Thermodynamics ch1

Thermodynamics and Heat Transfer Chapter 1
0HT121 Thermodynamic Properties/Concepts

Table of Contents

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

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

The Thermodynamic System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Working Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Thermodynamic Properties of a Working Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
State of a Working Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
Phase of a Working Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
Closed System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
Open System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Steady State, Steady Flow System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

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

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

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

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

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

Determination of Thermodynamic Data from Steam Tables . . . . . . . . . . . . . . . . . . . . 30
Specific Volume of Saturated Liquid Water and Saturated (Dry) Steam . . . . . . . 30
Specific Enthalpy of Saturated Liquid Water and Saturated (Dry) Steam . . . . . . 32
Specific Entropy of Saturated Liquid Water and Saturated (Dry) Steam . . . . . . . 35
Specific Volume, Enthalpy, and Entropy of Wet Steam . . . . . . . . . . . . . . . . . . . . 38
Thermodynamic Properties of Superheated Steam . . . . . . . . . . . . . . . . . . . . . . . 42
Thermodynamic Properties From the Mollier Diagram . . . . . . . . . . . . . . . . . . . . 44
Thermodynamic 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 System
unit of measurement:

C Sensible heat
C Latent heat
• Heat of vaporization/condensation
C Enthalpy
C Entropy

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

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 whether
the whether the water is subcooled (compressed), saturated, or superheated.

8. Given selected thermodynamic data for subcooled water, determine the water’s
subcooling margin.

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Introduction
Thermodynamics is the area of physics that deals with the processes associated with
the conversion of thermal energy into mechanical action. For example, the thermal
energy (heat) transferred to the Secondary Side of the steam generators from the
Reactor Coolant passing through steam generator tubes produces steam which does
mechanical 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 passage
of matter into or out of the system. Therefore, a thermodynamic system is a system in
which energy forms change within the system.

Working Fluid
The working fluid of a thermodynamic system is any fluid (including gases) which
receives, transports, and transfers energy in the system. For example, the Reactor
Coolant 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 the
energy across the tubes of the steam generators to the water on the other side of the
tubes.

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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 earlier
lessons:

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

Temperature: a measure of the average thermal energy of the molecules of a
substance; a direct indication of the average kinetic energy of the substance’s
individual molecules.

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

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

Internal energy: energy possessed by the fluid due to the average kinetic energy
of 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 its
thermodynamic properties. If the values of each of the thermodynamic properties of the
working fluid are known, or if these values can be determined from thermodynamic
properties 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 may
exist: solid, liquid, or gas. A substance may exist in a combination of these phases
such as a solid-liquid combination or a liquid-vapor (gaseous) combination. General
characteristics of the three phases of a substance can be summarized as shown in
Table 1:

Solid Liquid Gas
Definite volume Definite volume Indefinite volume
Definite shape Indefinite shape Indefinite shape
Molecules in fixed Molecules can move Molecules move
position and interact with each independently
other
Lowest energy per Intermediate energy per Highest energy per
molecule molecule molecule

Table 1: Solids, Liquids, and Gases

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

Closed System
A system is defined to be a closed system if matter does not cross the boundaries of
the system. Energy may or may not flow into or out of a closed system (may cross the
boundaries 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 of
mass.

Steady State, Steady Flow System
A steady state system is a system whose working fluid exists in a constant state at any
given location of the system. The system is a steady flow system if the mass flow rate
is 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. For
example, the gradual cooling of coffee in a thermos jug is a process because the
coffee’s temperature is decreasing (and other thermodynamic properties are changing,
too). In addition to designating processes by the properties that change, processes can
be characterized by the fact that certain properties do not change. For example, an
isothermal process is one in which temperature remains constant, but at least one of
its other thermodynamic properties change. A process is designated isobaric if
pressure remains constant, isometric if volume remains constant, and adiabatic if no
heat is transferred.

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

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Universal Gas Law

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

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

• 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:

1
For constant Tgas, Pgas %
Vgas

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

For constant Pgas, V % T

• When the volume of a gas is kept constant, the absolute pressure exerted by
a gas is directly proportional to its absolute temperature:

For constant Vgas, P % T

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

Trankine = Tfahrenheit + 460E (NOTE: 1RE = 1FE)

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

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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 quantity
of 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:

PV % T

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:

PV % mT

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

PV ' m Ri T

Ri is called the gas constant for the individual gas being considered:

PV Pν
Ri ' '
mT T

where ν is the specific volume of the particular gas.

