Download Thermodynamic Cycles

Revision 1
December 2014
Thermodynamic
Cycles
Student Guide
GENERAL DISTRIBUTION
GENERAL DISTRIBUTION: Copyright © 2014 by the National Academy for Nuclear Training. Not for
sale or for commercial use. This document may be used or reproduced by Academy members and
participants. Not for public distribution, delivery to, or reproduction by any third party without the prior
agreement of the Academy. All other rights reserved.
NOTICE: This information was prepared in connection with work sponsored by the Institute of Nuclear
Power Operations (INPO). Neither INPO, INPO members, INPO participants, nor any person acting
on behalf of them (a) makes any warranty or representation, expressed or implied, with respect to the
accuracy, completeness, or usefulness of the information contained in this document, or that the use of
any information, apparatus, method, or process disclosed in this document may not infringe on
privately owned rights, or (b) assumes any liabilities with respect to the use of, or for damages
resulting from the use of any information, apparatus, method, or process disclosed in this document.
ii
Table of Contents
INTRODUCTION .......................................................................................................................2
TLO 1 FIRST LAW OF THERMODYNAMICS ..............................................................................2
Overview ............................................................................................................................2
ELO 1.1 Thermodynamic Surroundings, Equilibrium, and Control Volume....................3
ELO 1.2 Thermodynamic Relationship Between Real and Ideal Processes .....................9
ELO 1.3 Applying the First Law of Thermodynamics for Open Systems ......................15
ELO 1.4 Identifying Cyclic Process Paths on a T-s Diagram Principle ..........................23
ELO 1.5 Thermodynamic Energy Balances on Major Components ...............................27
TLO 1 Summary ..............................................................................................................36
TLO 2 THE SECOND LAW OF THERMODYNAMICS .................................................................40
Overview ..........................................................................................................................40
ELO 2.1 Thermodynamic Entropy ..................................................................................41
ELO 2.2 Carnot’s Principle of Thermodynamics ............................................................45
ELO 2.3 Thermodynamics of Ideal and Real Processes ..................................................52
ELO 2.4 Thermodynamic Power Plant Efficiency ..........................................................54
TLO 2 Summary ..............................................................................................................79
THERMODYNAMIC PROCESSES SUMMARY ............................................................................83
iii
This page is intentionally blank.
iv
Thermodynamic Cycles
Revision History
Revision
Date
Version
Number
Purpose for Revision
Performed
By
11/7/2014
0
New Module
OGF Team
12/11/2014
1
Added signature of OGF
Working Group Chair
OGF Team
Rev 1
1
Introduction
Four basic laws, or principles, govern the study of thermodynamics and
define basic characterizations of the physical environment.
The zero law of thermodynamics is two bodies in thermal equilibrium are at
the same temperature. The first law of thermodynamics says energy can
never be created or destroyed, but only altered in form. The second law
states the total entropy of the universe must increase in every spontaneous
process. The third law of thermodynamics is that the entropy of a pure,
perfectly crystalline compound when the temperature is zero and there is no
disorder.
This module will focus on the first and second laws.
The first law of thermodynamics requires a balance of the various forms of
energy as they pertain to the specified thermodynamic system (control
volume) studied. The second law of thermodynamics does not take into
account the feasibility of the process or the efficiency of the energy
transformation studied. The second law of thermodynamics allows us to
determine the maximum efficiency of the operating system so that design
comparisons maximize the system’s efficiency.
Objectives
At the completion of this training session, the trainee will demonstrate
mastery of this topic by passing a written exam with a grade of 80 percent
or higher on the following Terminal Learning Objectives (TLOs):
1. Apply the First Law of Thermodynamics to analyze
thermodynamic systems and processes.
2. Apply the Second Law of Thermodynamics to analyze real and
ideal systems and components.
TLO 1 First Law of Thermodynamics
Overview
The First Law of Thermodynamics is a balance of the various forms of
energy as they pertain to the specified thermodynamic system (control
volume) studied. The operator must understand the First Law of
Thermodynamics and be able to apply it to the thermodynamic processes
occurring in the primary and secondary system, as well as other tertiary
systems throughout the plant.
Meeting these objectives ensures the operator has the required knowledge to
determine quickly when the plant thermodynamic parameters are not at the
proper values for the present plant operating conditions. The operator must
be able to work with system engineers who maintain the plant’s
thermodynamic efficiency. The operator must be able to determine when
2
Rev 1
equipment is not at its maximum efficiency and the best solution to return
maximum efficiency.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Relate the following terms to Open, Closed, and Isolated Systems:
a.
b.
c.
d.
Thermodynamic surroundings
Thermodynamic equilibrium
Control volume
Steady state
2. Describe the following processes:
a.
b.
c.
d.
e.
f.
g.
Thermodynamic process
Cyclic process
Reversible process
Irreversible process
Adiabatic process
Isentropic process
Throttling process (Isenthalpic)
3. Apply the First Law of Thermodynamics for open systems or cyclic
processes.
4. Identify the path(s) on a T-s diagram that represents the
thermodynamic processes occurring in a fluid system.
5. Given a defined system, perform energy balances on all major
components in the system.
ELO 1.1 Thermodynamic Surroundings, Equilibrium, and
Control Volume
Introduction
The First Law of Thermodynamics states that energy can be neither created
nor destroyed, but only altered in form. The energy forms may not always
be the same but the total energy in the system remains constant.
Rev 1
3
Figure: Energy Balance Equals the First Law of Thermodynamics
This energy balance for any system includes mass and energy crossing the
control boundary, external work and/or heat crossing the boundary, and a
change of stored energy within the control volume. The mass flow of
matter (typically fluid) is associated with the kinetic, potential, internal, and
flow energies that affect the overall energy balance of the system. The
exchange of external work and/or heat determines the stored energy change
and yields a change in stored energy in the control volume.
The principle of the First Law of Thermodynamics is that energy is always
conserved, meaning that energy can neither be created nor destroyed, but
rather transformed into various forms as the matter within the control
volume is being studied. The system is a region in space (control volume)
through which the matter passes. The various energies associated with the
matter are observed as they cross the boundaries of the system and the
balance is made.
Thermodynamic Systems
Defining an appropriate system can greatly simplify a thermodynamic
analysis. A thermodynamic system is any three-dimensional region of
space that is bounded by one or more surfaces. The bounding surfaces may
be real or imaginary and may be at rest or in motion. The boundary may
change its size or shape. The region of physical space that lies outside the
selected boundaries of the system is called the surroundings or the
environment.
A system in thermodynamics is nothing more than the collection of matter
that is being studied. A system could be the water within one side of a heat
exchanger, the fluid inside a length of pipe, or the entire lubricating oil
system for a diesel engine. Determining the boundary in order to solve a
thermodynamic problem for a system depends on what information is
known about the system and what question is asked about the system.
There are three types of systems to use in thermodynamic calculations:
isolated, open, and closed.
4
Rev 1
Figure: Types of Thermodynamic Systems
An isolated system is one that is not influenced in any way by the
surroundings. This means that no energy in the form of heat or work may
cross the boundary of the system. Additionally, no mass may cross the
boundary of the system. We could assume that the reactor coolant system
(RCS) is an isolated system if charging and letdown were secured and that
no heat was being transferred through the steam generators or to the
environment.
A closed system has no transfer of mass with its surroundings, but may
have a transfer of energy (either heat or work) with its surroundings. Again,
the RCS could function as a closed system if charging and letdown were
secured but heat was being transferred through the steam generator to the
secondary plant.
An open system is one that may have a transfer of both mass and energy
with its surroundings. The RCS can also be arranged as an open system
with mass and energy being transferred in and out by the charging and
letdown system, as well as heat is being lost the environment and being
transferred through the steam generator to the secondary system. The
reactor coolant system could be described as all three types of
thermodynamic systems under certain operational conditions.
Rev 1
5
Figure: Reactor Coolant System a Type of Thermodynamic System
The open system is the most general of the three types and indicates that
mass, heat, and external work are allowed to cross the control boundary of
the system. The energy balance for this type of system is expressed in
words as:
All energies into the system equal all energies leaving the system plus the
change in storage of energies within the system.
The analysis of these systems assumes a steady state condition in which
there is no accumulation of mass or energy within the control volume, and
the properties at any point within the system are independent of time.
When a system is in equilibrium with regard to all possible changes in state,
the system is in thermodynamic equilibrium.
Energy in thermodynamic systems is composed of kinetic energy (KE),
potential energy (PE), internal energy (U), and flow energy (PV); as well as
heat and work processes.
The energy input for a steady flow system equals the total energy of the
working fluid entering the system (Ein) plus the heat (Q) added to the
system. The energy output term equals the work (W) done by the system
plus the total energy of the working fluid leaving the system, (Eout). The
temperature and mass of the system is constant and the energy balance
equation for a steady flow system is shown below in the figure and written
as follows:
𝛴 (𝑎𝑙𝑙 𝑒𝑛𝑒𝑟𝑔𝑖𝑒𝑠 𝑖𝑛)
= 𝛴 (𝑎𝑙𝑙 𝑒𝑛𝑒𝑟𝑔𝑖𝑒𝑠 𝑜𝑢𝑡) + 𝛥 (𝑒𝑛𝑒𝑟𝑔𝑦 𝑠𝑡𝑜𝑟𝑒𝑑 𝑖𝑛 𝑠𝑦𝑠𝑡𝑒𝑚)
𝛴 𝐸𝑖𝑛 = 𝛴 𝐸𝑜𝑢𝑡 + 𝛥𝐸𝑠𝑡𝑜𝑟𝑒𝑑
6
Rev 1
If heat (Q) is exchanged or work is done (W) on or by the initial energy,
energy has been added or removed and it must also be included in the
balance. Most cycles do not store or accumulate energy so the balance
equation becomes:
𝑄 + 𝐸𝑖𝑛 = 𝑊 + 𝐸𝑜𝑢𝑡 or 𝑊 = 𝑄 + 𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡
Reactor Collant
Figure: Steady Flow Systems
Figure: Basic Energy Balance of the First Law of Thermodynamics
Therefore, in equation form, the balance appears as indicated on the figure
below:
Where:
•
𝑄̇ = heat flow into the system (British Thermal Units per hour
[BTUs/hr])
•
𝑚̇𝑖𝑛 = mass flow rate into the system (pound mass per hour [lbm/hr])
•
Uin = specific internal energy into the system (BTU/lbm)
•
Pinνin = pressure-specific volume energy into the system (feet poundforce per pound mass [ft-lbf/lbm])
Rev 1
7
•
2
𝑖𝑛
2𝑔𝑐
𝑉
= kinetic energy into the system (ft-lbf/lbm)
•
𝑉𝑖𝑛 = average velocity of fluid (feet per second [ft/sec])
•
gc = the gravitational constant (32.17 ft-lbm/lbf-sec2)
•
(𝑔 ) 𝑍𝑖𝑛 = potential energy of the fluid entering the system (ft-lbf/lbm)
𝑔
𝑐
Where:
•
𝑍𝑖𝑛 = height above reference level (ft)
•
g = acceleration due to gravity (ft/sec2)
•
gc = the gravitational constant (32.17 ft-lbm/lbf-sec2)
•
W = work flow out of the system (ft-lbf/hr)
•
𝑚̇𝑜𝑢𝑡 = mass flow rate out of the system (lbm/hr)
•
Uout = specific internal energy out of the system (BTU/lbm)
•
Poutνout = pressure-specific volume energy out of the system (ftlbf/lbm)
•
•
2
𝑜𝑢𝑡
2𝑔𝑐
𝑉
= kinetic energy out of the system (ft-lbf/lbm)
𝑔
(𝑔 ) 𝑍𝑜𝑢𝑡 = potential energy of the fluid exiting the system (ft-lbf/lbm)
𝑐
Figure: Continuity Equation for the First Law of Thermodynamics
The principle of the first law of thermodynamics is illustrated above in the
figure. All of the energies entering and leaving the control volume
boundary, and any work done on or by the control volume, and any heat
8
Rev 1
transferred into and out of the control volume boundaries are accounted for
in the energy balance equation.
Knowledge Check
A system that is not influenced in any way by its
surroundings is a(an) …
A.
open system
B.
closed system
C.
isolated system
D.
primary system
Knowledge Check
A system that has energy transferred but no mass
transferred across its boundaries is a(an) …
A.
open system
B.
closed system
C.
isolated system
D.
primary system
ELO 1.2 Thermodynamic Relationship Between Real and Ideal
Processes
Introduction
A thermodynamic process is the succession of states through which the
system passes. Here, a system state changes when one or more properties
of the system changes. One example of a thermodynamic process is
increasing the temperature of a fluid while maintaining a constant pressure.
Another example is increasing the pressure of a confined gas while
maintaining a constant temperature. We explore the six basic
thermodynamic processes of steady flow systems in this module, shown
below in the figure, including:
1. Transfer heat into the system.
2. Transfer heat out of the system.
3. Perform work on the system.
Rev 1
9
4. Perform work by the system.
5. Transfer mass into the system.
6. Transfer mass out of the system.
Figure: Six Basic Processes of Steady Flow Systems
Types of Thermodynamic Processes
When the system (the fluid studied) experiences work, heat, or internal
energy exchange, its properties (temperature, pressure, and volume) change
from one value to another and the fluid goes through a process. In some
processes, the relationships between pressure, temperature, and volume are
specified as the fluid goes from one thermodynamic state to another. The
most common processes are those in which the temperature, pressure, or
volume holds constant during the process. These are respectively classified
as isothermal, isobaric, or isovolumetric processes. Iso means a constant or
one. If the fluid passes through various processes and then eventually
returns to the same state in which it began, the system undergoes a cyclic
process.
Cyclic Process
The system has undergone a cyclic process or cycle when a system returns
to its initial values (state) after processes change its state. At the conclusion
of a cycle, all the properties have the same value they had at the beginning.
Steam or water circulating through a closed cooling loop undergoes a cycle.
Reversible Process
A reversible process for a system is a process that, once having taken place,
can be reversed without leaving any change in either the system or
surroundings. There are no truly reversible processes. However, for
analytic purposes, reversible processes are used to determine maximum
theoretical efficiencies. Engineering and calculation each use the reversible
process as a starting point on which to base system designs.
10
Rev 1
Reversible process can be approximated if a process is performed in a series
of small steps. Heat transfer may be considered reversible if it occurs due
to a small temperature difference between the system and its surroundings.
A temperature difference of 0.00001 °F (degrees Fahrenheit) appears to be
more reversible than transferring heat across a temperature difference of
100 °F.
Irreversible Process
An irreversible process cannot return both the system and the surroundings
to their original conditions if reversed. For example, an automobile engine
does not give back the fuel it took to drive up a hill as it coasts down the hill
to its original position.
There are factors that are present in real, irreversible processes that prevent
these processes from being reversible. Friction, unrestrained expansion of a
fluid, heat transfer through a finite temperature difference, and mixing of
two different substances are examples of these factors.
A piston and cylinder with large and small weights will be used to
demonstrate the difference between reversible and irreversible processes as
shown below in the figure. The cylinder is filled with a gas that results in a
pressure on the piston face adequate to prevent downward motion of the
piston. On the top of the piston a platform is mounted which can support a
number of small weights of mass (m). A series of shelves located next to
