thermo & heat transfer

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Thermodynamics and Heat Transfer Chapter 1

0HT121 Thermodynamic Properties/Concepts

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Table of Contents

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

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

The Thermodynamic System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Working Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Thermodynamic Properties of a Working Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . 2State of a Working Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2Phase of a Working Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3Closed System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

Open System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Steady State, Steady Flow System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

Universal Gas Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5Applications of the Universal Gas Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Specific Heat Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10Power Added to the Reactor Coolant in the Reactor Core . . . . . . . . . . . . . . . . . 13

Heat of Fusion and Heat of Vaporization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17General States of Water: Subcooled, Saturated, and Superheated . . . . . . . . . . 22

Specific Enthalpy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Specific Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

Determination of Thermodynamic Data from Steam Tables . . . . . . . . . . . . . . . . . . . . 30Specific Volume of Saturated Liquid Water and Saturated (Dry) Steam . . . . . . . 30Specific Enthalpy of Saturated Liquid Water and Saturated (Dry) Steam . . . . . . 32Specific Entropy of Saturated Liquid Water and Saturated (Dry) Steam . . . . . . . 35

Specific Volume, Enthalpy, and Entropy of Wet Steam . . . . . . . . . . . . . . . . . . . . 38Thermodynamic Properties of Superheated Steam . . . . . . . . . . . . . . . . . . . . . . . 42Thermodynamic Properties From the Mollier Diagram . . . . . . . . . . . . . . . . . . . . 44Thermodynamic Properties of Subcooled Liquid Water . . . . . . . . . . . . . . . . . . . . 49


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Determining The General State of Water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

Subcooling Margin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

Exercise Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73


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Lesson Objectives

1. Apply the Universal Gas Law to solve problems relating changes in mass, pressure,temperature and volume of a gas.

2. Define each of the following concepts, including the appropriate English Systemunit of measurement:

Sensible heatLatent heat

Heat of vaporization/condensationEnthalpyEntropy

3. Given the Specific Heat Equation, solve problems involving heat transferred, powertransferred, mass, mass flow rate, initial temperature, and final temperature ofwater.

4. List characteristics of each of the following:

subcooled water saturated liquid water wet steam saturated dry steam superheated steam

5. Explain the terms quality and percent moisture as they apply to saturated water.


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6. Given the Steam Tables and/or Mollier Diagram, determine thermodynamic

property values of water, including:

temperature pressure specific volume specific enthalpy heat of vaporization specific entropy quality moisture content degrees of superheat

7. Given the steam tables and temperature and pressure of water, determine whetherthe whether the water is subcooled (compressed), saturated, or superheated.

8. Given selected thermodynamic data for subcooled water, determine the waterssubcooling margin.


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Introduction

Thermodynamics is the area of physics that deals with the processes associated withthe conversion of thermal energy into mechanical action. For example, the thermalenergy (heat) transferred to the Secondary Side of the steam generators from theReactor Coolant passing through steam generator tubes produces steam which doesmechanical work on the Main Turbine blading. This chapter will introducecharacteristics of a thermodynamic system.

The Thermodynamic System

A system is any particular portion of the universe which we intend to study directly.Around the system are boundaries that the mind constructs. Thermodynamics is thestudy of the energy forms associated with a system, either with or without the passage

of matter into or out of the system. Therefore, a thermodynamic system is a system inwhich energy forms change within the system.

Working Fluid

The working fluidof a thermodynamic system is any fluid (including gases) whichreceives, transports, and transfers energy in the system. For example, the ReactorCoolant System water is the working fluid of this system because it receives energy

from the reactor core, transports this energy to the steam generators, and transfers theenergy across the tubes of the steam generators to the water on the other side of thetubes.


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Thermodynamic Properties of a Working Fluid

The condition of a working fluid is described in terms of its thermodynamic properties.Some of the thermodynamic properties of working fluids have been defined in earlierlessons:

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

Temperature: a measure of the average thermal energy of the molecules of asubstance; a direct indication of the average kinetic energyof the substancesindividual molecules.

Specific volume: the ratio of the volume occupied by the fluid to the mass itpossesses; it is volume per unit mass.

Density: the ratio of the mass possessed by the fluid to the volume it occupies; it ismass per unit volume.

Internal energy: energy possessed by the fluid due to the average kinetic energyof the individual molecules of the fluid, i.e., due to the temperature of the fluid.

Other thermodynamic properties will be introduced later in this lesson.

State of a Working Fluid

The state of a working fluid in a system is determined by the values of itsthermodynamic properties. If the values of each of the thermodynamic properties of theworking fluid are known, or if these values can be determined from thermodynamicproperties that are known, then the working fluid is at a unique state.


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Phase of a Working Fluid

Thephase of a working fluid is a generalcondition of the fluid described by its volume,shape, and energy level. There are three basicphases in which a substance mayexist: solid, liquid, or gas. A substance may exist in a combination of these phasessuch as a solid-liquid combination or a liquid-vapor (gaseous) combination. Generalcharacteristics of the three phases of a substance can be summarized as shown inTable 1:

Solid Liquid Gas

Definite volume Definite volume Indefinite volume

Definite shape Indefinite shape Indefinite shape

Molecules in fixedposition

Molecules can moveand interact with eachother

Molecules moveindependently

Lowest energy permolecule

Intermediate energy permolecule

Highest energy permolecule

Table 1: Solids, Liquids, and Gases

NOTE: A change of phase is always accompanied by a change of state, but a changeof state may or may not be accompanied by a change of phase. For example, there aremany unique states of water all of which would be classified as the liquid phase of thewater.

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.


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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.

Steady State, Steady Flow System

A steady state system is a system whose working fluid exists in a constant state at anygiven location of the system. The system is a steady flowsystem if the mass flow rateis constant at any given location in the system.

Process

The termprocess is used to describe a change in the state of a working fluid. Forexample, the gradual cooling of coffee in a thermos jug is a process because thecoffees temperature is decreasing (and other thermodynamic properties are changing,too). In addition to designating processes by the properties that change, processes canbe characterized by the fact that certain properties do not change. For example, anisothermal process is one in which temperature remains constant, but at least one of

its other thermodynamic properties change. A process is designated isobaricifpressure remains constant, isometricif volume remains constant, and adiabaticif no

heat is transferred.

Cycle

A cycle is a series of processes which periodically results in a final state of a systemwhich is identical to the initial state of the system before the series of processes wasbegun.


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For constant Tgas, Pgas1

Vgas

For constant Pgas, V T

For constant Vgas

, P T

Universal Gas Law

As shown in Table 1, the physical characteristics of the gaseous phase of a workingfluid are distinctly different from those of its liquid phase. In particular, gas moleculesare so widely spaced in comparison to those of a liquid that each molecule acts more orless independently of other molecules. Gases are easily compressed, and they tend toexpand freely to fill any closed container, regardless of the amount of gas placed in thecontainer.

The basic relationships between gas pressure, temperature, and volume are describedbelow:

When the temperature of a gas is kept constant, the volume of an enclosed

mass of gas varies inversely with the absolute pressure of the gas:

When the pressure of a gas is kept constant, the volume of a gas is directlyproportional to its absolute temperature:

When the volume of a gas is kept constant, the absolute pressure exerted by

a gas is directly proportional to its absolute temperature:

The relationships above apply only if pressure and temperature are expressed inabsolute units! Absolute temperature was discussed in Classical Physics:

Trankine = Tfahrenheit + 460 (NOTE: 1R = 1F )

Tkelvin = Tcelsius + 273 (NOTE: 1K = 1C )


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PV T

PV mT

PV mR i T

Ri

PV

mT

P

T

The three formulas on the previous page can be combined into a single, more general

law relating the absolute pressure, volume, and absolute temperature of a fixed quantityof gas. The relationship is called the Ideal Gas Law:

The product of the absolute pressure and the volume of a gas is directly

proportional to the absolute temperature of the gas:

If the quantity of gas (i.e., the mass of the gas) is also allowed to vary, the Ideal Gas

Law becomes the Universal Gas Law:

This joint variation relates all of the significant thermodynamic variables for gases. Itcan be made into an equation by inserting a constant of proportionality, designated Ri:

Ri is called the gas constantfor the individual gas being considered:

where is the specific volume of the particular gas.