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

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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 states
of the gas:
P1 V1 P2 V2
'
m1 T1 m2 T2

Pressure and temperature must be expressed in absolute units (psia and ER) for this
proportion 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, we
will assume the proportion shown above to be valid under all conditions.

The relationship between the thermodynamic properties of steam is predicted with
reasonable accuracy by the proportion above when the steam is at pressures below 2
psia or temperatures above 2400EF. Since typical steam pressures and temperatures
in our plant do not fall in these ranges, the Universal Gas Law does not produce
accurate results. The relationships between thermodynamic properties of steam have
been experimentally determined and are tabulated in Steam Tables (to be discussed
later).

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Applications of the Universal Gas Law
As stated earlier, the Universal Gas Law above can be used to solve problems involving
thermodynamic 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 70EF
water to the 80% level. Nitrogen gas at 14.7 psia is used as a cover gas above the
liquid level, and the pressurizer vent is closed. If the pressurizer is now drained to the
20% level, what will be the final pressure of the nitrogen in the pressurizer? Assume
nitrogen temperature remains constant during the draining process, and ignore any
effects 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, the
Universal Gas Law reduces to:

P1V1 ' P2V2

V1
i.e., P2 ' P1
V2

Therefore,
V1 0.20 (1800 ft 3)
P2 ' P1 ' (14.7 psia) ' 3.7 psia
V2 0.80 (1800 ft 3)

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

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

Solution
The mass and volume of the gas remain constant until the tank pressure exceeds 390
psig. Therefore, the Universal Gas Law simplifies to become:
P1 P2
'
T1 T2

P1 = 350 psi + 14.7 psi = 364.7 psia

P2 = 390 psi + 14.7 psi = 404.7 psia

T1 = 70E + 460E = 530ER

Therefore,
P2 404.7 psia
T2 ' T1 ' (530ER) ' 588ER
P1 364.7 psia

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

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Specific Heat Equation
As discussed in previous Classical Physics and Fluid Mechanics lessons, heat (Q) is
energy in transition. It is energy that moves from one location to another because a
temperature 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 is
defined to be the amount of heat required to raise the temperature of one pound mass
of 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 increase
of the mass. The amount of heat required to raise the temperature of one pound mass
of any substance by one EF is defined to be the specific heat capacity, cp, of the
substance. (The subscript (p) of the symbol cp indicates that the process of heat
addition 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 will
increase the temperature of that one pound mass to 40EF (by 1EF), the specific heat
capacity of water under these conditions is equal to 1 BTU/lbmEF.

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

Q ' m cp ∆ T ' m cp (T2 & T1)

where
Q ' Heat added to the mass (BTU)
m ' mass (lbm)
∆ T' Temperature change of the mass (EF)
BTU
cp ' Specific heat capacity of the mass
lbm EF

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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 is
flowing past a heat source, it is the flow rate of the water (m-dot), the temperature
change of the water, and the specific heat capacity of the water that determine the
power that is delivered to the water. The Specific Heat Equation becomes the Specific
Power Equation for these dynamic systems:

0 0 0
Q ' m cp ∆ T ' m cp (T2 & T1)

where

0 BTU
Q ' Power added to the flowing water
hr
lbm
0
m ' mass flowrate of the water through the heat source
hr
∆ T' Temperature change of the water as it passes through the heat source (EF)
BTU
cp ' Specific heat capacity of the water
lbm EF
T2 ' Outlet (hot) temperature of the water
T1 ' Inlet (cooler) temperature of the water

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

Solution

0 lbm BTU BTU
0
Q ' m cp ∆ T ' (50 ) 1.00 (68 & 60) EF ' 400
hr lbm EF hr

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

Example D
7,800 BTU/hr is added to water flowing at 200 lbm/hr through a heat source. Determine
the change in water temperature as it passes through the heat source. The specific
heat capacity of the water is 1.00 BTU/lbmEF.

Solution
0 0
Q ' m cp ∆ T
BTU
7,800
0
Q hr
∆T ' '
m cp lbm BTU
200 1.00
hr lbm EF
∆ T ' 39EF

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 the
specific heat capacity of water is 1.00 BTU/lbmEF. However, as water approaches
saturation conditions (i.e., water temperature is relatively close to Tsat for the given
pressure), the magnitude of its specific heat capacity increases.

