FluidsCh1

Fluid Mechanics Chapter 10FD121 Fluids Statics and Dynamics Concepts

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Table of Contents

Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii

Fluid Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Fluid Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Absolute Zero Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3Atmospheric Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Gauge, Vacuum, and Absolute Pressures . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Typical Pressures in SONGS Plant Components . . . . . . . . . . . . . . . . . . . . . . 8

Density (ρ) and Specific Volume (ν) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8Hydrostatic Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

Saturation Temperature and Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19Determining TSAT/PSAT From Steam Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

Subcooled Water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Saturated Liquid Water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

Density of Subcooled and Saturated Liquid Water . . . . . . . . . . . . . . . . . . . . . . . . . . . 23Use of Steam Tables to Determine Density of Saturated Liquid Water . . . . . . . 25Use of Steam Tables to Determine Density of Subcooled Liquid Water . . . . . . 26

Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28Working Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28Closed System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28Open System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

Fluid Flow Rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29


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Fluid Energy Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34Heat added, Qin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34Heat removed, Qout . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34Work done on the fluid, Won . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35Work done by the fluid, Wby . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Energies Possessed By the Flowing Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36Kinetic energy, KE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36(Gravitational) potential energy, PE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37Internal energy, U . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37Flow energy, FE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

The General Energy Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

The Basic Bernoulli Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

Effect of Friction in a Fluid Flow System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

Fluid Viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

Laminar and Turbulent Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

Buoyant Force and Drag Force on Reactor Core . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

Fluid Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

Water Hammer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62Causes/Methods of Prevention of Water Hammer . . . . . . . . . . . . . . . . . . . . . . . 62

Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

Exercise Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76


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Lesson Objectives

1. Given a gauge pressure reading or a vacuum pressure reading, convert thereading to an absolute pressure reading.

2. Define fluid density and fluid specific volume, and state the relationship betweenthem.

3. Given a table of conversion factors, perform conversions between pressuresexpressed in the following pressure units:

C Pounds per square inch (psi)C Inches of mercury (in. Hg.)C Feet of water (ft. H20)

4. State how the density and the specific volume of liquid water change as

C the temperature of the water changesC the pressure of the water changes

5. State and apply the mathematical formulas for:

C Volumetric Flow RateC Mass Flow Rate

6. Apply the General Energy Equation to determine the effect of system configurationchanges on system pressure.

7. Describe the concept of fluid viscosity.

8. Compare/contrast laminar flow and turbulent flow characteristics.

9. Describe the buoyant force exerted by a fluid on an object partially or totallysubmersed in that fluid.


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10. Identify characteristics of water hammer, including:

• its definition• fluid flow conditions that can trigger a water hammer event• potential consequences of a water hammer event• methods that will minimize the possibility of a water hammer event occurring


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Pressure 'ForceArea

P 'FA

Lesson Purpose

Fluid Characteristics

Fluid Mechanics is the area of physics which mathematically describes the behavior offluids at rest (fluid statics) and in motion (fluid dynamics).

Fluid: Any substance that flows; a substance which will always take on the shapeof its container.

A fluid may be classified as either a liquid or gas/vapor (the terms gas and vapor will beused interchangeably). Water can exist in the solid, liquid, or gas/vapor state. Usually,when the term “water” is used, it means water in the liquid state.

Fluid properties will now be defined.

Fluid Pressure

Fluid Pressure: the ratio of the total force exerted by a fluid to the total area towhich the force is applied; it is force per unit area.

The equation for pressure is given below:


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In the English system of measurement, pressure is normally expressed in units ofpounds force per square inch (lbf/in2), abbreviated psi.

For static fluids (i.e., fluids at rest) the pressure on or by any particle of the fluid is thesame in all directions. Figure 1 demonstrates for a static fluid under a pressure of 25lbf/in2:

Figure 1

Pressure is transmitted from particle to particle so that the pressure at the sameelevation in the fluid will be constant as shown by Figure 2.

Figure 2


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The pressure for a static fluid at a given elevation will always be constant regardless ofthe size or shape of the connecting pipe as long as the two locations are connected bythe same intervening fluid (see Figure 3).

Figure 3

Absolute Zero Pressure

A vacuum, by definition, is a volume that contains no mass. The lowest level ofpressure that can exist in nature is that which would be present in a vacuum, and it iscalled absolute zero pressure:

Absolute zero pressure: the pressure that would exist in a perfect vacuum;therefore, its magnitude is 0 psi.


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Atmospheric Pressure

Because the air that surrounds us has weight, it is exerting force onto the area of ourbodies. The force per unit area exerted by the atmosphere is called atmosphericpressure:

Atmospheric Pressure: the pressure that is produced by the weight of the air,which makes up the atmosphere, acting over the entire surface of the earth; it isthe force per unit area of contact exerted by the atmosphere.

All objects on the earth experience atmospheric pressure. Additionally, atmosphericpressure varies with changes in altitude. As the altitude increases, the atmosphericpressure decreases, because the weight of the air above this altitude is less. Areference pressure, call standard atmospheric pressure, is defined:

Standard sea level atmospheric pressure is 14.7 psi.

Gauge, Vacuum, and Absolute Pressures

Fluid pressures are normally expressed relative to designated reference pressurelevels. These reference pressures are atmospheric pressure and absolute zeropressure. Absolute pressure is defined relative to the absolute zero pressure level:

Absolute Pressure (psia): pressure expressed relative to absolute zero pressureas the zero reference; absolute pressure is pressure above that which would existin a perfect vacuum.

For example, standard atmospheric pressure, expressed as an absolute pressure,is 14.7 psia. The a attached to the psi unit indicates that the value given is anabsolute pressure. 14.7 psia, therefore, indicates that the pressure is 14.7 psiabove that which exists in a perfect vacuum.


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Gauge pressure is defined relative to the atmospheric pressure level:

Gauge Pressure (psig): pressure expressed relative to atmospheric pressure asthe zero reference; it is equal in magnitude to the number of psi aboveatmospheric pressure.

For example, if the pressure in a tank is 20 psia, and atmospheric pressure is 14.7psia, the tank pressure expressed as a gauge pressure would be 5.3 psig. The gattached to the psi unit indicates that the value given is a gauge pressure. Themagnitude of the pressure (5.3 psi) indicates that the pressure exceedsatmospheric pressure by 5.3 psi.

Vacuum pressure, like gauge pressure, is defined relative to the atmospheric pressurelevel:

Vacuum Pressure (psiv): pressure expressed relative to atmospheric pressure asthe zero reference; it is equal in magnitude to the number of psi belowatmospheric pressure, expressed as a positive number.

For example, if the pressure in a tank is 10 psia, and atmospheric pressure is 14.7psia, the tank vacuum pressure would be 4.7 psiv. The v attached to the psi unitindicates that the value given is a vacuum pressure. The magnitude of thepressure (4.7 psi) indicates that the pressure is 4.7 psi below atmosphericpressure.


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Pabs ' Patm % Pgauge

Pabs ' Patm & Pvac

Figure 4 illustrates the concepts of absolute, gauge, and vacuum pressures, and theirrelationships to their reference pressure levels:

Figure 4

With the aid of Figure 4, two equations can be written to relate gauge pressure andvacuum pressure to the equivalent absolute pressure:

To convert gauge pressure to absolute pressure, add the absolute atmosphericpressure to the pressure indicated by the pressure gauge:

To convert vacuum pressure to absolute pressure, subtract the pressure indicatedby the vacuum gauge from the absolute atmospheric pressure:


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Pabs ' Patm % Pgauge

Pabs ' 14.7 psi % 92 psi ' 106.7 psia

Pabs ' Patm & Pvac

Pabs ' 14.7 psi & 13.0 psi ' 1.7 psia

EXAMPLE A

A pressure gauge attached to the shell side of a feedwater heater reads 92 psig. Determine the equivalent absolute pressure.

SOLUTION

Atmospheric pressure must be added to the gauge reading to obtain the absolutepressure:

Therefore, 92 psig is equivalent to 106.7 psia.

EXAMPLE B

A vacuum gauge on a steam condenser indicates 13.0 psiv. Determine the equivalentabsolute pressure.

SOLUTION

The pressure indicated by the vacuum gauge must be subtracted from the atmosphericpressure to obtain the absolute pressure:

Therefore, 13.0 psiv is equivalent to 1.7 psia.


*These plant-specific pressures are provided as examples, only. You are not expected to knowthese pressure values for your pre-employment exam.

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ρ 'mV

Typical Pressures in SONGS Plant Components

In the SONGS Units*, pressures range from a high of 2,250 psia to a low of about 1psia. Typical values include:

Reactor Coolant System (Pressurizer ): 2,250 psia

Main Steam (Steam Generator Outlet), 100% power: 805 to 813 psia

Feedwater Pump Discharge, 100% power: 1030 to 1050 psig

Main Condenser (steam side), 100% power: 1 to 2 psia

Density (ρ) and Specific Volume (ν)

Fluid density (ρ): the ratio of the mass possessed by the fluid to the volume itoccupies. Therefore, density is mass per unit volume, with units lbm/ft3.

where:

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

m = mass of that volume, lbm

V = total volume of the substance, ft3


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ν 'Vm

ρ '1ν

and ν '1ρ

Fluid specific volume (ν): the ratio of the volume occupied by the fluid to the mass itpossesses. Thus, specific volume is volume per unit mass, with units of ft3/lbm:

where:

ν = (Greek letter nu) = specific volume, ft3/lbm

V = volume, ft3

m = mass, lbm

From the definitions of density and specific volume, it is obvious that these quantitiesare reciprocals of each other:

Both density and specific volume measure the same property, namely how close themolecules or atoms making up a substance are to each other:

A relatively high value of specific volume (or low value of density) implies that themolecules or atoms in the substance are relatively far apart. This is true of gasesand vapors such as hydrogen, oxygen, and steam.

Conversely, a relatively low value of specific volume (or high value of density)implies that the molecules are relatively close together. This is true of liquids andsolids such as water and ice.


