Thermodynamics ch1

Thermodynamics and Heat Transfer Chapter 10HT121 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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3Open 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 . . . . . . . 35Specific 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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51Subcooling 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:

C Sensible heatC Latent heat• Heat of vaporization/condensationC EnthalpyC Entropy

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 thermodynamicproperty 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 water’ssubcooling 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 introduce characteristics 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 the study of the energy forms associated with a system, either with or without the passageof 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 fluid of 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 energyfrom 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 energy of the substance’sindividual 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

The phase of a working fluid is a general condition of the fluid described by its volume,shape, and energy level. There are three basic phases 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 GasDefinite 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 flow system if the mass flow rateis constant at any given location in the system.

Process

The term process 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 thecoffee’s 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 ofits other thermodynamic properties change. A process is designated isobaric ifpressure remains constant, isometric if volume remains constant, and adiabatic if noheat 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, Pgas % 1Vgas

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 enclosedmass 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 bya 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 + 460E (NOTE: 1RE = 1FE)

Tkelvin = Tcelsius + 273E (NOTE: 1KE = 1CE)


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PV % T

PV % mT

PV ' mRi T

Ri 'PVmT

'PνT

The three formulas on the previous page can be combined into a single, more generallaw 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 directlyproportional 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 GasLaw 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 constant for the individual gas being considered:

where ν 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 32EF and 14.7 psia has aspecific volume of 12.393 ft3/lbm. Thus, the formula above yields Rair = 53.32ft@lbf/lbmER.


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P1V1

m1T1

'P2V2

m2T2

In most applications it is not necessary to know the value of the gas constant for the gasbeing 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 ER) for thisproportion to be valid!

The Universal Gas Law provides accurate results for almost all gases and vapors whenthey 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 2400EF. 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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P1V1 ' P2V2

i.e., P2 'V1

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 70EFwater 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? Assumenitrogen 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 (m1 = m2).The temperature of the gas remains constant at 70EF (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

P1

T1 '404.7 psia364.7 psia

(530ER) ' 588ER

Example B

A gaseous radwaste tank (volume = 750 ft3) contains gas at 350 psig and 70EF. 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 = 70E + 460E = 530ER

Therefore,

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


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Q ' mcp∆T ' mcp(T2 & T1)

Q ' Heat added to the mass (BTU)m ' mass (lbm)∆T' Temperature change of the mass (EF)

cp ' Specific heat capacity of the mass BTUlbmEF

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 usingany 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 EF under specified conditions of pressure (14.7 psia) and temperature(39EF).

If heat is supplied to a mass it generally (but not always) causes a temperature increaseof the mass. The amount of heat required to raise the temperature of one pound massof any substance by one EF 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 39EF willincrease the temperature of that one pound mass to 40EF (by 1EF), the specific heatcapacity of water under these conditions is equal to 1 BTU/lbmEF.

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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0Q ' 0mcp∆T ' 0mcp (T2 & T1)

0Q ' Power added to the flowing water BTUhr

0m ' mass flowrate of the water through the heat source lbmhr

∆T' Temperature change of the water as it passes through the heat source (EF)

cp ' Specific heat capacity of the water BTUlbmEF

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 asubstance 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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0Q ' 0mcp∆T ' (50 lbmhr

) 1.00 BTUlbmEF

(68 & 60)EF ' 400 BTUhr

0Q ' 0mcp∆T

∆T '0Q

mcp'

7,800 BTUhr

200 lbmhr

1.00 BTUlbmEF

∆T ' 39EF

Example C

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

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 specificheat capacity of the water is 1.00 BTU/lbmEF.

Solution

The water temperature increases by 39EF 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/lbmEF. 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 cp remainsessentially constant (near 1.0 BTU/lbmEF) 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 about566.5EF and RCS pressure is 2250 psia. Under these conditions, the value of cp isapproximately 1.4 BTU/lbmEF.


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Figure 1: Variation of cp with Temperature and Pressure


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0QCORE ' 0mcp (TH & TC)

0QCORE ' Power transferred to RCS

0m ' RCS flowrate through corecp ' 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 (thethermal 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: 539EF Hot Leg Outlet Temperature: 594.85EF

Total RCS core flow rate: 150 x 106 lbm/hrCoolant Specific Heat Capacity: 1.4 BTU/lbmEF

NOTE: 1 Mw = 3.412 x 106 BTU/hr


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0QCORE ' 0mcp∆T

' 150 x 106 lbmhr

1.4 BTUlbm&EF

594.85EF & 539EF ' 1.173 x 1010 BTUhr

0QCORE ' 1.173 x 1010 BTUhr

1 Mw

3.412 x 106 BTUhr

' 3,438 Mw

Solution

100% rated power for SONGS Units 2/3 cores is 1.173 x 1010 BTU/hr, equivalent to 3,438 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 212EF and atmospheric pressure,the water begins to boil. Its temperature remains constant as 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 32EF and 14.7psia to change the ice into the liquid phase at 32EF .

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 212EF and 14.7psia to change the water into the gaseous (vapor) phase at 212EF .