If careful measurements of gas pressure, volume, and temperature are made, the valueof Ri for that gas can be computed. For example, air at 32 F and 14.7 psia has aspecific volume of 12.393 ft3/lbm. Thus, the formula above yields Rair= 53.32ft lbf/lbm R.


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P1V1

m1T

1

P2V2

m2T

2

In most applications it is not necessary to know the value of the gas constant for the gas

being analyzed. Since PV mT, a proportion can be written relating two different statesof the gas:

Pressure and temperature must be expressed in absolute units (psia and R) for thisproportion to be valid!

The Universal Gas Law provides accurate results for almost all gases and vapors when

they are at low pressures or high temperatures. For all plant gases except steam, wewill assume the proportion shown above to be valid under all conditions.

The relationship between the thermodynamic properties of steam is predicted withreasonable accuracy by the proportion above when the steam is at pressures below 2psia or temperatures above 2400 F. Since typical steam pressures and temperaturesin our plant do not fall in these ranges, the Universal Gas Law does not produceaccurate results. The relationships between thermodynamic properties of steam havebeen experimentally determined and are tabulated in Steam Tables (to be discussedlater).


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P1V

1P

2V

2

i.e., P2V1

V2

P1

P2

V1

V2

P1

0.20(1800 ft 3)

0.80(1800 ft 3)(14.7 psia) 3.7 psia

Applications of the Universal Gas Law

As stated earlier, the Universal Gas Law above can be used to solve problems involvingthermodynamic properties of gases in our Units. The following examples illustrate.

Example A

While the plant is shutdown, the pressurizer (total volume 1800 ft3) is filled with 70 Fwater to the 80% level. Nitrogen gas at 14.7 psia is used as a cover gas above theliquid level, and the pressurizer vent is closed. If the pressurizer is now drained to the20% level, what will be the final pressure of the nitrogen in the pressurizer? Assume

nitrogen temperature remains constant during the draining process, and ignore anyeffects of water vapor on the gas space pressure.

Solution

The total mass of the nitrogen does not change as it is allowed to expand (m 1 = m2).The temperature of the gas remains constant at 70 F (T1 = T2). Therefore, theUniversal Gas Law reduces to:

Therefore,

The increase in nitrogen volume, therefore, decreases pressure to 3.7 psia.


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P1

T1

P2

T2

T2

P2

P1T

1404.7 psia

364.7 psia(530 R) 588 R

Example B

A gaseous radwaste tank (volume = 750 ft3) contains gas at 350 psig and 70 F. If arelief valve on the tank has a setpoint of 390 psig, at what tank temperature will therelief valve lift?

Solution

The mass and volume of the gas remain constant until the tank pressure exceeds 390psig. Therefore, the Universal Gas Law simplifies to become:

P1 = 350 psi + 14.7 psi = 364.7 psia

P2 = 390 psi + 14.7 psi = 404.7 psia

T1 = 70 + 460 = 530 R

Therefore,

Therefore, therelief valve will lift when T2 reaches 588 - 460 = 128 F


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Q mcpT mcp (T2 T1)

Q Heat added to the mass (BTU)

m mass (lbm)

T Temperature change of the mass ( F)

cp Specific heat capacity of the massBTU

lbm F

Specific Heat Equation

As discussed in previous Classical Physics and Fluid Mechanics lessons, heat (Q) isenergy in transition. It is energy that moves from one location to another because atemperature difference exists between the two locations. Heat can be quantified using

any energy unit; it is normally expressed in British Thermal Units (BTU). One BTU isdefined to be the amount of heat required to raise the temperature of one pound massof water by one F under specified conditions of pressure (14.7 psia) and temperature(39 F).

If heat is supplied to a mass it generally (but not always) causes a temperature increase

of the mass. The amount of heat required to raise the temperature ofone pound massof any substance by one F is defined to be the specific heat capacity, cp, of thesubstance. (The subscript (p) of the symbol cp indicates that the process of heataddition to the substance occurs under constant pressure conditions.) For example,since 1 BTU, by definition, added to one pound mass of water at 14.7 psia and 39 F willincrease the temperature of that one pound mass to 40 F (by 1 F), the specific heatcapacity of water under these conditions is equal to 1 BTU/lbm F.

In general, the relationship between the heat added to a mass (m) and thecorresponding change in temperature of the mass is given by the Specific HeatEquation:

where


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Q mcp T mcp (T2 T1)

Q Power added to the flowing water BTUhr

m mass flowrate of the water through the heat sourcelbm

hr

T Temperature change of the water as it passes through the heat source ( F)

cp Specific heat capacity of the waterBTU

lbm F

T2 Outlet (hot) temperature of the water

T1

Inlet (cooler) temperature of the water

In plant applications of the specific heat equation it is generally not the heat added to a

substance but the thermal power(Q-dot) added that is calculated. When water isflowing past a heat source, it is the flow rate of the water (m-dot), the temperaturechange of the water, and the specific heat capacity of the water that determine thepower that is delivered to the water. The Specific Heat Equation becomes the SpecificPower Equation for these dynamic systems:

where


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Q mcp T (50lbm

hr) 1.00

BTU

lbm F(68 60) F 400

BTU

hr

Q mcp T

TQ

mcp

7,800BTU

hr

200lbm

hr1.00

BTU

lbm F

T 39 F

Example C

How much thermal power (BTU/hr) is added to water passing through a heat exchangerat 50 lbm/hr if the waters temperature increases from 60 F to 68 F while passingthrough the heat exchanger? The specific heat capacity of the water is 1.00BTU/lbm F.

Solution

Therefore, the thermal power added to the water while in the heat exchanger is 400BTU/hr.

Example D

7,800 BTU/hr is added to water flowing at 200 lbm/hr through a heat source. Determinethe change in water temperature as it passes through the heat source. The specific

heat capacity of the water is 1.00 BTU/lbm F.

Solution

The water temperature increases by 39 F as a result of the power addition.


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Power Added to the Reactor Coolant in the Reactor Core

It was stated earlier that for water at standard temperature and pressure conditions thespecific heat capacity of water is 1.00 BTU/lbm F. However, as water approachessaturation conditions (i.e., water temperature is relatively close to Tsat for the givenpressure), the magnitude of its specific heat capacity increases.

The variation of cp for water as a function of pressure and temperature is illustrated inFigure 1. The Figure shows that for a given water pressure, the value of c p remainsessentially constant (near 1.0 BTU/lbm F) when the water is at a temperaturesufficiently less than saturation temperature, and its value begins to increase as thewater temperature approaches saturation temperature for the given pressure.

At Units 2/3 100% reactor power conditions, the average RCS temperature is about566.5 F and RCS pressure is 2250 psia. Under these conditions, the value of cp isapproximately 1.4 BTU/lbm F.


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Figure 1: Variation of cp with Temperature and Pressure


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QCORE mcp (TH TC)

QCORE Power transferred to RCS

m RCS flowrate through core

cp Specific heat capacity of RCS

TH RCS Hot Leg outlet temperature

TC RCS Cold Leg inlet temperature

The Specific Power Equation can be used to determine the heat transfer rate (the

thermal power delivered) from the reactor core to the reactor coolant:

where:

Example E

Calculate reactor core thermal power, in Mw units, given the following Unit 2 conditionsat 100% power:

Cold Leg Inlet Temperature: 539 FHot Leg Outlet Temperature: 594.85 FTotal RCS core flow rate: 150 x 106 lbm/hrCoolant Specific Heat Capacity: 1.4 BTU/lbm F

NOTE: 1 Mw = 3.412 x 106 BTU/hr


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QCORE mcp T

150 x 106lbm

hr1.4

BTU

lbm F594.85 F 539 F 1.173 x 1010

BTU

hr

QCORE 1.173 x 1010 BTU

hr

1 Mw

3.412 x 106BTU

hr

3,438 Mw

Solution

100% rated power for SONGS Units 2/3 cores is 1.173 x 1010 BTU/hr, equivalent to3,438 Mwth. This means that the heat generated by core fission is transferred at a rateof 1.173 x 1010 BTU/hr (3,438 Mw) to the Reactor Coolant passing through the core.