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ρ '1ν

ρ '1

0.016026 ft 3

lbm

ρ '1

0.016026lbmft 3

' 62.4 lbmft 3

EXAMPLE C

A container is filled with a water having a specific volume of 0.016026 ft3/lbm. Calculatethe density of the water.

SOLUTION

62.4 lbm/ft3 is called the standard density of water. Water at 14.7 psia and 60EF hasthis density.


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Figure 5 Hydrostatic Pressure

Hydrostatic Pressure

A column of fluid has a pressure at the bottom of the column which is equal to thepressure at the top of the column plus the pressure due to the weight of the fluid in thecolumn. The pressure due to the weight of a column of fluid is called hydrostaticpressure:

Hydrostatic pressure (psi): the pressure produced by the weight of a fluidcolumn; it is the difference between the pressure at the bottom of the fluid columnand the pressure at the top of the column.

In Figure 5, the pressure difference P2 - P1 represents the hydrostatic pressure due tothe fluid in the container:


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P2 ' P1 %WA

. Therefore,

Hydrostatic Pressure is P2 & P1 'WA

where: P1 ' pressure at location 1, lbfft 2

P2 ' pressure at location 2, lbfft 2

W ' weight of the fluid column, lbf

A ' cross&sectional area of the fluid column, ft 2

P2 & P1 '

mggc

Ahh

, or

P2 & P1 'mA

hh

ggc

The definition of hydrostatic pressure can be used to derive a formula which can beused to calculate the hydrostatic pressure of any fluid column. Using Figure 5, thepressure at Point 2 (P2) is equal to the pressure at Point 1 (P1) plus the pressureproduced by the weight of the fluid column. This can be expressed in equation form asshown below:

Recall from the Classical Physics lesson that weight is given by the formula W =mg/gc. Therefore, substituting mg/gc for W, and multiplying the right-hand side of theequation by h/h, where h is the height of the fluid column, results in the followingequation:


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P2 & P1 'mV

h ggc

∆Phydrostatic ' P2 & P1 ' ρh ggc

P2 & P1 ' Hydrostatic Pressure, lbfft 2

P1 ' Cover Pressure, lbfft 2

P2 ' Pressure at bottom of column, lbfft 2

ρ ' Fluid density, lbmft 3

h ' Height (or depth) of fluid column, ft

ggc

'

32.2 ftsec2

32.2 ft @ lbmlbf @sec2

'1 lbf1 lbm

The product of the cross-sectional area (A) and the height (h) is equal to the volume, V,of the fluid column (V = Ah). Therefore, the equation becomes

The equation is now free of the weight and area terms. Since m/V is the density (ρ) ofthe fluid in the fluid column, the final form of the hydrostatic pressure equation becomes:

Hydrostatic Pressure Equation:

where:

Note that the formula expresses the fluid’s hydrostatic pressure (P2 - P1) in terms ofquantities which can easily be determined, namely the fluid density (ρ) and the height(h) of the fluid column.


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144 in 2

1 ft 2or 1 ft 2

144 in 2

∆Phydrostatic ' ρh ggc

' 62.4 lbmft 3

33.92 ft 1 lbf1 lbm

1 ft 2

144 in 2

' 14.7 lbfin 2

, or, 14.7 psi

Also, the unit of measurement of the hydrostatic pressure formula ρhg/gc in fundamentalunits of measurement of the FPS measurement system is lbf/ft2. If the pressure termsof the hydrostatic pressure equation are expressed in psi (lbf/in2), the formula will not beunit consistent. Therefore, the conversion factor,

must be used where appropriate to maintain unit consistency. The examples whichfollow will illustrate this.

EXAMPLE D

Find the hydrostatic pressure, in psi, produced by a column of water 33.92 ft high. Thedensity of water is 62.4 lbm/ft3.

SOLUTION

The pressure can be found by direct substitution of the column height and the waterdensity into the hydrostatic pressure equation. The conversion factor 1 ft2/144 in2 isinserted on the right-hand side of the equation so that the unit of measurement is psi:

Therefore, a column of water 33.92 ft in height, with density 62.4 lbm/ft3, produces 14.7psi (“one atmosphere”) of pressure.


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33.92 ftH2O ' 14.7 lbfin 2

1 ftH2O ' 0.433 psi, for standard density 62.4 lbmft 3

water

Pressure (psi) ' Pressure ftH2O x 0.433 psi1 ftH2O

psi ' ftH2O x 0.433 psi1 ftH2O

Water pressure is sometimes expressed in “feet of water (ftH2O) .” Since 33.92 feetof water with density 62.4 lbm/ft3 produces a hydrostatic pressure equivalent to 14.7 psi,a conversion factor can be created between these two pressure units:

or, dividing both sides by 33.92,

Therefore, if pressure is given in feet of water, the equivalent pressure in psi units canbe obtained by as follows:

NOTE: This conversion factor is valid only when the density of the water column is 62.4lbm/ft3, i.e. the water is at its standard density. It will be shown later that the density ofwater will be significantly changed if the temperature of the water is changed, and thiswould invalidate the conversion factor above.


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∆P ' ρh ggc

(Hydrostatic pressure equation)

' 848.7 lbmft 3

1 ft 3

1728 in 3(29.93 in) 1 lbf

1 lbm

' 14.7 lbfin 2

' 14.7 psi

29.93 in Hg ' 14.7 psi

1 in Hg ' 0.491 psi

Water pressure is sometimes expressed in “inches of mercury (in Hg).” ExampleE shows that a column of mercury 29.93 inches high will produce a pressure equivalentto 14.7 psia (atmospheric pressure):

EXAMPLE E

Determine the hydrostatic pressure produced by a column of mercury that is 29.93inches in height. The density of mercury is 848.7 lbm/ft3.

SOLUTION

Since 29.93 inches of mercury produces a hydrostatic pressure equivalent to 14.7 psi,the conversion factor then becomes

or, dividing both sides by 29.93,


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Pressure (psi) ' Pressure (in Hg) x 0.491 psi1 in Hg

psi ' in Hg x 0.491 psi1 in Hg

Figure 6 Pressure Scale Relationships

Therefore, if pressure is given in inches of mercury, the equivalent absolute pressure inpsi can be obtained as follows:

Figure 6 shows relationships between the common pressure scales (vacuum, gauge,and absolute) in psi, ft H2O, and in Hg units. Note the inverse relationship betweeninches of Hgvac and inches of Hgabs (“in Hgabs” is referred to as “inches of mercurybackpressure”).

pounds per square inch: # inches: “ feet: ‘


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27 in Hgvac0.491 psi1 in Hg

' 13.3 psiv

EXAMPLE E

A pressure gauge on a condenser reads 27 in Hg vacuum. Which one of the followingbest approximates the absolute pressure corresponding to this vacuum?

A. 1.0 psia

B. 1.5 psia

C. 3.0 psia

D. 15.0 psia

SOLUTION

Converting 27 in Hg vacuum to psiv,

Therefore, the absolute pressure can be obtained:

Pabs = Patm - Pvac = 14.7 - 13.3 = 1.4 psia

Answer: Choice B.


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Saturation Temperature and Pressure

Consider a coffee pot on a stove open to the atmosphere at a location where the pressure is one standard atmosphere, 14.7 psia. Consider this coffee pot to be partiallyfilled with cold liquid water at 32EF.

If heat is added to the water in the coffee pot, its temperature rises until 212EF isreached.

If additional heat is now added, the temperature of the water no longer rises butsome of the water begins to boil.

This temperature at which the water first begins to boil while at a given pressure, suchas 14.7 psia, is known as the saturation temperature, Tsat:

Saturation temperature Tsat: The temperature at which a liquid will boil at a givenpressure if heat is added to the liquid.

Now suppose the coffee pot is in a location where the pressure is 12 psia, and thetemperature of the liquid water in the pot is 32EF.

If heat is added to this water the water temperature increases to 202EF.


Therefore, by the definition of saturation temperature, 202EF is the saturationtemperature that corresponds to a pressure of 12 psia.

Now suppose the outlet of the coffee pot is restricted so that the pressure above the32EF liquid water in the pot is 100 psia.

If heat is added to the water, its temperature increases to 328EF.


Thus, 328EF is the saturation temperature that corresponds to a pressure of 100 psia.


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The conclusion may now be stated:

For every pressure there is a corresponding temperature, called thesaturation temperature, at which water boils when heat is added.

The converse of this statement is also true:

For every temperature there is a pressure, called the saturation pressure, atwhich water boils when heat is added.

Saturation temperature and saturation pressure, therefore, are dependent quantities.We cannot specify a saturation temperature unless the corresponding (saturation)pressure is known, and we cannot specify a saturation pressure unless thecorresponding (saturation) temperature is known.

Determining TSAT/PSAT From Steam Tables

The ABB Steam Tables provide thermodynamic data concerning water. In particular,for any saturation temperature the corresponding saturation pressure can be obtained,and vice versa.

For the purpose of this and all subsequent lessons dealing with obtainingthermodynamic data from the Steam Tables, the values will be provided for you.In other words, you will NOT have to have a version of the Steam Tables for theselessons. These lessons will simply say such things as “from the Steam Tables,the value of the saturation pressure is...”. When the actual classroom training onthese materials begins, you will be taught how to use the Steam Tables to determinethermodynamic data concerning water.


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Subcooled Water

When water (H2O) exists at a temperature below saturation temperature, it is classifiedas subcooled liquid water:

Subcooled (liquid) water: Water that exists at a temperature below saturationtemperature for the pressure of the water.

If heat is added to a subcooled liquid, the temperature of the liquid willincrease.

For example, water at atmospheric pressure (14.7 psia) and at a temperature below212EF is subcooled water. If the temperature of the water was, say, 100EF, and heatwas added to it, its temperature would increase.

As another example, consider the water in the Reactor Coolant System (RCS):

The Pressurizer maintains the pressure in the RCS at 2250 psia.

From the Steam Tables, saturation temperature for 2250 psia is 653EF (liquidwater at 2,250 psia will not start to boil until its temperature reaches 653EF).

At 100% power, the water entering the reactor core through the Cold legs is at539EF, and the water leaving the reactor core through the Hot Legs is at 594EF.