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Figure 2 illustrates the effect of adding heat to water under atmospheric pressure (14.7psia) conditions:

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

2. Once the ice reaches 32EF, 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 (212EF) 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 (212EF), 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 212EF, 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 (0EF to 32EF)144 BTUs per pound ' Latent Heat added at 32EF to melt the ice180 BTUs per pound ' Sensible Heat added to raise liquid temperature (32EF to 212EF)970 BTUs per pound ' Latent Heat added at 212EF to vaporize the water

Further Heat Addition: Sensible Heat added to raise steam temperature (212EF to > 212EF

Figure 2 Temperature vs Heat 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 Fusion BTUlbm

hvaporization ' Heat of Vaporization BTUlbm

Q1 ' mcp∆T

' (5 lbm) 0.5 BTUlbmEF

(32 & 22)EF

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 22EF and 14.7 psia to steamat 212EF. The specific heat capacity of ice is 0.5 BTU/lbmEF.

Solution

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


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Q2 & Q1 ' mhfusion

' (5 lbm) 144 BTUlbm

Q2 & Q1' 720 BTU

Q3 & Q2' mcp∆T

' (5 lbm) 1 BTUlbmEF

(212 & 32)EF

Q3 & Q2 ' 900 BTU

Q4 & Q3 ' mhvaporization

' (5 lbm) 970 BTUlbm

Q4 & Q3 ' 4,850 BTU

Qtotal ' 25 BTU % 720 BTU % 900 BTU % 4,850 BTU

Qtotal ' 6,495 BTU

The latent heat addition (Q2 - Q1) necessary to melt the ice can be calculated using theLatent Heat Equation:

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

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

Therefore, the total heat added to change 5 lbm of ice at 22EF and 14.7 psia to steam at212EF 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 212EF if heat is added to it; P = PSAT = 14.7psia when T = TSAT = 212EF. 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 below saturation temperature for the pressureof 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 653EF. However,even at it’s hottest location in the RCS, the temperature of the water is normally nohigher than 594EF. Because 594EF is less than TSAT (653EF) 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 212EF/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 isadded 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, x 'Mass 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-steammixture 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 term percent 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 ofvaporization 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).


0HT121 Rev 1-1, 12/27/0025

x '0.8 lbm

0.2 lbm % 0.8 lbm'

0.8 lbm1 lbm

' 0.80

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


0HT121 Rev 1-1, 12/27/0026

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. Ithas 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.


0HT121 Rev 1-1, 12/27/0027

h ' u %PνJ

h ' specific enthalpy, BTUlbm

u ' specific internal energy BTUlbm

P ' pressure lbfft 2

ν ' specific volume ft 3

lbm

J ' conversion factor, 778 ft lbf1 BTU

Specific Enthalpy (h)

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


0HT121 Rev 1-1, 12/27/0028

∆s ' I dqT

∆s ' change in specific entropy, BTUlbmER

dq ' increment of heat added during the process, BTUlbm

T ' absolute temperature at which the dq is supplied, ER

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:


0HT121 Rev 1-1, 12/27/0029

Note that the unit of measurement of entropy is BTU/lbmER, i.e., BTUs per pound massper degree Rankine. Since the magnitude of 1ER is the same as the magnitude of 1EF,entropy can also be expressed with the unit BTU/lbmEF; 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/lbmER unit and alltemperatures used in that calculation must be in ER.

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 itpossesses into useful mechanical work. Briefly, an ideal turbine is one in which all ofthe 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 insteam 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 fluid’s 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.


0HT121 Rev 1-1, 12/27/0030

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 ofthe data that you would normally find in the Steam Tables will be provided to youhere. When the actual classroom 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 for saturated water (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 (EF), 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.


0HT121 Rev 1-1, 12/27/0031

The first specific volume column, labeled νf, provides the specific volume of saturatedliquid water at the given temperature and pressure. The third specific volume column,labeled νg, provides the specific volume of saturated (dry) steam at the giventemperature and pressure. The middle specific volume column, labeled νfg, representsthe change in specific volume as the water goes from saturated liquid to saturatedsteam at the given temperature and pressure. Example H illustrates.

Example H

Determine:

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

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

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

Solution

a) The water is saturated liquid at 528EF. Table 1 shows that νf, the specificvolume of saturated liquid water at 528EF, is 0.02112 ft3/lbm.

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

b) The water is saturated (dry) steam at 528EF. Table 1 shows that νg, thespecific volume of saturated dry steam at 528EF, is 0.51995 ft3/lbm.

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


0HT121 Rev 1-1, 12/27/0032

c) The change in specific volume as water goes from saturated liquid at 528EF tosaturated (dry) steam at 528EF is found in the νfg column of Table 1. Table 1shows that νfg = 0.49843 ft3/lbm.

Physical interpretation of this νfg value: The volume occupied by a pound massof saturated liquid water increases by 0.49843 ft3 as the water changes tosaturated (dry) steam at 528EF (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 SteamTables available or not:

Example H shows that:

νg = νf + νfg

This is true in all cases, because of how νg, νfg, and νg 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 of saturated liquid water.

hg: The specific enthalpy of saturated (dry) steam.

hfg: The change in specific enthalpy as saturated liquid water changes to saturated(dry) steam.