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Heat of Fusion and Heat of Vaporization

The Specific Heat Equation describes the relationship between the heat added to asubstance and the resultant change in temperature of the substance. Whenever heatthat is added to (or removed from) a substance results in a temperature change in thatsubstance, the heat is referred to as sensible heat:

Sensible heat: heat addition or removal which causes a temperature change.

Under certain circumstances, heat addition or removal does not cause a change in thetemperature of the substance; the Specific Heat Equation does not apply. For example,if heat is added to water initially in the liquid phase at 212 F and atmospheric pressure,the water begins to boil. Its temperature remains constantas the heat is added, but itsphase changes from liquid to vapor. Whenever heat that is added to (or removed from)a substance results in a phase change of the substance, the heat is referred to aslatent heat:

Latent heat: heat addition or removal which causes a phase change.

The (latent) heat required to change the phase of a substance from solid to liquid isreferred to as the heat of fusion. The (latent) heat required to change the phase of asubstance from liquid to gas (or vapor) is called the heat of vaporization.

Under atmospheric pressure conditions, the heat of fusion of water is equal to 144BTU/lbm. This means that 144 BTU must be added to each lbm of ice at 32 F and 14.7psia to change the ice into the liquid phase at 32 F .

The heat of vaporization of water under atmospheric pressure conditions is equal to 970BTU/lbm. Thus, 970 BTU must be added to each lbm of liquid water at 212 F and 14.7psia to change the water into the gaseous (vapor) phase at 212 F .


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Figure 2illustrates the effect of adding heat to water under atmospheric pressure (14.7

psia) conditions:

1. Beginning with ice at 0 F, any heat added results in a temperature increase of theice until its temperature reaches 32 F. This heat addition is sensible heat, becauseit results in a temperature increase in the ice.

2. Once the ice reaches 32 F, further heat addition does not change the temperatureof the ice. The ice changes phase as heat is added; the heat added is latent heat.

As stated earlier, any constant temperature process is referred to as an isothermalprocess; thus, the phase change (melting) process is isothermal.

3. After the ice has fully melted, further heat addition results in a temperature increaseof the liquid water until the water reaches saturation temperature (212 F) for itspressure of 14.7 psia. Because heat addition results in a temperature increase,this heat addition is sensible heat.

4. Once the water reaches saturation temperature (212 F), further heat additionresults in an (isothermal) phase change from liquid to gas (steam); the heat addedis latent heat.

5. After all of the liquid has been changed into steam at 212 F, further heat additionresults in an increase in steam temperature; the heat added is sensible heat.


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16 BTUs per pound Sensible Heat added to raise ice temperature (0 F to 32 F)

144 BTUs per pound Latent Heat added at 32 F to melt the ice

180 BTUs per pound Sensible Heat added to raise liquid temperature (32 F to 212 F)

970 BTUs per pound Latent Heat added at 212 F to vaporize the water

urther Heat Addition: Sensible Heat added to raise steam temperature (212 F to > 212 F

Figure 2 Temperature vs Head Added to H2O at Atmospheric Pressure

where:


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Q (to melt) mhfusion Q (to vaporize) mhvaporization

Q Heat added or removed (BTU)

m mass (lbm)

hfusion Heat of FusionBTU

lbm

hvaporization Heat of VaporizationBTU

lbm

Q1 mcp T

(5 lbm) 0.5BTU

lbm F(32 22) F

Q1 25 BTU

The amount of energy gained or lost during the transition between the solid and liquidphases or the liquid and vapor phases is given by the relationship

where

Example F

Determine the heat input required to change 5 lbm of ice at 22 F and 14.7 psia to steamat 212 F. The specific heat capacity of ice is 0.5 BTU/lbm F.

Solution

The sensible heat addition (Q1) necessary to raise the temperature of the ice to 32 Fcan be calculated using the Specific Heat Equation:


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Q2 Q1 mhfusion

(5 lbm) 144BTU

lbm

Q2 Q1 720 BTU

Q3 Q2 mcp T

(5 lbm) 1BTU

lbm F(212 32) F

Q3 Q2 900 BTU

Q4 Q3 mhvaporization

(5 lbm) 970BTU

lbm

Q4 Q3 4,850 BTU

Qtotal 25 BTU 720 BTU 900 BTU 4,850 BTU

Qtotal 6,495 BTU

The latentheat addition (Q2 - Q1) necessary to melt the ice can be calculated using the

Latent Heat Equation:

The sensible heat addition (Q3 - Q2) which raises the liquid temperature to 212 F is:

The latentheat addition (Q4 - Q3) which boils the water to steam is:

Therefore, the total heat added to change 5 lbm of ice at 22 F and 14.7 psia to steam at212 F and 14.7 psia is:


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General States of Water: Subcooled, Saturated, and Superheated

In Fluid Mechanics, Chapter 1, the concepts of saturation temperature and saturationpressure were defined:

Saturation temperature, TSAT: The temperature at which water at a given pressurewill boil if heat is added to the water.

Saturation pressure, PSAT: The pressure at which water at a given temperature willboil if heat is added to the water.

Saturation temperature and saturation pressure are referred to as dependentthermodynamic properties. Another way of saying this is that the saturationtemperature depends on the pressure of the water, and the saturation pressuredepends on the temperature of the water. For each temperature there is only onesaturation pressure, and for each pressure there is only one saturation temperature.

For example, water at 14.7 psia will boil at 212 F if heat is added to it; P = PSAT = 14.7psia when T = TSAT = 212 F. Liquid water under these conditions is referred to assaturated liquid water:

Saturated liquid water: Liquid water that exists at TSAT/PSAT conditions.

The Steam Tables, Tables 1 and 2, list TSAT/PSAT pairs in the first two columns of eachTable. When water exists as saturated liquid water it is said to be saturated with energy;it holds as much energy (BTU/lbm) as it can hold and still exist in the 100% liquid state.


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When liquid water exists at a temperature belowsaturation temperature for the pressure

of the water, the water is called subcooled liquid water:

Subcooled liquid water: Liquid water at a temperature below saturationtemperature for the pressure of the water.

For example, the Reactor Coolant System (RCS) is normally maintained at a pressureof 2,250 psia. The saturation temperature for water at this pressure is 653 F. However,even at its hottest location in the RCS, the temperature of the water is normally nohigher than 594 F. Because 594 F is less than TSAT (653 F) for 2,250 psia, this water issubcooled.

When liquid water is at saturation temperature for the pressure of the water, heataddition results in some of that water changing to steam. Eventually, if enough heat isadded to the water, it becomes 100% steam. The steam is still at saturation temperaturefor the pressure of the water (temperature did not change as heat was added), but thestate of the water is now referred to as saturated (dry) steam:

Saturated (dry) steam: water that exists as 100% steam at TSAT/PSAT conditions.

For example, if one pound of 212 F/14.7 psia saturated liquid water receives a heatinput of 970 BTU (the latent heat of vaporization for water at this TSAT/PSAT condition),the water changes to 100% saturated (dry) steam. If however, less than 970 BTU is

added to this pound of saturated liquid water, some of the water will become steam andsome will remain liquid. When water exists as a mixture of liquid water and steam atTSAT/PSAT conditions, the water is referred to as wet steam:

Wet steam: water existing as a mixture of liquid water and steam at TSAT/PSATconditions.