Since the hottest spot in the RCS is the Hot Legs, where the temperature is about594EF, the entire RCS is a subcooled system.

Any heat added to the water in the RCS causes its temperature to increase.

Since the reactor core adds heat to subcooled, 2,250 psia water as it passesthrough the core, the temperature of the water continuously increases (from ColdLeg core inlet temperature of 539EF to Hot Leg core outlet temperature of 594EF).


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Saturated Liquid Water

When liquid water exists at saturation temperature for the pressure of the water, thewater is called saturated liquid water:

Saturated liquid water: Water that exists at a temperature equal to saturationtemperature for the pressure of the liquid.

Since saturation temperature and saturation pressure are dependent quantities,saturated liquid water exists at saturation temperature/pressure conditions.

Since saturated liquid water exists at boiling temperature for the pressure of the water,heat added to saturated liquid water will cause the water to boil (change phase fromliquid to steam).

When heat is added to saturated liquid water at constant pressure, the temperature ofsaturated liquid water and the steam that is produced remain constant until the liquidwater has been converted totally into steam.

For example, water at atmospheric pressure (14.7 psia) and 212EF is saturated liquidwater.

Any heat added to this water will cause some of the water to change phase tosteam, i.e., to boil.

As another example, consider the liquid water that is in the Pressurizer under normaloperating conditions:

Its pressure is 2250 psia, and its liquid water temperature is 653EF.

Since 653EF is saturation temperature for 2250 psia, this liquid water exists atsaturation temperature/pressure conditions; it is saturated liquid water.

If heat is added to the water (by the Pressurizer heaters), some of the water willchange phase into steam.


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Density of Subcooled and Saturated Liquid Water

The density of subcooled liquid water varies with both the temperature and the pressureof the water:

As the temperature of subcooled water increases, the density of the waterdecreases. (As the temperature of subcooled water decreases, the density of thewater increases.)

As the pressure of subcooled water increases, the density of the waterincreases. (As the pressure of subcooled water decreases, the density of thewater decreases.)

For both temperature and pressure changes, the effect on the density of subcooledwater can be explained in terms of expansion or contraction of the volume occupied bythe water:

As the temperature of subcooled water increases, the water expands. Theexpansion occurs because the average thermal energy of the water moleculesincreases, and this increased vibrational energy of the individual moleculespushes them to greater separation distances from each other. The mass of thewater now occupies more volume (i.e., same mass m, but V8), and since density= m/V, this means that the density decreases.

As the pressure of subcooled water increases, the water contracts. Thecontraction occurs because the pressure increase “squeezes” the moleculescloser together. The mass of the water now occupies less volume, and sincedensity = m/V, this means that the density increases.


0FD121 24 Rev 1-3, 10/15/04

Figure 7 Subcooled Water Density Versus Temperature

The variation of subcooled water density with temperature (for two given pressures,1000 psia and 2000 psia) is shown in Figure 7:

Note that:

For a given pressure, the density of the water decreases at an increasing rateas its temperature increases, an important fact relative to numerousfundamentals concepts that will be discussed during this training.

For any given temperature, the density of 2000 psia water is slightly larger thanthe density of 1000 psia water. An increase of 1000 psi (from 1000 psia to 2000psia) has very little effect on the density of the subcooled water.


0FD121 25 Rev 1-3, 10/15/04

The following conclusions can be drawn:

The density of subcooled water is primarily determined by the temperature of thewater.

Even though subcooled water pressure changes do affect the density of the waterslightly, the effect of a pressure change on the density of subcooled water isessentially negligible (subcooled water is essentially “incompressible.”).

For precise engineering calculations, the minor changes in the specific volume ofsubcooled water with changes in pressure must be accounted for. The source used todetermine the precise specific volume of water at temperatures below saturationtemperature for the given pressure is called the Subcooled Liquid Tables.

Use of Steam Tables to Determine Density of Saturated Liquid Water

ABB Steam Tables provide values for the specific volume of saturated liquid water.When the symbol “ f” is used in this training, it means the value of the specific volume ofsaturated liquid water at the specified saturation temperature/saturation pressureconditions.

Example F illustrates how to use the Steam Tables to determine the specific volume(and density) of saturated liquid water.


0FD121 26 Rev 1-3, 10/15/04

νf ' 0.02146 ft 3

lbm

ρ '1νf

'1

0.02146 ft 3

lbm

' 46.6 lbmft 3

EXAMPLE F

Determine the specific volume and the density of saturated liquid water at 540EF(962.79 psia).

SOLUTION

From Steam Table 1, the specific volume of saturated liquid water at 540EF (962.79psia) is:

Therefore, since density is the reciprocal of specific volume,

Use of Steam Tables to Determine Density of Subcooled Liquid Water

Since the density of subcooled water depends primarily on the temperature of the water, using vf for the temperature of the water is sufficiently accurate. In other words,

To determine the specific volume of subcooled liquid water, use the SteamTables value for saturated liquid water (use f) at the given temperature of thewater.

Example G illustrates the acceptable method for determining the density of subcooledliquid water.


0FD121 27 Rev 1-3, 10/15/04

ρ . 1νf, 540EF

'1

0.02146 ft 3

lbm

' 46.6 lbmft 3

EXAMPLE G

At 80% power conditions, the water entering the suction of a Reactor Coolant Pump(RCP) is at 540EF and 2,250 psia. Determine the density of this water.

SOLUTION

Since the water is at 540EF, and this temperature is below saturation temperature forthe pressure of the water, the water entering the RCP is subcooled. Since the densityof subcooled water is primarily determined by the temperature of the water,

In other words, the answer is the same as it was in Example F.


0FD121 28 Rev 1-3, 10/15/04

Systems

A system is any particular portion of the universe which we intend to study directly. Around the system are boundaries that the mind constructs. The choice of the system,defined by its contents and its boundaries, is purely arbitrary.

Working Fluid

The working fluid of a system is any fluid (including gases) which receives, transports,and transfers energy in the system. For example, the Reactor Coolant System water isthe working fluid of this system because it receives energy from the reactor core,transports this energy to the steam generators, and transfers the energy across thetubes of the steam generators to the water on the other side of the tubes.

Closed System

A system is defined to be a closed system if matter does not cross the boundaries ofthe system. Energy may or may not flow into or out of a closed system (may cross theboundaries of the system), but mass may not.

Open System

If matter crosses a boundary of a system, the system is called an open system. Energy may cross the boundary of an open system either alone or with the flow ofmass.


0FD121 29 Rev 1-3, 10/15/04

0V 'Vt

0V ' Volumetric Flow Rate: ft 3

sec, gal

min, etc.

V ' Volume of fluid passing flowing past a given point

in a given time interval: ft 3, galt ' time interval: sec, min, etc.

0V ' Av

0V ' Volumetric Flow Rate: ft 3

sec

A ' Cross&sectional area of piping, ft 2

v ' fluid velocity: ftsec

Fluid Flow Rate

The concepts of volumetric flow rate and mass flow rate will now be defined:

Volumetric flow rate, V.

: the volume of fluid flowing past a given point in a fluidflow system per unit time.

where:

The dot notation ( V.

) is used to denote that the quantity under consideration ismeasured on a "per unit of time" basis (“volume per unit time”).

Volumetric flow rate can be determined if the fluid velocity andcross-sectional area of the piping through which the fluid isflowing are known:

where:


0FD121 30 Rev 1-3, 10/15/04

0m 'mt

0m ' Mass flow rate, lbmsec

(or lbmhr

, etc.)

m ' mass flowing past a given location in a given time interval, lbmt ' time interval, sec (or hr, etc.)

0m ' ρAv

0m ' mass flow rate, lbmsec

ρ ' fluid density , lbmft 3

A ' cross&sectional flow area of the pipe, ft 2

v ' magnitude of the fluid velocity, ftsec

Mass flow rate, m.

: the amount of fluid mass flowing past a given point in a fluidflow system per unit of time.

where

Mass flow rate can be determined if fluid density, fluid velocity, and cross-sectional areaof pipe through which the fluid is flowing are known:

Mass Flow Rate Equation

where:


0FD121 31 Rev 1-3, 10/15/04

0V ' Av and 0m ' ρAv Y 0m ' ρ 0V

0m1 ' 0m2

Therefore,ρ1A1v1 ' ρ2A2v2

The relationship between mass flow rate and volumetric flow rate is shown by the twodefining formulas on the previous page: mass flow rate is equal to fluid densitymultiplied by volumetric flow rate:

EXAMPLE H

For the constant mass flow rate system shown below, the fluid velocity in the 2" pipe is10 ft/sec. How does the change in pipe diameter from 2" at point 1 to 1" at point 2 affectthe fluid velocity? Assume that the fluid density is constant in the pipe.

SOLUTION

Since the mass flow rate is constant,


0FD121 32 Rev 1-3, 10/15/04

A1v1 ' A2v2

A1 ' πr 21 ' π (1 inch)2 ' 3.14 in 2

A2 ' πr 22 ' π (0.5 inch)2 ' 0.785 in 2

A1v1 ' A2v2

(3.14 in 2) 10 ftsec

' (0.785 in 2)v2

v2 '3.140.785

10 ftsec

' 40 ftsec

Since the fluid density is constant (ρ1 = ρ2):

Since the area A of a circle is A = πr2,

Therefore,

The velocity of the fluid flow increases to 40 ft/sec as a result of the reduction in the pipediameter.


0FD121 33 Rev 1-3, 10/15/04

0m ' ρAv

0m '1ν

Av

0m '1

1.7665 ft 3

lbm

π 1.5 in 2 20 ftsec

1 ft 2

144 in 2

0m 'π 1.5 2 (20)

(1.7665) (144)lbmft 3

in 2

1ft

secft 2

in 2

0m ' 0.556 lbmsec

0m ' 0.556 lbmsec

3,600 sec1hr

' 2001 lbmhr

EXAMPLE I

Steam flows through a 3-inch diameter pipe at 20 ft/sec. The specific volume of thesteam is 1.7665 ft3/lbm. What is the mass flow rate (in lbm/hr) of the steam?