The variation of cp for water as a function of pressure and temperature is illustrated in
Figure 1. The Figure shows that for a given water pressure, the value of cp remains
essentially constant (near 1.0 BTU/lbmEF) when the water is at a temperature
sufficiently less than saturation temperature, and its value begins to increase as the
water temperature approaches saturation temperature for the given pressure.

At Units 2/3 100% reactor power conditions, the average RCS temperature is about
566.5EF and RCS pressure is 2250 psia. Under these conditions, the value of cp is
approximately 1.4 BTU/lbmEF.

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Figure 1: Variation of cp with Temperature and Pressure
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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:

0 0
QCORE ' m cp (TH & TC)
where:

0
QCORE ' Power transferred to RCS
0
m ' RCS flowrate through core
cp ' Specific heat capacity of RCS
TH ' RCS Hot Leg outlet temperature
TC ' RCS Cold Leg inlet temperature

Example E
Calculate reactor core thermal power, in Mw units, given the following Unit 2 conditions
at 100% power:

Cold Leg Inlet Temperature: 539EF
Hot Leg Outlet Temperature: 594.85EF
Total RCS core flow rate: 150 x 106 lbm/hr
Coolant Specific Heat Capacity: 1.4 BTU/lbmEF

NOTE: 1 Mw = 3.412 x 106 BTU/hr

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Solution
0 0
QCORE ' m cp ∆ T

lbm BTU BTU
' 150 x 106 1.4 594.85EF & 539EF ' 1.173 x 1010
hr lbm&EF hr

0 BTU 1 Mw
QCORE ' 1.173 x 1010 ' 3,438 Mw
hr BTU
3.412 x 106
hr

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 rate
of 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 a
substance and the resultant change in temperature of the substance. Whenever heat
that is added to (or removed from) a substance results in a temperature change in that
substance, 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 the
temperature 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 its
phase 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 as
latent 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 is
referred to as the heat of fusion. The (latent) heat required to change the phase of a
substance 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 144
BTU/lbm. This means that 144 BTU must be added to each lbm of ice at 32EF and 14.7
psia to change the ice into the liquid phase at 32EF .

The heat of vaporization of water under atmospheric pressure conditions is equal to 970
BTU/lbm. Thus, 970 BTU must be added to each lbm of liquid water at 212EF and 14.7
psia 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 (14.7
psia) conditions:

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

2. Once the ice reaches 32EF, further heat addition does not change the temperature
of 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 isothermal
process; thus, the phase change (melting) process is isothermal.

3. After the ice has fully melted, further heat addition results in a temperature increase
of the liquid water until the water reaches saturation temperature (212EF) for its
pressure 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 addition
results in an (isothermal) phase change from liquid to gas (steam); the heat added
is latent heat.

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

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Figure 2 Temperature vs Heat Added to H2O at Atmospheric Pressure

where:

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 ice
180 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

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

Q (to melt) ' m hfusion Q (to vaporize) ' m hvaporization

where
Q ' Heat added or removed (BTU)
m ' mass (lbm)
BTU
hfusion ' Heat of Fusion
lbm
BTU
hvaporization ' Heat of Vaporization
lbm

Example F
Determine the heat input required to change 5 lbm of ice at 22EF and 14.7 psia to steam
at 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 32EF
can be calculated using the Specific Heat Equation:
Q1 ' m c p ∆ T
BTU
' (5 lbm) 0.5 (32 & 22) EF
lbm EF
Q1 ' 25 BTU

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The latent heat addition (Q2 - Q1) necessary to melt the ice can be calculated using the
Latent Heat Equation:
Q2 & Q1 ' m hfusion
BTU
' (5 lbm) 144
lbm
Q2 & Q1' 720 BTU

The sensible heat addition (Q3 - Q2) which raises the liquid temperature to 212EF is:
Q3 & Q2' m c p ∆ T
BTU
' (5 lbm) 1 (212 & 32) EF
lbm EF
Q3 & Q2 ' 900 BTU

The latent heat addition (Q4 - Q3) which boils the water to steam is:
Q4 & Q3 ' m hvaporization
BTU
' (5 lbm) 970
lbm
Q4 & Q3 ' 4,850 BTU