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Quality, xMass of steam

Mass of water Mass of steam

Mass of steam

Total mass of water/steam mixture

Moisture fraction 1 quality

Percent moisture (%) 100% (% quality)

Any liquid water-steam mixture is said to have a quality, x(the quality of a water-steam

mixture is one of its thermodynamic properties). The quality of a water/steam mixture,by definition, represents that fraction of the total mass of the mixture which is steam:

Quality is often expressed as a percent by multiplying by 100.

The termpercent moisture (or moisture fraction) is sometimes used instead of percentquality:

For example, saturated liquid water has quality = 0 (or 0%), and a moisture fraction = 1(or percent moisture = 100%). As latent heat is added to saturated liquid water steam isgenerated, quality increases, and moisture content decreases. When the latent heat of

vaporization has been added to the water the quality of the mixture becomes 1 (100%steam) and moisture content equals zero. Saturated steam is often referred to as drysteam because it contains no moisture (no liquid water).


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x0.8 lbm

0.2 lbm 0.8 lbm

0.8 lbm

1 lbm0.80

or, x 0.80(100) 80% (expressed as percent)

Example G

Determine the quality of a mixture of 0.2 lbm liquid water and 0.8 lbm steam.

Solution

Note that the moisture content of the mixture in this example is 1 - x = 1 - 0.80 = 0.20, or20%.

If heat is added to saturated (dry) steam at constant pressure, the temperature of thesteam will increase. Steam existing at a temperature above saturation temperature forthe pressure of the steam is called superheated steam:

Superheated steam: Steam existing at a temperature above TSAT for the pressureof the steam.

For example, if steam exists at 180 psia and 498 F, this steam is superheated becauseTSAT = 373 F when P = 180 psia. This steam is said to have 498 - 373 = 125 degrees ofsuperheat because it exists at a temperature 125 F above saturation temperature forits given pressure. Degrees of superheat is another thermodynamic property of water.


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Summarizing classifications of the general states of water:

Water in a particular state is either subcooled, saturated or superheated.

1. If the water is subcooled, then it is in the liquid phase and its temperature isless than saturation temperature for the pressure of the water.

2. If the water is saturated, then its temperature is equal to saturationtemperature for the pressure of the water. There are three classifications ofsaturated water (water at TSAT/PSAT conditions):

a. Saturated water that is 100% liquid water is called saturated liquid water. It

has a quality of 0% (a moisture content of 100%).

b. Saturated water that exists as a water-steam mixture is called wet steam.Because it contains both liquid water and steam, its quality is greater than0% but less than 100%.

c. Saturated water that is 100% steam is called saturated (dry) steam. It hasa quality of 100% and a moisture content of 0%.

Even though the water exists at TSAT/PSAT conditions for each case above, thestates of the water in each case are unique. Of the three conditions described,

saturated liquid water contains the least energy per pound, wet steam the nexthigher energy per pound (and energy per pound increases as qualityincreases), and saturated (dry) steam the most energy per pound.

3. If the water is superheated steam, then it exists as 100% steam and itstemperature is higher than saturation temperature for the pressure of thewater. Superheated steam contains more energy per pound mass than doessaturated (dry) steam at the same pressure.


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h uP

J

h specific enthalpy,BTU

lbm

u specific internal energyBTU

lbm

P pressurelbf

ft 2

specific volumeft 3

lbm

J conversion factor,778 ftlbf

1 BTU

Specific Enthalpy (h)

The thermodynamic property of a working fluid which accounts for both its specific

internal energy and its specific flow energy is called enthalpy (h). Enthalpy is definedas the sum of the internal energy and flow energy of the working fluid:

where:

Enthalpy is energy possessed by the working fluid that is available to do work. Becauseof the vibrational energy of the individual atoms/molecules of the working fluidassociated with the temperature of the fluid (i.e., because of its internal energy) andbecause of its energy associated with the fact that the fluid volume is pressurized (i.e.,its flow energy), the working fluid has the potential to do work.

For example, steam enters a turbine with a specific enthalpy determined by thetemperature and pressure of the steam. As this steam does work (i.e., applies a forcethrough a distance) on the turbine blades, the temperature and pressure of the steamdecrease (its enthalpy decreases). The amount of work done by each pound mass ofthe steam while it is in the turbine is equal to the change in the enthalpy of the steam.


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sdq

T

s change in specific entropy,BTU

lbm R

dq increment of heat added during the process,BTU

lbm

T absolute temperature at which the dq is supplied, R

Specific Entropy (s)

In any closed thermodynamic system, not all energy possessed by a working fluid canbe converted into useful mechanical work. The thermodynamic property of the workingfluid which quantitatively describes the unavailability of energy for the performance ofwork is called entropy (s). The entropy of a working fluid is a mathematicallydetermined number whose magnitude increases as the temperature and/or the pressureof the fluid decreases. This means that the fraction of the energy possessed that iscapable of being converted to useful work when water is at a relatively lowtemperature/pressure is less than the fraction of the energy possessed that is capableof being converted to useful work when water is at a relatively hightemperature/pressure.

The calculation of entropy involves calculus and is beyond the scope of this training. Asis the case with enthalpy, it is the change in entropy during a thermodynamic process,not the specific entropy values at each endpoint of the process, that is of interest inthermodynamic analysis. The mathematical formula for how an entropy change iscalculated is provided below for information, only:

where:


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Note that the unit of measurement of entropy is BTU/lbm R, i.e., BTUs per pound mass

per degree Rankine. Since the magnitude of 1 R is the same as the magnitude of 1 F,entropy can also be expressed with the unit BTU/lbm F; the entropy magnitude will bethe same regardless of which of these two units is used. Technically, however, whenentropy is used in calculations its value is expressed in the BTU/lbm R unit and alltemperatures used in that calculation must be in R.

The thermodynamic property of entropy will be used when discussing steam cycle andsteam turbine efficiency, i.e., how efficiently the cycle or turbine converts the energy it

possesses into useful mechanical work. Briefly, an idealturbine is one in which allofthe steam energy change from inlet to outlet of the turbine is converted into rotationalenergy of the turbine shaft (into useful work), and the maximum possible change in

steam energy occurs for the given inlet and outlet steam pressure conditions. This idealprocess results in no change in the entropy of the working fluid because all of its energychange produces useful mechanical work; none of the energy change is unavailable forthe performance of work.

Of course, no turbine is ideal; friction within its moving parts and internal friction withinthe steam itself will not allow all of the steam energy change to be converted to usefulshaft work. The working fluids entropy increase in any actual turbine process reflectshow efficiently the turbine converts available steam energy into useful work. For giventurbine inlet and outlet steam pressures, the greater the increase in steam entropyacross the turbine, the less efficiently the turbine is producing useful shaft work.

Entropy and its relationship to plant operation will be discussed in a later lesson.However, the method used to obtain entropy values, as well as values for otherthermodynamic properties of water in a given state, will be discussed next.


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Determination of Thermodynamic Data from Steam Tables

The ABB Steam Tables provide information that can be used to determine thethermodynamic properties of water in a given state. If enough specific information aboutthe state of the water is known, the unique values of each of its thermodynamicproperties can be determined using these Tables.

Again, a reminder: Because you do not have Steam Tables available to you, all of

the data that you would normally find in the Steam Tables will be provided to you

here. When the actualclassroom training occurs, you will learn how to obtain this datafrom the Steam Tables.

However, there will be particular symbols used in this and subsequent lessons that youmust recognize. For this reason, much of the description of how to use the SteamTables will be left here. Continuing now with the training....

The ABB Steam Tables contain three sets of Tables (Tables 1, 2, and 3). Each of theseTables contains temperature, pressure, specific volume, specific enthalpy, and specificentropy data.

Specific Volume of Saturated Liquid Water and Saturated (Dry)

Steam

Tables 1 and 2 provide thermodynamic data forsaturatedwater (i.e., saturated liquidwater, wet steam, and saturated (dry) steam data). As stated earlier, saturated waterexists at saturation temperature for the pressure of the water (the water is in a TSAT/PSATcondition). Column 1 of Table 1 lists temperatures ( F), with their correspondingsaturation pressures (psia) in Column 2. Column 1 of Table 2 lists pressures, with theircorresponding saturation temperatures in Column 2.