SOLUTION


0FD121 34 Rev 1-3, 10/15/04

Fluid Energy Forms

Recall from Classical Physics that energy is the capacity to do work. Anycharacteristic of a flowing fluid that can be converted into useful work is a form ofenergy possessed by that fluid.

There are six forms of energy that have importance in fluid flow analysis: heat, work,potential energy, kinetic energy, flow energy, and internal energy. Heat and workare energy forms that cross the boundaries of fluid flow systems; potential, kinetic, flowand internal energies are energy forms possessed by the flowing fluid.

As discussed in Classical Physics, heat (Q) is defined to be energy in transition. It isenergy that is transferred from one location to another because there is a temperaturedifference between the two locations. Heat is always transferred from a location ofhigher temperature to a location of lower temperature.

Heat added, Qin: Energy added to the system as a result of heat crossing theboundary of the system into the fluid. Heat is added any time the temperature onthe outside of the fluid system boundary is higher than the temperature on theinside of the fluid system boundary.

For example, the reactor core adds heat to the RCS.

This causes the coolant’s thermal energy (INTERNAL ENERGY) toincrease as it passes through the core.

Heat removed, Qout: Energy removed from the system as a result of heat crossingthe system boundary out of the fluid. Heat is removed from a system any time thetemperature on the outside of the fluid system boundary is less than thetemperature on the inside of the fluid system boundary.

For example, Steam Generators remove heat from the RCS.

The thermal energy (INTERNAL ENERGY) of the reactor coolantdecreases as it passes through the Steam Generator Tubes.

Also, heat losses through piping carrying a hot fluid remove heat from thefluid.


0FD121 35 Rev 1-3, 10/15/04

Also as discussed in Classical Physics, work (W) is defined as the application of aforce through a distance. Work done on a system by a force external to the systemincreases the total energy of the system.

Work done on the fluid, Won: Energy added to the system as a result of anexternally supplied force acting through a distance on the fluid.

For example, a pump does work ON the flowing fluid.

A pump increases the FLOW ENERGY and INTERNAL ENERGY of thefluid (increases the fluid pressure and temperature) as the fluid flowsthrough the pump.

On the other hand, a fluid can do work as it moves through a system; when it does thiswork, its total energy decreases.

Work done by the fluid, Wby: Energy removed from the system as a result of thefluid supplying a force through a distance to an object external to the system.

For example, work is done BY steam as it passes through a turbine becausethe steam applies a force through a distance to the turbine blades.

This process reduces the FLOW ENERGY and the INTERNAL ENERGYof the steam as it flows through the turbine.


0FD121 36 Rev 1-3, 10/15/04

KE 'mv 2

2gc

KE ' Kinetic energy (ft&lbf)

m ' Mass of the fluid in motion (lbm)

v ' Velocity of the fluid in motion ftsec

gc ' 32.2 ft @ lbmlbf @sec2

Energies Possessed By the Flowing Fluid

Potential energy (PE), kinetic energy (KE), internal energy (U) and flow energy (FE) arecharacteristics of a fluid in a fluid flow system. This means that, unlike heat and workwhich describe energies that cross fluid flow system boundaries, these four forms ofenergy are described in terms of and change with particular characteristics of the fluiditself.

Kinetic, potential, and internal energy were introduced in Classical Physics. Reviewingthese definitions:

1. Kinetic energy, KE: energy possessed by the fluid due to its mass and itsvelocity:

:

where:

If the fluid has a speed greater than zero (v > 0), then it possesses kinetic energy. Thespeed associated with the fluid’s kinetic energy is the average speed of the bulk flow ofthe fluid.


0FD121 37 Rev 1-3, 10/15/04

PE 'mghgc

PE ' potential energy, (ft @ lbf)m ' mass of the fluid, (lbm)h ' elevation of the fluid above the reference level (ft)ggc

'1 lbf1 lbm

2. (Gravitational) potential energy, PE: energy due to the elevation of the fluidabove/below a designated reference elevation level:

where:

If the flowing fluid undergoes an increase in elevation, its potential energy increases. Ifthe fluid undergoes an elevation decrease, its potential energy decreases.

3. Internal energy, U: energy due to the (random) molecular vibrational motion ofthe fluid molecules; i.e., due to the temperature of the fluid. Internal energy isthermal energy.

No specific formula is given here for Internal Energy as a function of temperature.Qualitatively, however, a “hot” fluid has more internal energy than “cool” fluid,because its temperature is higher. If the temperature of the fluid increases, itsinternal energy increases.


0FD121 38 Rev 1-3, 10/15/04

FE ' Flow Energy, ft @ lbf

P ' Fluid pressure, lbfft 2

V ' Fluid volume, ft 3

Finally, a flowing fluid also has the capacity to do work due to the fact that its moleculesare under pressure. Similar to a coiled spring, which can apply a force through adistance (do work) if the spring is allowed to unwind, a fluid under pressure can apply aforce through a distance if the pressure of the fluid is allowed to decrease. The “stored”energy associated with the fluid pressure is called flow energy:

4. Flow energy, FE: energy due to the fluid pressure and volume occupied by thefluid under that pressure:

FE = PV

where:

Differences in fluid pressures at two locations in a fluid flow system produce thedriving force that maintains fluid flow. This is why this form of energy is called“flow” energy; it is the fluid pressure difference, and the associated flow energydifference, that causes the fluid to flow.


0FD121 39 Rev 1-3, 10/15/04

The General Energy Equation

The Law of Conservation of Energy states that energy can be neither created nordestroyed. In its most general sense, this Law says that the total energy in the Universeis constant; any event which increases the amount of one form of energy must result inan equivalent decrease in the amount of one or more other forms of energy.

The General Energy Equation (GEE) is a mathematical statement of the Law ofConservation of Energy, applied to a fluid flow system. The GEE says that the totalenergy possessed by the fluid (potential, kinetic, internal, and flow energy) remainsconstant as long as no form of energy is allowed to cross the boundary into or out of thesystem. If work is done on the fluid by a force external to the system, or if heat istransferred across the boundary of the system into the fluid, the total energy possessedby the fluid will increase by the an amount equal to the amount which was supplied. Ifwork is done by the system, or heat is transferred out of the system, the total energypossessed by the fluid will decrease by an amount equal to the amount which wasremoved.

Specifically, the General Energy Equation expresses the relationship between the totalenergy of the fluid at two different locations in a steady flow system. It considers all ofthe energy forms possessed by the fluid and the forms of energy that can be transferredinto or out of the fluid between the two fluid flow locations. Figure 8 (next page)represents a fluid flow system, and it indicates that:

C At point 1, the fluid possesses potential, kinetic, flow, and internal energy.

C At point 2, the fluid possesses potential, kinetic, flow, and internal energy.

C Between points 1 and 2, energy may be ADDED to the fluid (by heat, Qin, crossingthe boundary into the fluid or by work being done on the fluid, Win, by someexternal force), and energy may be REMOVED from the fluid (by heat, Qout,crossing the boundary out of the fluid or work being done by the fluid, Wout).


0FD121 40 Rev 1-3, 10/15/04

Figure 8 Forms of Energy in a Fluid Flow System

For a fluid flow system, the General Energy Equation can be described verbally:

The total energy possessed by a fluid at a given point in a steady flow system (i.e.,KE1 + PE1 + FE1 + U1), plus the net energy added to the fluid between this firstpoint and a second point in the system (i.e., Win - Wout and Qin - Qout), is equal tothe total energy possessed by the fluid at the second point in the system (i.e., KE2+ PE2 + FE2 + U2).


0FD121 41 Rev 1-3, 10/15/04

KE1%PE1%FE1%U1% (Win&Wout)% (Qin&Qout) ' KE2%PE2%FE2%U2

OR

KE1%PE1%FE1%U1%Win%Qin ' KE2%PE2%FE2%U2%Wout%Qout

KE: Kinetic energy of the fluidPE: Potential energy of the fluidFE: Flow energy of the fluid

U: Internal energy of the fluidWin: Work done on the fluid as it moves from Point 1 to Point 2

Wout: Work done by the fluid as it moves from Point 1 to Point 2Qin: Heat transferred into the fluid as it moves from Point 1 to Point 2

Qout: Heat transferred out of the fluid as it moves from Point 1 to Point 2

The mathematical formula equivalent to the statement of the General Energy Equationis:

where:

Therefore, the General Energy Equation accounts for the total energy possessed by afluid at a given location (KE1 + PE1 + FE1 + U1), the energies added or lost between thegiven location and another location in the system (Win - Wout and Qin - Qout), and theresulting total energy possessed by the fluid at the second location in the system (KE2 +PE2 + FE2 + U2).

The General Energy Equation (GEE) is the basis for most energy analysis calculationsused in power plant fundamentals. In particular, a significant portion of the formulas thatare used for calculations of heat transfer, pump work, core power, etc., are derived fromthe GEE. The GEE has numerous forms, each one applicable to particular analyticalsituations. Variations to the GEE shown above will be discussed on an as-needed basis.


0FD121 42 Rev 1-3, 10/15/04

KE1 % PE1 % FE1 ' KE2 % PE2 % FE2

The Basic Bernoulli Equation

Consider a fluid flow system having:

C No energy transfers across the system boundaries, that is

Win = Won = Qin = Qout = 0

C No change in fluid temperature (i.e, no change in fluid internal energy), that is

U2 = U1

The General Energy Equation applied to this special set of circumstances reduces to:

The equation above is called the Basic Bernoulli Equation for frictionless,incompressible fluid flow in a steady flow system.

This Bernoulli Equation can be used to qualitatively explain how a change in fluidvelocity (v), fluid elevation (h), or fluid pressure (P) will affect the value of the other twoparameters. The Examples which follow will illustrate.


0FD121 43 Rev 1-3, 10/15/04

KE1 % PE1 % FE1 ' KE2 % PE2 % FE2

KE1 % FE1 ' KE2 % FE2

EXAMPLE J

For the constant mass flow rate system shown below, what is the effect on fluidpressure by the reduction in pipe diameter from 2" at point 1 to 1" at point 2? Assumefluid temperature remains constant from point 1 to point 2.