the piston platform is arranged to store the weights.
Figure: Real Piston and Cylinder
No heat can flow into or out of the system because the cylinder is perfectly
insulated from the surroundings (adiabatic enclosure). An adiabatic process
will be described since there will be no external source or sink of heat
energy.
Rev 1
11
A weight is removed from the piston's platform and placed on an adjacent
shelf. Due to the friction between the piston and cylinder that occurs in any
real process, the piston will not move initially. Moving additional weights
from the platform to the shelf will cause the piston to break free and
overcome the restraining friction forces. The piston will then accelerate
upward since the pressure applied by the gas is now more than sufficient to
balance the weight against it. As the platform moves upward, useful work
is produced to move the remaining weights against the gravitational field.
The piston will decelerate to some final position, oscillations will eventually
be damped out due to friction between the piston and cylinder, and viscous
friction of the gas itself. These effects prevent the piston from rising as far
as it would have if they had not occurred. The amount of useful work
produced is not as great as what would have been produced in an ideal
process. The process is not reversible since it would be impossible to
retrace the oscillations in a compression process. Because of the same type
of frictional dissipative effects, replacing more than the original weights
removed is required to return the piston to its original position.
In an ideal or reversible process shown below the movement between the
piston and cylinder would be completely frictionless. The weights are
replaced with very small, almost infinitesimal masses Δm (change in mass),
such as metal shavings. Once again, the gas pressure is sufficient to balance
exactly the downward force applied by the weight of the piston and the
metal shavings.
Figure: Ideal Piston and Cylinder
Since no friction exists between the piston and cylinder, if a single metal
shaving is moved from the platform onto the shelf, the piston responds
immediately by moving upward by a small amount, ΔL (change in length).
Since the removed mass was small and no friction exists to retard
12
Rev 1
movement, the piston will not accelerate an appreciable amount, so no
overshoot and oscillation occur. As additional shavings are removed, the
piston continues to move upward in small and well-defined increments.
Work is done each time the piston moves upward.
At any time, this process could be reversed by placing weights from the
shelves back onto the platform, which would move the piston downward.
For example, to return the piston to its original position we would have to
raise the shaving on the lowest shelf a height of ΔL and place it back on the
platform. It would require exactly as much work to raise the small mass,
Δm, and a height of ΔL as was produced initially in the gas expansion
process when the weight was removed. This process is reversible since it
can be undone by an infinitesimal change in external conditions (replacing
the mass, Δm), and no permanent change resulted in either the system or its
surroundings.
Real processes are not reversible. However, it is often useful to idealize a
process and examine it as if it was. Then the real process can be compared
to the ideal to see how well the process or system is designed and built.
This comparison is often given in terms of efficiency. The more efficient a
process is, the closer it approaches its reversible ideal. Since a reversible
process represents a maximum work output or maximum desired effect for a
minimum work or heat input, a great deal is gained in optimizing
efficiencies.
Adiabatic Process
An adiabatic process is one in which there is no heat transfer into or out of
the system. The system can be considered perfectly insulated.
Isentropic Process
Entropy of the fluid remains constant. If the process is reversible and
adiabatic, entropy can remain unchanged. An isentropic process can also be
called a constant entropy or an ideal process. Real processes always result
in a change in entropy.
Throttling Process
A throttling process is defined as a process in which there is no change in
enthalpy from state one to state two (h1= h2), no work is done (W = 0), and
the process is adiabatic (Q = 0). For a better understanding of the theory of
the ideal throttling process, we can compare what we can observe with the
above theoretical assumptions.
An example of a throttling process is an ideal gas flowing through a valve
in mid-position.
From experience, we can observe that:
Rev 1
13
𝑃𝑖𝑛 > 𝑃𝑜𝑢𝑡 , 𝑣𝑒𝑙𝑖𝑛 < 𝑣𝑒𝑙𝑜𝑢𝑡 (Where P = pressure and vel =
velocity)
These observations confirm the theory that:
ℎ𝑖𝑛 = ℎ𝑜𝑢𝑡
Remember:
ℎ = 𝑢 + 𝑃𝑣 (v = specific volume)
If pressure decreases, then specific volume must increase if enthalpy is to
remain constant (assuming u is constant). The change in specific volume is
observed as an increase in gas velocity and this is verified by our
observations.
The theory also states W = 0. Our observations again confirm this as true
because clearly the throttling process does no work. Finally, the theory
states that an ideal throttling process is adiabatic. This cannot be proven by
observation because a real throttling process is not ideal and involves some
heat transfer.
Knowledge Check
In what type of process does the fluid pass through
various system processes and then returns to the same
state it began?
14
A.
throttling process
B.
isentropic process
C.
adiabatic process
D.
cyclic process
Rev 1
Knowledge Check
Steam flowing through the main turbine control valve is
a(n) …?
A.
isenthalpic process
B.
isentropic process
C.
adiabatic process
D.
reversible process
ELO 1.3 Applying the First Law of Thermodynamics for Open
Systems
Introduction
We learned that under defined conditions we can use the principle of the
first law of thermodynamics to determine the state of a system. The first
law of thermodynamics states that energy can be neither created or
destroyed, but only altered in form. We can account for the types and
quantities of energy entering our system and compare with the types and
quantities of energy exiting our system.
Analyzing Systems Using the First Law of Thermodynamics
The control volume approach is one in which a fixed region in space is
established with specified control boundaries and the energies that cross the
boundary are studied and an energy balance performed. For example, a
pump can have boundaries set at the suction and discharge. We can analyze
the change in the various energies as fluid progresses through and exits the
pump. By comparing the enthalpy into the pump and the enthalpy out of
the pump, we determine how much work the pump did on the fluid.
First Law of Thermodynamics Principle Statement
The principle of the first law of thermodynamics is demonstrated below in
the examples. All of the energies entering and leaving the control volume
boundary as well as any work done on or by the control volume and any
heat transferred into and out of the control volume boundaries are accounted
for in the energy balance equation.
Rev 1
15
Analyzing Open Systems and Cyclic Processes Guidelines
Open System Control Volume Analysis
Heat and/or work can be directed into or out of the control volume. For
convenience and as a standard convention, the net energy exchange is
presented here with the net heat exchange assumed to be into the system
and the net work assumed to be out of the system. If no mass crosses the
boundary but work and/or heat do, then the system is called a closed
system. If mass, work, and heat do not cross the boundary (that is, the only
energy exchanges taking place are within the system), then the system is
called an isolated system. Isolated and closed systems are nothing more
than specialized cases of the open system. In this text, the open system
approach to the first law of thermodynamics will be emphasized because it
is more general. Finally, almost all practical applications of the first law
require an open system analysis.
Control Volume Approach
Two basic approaches exist in studying thermodynamics: the control mass
approach and the control volume approach. Thermodynamic problems are
usually analyzed using the control volume approach.
The figure below shows the control volume approach that is a fixed region
in space with specified control boundaries. The energies that cross the
boundary of this control volume, including those with the mass, are then
studied and the energy balance performed. The control volume approach is
often used today in analyzing thermodynamic systems. It is more
convenient and efficient in tracking the energy balances.
Figure: Open System Control Volume Concept for a Pump
Boundaries can be set around any component, such as the suction and
discharge of the pump as shown above in the graphic. The inlet energies
are compared to exit energies to determine how the pump altered the
16
Rev 1
energies while doing work on the system to perform its function of
increasing system pressure.
In the pump example, we would see that the pump altered the energies by
doing work on the fluid and increasing its flow energy by raising the
discharge pressure. A detailed explanation of this energy conversion is
given in the pumps module.
It is often necessary to show multiple processes inside of our boundary to
account for all the energy transformations taking place. The two figures
below illustrate that different processes occur inside the boundary to
produce different states of the fluid. We can determine the state of the fluid
at any given point and the state of the fluid exiting the boundaries as well as
the work done by the system by analyzing these processes.
Figure: Open System Control Volume for Multiple Processes
Figure: Multiple Control Volumes in the Same System
Rev 1
17
When mass crosses the boundary, it carries potential energy, kinetic energy,
and internal energy with it. The mass possesses another form of energy
associated with the fluid pressure referred to as flow energy since the flow
is normally supplied by some driving power (for example, a pump).
Therefore, the various forms of energy crossing the control boundary with
the mass are (U + PΝ + KE + PE).
Enthalpy has been previously defined as ℎ = 𝑈 + 𝑃𝑉. Therefore, the above
expression can be written as m (h + KE + PE). In addition to the mass and
its energies, externally applied work (W), usually designated as shaft work,
is another form of energy that may cross the system boundary.
Energy that is not caused by mass or shaft work is classified as heat energy
(Q). In order to complete and satisfy the conservation of energy
relationship, we must describe the relationship in equation form as follows:
𝑚̇(ℎ𝑖𝑛 + 𝑃𝐸𝑖𝑛 + 𝐾𝐸𝑖𝑛 ) + 𝑄̇ = 𝑚̇(ℎ𝑜𝑢𝑡 + 𝑃𝐸𝑜𝑢𝑡 + 𝐾𝐸𝑜𝑢𝑡 ) + 𝑊̇
Where:
•
𝑚̇ = mass flow rate of working fluid (lbm/hr)
•
hin = specific enthalpy of the working fluid entering the system
(BTU/lbm)
•
hout = specific enthalpy of the working fluid leaving the system
(BTU/lbm)
•
PEin = specific potential energy of working fluid entering the system
(ft-lbf/lbm)
•
PEout = specific potential energy of working fluid leaving the system
(ft-lbf/lbm)
•
KEin = specific kinetic energy of working fluid entering the system (ftlbf/lbm)
•
KEout = specific kinetic energy of working fluid leaving the system (ftlbf/lbm)
•
𝑊̇ = rate of work done by the system (ft-lbf/hr)
•
𝑄̇ = heat rate into the system (BTU/hr)
The heat transferred and the work done through the boundaries must be
accounted for; they are assigned a positive or a negative number depending
on their interaction with the control volume.
18
Rev 1
Figure: Heat and Work in a System
It is important to understand the concept of positive and negative heat and
work while solving thermodynamic problems. For example, heat added to
feedwater is positive, but heat lost to ambient is negative. Work done by
the feedwater pump to increase the feedwater pressure is negative work, but
the work done to turn the main turbine is positive work. The convention for
positive or negative work is the opposite of the convention for heat transfer.
For instance, work is negative when transferred into the system, such as the
work of a pump on the fluid. Work is positive when transferred out of the
system, such as the work in rotating the main turbine.
Rev 1
19
Tip
Example 1: Open System Control Volume
This example illustrates the use of the control volume
concept while solving a first law problem involving
energy terms mentioned previously.
The enthalpies of steam entering and leaving a steam
turbine are 1,349 BTUs/lbm and 1,100 BTUs/lbm,
respectively. The estimated heat loss is 5 BTUs/lbm of
steam. The flow enters the turbine at 164 ft/sec at a
point 6.5 ft above the discharge and leaves the turbine at
262 ft/sec. Determine the work of the turbine.
Always ensure you have the proper mental picture of the
system being analyzed by making a drawing showing
the boundaries for the analysis.
Figure: Open System Control Volume Concept
𝑚𝑖𝑛 (ℎ𝑖𝑛 + 𝑃𝐸𝑖𝑛 + 𝐾𝐸𝑖𝑛 ) + 𝑞 = 𝑚𝑜𝑢𝑡 (ℎ𝑜𝑢𝑡 + 𝑃𝐸𝑜𝑢𝑡 + 𝐾𝐸𝑜𝑢𝑡 ) + 𝑤
1. Divide by m since: 𝑚𝑖𝑛 = 𝑚𝑜𝑢𝑡 = 𝑚
Where:
•
q = heat added to the system per pound (BTU/lbm)
•
w = work done by the system per pound (ft-lbf/lbm)
2. Use Joule’s constant, 𝐽 = 778
known values.
𝑓𝑡–𝑙𝑏𝑓
𝐵𝑡𝑢
for conversions and substitute
1,349
(164)2
𝐵𝑇𝑈𝑠
6.5 𝐵𝑇𝑈𝑠
+ (
)
+ [
]
𝑙𝑏𝑚
778 𝑙𝑏𝑚
2(32.17)(778)
1,100
(262)2
𝐵𝑇𝑈𝑠
𝐵𝑇𝑈𝑠
+ 0 𝑃𝐸𝑜𝑢𝑡 + [
]
+𝑤
𝑙𝑏𝑚
2(32.17)(778) 𝑙𝑏𝑚
Note: The minus sign indicates heat out of the turbine.
3. Solve for work (w):
20
Rev 1
1,349
𝐵𝑇𝑈𝑠
𝐵𝑇𝑈𝑠
𝐵𝑇𝑈𝑠
𝐵𝑇𝑈𝑠
+ 8.3548 × 10−3
+ 0.5368
–5
=
𝑙𝑏𝑚
𝑙𝑏𝑚
𝑙𝑏𝑚
𝑙𝑏𝑚
1,100
𝐵𝑇𝑈𝑠
𝐵𝑇𝑈𝑠
= 1.37
+ 𝑤
𝑙𝑏𝑚
𝑙𝑏𝑚
1,344.54
𝐵𝑇𝑈𝑠
𝐵𝑇𝑈𝑠
= 1,101.37
+𝑤
𝑙𝑏𝑚
𝑙𝑏𝑚
𝑤 + 1,344.54
𝐵𝑇𝑈𝑠
𝐵𝑇𝑈𝑠
– 1,101.37
𝑙𝑏𝑚
𝑙𝑏𝑚
𝑤 = 243.17
Tip
𝐵𝑇𝑈𝑠
𝑙𝑏𝑚
Example 2: Gas Filled Piston
A system comprised of a certain mass of air is contained
in a cylinder fitted with a piston. The air expands from
an initial state for which E1 = 70 BTUs to a final state of
E2 = 20 BTUs. The air does 60 BTUs of work on its
surroundings during the expansion. Find the amount of
heat transferred to or from the system during the
process.
Draw figure:
Figure: Gas Filled Piston
Using equation:
𝑄 + 𝐸1 = 𝑊 + 𝐸2
𝑄 = 𝑊 + 𝐸2 − 𝐸1
𝑄 = 60 + 20 − 70 = 10 𝐵𝑇𝑈
The positive sign of the result indicates that 10 BTUs of heat is added to the
system.
In this example, notice that heat was added to the system and work was
taken from the system. The total decrease in the stored energy of the
system equals the difference between the energy added as heat and the
energy removed as work.
Rev 1
21
Tip
Example 3: Heat and Work
Calculate the final E value of a mass of water that is at
an initial state 20 BTUs of energy and then undergoes a
process during which 7,780 ft-lbf of work is done on the
water and 3 BTUs of heat is removed from it.
Figure: Heat and Work in a Closed System
Applying the First Law of Thermodynamics to the system (comprised of the
mass of water), the final amount of stored energy E2 can be determined.
Note
Heat removed is a negative value and work done on a
system is a negative value.
𝑄 + 𝐸1 = 𝑊 + 𝐸2
𝐸2 = 𝑄 + 𝐸1 − 𝑊
We must convert all of the energy values to the same
units.
1 𝐵𝑇𝑈
𝑊 = (7,780 𝑓𝑡– 𝑙𝑏𝑓) 778 𝑓𝑡–𝑙𝑏𝑓 = 10 𝐵𝑇𝑈Now,
substituting into and solving the equation gives:
𝐸2 = (−3) + 20 – (−10)
𝐸2 = 27 𝐵𝑇𝑈
Knowledge Check
Fill in the blanks for the polarity of heat and work. Heat
added to feedwater is ______ but heat lost to ambient is
_______ while work done by the feedwater pump to
increase the feedwater pressure is ________ work but the
work done to turn the main turbine is __________work.
22
A.
positive, negative, negative, positive
B.
negative, positive, positive, negative
Rev 1
Knowledge Check
In an Open Steady Flow System, choose the energies that
are associated with the mass crossing the system
boundary.
A.
kinetic energy, potential energy, internal energy, flow
energy
B.
work, kinetic energy, potential energy, heat
C.
kinetic energy, heat, internal energy, flow energy
D.
work, potential energy, internal energy, flow energy
ELO 1.4 Identifying Cyclic Process Paths on a T-s Diagram
Principle
Introduction
In some processes, the relationships between pressure, temperature, and
volume are specified as the fluid goes from one thermodynamic state to
another. We will analyze some of the many processes that occur in the
plant.
Going Through a Process
Glossary
The system (the fluid studied) changes its properties
(such as temperature, pressure, and/or volume) from one
value to another when exchanging work, heat, or internal
energy.
Analyzing Cyclic Processes Guidelines
The most common processes are those in which the temperature, pressure,
or volume are held constant during the process. These classify as
isothermal, isobaric, or isovolumetric processes. If the fluid passes through
various processes and then eventually returns to the same state in which
began, the system undergoes a cyclic process.
Iso
Glossary
Rev 1
Iso is a constant or one.
23
Figure: T-s Diagram With Rankine Cycle
One such cyclic process used is the Rankine cycle, two examples of which
are shown above in the figures. The Rankine cycle is an ideal cycle where
no increase in entropy occurs as work is done on and by the system. The
Rankine cycle is ideal and cannot be constructed because it is 100 percent
efficient, but it has value by setting a maximum efficiency to which real
cycles are compared. These comparisons will be analyzed later.
The Rankine cycle processes are described below:


ab: Liquid is compressed with no change in entropy (by ideal pump).
bc: Constant pressure transfer of heat in the boiler. Heat is added to
the compressed liquid, into the two-phase and superheat states.
 cd: Constant entropy expansion with shaft work output (in an ideal
turbine).
 da: Constant pressure transfer of heat in the sink. Unavailable heat is
rejected to the heat sink (condenser).
These are the individual processes that the fluid undergoes while
completing the entire cycle. Rankine cycles will be discussed in greater
detail later in this module.
Typical Steam Plant Cycle
The figure below shows a typical steam plant cycle. Heat is supplied to the
steam generator (boiler) where liquid converteds to steam (vapor). The
vapor expands adiabatically in the turbine to produce a work output. Vapor
leaving the turbine enters the condenser where heat is removed and the
vapor condenses into the liquid state. The condensation process is the heatrejection mechanism for the cycle. The liquid is delivered to the condensate
pump and then the feed pump, where its pressure is raised to the saturation
pressure, corresponding to the steam generator temperature, and the highpressure liquid is delivered to the steam generator where the cycle is
repeated.
24
Rev 1
Figure: Typical Steam Plant Cycle
A typical steam plant system consists of the following:

A heat source (steam generator) converts the thermal energy from the
reactor into steam (5-1 above and below on the Rankine diagram).
 A steam turbine converts the steam energy into work through a
constant entropy expansion (1-2 above and below on the Rankine
diagram below).
 A condenser converts the turbine exhaust back to liquid and rejects
the heat (2-3 above and below on the Rankine diagram).
 Pumps raise the fluid pressure to transfer the fluid back to the heat
source (3-5 above and below on Rankine diagram as both the
condensate and feedwater pumps are in series and shown as one
pumping process) to repeat the cycle.
Figure: Rankine Cycle for a Typical Steam Plant
The steam plant in its entirety is a large closed system. However, each
component of the system may be analyzed thermodynamically as an open
system as the fluid passes through it. We analyze each major component in
the system during this course.
Rev 1
25
Knowledge Check
Which one of the following will cause overall nuclear
power plant thermal efficiency to increase?
A.
increasing total steam generator blowdown from 30
gallons per minute (gpm) to 40 gpm.
B.
changing steam quality from 99.7 percent to 99.9
percent.
C.
bypassing a feedwater heater during normal plant
operations.
D.
increasing condenser pressure from 1 pound or pounds
per square inch absolute (psia) to 2 psia.
Knowledge Check
What type of property diagram is frequently used to
analyze Rankine cycles?
A.
P-T diagram
B.
P-ν diagram
C.
h-T diagram
D.
T-s diagram
Knowledge Check
On a T-s diagram, if the temperature of the heat sink is
lowered, the efficiency of the cycle will _______ because
________.
26
A.
increase, reject more heat
B.
increase, reject less heat
C.
decrease, must add more heat
D.
decrease, must add less heat
Rev 1
ELO 1.5 Thermodynamic Energy Balances on Major
Components
Introduction
Thermodynamic energy balances can be used on a variety of major
components and systems, including but not limited to the steam generators,
pumps, heat exchangers, and condensers. The thermodynamic energy
balance also depends on whether the system is open or closed.
Steam Generator Analysis
The steam generator is a two-phase heat generator that acts as the heat sink
for the reactor and the heat source for the secondary system. The hot fluid
(TH) from the reactor passes through the primary side of the steam
generator, then through tubes. Some of its energy transfers to the secondary
side of the heat exchanger where lower pressure water vaporizes. The
previously hot fluid leaves the steam generator at a lower temperature (TC)
and is pumped back to the heat source to be reheated.
Each major component of a steam plant can be treated as a separate open
system. A thermodynamic analysis using the various forms of energies
discussed, can be applied to any particular component in studying its
behavior. The steps in the thermodynamic analysis of the steam generator
are shown in the following table:
Steam Generator Analysis Table
Step
Action
1.
Draw the system with the boundaries.
2.
Write the general energy equation and solve for the required
information.
3.
Determine which energies can be ignored to simplify the equation.
4.
Make substitutions to ensure correct units are obtained.
Steam Plant Component Demonstration In An Open System
Primary Side Of Steam Generators
Fluid from the heat source at a steam generating facility enters the steam
generator (heat exchanger) of the facility at 610 °F and leaves at 540 °F.
Rev 1
27
The flow rate is approximately 1.38 x 108 lbm/hr. If the specific heat of the
fluid is taken as 1.5 BTUs/lbm-°F, what is the heat transferred out of the
steam generator?
Solution:
Step 1:
Draw the system with the boundaries. As can be seen in the figure below,
the Steam Generator (SG) System can be analyzed on either the primary
side or the secondary side, assuming all the heat transferred from the
primary system converts to steam in the secondary system. The energy
transferred out of the primary system should be the same as the energy
transferred into the secondary system. Two system boundaries can be
drawn. The red circle represents energies to be considered on the primary
side balance. The green circle represents energies to analyze for the
secondary side balance. Please keep in mind that the energy transferred out
of the primary system equals the energy transferred into the secondary
system.
Figure: Steam Generation System
Step 2:
Write the general energy equation simplified for this component.
𝑚̇𝑖𝑛 (ℎ𝑖𝑛 + 𝑃𝐸𝑖𝑛 + 𝐾𝐸𝑖𝑛 ) + 𝑄̇ = 𝑚̇𝑜𝑢𝑡 (ℎ𝑜𝑢𝑡 + 𝑃𝐸𝑜𝑢𝑡 + 𝐾𝐸𝑜𝑢𝑡 ) + 𝑊̇
28
Rev 1
Step 3:
Simplify the equation by eliminating the energies that are insignificant to
this process.
Neglecting PE and KE, and assuming no work is done on the system:
𝑚̇(ℎ𝑖𝑛 ) + 𝑄̇ = 𝑚̇(ℎ𝑜𝑢𝑡 )
𝑄̇ = 𝑚̇(ℎ𝑜𝑢𝑡 – ℎ𝑖𝑛 )
Step 4:
Substituting 𝑄̇ = 𝑚̇𝑐𝑝 𝛥𝛵, where cp = specific heat capacity (BTU/lbm-°F):
= 𝑚̇(𝑐𝑝 )(𝑇𝑜𝑢𝑡 – 𝑇𝑖𝑛 )
= (1.38 × 108
𝑙𝑏𝑚
𝐵𝑇𝑈
) (1.5
) (540 − 610 ℉)
ℎ𝑟
𝑙𝑏𝑚-℉
𝑄̇ = −1.45 × 1010
𝐵𝑇𝑈
ℎ𝑟
The minus sign indicates heat out of the heat exchanger, which is consistent
with the physical case. This example demonstrates that for a heat
exchanger, the heat transfer rate can be calculated using the equation:
𝑄̇ = 𝑚̇(ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 ) 𝑜𝑟 𝑄̇ = 𝑚̇𝑐𝑝 𝛥𝑇
Pump Analysis
It is important to note that the latter equation can only be used when no
phase change occurs since ΔT = 0 during a phase change (ΔT is change in
temperature). The first equation can be used for a phase change heat
transfer process as well as for latent heat calculations because it relies on
the change in enthalpy rather than temperature.
The pumps used for returning the fluid to the heat source can be analyzed as
a thermodynamic system.
Pumps In An Open System
A pump returns the fluid from the heat exchanger back to the core. The
flow rate through the pump is approximately 3.0 x 107 lbm/hr with the fluid
entering the pump as saturated liquid at 540 °F. The pressure rise across the
pump is 90 psia. What is the work of the pump, neglecting heat losses and
changes in potential and kinetic energy?
Rev 1
29
Solution:
Step 1:
Draw the system with boundaries.
Figure: Pump Returns Fluid from Heat Exchanger to Core
Step 2:
Write the general energy equation:
𝑚̇(ℎ𝑖𝑛 + 𝑃𝐸𝑖𝑛 + 𝐾𝐸𝑖𝑛 ) + 𝑄̇ = 𝑚̇(ℎ𝑜𝑢𝑡 + 𝑃𝐸𝑜𝑢𝑡 + 𝐾𝐸𝑜𝑢𝑡 ) + 𝑊̇
Step 3:
Simplify the equation.