Therefore, the total heat added to change 5 lbm of ice at 22EF and 14.7 psia to steam at
212EF and 14.7 psia is:

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

Qtotal ' 6,495 BTU

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General States of Water: Subcooled, Saturated, and Superheated
In Fluid Mechanics, Chapter 1, the concepts of saturation temperature and saturation
pressure were defined:

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

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

Saturation temperature and saturation pressure are referred to as dependent
thermodynamic properties. Another way of saying this is that the saturation
temperature depends on the pressure of the water, and the saturation pressure
depends on the temperature of the water. For each temperature there is only one
saturation 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 = 14.7
psia when T = TSAT = 212EF. Liquid water under these conditions is referred to as
saturated 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 each
Table. 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 pressure
of the water, the water is called subcooled liquid water:

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

For example, the Reactor Coolant System (RCS) is normally maintained at a pressure
of 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 no
higher than 594EF. Because 594EF is less than TSAT (653EF) for 2,250 psia, this water is
subcooled.

When liquid water is at saturation temperature for the pressure of the water, heat
addition results in some of that water changing to steam. Eventually, if enough heat is
added to the water, it becomes 100% steam. The steam is still at saturation temperature
for the pressure of the water (temperature did not change as heat was added), but the
state 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 heat
input 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 and
some will remain liquid. When water exists as a mixture of liquid water and steam at
TSAT/PSAT conditions, the water is referred to as wet steam:

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

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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:

Mass of steam
Quality, x '
Mass of water % Mass of steam

Mass of steam
'
Total mass of water/steam mixture

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

The term percent moisture (or moisture fraction) is sometimes used instead of percent
quality:
Moisture fraction ' 1 & quality

Percent moisture (%) ' 100% & (% quality)

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 is
generated, 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 dry
steam because it contains no moisture (no liquid water).

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

Solution

0.8 lbm 0.8 lbm
x ' ' ' 0.80
0.2 lbm % 0.8 lbm 1 lbm

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

Note that the moisture content of the mixture in this example is 1 - x = 1 - 0.80 = 0.20, or
20%.

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

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

For example, if steam exists at 180 psia and 498EF, this steam is superheated because
TSAT = 373EF when P = 180 psia. This steam is said to have 498 - 373 = 125 “degrees of
superheat” because it exists at a temperature 125EF above saturation temperature for
its 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 is
less than saturation temperature for the pressure of the water.

2. If the water is saturated, then its temperature is equal to saturation
temperature for the pressure of the water. There are three classifications of
saturated 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 than
0% but less than 100%.

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

Even though the water exists at TSAT/PSAT conditions for each case above, the
states 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 next
higher energy per pound (and energy per pound increases as quality
increases), and saturated (dry) steam the most energy per pound.

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

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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 defined
as the sum of the internal energy and flow energy of the working fluid:

Pν
h ' u %
J

where:

BTU
h ' specific enthalpy,
lbm
BTU
u ' specific internal energy
lbm
lbf
P ' pressure
ft 2
ft 3
ν ' specific volume
lbm
778 ft lbf
J ' conversion factor,
1 BTU

Enthalpy is energy possessed by the working fluid that is available to do work. Because
of the vibrational energy of the individual atoms/molecules of the working fluid
associated with the temperature of the fluid (i.e., because of its internal energy) and
because 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 the
temperature and pressure of the steam. As this steam does work (i.e., applies a force
through a distance) on the turbine blades, the temperature and pressure of the steam
decrease (its enthalpy decreases). The amount of work done by each pound mass of
the steam while it is in the turbine is equal to the change in the enthalpy of the steam.

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Specific Entropy (s)
In any closed thermodynamic system, not all energy possessed by a working fluid can
be converted into useful mechanical work. The thermodynamic property of the working
fluid which quantitatively describes the unavailability of energy for the performance of
work is called entropy (s). The entropy of a working fluid is a mathematically
determined number whose magnitude increases as the temperature and/or the pressure
of the fluid decreases. This means that the fraction of the energy possessed that is
capable of being converted to useful work when water is at a relatively low
temperature/pressure is less than the fraction of the energy possessed that is capable
of being converted to useful work when water is at a relatively high
temperature/pressure.