SOLUTION

From the Bernoulli Equation,

Because there is no fluid elevation change between point 1 and point 2, thepotential energy at point 1 is equal to the potential energy at point 2, that is, PE1 =PE2. Therefore the Bernoulli Equation reduces to:

Example H, earlier, showed that the reduction in piping diameter causes the fluidvelocity to increase from 10 ft/sec to 40 ft/sec. Therefore, since v2 > v1, and kineticenergy depends on velocity, this means that the reduced diameter piping causesthe fluid kinetic energy to increase, that is, KE2 > KE1.

Since KE2 > KE1 and KE1 + FE1 = KE2 + FE2, this means that fluid flow energymust decrease as a result of the reduction in pipe diameter, that is, FE2 < FE1.Therefore, since flow energy depends on pressure, this means that the fluidpressure decreases as a result of the reduction in pipe diameter, that is P2 < P1.

Conclusion: A pipe diameter reduction increases the kinetic energy of theflowing fluid (increases the fluid speed) and a consequently decreases thefluid flow energy (decreases the fluid pressure).


0FD121 44 Rev 1-3, 10/15/04

KE1 % PE1 % FE1 ' KE2 % PE2 % FE2

PE1 % FE1 ' PE2 % FE2

EXAMPLE K

For the constant mass flow rate system shown below, what is the effect on fluid velocityand fluid pressure by the 12" increase in fluid elevation from point 1 to point 2? Thediameter of the piping at point 2 is the same as it is at point 1.

SOLUTION

From the Bernoulli Equation,

Because there is no diameter change between point 1 and point 2, the velocity of thefluid at point 1 is equal to the velocity of the fluid at point 2, that is v1 = v2. Therefore,since the velocity of the fluid does not change, the kinetic energy of the fluid does notchange, that is, KE1 = KE2.

Therefore the Bernoulli Equation reduces to:

Because the elevation of the fluid increases, the potential energy of the fluid at point 2 isgreater than the potential energy of the fluid at point 1, that is, PE2 > PE1.


0FD121 45 Rev 1-3, 10/15/04

Since PE2 > PE1 and PE1 + FE1 = PE2 + FE2, this means that flow energy mustdecrease as a result of the increase in pipe elevation, that is, FE2 < FE1. Therefore,since flow energy depends on pressure, this means that the fluid pressure decreasesas a result of the increase in pipe elevation, that is P2 < P1.

Conclusion: A fluid flow elevation increase causes an increase in thepotential energy of the flowing fluid and a consequent decrease in the fluidflow energy (a decrease in the fluid pressure).


0FD121 46 Rev 1-3, 10/15/04

Effect of Friction in a Fluid Flow System

One of the assumptions of the Basic Bernoulli Equation was that the internal energy offlowing fluid remained constant as the fluid flowed through a system. Because internalenergy is directly related to the temperature of the fluid, this means that the BernoulliEquation assumed fluid temperature remained constant as the fluid flowed.

However, independent of whether heat is transferred into or out of a system by a heatsource or a heat sink, fluid friction will cause the temperature and consequently theinternal energy of the flowing fluid to increase:

Water molecules experience frictional forces as they collide with each other andslide over each other.

This results in an increase in the vibrational energy of the molecules, i.e., thetemperature of the water increases.

Since the water’s internal energy depends on its temperature, the increase in thetemperature of the water due to the friction causes the internal energy of the waterto increase.

EXAMPLE L

Using the General Energy Equation, determine the effect of fluid friction on the fluid’sflow energy (consequently, the fluid’s pressure) in the fluid flow system shown below.Assume the system is perfectly insulated.


0FD121 47 Rev 1-3, 10/15/04

KE1 % PE1 % FE1 % U1 % Win % Qin ' KE2 % PE2 % FE2 % U2 % Wout % Qout

FE1 % U1 ' FE2 % U2

SOLUTION

The GEE for any steady flow process is:

The following assumptions apply for the fluid flow system shown:

Win = Wout = 0, because no work is done or by the fluid.

Qin = Qout = 0, because the pipe is perfectly insulated.

KE1 = KE2, because the pipe diameter (and hence the fluid velocity) does notchange.

PE1 = PE2, because there is no elevation change.

Thus, the GEE reduces to:

Since fluid friction results in an increase in the temperature of the water, theinternal energy of the water increases:

U2 > U1

Since U2 > U1, and FE1 + U1 = FE2 + U2, this means that:

FE2 < FE1

Since the fluid’s flow energy decreases, the fluid’s pressure drops.

Conclusion: Fluid friction in a flowing fluid causes an increase in the fluidtemperature and therefore an increase in the internal energy of the fluid. Consequently, the fluid flow energy decreases (fluid pressure drops) due tofluid friction.


0FD121 48 Rev 1-3, 10/15/04

Fluid Viscosity

Viscosity is a property of a fluid that describes the readiness with which a fluid flows. The greater the viscosity of a fluid is, the more resistant that fluid is to flow. Viscosity isan indication of the cohesive properties and the relative distances between theindividual molecules of the fluid.

The coefficient of viscosity, designated by the symbol µ (mu), is a measure of the fluid'sresistance to flow. The English Engineering System unit of viscosity is lbf-sec/ft2. Thegreater the magnitude of µ is, the more resistant the fluid will be to flow. The conceptsof viscosity and coefficient of viscosity are often used interchangeably.

The average distance between adjacent molecules of the fluid is reflected by the densityof the fluid. The greater the density of the fluid is, the closer the molecules of the fluidare to each other. For example, water at 50EF has greater density than water at 200EF;the average distance between adjacent molecules of the 50EF water is less than that ofthe 200EF water. Consequently, the 50EF water is more likely to experienceinteractions between its molecules that result in more resistance to flow of the bulk fluidthan is the case for 200EF water. The 50EF water is said to be more viscous than the200EF water is; it has a greater viscosity.

In a gas, viscosity is a result of intermolecular vibration and molecular collisions. Viscosity in liquids is caused predominantly by intermolecular friction: the resistance torelative motion between two bodies in contact.

In liquids, increased molecular vibration caused by rising temperature tends to weakenthe attraction between adjacent molecules, reducing the viscosity of the liquid. Ingases, however, molecular attraction is very weak. Molecular vibration, increasing withtemperature, results in more collisions and interference between molecules moving pasteach other. The viscosity of gases increases with temperature.


0FD121 49 Rev 1-3, 10/15/04

Table 1 shows the effect of temperature and pressure changes on the viscosity of waterand steam.

Note: Steam values are italicized.

1 10 100 500 1000 2000 5000EF psia psia psia psia psia psia psia

1500 8.6 8.6 8.6 8.7 8.7 8.7 9.21000 6.3 6.3 6.3 6.3 6.4 6.6 7.4500 3.9 3.9 3.9 3.8 21.6 22.0 23.1200 2.5 2.5 62.6 62.8 62.9 63.2 64.0100 142.0 142.0 142.0 142.0 142.1 142.1 142.050 271.4 271.4 271.4 271.2 271.0 270.6 269.332 366.1 366.1 366.1 365.7 365.3 364.5 361.9

Absolute Viscosity µ (lbf-sec/ft2 x 10-7)

Table 1 Fluid Viscosity versus Temperature/Pressure

Table 1 indicates that water maintained at the constant temperature experiencesminimal changes in viscosity with increasing or decreasing pressure. However,temperature changes cause more significant changes in fluid viscosity. The viscosity ofliquid water decreases with increasing temperature; the viscosity of steam increaseswith temperature increase. Therefore, liquid water becomes less resistant to flow as itstemperature increases, and steam becomes more resistant with temperature increases.


0FD121 50 Rev 1-3, 10/15/04

The viscosity of a flowing fluid is a direct indication of the work which must be done tomaintain that fluid in steady flow (i.e., constant mass flow rate). For example, ifsomething occurs (such as a temperature decrease of flowing water) that results in anincrease in its viscosity, the fluid now offers more resistance to flow at the desired flowrate. This resistance to flow is the frictional effect which must be overcome by appliedexternal forces (such as the forces exerted by the impeller of a pump) in order tomaintain the desired flow rate. Thus, the viscosity of water is a significant factoraffecting fluid flow characteristics.

Viscosity is also a significant factor in the choice of lubricating fluids put betweenmoving and stationary parts of plant components, such as rotating shafts in stationaryshaft sleeves. A lubricant is chosen based on its viscosity and its expected temperaturerange within the component parts it lubricates, as well as the rotational speeds andclearances in the component. If a lubricant's temperature is either excessively high orexcessively low relative to the normal operating temperature range, a loss of adequatelubrication will result.

The reasons for the loss of lubrication can be briefly summarized as follows:

C At excessively high temperatures, the decreased viscosity (decreased cohesion)and the chemical breakdown of the lubricant result in the loss of a sufficientlubricant layer between moving parts.

C At excessively low temperatures, high cohesive forces of the lubricant can preventthe lubricant from establishing the adequate lubricant layer between moving parts.

Laminar and Turbulent Flow

There are three basic classifications of fluid flow in a pipe: Laminar flow, turbulent flowand mixed flow.

Laminar flow is characterized by the fluid flowing in smooth, orderly layers. In laminarflow one layer of fluid glides smoothly over the adjacent layer as shown in Figure 9. There is minimal transfer of fluid particles between adjacent layers. The further aparticular fluid layer is from the fixed piping surface, the greater is the speed of the layerin the direction of bulk flow.


0FD121 51 Rev 1-3, 10/15/04

Turbulent flow is characterized by the irregular, random motion of the fluid particles ina transverse direction to the direction of bulk flow. This is also shown in Figure 9. Themajority of the bulk flow is moving at essentially constant speed. Near the stationarypiping wall, however, the speed of the fluid is much slower because of the combinedresistance effects created by the roughness of the pipe surface and the viscositycharacteristics of the fluid.