Assume 𝑄̇ = 0, and neglect changes in PE and KE
𝑚̇(ℎ𝑖𝑛 ) = 𝑚̇(ℎ𝑜𝑢𝑡 ) + 𝑊̇

𝑊̇ = 𝑚̇(ℎ𝑖𝑛 + ℎ𝑜𝑢𝑡 ) where 𝑊̇ is the rate of doing work by the pump
(vP)
ℎ𝑖𝑛 = 𝑈𝑖𝑛 + ѵ𝑃𝑖𝑛
ℎ𝑜𝑢𝑡 = 𝑈𝑜𝑢𝑡 + ѵ𝑃𝑜𝑢𝑡
(ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 ) = (𝑈𝑖𝑛 − 𝑈𝑜𝑢𝑡 ) + (ѵ𝑃𝑖𝑛 − ѵ𝑃𝑜𝑢𝑡 ) = 𝛥𝑈 + (ѵ𝑃𝑖𝑛 − ѵ𝑃𝑜𝑢𝑡 )

Since no heat is transferred, ΔU = 0, (ΔU is change in internal energy)
and the specific volume out of the pump is the same as the specific
volume entering since water is incompressible.
(ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 ) = ѵ(𝑃𝑖𝑛 − 𝑃𝑜𝑢𝑡 )
Step 4:
Arrange the correct terms:

Substituting the expression for work, 𝑊̇ = 𝑚̇(ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 ) we have:
𝑊̇ = 𝑚̇ѵ(𝑃𝑖𝑛 − 𝑃𝑜𝑢𝑡 )
30
Rev 1

Using 0.01246 for specific volume:
𝑓𝑡 3
𝑙𝑏𝑚
𝑖𝑛2
107
(0.01246
) (−90 𝑝𝑠𝑖𝑎) (144 2 )
ℎ𝑟
𝑙𝑏𝑚
𝑓𝑡
𝑊̇ = 3.0 ×
𝑓𝑡– 𝑙𝑏𝑓
778 𝐵𝑇𝑈
𝐵𝑇𝑈
𝑊̇ = −6.23 × 106 ℎ𝑟 or -2,446 hp
The minus sign indicates that work is put into the fluid by the pump. For
example, 1 horsepower (hp) = 2,545 BTUs/hr.
Thermodynamic Balance Across Heat Source
In a particular facility, the temperature leaving the heat source is 612 °F,
while that entering the heat source is 542 °F. The coolant flow through the
heat source is 1.32 x 108 lbm/hr. The cp of the fluid averages 1.47
BTUs/lbm-°F. How much heat is removed from the heat source?
Figure: Heat Exchanger Analysis Shows Thermodynamic Balance
Solution:
Step 1:
Draw the system with the boundaries.
Step 2:
Make needed substitutions to ensure correct units are obtained.
Substituting 𝑄̇ = 𝑚̇𝑐𝑝 𝛥𝑇, where cp = specific heat capacity:
 𝑄̇ = 𝑚̇(𝑐𝑝)(𝑇𝑜𝑢𝑡 – 𝑇𝑖𝑛 ) + 𝑊̇
𝑙𝑏𝑚
𝐵𝑇𝑈
𝑄̇ = 1.32 × 108
. (1.47
) (612 – 542℉) + 0
ℎ𝑟
𝑙𝑏𝑚-℉
𝑄̇ = 1.36 𝑥 1010 𝐵𝑇𝑈/ℎ𝑟

For this example 𝑄̇ = 𝑚̇𝑐𝑝 𝛥𝑇 has been used to calculate the heat transfer
rate since no phase change has occurred. However, 𝑄̇ = 𝑚̇(ℎ𝑜𝑢𝑡 −
Rev 1
31
ℎ𝑖𝑛 ) could also have been used if the problem data included inlet and outlet
enthalpies.
In the above examples, the individual principal components of a steam
generating system have been thermodynamically analyzed. If all of these
components combine into an overall system, the system could be analyzed
as a closed system problem. Such an analysis is illustrated in the following
example.
Primary Side Thermodynamic Balance
A steam generating facility is studied as a complete system. The heat
produced by the heat source is 1.36 x 1010 BTUs/hr. The heat removed by
the heat exchanger (steam generator) is 1.361 x 1010 BTUs/hr. What is the
required pump power to maintain a stable temperature?
Solution:
Step 1:
Draw the system with the boundaries.
Figure: Pump Power of a Steam Generating Facility
𝑊̇𝑝 = pump work
𝑄̇𝑐 = heat produced by the heat source
𝑄̇𝑆𝐺 = heat transferred into steam generator
32
Rev 1
Steps 2 & 3: Write and simplify the equation:
𝑚̇(ℎ + 𝑃𝐸 + 𝐾𝐸) + 𝑊̇𝑝 + 𝑄̇𝑐 = 𝑄̇𝑆𝐺 + 𝑚̇ + 𝑃𝐸 + 𝐾𝐸
Step 3:
For a closed system, the mass entering and leaving the system is zero (0);
therefore, ṁ is constant. The energy entering and leaving the system is
zero, and you can assume that the KE and PE are constant so that:
𝑄̇𝑐 + 𝑊̇𝑝 = 𝑄̇𝑆𝐺
𝑊̇𝑝 = 𝑄̇𝑆𝐺 − 𝑄̇𝑐
= 1.361 × 1010
= 1.0 × 107
𝐵𝑇𝑈
𝐵𝑇𝑈
– 1.36 × 1010
ℎ𝑟
ℎ𝑟
𝐵𝑇𝑈
ℎ𝑟
Step 4:
Arrange equation for the proper units.
Recall that 1 hp = 2,545 BTUs/hr. Therefore, converting to hp:
𝑊̇𝑝 = 3,929 ℎ𝑝
Both the primary side and the secondary side have their own energy
balances as the heat energy is transferred from one fluid to the other during
the heat exchanger analysis. In calculating heat exchanger heat transfer
rates, we found that we could use the equation below:
𝑄̇ = 𝑚̇𝑐𝑝 𝛥ℎ (Δh is change in enthalpy)
A short analysis of the secondary side of the heat exchanger helps in
understanding the heat exchanger's importance in the energy conversion
process.
Thermodynamic Balance of Overall Secondary Side
Example 6:
Steam flows through a condenser at 4.4 x 106 lbm/hr, entering as saturated
vapor at 104 °F (h = 1,106.8 BTUs/lbm), and leaving at the same pressure
as subcooled liquid at 86 °F (h = 54 BTUs/lbm). Cooling water is available
at 64.4 °F (h = 32 BTUs/lbm). Environmental requirements limit the exit
temperature to circulating water inlet 77 °F (h = 45 BTUs/lbm)
Determine the required cooling water flow rate.
Rev 1
33
Solution:
Step 1:
Draw the system with the boundaries.
Figure: Typical Single-Pass Condenser End View
Step 2:
Write the equation.
Thermal balance gives the following:
𝑄̇𝑠𝑡𝑚 = −𝑄̇𝑐𝑤
Steps 3 and 4:
Simplify the equation and arrange for required units.
𝑚̇𝑠𝑡𝑚 (ℎ𝑜𝑢𝑡 – ℎ𝑖𝑛 )𝑠𝑡𝑚 = 𝑚̇(ℎ𝑜𝑢𝑡 – ℎ𝑖𝑛 )𝑐𝑤
𝑚̇𝑐𝑤 =
𝑚̇𝑐𝑤
𝑚̇𝑠𝑡𝑚 (ℎ𝑜𝑢𝑡 – ℎ𝑖𝑛 )𝑠𝑡𝑚
(ℎ𝑜𝑢𝑡 – ℎ𝑖𝑛 )𝑐𝑤
𝐵𝑇𝑈𝑠
54– 1,106.8
𝑙𝑏𝑚
𝑙𝑏𝑠
= 4.4 × 106
ℎ𝑟 (45 − 32 𝐵𝑇𝑈𝑠)
𝑙𝑏𝑚
𝑚̇𝑐𝑤 = 3.67 × 108
𝑙𝑏𝑚
ℎ𝑟
All of the heat removed from the steam (stm) condensing and subcooling is
transferred to the circulating water (cw) system.
𝑄̇ = 𝑚̇𝛥ℎ was required in this example since a phase change occurred.
when the steam was condensed to water. Since ΔT = 0 for a phase
34
Rev 1
change, Q = ṁcp𝛥T would not have worked. Had we attempted to
solve the problem using Q = ṁcp𝛥T, we would have discovered that an
error occurs since the 18°F ΔT is the ΔT needed to subcool the liquid
from saturation at 104°F to a subcooled value of 86°F. This change in
temperature does not account for the much larger heat transfer process
necessary to condense the steam to a saturated liquid which must also
be taken into account.
Knowledge Check
Why can’t a formula be used in all heat transfer systems
in the plant?
A.
Cp is too hard to measure.
B.
The temperature measurements are not accurate.
C.
A phase change occurs without a change in temperature.
D.
A phase change results in the ΔT being too large.
Knowledge Check
Reactor coolant enters a reactor core at 545 °F and leaves
at 595 °F. The reactor coolant flow rate is 6.6 x 107
lbm/hour and the specific heat capacity of the coolant is
1.3 BTUs/lbm-°F. What is the reactor core thermal
power?
Rev 1
A.
101 Megawatts (Mw)
B.
126 Mw
C.
1,006 Mw
D.
1,258 Mw
35
TLO 1 Summary
During this lesson, you learned about the First Law of Thermodynamics,
which states that energy can be neither created nor destroyed, but only
altered in form. The energy forms may not always be the same but the total
energy in the system remains constant. You learned about open, closed,
isolated, and steady flow systems. You studied processes including
thermodynamic, cyclic, reversible, irreversible, adiabatic, isentropic, and
isenthalpic. The listing below provides a summary of sections in this TLO.
1. Review ELO 1.1 by asking students the following:







•
The first law of thermodynamics states energy can be neither
created nor destroyed, but only altered in form.
The law to energy balance states that energy into the system
equals the energy leaving the system.
During a cyclic process, fluid passes through various processes
and states then returns to the same state in which it began.
Have a student explain the concept of heat transfer in a heat
exchanger.
— The amount of energy transferred across a heat exchanger
depends on the temperature of the fluid entering the heat
exchanger from both sides and the flow rates of these fluids.
Have students define: A thermodynamic system is a collection of
matter and space with its boundaries defined in such a way that the
energy transfer across the boundaries can be best understood.
Have students decide which diagram is usually used to
demonstrate a thermodynamic system (T-s).
Have students draw a simple representation of system groups
showing the following:
— Isolated system is a system where neither mass nor energy
can cross the boundaries.
— Closed system is a system where only energy can cross the
boundaries.
— Open system is a system where both mass and energy can
cross the boundaries.
Define the following terms:
— A control volume is a fixed region of space studied as a
thermodynamic system.
— A steady state is a condition where the properties at any
given point within the system are constant over time. Both
mass and energy are stable within the system.
2. Review ELO 1.2:


36
Define thermodynamic process is the succession of states through
which that a system passes.
Define the following processes, which can be described by any of
the following terms:
Rev 1
— A cyclic process is a series of processes that result in the
system returning to its original state.
— An reversible process is a process that can be reversed
resulting in no change in the system or surroundings.
— An irreversible process is a process that if reversed results in
a change to the system or surroundings.
— An adiabatic process is a process in which there is no heat
transfer across the system boundaries.
— An isentropic process is a process in which the entropy of the
system remains unchanged.
— A throttling process is a process in which enthalpy is
constant (h1 = h2), work = 0, and which is adiabatic, Q = 0.
3. Review ELO 1.3:


Define thermodynamic cycle, which is a continuous series of
thermodynamic processes transferring heat and work, while
varying pressure, temperature, and other state variables, eventually
returning a system to its initial state.
Describe the four basic processes in any thermodynamic cycle:
— Energy is supplied from a source (steam generator).
— Some of the energy is converted to work in a turbine.
— Most of the remaining steam energy is rejected to a heat sink
(condenser).
— Condensed steam (liquid water) is pumped back to the
source to restart the cycle.
4. Review ELO 1.4:
•




Describe the two primary classes of thermodynamic cycles:
— Power cycles
— Heat pump cycles
Thermodynamic cycle efficiency is the ratio of net work or energy
output of the system divided by the heat energy added to the
system.
Review reversible cycle. A Carnot cycle is an ideal heat engine
that converts heat into work through reversible processes.
The most efficient existing cycle is one that converts a given
amount of thermal energy into the greatest amount of work or,
conversely, creates a temperature difference by accomplishing a
given amount of work.
Heat engine is an engine that converts heat energy to mechanical
work by exploiting the temperature gradient between a hot source
and a cold sink.
5. Review ELO 1.5:


Rev 1
Review the steps for solving energy balance problems.
Review the use of the continuity equation:
37
•
•
•
𝑚̇(ℎ + 𝑃𝐸 + KE) + 𝑊̇𝑝 + 𝑄̇𝑐 = 𝑄̇𝑆𝐺 + 𝑚̇ + 𝑃𝐸 + 𝐾𝐸
Review assumptions normally made to simplify the equation.
No change in PE, KE, or U in most applications.
Typically, changes in enthalpy are used to simplify the equation.
Objectives
Now that you have completed this lesson, you should be able to do the
following:
1. Relate the following terms to Open, Closed, and Isolated Systems:
a. Thermodynamic surroundings
b. Thermodynamic equilibrium
c. Control volume
d. Steady state
2. Describe the following processes:
a. Thermodynamic process
b. Cyclic process
c. Reversible process
d. Irreversible process
e. Adiabatic process
f. Isentropic process
g. Throttling process (Isenthalpic)
3. Analyze an open system or cyclic processes, including all energy
transfer processes crossing the boundaries using the first law of
thermodynamics.
4. Identify the path(s) on a T-s diagram that represents the
thermodynamic processes occurring in a fluid system.
5. Given a defined system, perform energy balances on all major
components in the system.
Knowledge Check
To achieve maximum overall nuclear power plant
thermal efficiency, feed water should enter the steam
generator (SG) _____________ and the pressure
difference between the SG and the condenser should be
as _____________ as possible.
38
A.
as subcooled as practical; great
B.
as subcooled as practical; small
C.
close to saturation; great
D.
close to saturation; small
Rev 1
Knowledge Check
Feed water heating increases overall nuclear power plant
thermal efficiency because…
A.
The average temperature at which heat is transferred in
the steam generators is increased.
B.
Less steam flow passes through the turbine, thereby
increasing turbine efficiency.
C.
Increased feed water temperature lowers the temperature
at which heat is rejected in the condenser.
D.
Less power is required by the feed water pumps to pump
the warmer feed water.
Knowledge Check
A nuclear power plant is operating at full power with 0
°F of condensate subcooling. If main condenser cooling
water inlet temperature increases by 3 °F, the overall
nuclear power plant thermal efficiency will…
A.
decrease due to a degraded main condenser vacuum.
B.
increase due to an improved main condenser vacuum.
C.
decrease due to increased main condenser heat rejection.
D.
increase due to decreased main condenser heat rejection.
Knowledge Check
A nuclear power plant is operating at 90 percent of rated
power. Main condenser pressure is 1.7 psia and hotwell
condensate temperature is 120 °F. If main condenser
cooling water flow rate is reduced by 5 percent, overall
steam cycle efficiency will…
Rev 1
A.
Increase because condensate depression will decrease.
B.
Decrease because condensate depression will increase.
39
C.
Increase because the work output of the main turbine will
increase.
D.
Decrease because the work output of the main turbine
will decrease.
Knowledge Check
Which one of the following actions result in a decrease in
overall nuclear power plant thermal efficiency?
A.
increasing steam quality by adding additional heat to the
steam prior to entering the turbine
B.
increasing the temperature of the feed water entering the
steam generator
C.
decreasing the amount of condensate depression in the
main condenser
D.
decreasing the amount of turbine steam extracted for feed
water heating
TLO 2 The Second Law of Thermodynamics
Overview
The first law of thermodynamics requires a balance of the various forms of
energy as they pertain to the specified thermodynamic system and/or
control volume studied. However, this first law of thermodynamics does
not take into account the feasibility of the process or the efficiency of the
energy transformation studied. The second law of thermodynamics allows
us to determine the maximum efficiency of the operating system so that
design comparisons maximize the system’s efficiency.
The second law of thermodynamics states that it is impossible to construct a
device that operates within a cycle that can convert all the heat supplied it
into mechanical work. Recognizing that even the most thermally and
mechanically perfect cycles must reject some heat defines thermodynamic
power cycle efficiency. Maximizing the power cycle efficiency is a major
part of the operator's job. There are many important and interdependent
plant parameters affecting plant efficiency.
To produce the maximum electrical power output for the allowed core
thermal power input, the operator must continually monitor these
parameters and adjust plant conditions as necessary. Advancements in
40
Rev 1
sensing these key plant parameters give the operator real time data of actual
plant efficiency and core power levels, so the operator is able to make
immediate adjustments to maintain the plant within licensed limitations, set
by the Nuclear Regulatory Commission (NRC). The ability to relate current
plant conditions to the plant’s thermal efficiency is a fundamental operator
attribute. Precise control of plant parameters requires continual oversight
by the operator and adjustments as needed.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Explain the second law of thermodynamics using the term entropy.
2. Given a thermodynamic system, determine the:
a. Maximum efficiency of the system
b. Efficiency of the components within the system
3. Differentiate between the path for an ideal process and that for a real
process that for a real process on a T-s or h-s diagram.
4. Describe how individual factors affect system or component
efficiency.
ELO 2.1 Thermodynamic Entropy
Introduction
We must first understand that the first law of thermodynamics governs all
cycles when we examine thermodynamic cycles. The first law states energy
can be neither created nor destroyed, but only altered in form. This means
that all of the energy added to a cycle must be accounted for in its entirety.
The first law of thermodynamics places no restrictions on how conversions
from heat to work or vice versa take place or to what extent these
conversions may proceed, which is addressed by the second law of
thermodynamics.
The second law is based on experimental evidence and observations of
actual processes. It suggests that processes proceed in a certain direction
but not in the opposite direction. The second law, which implies that all
real processes are irreversible, governs all real processes. With the second
law of thermodynamics, the limitations imposed on any process can be
studied to determine the maximum possible efficiencies of such a process
then a comparison can be made between the maximum possible efficiency
and the actual efficiency achieved.
Energy-Conversion Systems
One of the areas of application of the second law is the study of energyconversion systems. For example, it is not possible to convert all the energy
obtained from a nuclear reactor into electrical energy. There must be losses
in the conversion process. As shown the figure below on the left as heat
flows from the heat source to the heat sink, it is capable of doing work. As
Rev 1
41
shown below in the figure on the right, not all of the heat transfers into
work. This is the second law: there must be rejected heat.
Figure: Second Law of Thermodynamics for a Heat Engine
Considering those losses, the second law of thermodynamics can be used to
derive an expression for the maximum possible energy conversion
efficiency. The second law denies the possibility of completely converting
into work all of the heat supplied to a system operating in a cycle, no matter
how perfectly designed the system may be. The restriction placed by the
second law requires that some of the heat supplied (QS) to the engine must
be rejected as heat (QR). The difference between the heat supplied and the
heat rejected is the net amount of work produced in the cycle (WNET). The
cycle efficiency is the percentage of energy input to a cycle that is
converted to net work output. The concept of the second law is best stated
using Max Planck's description:
Figure: Kelvin-Planck's Second Law of Thermodynamics
The first law of thermodynamics does not define the energy conversion
process completely. The first law relates to and evaluates the various
energies involved in a process. However, no information about the
direction of the process can be obtained by the application of the first law.
Early in the development of the science of thermodynamics, investigators
noted that while work could be converted completely into heat, the converse
was never true for a cyclic process. Certain natural processes were
42
Rev 1
observed always to proceed in a certain direction; for example, heat transfer
occurs from a hot to a cold body. The second law was developed as an
explanation of these natural phenomena.
Figure: Heat Flow Direction
Entropy
The physical property of matter called entropy (S) explains the second law
of thermodynamics. The change in entropy determines the direction in
which a given process proceeds. Entropy also measures the unavailability
of heat to perform work in a cycle. The second law predicts that not all heat
provided to a cycle can be transformed into an equal amount of work; some
heat rejection must take place. The change in entropy is the ratio of heat
transferred during a reversible process to the absolute (abs) temperature of
the system.
∆𝑆 =
∆𝑄
(𝑓𝑜𝑟 𝑎 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠)
𝑇𝑎𝑏𝑠
Where:
•
∆𝑆 = the change in entropy of a system during some process
(BTU/°R [degrees Rankine])
•
∆𝑄 = the amount of heat added to the system during the process
(BTU) (∆Q is change in heat)
•
𝑇𝑎𝑏𝑠 = the absolute temperature at which the heat was transferred
(°R)
Rev 1
43
Figure: Entropy
Entropy (S) is a natural process that starts in one equilibrium state, ends in
another state, and will go in the direction that causes the entropy of the
system plus the environment to increase for an irreversible process and to
remain constant for a reversible process. Therefore, Sf = Si (reversible) and
Sf > Si (irreversible).
The second law of thermodynamics is also expressed as ΔS ≥ 0 for a closed
cycle. In other words, entropy must increase or stay the same for a cyclic
system; it can never decrease.
Entropy is an extensive property of a system and like the total internal
energy or total enthalpy, may be calculated from specific entropies based on
a unit mass quantity of the system, so that S = ms. Values of the specific
entropy are tabulated along with specific enthalpy, specific volume, and
other thermodynamic properties of interest in the steam tables described in a
previous module.
The property of specific entropy is used advantageously as one of the
coordinates when representing a reversible process graphically. The area
under a reversible process curve on the T-s diagram represents the quantity
of heat transferred during the process.
Figure: T-s Diagram With Rankine Cycle
Reversible processes are often used in thermodynamic problems by
comparison to the real irreversible process to aid in a second law analysis.
Reversible processes can be depicted on diagrams such as h-s and T-s,
shown below in the figure.
44
Rev 1
Actual or irreversible processes cannot be drawn because they are not a
succession of equilibrium conditions. Since only the initial and final
conditions of irreversible processes are known, they are represented in the
figure below by dotted or dashed lines shown on the diagram. The dashes
represent the intermediate state of the fluid is not determined, only the
beginning and end states are known. In both real processes shown below,
entropy increases, the dashed lines slant to the right. The compression of
the fluid by the pump and the expansion of the fluid through the turbine
both increase entropy compared to the ideal case.
Figure: T-s and h-s Diagrams for Expansion and Compression Processes
Knowledge Check
The second law of thermodynamics can also be
expressed as ________ for a closed cycle.
A.
Sf = S i
B.
ΔS ≥ 0
C.
ΔT < 0
D.
ΔS < 0
ELO 2.2 Carnot’s Principle of Thermodynamics
Introduction
In 1824 Nicolas Léonard Sadi Carnot, a French military engineer and
physicist known as the father of thermodynamics, advanced the study of the
second law by using reversible processes that disclosed a principle
consisting of the following:

Rev 1
No engine can be more efficient (η) than a reversible (ideal) engine
operating between the same high temperature heat source and low
temperature heat sink.
45

Efficiencies of all reversible engines operating between the same
constant temperature reservoirs are the same.
 Efficiency of a reversible engine depends only upon the temperatures
of the heat source and heat receiver.
Figure: Carnot's Efficiency Principle
Carnot Cycle Guidelines
The Carnot cycle can best be described using an ideal frictionless thermally
isolated piston operating between a constant heat source and heat sink. A
P-v and T-s diagram shown below illustrates the cycle as the heat source is
applied to the piston, causing a reversible isothermal expansion between
point 1 and 2. The piston then moves doing an amount of work (w1-2) due
to the isothermal (constant temperature) expansion of the gas, shown on the
figure as the line between points 1-2.
The gas is allowed to finish expanding adiabatically between point 2 and 3
and an amount of work (w2-3) is done, shown as the line between points 2-3.
Next, the heat sink is applied to the piston, and a reversible isothermal
compression of the gas occurs between points 3 and 4. The piston is used to
compress the gas and an amount of heat (q3-4) is transferred to the heat sink
through the cylinder head. This isothermal compression requires some
amount of work (w3-4) to be done on the piston, shown as the line between
points 3-4.
In process 4, the cylinder is removed from the heat sink. The piston returns
to its initial state by undergoing adiabatic compression requiring some
amount of work (w4-1). The cycle is completed when the cylinder is again
placed in contact with the heat source, shown in the figure below as the line
between points 4-1:
46
Rev 1
Figure: Single Piston Carnot Engine Cycle
During the Carnot cycle just described, a certain amount of heat and work
were added or removed from the system. Work is done on the system when
the piston travels into the cylinder and compresses the gas. Work is done
by the system as the gas expands to force the piston out of the cylinder.
Heat is added to the system to cause the piston to move outward (𝑄𝐴 =
𝑞1−2 ). Heat is added, entropy increases, and the process line goes from left
to right. Heat is removed from the system as the piston is compressed in the
final portion of the cycle (𝑄𝑅 = 𝑞3−4 ). Entropy decreases, and the process
line goes from right to left as heat is removed from the system. Two ideal
assumptions are made that result in the Carnot cycle having the highest
possible efficiency:
1. Both work processes occur with no friction and thus there is no
change in entropy.
2. The heat addition and heat rejection occur with no change in the
temperature of the working fluid. Therefore, the temperature
difference (T) between the working fluid and the heat source and the
heat sink remains constant.
We define thermodynamic cycle efficiency by analyzing the energy output
or work (W) produced compared to the energy input (QA). The greater the
percentage of energy input converted to work, the greater the cycle
efficiency.
The figure below shows a Carnot cycle representation. The heat input (QH)
is the area under line 2-3. The heat rejected (QC) is the area under line 1-4.
The difference between the heat added and the heat rejected is the net work
(sum of all work processes), represented as the area of rectangle 1-2-3-4.
Rev 1
47
Figure: Carnot Cycle Representation
In a perfectly efficient cycle, all of the energy put into the cycle converts to
a useful work output. However, as stated previously, heat must be rejected
for the cycle to be continuous.
The efficiency (η) of the cycle is the ratio of the net work of the cycle to the
heat input to the cycle. This ratio can be expressed by the following
equation:
𝜂=
𝑊𝑛𝑒𝑡
ℎ𝑒𝑎𝑡 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑
𝜂=
(𝑄𝐴 − 𝑄𝑅 )
𝑄𝐴
𝜂=
(𝑄𝐻 − 𝑄𝐶 ) (𝑇𝐻 − 𝑇𝐶 )
=
𝑄𝐻
𝑇𝐻
𝑇
= 1 − (𝑇 𝐶 )Where:
𝐻
•
η = cycle efficiency
•
TC = designates the low-temperature reservoir (°R)
•
TH = designates the high-temperature reservoir (°R)
This equation shows that the maximum possible efficiency exists when TH
is at its highest possible value or when TC is at its lowest value. The above
represents an upper limit of efficiency for any given system operating
between the same two temperatures since all practical systems and
processes are irreversible. The system's maximum possible efficiency
would be that of a Carnot cycle, but because Carnot cycles represent
reversible processes, the real system cannot reach the Carnot efficiency
value. Thus, the Carnot efficiency serves as an unattainable upper limit for
48
Rev 1
any real system's efficiency. The following example demonstrates the
above principles.
Example 1: Carnot Efficiency
An inventor claims to have an engine that receives 100 BTUs of heat and
produces 25 BTUs of useful work when operating between a source at 140
°F and a receiver at 0 °F. Is the claim a valid claim?
Solution 1:
𝑇ℎ = 140 ℉ + 460 = 600 °𝑅
𝑇𝑐 = 0 ℉ + 460 = 460 °𝑅
𝜂=
600 − 460
× 100 = 23.3%
600
Claimed efficiency = 25/100 = 25 percent; this exceeds the Carnot
efficiency value.
Therefore, the claim is invalid.
The second law determines the maximum possible efficiencies obtained
from a power system. Actual efficiencies will always be less than this
maximum. Real systems have losses, such as friction, that are not
reversible and that preclude real systems from obtaining the maximum
possible efficiency. An illustration of the difference that may exist between
the ideal and actual efficiency is presented in the figure below and in the
following example:
Example 2: Actual Versus Ideal Efficiency
The actual efficiency of a steam cycle is 18.0 percent. The facility operates
from a steam source at 340 °F and rejects heat to atmosphere at 60 °F.
Compare the Carnot efficiency to the actual efficiency.
Figure: Real Process Cycle Compared to Carnot Cycle
Rev 1
49
Solution:
Solve for the Carnot maximum efficiency:
𝑇𝑐
𝜂 = 1−( )
𝑇ℎ
𝜂 = 1−
60 + 460
340 + 460
𝜂 = 1−
520
800
𝜂 = 1 − 0.65
𝜂 = 35%
= 35 percent compared to 18.0 percent actual efficiency. .The second law
equations are treated in much the same manner as the first law equations.
An isolated, closed, or open system used in the analysis depends on the
types of energy that cross the boundary. The open system analysis is still
the more general case, with the closed and isolated systems being special
cases of the open system. The approach used to solve second law problems
is similar to that used in the first law analysis.
A control volume using the second law is shown below in the figure. In this
diagram, the fluid moves through the control volume from the inlet section
to the outlet section while work is delivered externally to the control
volume. We assume that the boundary of the control volume is at some
environmental temperature and that all of the heat transfer (Q) occurs at this
boundary.
Entropy is a property that may be transported with the flow of the fluid into
and out of the control volume, just like enthalpy or internal energy. The
entropy flow into the control volume resulting from mass transport, 𝑚̇𝑖𝑛 𝑠𝑖𝑛 ,
and the entropy flow out of the control volume is 𝑚̇𝑜𝑢𝑡 𝑠𝑜𝑢𝑡 , assuming that
the properties are uniform at sections in and out. Entropy may also be
added to the control volume because of heat transfer at the boundary of the
control volume.
50
Rev 1
Figure: Control Volume for Second Law Analysis
A simple demonstration of the use of this form of system in second law
analysis gives the student a better understanding of its use.
Example 3: Open System Second Law
Steam enters the nozzle of a steam turbine with a velocity of 10 ft/sec at a
pressure of 100 psia and temperature of 500 °F. At the nozzle discharge,
the pressure and temperature are one (1) atmospheric pressure (atm) at 300
°F. What is the increase in entropy for the system if the mass flow rate is
10,000 lbm/hr?
Solution:
𝑚̇𝑠𝑖𝑛 + 𝑝̇ = 𝑚̇𝑠𝑜𝑢𝑡
Where:
•
𝑝̇ = entropy added to the system
•
𝑝̇ = 𝑚̇(𝑠𝑜𝑢𝑡 − 𝑠𝑖𝑛 )
•
sin = 1.7088 BTUs/lbm-°R (from steam tables)
•
sout = 1.8158 BTUs/lbm-°R (from steam tables)
•
𝑝̇
•
𝐵𝑇𝑈
= 𝑠𝑜𝑢𝑡 − 𝑠𝑖𝑛 = 1.8158 − 1.7088 𝑙𝑏𝑚-°𝑅
𝑚̇
𝑝̇
𝑚̇
𝐵𝑇𝑈
= 0.107 𝑙𝑏𝑚-°𝑅
•
𝑝̇ = 10,000(0.107)
•
𝑝̇ = 1,070
Rev 1
𝐵𝑇𝑈
𝑙𝑏𝑚-°𝑅
= 𝑒𝑛𝑡𝑟𝑜𝑝𝑦 𝑎𝑑𝑑𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚
51
The second law of thermodynamics gives a maximum efficiency limit
(which is never reached in physical systems) that an ideal thermodynamic
system can perform. The efficiency is determined by knowing the inlet and
exit absolute temperatures of the overall system (one that works in a cycle)
and applying Carnot's efficiency equation.
Knowledge Check
The steam generators produce dry saturated steam at
1,000 psig. The main condenser is operating with zero
subcooling at 1 psia. What is the maximum efficiency
obtainable?
A.
44 percent
B.
34 percent
C.
28 percent
D.
23 percent
Knowledge Check
Determine the Carnot Efficiency of a steam engine that is
supplied with saturated steam at 300 psia and exhausts to
atmosphere…
A.
44 percent
B.
56 percent
C.
42 percent
D.
35 percent
ELO 2.3 Thermodynamics of Ideal and Real Processes
Introduction
It is convenient to arrange the various thermodynamic processes on a
property diagram in evaluating the various cycles present in a nuclear power
plant. The most common set of coordinates used is a plot of temperature
versus specific entropy, a T-s diagram. Using this type of diagram, we can
analyze the various processes that take place and how these processes effect
the entire cycle as well as the amount of heat and work, both of which occur
during the processes. Any ideal thermodynamic process can be drawn as a
52
Rev 1
path on a property diagram. A real process that approximates the ideal
process can also be represented on the same diagrams, usually by dashed
lines.
Carnot Cycle Guidelines
Entropy is constant in an ideal expansion or compression process.
Isentropic processes are represented by vertical lines on T-s and h-s
diagrams, shown below in the figures. A real expansion and real
compression process operating between the same pressures as the ideal
process are shown by dashed lines and will slant slightly toward the right,
since the entropy increases from the start to the end of the real process.
All real processes are irreversible. It is helpful to compare real processes to
ideal processes in system design. The reversible process indicates a
maximum work output for a given input, which compares to real work
output for efficiency purposes. The h-s diagram clearly shows that the real
expansion process (turbine) results in a smaller change in enthalpy, meaning
less energy is extracted in the real turbine than the ideal turbine. The figure
below also shows that more enthalpy must be added during the compression
(pump) process, meaning more energy must be added to the real system
than the ideal system. These differences in enthalpy across various
components tell us how efficient the real process compares to the ideal
process of the Carnot cycle.
Figure: T-s and h-s Diagrams for Expansion and Compression Processes
Rev 1
53
Knowledge Check
Why are real processes shown with dotted lines on
property diagrams?
A.
They occur faster than real processes.
B.
The value of entropy during the process is not
