressure everyday with a straw. Find some flexible tubing and try the experiment shown on Figure 1-20 and Figure 1-24. Try the following simple experiment. Get a small container and a short length of clear plastic tube. Our goal will be to put some water on a shelf so to speak. 1. Suction is applied to the tube and the liquid is lifted up to point 4. 2. Bend the tube as you apply suction to get the fluid past point 5. At this point a siphon (see section 1.8) is established. 3. The tube is bent at points 7 and 8 and the liquid level establishes itself at point 9, which is the same level as point 1. The liquid in the tube remains stable and suspended at the level of point 4 and 5. Liquid has been raised from a lower elevation at point 1 to a higher one at point 4, like putting a book on a shelf. If the tube was punctured at point 4 or 5, what would happen? Air would enter the tube and the liquid would drop to its lowest level. We have managed to create negative relative pressure at point 4, which is easily maintained without further intervention.

Figure 1-20 Creating negative pressure.

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

1 • 15

Use a longer tube this time, fill it with water and put your finger on the end of the tube sealing it off. The lower end of the tube is open (see Figure 1-21a). It is possible to suspend a column of water 34 feet high by sealing the top. This creates a volume of zero pressure at the top end, between the finger and the fluid surface. At the bottom end, which is open, atmospheric pressure is pushing on the fluid in an upward direction. The weight of the liquid column is balanced by the force generated by atmospheric pressure.

Figure 1-21b Difference in pressure in a water column suspended from an open tube 34 feet high.

Figure 1-21a Water suspended from an open tube 34 feet high.

1 • 16

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

A common unit in North America for measuring pressure is the psig (or pound per square inch gauge). Zero psig corresponds to the level of pressure in the local atmosphere. The “g” stands for gauge, meaning dial gauge. The equivalent to 0 psig in pressure head is 0 feet of fluid. These units are relative to the local atmospheric pressure or atmospheric pressure head. Pressure in certain parts of the system can drop below the local atmospheric pressure, and become negative. An area of negative pressure is under a vacuum. A breach of the containment wall, in an area under vacuum, will cause air to be drawn into the system. This is what would happen if the tube were punctured at point 4 in Figure 1-20. The units often used to express low pressure are psia (pounds per square inch absolute). The levels of pressure head are expressed in feet of fluid absolute or in Hg (inch of Mercury). These units are absolute, and are therefore not relative to any other pressure. A perfect vacuum corresponds to 0 psia. Figure 1-22 shows graphically the relationship between absolute and relative pressure. The atmospheric pressure at sea level is 14.7 psia. Not all plants are at sea level, for example, Johannesburg is 5,200 feet above sea level and the local atmospheric pressure is 12 psia. This effect has to be considered when calculating the available N.P.S.H. (Net Positive Suction Head ) at the pump suction (see Chapter 3).

Figure 1-22 Absolute vs. relative pressure scales.

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

1 • 17

The relationship between an absolute (psia) and a relative (psig) pressure measurement is:

p ( psia ) = p ( psig ) + p A ( psia ) where pA is the local barometric or atmospheric pressure in psia (i.e. pA = 14.7 psia at sea level). 1.8

THE SIPHON EFFECT At first glance, a fluid moving vertically upwards without assistance creates a surprising effect. Figure 1-23 compares the movement of a rope with that of a ball. Both objects are solid, however the rope can emulate the behavior of a fluid where a ball cannot. A ball moves toward an incline and encounters a rise before it gets to a sharp drop; can it get over the hump without any intervention? No, not if it has a low velocity. Imagine the ball stretched into the shape of a rope, lying on a smooth surface, and draped across the hump. Even when starting from rest, the rope will slide down and drag the overhung part along with it. A fluid in a tube will behave in the same way as the rope.

Figure 1-23 A siphon as a rope. A volume of fluid is formless. A fluid will take on the shape of its container, whatever shape it may be. Its natural tendency is to lie flat. Gravity (or potential energy) is responsible for this behavior. Gravity is also responsible for the siphon effect.

[1-4]

1 • 18

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

Figure 1-24 shows an experiment that requires a few feet of flexible tube. Fill the tube with water and keep the bottom end sealed in some manner. Pinch the end of the tube at point 1, turn it around l80° to form a U shape as in position C. Release the end of the tube at point 1. What happens? No fluid leaves the tube. Why? If the water were to come out on the short side then a void would be created in the tube on the level of point 2. A void would produce a volume under low pressure. Low relative pressure acts as a pulling force on the short fluid column. Actually, it is not low pressure that pulls the fluid up but atmospheric pressure that pushes the short column of fluid upward. Therefore, fluid coming out at point 1 would create a vacuum at point 2 and a vacuum stops the movement so that there can be no movement out at point 1. What this Figure 1-24 Water suspended in an open means is that the pressure at point 1 tube. (atmospheric pressure) is sufficient to prevent the short column of fluid from moving downward. Why, because if it did not, a void would have to be created at point 2. The effect of creating a void at point 2 would reduce the pressure to 0 psia. This low pressure can suspend a column of water 34 feet high. This is clearly not necessary for the short column of fluid to the left of 2. Something less than a void needs to be created to suspend the short column of fluid. In Figure 1-24 E, a reduced pressure is present at point 2 which suspends the short column. There is no void required to suspend the short column. The following experiment will show how easy it is to create low pressure. The effect produced is surprising and shows up an unusual property of fluids, the ability of fluids to be suspended in mid air without apparent means of support.

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

1 • 19

Experiment No.1

The experiment consists of a vertical tube, full of water, and closed at the bottom end. We take the top end and turn it around vertically downward as shown in Figure 1-25. What happens to the fluid in the tube? Is it in equilibrium, is it suspended, or will some portion of it drop out of the tube? If the fluid is suspended, there must be a balance of forces that holds it in place, preventing it from falling out of the short end of the tube. The pressure p0 within the fluid produces a force F0 at the section of point 0 (see Figure 125). Similarly, the atmospheric pressure pA produces a force FA at the end of the tube. W is the weight of that portion of fluid between point 0 and A.

Figure 1-25 Balance of forces between points 0 and A.

The balance of forces is:

FA = F0 + W also F = p × A then

Therefore

p A × A = p0 × A + γz A

p A = p0 + γz

For the fluid to be in equilibrium, p0 must be smaller than pA. This means that p0 will be negative with respect to the atmospheric pressure.

1 • 20

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

There are two principles in the following argument that should be made clear: •

In a static system, the fluid particles on the same level are at the same pressure.

•

Fluids are incompressible.

We know that for the fluid to be stable there must be a balance of forces. How does this balance of forces come about? Let’s see if we can duplicate this experiment without bending any tubes. Experiment No.2

Let’s start with a vertical tube full of water with one end closed as shown in Figure 1-26. If we apply a source of vacuum, the fluid will remain exactly where it is since it is incompressible. In other words, if the pressure is lowered the fluid does not expand and alternatively, if the pressure is raised it does not contract.

Figure 1-26 The level in the tube does not change when we apply a vacuum. Attach a tee to the top end of the tube, then a short leg to the tee and connect one branch to a reservoir. Reattach the vacuum source to the tee as in Figure 1-27. The fluid is pulled up the short leg causing it to be in balance and therefore suspended. If we shut off the vacuum source and remove the reservoir, we have exactly the exact same system we started out with in Figure 1-25.

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

Figure 1-27 Lifting fluid with a source of vacuum.

1 • 21

Figure 1-28 The vacuum source is disconnected and the fluid is suspended.

We managed to create a similar system to the original one (Figure 1-25) without bending the tube. We know that the pressure in the tube drops as we raise a fluid vertically from the outlet. It drops just sufficiently to suspend the column of fluid. This was made apparent as we applied a vacuum to lift the fluid. The bending of the tube in Figure 1-25 disguises this effect, giving us a suspended column of fluid without a clear indication of the mechanism. Experiment no. 2 shows the mechanism by which the fluid is suspended.

A typical siphon situation is shown in Figure 129. The graph shows how the pressure head varies within the system. In the next chapter, we will develop the method required to calculate the pressure head at any location in the system.

Figure 1-29 The pressure variation within a simple siphon system.

1 • 22

1.9

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

SPECIFIC GRAVITY We often need to calculate the pressure head that corresponds to the pressure. Pressure can be converted to pressure head or fluid column height for any fluid. However, not all fluids have the same density. Water for example has a density of 62.34 pounds per cubic foot whereas gasoline has a density of 46.75 pounds per cubic foot. Specific gravity is the ratio of the fluid density to water density at standard conditions. By definition water has a specific gravity (SG) of 1. To convert pressure to pressure head, the specific gravity SG of the fluid must be known. The specific gravity of a fluid is: S G

ρ F ρW

=

where ρF is the fluid density and ρW is water density at standard conditions. Since

p =γF z =

ρF g z gc

ρF = SG ρW

and

therefore

p = SG

ρW g z gc

where γF is the fluid density in terms of weight per unit volume and ρF is the density in terms of mass per unit volume. The constant gc is required to provide a relationship between mass in lbm and force in lbf (see Appendix D). The quantity ρWg/gc ( ρW = 62.34 lbm/ft3 for water at 60 °F) is:

ρW g 62.34(lbm / ft 3) × 32.17(ft / s 2) gc

=

32.17(lbm− ft / lbf − s ) 2

×

 lbf  1(ft) = 1  2  2 144(in ) 2.31 in − ft 

After simplification, the relationship between the fluid column height and the pressure at the bottom of the column is:

p( psi ) =

1 SG z ( ft of fluid ) 2.31

Figure 1-30 Pressure vs. elevation in a fluid column.

[1-5]

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

1 • 23

Here he goes again with the siphon. OK, maybe not everyone has the same interest in siphons as I do. The explanation for the behavior of a siphon is quite simple: a siphon is like a rope: if you pull on one end the other will follow. Why is it important to know how a siphon works? Siphons by their very nature produce an area of low pressure. A feed pipe with top entry to a tank behaves like a siphon. Any area in a piping system that is higher than the discharge point will likely be under low pressure.

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS 2.0

ENERGY AND THERMODYNAMIC PROPERTIES This chapter requires some introduction to thermodynamic properties and states. No need to panic, we will use only what we need to know to calculate Total Head. Several measurable quantities are used to define the state of a substance: temperature (T), pressure (p), velocity (v), and elevation (z). A specific type of energy corresponds to each of these quantities. Kinetic Energy

The formula for kinetic energy is:

KE =

1 2 mv 2

The above equation states that the kinetic energy of a body is equal to one half the mass times the velocity squared. Potential Energy

Another form of energy is potential energy:

PE = mg z which means that the potential energy of a body is equal to the weight (mg) times the vertical distance (z) above a surface upon which the object would come to rest. These two types of energies interact. For example, an object at the top of an incline has a certain amount of potential energy. After it's release, potential energy is gradually converted to kinetic energy. When it reaches the bottom, all the potential energy has converted to kinetic energy.