Figure 9 Laminar Flow and Turbulent Flow Characteristics

Classic turbulent flow is actually a combination of turbulent flow and laminar flow. Sinceflow very near the piping surface moves slowly compared to the bulk flow in the centerof the pipe, it has characteristics of laminar flow. This laminar region is called theboundary layer. The size of the boundary layer depends on the fluid's viscosity, surfaceroughness, surface geometry, and relative velocity of the fluid. Higher average fluidvelocity results in a thinner boundary layer; a higher fluid viscosity results in a thickerboundary layer.


0FD121 52 Rev 1-3, 10/15/04

Regardless of the nature of the fluid flow (laminar, turbulent, or mixed), there exists alayer of fluid immediately adjacent to the piping surface whose velocity is zero. Thethickness of this layer is determined primarily by the relative roughness of the pipingsurface. The fluid layer immediately adjacent to the fixed layer has nonzero velocity,and relative layer velocities increase rapidly with their distance from the surface. Whenfluid flow is laminar, the fastest moving layer is the one found at the center of the pipe. When fluid flow is turbulent, the fastest moving layer is the last layer of the boundarylayer, the layer closest to the center of the pipe. Because of the turbulence in the regionoutside of the boundary layer, the bulk fluid velocity at any location in this area isessentially constant.

Most of the flow in SONGS systems is highly turbulent. Turbulent flow versus laminarflow results in both a positive and negative operational effect:

• The more turbulent the flow is through a heat source/heat exchanger, the moreefficiently the flow will pick up/give up heat. That is turbulence enhances heattransfer. Factors affecting heat transfer will be discussed in detail in a laterchapter.

• The more turbulent the flow is through any system, the greater are the frictionaleffects of the flow. More friction means larger pressure drops, which means largerflow energy input requirements from the pump(s) supplying the flow. The effects ofturbulence on system pressure will be discussed in detail in Chapter 3.


0FD121 53 Rev 1-3, 10/15/04

Figure 10

Buoyant Force and Drag Force on Reactor Core

The principle of hydrostatic pressure can be used to explain why objects partially ortotally submerged in a liquid experience an upward force, called the buoyant force,exerted on the object due to the fluid. For objects which float, the buoyant force is equalin magnitude but opposite in direction to the weight of the object. Objects which aresubmerged also experience buoyant force producing the apparent effect of decreasingthe object's weight. The following development quantifies this buoyant force.

Consider the block submerged in the liquid shown in Figure 10 below:

The liquid exerts forces on all six surfaces of the block. The horizontal forces exertedon opposite sides of the block cancel each other. However, the upward force F1 due tothe fluid pressure below the block and the downward force F2 due to fluid pressureabove the block do not cancel, because the pressure of the liquid at elevation 1 isgreater than the pressure of the liquid at elevation 2. The difference F1 - F2 is thebuoyant force exerted on the block.


0FD121 54 Rev 1-3, 10/15/04

F1 & F2 ' P1A & P2A ' P1 & P2 A

P1 & P2 ' ρh ggc

F1 & F2 ' ρh ggc

A

F1 & F2 ' ρVfggc

,

Where Vf ' Volume of Fluid Displaced

B ' ρVfggc

Since P = F/A, we know that F1 = P1A and F2 = P2A. Therefore,

(1)

P1 - P2 is the hydrostatic pressure due to a column of the fluid h feet high, thus,

Substituting this equation into (1),

The product of h and A is just the volume of the block. This volume also represents thevolume of the fluid displaced by the block. Thus,

Since ρ is the fluid density and Vf is the volume of the fluid displaced, ρVf represents themass (in lbm) of the fluid displaced. Therefore, ρVf(g/gc) represents the weight of thefluid displaced. The buoyant force has now been quantified:

The buoyant force B exerted by a fluid on an object in the fluid is equal to themagnitude of the weight of the fluid displaced by the object. If the fluid has densityρ and volume Vf, then the buoyant force is


0FD121 55 Rev 1-3, 10/15/04

B ' ρVfggc

'1

0.02221 ft 3

lbm

(900 ft 3) 1 lbf1 lbm

' 40,522 lbf

Buoyant force acts on Unit 2/3 reactor cores. The Combustion Engineering System 80reactor core sits in the reactor vessel, unsupported by any restraining devices otherthan its lower support. Since the vessel is full of water (the core is totally covered), thecore experiences an upward force equal in magnitude to the weight of the water itdisplaces in the vessel. Example M below demonstrates the effect of buoyant force onthe apparent weight of a reactor core.

Example M

A reactor core sits in a reactor vessel, unsupported by any restraining devices otherthan its lower support. The total volume of the core is 900 ft3. The weight of the core is407,430 lbf.

a. Determine the effective weight of the core if it is totally submersed in water at564EF and 2250 psia.

b. Determine the effective weight of the core if it is totally submersed in water at70EF and 14.7 psia.

Solution

The weight of the core will be effectively reduced by an amount equal to the upwardbuoyant force caused by the water it displaces.

a. The specific volume of the water is obtained from the Steam Tables: ν = 0.02221ft3/lbm. The volume of the water displaced is 900 ft3. Therefore,


0FD121 56 Rev 1-3, 10/15/04

Core Effective Weight ' Actual Core Weight & Buoyant Force' 407,430 lbf & 40,522 lbf' 366,908 lbf

B ' ρVfggc

'1

0.016050 ft 3

lbm

(900 ft 3) 1 lbf1 lbm

' 56,075 lbf

Core Effective Weight ' Actual Core Weight & Buoyant Force' 407,430 lbf & 56,075 lbf' 351,355 lbf

The effective weight of the core under these conditions will be

Therefore, the buoyant force of water under these conditions effectively reducesthe weight of the core to 90% (366908/407430 X 100) of its actual value.

b. The specific volume of the water is obtained from the Steam Tables: v = 0.016050ft3/lbm. The volume of the water displaced is 900 ft3. Therefore,

The effective weight of the core under these conditions will be

Therefore, the buoyant force of the water under these conditions effectivelyreduces the weight of the core to 86% of its actual value.

Example M shows that the downward force which holds the core in its reactor vesselposition decreases as the temperature of the water decreases. This occurs becausethe buoyant force of the water is directly proportional to its density; cooler water exertsgreater upward force on a totally submersed body than does hotter water.


0FD121 57 Rev 1-3, 10/15/04

Buoyant force is not the only upward force acting on the core. The flow of the reactorcoolant (water) through the core enters at the bottom of the core and flows through itsfuel assemblies in the upward direction. The frictional drag forces caused by thisupward coolant flow are also in the upward direction. If the magnitude of the buoyantforce and the frictional drag forces combined is great enough, the core can be liftedfrom its normal position. Core uplift can result in consequent damage to core internals.

The probability that a core uplift event will occur varies with the coolant flow rate andwith the temperature of the reactor coolant:

C If coolant flow rate through the core increases, the velocity of the coolantincreases. Therefore, the upward drag force of the coolant on core surfaceswill be greater.

C If coolant temperature decreases, the density of the coolant increases. Therefore, since the buoyant force on the core is directly proportional to thedensity of the water it displaces, lower temperature coolant will exert a greaterupward force on the core.

C If coolant temperature decreases, the viscosity of the coolant increases (Table 1). Therefore, since the drag force of upward-flowing coolant variesdirectly with the viscosity (cohesiveness) of the water, a decrease in coolanttemperature will increase the drag force on the core surfaces.

Combustion Engineering reactor vessels have an operating restriction that is based onboth buoyant force and the coolant viscosity. Specifically, procedure restrictionsprevent starting the fourth Reactor Coolant Pump with coolant temperature less than500EF. Starting the fourth RCP increases coolant flow rate through the core, increasingthe upward drag force on the core. If this is done at temperatures less than 500EF,when both bouyant force and viscous drag forces are greater than they would be athigher temperatures, the total upward force could be sufficient to lift fuel assemblies outof their lower support structure. This can result in damage to the fuel assemblies, aswell as misalignment of the control element assembly (CEA) drive mechanisms.


0FD121 58 Rev 1-3, 10/15/04

6p ' m6v

6p ' momentum

m ' mass

6v ' velocity

Fluid Momentum

A flowing fluid has both mass and velocity, and each of these characteristics contributesto the fluid’s resistance to changes to its present state of motion. The property ofmoving fluid described by the product of its mass and its velocity is called fluidmomentum, and is denoted by the symbol

6 p :

Momentum: the product of the mass and velocity of a flowing fluid.

where:

Momentum results when an unbalanced force exerted on a fluid causes it to accelerateit to some velocity ( 6v ), in accordance with Newton’s Second Law of Motion. Therefore,in order to change a fluid’s momentum, a force must be exerted for a given amount oftime.


0FD121 59 Rev 1-3, 10/15/04

6I ' 6Ft

6I ' impulse

6F ' force

t ' time

Newton’s Second Law (discussed in the Classical Physics lesson) tells us that whenan unbalanced force is applied to a mass over a time interval, the mass accelerates;i.e., its velocity changes. The product of the applied force and the time over which theforce acts is called impulse:

Impulse (I): the product of the force (F) applied to a mass and the time (t) throughwhich the force acts:

where:


0FD121 60 Rev 1-3, 10/15/04

Impulse ' Change in momentum

Ft ' mvf & mvi

Ft ' ∆p

mvf ' object )s momentum after the force has been applied (final momentum)

mvi ' object )s momentum before the force was applied (initial momentum)

∆p ' change in the object )s momentum ())∆)) is the symbol meaning ))change in )))

Therefore, because the velocity changes, this applied force results in a change inmomentum of the mass. The impulse imparted to the object is equal to the change inthe momentum of the mass:

where:

For example, consider the fluid flow situation shown in Figure 11. The diameter of thepipe shown is constant, and the speed of the fluid flowing through the pipe is alsoconstant. The velocity of the fluid, however, changes at the bend of the pipe, notbecause the speed of the flow changes, but because the direction of flow changes. Thebend in the pipe forces the fluid velocity to change, i.e., the pipe is applying a force tothe fluid at the location of the bend which causes the direction of fluid flow, but not itsspeed, to change. Since the pipe is not moving (not accelerating), this means that thefluid is exerting a force of equal magnitude but opposite direction on the pipe bend as itsdirection is changed.