The remaining columns of Table 1 are formatted exactly as they are in Table 2. In eachof these Tables, there are three specific volume columns, three specific enthalpycolumns, and three specific entropy columns.


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The first specific volume column, labeled f, provides the specific volume ofsaturated

liquid waterat the given temperature and pressure. The third specific volume column,labeled g, provides the specific volume ofsaturated (dry) steam at the given

temperature and pressure. The middle specific volume column, labeled fg, representsthe change in specific volume as the water goes from saturated liquid to saturatedsteam at the given temperature and pressure. Example Hillustrates.

Example H

Determine:

a) The specific volume of saturated liquid water at 528 F

b) The specific volume of saturated (dry) steam at 528 F

c) The change in specific volume as water goes from saturated liquid at 528 F tosaturated (dry) steam at 528 F due to addition of the latent heat of vaporizationto this water.

Solution

a) The water is saturated liquid at 528 F. Table 1 shows that vf, the specific

volume of saturated liquid water at 528 F, is 0.02112 ft3

/lbm.

Physical interpretation of this specific volume: Each pound mass of saturatedliquid water at 528 F (and 870.31 psia) occupies 0.02112 ft3.

b) The water is saturated (dry) steam at 528 F. Table 1 shows that vg, the

specific volume of saturated dry steam at 528 F, is 0.51995 ft3/lbm.

Physical interpretation of this specific volume: Each pound mass of saturated(dry) steam at 528 F (and 870.31 psia) occupies 0.51995 ft3.


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c) The change in specific volume as water goes from saturated liquid at 528 F to

saturated (dry) steam at 528 F is found in the vfg column ofTable 1. Table 1shows that vfg = 0.49843 ft

3/lbm.

Physical interpretation of this vfg value: The volume occupied by a pound massof saturated liquid waterincreases by 0.49843 ft3 as the water changes tosaturated (dry) steam at 528 F (and 870.31 psia), i.e., as the latent heat ofvaporization is added to this water.

KEY POINT YOU MUST REMEMBER, regardless of whether you have Steam

Tables available or not:

Example Hshows that:

vg = vf + vfg

This is true in all cases, because of how vg, vfg, and vg are defined. The equation simplystates that the specific volume of saturated liquid water, plus the increase in specificvolume as that liquid volume boils to steam, is equal to the specific volume of thesaturated steam which results. The temperature and pressure remain constant as thephase change occurs.

Specific Enthalpy of Saturated Liquid Water and Saturated (Dry)Steam

The three specific enthalpy columns of C-E Steam Tables 1 and 2 are described in amanner analogous to the previous specific volume columns discussion:

hf: The specific enthalpy ofsaturated liquid water.

hg: The specific enthalpy ofsaturated (dry) steam.

hfg: The change in specific enthalpy as saturated liquid water changes to saturated

(dry) steam.


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Based on earlier discussions of the concept of latent heat of vaporization, it is clear that

the definition above forhfg describes the latent heat of vaporization:

hfg: The latent heat of vaporization for saturated liquid water at the given TSAT/PSATcondition.

Numbers provided in the Steam Tables do not represent absolute enthalpy values; i.e.,they are not calculated values of u + Pv/J for the given temperature/pressure conditions.The enthalpy of saturated liquid water at a temperature of approximately 32 F wasarbitrarily assigned the value of zero (its absolute enthalpy, as determined by thedefining formula, is actually greater than zero). All values listed in the Tables areenthalpy values relative to this zero reference; i.e., the values listed represent the

change in enthalpy relative to the zero reference.

Thus, any hfvalue listed in the Steam Tables can be interpreted as the increase inenergy of a pound mass of the water as it changes from saturated liquid at 32 F tosaturated liquid at the given temperature/pressure conditions. Any hg value representsthe increase in energy of a pound mass of water as it changes from saturated liquid at32 F to saturated (dry) steam at the given temperature/pressure conditions.

Example I

Determine:

a) The specific enthalpy of saturated liquid water at 850 psia

b) The specific enthalpy of saturated (dry) steam at 850 psia

c) The change in specific enthalpy as water goes from saturated liquid at 850 psiato saturated (dry) steam at 850 psia due to addition of the latent heat ofvaporization to this water.


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Solution

a) The water is saturated liquid at 850 psia. Table 2 shows that hf, the specific

enthalpy of saturated liquid water at 850 psia, is 518.4 BTU/lbm.

Physical interpretation of this specific enthalpy: Each pound mass of saturatedliquid water at 850 psia (and 525.24 F) contains 518.4 BTUs.1

b) The water is saturated (dry) steam at 850 psia. Table 2 shows that hg, thespecific enthalpy of saturated dry steam at 850 psia, is 1198.0 BTU/lbm.

Physical interpretation of this specific enthalpy: Each pound mass of saturated(dry) steam at 850 psia (and 525.24 F) contains 1198.0 BTUs.1

c) The change in specific enthalpy as water goes from saturated liquid at 850 psiato saturated (dry) steam at 850 psia is found in the hfg column ofTable 2.Table 2 shows that hfg = 679.5 BTU/lbm.

Physical interpretation of this hfg value: The energy addition required to changeone pound of saturated liquid water at 850 psia to saturated dry steam at 850psia is 679.5 BTUs, that is, the latent heat of vaporization of water at 850 psiais 679.5 BTU/lbm.



Example Ishows that:hg = hf+ hfg

This is true in all cases, because of how hg, hfg, and hg are defined. The equation simplystates that the specific enthalpy of saturated liquid water, plus the increase in specificenthalpy as that liquid volume boils to steam, is equal to the specific enthalpy of thesaturated steam which results. The temperature and pressure remain constant as the

phase change occurs.


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Specific Entropy of Saturated Liquid Water and Saturated (Dry)

Steam

The three specific entropy columns of ABB Steam Tables 1 and 2 are described in amanner analogous to the previous specific volume and enthalpy columns discussion:

sf: The specific entropy ofsaturated liquid water.

sg: The specific entropy ofsaturated (dry) steam.

sfg: The change in specific entropy as saturated liquid water changes to saturated

(dry) steam.

Specific entropy values provided in the Steam Tables do not represent the absoluteentropy values; i.e., they are not calculated values of the mathematical formula shownearlier. The entropy of saturated liquid water at 32 F was arbitrarily assigned the valueof zero (its absolute entropy, as determined by the defining formula, is actually greaterthan zero). All values listed in the Tables are entropy values relative to this zeroreference; i.e., the values listed represent the change in entropy relative to the zeroreference.

Thus, any sf value listed in the Steam Tables can be interpreted as the increase in the

entropy of a pound mass of the water as it changes from saturated liquid at 32 F tosaturated liquid at the given temperature/pressure conditions. Any sg value representsthe increase in entropy of a pound mass of water as it changes from saturated liquid at32 F to saturated (dry) steam at the given temperature/pressure conditions.


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Example J

Determine: a) The specific entropy of saturated liquid water at 500 F

b) The specific entropy of saturated (dry) steam at 500 F

c) The change in specific entropy as water goes from saturated liquid at500 F to saturated (dry) steam at 500 F due to addition of the latentheat of vaporization to this water.

Solution

a) The water is saturated liquid at 500 F. Table 1 shows that sf, the specificentropy of saturated liquid water at 500 F, is 0.6890 BTU/lbm R .

Physical interpretation of this specific entropy: Specific entropy = 0.6890BTU/lbm R is a direct indication of that portion of the waters energy thatCANNOT be converted into useful work.

b) The water is saturated (dry) steam at 500 F. Table 1 shows that sg, thespecific entropy of saturated dry steam at 500 F 850, is 1.4333 BTU/lbm R.