The calculation of entropy involves calculus and is beyond the scope of this training. As
is 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 in
thermodynamic analysis. The mathematical formula for how an entropy change is
calculated is provided below for information, only:

dq
∆s ' I
T
where:

BTU
∆s ' change in specific entropy,
lbm ER

BTU
dq ' increment of heat added during the process,
lbm

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

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Note that the unit of measurement of entropy is BTU/lbmER, i.e., BTUs per pound mass
per 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 be
the same regardless of which of these two units is used. Technically, however, when
entropy is used in calculations its value is expressed in the BTU/lbmER unit and all
temperatures used in that calculation must be in ER.

The thermodynamic property of entropy will be used when discussing steam cycle and
steam turbine efficiency, i.e., how efficiently the cycle or turbine converts the energy it
possesses into useful mechanical work. Briefly, an ideal turbine is one in which all of
the steam energy change from inlet to outlet of the turbine is converted into rotational
energy 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 ideal
process results in no change in the entropy of the working fluid because all of its energy
change produces useful mechanical work; none of the energy change is unavailable for
the performance of work.

Of course, no turbine is ideal; friction within its moving parts and internal friction within
the steam itself will not allow all of the steam energy change to be converted to useful
shaft work. The working fluid’s entropy increase in any actual turbine process reflects
how efficiently the turbine converts available steam energy into useful work. For given
turbine inlet and outlet steam pressures, the greater the increase in steam entropy
across 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 other
thermodynamic 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 the
thermodynamic properties of water in a given state. If enough specific information about
the state of the water is known, the unique values of each of its thermodynamic
properties 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 actual classroom training occurs, you will learn how to obtain this data
from the Steam Tables.

However, there will be particular symbols used in this and subsequent lessons that you
must recognize. For this reason, much of the description of “how to use the Steam
Tables” 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 these
Tables contains temperature, pressure, specific volume, specific enthalpy, and specific
entropy data.

Specific Volume of Saturated Liquid Water and Saturated (Dry)
Steam
Tables 1 and 2 provide thermodynamic data for saturated water (i.e., saturated liquid
water, wet steam, and saturated (dry) steam data). As stated earlier, saturated water
exists at saturation temperature for the pressure of the water (the water is in a TSAT/PSAT
condition). Column 1 of Table 1 lists temperatures (EF), with their corresponding
saturation pressures (psia) in Column 2. Column 1 of Table 2 lists pressures, with their
corresponding saturation temperatures in Column 2.

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

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The first specific volume column, labeled νf, provides the specific volume of saturated
liquid water at the given temperature and pressure. The third specific volume column,
labeled νg, provides the specific volume of saturated (dry) steam at the given
temperature and pressure. The middle specific volume column, labeled νfg, represents
the change in specific volume as the water goes from saturated liquid to saturated
steam 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 to
saturated (dry) steam at 528EF due to addition of the latent heat of vaporization
to this water.

Solution

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

Physical interpretation of this specific volume: Each pound mass of saturated
liquid 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, the
specific 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.

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

Physical interpretation of this νfg value: The volume occupied by a pound mass
of saturated liquid water increases by 0.49843 ft3 as the water changes to
saturated (dry) steam at 528EF (and 870.31 psia), i.e., as the latent heat of
vaporization is added to this water.

KEY POINT YOU MUST REMEMBER, regardless of whether you have Steam
Tables 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 simply
states that the specific volume of saturated liquid water, plus the increase in specific
volume as that liquid volume boils to steam, is equal to the specific volume of the
saturated steam which results. The temperature and pressure remain constant as the
phase 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 a
manner 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.

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Based on earlier discussions of the concept of latent heat of vaporization, it is clear that
the definition above for hfg describes the latent heat of vaporization:

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

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 was
arbitrarily assigned the value of zero (its absolute enthalpy, as determined by the
defining formula, is actually greater than zero). All values listed in the Tables are
enthalpy values relative to this zero reference; i.e., the values listed represent the
change in enthalpy relative to the zero reference.

Thus, any hf value listed in the Steam Tables can be interpreted as the increase in
energy of a pound mass of the water as it changes from saturated liquid at 32EF to
saturated liquid at the given temperature/pressure conditions. Any hg value represents
the increase in energy of a pound mass of water as it changes from saturated liquid at
32EF 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 psia
to saturated (dry) steam at 850 psia due to addition of the latent heat of
vaporization 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 saturated
liquid 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, the
specific 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 psia
to 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 change
one pound of saturated liquid water at 850 psia to saturated dry steam at 850
psia is 679.5 BTUs, that is, the latent heat of vaporization of water at 850 psia
is 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 simply
states that the specific enthalpy of saturated liquid water, plus the increase in specific
enthalpy as that liquid volume boils to steam, is equal to the specific enthalpy of the
saturated steam which results. The temperature and pressure remain constant as the
phase change occurs.