0HT121 Rev 1-1, 12/27/0033

Based on earlier discussions of the concept of latent heat of vaporization, it is clear thatthe definition above for hfg 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 32EF 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 thechange in enthalpy relative to the zero reference.

Thus, any hf value 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 32EF 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 at32EF 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.


1This is the enthalpy relative to the zero reference enthalpy (h / 0 at . 32EF)

0HT121 Rev 1-1, 12/27/0034

Solution

a) The water is saturated liquid at 850 psia. Table 2 shows that hf, the specificenthalpy 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 525.24EF) 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 525.24EF) 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 of Table 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 I shows 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 thephase change occurs.


0HT121 Rev 1-1, 12/27/0035

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 of saturated liquid water.

sg: The specific entropy of saturated (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 32EF 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 theentropy of a pound mass of the water as it changes from saturated liquid at 32EF 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 at32EF to saturated (dry) steam at the given temperature/pressure conditions.


0HT121 Rev 1-1, 12/27/0036

Example J

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

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

c) The change in specific entropy as water goes from saturated liquid at500EF to saturated (dry) steam at 500EF due to addition of the latentheat of vaporization to this water.

Solution

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

Physical interpretation of this specific entropy: Specific entropy = 0.6890BTU/lbmER is a direct indication of that portion of the water’s energy thatCANNOT be converted into useful work.

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

Physical interpretation of this specific entropy: Specific entropy = 1.4333BTU/lbmER is a direct indication of that portion of the water’s 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 smaller than the fraction of the energy possessed by the saturatedliquid that is available to do work. The key word here is fraction; saturatedsteam at 500EF can obviously do more total work than can saturated liquidwater at 500EF. However, the fraction of the total energy of the saturatedsteam that is available to do work is smaller than the fraction of the saturatedliquid water’s total energy available to do work.


0HT121 Rev 1-1, 12/27/0037

c) The change in specific entropy as water goes from saturated liquid at 500EF tosaturated (dry) steam at 500EF is found in the sfg column of Table 1. Table 1shows that sfg = 0.7443 BTU/lbmER.

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


Example J shows 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.


0HT121 Rev 1-1, 12/27/0038

νwet steam ' νf % xνfg

hwet steam ' hf % xhfg

swet steam ' sf % xsfg

ν ' specific volume of wet steamh ' specific enthalpy of wet steams ' specific entropy of wet steamνf ' specific volume of saturated liquid

hf ' specific enthalpy of saturated liquid

sf ' specific entropy of saturated liquid

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

For a saturated water/steam mixture,

where:


0HT121 Rev 1-1, 12/27/0039

ν ' νf % xνfg

' 0.017274 ft 3

lbm% 0.93 8.4967 ft 3

lbm

ν ' 7.9192 ft 3

lbm

ρ '1ν

'1

7.9192 ft 3

lbm

ρ ' 0.1263 lbmft 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 =281.02EF (saturation temperature for 50 psia).


0HT121 Rev 1-1, 12/27/0040

h ' hf % xhfg

' 250.2 BTUlbm

% 0.93 923.9 BTUlbm

h ' 1109.4 BTUlbm

s ' sf % xsfg

' 0.4112 BTUlbmER

% 0.93 1.2474 BTUlbmER

s ' 1.5713 BTUlbmER

d) From Table 2,

e) From Table 2,


0HT121 Rev 1-1, 12/27/0041

ν88% quality steam ' νf % xνfg ' 0.016136 ft 3

lbm% 0.88 333.59 ft 3

lbm' 293.6 ft 3

lbm

h88% quality steam ' hf % xhfg ' 69.73 BTUlbm

% 0.88 1036.1 BTUlbm

' 981.5 BTUlbm

s88% quality steam ' sf % xsfg ' 0.1326 BTUlbmER

% 0.88 1.8455 BTUlbmER

' 1.7566 BTUlbmER

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:


0HT121 Rev 1-1, 12/27/0042

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, butformatted differently. The first column of Table 3 contains steam pressure values, withsaturation temperatures listed in parentheses just below each pressure value. The nexttwo 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 bywhich Tstm exceeds Tsat is referred to as the degrees of superheat of the steam, orsimply the superheat. Table 3 also provides information about the steam superheat,found in rows designated (Sh).


0HT121 Rev 1-1, 12/27/0043

ν ' 0.5615 ft 3

lbm

h ' 1379.7 BTUlbm

s ' 1.5415 BTUlbmER

sh ' 232.81EF

Example M

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

Solution

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


0HT121 Rev 1-1, 12/27/0044

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 hasa 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 860EF 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:

C enthalpy

C entropy

C pressure

C temperature

C degrees of superheat of superheated steam

C moisture content (and consequently, quality) of wet steam


0HT121 Rev 1-1, 12/27/0045

Concerning the Mollier Diagram:

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 of saturated (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 212EF.