Physical interpretation of this specific entropy: Specific entropy = 1.4333BTU/lbm R is a direct indication of that portion of the waters energy thatCANNOT be converted into useful work. Also, because the specific entropy ofthis saturated steam has a magnitude that is greater than the magnitude of thespecific entropy of saturated liquid water (1.4333 versus 0.6890 determined inpart a), the fraction of the energy possessed by this steam that is available todo work is smallerthan the fraction of the energy possessed by the saturatedliquid that is available to do work. The key word here is fraction; saturated

steam at 500 F can obviously do more total work than can saturated liquidwater at 500 F. However, the fraction of the total energy of the saturatedsteam that is available to do work is smaller than the fraction of the saturated

liquid waters total energy available to do work.


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c) The change in specific entropy as water goes from saturated liquid at 500 F to

saturated (dry) steam at 500 F is found in the sfg column ofTable 1. Table 1shows that sfg = 0.7443 BTU/lbm R.

Physical interpretation of this sfg value: When energy that is added to saturatedliquid water at 500 F, taking the water to saturated (dry) steam at 500 F, someof the energy that was added is now unavailable to do useful work.



Example Jshows that:sg = sf+ sfg

This is true in all cases, because of how sg, sfg, and sg are defined. The equation simplystates that the specific entropy of saturated liquid water, plus the increase in specificentropy as that liquid boils to steam, is equal to the specific entropy of the saturatedsteam which results. The temperature and pressure remain constant as the phasechange occurs.


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wet steam f xfg

hwet steam hf xhfg

swet steam sf xsfg

v specific volume of wet steam

h specific enthalpy of wet steam

s specific entropy of wet steam

vf specific volume of saturated liquid

hf specific enthalpy of saturated liquid

sf specific entropy of saturated liquid

vfg specific volume of vaporization

hfg specific enthalpy of vaporization

sfg specific entropy of vaporization

x quality of the water/steam mixture

Specific Volume, Enthalpy, and Entropy of Wet Steam

Since wet steam is a mixture of saturated steam and saturated liquid water, the values

of its thermodynamic parameters depend on the quality (x) of the water/steam mixture.

For a saturated water/steam mixture,

where:


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v vf xvfg

0.017274

ft 3

lbm 0.93 8.4967

ft 3

lbm

v 7.9192ft 3

lbm

1

v

1

7.9192ft 3

lbm

0.1263lbm

ft 3

Example K

For wet steam at 50 psia and 93 percent quality, determine: (a) its specific volume, (b)its density, (c) its temperature, and (d) its enthalpy, and (e) its entropy.

Solution

a) Table 2 of the Steam Tables lists integral values for pressure. Using this Table,

b) Density is the reciprocal of specific volume:

c) Because wet steam exists at saturation temperature/pressure conditions, T =281.02 F (saturation temperature for 50 psia).


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h hf xhfg

250.2BTU

lbm0.93 923.9

BTU

lbm

h 1109.4BTU

lbm

s sf xsfg

0.4112BTU

lbm R0.93 1.2474

BTU

lbm R

s 1.5713BTU

lbm R

d) From Table 2,

e) From Table 2,


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88% quality steam f xfg 0.016136ft 3

lbm0.88 333.59

ft 3

lbm293.6

ft 3

lbm

h88% quality steam hf xhfg 69.73BTU

lbm0.88 1036.1

BTU

lbm981.5

BTU

lbm

s88% quality steam sf xsfg 0.1326BTU

lbm R0.88 1.8455

BTU

lbm R1.7566

BTU

lbm R

Example L

Wet steam with a quality of 88% leaves the Low Pressure Turbines and enters the MainCondenser at a pressure of 1 psia. Using the C-E Steam Tables, determine a) thespecific volume, b) the enthalpy, and c) the entropy of steam leaving the Low PressureTurbines.

Solution

From Table 2:


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Thermodynamic Properties of Superheated Steam

The thermodynamic properties of superheated steam can be obtained from Table 3 ofthe C-E Steam Tables. If two of the thermodynamic properties of the superheatedsteam are known, Table 3 can be used to determine values for the remainingproperties.

Similar to saturated water/steam Tables 1 and 2, Table 3 provides temperature,pressure, specific volume, enthalpy and entropy data for superheated steam, but

formatted differently. The first column ofTable 3 contains steam pressure values, withsaturation temperatures listed in parentheses just below each pressure value. The next

two columns contain saturated liquid and saturated steam thermodynamic propertiesassociated with the pressure listed in the first column. Therefore, the first three columnsof the Superheated Steam Table 3 are redundant with the information provided inTables 1 and 2.

Thermodynamic data for superheated steam begins in the fourth column. Becausesuperheated steam, by definition, exists at a temperature above Tsat for the pressure ofthe steam, several temperature columns are provided. The number of degrees by

which Tstm exceeds Tsat is referred to as the degrees of superheatof the steam, orsimply the superheat. Table 3 also provides information about the steam superheat,found in rows designated (Sh).


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v 0.5615ft 3

lbm

h 1379.7BTU

lbm

s 1.5415BTU

lbm R

sh 232.81 F

Example M

Determine the specific volume, enthalpy, entropy, and degrees of superheat of steam at1200 psia and 800 F.

Solution

For water at 1200 psia, Tsat = 567.19 F. Since the temperature of the steam, 800 F, isgreater than Tsat, we know that the steam is superheated. Since the steam is

superheated, Table 3 must be used to obtain thermodynamic data (Tables 1 and 2provide information about saturated water, only).

From Table 3, the desired data is found in the row corresponding to 1200 psia and thecolumn corresponding to 800 F:


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Example N

Determine the specific volume of 200 psia steam having an enthalpy of 1477.0BTU/lbm.

Solution

Table 2 shows that saturated steam at 200 psia has an enthalpy of 1198.3 BTU/lbm.

Since 1477.0 BTU/lbm is greater than hg for 200 psia, we know that the steam issuperheated and therefore Table 3 must be used to obtain its thermodynamicproperties. Table 3 indicates that steam at 200 psia with enthalpy 1477.0 BTU/lbm has

a specific volume v = 4.0008 ft3

/lbm.

Thermodynamic Properties From the Mollier Diagram

The Mollier Diagram is a graphical representation of thermodynamic property data forwet steam with at least 44% quality up to superheated steam with as much as 860 F ofsuperheat.

Figure 3 is a simplified facsimile of the Mollier Diagram. The detailed Mollier Diagram isattached to the end of the C-E Steam Tables.

For a given state of superheated, saturated, or wet steam, the Mollier Diagram can beused to obtain values for:

enthalpy

entropy

pressure

temperature

degrees of superheat of superheated steam

moisture content (and consequently, quality) of wet steam

Concerning the Mollier Diagram:


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1. The horizontal axis represents specific entropy values.

2. The vertical axis is represents specific enthalpy values.

3. Points of the saturation line represent unique states ofsaturated (dry) steam.

4. The region below the saturation line is the wet steam region; each point in thisregion represents a unique state of wet steam.

Constant percent moisture lines are drawn in this wet steam region.

5. The region above the saturation line is the superheated steam region.

Each point on a given constant steam temperature line represents a unique stateof superheated steam existing at the given steam temperature.

Each point on a given constant superheat line represents a unique state ofsuperheated steam existing with the given degrees of superheat.

6. Constant pressure lines run upward diagonally across the diagram.

When the constant pressure line is a solid line, the unit of pressure for that line ispsia.

When the constant pressure line is a dotted line, the unit of pressure is in Hgabs (notshown on Figure 2.4).

7. Constant pressure lines in the wet steam region are also constant temperaturelines. For example, any point on the "Standard Atmosphere (14.696 psia)" constantpressure line in the wet steam region represents a state of wet steam existing at atemperature of 212 F.