1
This is the enthalpy relative to the zero reference enthalpy (h / 0 at . 32EF)
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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 a
manner 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 absolute
entropy values; i.e., they are not calculated values of the mathematical formula shown
earlier. The entropy of saturated liquid water at 32EF was arbitrarily assigned the value
of zero (its absolute entropy, as determined by the defining formula, is actually greater
than zero). All values listed in the Tables are entropy values relative to this zero
reference; i.e., the values listed represent the change in entropy relative to the zero
reference.

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 32EF to
saturated liquid at the given temperature/pressure conditions. Any sg value represents
the increase in entropy of a pound mass of water as it changes from saturated liquid at
32EF 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 500EF

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

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

Solution

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

Physical interpretation of this specific entropy: Specific entropy = 0.6890
BTU/lbmER is a direct indication of that portion of the water’s energy that
CANNOT be converted into useful work.

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

Physical interpretation of this specific entropy: Specific entropy = 1.4333
BTU/lbmER is a direct indication of that portion of the water’s energy that
CANNOT be converted into useful work. Also, because the specific entropy of
this saturated steam has a magnitude that is greater than the magnitude of the
specific entropy of saturated liquid water (1.4333 versus 0.6890 determined in
part a), the fraction of the energy possessed by this steam that is available to
do work is smaller than the fraction of the energy possessed by the saturated
liquid that is available to do work. The key word here is fraction; saturated
steam at 500EF can obviously do more total work than can saturated liquid
water at 500EF. However, the fraction of the total energy of the saturated
steam that is available to do work is smaller than the fraction of the saturated
liquid water’s total energy available to do work.

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

Physical interpretation of this sfg value: When energy that is added to saturated
liquid water at 500EF, taking the water to saturated (dry) steam at 500EF, some
of 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 simply
states that the specific entropy of saturated liquid water, plus the increase in specific
entropy as that liquid boils to steam, is equal to the specific entropy of the saturated
steam which results. The temperature and pressure remain constant as the phase
change occurs.

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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,

νwet steam ' νf % x νfg

hwet steam ' hf % x hfg

swet steam ' sf % x sfg

where:
ν ' specific volume of wet steam
h ' specific enthalpy of wet steam
s ' specific entropy of wet steam
νf ' 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

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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,
ν ' νf % x νfg

ft 3 ft 3
' 0.017274 % 0.93 8.4967
lbm lbm

ft 3
ν ' 7.9192
lbm

b) Density is the reciprocal of specific volume:
1
ρ '
ν

1
'
ft 3
7.9192
lbm

lbm
ρ ' 0.1263
ft 3

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

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d) From Table 2,
h ' hf % x hfg

BTU BTU
' 250.2 % 0.93 923.9
lbm lbm

BTU
h ' 1109.4
lbm

e) From Table 2,

s ' sf % x sfg

BTU BTU
' 0.4112 % 0.93 1.2474
lbm ER lbmER

BTU
s ' 1.5713
lbmER

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

Solution
From Table 2:
ft 3 ft 3 ft 3
ν88% quality steam ' νf % x νfg ' 0.016136 % 0.88 333.59 ' 293.6
lbm lbm lbm

BTU BTU BTU
h88% quality steam ' hf % x hfg ' 69.73 % 0.88 1036.1 ' 981.5
lbm lbm lbm

BTU BTU BTU
s88% quality steam ' sf % x sfg ' 0.1326 % 0.88 1.8455 ' 1.7566
lbmER lbmER lbmER

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

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 of Table 3 contains steam pressure values, with
saturation temperatures listed in parentheses just below each pressure value. The next
two columns contain saturated liquid and saturated steam thermodynamic properties
associated with the pressure listed in the first column. Therefore, the first three columns
of the Superheated Steam Table 3 are redundant with the information provided in
Tables 1 and 2.