0HT121 Rev 1-1, 12/27/0046

Figure 3 Mollier Diagram


0HT121 Rev 1-1, 12/27/0047

h . 1081 BTUlbm

s . 1.532 BTUlbmER

T . 283EF

Example Q

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

C EnthalpyC EntropyC Temperature

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 280EFconstant temperature line, and well below the next indicated constant temperature line(320EF). 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 isTSAT for 50 psia. Table 2 provides this temperature to be 281.02EF.


0HT121 Rev 1-1, 12/27/0048

Example Q demonstrates advantages and disadvantages of using the Mollier Diagramto 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 (281.02EF) 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.


0HT121 Rev 1-1, 12/27/0049

Thermodynamic Properties of Subcooled Liquid Water

As stated earlier, liquid water at a temperature below the saturation temperature for theexisting water pressure is called subcooled liquid water. Another way of describingsubcooled water is to say that the pressure of the water is greater than the saturationpressure associated with the existing temperature of the water.

The C-E Steam Tables list thermodynamic data for saturated and superheated water;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 tothe thermodynamic properties of saturated liquid water at the SAMETEMPERATURE 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 300EFand 500 psia.

Solution

Water at 300EF has a saturation pressure of 67.005 psia (Table 1). Since the water’spressure 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 300EF saturated liquid water:

ν (300EF, 500 psia) . νf (300EF, 67.005 psia) = 0.01745 ft3/lbmh (300EF, 500 psia) . hf (300EF, 67.005 psia) = 269.7 BTU/lbm

s (300EF, 500 psia) . sf (300EF, 67.005 psia) = 0.4372 BTU/lbmER


0HT121 Rev 1-1, 12/27/0050

ν (544EF, 2250 psia) . νf (544EF, 995.22 psia) ' 0.02157 ft 3

lbm

h(544EF, 2250 psia) . hf (544EF, 995.22 psia) ' 541.8 BTUlbm

NOTE: If the Subcooled Liquid Tables had been used to determine more precisevalues for the thermodynamic properties of this subcooled water, the following valueswould be obtained:

ν (300EF, 500 psia) = 0.01742 ft3/lbm

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

s (300EF, 500 psia) = 0.4364 BTU/lbmER

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 Tablesare the only source 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 544E and 2250 psia.

Solution

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

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


0HT121 Rev 1-1, 12/27/0051

s (544EF, 2250 psia) . sf (544EF, 995.22 psia) ' 0.7427 BTUlbmER

c. The value of sf for saturated water at 544EF is used as the approximation of theentropy 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 100EF and a pressure of 1 psia. Is thewater subcooled, saturated, or superheated?


0HT121 Rev 1-1, 12/27/0052

Solution

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

1) Find 100EF 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 101.74EF. Since 100EF < 101.74EF, the water is subcooled.

Example S

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

Solution

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

OR:

From Steam Table 2, TSAT = 499.80EF for P = 680 psia (by interpolation). Since 500EF >499.80EF, 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 S shows thatthe Diagram is insufficient for absolute determination that the water is superheated.


0HT121 Rev 1-1, 12/27/0053

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 or Table 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 superheatedsteam.

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

where


0HT121 Rev 1-1, 12/27/0054

SCM ' TSAT & TACTUAL

TSAT (2250 psia) ' 653EF (interpolating from Table 2)

Therefore, SCM ' 653EF & 607EF ' 46EF

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

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 607EF.

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

Solution

a.


0HT121 Rev 1-1, 12/27/0055

SCM ' TSAT & TACTUAL

TSAT (2100 psia) ' 642.76EF (Table 2)

Therefore, SCM ' 642.76EF & 575EF ' 67.76EF

b.


0HT121 Rev 1-1, 12/27/0056

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 1EF 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 theinitial state of the system before the series of processes began.

Density (ρ) - 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/lbmER

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 of the individual molecules of a substance. Unit: BTU/lbm


0HT121 Rev 1-1, 12/27/0057

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 pressureconditions.

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 1EF. Unit: BTU/lbmEF

Specific Volume (ν) - Volume occupied per unit mass of a substance. Unit: ft3/lbm


0HT121 Rev 1-1, 12/27/0058

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 asubstance. Unit: EF, ER, EC, EK

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.


0HT121 Rev 1-1, 12/27/0059

Exercises

Exercise 1

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

1. a. A tank partially filled with water has an air volume above the water surface of 4.0 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.


0HT121 Rev 1-1, 12/27/0060

2. Containment is sealed leak tight while the inside temperature is 80EF and thepressure is atmospheric (14.7 psia). If the temperature inside containmentuniformly increases to 115EF, 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 pressurein the tank?


0HT121 Rev 1-1, 12/27/0061

Exercise 2:

Objective 2: Define the following concepts, including the appropriate EnglishSystem 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 water’s 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 .


0HT121 Rev 1-1, 12/27/0062

Exercise 3:

Objective 3: Given the Specific Heat Equation, solve problems involving heattransferred, power transferred, mass, mass flow rate, initial temperature, and finaltemperature 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 47EF to 75EFas the water passes through the heat source. The specific heat capacity of thewater is 1.00 BTU/lbmEF.