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Figure 3 Mollier Diagram


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h 1081BTU

lbm

s 1.532BTU

lbm R

T 283 F

Example Q

For steam with quality 90% and pressure 50 psia, use the Mollier diagram to estimateeach of the following thermodynamic properties:

EnthalpyEntropyTemperature

Solution

The unique point on the Mollier diagram corresponding to the given information is foundwhere the 10% constant moisture line intersects the 50 psia constant pressure line.Reading horizontally back from this point to the vertical axis, the enthalpy can bedetermined:

Reading vertically down from this point to the horizontal axis, the entropy can bedetermined:

The temperature of wet steam at 50 psia, regardless of its quality, is exactly the sameas the temperature of saturated steam at 50 psia. Therefore, if the 50 psia constantpressure line is followed up to its intersection point with the saturation line, the constanttemperature lines which end at the saturation line can be used to estimate thetemperature of the 50 psia wet steam. The intersection point is slightly above the 280 Fconstant temperature line, and well below the next indicated constant temperature line(320 F). Therefore, an estimate of the temperature of the wet steam could be:

However, it is clear that because the steam is wet steam at 50 psia, its temperature is

TSAT for 50 psia. Table 2 provides this temperature to be 281.02 F.


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Example Qdemonstrates advantages and disadvantages of using the Mollier Diagram

to obtain thermodynamic information. Without the Mollier diagram, the wet steamformulas (h = hf+ xhfg; s = sf + xsfg) would have to be applied to find the steam enthalpyand entropy. The Mollier diagram allows this data to be read directly from the graph.However, when the enthalpy and entropy values are read from the graph, the precisionof their values is limited. Finally, the process described above for use of the Mollierdiagram to find the temperature of the the wet steam is tedious and the accuracy of theanswer is questionable. It is much easier (and more accurate) to go to the SteamTables to obtain the saturation temperature (281.02 F) for 50 psia.

Summarizing:

The state of water is uniquely defined if two independent thermodynamic properties ofthe water are known. For example, if the enthalpy and the entropy of steam are known,these two values can be used to locate a unique point on the Mollier diagram. Once thepoint is located, the values of the other thermodynamic properties can be read from thislocation.

The Mollier Diagram is simply another option for obtaining thermodynamic data; theSteam Tables provide the same information. If precise thermodynamic data is needed,the Steam Tables should be used. When estimation of thermodynamic data isacceptable, the Mollier Diagram produces the data more efficiently.

If thermodynamic data for wet steam with a moisture content greater than 56% isneeded, or if saturated liquid or subcooled water data is needed, the Mollier Diagramdoes NOT provide the information; the Steam Tables are the only option to obtain it.


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Thermodynamic Properties of Subcooled Liquid Water

As stated earlier, liquid water at a temperature below the saturation temperature for the

existing water pressure is called subcooled liquidwater. Another way of describingsubcooled water is to say that the pressure of the water is greaterthan the saturationpressure associated with the existing temperature of the water.

The C-E Steam Tables list thermodynamic data forsaturatedand superheatedwater;they do NOT list thermodynamic property values for subcooled water. For precisethermodynamic property value information, Subcooled Water Tables must be used.

However, the C-E Steam Tables can be used to obtain reasonable approximations ofthe thermodynamic properties of subcooled water. The following rule apply for use ofthe C-E Steam Tables to determine thermodynamic properties of subcooled water:

The thermodynamic properties of subcooled water are approximately equal to

the thermodynamic properties of saturated liquid water at the SAME

TEMPERATURE as the temperature of the subcooled water.

The Examples which follow will illustrate.

Example O

Determine a) the specific volume, b) the enthalpy, and c) the entropy of water at 300 Fand 500 psia.

Solution

Water at 300 F has a saturation pressure of 67.005 psia (Table 1). Since the waterspressure is 500 psia, we know that the water is subcooled. Therefore, per the rulestated above, the values of the specific volume, enthalpy, and entropy of this water areapproximated by the values listed for 300 F saturated liquid water:

(300 F, 500 psia) f (300 F, 67.005 psia) = 0.01745 ft3

/lbmh (300 F, 500 psia) hf (300 F, 67.005 psia) = 269.7 BTU/lbm

s (300 F, 500 psia) sf (300 F, 67.005 psia) = 0.4372 BTU/lbm R


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(544 F, 2250 psia) f(544 F, 995.22 psia) 0.02157ft 3

lbm

h(544 F, 2250 psia) hf(544 F, 995.22 psia) 541.8BTU

lbm

NOTE: If the Subcooled Liquid Tables had been used to determine more precise

values for the thermodynamic properties of this subcooled water, the following valueswould be obtained:

(300 F, 500 psia) = 0.01742 ft3/lbm

h (300 F, 500 psia) = 270.5 BTU/lbm

s (300 F, 500 psia) = 0.4364 BTU/lbm R

The numbers above show that small differences from the actual values. The effects of500 psia versus 67.005 psia on the values are minimal. Since the ABB Steam Tables

are the onlysource of thermodynamic data the NRC allows the student to use duringthe Generic Fundamentals (GFE) Exam, the method shown above must be used toobtain thermodynamic data for subcooled water.

Example P

Using the C-E Steam Tables, determine the a) specific volume, b) enthalpy, and c) theentropy of liquid water at 544 and 2250 psia.

Solution

a. The value offfor saturated water at 544 F is used as the approximation of thespecific volume of this subcooled liquid. Therefore,

b. The value ofhf for saturated water at 544 F is used as the approximation of theenthalpy of this subcooled liquid. Therefore,


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s(544 F, 2250 psia) sf(544 F, 995.22 psia) 0.7427BTU

lbm R

c. The value of sf for saturated water at 544 F is used as the approximation of the

entropy of this subcooled liquid. Therefore,

Determining The General State of Water

If the temperature and pressure of water are known, the Steam Tables or Mollierdiagram can be used to determine whether the water is subcooled (compressed),saturated, or superheated.

Recall that a subcooled liquid is one whose temperature is below the saturationtemperature for the existing pressure or whose pressure is above the saturationpressure for the existing temperature. When water is at saturation conditions, itstemperature is the saturation temperature for the existing pressure or, equivalently, itspressure is the saturation pressure for the existing temperature. Finally, when steam issuperheated, its temperature is above the saturation temperature for the existingpressure or, equivalently, its pressure is below the saturation pressure for the existingtemperature.

Example R

Water leaves the condenser at a temperature of 100 F and a pressure of 1 psia. Is thewater subcooled, saturated, or superheated?


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Solution

Since temperature and pressure are known, Table 1 orTable 2 can be used todetermine the general state of the water:

1) Find 100 F in Table 1. The corresponding saturation pressure is 0.94924 psia.Since 1 psia > 0.94924 psia, the water is subcooled.

OR:

2) Find 1 psia in Table 2. The corresponding saturation temperature is 101.74 F.Since 100 F < 101.74 F, the water is subcooled.

Example S

What is the general state (subcooled, saturated, or superheated) of water at 500 F and680 psia?

Solution

From Steam Table 1, PSAT = 680.86 psia for 500 F. Since 680 psia < PSAT, the watermust be superheated steam.

OR:

From Steam Table 2, TSAT = 499.80 F for P = 680 psia (by interpolation). Since 500 F >499.80 F, the water must be superheated steam.

NOTE: Using Steam Table 1 information is clearly preferable to using Steam Table2 information, because no interpolation was necessary with Table 1. The point hereis that you should weigh all options Talbes 1,2,3, and the Mollier Diagram) whendetermining the general state of water from the available data; usually, one of thoseoptions will prove to be clearly the most efficient option!

The Mollier Diagram is another option for determining the general state of water.However, examination of the Mollier for the data provided in Example Sshows thatthe Diagram is insufficient for absolute determination that the water is superheated.


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Subcooling Margin (SCM) TSAT TACTUAL

TSAT Saturation temperature for the water pressure

TACTUAL Actual Temperature of the water

Example T

What is the general state (subcooled, saturated, or superheated) of water at 175 psiawith enthalpy 1270 BTU/lbm?

Solution

The Mollier Diagram shows that water with enthalpy 1270 BTU/lbm is superheatedsteam, regardless of the pressure of the water. More specifically, it is clear that theintersection of the constant enthalpy 1270 BTU/lbm line and the constant pressure 175psia line occurs in the superheated region of the Diagram.