Figure 2-1 Transformation of potential energy to kinetic energy.

2•2

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

The principle of conservation of energy states that energy can neither be created or lost. Consequently, if one form of energy decreases then another form must increase. This allows us to make an energy balance that describes the energy variation of the object in Figure 2-1.

∆KE + ∆PE = 0

[2-1]

In other words, equation [2-1] reads: the sum of the potential and kinetic energy variation between points 1 and 2 is equal to zero. Energy levels are associated with positions within a system, position 1 is the top of the slope and position 2 the bottom.

∆ PE = PE 1 − PE 2 and ∆ KE = K E 1 − KE 2 mg ( z 1 − z 2 ) +

1 m( v1 2 − v 2 2 ) = 0 2

v 2 2 − v1 2 = 2 g ( z1 − z 2 )

[2-2]

Equation [2-1] expresses the principle of conservation of energy for this system. This leads to equation [2-2] that describes how velocity changes with respect to height. The principle of conservation of energy makes it possible to account for all forms of energy in a system. The energies present in the system are in continuous change and the term delta (∆) is used in equation [2-1] to indicate a change or variation in energy. Thermodynamic properties

Thermodynamic properties are the different types of energies associated with a body (for example, potential, kinetic, internal or external). It is characteristic of a thermodynamic property is that its value is independent of the method or path taken to get from one value to another or from an initial state 1 to another state 2. The potential energy (PE) and the kinetic energy (KE) are thermodynamic properties. We require other energy quantities (work and heat) to account for real world conditions. These quantities are dependent on the path or method used to get from state 1 to state 2 (more about paths later). 2.1

CLOSED SYSTEMS AND INTERNAL ENERGY The system shown in Figure 2-2 is a fluid contained in a sealed tube with its inlet connected to its outlet forming a closed system.

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

2•3

Internal Energy

All fluids have internal energy (U). If we apply a heat source to the system, the temperature, pressure and internal energy of the fluid will increase. Internal energy is the energy present at the molecular level of the substance.

Figure 2-2 The relationship between internal energy and heat gain within a closed system. Closed Systems

In a closed system no mass enters or leaves the boundary. What happens if we put a fluid within a closed environment such as a sealed container and apply heat? Pressure and temperature in the fluid will increase. The temperature rise indicates that the internal energy of the fluid has increased. The heat source increases the internal energy of the fluid from U1 to U2 . The energy quantities present in this system are the internal energy (U) and the heat loss (Q). Therefore, the energy balance is:

Q = QC − QE = ∆U = U 1 − U 2 QC is the quantity of heat absorbed by the fluid from the source and QE is the heat loss of the fluid to the environment. The internal energy has changed from its level at instant 1 to another level at instant 2 (after heat is applied). The internal energy (U) is a thermodynamic property. 2.2

CLOSED SYSTEMS, INTERNAL ENERGY AND WORK Another way to increase the internal energy of a fluid is to do work on it by means of a pump. In this way, without applying heat, the work done on the fluid through the pump raises the internal energy of the fluid from U1 to U2. This causes the temperature of the fluid to increase, producing a heat loss QE to the environment.

Figure 2-3 The relationship between internal energy, work and heat loss within a closed system.

2•4

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

In this situation, the energy balance is:

QE − W = ∆U = U 1 − U 2 The sign convention is positive energy for heat leaving the system and negative energy for heat or work entering the system. 2.3

OPEN SYSTEMS AND ENTHALPY

Figure 2-4 The relationship between internal energy, heat loss, work, and a pressurized environment for an open system. An open system is one where mass is allowed to enter and exit. The mass entering the system displaces an equal amount of mass that exits. A clearly defined boundary, called the control volume, envelops the system and intersects the inlet and outlet. The control volume allows us to apply the principle of conservation of energy. Each unit of mass (m1) and volume (V1) that enters the system is subject to a pressure (p1) at the inlet. The same is true of the mass leaving the system (m2) of volume (V2) and subjected to a pressure (p2). The principle of conservation of mass requires that m1 = m2. In this book, we are dealing only with incompressible fluids where V1 = V2. The pressure at the inlet produces a certain quantity of work that helps push the fluid through the system. The pressure at the outlet produces work opposing that movement. The difference between these two work components is the work associated with pressure at the inlet and outlet of the system. This difference is the energy term ∆(pV) and is the only difference between the systems shown in Figure 2-3 vs. the system shown in Figure 2-4.

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

2•5

The energy balance is:

Q E − W = ∆ U − ∆ ( pV ) = ∆ En ∆U and ∆(pV) are terms that always occur together in an open system. For that reason their sum has been given the name enthalpy (En). Enthalpy is also a thermodynamic property. 2.4

OPEN SYSTEMS, ENTHALPY, KINETIC AND POTENTIAL ENERGY In closed systems, the fluid particles move from one end of the container to another but they do not leave the container. It is impossible to have a net effect on the velocity and therefore on the kinetic energy of the system. In open systems, fluid particles move from the inlet to the outlet and leave the system. If the velocity varies between inlet and outlet, the kinetic energy varies. The pump is the energy source that compensates for the variation. The same is true for potential energy in a closed system. A fluid particle that goes from the bottom of a container to the top will eventually return to the bottom. There can be no net effect on the potential energy of this system. In an open system, fluid particles can leave the system at a different elevation than they enter producing a variation in potential energy. The pump also compensates for this energy variation. Potential Energy

The difference in vertical position at which the fluid enters and leaves the system causes the potential energy of a fluid particle to increase or decrease. The potential energy increases when the fluid leaves at a higher elevation than it enters. The pump provides energy to compensate for the increase in potential energy. If the fluid leaves the system at a lower elevation than it enters, the potential energy decreases and it is possible to convert this energy to useful work. This is what happens in a hydroelectric dam. A channel behind the dam wall brings water to a turbine located beneath the dam structure. The flow of water under pressure turns a turbine that is coupled to a generator producing electricity. The difference in elevation or potential energy of the fluid is the source of energy for the generator. Kinetic Energy

There is often a velocity difference between the inlet and outlet of a system. Usually the velocity is higher at the outlet versus the inlet; this produces an increase in kinetic energy. The pump compensates for this difference in kinetic energy levels. The complete energy balance for an open system is:

Q E − W = ∆En + ∆KE + ∆PE

[2-3]

2•6

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

Equation [2-3] has the thermodynamic properties on the right-hand side. The heat loss (QE) and the work done on the system (W), which are not thermodynamic properties, are on the left-hand side of the equation. To determine QE and W, it is necessary to know how the heat is produced (i.e. path that the fluid must take between inlet and outlet) and how the work is done. 2.5

WORK DONE BY THE PUMP The role of a pump is to provide sufficient pressure to move fluid trough the system at the required flow rate. This energy compensates for the energy losses due to friction, elevation, velocity and pressure differences between the inlet and outlet of the system. This book does not deal with the method by which a pump can pressurize a fluid within its casing. The reader is referred to reference 15, a very detailed treatment of this subject. For our purpose, the Figure 2-5 The work provided by the pump. pump is a black box whose function is to increase the fluid pressure at a given flow rate.

2.6

FLUID AND EQUIPMENT FRICTION LOSS The total heat loss (QE) is the sum of fluid friction loss and equipment friction loss.

Q E = Q F + Q EQ

Fluid Friction Loss (QF)

How is it possible to go from a nebulous concept such as fluid friction to the pressure drop due to friction? Friction occurs within the fluid and between the fluid and the walls.. A detailed method of fluid friction calculation is presented in Chapter 3. Friction in fluids produces heat (QF ) which dissipates to the environment. The temperature increase is negligible. If we can estimate the friction force, we know that we must supply an equal and opposite force to overcome it.

Figure 2-6 Heat loss due to fluid friction within pipes.

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

2•7

The energy corresponding to the heat loss (QF) must be supplied by the pump. The net force (FF) required to balance the friction force is F1-F2 These forces are the result of the action of pressures p1 and p2 (see Figure 2-6). The difference between p1 and p2 is the pressure differential or pressure drop due to friction for a given length of pipe. Many publications (reference 1 and 8 for example) provide pressure head drop values in the form of tables. The pressure head drop can also be calculated by the Colebrook equation, which we will discuss in chapter 3. Equipment Friction Loss (QEQ)

Many different types of equipment are present in typical industrial systems. Control valves, filters, etc., are examples of equipment that have an effect on the fluid. Because of the great variety of different equipment that can be installed, it is not possible to analyze everyone of them. However we can quantify the effect of the equipment by determining the pressure drop across the equipment much as we did with the pressure drop across the pump. Again it is not necessary to know exactly what happens inside the equipment to be able estimate this effect. Pressure drop (difference between inlet and outlet pressure) is the outward manifestation of the effect of equipment. This pressure drop produces heat (QEQ) which is then lost through the fluid to the environment. The heat loss is calculated in the same manner as the pump work (W).

Q EQ = ∆ p EQ

m

ρ

The product supplier normally makes available the amount of pressure drop for the equipment at various flow rates in the form of charts or tables. 2.7

THE CONTROL VOLUME Up to now we have been using control volumes without defining their purpose. A control volume is an imaginary boundary that surrounds a system and intersects all inlets and outlets. This makes it possible to apply the principle of conservation of mass. The boundary of the system is determined by appropriately locating the inlet and the outlet. The term “appropriate” refers to locating the boundary at points where the conditions (pressure, elevation and velocity) are known. The control volume encompasses all internal and external the energy sources affecting the system. This makes it possible to apply the principle of conservation of energy. How is the control volume positioned in real systems? Figure 2-7 shows a typical industrial process. Suppose that we locate the inlet (point 1) on the inlet pipe as shown in Figure 2-7. Is this reasonable? Where exactly should we locate it, will any position due? Clearly no. Point l could be located anywhere on the feed line to the suction tank. The problem with locating point 1 on the inlet line is precisely that the location is indeterminate.

2•8

.

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

Figure 2-7 Incorrect positioning of the control volume in a pumping system. However, consider Figure 2-8. The pressure head at the inlet of the pump is proportional to the elevation of the suction tank fluid surface. We locate the inlet of the system (point 1) on the suction tank fluid surface since any change in the elevation of point 1 will affect the pump. The inlet pipe is the means by which we supply the pump with enough fluid to operate. As far as this system is concerned, the inlet feed pipe is irrelevant. This is because the pressure head at the inlet of the pump (or any point within the system) is dependent on the level of the suction fluid tank surface. The other reason that requires us to locate point 1 on the liquid surface of the suction tank is that the control volume must contain all the energy sources that affect the system and if the suction and discharge tanks are open to atmosphere than this pressure will be present at points 1 and 2 (H1 = H2 = 0 = atmospheric pressure).