2. A heat source transfers 10 Mw to water passing through it at 2 x 106 lbm/hr. If thewater enters the heat source at 72EF, what is the outlet temperature of the water?The specific heat capacity of the water is 1.00 BTU/lbmEF.

3. 3000 BTU of heat is added to 125 lbm of water initially at 78EF. The specific heatcapacity of water is 1.00 BTU/lbmEF. Determine the final temperature of the water.

4. Determine the power (in BTU/hr) delivered to 100 psia water flowing through a heatsource at 6,000 lbm/hr if the water enters the source at 56EF and leaves at 179EF.


0HT121 Rev 1-1, 12/27/0063

Exercise 4:

Objective 4: List characteristics of each of the following:

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

Objective 5: Explain the terms quality and percent moisture as they apply tosaturated water.

Select ALL items from Column 2 that apply to each item in Column 1:

Column 1 Column 2

Saturated liquid watera. Temperature is greater than TSAT for

the existing pressure

b. Temperature is less than TSAT for theexisting pressure

Wet steam c. Temperature is equal to TSAT for theexisting pressure

d. Quality is 100%

Superheated steam e. Moisture content is 100%

f. Quality is 0%

Subcooled liquid water g. Moisture content is 0%

h. 0% < Quality < 100%

Saturated (dry) steami. Enthalpy is equal to hf for the given

temperature

j. Enthalpy is equal to hg for the givenpressure

k. Enthalpy is greater than hg for thegiven pressure


0HT121 Rev 1-1, 12/27/0064

NOTE: Because Exercises 5 and 6 require that you use the Steam Tables and theseSteam Tables are not available to you, you will NOT be held responsible for these typesof questions on your SONGS Pre-Employment Exam!!

Exercise 5

Objective 5: Explain the terms quality and percent moisture as they apply tosaturated water.

Objective 6: Given the Steam Tables and/or Mollier Diagram and two independentthermodynamic properties, determine the following thermodynamic properties ofwater:

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

1. Determine the specific volume of a) saturated liquid water and b) saturated (dry)steam at 224EF.

2. Determine the specific volume of a) saturated liquid water and b) saturated (dry)steam at 180 psia.

3. Determine the specific enthalpy of a) saturated liquid water and b) saturated drysteam at 652EF.

4. Determine the specific enthalpy of a) saturated liquid water and b) saturated drysteam at 70 psia.


0HT121 Rev 1-1, 12/27/0065

5. Determine the specific entropy of a) saturated liquid water and b) saturated drysteam at 112EF.

6. Determine the specific entropy of a) saturated liquid water and b) saturated drysteam at 1 psia.

7. Determine the a) temperature, b) specific volume, c) specific enthalpy, and d)specific entropy of 280 psia wet steam with a quality of 92%.

8. Determine the a) pressure, b) specific volume, c) specific enthalpy, and d) specificentropy of 280EF wet steam with a moisture content of 14%.


0HT121 Rev 1-1, 12/27/0066

9. Determine the temperature of 180 psia steam with an enthalpy of 1271.2 BTU/lbm.

10. Determine the enthalpy and specific volume of 220 psia steam at a temperature of550EF.

11. Steam at 240 psia has 52.61EF of superheat. What is its enthalpy?

12. Determine the a) specific volume, b) specific enthalpy, and c) specific entropy ofwater at 280EF and 50 psia.

13. Determine the a) specific volume, b) specific enthalpy, and c) specific entropy ofwater at 280 psia and 50EF.


0HT121 Rev 1-1, 12/27/0067

1278 Btulbm

1120 Btulbm

990 Btulbm

1.6 BtulbmER

1.39 BtulbmER

14. Using the Mollier Diagram to complete the chart below:

Enthalpyh

Entropys

PressureP

TemperatureT

Qualityx*

Degrees ofSuperheat*

1. 1100 psia (Saturated steam)

2.3.5 in Hg 0.87

3.180 psia

4.400EF

5.

6.140 psia

* if applicable


0HT121 Rev 1-1, 12/27/0068

15. Determine the specific volume, specific enthalpy, and specific entropy, of 122EFwet steam with quality = 0.618.

16. Wet steam at 300 psia has a specific volume of 1.45 ft3/lbm. Determine: (a)temperature in degrees F, (b) quality in percent, (c) enthalpy in BTU/lbm.

17. Determine the following:

(a) enthalpy of 100 psia steam with 70% quality

(b) entropy of 50EF/0.178 psia liquid water

(c) enthalpy of 300 psia/500EF steam

(d) temperature of 600 psia steam with 87% quality

(e) enthalpy of 300 psia steam with specific volume 2.86 ft3/lbm.


0HT121 Rev 1-1, 12/27/0069


Specific volume of liquid water at: (a) 100EF and 0.95 psia,(b) 200EF and 100 psia,(c) 100EF and 200 psia.

Enthalpy of liquid water at: (d) 200EF and 11.5 psia,(e) 50EF and 100 psia.