determined.
C.
The entropy values during the process are the same as the
real process until the outlet from the process.
D.
You would not be able to distinguish between real and
ideal processes if the real process was a solid line.
ELO 2.4 Thermodynamic Power Plant Efficiency
Introduction
The actual construction of the steam cycle in nuclear power plants is
considerably more complex than the basic cycles that we have covered so
far, including numerous required auxiliary systems and instrumentation and
control equipment. However, adding this detail to the steam cycle does not
contribute significantly to an understanding of the energy transfer
characteristics of the overall plant. A simplified steam cycle for a typical
steam-electric plant is shown below in the figure. Notice that the cycle
shown includes two components we have not yet discussed, the Moisture
Separator Reheater (MSR) and the Feedwater Preheater. We discuss both
of these components and their effect on cycle efficiency later in this
module.
To analyze a complete power plant steam cycle, it is first necessary to
analyze the elements that make up the cycle. Although specific designs
differ, there are three basic types of elements in power cycles: (1) turbines,
(2) pumps, and (3) heat exchangers. Each of these three elements imparts a
characteristic change in the properties of the working fluid.
Typical Steam Cycle
Previously we calculated system efficiency by knowing the temperature of
the heat source and the heat sink. It is also possible to calculate the
efficiencies of each individual component by comparing the actual work
produced by the component to the work that produced by an ideal
component operating isentropically between the same inlet and outlet
conditions.
54
Rev 1
Figure: Typical Steam Cycle
Steam Turbine Efficiency Guidelines
A steam turbine extracts energy from the working fluid (steam) to do work
in the form of rotating the turbine shaft. The steam works as it expands
through the turbine. The shaft work converts to electrical energy by the
generator. In the application of the first law general energy equation to a
simple turbine under steady flow conditions, demonstrates that the decrease
in the enthalpy of the working fluid Hin - Hout equals the work done by the
working fluid in the turbine (Wt).
Figure: Turbine Work
𝐻𝑖𝑛 − 𝐻𝑜𝑢𝑡 = 𝑊𝑡
Rev 1
55
𝑚̇(ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 ) = 𝑤̇𝑡
Where:
•
Hin = enthalpy of the working fluid entering the turbine (BTU)
•
Hout = enthalpy of the working fluid leaving the turbine (BTU)
•
Wt = work done by the turbine (ft-lbf)
•
𝑚̇ = mass flow rate of the working fluid (lbm/hr)
•
hin = specific enthalpy of the working fluid entering the turbine
(BTU/lbm)
•
hout = specific enthalpy of the working fluid leaving the turbine
(BTU/lbm)
•
𝑤̇𝑡 = power of the turbine (BTU/hr)
Ideal Versus Real Turbine
The calculation of turbine work using only the enthalpy change is valid
because the change of kinetic and potential energy and the amount of heat
lost by the working fluid while in the turbine are negligible. These
assumptions are valid for most practical applications. However, to apply
these relationships, one additional definition is necessary. In any ideal case,
the working fluid does work reversibly by expanding at constant entropy.
In an ideal turbine, the entropy of the working fluid entering the turbine Sin
equals the entropy of the working fluid leaving the turbine.
𝑆𝑖𝑛 = 𝑆𝑜𝑢𝑡 or 𝑠𝑖𝑛 = 𝑠𝑜𝑢𝑡
Where:
•
Sin = entropy of the working fluid entering the turbine (BTU/°R)
•
Sout = entropy of the working fluid leaving the turbine (BTU/°R)
•
sin = specific entropy of the working fluid entering the turbine
(BTU/lbm-°R)
•
sout = specific entropy of the working fluid leaving the turbine
(BTU/lbm-°R)
An ideal turbine performs the maximum amount of work theoretically
possible, and therefore provides a basis for analyzing the performance of
real turbines.
Because of friction losses in the blades, steam leakage past the blades and to
a lesser extent mechanical friction, a real turbine does less work than an
56
Rev 1
ideal turbine. Turbine efficiency (ηt), is defined as the ratio of the actual
work done by the turbine (Wt.actual) to the work that would be done by the
turbine if it were an ideal turbine (Wt.ideal). It is shown that Wi is larger than
Wa, meaning more enthalpy is extracted from the steam in an ideal turbine
operating between the same temperatures (pressures) than in a real turbine.
Figure: h-s Diagram for Ideal and Real Turbines
𝜂𝑡 =
𝑊𝑡.𝑎𝑐𝑡𝑢𝑎𝑙
𝑊𝑡.𝑖𝑑𝑒𝑎𝑙
𝜂𝑡 =
(ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 )𝑎𝑐𝑡𝑢𝑎𝑙
(ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 )𝑖𝑑𝑒𝑎𝑙
Where:
•
𝜂𝑡 = turbine efficiency (no units)
•
Wt.actual = actual work done by the turbine (ft-lbf)
•
Wt.ideal = work done by an ideal turbine (ft-lbf)
•
(hin – hout)actual = actual enthalpy change of the working fluid
(BTU/lbm)
•
(hin – hout)ideal = actual enthalpy change of the working fluid in an ideal
turbine (BTU/lbm)
A vertical line on the T-s diagram is a constant entropy ideal process.
Entropy increases in the actual turbine process. The smaller the increase in
entropy, the closer the turbine efficiency (ηt) is to 1.0 or 100 percent.
If an ideal turbine became a real turbine, its output would decrease due to
losses such as friction, windage, moisture, and tip leakage. To raise the
turbine back to its original output, the turbine steam supply valves would
open, increasing the mass flow rate of the steam going into the turbine. The
work of the turbine increases to overcome the losses. Opening the steam
Rev 1
57
supply valves decreases steam generator pressure and adding more heat
raises the steam generator pressure back to its original value. However,
since the heat added is greater than the increase in work from the turbine,
the cycle efficiency decreases.
Pump Efficiency Guidelines
A pump performs work on a system's working fluid to overcome the head
loss and keep the fluid moving. Like the turbine, the application of the first
law general energy equation to a simple pump under steady flow conditions
results in the work of the pump (Wp), equals the change in working fluid
enthalpy across the pump ( Hout - Hin).
𝐻𝑜𝑢𝑡 − 𝐻𝑖𝑛 = 𝑊𝑝
𝑚̇(ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 ) = 𝑊̇𝑝
Where:
•
Hout = enthalpy of the working fluid leaving the pump (BTU)
•
Hin = enthalpy of the working fluid entering the pump (BTU)
•
Wp = work done by the pump on the working fluid (ft-lbf)
•
𝑚̇ = mass flow rate of the working fluid (lbm/hr)
•
hout = specific enthalpy of the working fluid leaving the pump
(BTU/lbm)
•
hin = specific enthalpy of the working fluid entering the pump
(BTU/lbm)
•
𝑊̇𝑝 = power of pump (BTU/hr)
Real Versus Ideal Pump
As in the turbine, the kinetic and potential energy changes and the heat lost
by the working fluid while in the pump are negligible. These are valid
assumptions along with the assumption that the working fluid is
incompressible. For the ideal case, it can be shown that the work done by
the pump (Wp) equals the change in enthalpy across the ideal pump.
𝑊𝑝.𝑖𝑑𝑒𝑎𝑙 = (𝐻𝑜𝑢𝑡 − 𝐻𝑖𝑛 )𝑖𝑑𝑒𝑎𝑙
𝑊̇𝑝.𝑖𝑑𝑒𝑎𝑙 = 𝑚̇(ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 )𝑖𝑑𝑒𝑎𝑙
Where:
•
58
Wp = work done by the pump on the working fluid (ft-lbf)
Rev 1
•
Hout = enthalpy of the working fluid leaving the pump (BTU)
•
Hin = enthalpy of the working fluid entering the pump (BTU)
•
𝑊̇𝑝 = power of pump (BTU/hr)
•
𝑚̇ = mass flow rate of the working fluid (lbm/hr)
•
hout = specific enthalpy of the working fluid leaving the pump
(BTU/lbm)
•
hin = specific enthalpy of the working fluid entering the pump
(BTU/lbm)
The ideal pump provides a basis for analyzing the performance of actual
pumps, which requires more work because of unavoidable losses due to
friction and fluid turbulence. The work done by a pump Wp equals the
change in enthalpy across the actual pump.
𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙 = (𝐻𝑜𝑢𝑡 − 𝐻𝑖𝑛 )𝑎𝑐𝑡𝑢𝑎𝑙
𝑊̇𝑝.𝑎𝑐𝑡𝑢𝑎𝑙 = 𝑚̇(ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 )𝑎𝑐𝑡𝑢𝑎𝑙
Pump Efficiency
Now, consider a real pumping process opposed to the ideal pumping
process. In the real pumping process with friction taken into account, the
entropy increases across the pump. Recall that entropy increases in all real
processes. Because the work of the pump (WPUMP) increases to make up for
the frictional losses, the net work (WNET) decreases. Since the net work
decreases, the overall cycle efficiency also decreases.
Pump efficiency (ηp) is the ratio of the work required by the pump if it were
an ideal pump (Wp.ideal) to the actual work required by the pump (Wp.actual).
𝜂𝑝 =
𝑊𝑝.𝑖𝑑𝑒𝑎𝑙
𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙
Example:
A pump operating at 75 percent efficiency has an inlet specific enthalpy of
200 BTU/lbm. The exit specific enthalpy of the ideal pump is 600
BTU/lbm. What is the exit specific enthalpy of the actual pump?
Solution:
Using the equation above:
𝜂𝑝 =
𝑊𝑝.𝑖𝑑𝑒𝑎𝑙
𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙
Rev 1
59
𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙 =
𝑊𝑝.𝑖𝑑𝑒𝑎𝑙
𝜂𝑝
(ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 )𝑎𝑐𝑡𝑢𝑎𝑙 =
ℎ𝑜𝑢𝑡.𝑎𝑐𝑡𝑢𝑎𝑙 =
ℎ𝑜𝑢𝑡.𝑎𝑐𝑡𝑢𝑎𝑙 =
(ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 )𝑖𝑑𝑒𝑎𝑙
𝜂𝑝
(ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 )𝑖𝑑𝑒𝑎𝑙
+ ℎ𝑖𝑛.𝑎𝑐𝑡𝑢𝑎𝑙
𝜂𝑝
(600
𝐵𝑇𝑈
𝐵𝑇𝑈
− 200
)
𝑙𝑏𝑚
𝑙𝑏𝑚 + 200 𝐵𝑇𝑈
75
𝑙𝑏𝑚
ℎ𝑜𝑢𝑡.𝑎𝑐𝑡𝑢𝑎𝑙 = 533.3
𝐵𝑇𝑈
𝐵𝑇𝑈
+ 200
𝑙𝑏𝑚
𝑙𝑏𝑚
ℎ𝑜𝑢𝑡.𝑎𝑐𝑡𝑢𝑎𝑙 = 733.3 𝐵𝑇𝑈/𝑙𝑏𝑚
Pump efficiency (ηp) relates the minimum amount of work theoretically
possible to the actual work required by the real pump. However, the work
required by a pump is normally only an intermediate form of energy.
Usually, a motor or turbine runs the pump. Pump efficiency does not
account for losses in this motor or turbine. An additional efficiency factor,
motor efficiency (ηm) is the ratio of the actual work required by the pump to
the electrical energy input to the pump motor, when expressing both in the
same units.
𝜂𝑚 =
𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙
𝑊𝑚.𝑖𝑛 𝐶
Where:
•
𝜂𝑚 = motor efficiency (no units)
•
Wp.actual = actual work required by the pump (ft-lbf)
•
Wm.in = electrical energy input to the pump motor per kilowatt hour
(kWh)
•
C = conversion factor = 2.655 x 106 ft-lbf/kWh
Motor Efficiency Guidelines
Like pump efficiency, motor efficiency is always less than 1.0 or 100
percent for an actual pump motor. The combination of pump efficiency and
motor efficiency relates the ideal pump to the electrical energy input to the
pump motor.
𝜂𝑚 𝜂𝑝 =
60
𝑊𝑝.𝑖𝑑𝑒𝑎𝑙
𝑊𝑚.𝑖𝑛 𝐶
Rev 1
Where:
•
𝜂𝑚 = motor efficiency (no units)
•
𝜂𝑝 = pump efficiency (no units)
•
𝑊𝑝.𝑖𝑑𝑒𝑎𝑙 = ideal work required by the pump (ft-lbf)
•
𝑊𝑚.𝑖𝑛 = electrical energy input to the pump motor (kWh)
•
C = conversion factor = 2.655 x 106 ft-lbf/kWh
Heat Exchangers Guidelines
Several types of heat exchangers transfer heat between two working fluids
in power plant steam cycles. The steam generator or boiler uses reactor
coolant to heat and vaporize the feedwater. The main condenser uses
circulating water to cool and condense the turbine exhaust steam before
returning it to the steam generator. Numerous smaller heat exchangers
throughout the steam cycle heat or cool support systems. Two primary
factors—first, the mass flow rates of the fluids flowing through the heat