Thermodynamic data for superheated steam begins in the fourth column. Because
superheated steam, by definition, exists at a temperature above Tsat for the pressure of
the steam, several temperature columns are provided. The number of degrees by
which Tstm exceeds Tsat is referred to as the degrees of superheat of the steam, or
simply the superheat. Table 3 also provides information about the steam superheat,
found in rows designated (Sh).

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Example M
Determine the specific volume, enthalpy, entropy, and degrees of superheat of steam at
1200 psia and 800EF.

Solution
For water at 1200 psia, Tsat = 567.19EF. Since the temperature of the steam, 800EF, is
greater 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 2
provide information about saturated water, only).

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

ft 3
ν ' 0.5615
lbm
BTU
h ' 1379.7
lbm
BTU
s ' 1.5415
lbmER
sh ' 232.81EF

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Example N
Determine the specific volume of 200 psia steam having an enthalpy of 1477.0
BTU/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 is
superheated and therefore Table 3 must be used to obtain its thermodynamic
properties. 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 for
wet steam with at least 44% quality up to superheated steam with as much as 860EF of
superheat.

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

For a given state of superheated, saturated, or wet steam, the Mollier Diagram can be
used 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

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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 this
region 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 state
of superheated steam existing at the given steam temperature.

Each point on a given constant superheat line represents a unique state of
superheated 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 is
psia.

When the constant pressure line is a dotted line, the unit of pressure is in Hgabs (not
shown on Figure 2.4).

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

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Figure 3 Mollier Diagram

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

C Enthalpy
C Entropy
C Temperature

Solution
The unique point on the Mollier diagram corresponding to the given information is found
where 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 be
determined:
BTU
h . 1081
lbm

Reading vertically down from this point to the horizontal axis, the entropy can be
determined:
BTU
s . 1.532
lbm ER

The temperature of wet steam at 50 psia, regardless of its quality, is exactly the same
as the temperature of saturated steam at 50 psia. Therefore, if the 50 psia constant
pressure line is followed up to its intersection point with the saturation line, the constant
temperature lines which end at the saturation line can be used to estimate the
temperature of the 50 psia wet steam. The intersection point is slightly above the 280EF
constant temperature line, and well below the next indicated constant temperature line
(320EF). Therefore, an estimate of the temperature of the wet steam could be:
T . 283EF

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 281.02EF.

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47


Example Q demonstrates advantages and disadvantages of using the Mollier Diagram
to obtain thermodynamic information. Without the Mollier diagram, the wet steam
formulas (h = hf + xhfg; s = sf + xsfg) would have to be applied to find the steam enthalpy
and 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 precision
of their values is limited. Finally, the process described above for use of the Mollier
diagram to find the temperature of the the wet steam is tedious and the accuracy of the
answer is questionable. It is much easier (and more accurate) to go to the Steam
Tables to obtain the saturation temperature (281.02EF) for 50 psia.

Summarizing:

The state of water is uniquely defined if two independent thermodynamic properties of
the 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 the
point is located, the values of the other thermodynamic properties can be read from this
location.

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

If thermodynamic data for wet steam with a moisture content greater than 56% is
needed, or if saturated liquid or subcooled water data is needed, the Mollier Diagram
does 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 liquid water. Another way of describing
subcooled water is to say that the pressure of the water is greater than the saturation
pressure 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 precise
thermodynamic property value information, Subcooled Water Tables must be used.

However, the C-E Steam Tables can be used to obtain reasonable approximations of
the thermodynamic properties of subcooled water. The following rule apply for use of
the 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 300EF
and 500 psia.

Solution

Water at 300EF has a saturation pressure of 67.005 psia (Table 1). Since the water’s
pressure is 500 psia, we know that the water is subcooled. Therefore, per the rule
stated above, the values of the specific volume, enthalpy, and entropy of this water are
approximated by the values listed for 300EF saturated liquid water:

ν (300EF, 500 psia) . νf (300EF, 67.005 psia) = 0.01745 ft3/lbm
h (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

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49


NOTE: If the Subcooled Liquid Tables had been used to determine more precise
values for the thermodynamic properties of this subcooled water, the following values
would 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 of
500 psia versus 67.005 psia on the values are minimal. Since the ABB Steam Tables
are the only source of thermodynamic data the NRC allows the student to use during
the Generic Fundamentals (GFE) Exam, the method shown above must be used to
obtain thermodynamic data for subcooled water.