Entropy of liquid water at: (f) 100EF and 0.94924 psia,(g) 300EF and 400 psia,(h) 300EF and 200 psia.

Specific volume of steam at: (i) 200 psia and 100% quality(j) 100 psia and 95% quality.

Entropy of wet steam at: (k) 185.3 psig and 92% quality.


0HT121 Rev 1-1, 12/27/0070


(a) Specific volume of 200EF/88% quality steam.

(b) Enthalpy of 400 psia/500EF steam.

(c) Degrees of superheat 300 psia/800EF steam.

(d) Entropy of 300 psia/1000EF steam.

(e) Pressure of superheated steam with specific volume 1.2841 ft3/lbm and temperature 500EF.

(f) Temperature of steam at 885 psia and 92% quality.

(g) Temperature of saturated steam at 28.90 in Hgvac.

(h) Quality of wet steam with entropy 1.350 BTU/lbm ER and temperature400EF.

(i) Enthalpy of superheated steam with specific volume 4.0 ft3/lbm and pressure200 psia.


0HT121 Rev 1-1, 12/27/0071

Exercise 6

Objective 7: Given the temperature and pressure of water, determine whether thewater is

• subcooled• saturated• superheated

1. Determine the general state of the water, given the following temperatures andpressures:

Temperature (EF) Pressure (psia) General State

a. 212 14.696

b. 620 2250

c. 517 200

d. 192 10


0HT121 Rev 1-1, 12/27/0072

Exercise 7:

Objective 8: Given selected thermodynamic data for subcooled water, determinethe water’s subcooling margin.

1. Following a loss of offsite power and subsequent reactor trip, minimum indicatedRCS pressure is 2000 psia and its maximum indicated temperature is 600EF. Whatis the RCS subcooling margin?


0HT121 Rev 1-1, 12/27/0073

P1V1

m1T1'

P2V2

m2T2Y P1V1 ' P2V2 Y P2 '

V1

V2P1

P2 '4 x 105 in 3

6 x 105 in 315 psia ' 10 psia

P 'FA

Y F ' PA Y F ' 14.7 lbfin 2

& 10 lbfin 2

3 x 105 in 2 ' 1,410,000 lbf

P1V1

m1T1'

P2V2

m2T2Y

P1

T1'

P2

T2Y P2 '

T2

T1P1

P2 '575 ER540 ER

14.7 psia ' 15.7 psia

Exercise Solutions

Exercise 1:

1a.

1b.

2.


0HT121 Rev 1-1, 12/27/0074

P1V1

m1T1'

P2V2

m2T2Y

P1

m1'

P2

m2Y P2 '

m2

m1P1

P2 '325 lbm250 lbm

300 psia ' 390 psia

3.

Exercise 2:

1. Sensible

2. Enthalpy

3. latent

4. Entropy

5. Heat of vaporization

6. Heat of condensation


0HT121 Rev 1-1, 12/27/0075

0Q ' 0mcp∆T

' 1,000 lbmhr

1.00 BTUlbmEF

(75EF & 47EF)

' 28,000 BTUhr

0Q ' 0mcp∆T

10 Mw3.41 x 106 BTU

hr1 Mw

' 2 x 106 lbmhr

1.00 BTUlbmEF

(TOUT & 72EF)

3.41 x 107 BTUhr

' 2 x 106 BTUhrEF

TOUT & 72EF

TOUT & 72EF '

3.41 x 107 BTUhr

2 x 106 BTUhrEF

TOUT & 72EF ' 17.05EF

TOUT ' 89.05EF

Exercise 3:

1.

2.


0HT121 Rev 1-1, 12/27/0076

Q ' mcp (Tf & Ti)

Tf 'Q

mcp% Ti

Tf '3000 BTU

(125 lbm) 1 BTUlbmEF

% 78EF

Tf ' 102EF

0Q ' 0mcp Tout & Tin ' 6000 lbmhr

1 BTUlbmEF

179EF & 56EF ' 8.46 x 105 BTUhr

3.

4.

Exercise 4

Saturated liquid water: c, e, f, i

Wet steam: c, h

Superheated steam: a, d, g, k

Subcooled liquid water: b, e, f, i (because we must use hf)

Saturated dry steam: c, d, g, j


0HT121 Rev 1-1, 12/27/0077

Exercise 5

REMEMBER: You are not responsible for these types of questions on your pre-employment exam!

1. a. From Steam Table 1, the specific volume of saturated liquid water at 224EF is νf= 0.016805 ft3/lbm.

b. From Steam Table 1, the specific volume of saturated (dry) steam at 224EF is νg= 0.21.545 ft3/lbm.

2. a. From Steam Table 2, the specific volume of saturated liquid water at 180 psia isνf = 0.01827 ft3/lbm.

b. From Steam Table 2, the specific volume of saturated (dry) steam at 180 psia isνg = 2.5312 ft3/lbm.

3. a. From Steam Table 1, the specific enthalpy of saturated liquid water at 652EF is hf= 700.0 BTU/lbm.

b. From Steam Table 1, the specific enthalpy of saturated (dry) steam at 652EF ishg = 1118.7 BTU/lbm.