Figure 2-8 Proper positioning of the intersections of the control volume with a pumping system.

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

2•9

This leads to a representation of a generalized system that we will use from now on. A general system must take into account the possibility that the suction and discharge tanks are pressurized. The terms H1 and H2 in Figure 2-8 represent respectively the suction and discharge pressure head at the fluid surfaces of these tanks. A general system must take into account the possibility that the system contains no tanks and that points 1 and 2 are connections to pipe lines of other systems. The only difference between a tank and a pipe is that a tank can contain a much greater volume of fluid. Therefore, whether the system has tanks or not is immaterial. However the conditions (i.e. pressure, velocity and elevation) of the inlet and outlet of the system must be known. It will always be possible to disconnect our system from another if the pressure head, velocity and elevation (H , v and z) at the connection point is known (more about disconnecting interconnected systems later in this chapter). Therefore, Figure 2-8 can represent all possible situations. 2.8

THE DETERMINATION OF TOTAL HEAD FROM THE ENERGY BALANCE We now have all the background information necessary to determine the work required of the pump and therefore the Total Head. First, we calculate the energy lost in friction due to fluid movement through pipes (QF) and equipment (QEQ.). Next, calculate the energy variations due to conditions at the inlet and outlet. 1. Potential energy variation (∆PE ) is associated with the difference in height of a fluid element between the outlet and inlet of the system. 2. Kinetic energy variation (∆KE) is associated with the variation in velocity of a fluid element between the outlet and inlet of the system. 3. Enthalpy variation (∆En) consists of two components ∆En = ∆U + ∆(pV). The first is the internal energy (∆U) of the fluid. In most common fluid transfer situations, there is no significant temperature change and therefore no significant internal energy change (∆U = U1 - U2 = 0). The other component of enthalpy ∆(pV) is the energy associated with the difference in pressure at the inlet and outlet of the system. The difference between the inlet and outlet energies added to the total friction loss (QE) gives the resultant work W that must be supplied by the pump.

2.9

SYSTEM OR TOTAL HEAD EQUATION FOR A SINGLE INLET - SINGLE OUTLET SYSTEM The simplest case we can consider is a system with one inlet and one outlet. The first law of thermodynamics or the principle of conservation of energy for open systems states:

Q& E − W& = ∆ E& n + ∆ K& E + ∆ P& E

[2-4]

2 • 10

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

Equation [2-4] expresses the rate of variation of the energy terms instead of the energy variation as in equation [2-3]. This allows us to balance mass flow rates instead of mass between the inlet and outlet of the system. In Figure 2-8, the box with the letters EQ represents the effect of all the equipment present in the line between points 1 and 2. H1 and H2 are the pressure heads present at the suction tank and the discharge tank fluid surface respectively. z1 and v1 are respectively the elevation and the velocity of a fluid particle at the inlet of the system, and z2 and v2 are the same variables for a fluid particle at the outlet.

ENERGY RATE = v ∆F = V&∆p =

m&

ρ

∆p

The following equations are all based on the above which is the work or energy rate & are respectively the required to move a body of mass m at a velocity v. V& and m volumetric and mass flow rate. ∆p is the pressure differential and ρ is the fluid density.

&E ) The energy rate of heat loss ( Q

The rate of heat transfer (Q& E ) is the heat generated in the system by fluid friction in the pipes and through the equipment:

( ρ

m& Q& E = ∆ p F 1− 2 + ∆ p EQ 1− 2

)

[2-5]

where ∆pF1-2 is the pressure drop associated with fluid friction for a fluid particle traveling between points 1 and 2. ∆pEQ1-2 is the sum of all pressure drops produced by all of the equipment between the same points.

&) The rate of mechanical work (W Similarly, the rate of mechanical energy introduced into the system, such as supplied by a pump is: [2-6] m&

W& =

ρ

∆p P

where ∆pP is the difference in pressure between the discharge and the suction of the pump (points P and S in Figure 2-8). The rate of enthalpy variation ( ∆E& )

The rate of enthalpy variation is composed of the rate of internal energy variation ( ∆U& ), and the difference in pressure energy between the inlet or outlet of the system ∆ ( pV& ) .

∆U& is normally zero or negligible in most fluid transfer situations, therefore: ∆E& = ∆U& + ∆ ( pV& ) = U& 1 − U& 2 + ( p1V&1 − p2V&2 ) = p1V&1 − p2V&2 since U& 1 − U& 2 = 0

2 • 11

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

also for incompressible fluids V&1 = V&2 , therefore [2-7]

m& ∆ E& n = V& ( p1 − p 2 ) = ( p1 − p 2 )

ρ

where p1 is the pressure at the suction tank liquid surface and p2 is the pressure at the discharge tank liquid surface.

& ) The rate of kinetic energy variation ( ∆KE Kinetic energy is the energy associated with the velocity (v) of a body of mass (m). The rate of kinetic energy variation of the system is:

& = ∆KE

[2-8]

1 m& (v12 − v2 2 ) 2 gc

where v1 is the velocity of a particle at the surface of the suction tank or the system inlet velocity. v2 is the velocity of a particle at the surface of the discharge tank or the system outlet velocity. The constant (gc) is required to make the units consistent in the FPS system.

& ) The rate of potential energy variation ( ∆PE Potential energy is the energy associated with the vertical position z2 or z1 of a mass (m) subject to the influence of a gravity field. The change of potential energy of fluid particles in the system is:

& = m& ∆PE

g (z − z ) gc 1 2

[2-9]

where z1 and z2 are respectively the elevation of a particle on the fluid surface of the suction and discharge tank. Elevations such as z1 and z2 are often taken with respect to a DATUM or reference plane.

& By substituting equations [2-5] to [2-9] in equation [2-4] and dividing by m ( ∆ p F 1− 2 + ∆ p EQ 1− 2 ) ∆ p P ( p1 − p 2 ) 1 − = + ( v 2 − v 2 2 ) + ( z1 − z 2 ) g g g 2g 1 ρ ρ ρ gc gc gc

g , we obtain: gc [2-10]

Equation [2-10] is Bernoulli’s equation with the pump pressure increase (∆pP) and fluid (∆pF1-2) and equipment (∆pEQ1-2) friction terms added. Pressure can be expressed in terms of fluid column height or pressure head as demonstrated in chapter 1.

p=

ρgH gc

[2-11]

2 • 12

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

All pressure terms are replaced by their corresponding fluid column heights with the use of equation [2-11], for example p1 =

ρgH1 gc

, ∆p F 1− 2 =

ρg∆H F 1− 2 gc

, etc.). The constant gc

cancels out. The Total Head is:

∆H P ( ft fluid ) = (∆H F 1− 2 + ∆H EQ1− 2 ) +

1 2 2 (v 2 − v1 ) + z 2 + H 2 − ( z1 + H 1 ) 2g

[2-12]

The unit of Total Head (∆HP) is feet of fluid. The pump manufacturers always express the Total Head (∆HP) in feet of water, is a correction required if the fluid is other than water? Do we need to convert feet of fluid to feet of water? The terms in equations [2-10] and [212] are in energy per pound of fluid, or lbf-lbf/lbf; which is the same as feet. Since head is really energy per unit weight, the density of the fluid becomes irrelevant (or in other words 1 pound of water is the same as 1 pound of mercury). However, we will see later that the motor power required to move the fluid at a certain rate does required that the density of the fluid be considered (see Chapter 4). The pump manufacturers test the performance of their pumps with water. The capacity of a centrifugal pump is negatively affected by the fluid’s viscosity; therefore the three major performance parameters of a pump (total head, flow rate and efficiency) will have to be corrected for fluids with a viscosity higher than water ( see reference 1 and the web site www.fluidedesign.com).

2 • 13

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

EXAMPLE 2.1 – CALCULATE THE TOTAL HEAD FOR A TYPICAL PUMPING SYSTEM

Figure 2-9 An example of a calculation for Total Head of a typical pumping system. Our first example will take us through the ins and outs of a Total Head calculation for a simple pumping system. The fluid is water at 60 °F. The equation for Total Head is:

∆HP = ∆HF1−S + ∆HFP−4 + ∆HF 4−2 + ∆HEQ1−S + ∆HEQP−4 + ∆HEQ4−2 +

1 2 2 (v2 − v1 ) + z2 + H2 − (z1 + H1) 2g

Point 4 is not essential to solve this problem but required for comparison purposes in example 2.2 where a branch is added at point 4. Friction Head Difference Pipe & Fittings Since the pipe diameter changes between points 1 and 2, it is necessary to determine the friction occurring between points 1 and S and points P and 2. POINT 1 TO S Pipe friction loss between points 1 and S.

∆H F 1− 2 = ∆H F 1− S + ∆H FP − 2 The friction loss term ∆HF1-S is made up of fluid friction and fittings friction.

2 • 14

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

Pipe friction = Fluid friction (pipe) + Fluid friction (fittings)

∆ H F 1− S = ∆ H FP 1− S + ∆ H FF 1− S and ∆ H FP − 2 = ∆ H FPP − 2 = ∆ H FFP − 2 where ∆HFP1-S is the fluid friction between points 1 and S and ∆HFF1-S is the friction produced by fittings (for example elbows, isolation valves, etc.) between the same two points. From the tables (reference 1 or 8). for a 8" dia. Pipe and a flow rate of 500 USGPM: ∆HFP1-S/L = 0.42 ft/100 ft of pipe.

∆H FF 1− S =

∆H FP L1− S 8 × = 0.42 × = 0.03 ft L 100 100

Fittings Friction Loss between points 1 and S In this portion of the line there is (1) 8" butterfly valve, from the tables (reference 1 or 8) K = 0.25

∆H FF 1− S

Kv S 2 0.25 × 319 . 2 = = = 0.04 ft 2g 2 × 32.17

The pipe and fittings friction loss between points 1 and S is: ∆HF1-S = ∆HFP1-S + ∆HFF1-S = 0.03 + 0.04 = 0.07 ft POINT P TO 4. Pipe friction loss between points P and 4. From the tables (reference 1 or 8), for a 4" dia. Pipe at a flow rate of 500 USGPM: ∆HFPP-4 /L= 13.1 ft/100 ft of pipe.