Example P
Using the C-E Steam Tables, determine the a) specific volume, b) enthalpy, and c) the
entropy 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 the
specific volume of this subcooled liquid. Therefore,

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

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

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

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50


c. The value of sf for saturated water at 544EF is used as the approximation of the
entropy of this subcooled liquid. Therefore,

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

Determining The General State of Water
If the temperature and pressure of water are known, the Steam Tables or Mollier
diagram 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 saturation
temperature for the existing pressure or whose pressure is above the saturation
pressure for the existing temperature. When water is at saturation conditions, its
temperature is the saturation temperature for the existing pressure or, equivalently, its
pressure is the saturation pressure for the existing temperature. Finally, when steam is
superheated, its temperature is above the saturation temperature for the existing
pressure or, equivalently, its pressure is below the saturation pressure for the existing
temperature.

Example R
Water leaves the condenser at a temperature of 100EF and a pressure of 1 psia. Is the
water subcooled, saturated, or superheated?

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51


Solution
Since temperature and pressure are known, Table 1 or Table 2 can be used to
determine 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 and
680 psia?

Solution
From Steam Table 1, PSAT = 680.86 psia for 500EF. Since 680 psia < PSAT, the water
must 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 Table
2 information, because no interpolation was necessary with Table 1. The point here
is that you should weigh all options Talbes 1,2,3, and the Mollier Diagram) when
determining the general state of water from the available data; usually, one of those
options 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 that
the Diagram is insufficient for absolute determination that the water is superheated.

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52


Example T

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

Solution
The Mollier Diagram shows that water with enthalpy 1270 BTU/lbm is superheated
steam, regardless of the pressure of the water. More specifically, it is clear that the
intersection of the constant enthalpy 1270 BTU/lbm line and the constant pressure 175
psia 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 superheated
steam.

Subcooling Margin
Subcooling margin (SCM) is the term used describe the amount by which water is
subcooled. The subcooling margin of subcooled water is defined as the difference
between the saturation temperature for the existing water pressure and the liquid's
actual temperature:

Subcooling Margin (SCM) ' TSAT & TACTUAL

where
TSAT ' Saturation temperature for the water pressure
TACTUAL 'Actual Temperature of the water

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53


Subcooling margin is an important Reactor Coolant System (RCS) parameter. As long
as the RCS subcooling margin is maintained above specified minimum values, control
of RCS pressure, inventory, and core heat removal is assured (to be discussed in detail
later).

Units 2/3 have Control Room indications of subcooling margins in the reactor vessel
upper head, just above the reactor core, and in RCS coolant system piping. In addition,
the Qualified Safety Parameter Display System (QSPDS) has two additional subcooled
margin indications that can be retrieved if necessary. QSPDS computes subcooled
margin in terms of temperature (TSAT - TACTUAL) and in terms of pressure. Pressure
subcooled 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 determine
the subcooling margin. Plant procedures specify using the highest available RCS hot
leg temperature (Th) indication and the lowest pressurizer pressure (Ppzr) indication. An
example 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 indicated
RCS hot leg temperature is 607EF.
b. the lowest indicated pressurizer pressure is 2100 psia, and the highest indicated
RCS hot leg temperature is 575EF.

Solution
a.

SCM ' TSAT & TACTUAL

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

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

0HT121 Rev 1-1, 12/27/00
54


b.
SCM ' TSAT & TACTUAL

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

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

0HT121 Rev 1-1, 12/27/00
55


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

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

Closed system - A system in which energy may cross the boundaries, but matter
cannot 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 (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 performance
of 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 because
a temperature difference exists between those locations. Unit: BTU

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

Heat of vaporization - Heat required at vaporization temperature to change one pound
mass 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

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56


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 in
the phase of the substance (but no change in temperature).

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

Open system - A system in which matter cross the boundaries (energy may or may not
cross 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/saturation
pressure conditions.

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

Specific Heat Capacity (cp) - Heat required to raise the temperature of one pound
mass of a material by 1EF. Unit: BTU/lbmEF

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

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Definitions, continued

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

Steady state, steady flow system - A system in which any given location remains in a
particular thermodynamic condition (the state of the working fluid at that location is not
changing), and the rate that mass enters the system is equal to the rate mass leaves
the system.

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

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

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

Temperature (T) - A measure of the average kinetic energy of the molecules of a
substance. Unit: EF, ER, EC, EK

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

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

0HT121 Rev 1-1, 12/27/00
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