OR:

Steam Table 1 orTable 2 can be used to show that 1270 BTU/lbm is greater than hg,the enthalpy of saturated (dry) steam at 175 psia. Therefore, the water is superheated

steam.

Subcooling Margin

Subcooling margin (SCM) is the term used describe the amount by which water issubcooled. The subcooling margin of subcooled water is defined as the difference

between the saturation temperature for the existing water pressure and the liquid'sactual temperature:

where


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SCM TSAT TACTUAL

TSAT (2250 psia) 653 F (interpolating from Table 2)

Therefore, SCM 653 F 607 F 46 FSCM TSAT TACTUAL

TSAT (2100 psia) 642.76 F (Table 2)

Therefore, SCM 642.76 F 575 F 67.76 F

Subcooling margin is an important Reactor Coolant System (RCS) parameter. As long

as the RCS subcooling margin is maintained above specified minimum values, controlof RCS pressure, inventory, and core heat removal is assured (to be discussed in detaillater).

Units 2/3 have Control Room indications of subcooling margins in the reactor vesselupper head, just above the reactor core, and in RCS coolant system piping. In addition,the Qualified Safety Parameter Display System (QSPDS) has two additional subcooledmargin indications that can be retrieved if necessary. QSPDS computes subcooledmargin in terms of temperature (TSAT - TACTUAL) and in terms of pressure. Pressuresubcooled margin is defined as PACTUAL - PSAT, where PACTUAL is pressurizer pressure(psia) and PSAT is saturation pressure (psia) for the hottest temperature of the reactor

coolant.

During accident conditions, licensed operators may be required to manually determinethe subcooling margin. Plant procedures specify using the highest available RCS hotleg temperature (Th) indication and the lowest pressurizer pressure (Ppzr) indication. Anexample of such a calculation follows.

Example U

Calculate the RCS Subcooling Margin if:

a. the lowest indicated pressurizer pressure is 2250 psia, and the highest indicatedRCS hot leg temperature is 607 F.

b. the lowest indicated pressurizer pressure is 2100 psia, and the highest indicatedRCS hot leg temperature is 575 F.

Solutiona.

b.


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Definitions

Adiabatic process - A process in which no heat is transferred.

British Thermal Unit BTU- Amount of heat required to increase the temperature ofone pound mass of water by 1 F under standard temperature and pressure conditions.

Closed system - A system in which energy may cross the boundaries, but mattercannot cross the boundaries.

Cycle - A series of processes which result in a final state of a system identical to the

initial state of the system before the series of processes began.

Density () - Mass of a substance per unit volume. Unit: lbm/ft3

Pressure (P) - Force per unit area. Unit: lbf/ft2, lbf/in2

Enthalpy (h) - Sum of the internal energy and flow energy of a substance. Unit:BTU/lbm

Entropy (s) - Quantitative description of the unavailability of energy for the performanceof work. Unit: BTU/lbm R

Fluid- Any substance that conforms to the shape of its container.

Heat (Q) - Energy in transition; energy that flows from one location to another becausea temperature difference exists between those locations. Unit: BTU

Heat of fusion - Heat required at melting temperature to change one pound mass of asolid into liquid form. Unit: BTU/lbm

Heat of vaporization - Heat required at vaporization temperature to change one poundmass of a liquid into gaseous form. Unit: BTU/lbm

Internal Energy (u) - Energy possessed by a substance due to the random motion ofthe individual molecules of a substance. Unit: BTU/lbm


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Definitions, continued

Isobaric process - A constant pressure process.

Isothermal process - A constant temperature process.

Latent heat- Heat added to or removed from a substance that produces a change inthe phase of the substance (but no change in temperature).

Moisture Fraction - The fraction of the total mass of a water/steam mixture that is liquidwater.

Open system - A system in which matter cross the boundaries (energy may or may notcross its boundaries).

Phase - The condition of a substance as defined by its fluidity.

Process - Series of state changes that occur when the properties of a system change.

Quality(x) - the fraction of the total mass of a water/steam mixture that is steam.

Saturated (Dry) Steam - 100% steam at saturation temperature/saturation pressure

conditions.

Saturated Liquid Water- 100% liquid water at saturation temperature/saturationpressure conditions.

Sensible heat- Heat added to or removed from a substance that produces a change inits temperature (but no change in phase).

Specific Heat Capacity (cp) - Heat required to raise the temperature of one poundmass of a material by 1 F. Unit: BTU/lbm F

Specific Volume () - Volume occupied per unit mass of a substance. Unit: ft

3

/lbm


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Definitions, continued

State - A particular condition of a substance described by the (unique) values of itsthermodynamic properties.

Steady state, steady flow system - A system in which any given location remains in aparticular thermodynamic condition (the state of the working fluid at that location is notchanging), and the rate that mass enters the system is equal to the rate mass leavesthe system.

Subcooled Liquid Water- Water that exists at a temperature below saturationtemperature for the pressure of the water, or, equivalently, exists at a pressure abovesaturation pressure for the temperature of the water.

Subcooling Margin (SCM) - The difference between the saturation temperature andthe actual temperature of subcooled water; SCM = TSAT - TACTUAL.

Superheated Steam - Steam existing at a temperature above saturation temperaturefor the pressure of the steam.

Temperature (T) - A measure of the average kinetic energy of the molecules of a

substance. Unit: F, R, C, K

Wet Steam - a mixture of liquid water and steam, both at saturationtemperature/saturation pressure conditions.

Working Fluid - Any fluid which receives, transports and transfers energy in a fluid flowsystem.


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Exercises

Exercise 1

Objective 1: Apply the Universal Gas Law to solve problems relating changes in

mass, pressure, temperature and volume of a gas.

1. a. A tank partially filled with water has an air volume above the water surface of4.0 x 105 in3 and a pressure of 15 psia. The tank is drained with the ventclosed until the air volume reaches 6.0 x 105 in3. What is the air pressure inthe tank? Ignore any effects of water vapor on the gas space pressure, andassume tank temperature remains constant during the process.

b. After the water is drained, the surface area of the tank above the liquid level is3.0 x 105 in2. What is the total (net) force acting to collapse the tank acting ofthe air space region? Assume pressure outside the tank is atmosphericpressure, 14.7 psia.


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2. Containment is sealed leak tight while the inside temperature is 80 F and the

pressure is atmospheric (14.7 psia). If the temperature inside containmentuniformly increases to 115 F, what will the containment pressure be?

3. A gaseous radwaste tank has a volume 700 ft3 and is filled to a pressure of 300psia. The mass of the gas in the tank is 250 lbm. If an additional 75 lbm of gasenters the tank, with no change in tank temperature, what is the final gas pressure

in the tank?


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Exercise 2:

Objective 2: Define the following concepts, including the appropriate English

System unit of measurement:

Sensible heat

Latent heat

Heat of vaporization/condensation

Enthalpy

Entropy

1. Heat addition to a substance that results in a temperature increase in thatsubstance is called heat.

2. The sum of a waters internal energy and flow energy is called .

3. Heat addition to a substance that results in a (constant temperature) phase changein that substance is called heat.

4. The mathematical quantity used to describe the fraction of energy possessed by asubstance that is unavailable to do useful work is called .

5. The energy added to saturated liquid water to produce saturated dry steam is called.

6. The energy removed from saturated dry steam to produce saturated liquid water iscalled .


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Exercise 3:

Objective 3: Given the Specific Heat Equation, solve problems involving heat

transferred, power transferred, mass, mass flow rate, initial temperature, and final

temperature of water.

1. Determine the thermal power (BTU/hr) transferred to water flowing at 1000 lbm/hrthrough a heat source if the temperature of the water increases from 47 F to 75 Fas the water passes through the heat source. The specific heat capacity of thewater is 1.00 BTU/lbm F.

2. A heat source transfers 10 Mw to water passing through it at 2 x 106 lbm/

thermo & heat transfer

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