∆ H FPP − 4 =

L ∆ H FP 18 × P − 4 = 131 = 2.3 ft . × L 100 100

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

2 • 15

Fittings Friction Loss between points P and 4 In this portion of the line there is: 4” dia. elbow, from the tables (reference 1 or 8) K = 0.25; (1) 4”Φ isolation (butterfly) valve; from the tables (reference 1 or 8) KVAL = 0.25

∆H FFP − 4

( K ELB + KVAL ) × v 4 2 ( 0.25 + 0.25) × 12.76 2 = = = 1.3 ft 2g 2 × 32.17

The pipe and fittings friction between points P and 4 is: ∆HFP-4 = ∆HFPP-4 + ∆HFFP-4 = 2.3 + 1.3 = 3.6 ft POINT 4 TO 2 Pipe Friction Loss between points 4 and 2. From the tables (reference 1 or 8), for a 4"dia. pipe at a flow rate of @ 500 USGPM: ∆HFPP-2 /L= 13.1 ft/100 ft of pipe.

∆ H FP 4 − 2 =

∆ H FP L4 − 2 114 × = 131 . × = 14.9 ft 100 100 L

Fittings Friction Loss between points 4 and 2 In this portion of the line there is (2) 4"dia. elbows, from the tables (reference 1 or 8) KELB = 0.25.

∆H FF 4 − 2

2 × K ELB × v 2 2 2 × 0.25 × 12 .76 2 = = = 1.3 ft 2g 2 × 32 .17

The pipe and fittings friction between points 4 and 2 is: ∆HF4-2 = ∆HFP4-2 + ∆HFF4-2 = 14.9 + 1.3 = 16.2 ft

2 • 16

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

Equipment Pressure Head Difference POINT 1 TO S There is no equipment in this portion of the line. ∆HEQ1-S = 0 POINT P TO 4 The filter between point P and 4 is considered a piece of equipment. The filter manufacturer's technical data states that this filter will produce a 5 psi pressure drop at a flow rate of 500 USGPM. For water 5 psi is equivalent to 11.5 ft of head (5 X 2.31, see equation [1-5]). ∆HEQP-4 = ∆HEQFIL = 11.5 ft POINT 4 TO 2 The control valve between point P and 4 is considered a piece of equipment. The control valve will have a head drop of 10 ft of fluid. The justification for this will be given later in Chapter 3. ∆HEQ4-2 = ∆HEQCV = 10 ft Velocity Head Difference The velocity head difference between points l and 2 is zero, since v1 and v2 are very small.

∆H v =

1 ( v 2 − v1 2 ) = 0 2g 2

Static Head Both tanks are not pressurized therefore H1 = H2= 0. The elevation difference z2 - z1 = 137 - 108 = 29 ft

2 • 17

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

Total Head Once again the equation for Total Head is:

∆HP = ∆HF1−S + ∆HFP−4 + ∆HF4−2 + ∆HEQ1−S + ∆HEQP−4 + ∆HEQ4−2 +

1 2 2 (v2 − v1 ) + z2 + H2 − (z1 + H1) 2g

By substituting the values previously calculated into the above equation: ∆HP= 0.07 + 3.6 + 16.2 + 0 + 11.5 + 10+ 0 + 137 + 0 - (108 + 0 ) = 70.4 ft The result of the calculation indicates that the pump will require a Total Head of 70.4 ft to deliver 500 USGPM. How can we guarantee that the pump will deliver 500 USGPM at 70 ft of Total Head? The answer is in Chapter 4, section 4.5.

2 • 18

2.10

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

METHOD FOR DETERMINING THE PRESSURE HEAD AT ANY LOCATION The system configuration in Figure 2-10 illustrates how drastically the pressure can vary in a simple pumping system. Why do we need to calculate the pressure at a specific location in the system? One reason is the pressure head just before the control valve is a parameter required to size the valve. It is calculated by the method described below. In addition, the system may have other equipment where the inlet pressure must be known to ensure proper operation.

Figure 2-10 The pressure head variation within a typical pumping system. A. The pressure head at any location on the discharge side of the pump. First, calculate the Total Head (∆HP) for the complete system using equation [2-12]. Then draw a control volume (see Figure 2-11) to intersect point X and point 1. Point X can be located anywhere between P and 2. The system equation [2-12] is then applied with point X as the outlet instead of point 2 and solved for the unknown variable Hx instead of ∆HP.

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

2 • 19

Figure 2-11 Using control volumes to determine the pressure head anywhere after the pump discharge From equation [2-12], the Total Head for the complete system is:

∆HP = ∆HF1−2 + ∆HEQ1−2 +

1 2 2 (v2 − v1 ) + z2 + H2 − (z1 + H1) 2g

Equation [2-12] is applied with the following changes: all terms with subscript 2 are replaced with the subscript X, H2=HX, ∆HF1-2 = ∆HF1-X , ∆HEQ1-2 = ∆HEQ1-X, v2 = vX, z2 = zX. The unknown term Hx is isolated on one side of the equation:

H X = ∆H P − (∆H F1− X + ∆H EQ1− X ) +

1 2 2 (v1 − vX ) + z1 + H1 − z X 2g

[2-13]

The unknown pressure head Hx can also be determined by using that part of the system defined by control volume 2 (see Figure 2-11) by using the same reasoning as above, in equation [2-12], ∆HP = 0 and all terms with subscripts 1 replaced with X. Therefore, H1=HX, ∆HF1-2 = ∆HFX-2 , ∆HEQ1-2 = ∆HEQX-2, v1= vX, z1 = zX

H X = (∆H FX −2 + ∆H EQX −2 ) +

1 2 2 (v2 − vX ) + z2 + H2 − zX 2g

[2-14]

2 • 20

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

We now have two methods for determining the pressure head at any location. We can use one equation to verify the results of the other. The calculation for HX is often quicker using equation [2-14]. Note that in equation [2-14] the value of ∆HP is not required to calculate Hx. B. The pressure head at any location on the suction side of the pump

Figure 2-12 Using control volumes to determine the pressure head anywhere before the pump inlet. We can determine the pressure head anywhere on the suction side of the pump by the same method. In this case, ∆HP = 0 since there is no pump within the control volume (see Figure 2-12). ∆HP = 0 and subscript 2 is replaced with subscript X in equation [2-12]. We obtain:

H X = −(∆H F 1− X + ∆H EQ1− X ) +

1 2 2 (v1 − v X ) + z1 + H1 − z X 2g

[2-15]

The velocity vX must be the same as would occur at point X in the complete system. In chapter 3, we will use equation [2-15] to calculate the available Net Positive Suction Head of the pump.

2 • 21

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

EXAMPLE 2.2 – CALCULATE THE PRESSURE HEAD AT THE INLET OF A CONTROL VALVE

Figure 2-13 Example of the calculation of the pressure head at the inlet of a control valve. It is often required to know the pressure head just ahead of a piece of equipment such as a control valve. We will apply the methods in the previous section to calculate the pressure head just ahead of the control valve in example 2.1. CALCULATE THE PRESSURE HEAD AT POINT 7 USING EQUATION [2-13] Applying equation [2-13] where point X is now point 7 (X = 7) and ∆HP = 70.4 from the calculation in example 2.1.

H 7 = 70.4 − (∆H F1−S + ∆H EQ1− S ) − (∆H FP−7 + ∆H EQP−7 ) + Friction Head Difference - Pipe and Fittings POINT 1 TO S From example 2.1: ∆HF1-S = 0.07 ft

1 2 2 (v1 − v7 ) + z1 + H 1 − z 7 2g

2 • 22

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

POINT P TO 7 ∆HFP-7 = ∆HFPP-7 + ∆HFFP-7 Pipe friction loss between point P and 7. From the tables (reference 1 or 8), for a 4"dia. at a flow rate of 500 USGPM: ∆HFPP-7 /L= 13.1 ft/100 ft of pipe.

∆ H FPP − 7 =

∆ H FP L P − 7 125 . × × = 131 = 16.4 ft L 100 100

Fittings Friction Loss between points P and 7 In this portion of the line there is: (3) 4” dia. elbow, KELB = 0.25; (1) 4” dia. isolation (butterfly) valve, KVAL = 0.25.

∆H FFP − 7

( 3 × K ELB + KVAL ) × v 7 2 ( 3 × 0.25 + 0.25) × 12.6 2 = = = 2.5 ft 2g 2 × 32.17

The pipe and fittings friction between points P and 7 is: ∆HFP-7 = ∆HFPP-7 + ∆HFFP-7 = 16.4 + 2.5 = 18.9 ft. Equipment Pressure Head Difference POINT 1 TO S ∆HEQ1-S = 0 POINT P TO 7 The filter between point P and 4 has a pressure head drop of: ∆HEQP-7 = ∆HEQFIL = 11.5

2 • 23

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

Velocity Head Difference The velocity head difference between points 1 and 7 is:

∆H v =

1 1 ( v1 2 − v 7 2 ) = (0 − 12.76 2 ) = −2.5 ft 2g 2g

Static Head The suction tank is not pressurized, therefore H1 = 0. The elevation difference z1 - z7 = 108 - 140 = -32 ft The pressure head at point 7 Again the equation for the pressure head at point 7 is:

H 7 = 70.4 − ( ∆H F 1− S + ∆H EQ1− S ) − ( ∆H FP − 7 + ∆H EQP − 7 ) +

1 2 2 ( v1 − v 7 ) + z 1 + H 1 − z 7 2g

By substituting the values calculated into the above equation we obtain: H7 = 70.4 - 0.07-(18.9 + 11.5) - 2.5 + 108 + 0 -140 = 5.4 ft CALCULATE THE PRESSURE HEAD AT POINT 7 USING EQUATION [2-14] This is a good opportunity to verify our calculation for H7 using equation [2-14]. Substituting 7 for X, the equation is:

H 7 = ( ∆H F 7 − 2 + ∆H EQ 7 − 2 ) +

Friction Head - Pipe and Fittings POINT 7 TO 2 ∆HF7-2 = ∆HFP7-2 + ∆HFF7-2

1 2 2 (v 2 − v 7 ) + z 2 + H 2 − z 7 2g

2 • 24

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

Pipe friction loss between point 7 and 2. From the tables (reference 1 or 8), for a 4" dia. at a flow rate of 500 USGPM: ∆HFP7-2 /L= 13.1 ft/100 ft of pipe.

∆ H FP 7 − 2 =

∆ H FP L7 − 2 7 × = 131 . × = 0.92 ft L 100 100

Fittings Friction Loss between points 7 and 2 In this portion of the line, there is no fittings. Therefore ∆HFF7-2 = 0 The pipe and fittings friction between points 7 and 2 is: ∆HF7-2 = ∆HFP7-2 + ∆HFF7-2 = 0.92 + 0 = 0.92 ft. Equipment Pressure ahead Difference POINT 7 TO 2 As previously stated, we allow a pressure head drop of 10 ft for the control valve. ∆HEQ7-2 = 10 ft Velocity Head Difference The velocity head difference between points 7 and 2 is:

∆Hv =

1 1 (v22 − v72 ) = ( 0 − 1 2 .7 6 2 ) = − 2 .5 f t 2g 2g

Static Head The discharge tank is not pressurized, therefore H2 = 0. The elevation difference z2 - z7 = 137 - 140 = -3 ft

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

The pressure head at point 7 Again the equation for the pressure head at point 7 is:

H 7 = ( ∆H F 7 − 2 + ∆H EQ 7 − 2 ) +

1 2 2 (v 2 − v 7 ) + z 2 + H 2 − z 7 2g

By substituting the values calculated into the above equation: H7 = 0.9 + 10 - 2.5 + 137 + 0 - 140 = 5.3 ft - the same result as previously.