Figure 11


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PFt ' m Pv2 & Pv1

PF 'mtPv2 & Pv1

The impulse equation can be used to determine the magnitude and direction of theforce exerted by the pipe which is forcing the direction change of the velocity.

The term m/t above represents the mass flow rate of the fluid, which is constant. Thespeed of the fluid is the same at location 1 as it is at location 2, but the directions of floware different at these two locations. In other words, the velocity changed from location 1to location 2. The force exerted by the fluid on the pipe is constant, and it is directed inthe same direction as the difference of the final and initial velocity vectors.

If the flow rate m/t were to suddenly increase, the equation above shows that themagnitude of the force applied by the fluid on the pipe through the pipe bend would alsosuddenly increase. The piping structure could be seriously damaged by this. In fluidflow systems where the flow rate is variable and fluid flow must change direction atsome location we frequently see piping snubbers installed to lessen the impact of thechanging force.


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Impulse ' Change in momentum

Ft ' mvf & mvi

Water Hammer

Water hammer is the term used to describe the overall effect of severe pressuretransients in a piping system. More specifically, it is the formation of a fluid pressureshock wave resulting from sudden changes in fluid flow velocity, and the subsequentdissipation of shock wave energy at the locations of flow obstructions.

As shown in the previous section, anytime the velocity of the fluid changes, themomentum of the fluid changes. If the speed of the flow is suddenly reduced, themomentum of the flow is also reduced. This change in momentum exerts an impulse atthe location where the speed change occurred:

The magnitude of the force (F) that will be applied at the location depends on the time(t) it took for the speed of the fluid to change. If the speed change occurs in a veryshort time, the magnitude of the force will be much larger.

Large forces exerted by “quick” fluid momentum changes can cause severe damage topiping and system components. The pressure shock wave produced by the quick dropin fluid kinetic energy moves at the speed of sound through the fluid, and when it hits anobstruction it is forced to change direction. A force each time the direction is changed,and the magnitude of the force is very large because the direction change occurs in avery short time. This is the “hammering” effect that is heard; every time the shock wavechanges direction the “bang” of the force is heard. The hammering continues until theenergy of the shock wave is totally dissipated. Causes/Methods of Prevention of Water Hammer

Cause 1: Rapid valve closure. The more quickly the valve is closed, the morequickly the fluid momentum is changed. Quick changes in fluid momentum resultin large forces applied at the location of the momentum change, and a high energyshock wave which must dissipate its energy.

Method of Prevention: Valves should be closed slowly.


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Cause 2: Introduction of flow to empty piping. An empty pipe offers very littleresistance to flow through it. Therefore, the flow accelerates to high velocities veryquickly. As soon as the flow encounters an obstruction to flow, such as an angledpipe bend, a large momentum change occurs in a short period of time; waterhammer occurs.

Method of Prevention: Piping systems should be filled and vented prior tostarting a pump supplying the system.

Cause 3: Introduction of low temperature liquid water into piping containingsteam. Cool water entering a pipe containing steam will cause rapid condensationof the steam. The steam volume collapses quickly, and the water accelerates intothis voided area. When the water runs into itself or a pipe wall at this location, alarge fluid momentum change occurs in a very short time frame; water hammeroccurs.

Method of Prevention: Water flow systems should be vented to preventsteam pockets from forming, before flow of water is introduced to thesesystems.

Cause 4: Introduction of steam flow into piping partially filled with liquidwater. When steam comes in contact with the cooler water, the steam will rapidlycondense. The steam volume collapses quickly, and the water accelerates intothis voided area. When the water runs into itself or a pipe wall at this location, alarge fluid momentum change occurs in a very short time frame; water hammeroccurs.

Method of Prevention: Steam flow systems should be drained before flow ofsteam is introduced to these systems.


0FD121 64 Rev 1-3, 10/15/04

Definitions

Absolute pressure (Pabs) - pressure expressed relative to a perfect vacuum as the zeroreference. English unit: psia.

Absolute Zero Temperature - the temperature at which all molecular motion of asubstance ceases; it is the temperature of the substance at which its thermal energy isequal to zero.

Atmospheric pressure (Patm) - pressure produced by the weight of the atmosphere.

Buoyant Force - the upward force exerted by a fluid on an object partially or totallysubmersed in the fluid. The buoyant force on an object is always equal to the weight ofthe fluid displaced by the object.

Density (ρ) - ratio of the mass, m, possessed by a substance to the volume, V, itoccupies. ρ = m/V. English unit: lbm/ft3.

Fluid - any substance which flows.

Fluid momentum: the product of the mass and velocity of a flowing fluid.

Flow Energy - energy due to the fluid pressure and the volume occupied by the fluidunder that pressure.

Gauge pressure (Pgauge) - pressure expressed relative to atmospheric pressure as thezero reference. English unit: psig.

Hydrostatic pressure (∆P) - pressure exerted by the weight of a fluid column. Englishunit: psi.

Impulse (I): the product of the force (F) applied to a mass and the time (t) throughwhich the force acts.

Internal Energy - energy due to the (random) molecular vibrational motion of the fluidmolecules, i.e., due to the temperature of the fluid.

Kinetic Energy - energy due to the mass and velocity of a fluid.


0FD121 65 Rev 1-3, 10/15/04

Definitions, continued:

Laminar flow - fluid flow in smooth, orderly layers

Pressure (P) - ratio of the total force exerted by a substance to the total area to whichthe force is applied. P = F/A. English unit: lbf/in2 (psi).

Potential Energy - energy possessed by a substance due to its elevation above areference elevation level.

Saturated liquid water - liquid water that is at saturation temperature for the pressureat which it exists. Any heat added to this water will cause the water to boil.

Saturation pressure (Psat) - the pressure at which water will boil at a given temperatureif heat is added to the water.

Saturation temperature (Tsat) - the temperature at which water will boil at a givenpressure if heat is added to the water.

Specific volume (ν) - ratio of the volume, V, occupied by a substance to mass, m, itpossesses. = V/m. English unit: ft3/lbm.

Subcooled (liquid) water - water that exists at a temperature below saturationtemperature for the pressure of the water. Any heat added to this water will cause thetemperature of the water to increase.

Turbulent flow - fluid flow with irregular, random motion of the individual fluid particleswith respect to the direction of bulk flow.

Vacuum pressure (Pvac) - pressure expressed relative to atmospheric pressure as thezero reference; it is the amount by which pressure is less than atmospheric pressure,expressed as a positive number. English unit: psiv.

Viscosity - Property of a fluid which quantifies the readiness with which the fluid flows.The larger the magnitude of a fluid’s viscosity is, the more resistance the fluid offers toflow.

Water hammer - a rapid change in the momentum of a flowing fluid that creates apressure shock wave which damages components of the fluid flow system


0FD121 66 Rev 1-3, 10/15/04

Exercises

Exercise 1

Objective 2: Define fluid density and fluid specific volume, and state therelationship between them.

1. The specific volume of a certain liquid is 0.0190 ft3/lbm. Find the density of theliquid.

2. The density of a certain liquid is 59.3 lbm/ft3. Find the specific volume of the liquid(to 4 decimal places).


0FD121 67 Rev 1-3, 10/15/04

Exercise 2

Objective 1: Given a gauge pressure reading or a vacuum pressure reading,convert the reading to an absolute pressure reading.

Objective 3: Given a table of conversion factors, perform conversions betweenpressures expressed in the following pressure units:

C Pounds per square inch (psi)C Inches of mercury (in. Hg.)C Feet of water (ft. H20)

1. A pressure gauge indicates 40 psig. Calculate the equivalent absolute pressure. The atmospheric pressure is 14.7 psia.

2. The pressure of a fluid in a rigid container is 8 psiv. Determine the equivalentabsolute pressure. The atmospheric pressure is 14.7 psia.

3. A gauge reads 30 psig. Calculate the absolute pressure.

4. A vacuum gauge reads 24 inches Hgvac. Find the absolute pressure in psia.

5. How many ft H2O is equivalent to a pressure difference of 200 psi? Assume thewater is at standard density (62.4 lbm/ft3).


0FD121 68 Rev 1-3, 10/15/04

Exercise 3

(Note: Because you have no Steam Tables available, you will NOT be tested bythis type of question!!)

Using the C-E Steam Tables:

1. At TSAT = 364EF, determine PSAT

2. At PSAT = 1150 psia, determine TSAT

3. At TSAT = 101.74EF, determine PSAT

4. At PSAT = 1028.49 psia, determine TSAT

5. At TSAT = 609EF, determine PSAT

6. At PSAT = 1410 psia, determine TSAT


0FD121 69 Rev 1-3, 10/15/04

Exercise 4

Objective 4: State how the density and the specific volume of subcooled liquidwater change as

C the temperature of the water changes

C the pressure of the water changes

1. TRUE or FALSE:

a. If the temperature of subcooled liquid water increases from 140EF to160EF, then the density of the water decreases and the specific volume ofthe water increases.

b. A 10EF decrease in the temperature of subcooled liquid water will causeapproximately the same increase in the water density as will a 10 psipressure increase.

c. For subcooled liquid water, the temperature of the water (not the pressure)is the primary factor determining the density of the water.


0FD121 70 Rev 1-3, 10/15/04

Exercise 5

Objective 5: State and apply the mathematical formulas for:

C Volumetric Flow Rate

C Mass Flow Rate

1. 120EF, 400 psia water is flowing through an 18" diameter pipe at 12 ft/sec.

a. Determine the flow rate, in gallons per minute.

b. The Steam Tables say that the specific volume of 120EF, 400 psia water is0.016204 ft3/lbm. Determine the flow rate, in pounds mass per hour.

2. A Reactor Coolant Pump delivers coolant flow through 30" diameter cold legpiping at 45 ft/sec. If the Reactor Coolant is at 540EF and 2,250 psia,

a. Determine the flow rate, in gallons per minute.

b. The Steam Tables say that the specific volume of 540EF, 2,250 psia water is0.02146 ft3/lbm. Determine the flow rate, in pounds mass per hour.