4. a. From Steam Table 2, the specific enthalpy of saturated liquid water at 70 psia ishf = 272.7 BTU/lbm.

b. From Steam Table 2, the specific enthalpy of saturated (dry) steam at 70 psia ishg = 1180.6 BTU/lbm.

5. a. From Steam Table 1, the specific entropy of saturated liquid water at 112EF is sf= 0.01507 BTU/lbmER .

b. From Steam Table 1, the specific enthalpy of saturated (dry) steam at 112EF issg = 1.9528 BTU/lbmER .

6. a. From Steam Table 2, the specific enthalpy of saturated liquid water at 1 psia is sf= 0.1326 BTU/lbmER .


0HT121 Rev 1-1, 12/27/0078

ν ' νf % xνfg ' 0.01880 ft 3

lbm% 0.92 1.63169 ft 3

lbm' 1.51995 ft 3

lbm

h ' hf % xhfg ' 387.1 BTUlbm

% 0.92 815.1 BTUlbm

' 1137.0 BTUlbm

s ' sf % xsfg ' 0.5805 BTUlbmER

% 0.92 0.9361 BTUlbmER

' 1.4417 BTUlbmER

ν ' νf % xνfg ' 0.017264 ft 3

lbm% 0..86 8.627 ft 3

lbm' 7.436 ft 3

lbm

b. From Steam Table 2, the specific enthalpy of saturated (dry) steam at 1 psia is sg= 1.9781 BTU/lbmER.

7. a. The temperature of wet steam at 280 psia is saturation temperature for 280 psia.Therefore, from Steam Table 2, T = TSAT = 411.07EF.

b. Specific volume:

c. Specific enthalpy:

d. Specific entropy:

8. a. The pressure of wet steam at 280EF is saturation pressure for 280EF. Therefore,from Steam Table 1, P = PSAT = 49.200 psia.

b. Specific volume:


0HT121 Rev 1-1, 12/27/0079

h ' hf % xhfg ' 249.17 BTUlbm

% 0.86 924.6 BTUlbm

' 1044.3 BTUlbm

s ' sf % xsfg ' 0.4098 BTUlbmER

% 0.86 1.2501 BTUlbmER

' 1.4849 BTUlbmER

c. Specific enthalpy:

c. Specific entropy:

9. Since h = 1271.2 BTU/lbm is greater than hg for saturated dry steam at 180 psia,we know the steam is superheated. Therefore, from Steam Table 3, h = 1271.2BTU/lbm corresponds to steam at T = 500EF.

10. Since 550EF is greater than saturation temperature for 220 psia, we know thesteam is superheated. Therefore, from Steam Table 3, enthalpy (h) = 1294.5BTU/lbm and specific volume ν = 2.6199 ft3/lbm.

11. From Steam Table 3, 240 psia steam with 52.61EF of superheat has enthalpy(h) = 1234.9 BTU/lbm.

12. Per Steam Table 2, TSAT = 281.02EF for 50 psia. Since the water temperature is280EF (< TSAT), the water is subcooled. (OR: Per Steam Table 1, PSAT = 49.200psia for 280EF. Since the water pressure is 50 psia (> PSAT), the water issubcooled.) Since the water is subcooled,

ν = νf, 280EF = 0.017264 ft3/lbm (Table 1)

h = hf, 280EF = 249.17 BTU/lbm (Table 1)

s = sf, 280EF = 0.4098 BTU/lbmER (Table 1)


0HT121 Rev 1-1, 12/27/0080

13. Per Steam Table 2, TSAT = 411.07EF for 280 psia. Since the water temperature is50EF (< TSAT), the water is subcooled. (OR: Per Steam Table 1, PSAT = 0.17796psia for 50EF. Since the water pressure is 280 psia (> PSAT), the water issubcooled.) Since the water is subcooled,

ν = νf, 50EF = 0.016023 ft3/lbm (Table 1)

h = hf, 50EF = 18.054 BTU/lbm (Table 1)

s = sf, 50EF = 0.0361 BTU/lbmER (Table 1)


0HT121 Rev 1-1, 12/27/0081

1189 Btulbm

1.378 BtulbmER

980 Btulbm

1.7039 BtulbmER

1278 Btulbm

1.645 BtulbmER

1120 Btulbm

1.43 BtulbmER

990 Btulbm

1.6 BtulbmER

1042 Btulbm

1.39 BtulbmER

14.