2 • 25

2 • 26

2.11

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

SYSTEM OR TOTAL HEAD EQUATION FOR A SINGLE INLET - DOUBLE OUTLET SYSTEM

Figure 2-14 The use of control volumes to determine the Total Head for a single inlet – double outlet system. Industrial systems often have more than one outlet (see Figure 2-14). Equation [2-4] is used as the basis for the analysis of the system using the principle of conservation of mass flow rate for a steady state system.

Q& E − W& = ∆ E& n + ∆ K& E + ∆ P& E

m& 1 = m& 2 + m& 3 The same logic used to deduce equation [2-12] can be used for a system with multiple inlets and outlets. Q& E is the sum of all the friction heat losses and equipment losses. W& is the work supplied by the pump. ∆E& n is the sum of the inlet minus the sum of the outlet

& is the sum of the inlet minus the sum of the outlet kinetic energies. enthalpies. ∆KE & is the sum of the inlet minus the sum of the outlet potential energies. The energy ∆PE balance is:

(∆p ρ

m& 1

F 1− 4

)

+ ∆p EQ1− 4 +

( ∆p ρ

m& 2

F 4−2

)

+ ∆p EQ 4 − 2 +

( ∆p ρ

m& 3

F 4−3

)

+ ∆p EQ 4 − 3 −

m& 1

ρ

∆p =

 m&  m&  m& g  m& m& m& 1  m& g  m& p1 −  2 p2 + 3 p3  + 1 v1 2 −  2 v 2 2 + 3 v 3 2  + 1 z1 −  2 z 2 + 3 z 3  gc  ρ ρ ρ  ρg c ρ ρ  gc  ρ  ρ  ρg c

m& 1

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

2 • 27

replacing all pressure terms by their corresponding fluid column height (for example

p1 =

ρgH1 gc

, ∆p F 1− 4 =

ρg∆H F 1− 4

, etc.), and all mass flow terms by their corresponding

gc & 1 = ρq1 etc., we obtain: volumetric flow rate q, m

q1 ∆ H P = q1 ( ∆ H F 1− 4 + ∆ H EQ 1− 4 ) + q 2 ( ∆ H F 4 − 2 + ∆ H EQ 4 − 2 ) + q 3 ( ∆ H F 4 − 3 + ∆ H EQ 4 − 3 ) − q1 H 1 + ( q 2 H 2 + q 3 H 3 ) +

1 2 2 2 (−q1 v1 + (q 2 v 2 + q 3 v 3 )) − q1 z1 + (q 2 z 2 + q 3 z 3 ) 2g

and since q1 = q2 + q3, the Total Head is:

∆H P = (∆H F 1− 4 + ∆H EQ1− 4 ) + (∆H F 4 − 2 + ∆H EQ 4 − 2 ) +

1 2 2 (v 2 − v1 ) + ( z 2 + H 2 − ( z1 + H 1 )) + 2g

q3 1 2 2 (∆HF 4−3 + ∆HEQ4−3) − (∆HF4−2 + ∆HEQ4−2 ) + (v3 − v2 ) + (z3 + H3 − (z2 + H2 )) q1 2g

[2-16]

For equation [2-16] to be true, the friction loss in the major branches has to be adjusted by closing the manual valves to balance the flows. The term EQ2 represents equipment on branch 2 such as a valve (see Figure 2-14). Closing the valve, reduces the flow at branch 2 but increase it at branch 3. The conditions in one branch affects the other because of the common connection point 4. The correct flow is obtained at point 3 by adjusting components in branch 2 (for example ∆HF4-3, ∆HEQ4-3 and H4). One would think that the main branch 2, the branch with the highest head and flow requirement, should determine the pump Total Head. So, why is it that the head requirements for branch 3 appear in equation [2-16]? All of the terms associated with branch 3 can be made to disappear after simplification. Determine the value of H4 using equation [2-14] and a control volume that surrounds branch 3:

H 4 = ( ∆ H F 4 − 3 + ∆H EQ 4 − 3 ) +

1 (v 2 − v 4−3 2 ) + ( z3 + H 3 − z4 ) 2g 3

or

( ∆H F 4 − 3 + ∆H EQ 4 − 3 ) = H 4 −

1 (v 3 2 − v 4 − 3 2 ) − ( z3 + H 3 − z4 ) 2g

We can also determine the value of H4 with equation [2-14] and a control volume surrounding branch 2:

[2-17]

2 • 28

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

H4 = ( ∆H F 4−2 + ∆H EQ4−2 ) +

1 (v2 2 − v4−2 2 ) + ( z2 + H2 − z4 ) 2g

or

( ∆H F 4 − 2 + ∆H EQ 4 − 2 ) = H 4 −

1 (v 2 − v 4 − 2 2 ) − ( z2 + H 2 − z4 ) 2g 2

[2-18]

v4-3 and v4-2 are respectively the velocity at point 4 for branch 3 and branch 2, substituting equations [2-17] and [2-18] into equation [2-16] we obtain:

∆H P = ( ∆H F 1− 4 + ∆ H EQ1− 4 ) + ( ∆ H F 4 − 2 + ∆ H EQ 4 − 2 ) +

+

1 ( v 2 2 − v1 2 ) + ( z 2 + H 2 − ( z1 + H 1 )) 2g

q3 1 2 2 ( v 4 − 3 − v 4 − 2 )) ( q1 2 g

[2-19]

Point 4 is the point at which the flow begins to divide and the fluid particles move towards the direction of branch 2 or branch 3. All fluid particles have the same velocity at point 4 although their trajectory is about to change.

v4 = v4 −3 = v4−2

[2-20]

When we replace equation [2-20] into [2-19] then the last term in [2-19] disappears and we obtain equation [2-21] which is the equation for a single branch system. This means that the branch has no impact on the Total Head as long as branch 2 represents the path requiring the highest energy. Figure 2-15 Location of H4.

∆HP = ∆HF1−2 + ∆HEQ1−2 +

1 2 2 (v − v ) + z2 + H2 − (z1 + H1 ) 2g 2 1

[2-21]

In many cases the path requiring the highest energy is obvious. What happens if the flow in branch 3 is equal to the flow in branch 2? Alternatively, if the flow is much smaller in branch 3, but the elevation of point 3 is higher than point 2?

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

2 • 29

These situations make judging which branch will produce the higher head difficult to do at a glance. To resolve this, calculate the Total Head for the fluid traveling through either branch and compare the results. The Total Head for the pump as calculated considering the flow through branch 3 is:

∆HP = ∆HF1−3 + ∆HEQ1−3 +

1 (v3 2 − v12 ) + z3 + H3 − (z1 + H1 ) 2g

[2-22]

For the system to work as intended, the highest Total Head as calculated from equations [2-21] or [2-22] will be used for sizing the pump. 2.12

GENERAL METHOD FOR DETERMINING TOTAL HEAD IN A SYSTEM WITH MULTIPLE INLETS AND OUTLETS The energy rate balance for any system is: Q& E − W& = ∆E& n + ∆K& E + ∆P& E Figure 2-16 represents a system with (n) inlets and (m) outlets.

Figure 2-16 A general system with a single pump and multiple inlets and outlets. In the system shown in Figure 2-16 there are 1 to p branches. These subscripts are used to identify the flow rates (for example q1, q2, ..qp) There are li to ni inlets, which are used to identity the inlet properties (for example v1i, H1i, z1i, v2i, H2i, z2i, ... vni, Hni, zni). Similarly, there are 1o to mo outlets that identify the outlet properties.

2 • 30

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

Figure 2-17 The use of the control volume with a generalized system representation.

There is one path, which for a given flow rate, will require the highest Total Head. Often, this path is obvious, but several paths may have to be checked to find the most critical. The following equations are all based on the general principle for energy rates in a fluid medium:

m& ENERGY RATE = v ∆ F = V&∆ P = ∆p ∝ q∆H

ρ

The friction term Q& E is:

q BR 1 ( ∆ H F 1 + ∆ H EQ 1 ) + q BR 2 ( ∆ H F 2 + ∆ H EQ 2 ) + ...+ q BRp ∆ H P =

t= p

∑ t =1

q t ( ∆ H Ft + ∆ H EQt

where the subscripts 1 to p stand for branch no.1, 2, etc,. up to branch no. p. The work rate term W& is:

q P ∆H P

2 • 31

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

The enthalpy rare term ∆E& n is: t =n

t =m

t =1

t =1

q1i H1i + q2i H 2i + ...+ qni H ni − (q1o H1o + q20 H 2o +...+ qmo H mo ) = ∑ qti H ti − ∑ qto H to ) & is: The kinetic energy rate term ∆KE t =m 1 1  t =n  (q1i v 21i + q2i v 2 2i + ...+ qni v 2 ni − (q1o v 21o + q20v 2 2o +...+ qmov 2 mo ) =  ∑ qti v 2 ti − ∑qtov 2 to )  2g 2g  t =1 t =1 

(

)

& is: The potential energy rate term ∆PE t =n

t =m

t =1

t =1

q1i z1 i + q 2i z 2i + ...+ q ni z ni − ( q1o z1o + q 20 z 2 o +...+ q mo z mo ) = ∑ q ti z ti − ∑ q to z to The Total Head for the system ∆H P is: t=p

qP∆HP = ∑qt (∆HFt + ∆HEQt) + t =1

t =m 1  t =n 2 t =m 2  t =n ∑qtiv ti − ∑qtov to  + ∑qti (zti + Hti ) − ∑qto(zto + Hto) 2g  t =1 t =1 t =1  t =1

[2-23]

This is a similar equation to [2-16] with the difference that there are many inlet and outlet branches. As in equation [2-16], this equation requires that the system parameters that satisfy the flow balance be known. 2.13

GENERAL METHOD FOR DETERMINING TOTAL HEAD IN A SYSTEM WITH MULTIPLE PUMPS, INLETS AND OUTLETS Inlets and Outlets vs. Internal Connection Points

One of the purposes of an open system is to move material from one point to another. A more complex system may use several inlets to move material to several outlets. Only the inlets and outlets of a system have thermodynamic properties, therefore the velocity, elevation and pressure at these points are properties critical to the system. Any internal connection points with branches used to re-circulate fluid within the system do not have thermodynamic properties as regards to the system. Figure 2-18 The effect of internal branches within a system.