0FD121 71 Rev 1-3, 10/15/04

Exercise 6

Objective 6: Apply the General Energy Equation to determine the effect of systemconfiguration changes on system pressure.

1. A fluid flow system has a constant mass flow rate (maintained by a pump in thesystem). For each of the system segments show below, determine:

C How system pressure at point 2 compares to system pressure at point 1, and

C Explain why system pressure changes as it does.

a. h2 = h1, d2 = d1, and point 2 is 40 feet downstream of point 1

b. d1 = d2, point 1 is 8 inches higher than point 2, and frictional effects are negligible.

.


0FD121 72 Rev 1-3, 10/15/04

Exercise 6, continued

C How system pressure at point 2 compares to system pressure at point 1, and

C Explain why system pressure changes as it does.

c. d1 = 2 inches, d2 = 6 inches, h1 = h2, and frictional effects are negligible.

d. d1 = 5 inches, d2 = 2 inches, point 1 is 8 inches lower than point 2.

e. d1 = 5 inches, d2 = 7 inches, point 1 is 8 inches higher than point 2, and frictionaleffects are negligible.


0FD121 73 Rev 1-3, 10/15/04

Exercise 7

Objective 7: Describe the concept of fluid viscosity.

1. How does the viscosity of a liquid vary with the temperature of the liquid?

2. How does the viscosity of a fluid vary with the pressure of the liquid?

Exercise 8

Objective 8: Compare/contrast laminar flow and turbulent flow characteristics.

TRUE or FALSE:

1. Turbulent flow through a given length of pipe results in a greater pressureloss than does laminar flow through the same section of pipe.

2. Laminar flow through a heat exchanger results in better heat transfer thandoes turbulent flow through the same heat exchanger.


0FD121 74 Rev 1-3, 10/15/04

Exercise 9

Objective 9: Describe the buoyant force exerted by a fluid on an object partially ortotally submersed in that fluid.

1. What happens to the buoyant force exerted by the water in the reactor vesselduring the time interval following a reactor trip (100% power to 0% power)? Whydoes this happen?

2. If an object submersed in a liquid is freely suspended in the liquid at a particulardepth, what can be said about the magnitude of the fluid’s buoyant force acting onthe object?


0FD121 75 Rev 1-3, 10/15/04

Exercise 10

Objective 10: Identify characteristics of water hammer, including:

• its definition• fluid flow conditions that can trigger a water hammer event• potential consequences of a water hammer event• methods that will minimize the possibility of a water hammer event

occurring

1. For each of the following fluid flow system changes, identify whether the changewould increase or decrease the possibility of a water hammer event occurring:

a. Venting a system prior to starting a pump in the system as opposed tostarting the pump prior to venting.

b. Quickly closing a valve in a flowing system as opposed to closing it moregradually.

c. Maintaining the discharge line completely filled for an automaticallystarting pump as opposed to allowing it to be partially or completely empty.

d. Draining all liquid out of a steam line before steam flow is introduced intothe line as opposed to introducing the steam flow with liquid water in theline.

e. Temperature of the flowing water increases above saturation temperaturefor the pressure of the water as opposed to staying constant belowsaturation temperature.

f. Steam flow system lines are warmed prior to initiating steam flow asopposed to introducing steam flow into cold lines.


0FD121 76 Rev 1-3, 10/15/04

ρ '1ν

'1

.0190 ft 3

lbm

' 52.6 lbmft 3

ν '1ρ

'1

59.3 lbmft 3

' 0.0169 ft 3

lbm

Exercise Solutions

Exercise 1

1.

2.


0FD121 77 Rev 1-3, 10/15/04

1. Pabs ' Patm % Pgauge

Pabs ' 14.7 psi % 40 psi

Pabs ' 54.7 psia

2. Pabs ' Patm & Pvac

Pabs ' 14.7 psi & 8 psi

Pabs ' 6.7 psia

3. Pabs ' Patm % Pgauge

Pabs ' 14.7 psi % 30 psi

' 44.7 psia

4. 24 inHgvac0.491 psi1 inHg

' 11.8 psiv

Pabs ' Patm & Pvac ' 14.7 psi & 11.8 psi ' 2.9 psia

5. 200 psi1 ft H2 O0.433 psi

' 462 ft H2 O

Exercise 2


0FD121 78 Rev 1-3, 10/15/04

14

'x

1686.1 & 1637.3Y x ' 12.2 Y P ' 1637.3 % 12.2 ' 1649.5 psia

Exercise 3

1. 160.903 psia

2. 561.82EF

3. 1.0 psia

4. 548EF

5. From Table 2, T = 609.05EF is close to the given 609EF value. Therefore, the PSAT. 1650 psia.

OR: Linear Interpolation from Table 1:

608EF -----------> 1637.3 psia609EF -----------> ?612EF -----------> 1686.1 psia

6. From Table 1, 1410 psia is saturation pressure for 588EF.


0FD121 79 Rev 1-3, 10/15/04

1a. 0V ' Av ' π 9 in 2 12 ftsec

1 ft 2

144 in27.48 gal

1 ft 360 sec1 min

'π (92)(12)(7.48)(60)

144in 2

1ft

secft 2

in 2galft 3

secmin

' 9,517 galmin

0m ' ρ 0V

'1

0.016204 ft 3

lbm

9517 galmin

1 ft 3

7.48 gal60 min

1 hr

' 4.71E6 lbmhr

Exercise 4

a. True

b. False. Liquid water density changes significantly as water temperature changes.Pressure changes have a negligible effect on the density of liquid water.

c. True. The major factor determining the density of liquid water is its temperature,not its pressure.

Exercise 5

1b. The specific volume of 120EF, 400 psia water is determined from the SteamTables, Table 1:

= 0.016204 ft3/lbm.

Therefore,


0FD121 80 Rev 1-3, 10/15/04

2a. 0V ' Av ' π 15 in 2 45 ftsec

1 ft 2

144 in 27.48 gal

1 ft 360 sec1 min

'π (15)2(45)(7.48)(60)

144in 2

1ft

secft 2

in 2galft 3

secmin

' 99,137 galmin

0m ' ρ 0V

'1

0.02146 ft 3

lbm

99,137 galmin

1 ft 3

7.48 gal60 min

1 hr

' 3.71E7 lbmhr

Exercise 5, continued

2b. The specific volume of 540EF and 2,250 psia water is determined from the SteamTables, Table 1:

= 0.02146 ft3/lbm.

Therefore,

Exercise 6

1. a. P2 is less than P1

Friction causes internal energy at point 2 to be higher than at point 1. Kineticenergy and potential energy did not change (same pipe diameter and sameelevation). Since total fluid energy did not change, but internal energy increased,the flow energy had to decrease. If flow energy decreases, pressure decreases.


0FD121 81 Rev 1-3, 10/15/04

Exercise 6, continued:

b. P2 is greater than P1

The 8 inch elevation drop causes potential energy to decrease. Kineticenergy did not change, and internal energy did not change (same pipediameter and frictional effects negligible). Since total fluid energy did notchange, but potential energy decreased, the flow energy had to increase. If flowenergy increases, pressure increases.

c. P2 is greater than P1

The increase in pipe diameter causes kinetic energy to decrease. Potentialenergy and internal energy did not change (same pipe elevation andfrictional effects negligible). Since total fluid energy did not change, but kineticenergy decreased, the flow energy had to increase. If flow energy increases,pressure increases.

d. P2 is less than P1

The 8 inch elevation increase causes potential energy to increase. Thesmaller pipe diameter causes kinetic energy to increase. Friction causesinternal energy to increase. Since total fluid energy did not change, butpotential, kinetic, and internal energies all increased, the flow energy had todecrease. If flow energy decreases, pressure decreases.

e. P2 is greater than P1

The 8 inch elevation drop causes potential energy to decrease. The largerpipe diameter causes kinetic energy to decrease. Internal energy did notchange (frictional effects negligible). Since total fluid energy did not change,but potential energy and kinetic energy decreased, the flow energy had toincrease. If flow energy increases, pressure increases.


0FD121 82 Rev 1-3, 10/15/04

Exercise 7

1. As the temperature of a liquid increases, the density of the liquid decreases. Thenow greater distances between the molecules of the liquid reduces the frictionalinteractions between the molecules. The liquid, therefore, flows more readily; itoffers less resistance to flow. This means the viscosity of the liquid hasdecreased.

2. As the pressure of a liquid increases, the density of the liquid increases slightly.The now smaller distances between the molecules of the liquid increases thefrictional interactions between the molecules. The liquid, therefore, flows lessreadily; it offers more resistance to flow. This means the viscosity of the liquid has increased. (Note: the effect of pressure changes on the viscosity of a liquid aremuch less dramatic than is the effect of temperature changes on the liquid’sviscosity.

Exercise 8

1. TRUE. The more turbulent the flow is, the more friction there is between fluidparticles. The more friction there is, the more the internal energy of the fluidincreases. The amount by which the internal energy increases is the same as theamount by which the flow energy decreases. Therefore, since internal energyincreases more when the flow is turbulent, flow energy decreases more when theflow is turbulent, i.e., pressure drops more in a section of turbulent flow through apipe than it does through the same pipe section when the flow is laminar.

2. FALSE. The more turbulent the flow is through the heat exchanger, the moreeffective the heat exchanger is at transferring heat between the two fluids passingthrough the heat exchanger. More detail on this subject will be presented in a laterchapter.


0FD121 83 Rev 1-3, 10/15/04

Exercise 9

1. The buoyant force increases. The reason for this is the rapid decrease in thetemperature of the reactor coolant in the core caused by the drop in reactor powerlevel (fission rate) from 100% to 0%. Since the temperature of the waterdecreases, its density increases. Since buoyant force on an object is directlyproportional to density of the liquid the object is submersed in, this means that thebuoyant force increases.

2. Since the object is not moving, this means the downward force due to the weightof the object is exactly canceled by the upward buoyant force exerted on theobject by the fluid it is in. That is, the buoyant force on this object equals theweight of the object.

Exercise 10

1a. Decrease

1b. Increase

1c. Decrease

1d. Decrease

1e. Increase

1f. Decrease

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