Enthalpyh

Entropys

PressureP

TemperatureT

Qualityx

Degrees ofSuperheat

1. 1100 psia (Saturated steam)

556.28EF (Steam Table 2)

-------- --------

2.3.5 in Hg 120EF

(Steam Table 1)

0.87--------

3.180 psia 515EF

-------- 142

4.247.259 psia (Steam Table 1)

400EF 0.902--------

5.6 psia 170EF

(Steam Table 1)

0.856--------

6.140 psia 353.04EF

(Steam Table 2)

0.825--------


0HT121 Rev 1-1, 12/27/0082

15. a) v = vf + xvfg = 0.016213 + 0.618(192.94) = 119.25 ft3/lbm

b) h = hf + xhfg = 89.96 + 0.618(1024.5) = 723.10 Btu/lbm

c) s = sf + xsfg = 0.1680 + 0.618(1.7613) = 1.2565 Btu/lbmER

16. a) T = TSAT, 300 psia = 417.35EF

b) v = vf + xvfg. Therefore, x = (v - vf)/vfg = (1.45 - 0.01889)/1.52384) = 0.94 =94%

c) h = hf + xhfg = 394.0 + 0.94(808.9) = 1154.4 BTU/lbm

17. a) h = hf + xhfg = 298.5 + 0.70(888.6) = 920.5 BTU/lbm

b) Since P = PSAT for 50EF, s = sf, 50EF = 0.0361 BTU/lbmER

c) Since 500EF > TSAT for 300 psia, the steam is superheated. Therefore, fromSteam Table 3, h = 1257.7 BTU/lbm

d) Since the steam is wet steam at 600 psia, T = TSAT, 600 psia = 486.2EF

e) Since 2.86 ft3/lbm > vg, 600 psia, the steam is superheated steam. From SteamTable 3, the temperature of the steam is 1,000EF. Therefore, the enthalpy ofthe steam (Table 3) is 1526.2 BTU/lbm

18. a) Since 0.95 psia = PSAT for 100EF, v = vf, 100EF = 0.016130 ft3/lbm

b) Since 200EF < TSAT for 100 psia, the water is subcooled. Therefore, v . vf, 200EF = 0.016637 ft3/lbm.

c) Since 100EF < TSAT for 200 psia, the water is subcooled. Therefore, v . vf, 100EF = 0.016130 ft3/lbm.

d) Since 11.5 psia = PSAT for 200EF, h = hf, 200EF = 168.09 BTU/lbm


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18. e) Since 50EF < TSAT for 100 psia, the water is subcooled. Therefore, h . hf, 50EF = 18.054 BTU/lbm.

f) Since 0.94924 psia = PSAT for 100EF, s = sf, 100EF = 0.1295 BTU/lbmER

g) Since 300EF < TSAT for 400 psia, the water is subcooled. Therefore, s . sf, 300EF = 0.4372 BTU/lbmER

h) Since 300EF < TSAT for 200 psia, the water is subcooled. Therefore, s . sf, 300EF = 0.4372 BTU/lbmER

i) v = vg, 200 psia = 2.2873 ft3/lbm.

j) v = vf + xvfg = 0.017740 + 0.95(4.4133) = 4.2104 ft3/lbm

k) 185.3 psig = 200 psia. Therefore, s = sf + xsfg = 0.5438 + 0.92(1.0016) =1.4653 BTU/lbmER

19. a) v = vf + xvfg = 0.016637 + 0.88(33.622) = 29.604 ft3/lbm

b) Since 500EF > TSAT for 400 psia, the steam is superheated. Therefore, fromSteam Table 3, h = 1245.1 BTU/lbm.

c) From Steam Table 3, Superheat = 382.65EF

d) Since 1000EF > TSAT for 300 psia, the steam is superheated. Therefore, fromSteam Table 3, s = 1.7964 BTU/lbmER.

e) From Steam Table 3, 500EF/1.2841 ft3/lbm steam has P = 400 psia


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3550

'x

531.95 & 525.24Y x ' 4.70 Y T ' 525.24 % 4.70 ' 529.94EF

28.90 inHgVAC0.491 psi1 inHg

' 14.2 psiv

PABS ' PATM & PVAC ' 14.7 psi & 14.2 psi ' 0.5 psia

T ' TSAT, 0.5 psia ' 79.586EF (Table 2)

19. f) Because it is wet steam, the temperature is TSAT for 885 psia. Interpolation isrequired:

P T850 ----------> 525.24 885900 ----------> 531.95

g)

h) s = sf + xsfg. Therefore, x = (s - sf)/sfg = (1.350 - 0.5667)/0.9607 = 0.815 =81.5%

i) From Steam Table 3, P = 200 psia and v = 4.0 ft3lbm yields h = 1477.0BTU/lbm


0HT121 Rev 1-1, 12/27/0085

Exercise 6

1. a. From Steam Table 1, the temperature and pressure listed represent TSAT/PSATconditions. Therefore, the general state of the water is saturated.

b. From Steam Table 1, PSAT = 1786.9 psia for 620EF water. Since 2250 psia > PSAT for the temperature of the water, the general state of the water is subcooled.

c. From Steam Table 2, TSAT = 381.80EF for 200 psia water. Since 517EF > TSAT forthe pressure of the water, the general state of the water is superheated.

d. From Steam Table 1, PSAT = 9.747 psia for 112EF water. Since 10 psia > PSAT forthe temperature of the water, the general state of the water is subcooled.

Exercise 7

For 2000 psia, TSAT = 635.80EF (Table 2). Since highest indicated RCS temperature is600EF,

Subcooling margin = TSAT - TACTUAL = 635.80EF - 600EF = 35.80EF,

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Thermodynamics ch1