2 • 32

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

These branches are required for the system to function properly. For example, take a single inlet-single outlet system with a re-circulation line from the discharge to the suction of the pump (see Figure 2-18). How does this re-circulation branch affect the system if the same flow is required at point 2 with or without the re-circulation branch? The energy rate balance is: Q& E − W& = ∆E& n + ∆K& E + ∆P& E

Q& E is affected in the above equation since there is more friction due to the extra branch. W& is also affected since the pump will require a greater capacity to compensate for the extra fluid that must go through the branch. The effect of the extra branch is to increase the total friction of the system and therefore the work to be provided by the pump. Multiple Pumps

The system in Figure 2-19 appears at first glance impossible to solve or at the least challenging. This is the same system as in Figure 2-17 with additional pumps and branches. Before doing any Total Head calculations, determine the flow rates for each branch. The flow through each one of the branches and pumps must be known and this requires knowledge of the purpose of each pump. Each pump will move fluid through its respective branches at its design flow rate. To determine the Total Head of pump A in Figure 2-19, in your mind disconnect all pumps and branches that are not intended to be powered by pump A and apply equation [2-23].

Figure 2-19 The use of the control volume in a multiple pump generalized system. We now have a much simpler system as shown in Figure 2-20. This is a similar system to the one shown in Figure 2-17 that can be resolved with equation [2-23]. All the connecting points (c1 to c5) are points where additional flow enters or leaves some part of the system. These points are internal to the system, and not inlet or outlet points as point

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

2 • 33

1i..ni and lo...mo. The branches left behind with pump A contain all the inlets and outlets of the system. In a way, we could say that pump A is essential to the system since all the inlets and outlets are attached to pump A. The other pumps and their branches form subsystems of the main system.

Figure 2-20 Simplifying complex systems. To determine the Total Head of pump B in Figure 2-20, first calculate the head at points c2 and c4 (see equation [2-15] for example) and then apply equation [2-23]. The Total Heads of pump C and D are similarly determined.

2 • 34

2.14

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

GENERAL METHOD FOR DETERMINING THE PRESSURE HEAD ANYWHERE IN A SYSTEM WITH MULTIPLE PUMPS, INLETS AND OUTLETS First, isolate each pump and its respective branches as in the previous section and calculate the Total Head of each pump. Assuming point X to be located somewhere after pump A in pump A's system, then the pressure head at point X is: t =BRp

t =n 1 t=n 2 t=m 2 QX∆HX =−QPA∆HPA + ∑Qt (∆HFt + ∆HEQt) + ∑Qtivti −∑QXvX  +∑Qti(zti + Hti) −QX zX 2gt=1 t =BR1 t =1  t=1

[2-24]

if point X is before pump A then the same equation applies with ∆HPA = 0. To calculate the pressure head at points w, y, and z, identify their respective sub-systems and apply equation [2-13] if they are single inlet- single outlet systems or equation [2-24]. This chapter was a mouthful. The shock of dealing with thermodynamic properties is compensated by the fact that they are quite easily determined. The thermodynamic properties at the inlet of a system are the enthalpy, the kinetic and potential energies. These are respectively proportional to: pressure, velocity and elevation. The heat loss is the sum of all the losses that occur due to fluid movement through the system and the effect of equipment. Whatever is left over between the total thermodynamic energy inletoutlet difference and the heat loss is the work required by the pump. When, we have calculated the work required of the pump or its Total Head. A general method for determining the Total Head for a single inlet-outlet system, leads to a method for determining the pressure head anywhere within the system. So that now, when the instrumentation guy asks what is the pressure before the control valve and what is the delta p, give it to him. A general method for calculating the Total Head in a single inlet-outlet system leads to a method for calculating the Total Head in a multiple inlet and outlet system. The heat or friction loss is calculated the same way no matter how many branches there is in a system. The difference between the heat loss and the net thermodynamic energy is of course the work required by the pump. Once more on to the breech… Since we came this far, might as well go all out. The next step was to find a method to solve multiple inlet and outlet systems with multiple pumps, because when pumps are added to a system, they are put there for a reason, Oh, nooooh, you say. It is always possible to convert a multiple pump system to a single pump system. The pumps and branches that are extracted can be analyzed separately without affecting the single pump system assuming that you know or calculate the pressure head, velocity and elevation of the connecting points.

THE COMPONENTS OF TOTAL HEAD This chapter will introduce some of the terminology used in pumping systems. The components of Total Head will be examined one by one. Some of the more difficult to determine components, such as equipment and friction head, will be examined in more detail. I hope this will help get our heads together. 3.0

THE COMPONENTS OF TOTAL HEAD Total Head is the measure of a pump's ability to push fluids through a system. Total Head is proportional to the difference in pressure at the discharge vs. the suction of the pump. It is more useful to use the difference in pressure vs. the discharge pressure as a principal characteristic since this makes it independent of the pressure level at the pump suction and therefore independent of a particular system configuration. For this reason, the Total Head is used as the Y-axis coordinate on all pump performance curves (see Figure 4-3). The system equation for a typical single inlet — single outlet system (see equation [212]) is:

∆HP = ∆HF1−2 + ∆HEQ 1−2 + 1 (v 2 −v1 ) + z2 + H2 − (z1 + H1) 2g 2

2

[3-1]

∆H P = ∆H F + ∆H EQ + ∆Hv + ∆HTS

[3-1a]

∆H P = ∆H F + ∆H EQ + ∆Hv + ∆H DS + ∆H SS

[3-1b]

Equations [3-1a] and [3-1b] represent different ways of writing equation [3-1], using terms that are common in the pump industry. This chapter will explain each one of these terms in details. 3.1

TOTAL STATIC HEAD (∆HTS) The total static head is the difference between the discharge static head and the suction static head, or the difference in elevation at the outlet including the pressure head at the outlet, and the elevation at the inlet including the pressure head at the inlet, as described in equation [3-2a].

∆HTS = ∆H DS − ∆H SS = z 2 + H 2 − ( z 1 + H 1)

[3-2] [3-2a]

H2 and H1 are the pressure heads at points 2 and 1 respectively. Some people include these pressure heads with the elevation head, others do not. I will be using the former approach, but either way the pressure heads have to be considered. The discharge static head (∆HDS) is normally positive (assuming H2 = 0 ), since fluid is usually pumped

3•2

THE COMPONENTS OF TOTAL HEAD

to a higher elevation. However, it may on occasion be negative (for example, lower discharge pipe end than pump centerline) and the complete system must be evaluated before it is known if a pump is required or not. The suction static head (∆HSS) can either be negative or positive, depending on whether the pump centerline is below the suction fluid surface or above, and the value of suction tank fluid surface pressure head (H1).

Figure 3-1 Relationship between the various heads in a pumping system. 3.2

SUCTION STATIC HEAD (∆HSS) The Suction Static head is the sum of the elevation and pressure head at the inlet of the system, minus the elevation of the pump center line, as stated in equation [3-3]. The inlet of the system is located at point 1, which is the surface of the suction tank fluid. H1 is the pressure head at the suction tank fluid surface. If the tank is open to atmosphere then H1 = 0. Typical pumping configurations are shown in Figure 3-2, the pump suction is under positive pressure in A, and possibly under negative relative pressure in B.

3•3

THE COMPONENTS OF TOTAL HEAD

The Suction Static head in these two cases is:

∆ H SS = z 1 + H 1 − z S

[3-3]

Figure 3-2B presents a situation where the pump has to lift the fluid up to the pump suction. The head at the suction is described as suction lift. This head is normally negative with respect to atmospheric pressure since the term z1 – zS is negative (assuming H1 = 0 or the same as the atmospheric pressure).

Figure 3-2 Suction static head and suction static lift.

3.3

NET POSITIVE SUCTION HEAD AVAILABLE (N.P.S.H.A.) The Net Positive Suction Head Available (N.P.S.H.A.) is the total energy per unit weight, or head, at the suction flange of the pump less the vapor pressure head of the fluid. This is the accepted definition which is published by the Hydraulic Institute’s Standards books (see the HI web site at www. pumps.org). The Hydraulic Institute is the organization that formulates and promotes the use of common standards used throughout the pump industry in the United States. The term "Net" refers to the actual head at the pump suction flange, since some energy is lost in friction prior to the suction. Why do we need to calculate the N.P.S.H.A.? This value is required to avoid cavitation of the fluid. Cavitation will be avoided if the head at the suction is higher than the vapor pressure head of the fluid. In addition, the pump manufacturers require a minimum N.P.S.H. to guarantee proper operation of the pump, they call this the N.P.S.H.R., where “R” stands for required.

3•4

THE COMPONENTS OF TOTAL HEAD

To determine N.P.S.H.A., first we calculate the pressure head HS at point S. A control volume is positioned (see Figure 3-3) to intersect the suction inlet of the pump and the fluid surface of the suction tank. The pressure head at any point on the suction side of the pump is given by equation [2-15] where the subscript X is replaced by S:

(v1 −vS ) HS( ft of fluid ) = − (∆HF1−S +∆HEQ 1−S ) + + (z1 − z S +H1) 2g 2

2

[3-4]

Figure 3-3 Using the control volume for calculating the pressure head at point S.

The specific energy or head E for any point in the system is the sum of the elevation (potential) energy, the velocity (kinetic) energy and the pressure energy. E is given by: 2

E =H + v + z 2g

[3-5]

By definition, the N.P.S.H available at the pump suction (point S) is based on a reference plane located at the pump suction centerline (z = 0 ). We can understand why since using any other reference will increase or decrease the energy level at point S which is obviously incorrect (see Figure 3-3). Therefore equation [3-5] becomes: 2

ES = H S + v S 2g

[3-6]

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THE COMPONENTS OF TOTAL HEAD

The head ES is given in equation [3-6], the barometric head (HB) is added to HS to convert ES from feet of fluid to feet of fluid absolute. Therefore equation [3-6] becomes: 2

ES ( ft of fluid absol .) = H S + v S + H B 2g

[3-7]

The value of HS in equation [3-4] is substituted in equation [3-7] to give: 2

v1 ES ( ft of fluid absol. ) = − (∆HF1−S +∆HEQ1−S ) + + ( z1 − z S +H1) + HB 2g

[3-8]

BOILING LIQUIDS

Different liquids boil at different temperatures for a fixed pressure; also, different liquids boil at different pressures at a fixed temperature. The temperature required to vaporize a liquid varies as the pressure in the surrounding environment. For example, water boils at a temperature of 212 0F at a surrounding pressure of 14.7 psia (the air pressure at sea level). However, a temperature of 189 0F is required to boil water at a pressure of 11 psia which is the atmospheric pressure at 8,500 feet of elevation above sea level (or the altitude of Mexico city). A short digression is in order. Since water boils at a lower temperature in, say, Mexico City than a city which is close to sea level, does this mean that it takes a longer time to boil an egg in Mexico city? Yes, it will take longer in Mexico City. Why? Because the same amount of heat transfer is required to get the egg to the right consistency regardless of the water temperature. It will take longer to transfer the amount of heat required to cook the egg if the water is boiling at a lower temperature. For most of us water boils at the high temperature of 212 0F (100 0C), it is very surprising to find that it takes 4 minutes to boil a “3-minute” egg in Mexico City. The pressure at which a liquid boils is called the vapor pressure and is always associated with a specific temperature. When pressure decreases in the fluid's environment, the boiling temperature drops. Many liquids (i.e. acetone, methyl alcohol, benzene, etc.) have a lower vapor pressure than water at the same temperature. Since the pressure throughout a system can vary drastically, it is important to consider the vapor pressure of the liquid in order to avoid vaporization. Data on vapor pressure vs. temperature for many liquids is readily available (see reference 1, 2 and 8).

3•6

THE COMPONENTS OF TOTAL HEAD

VAPOR PRESSURE AND CAVITATION

The pressure near the impeller eye is lower than the pressure at the pump suction flange, and depending on the kind of fluid and temperature, may be low enough for vaporization to occur. When this happens, both, vapor and liquid, will enter the pump, and the capacity of the pump will be reduced. The point of lowest pressure is near the eye of the impeller on the underside of the vane (see Figure 3-6), where bubbles can form. Only small bubbles are formed because the fluid is rapidly compressed as it travels from the start of the impeller vane to its tip. The rapid compression of bubbles causes small pieces of metal to be dislodged from the surface. These bubbles collapse rapidly in the high pressure near the tip of the vane causing noise and vibration. This rapid collapse of vapor bubbles is known as cavitation and is accompanied by a distinct gravely sound similar to the sound made by a cement mixer. The system should be designed in such a way as to provide sufficient N.P.S.H. available to avoid cavitation under normal running conditions.

Figure 3-4 Vapor pressure vs. temperature.

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THE COMPONENTS OF TOTAL HEAD

In order for the liquid to stay in a fluid state and not vaporize, the head at the inlet of the pump must be above the vapor pressure head of the fluid:

E S ≥ H va where Hva is the vapor pressure head of the liquid. The Net Positive Suction Head Available (N.P.S.H.A.) is the difference between the head ( ES ) at the pump suction and the vapor pressure head (Hva).

N.P.S.H.avail. = ES − H va

[3-9]

By substituting the value of ES from equation [3-8] into equation [3-9] then: 2

N.P.S.H.avail.(ft of fluid absol. ) =− (∆H F1− S +∆H EQ1− S )+ v1 +(z 1 − z S + H1)+ H B −H va 2g

[3-10]

where HB and Hva are in feet of fluid. Vapor and barometric pressures are often given in pounds per square inch absolute (psia). The conversion to feet of fluid absolute is: H ( ft of fluid ) =

2 .31 × p ( psia ) SG

by substitution into equation [3-10]: 2

N.P.S.H.avail.(ft of fluid absol. ) =− (∆H F1− S +∆H EQ1− S )+ v1 +(z 1 − z S + H1)+ H B − Hva + 2.31(p B (psia)− pva(psia)) 2g SG

[3-11]

The N.P.S.H. in equation [3-10] and [3-11] is in feet of fluid absolute and is a head term, which is independent of fluid density. Since the pump manufacturers use water as the fluid, the N.P.S.H. value they provide is in feet of water absolute. The pump requires a minimum suction pressure head in order to function properly and avoid cavitation. This is known as the N.P.S.H. required, which the pump manufacturer gives for a specific pump model, impeller diameter, speed and flow rate. In order to satisfy the pump manufacturer's requirements:

N.P.S.H.avail. ≥ N.P.S.H.req.

3•8

THE COMPONENTS OF TOTAL HEAD

Figure 3-5 shows typical relative proportions of the terms in equation [3-10].

Figure 3-5 Relative size of N.P.S.H. components.

THE COMPONENTS OF TOTAL HEAD

3•9

N.P.S.H. REQUIRED

The N.P.S.H. required provides us with the level of head in terms of feet of water absolute required at the pump suction flange. When that level of head is insufficient the capacity and head of the pump will drop and cavitation will occur. Figure 3-6 shows how the pressure varies between the pump suction flange and the discharge flange.

Figure 3-6 The pressure variation within a pump (source: the Durco company). Let’s follow the path of the fluid particles form point A to E in Figure 3-6 to see where and why the pressure is varying. There is a small friction loss in the short piece of piping of the pump suction. Both, friction loss and turbulence, accompany the right angle turn that the liquid makes in going from horizontal flow to outward radial flow in the impeller. The leading edges of the vanes act as obstruction in the path of the liquid, and entrance loss occurs at each vane edge. Any pre-rotation of the liquid as it enters the impeller, changes the inlet angle and results in more turbulence. All of these losses occur before the liquid is acted upon positively by the impeller vane. The combination of these losses, plus the losses that occurred in the suction line, can be great enough to lower the pressure of the fluid sufficiently to reach the vapor pressure, at which point the fluid will vaporize. This condition must be avoided. Once the liquid is in the impeller, with the

3 • 10

THE COMPONENTS OF TOTAL HEAD

vanes pushing from behind, the pressure starts to increase and eventually reaches the full discharge pressure head. HOW DO THE PUMP MANUFACTURERS MEASURE N.P.S.H. REQUIRED?

The pump manufacturers measure the N.P.S.H. required in a test rig similar to that shown in Figure 3-7. The system is run in a closed loop where flow, total head and power consumed is measured. In order to provide a low N.P.S.H., a vacuum pump is used to lower the pressure in the suction tank which will provide a low head at the pump suction. The pressure in the suction tank is lowered until a drop of 3% of the total head is measured (see Figure 3-8). When that occurs the N.P.S.H. is calculated and recorded as the N.P.S.H. required for that operating point. Heating coils are also used which increase the water temperature thereby increasing the vapor pressure and further lowering the N.P.S.H. as needed.

Figure 3-7 Test rig used to measure the N.P.S.H. required (courtesy of the Hydraulic Institute www.pumps.org).

3 • 11

THE COMPONENTS OF TOTAL HEAD

Figure 3-8 Measuring the drop in total head to define the N.P.S.H. required at the operating point. GUIDELINE FOR THE LEVEL OF N.P.S.H. AVAILABLE

As stated, a total head drop of 3% is the criteria for setting the level of N.P.S.H. required. Since this results in a performance drop then the user should ensure that there is a higher N.P.S.H. available. The recommendation that you will find in the literature is to have 5 ft of water absolute or a 15% margin above the N.P.S.H. required whichever is greatest. HOW CAN THE N.P.S.H. AVAILABLE BE INCREASED AND CAVITATION AVOIDED

Table 1 gives the major components of N.P.S.H. available and how they affect the level of N.P.S.H. available.

The N.P.S.H. available depends on: 1. The friction loss in the pump suction line 2. The height of the suction tank fluid surface with respect to the pump suction 3. The pressure in the suction tank

4. The atmospheric pressure

5. Fluid temperature

Table 3-1. How to affect the N.P.S.H. available.

Effect on N.P.S.H.A. The higher the friction loss, the lower the N.P.S.H. The lower the height of the fluid surface, the lower the N.P.S.H. This cannot be changed for atmospheric tanks. For tanks that are pressurized, the lower the pressure, the lower the N.P.S.H. available. This cannot be changed. The lower the atmospheric pressure, the lower the N.P.S.H. available. An increase in fluid temperature, increases the vapor pressure of the fluid which decreases the N.P.S.H. available.

3 • 12

THE COMPONENTS OF TOTAL HEAD

EXAMPLE 3.1 – CALCULATE THE NET POSITIVE SUCTION HEAD AVAILABLE Water is to be pumped at a rate of 500 USGPM from a sump. The owner prefers to use a self-priming centrifugal pump rather than a submersible pump. Determine the N.P.S.H. available.

Figure 3-9 Typical example of a pump with low suction head.

Equation [3-10] is applied and SG = 1, H1 = 0, v1 = 0:

N.P.S.H.avail.( ft fluid abs.) = − (∆HF1−S +∆H EQ1−S ) + z1 − z S + 2.31 × (p B ( psia)− pva( psia)) 1 The check valve has a pressure head drop of 5 feet, ∆HEQ1-S = 5 ft. The suction line total friction loss (∆HF1-S) is 0.54 ft of fluid. The barometric pressure is pva = 0.25 psia and the 0 vapor pressure for water at 60 F is pB = 14.7 psia. The height z1 - zS between the surface level of the tank and the pump centerline is -14 ft.

N.P.S.H.avail (ft fluid abs.) = −(0.54 + 5) −14 + 2.31× (14.7−0.25) =13.8 ft 1 The N.P.S.H.A. is 14 ft of fluid absolute. There should be no problem finding a suitable pump with 14 ft fluid absolute N.P.S.H.A..

THE COMPONENTS OF TOTAL HEAD

3.4

3 • 13

PUMP INTAKE SUCTION SUBMERGENCE The pump suction intake must be submerged sufficiently to avoid the formation of vortexes on the liquid surface of the suction tank. These vortexes can take many shapes and forms (see Figure 3-10).

Figure 3-10 Vortex shapes (reprinted with permission of the Hydraulic Institute). The formation of vortexes between the pump suction intake and the suction tank fluid surface causes air to enter the pump suction. This mixture of air and water in the pump reduces the pump capacity. The formation of such vortexes must therefore be avoided. There is a relationship between the intake velocity at the suction intake, and the submergence (S) of the intake. (For more information consult the Hydraulic Institute's Pump Intake Design manual ANSI/HI9.8-1998).

3 • 14

THE COMPONENTS OF TOTAL HEAD

There are many possible intake design geometries, a few are shown in Figure 3-10; they all have in common a minimum requirement for submergence to avoid the formation of vortexes. The minimum value for submergence (S) to avoid vortex formation is given in Figure 3-12.

Figure 3-11 Examples of different intake designs (reprinted with permission of the Hydraulic Institute).

3 • 15

THE COMPONENTS OF TOTAL HEAD

Figure 3-12 Minimum submergence requirements vs. flow and various velocities at the suction pipe intake (reprinted with permission of the Hydraulic Institute). The values for submergence that are given in the chart above can be calculated with the following equation:

S (in) = D(in) + 0.574 ×

q(USgpm) D (in) 1.5

3 • 16

3.5

THE COMPONENTS OF TOTAL HEAD

DISCHARGE STATIC HEAD (∆HDS) The Discharge Static Head is the sum of the elevation and pressure head at the outlet of the system, minus the elevation of the pump center line as stated in equation [3-12]. Its value depends on the elevation and pressure head at the outlet (point 2) of the system. When the discharge pipe end is submerged, the outlet of the system is located at the discharge fluid surface (see Figure 3-13A). When the discharge pipe end is not submerged, the outlet of the system is located at the discharge pipe end (see Figure 313B).

Figure 3-13 Discharge static head with submerged and open discharge pipe end.

H2 is the pressure head at the discharge tank fluid surface. If the tank is open to atmosphere then H2 = 0. In all cases, the discharge static head is:

∆H DS = ( z 2 + H 2 − z S )

[3-12]

Very often pipes enter a tank from the top and finish lower than the liquid surface of the tank as in Figure 3-13A. The outlet of the system remains at point 2 since the fluid particles eventually have to get to point 2.

3 • 17

THE COMPONENTS OF TOTAL HEAD

EXAMPLE 3.2 – CALCULATE THE SUCTION & DISCHARGE STATIC HEAD

A paper machine uses large volumes of steam in the paper drying process. This steam is condensed and recovered in a final stage under vacuum conditions and the condensate is pumped to a flash tank. The flash tank is kept under controlled pressure conditions and receives condensate from other areas of the plant. Its purpose is to collect condensate from various sources and return it to the boilers. What is the suction and discharge static head of the vacuum receiver system if p1 = -10 in of Hg and p2 = 2 psig?

Figure 3-14 An example of a suction tank with low pressure and a discharge tank under pressure.

Suction Static Head

∆HSS = z1 + H1 − z S

[3-13]

The pressure p1 is converted to psia:

p1 ( psia ) = 14.7 − ( p1 (in Hg ) × 0.4918) = 9.8 psia

The specific gravity of condensate at the temperature corresponding to 9.8 psia is very close to 1 (SG=1). To convert from psia to feet of fluid relative: (14 . 7 − p1(psia ))× 2 .31 (14 . 7 −9 . 8 )×2 . 31 H1= = = − 11 .3 ft of fluid SG 1 The above values are substituted into the equation [3-13]: ∆HSS = (102 - 11 - 96) = -5.3 ft of fluid. The suction static head is -5.3 feet of fluid. Discharge Static Head

∆H DS = z 2 + H 2 − z S Convert the pressure p2 to feet of fluid: p × 2 .31 2 × 2 .31 H2 = 2 = = 4.6 ft of fluid SG 1 The above values are substituted into the equation [3-14]: ∆HDS = (108 + 4.6 - 96) = 16.6 feet of fluid. The discharge static head is 16.6 feet of fluid.

[3-14]

3 • 18

3.6

THE COMPONENTS OF TOTAL HEAD

VELOCITY HEAD DIFFERENCE (∆Hv) The Velocity Head Difference is the head (expressed in ft of fluid) corresponding to the kinetic energy variation between the outlet and the inlet of the system.

∆H v =

1 ( v 2 2 − v1 2 ) 2g

The Velocity Head is defined as v 2/2g. In the case where a suction line is connected to a tank, point 1 is on the fluid surface of the suction tank (see Figure 3-15). This surface will be moving downwards very slowly, so that v is very small and therefore v1 2/2g ≈ 0. If point 1 is in fact a connection on another line, the velocity will have to be considered. Identical reasoning can be applied to the exit point 2. The Velocity Head is not normally a significant part Figure 3-15 Inlet and outlet velocities of a of the Total Head, typically 1 or 2 ft of typical system. fluid. However, some systems are designed with nozzles at the exit point, in order to accelerate the fluid and generate a high velocity. A good example of this is a paper machine head-box. The fiber slurry is ejected from the wide and narrow opening (slice lip) of the head-box at the same velocity as the paper machine, which in certain cases is as high as 4000 ft/min (45 miles per hour). In this case, the velocity head may represent as much as half of the pump Total Head. 3.7

EQUIPMENT PRESSURE HEAD DIFFERENCE (∆HEQ) The Equipment Pressure Head difference is pressure head loss due to friction imposed by the equipment (for example, control valve. filter, etc.). Isolation valves and pipe fittings are not considered equipment. The pressure head loss for equipment vs. flow rate is usually obtained from tables, charts or graphs provided by the supplier of the equipment. The total Equipment Pressure Head Difference between the inlet and the outlet of a system is made up of the sum of the pressure head drops across each piece of equipment.

∆H EQ1− 2 = ∆H EQ1 + ∆H EQ2 +...

3 • 19

THE COMPONENTS OF TOTAL HEAD

CONTROL VALVES

A typical pumping system has at least one control valve in the circuit. Depending on the type of valve, its opening, the upstream pressure, and the flow, it is possible to calculate the pressure head drop by consulting the appropriate supplier data tables. This pressure head drop is then added to the Total head of the pump to ensure that there is enough energy to move the fluid through the system at the design flow rate. During the design stage, a simple way to account for this is to assume or fix the pressure head drop of the control valve. If we assume a pressure head drop across the valve of 10 ft of fluid, then it will be generally possible to select a valve that will give this pressure head drop at a reasonable opening of say 90%. In other words, by using a ∆p of 10 ft for the pressure head drop, we have fixed one of the parameters required to size a valve, without unduly restricting the task of sizing the valve. Ten (10) ft of pressure drop is a common value used in designing systems with control valves. This criterion will generally result in a valve size that is one size smaller than the line. A more practical approach is required for existing systems with a control valve. We will need the manufacturer’s tables for the valve which gives a CV coefficient which is proportional to the pressure drop and flow rate for a given valve opening (see equation [3-15]). Obtain the valve opening, while the system is running normally. With this information, using the manufacturer’s catalog you can obtain a CV value. Calculate the pressure drop and convert to pressure head drop. Use this value in the calculations for Total Head. The definition of the CV coefficient is:  USgpm CV  1/ 2  psi

 q ( USgpm )  = ∆ p ( psi )  SG

[3-15]

where q and ∆p are respectively the flow rate and pressure drop across the valve for a given valve model and opening. EQUIPMENT

Any equipment in the line, such as filters, nozzles, etc., will have a specified pressure drop at a certain flow rate that is available in the literature or from the equipment supplier. Occasionally, certain types of equipment require a specific upstream pressure in order to operate properly. To accommodate this, we need to determine what the pressure is at the proposed equipment location. If the calculation shows the upstream pressure to be lower than required, it will have to be raised artificially by closing a manual valve on the downstream side of the equipment. The Total Head will then have to be increased to accommodate this added pressure head. Another option is to move the equipment closer to the pump where the line pressure is higher. In the event that the calculation shows the pressure immediately upstream of the equipment to be higher than required, it will have to be lowered artificially by closing a manual valve on the upstream side of the equipment. The Total Head will then have to

3 • 20

THE COMPONENTS OF TOTAL HEAD

be increased to accommodate this added pressure head. Alternatively, we can try moving the equipment closer to the discharge end where the line pressure is usually lower. 3.8

PIPE FRICTION HEAD DIFFERENCE FOR NEWTONIAN FLUIDS (∆HFP) The Friction Head is the friction due to the movement of fluid in a piping system and is proportional to flow rate, pipe diameter and viscosity. Tables of values for Friction Head are available in references 1 & 8. The Friction Head, as defined here, is made up of the friction loss due to the fluid movement and the friction loss due to the effect of pipe fittings (for example, 90° elbows, 45° bends, tees, etc.):

∆HF = ∆HFP +∆HFF the subscript FP refers to pipe friction loss and the subscript FF to fittings friction loss. NEWTONIAN FLUIDS

Newtonian fluids are a large class of fluids, whose essential property VISCOSITY, was first defined by Newton (see Appendix A for a list of Newtonian and non-Newtonian fluids). Viscosity is the relationship between the velocity of a given layer of fluid and the force required to maintain that velocity. Newton theorized that for most pure fluids, there is a direct relationship between force required to move a layer and its velocity. Therefore, to move a layer at twice the velocity, required twice the force. His hypothesis could not be tested at the time, but later the French researcher, Poiseuille, demonstrated its validity. This resulted in a very practical definition for viscosity (see Appendix A for more details). The Darcy-Weisbach formula expresses the resistance to movement of any fluid in a pipe:

∆ H FP v2 = f L D × 2g where f is a non dimensional friction factor. Often, the tables give values for friction loss in terms of ft of fluid per 100 ft of pipe. When the appropriate units are used (Imperial system), the Darcy-Weisbach equation becomes:

∆ H F P  ft o f fluid  v 2 ( ft / s ) 2 = 1 2 0 0 f   L  1 0 0 ft o f pipe  D ( in ) × 2 g ( ft / s 2 ) The friction factor is proportional to the Reynolds number which is defined as:

[3-16]

3 • 21

THE COMPONENTS OF TOTAL HEAD

Re = 7745.8

v(ft / s) D(in) ν (cSt)

[3-17]

The Reynolds number is proportional to the kinematic viscosity, the average velocity, and the pipe inside diameter. It is a non dimensional number. The kinematic viscosity (ν) is the ratio of the absolute viscosity (µ) to the fluid specific gravity (SG).

ν ( cSt ) =

µ ( cP ) SG

Values of viscosity for many fluids can be found in reference 8. Laminar flow - RE < 2000

Distinct flow regimes can be observed as the Reynolds number is varied. In the range of 0 to 2000, the flow is uniform and is said to be laminar. The term laminar refers to successive layers of fluid immediately adjacent to one another, or laminated. Looking at a longitudinal section of the pipe, the velocity of individual fluid particles is zero close to the wall and increases to a maximum value at the center of the pipe with every particle moving parallel to its neighbor. If we inject dye into the stream, we would notice that the dye particles maintain their cohesion for long distances from the injection point.

Figure 3-16 Laminar and turbulent flow velocity profiles. The friction loss is generated within the fluid itself. Figure 3-16 shows that each layer (in this case each ring) of fluid is moving progressively faster as we get closer to the center. The difference in velocity between each fluid layer causes the friction loss.

3 • 22

THE COMPONENTS OF TOTAL HEAD

The friction factor f is given by:

f =

64 Re

[3-18]

For viscous fluids (i.e.: ν ≥ 50 SSU), the combination of velocity and viscosity usually produces a low Reynolds number and therefore laminar flow. Pumping viscous fluids at a faster rate may cause the fluid to become turbulent resulting in high friction losses. The tables for viscous fluid friction loss given in references 1 & 8 are based on the equation for laminar flow, equation [3-18]. This equation can be theoretically derived and is found in most fluid dynamic volumes (see reference 11). An interesting aspect of laminar flow is that pipe roughness is not a factor in determining friction loss. Unstable flow - 2000