exchanger and second, the temperature difference between the two fluids—
determine the heat transfer rate of a heat exchanger.
Figure: Typical Parallel and Counter-Flow Heat Exchangers. Different
Flow Regimes and Associated Temperature Profiles In a Double-Pipe Heat
Exchanger
Applying the first law general energy equation to a simple heat exchanger
under steady flow conditions shows that the mass flow rates and enthalpies
of the two fluids are related as follows:
Rev 1
61
𝑚̇1 (ℎ𝑜𝑢𝑡.1 − ℎ𝑖𝑛.1 ) = −𝑚̇2 (ℎ𝑜𝑢𝑡.2 − ℎ𝑖𝑛.2 )
Where:
•
𝑚̇1 = mass flow rate of the working fluid 1 (lbm/hr)
•
𝑚̇2 = mass flow rate of the working fluid 2 (lbm/hr)
•
hout.1 = specific enthalpy of the working fluid 1 leaving the heat
exchanger (BTU/lbm)
•
hin.1 = specific enthalpy of the working fluid 1 entering the heat
exchanger (BTU/lbm)
•
hout.2 = specific enthalpy of the working fluid 2 leaving the heat
exchanger (BTU/lbm)
•
hin.2 = specific enthalpy of the working fluid 2 entering the heat
exchanger (BTU/lbm)
Condenser Guidelines
The most common condenser design is the single-pass condenser, shown
below in the illustration. This design provides cooling water flow through
straight tubes from the inlet water box on one end to the outlet water box on
the other end (single-pass), and is a cross-flow heat exchanger because the
steam flows across the heat transfer surface. Tube sheets with cooling
water tubes attached separate the water box area and the steam condensing
area. Long cooling water tubes are supported within the condenser by the
tube support sheets.
Figure: Typical Single-Pass Condenser
62
Rev 1
Condensers normally have a series of baffles that redirect the steam to
minimize direct impingement on the cooling water tubes as shown above in
the figure. The bottom area of the condenser is the hotwell. The
condensate collects in the hotwell, which in turn is the condensate pump's
suction source.
The condenser performs two major functions in the steam cycle:
1. It is the closed space where the wet steam exits the turbine, gives
up its latent heat of condensation, and condenses to liquid for
return to the steam generator or boiler as feedwater. This lowers
the plant’s operational cost by allowing reuse of the clean, treated
condensate and pumped back to the boiler.
2. It increases the cycle's efficiency by providing the lowest
temperature heat sink resulting in the largest possible ΔT and ΔP
(change in pressure) between the source (boiler) and the heat sink
(condenser).
Because condensation is taking place, the term latent heat of condensation
is used instead of latent heat of vaporization. The steam's latent heat of
condensation passes to the water flowing through the tubes of the
condenser. The specific volume decreases and a vacuum forms as steam
passes into the closed condenser and condenses to liquid. The vacuum
increases the plant’s efficiency by extracting more work from the turbine.
Figure: Typical Single-Pass Condenser End View
After the steam condenses, the saturated liquid continues transferring heat
to the cooling water as it falls to the bottom of the condenser, or hotwell, in
a process called subcooling. Some subcooling is desirable. Condensate
depression is the difference between the saturation (sat) temperature for the
existing condenser vacuum and the temperature of the condensate,
Rev 1
63
expressed as number of degrees condensate depression or degrees
subcooled. A few degrees of subcooling is necessary to prevent cavitation
in the condensate pumps. Cavitation is the formation of vapor bubbles in
the low-pressure region of the pump impeller (or eye) and the subsequent
collapse of the bubbles along the impeller vanes. Cavitation causes
excessive vibration, erosion of the impeller vanes, and increased bearing
wear that results in damaged pumps.
𝐶𝑜𝑛𝑑𝑒𝑛𝑠𝑎𝑡𝑒 𝐷𝑒𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 = 𝑇𝑠𝑎𝑡 − 𝑇𝑎𝑐𝑡𝑢𝑎𝑙
As can be seen on the T-v diagram below, condensate depression decreases
the plant’s operating efficiency because the subcooled condensate is
reheated in the boiler, requiring more heat or energy from the heat source.
Condensate depression increases the heat rejected from the cycle,
decreasing overall efficiency. Excessive condensate depression also allows
increased absorption of air by the condensate and accelerated oxygen
corrosion of plant materials.
Figure: T-v Diagram for Typical Condenser
A buildup of non-condensable gasses in the condenser decreases vacuum
and increases the saturation temperature where the steam condenses.
Accumulating non-condensable gasses also blankets the condenser’s tubes,
of the condenser reducing the heat transfer surface. Allowing the
condensate level to rise over the condenser’s lower tubes also reduces the
surface area. Reducing the heat transfer surface has the same effect as a
reduction in cooling water flow. Reducing the effective surface area results
in difficulty maintaining condenser vacuum if the condenser is operating
near its design capacity.
The temperature and flow rate of the cooling water through the condenser
controls the temperature of the condensate, which also regulates the
saturation pressure (vacuum) of the condenser.
Operators should maintain condenser vacuum as close to 29 inches of
Mercury (Hg) as practical. This allows maximum expansion of the steam,
64
Rev 1
and the maximum work. If the condenser was perfectly airtight and no air
or non-condensable gasses were present in the exhaust steam, it would only
be necessary to condense the steam and remove the condensate to create and
maintain a vacuum. The sudden reduction in steam volume as it condenses
maintains the vacuum. However, it is impossible to prevent the entrance of
air and other non-condensable gasses into the condenser. In addition, some
method must exist to initially create a vacuum in the condenser. Using an
air injector or vacuum pump establishes and maintains the condenser
vacuum.
Example:
Determine the quality of the steam entering a condenser operating at a 1
psia vacuum with 4 °F of condensate depression and circulating water Tin =
75 °F and Tout = 97 °F. Assume cp for the condensate and the circulating
water is 1 BTU/lbm-°F. The steam mass flow rate in the condenser is 8 x
106 lbm/hr and the circulating water is 3.1 x 108 lbm/hr.
Find 𝑄̇ of the circulating water:
𝑄̇ = 𝑚̇𝑐𝑝 𝛥𝑇
1 𝐵𝑇𝑈
= 𝑚̇ (
) (97 ℉ − 75 ℉)
𝑙𝑏𝑚 ℉
= (3.1 ×
108 𝑙𝑏𝑚
𝐵𝑇𝑈
) (22
)
ℎ𝑟
𝑙𝑏𝑚
= 6.82 × 109
𝐵𝑇𝑈
ℎ𝑟
6.82 x 109 BTUs/hr represents the 𝑄̇ necessary to condense the steam (stm)
and subcool it to 4 °F below saturation temperature. Therefore, the 𝑄̇
necessary to subcool the condensate (cond) is:
𝑄̇ = 𝑚̇𝑐𝑝 𝛥𝑇
106 𝑙𝑏𝑚 1 𝐵𝑇𝑈
= (8 ×
)(
) (4 ℉)
ℎ𝑟
𝑙𝑏𝑚 ℉
= 3.2 × 107
𝐵𝑇𝑈
ℎ𝑟
This number is insignificant compared to the total 𝑄̇ , and therefore, will not
be considered. From the steam tables, saturated liquid at 1 psia:
ℎ𝑓 = 69.73
Rev 1
𝐵𝑇𝑈𝑠
𝑙𝑏𝑚
65
ℎ𝑓𝑔 = 1,036.1
ℎ𝑔 = 1,105.8
𝐵𝑇𝑈𝑠
𝑙𝑏𝑚
𝐵𝑇𝑈𝑠
𝑙𝑏𝑚
Using:
𝑄̇ = 𝑚̇𝛥ℎ
Solve for 𝛥ℎ:
𝛥ℎ =
𝑄̇
𝑚̇
Therefore:
(ℎ𝑠𝑡𝑚 − ℎ𝑐𝑜𝑛𝑑 ) =
𝑄̇
𝑚̇
Solving for hstm:
ℎ𝑠𝑡𝑚 =
ℎ𝑠𝑡𝑚 =
𝑄̇
+ ℎ𝑐𝑜𝑛𝑑
𝑚̇
𝐵𝑇𝑈𝑠
ℎ𝑟 + 69.73 𝐵𝑇𝑈𝑠
𝑙𝑏
𝑙𝑏𝑚
8 × 106 𝑚
ℎ𝑟
6.82 × 109
= 922.23
𝐵𝑇𝑈𝑠
𝑙𝑏𝑚
Using:
ℎ𝑠𝑡𝑚 = ℎ𝑓 + 𝑋ℎ𝑓𝑔
Solving for X:
𝑋=
ℎ𝑠𝑡𝑚 − ℎ𝑓
ℎ𝑓𝑔
922.23
𝑋=
𝐵𝑇𝑈𝑠
𝐵𝑇𝑈𝑠
− 69.73
𝑙𝑏𝑚
𝑙𝑏𝑚
𝐵𝑇𝑈𝑠
1,036.1
𝑙𝑏𝑚
𝑋 = 0.823 𝑜𝑟 82.3% 𝑠𝑡𝑒𝑎𝑚 𝑞𝑢𝑎𝑙𝑖𝑡𝑦
66
Rev 1
Real Versus Ideal Cycle Efficiency Guidelines
In the preceding sections, we discussed the Carnot cycle, cycle efficiencies,
and component efficiencies. In this section, we apply this information to
compare and evaluate various ideal and real cycles. This determines how
modifying a cycle affects the cycle's available energy that can be extracted
for work.
A Carnot cycle's efficiency depends solely on the temperature of the heat
source and the heat sink. To improve a cycle's efficiency, all we have to do
is increase the temperature of the heat source and decrease the temperature
of the heat sink. In the real world, the ability to do this is limited by the
following constraints.
For a real cycle, the heat sink is limited by the fact that the earth is our final
heat sink, and therefore is fixed at about 60 °F (520 °R).
The heat source is limited to the combustion temperatures of the fuel burned
or the maximum limits placed on nuclear fuels by their structural
components (pellets, cladding, etc.). In the case of fossil fuel cycles, the
upper limit is ~ 3,040 °F (3,500 °R). However, even this temperature is not
attainable due to the metallurgical restraints of the boilers; and therefore,
they are limited to about ~ 1,500 °F (~ 1,960 °R) for a maximum heat
source temperature.
Using these limits to calculate the maximum efficiency attainable by an
ideal Carnot cycle gives the following:
𝜂=
𝑇𝑆𝑂𝑈𝑅𝐶𝐸 − 𝑇𝑆𝐼𝑁𝐾 1,960 °𝑅 − 520 °𝑅
=
= 73.5%
𝑇𝑆𝑂𝑈𝑅𝐶𝐸
1,960 °𝑅
This calculation indicates that the Carnot cycle, operating with ideal
components under real world constraints, should convert almost threequarters of the input heat into work. This ideal efficiency is beyond the
present capabilities of any real systems.
Heat Rejection
By analyzing a T-s diagram, we will understand why an efficiency of 73
percent is not possible for real components.
The energy added to a working fluid during the Carnot isothermal
expansion is given by qs. Not all of this energy is available for use by the
heat engine since a portion of it (qr) must be rejected to the environment.
This is given by:
𝑞𝑟 = 𝑇𝑜 𝛥𝑠 in units of BTU/lbm
Where:
Rev 1
67
•
To = the average heat sink temperature of 520 °R
The available energy (A.E.) for the Carnot cycle may be given as:
•
𝐴. 𝐸. = 𝑞𝑠 − 𝑞𝑟
Substituting the above equation for qr gives:
𝐴. 𝐸. = 𝑞𝑠 − 𝑇𝑜 𝛥𝑠 in units of BTU/lbm, (Δs is change in absolute entropy)
and equals the area of the shaded region labeled available energy in the
figure below between the temperatures 1,962 °R and 520 °R.
Figure: Carnot Cycle Versus Typical Power Cycle Available Energy
Typical Power Cycle
A typical power cycle employed by a fossil fuel plant as shown on the
above figure. The working fluid is water, which places certain restrictions
on the cycle. If we wish to limit ourselves to operation at or below 2,000
psia, it is readily apparent that constant heat addition at our maximum
temperature of 1,962 °R is not possible (2 to 4). The nature of water and
certain elements of the process controls require us to add heat in a constant
pressure process instead (1-2-3-4). Because of this, the average temperature
where we add heat is far below the maximum allowable material
temperature.
68
Rev 1
The actual available energy (area under the 1-2-3-4 on the above curve) is
less than half of what is available from the ideal Carnot cycle (area under 12'-4) operating between the same two temperatures. Typical thermal
efficiencies for fossil plants are 40 percent while nuclear plants have
efficiencies of 31 percent. These numbers are less than half the maximum
thermal efficiency of the ideal Carnot cycle calculated earlier.
The figure below shows a proposed Carnot steam cycle superimposed on a
T-s diagram. Several problems make it undesirable as a practical power
cycle. Significant pump work is required to compress a two-phase mixture
of water and steam from point 1 to the saturated liquid state at point 2.
Cavitation in the pump would occur if the inlet fluid was at saturation.
Third, a condenser designed to produce a two-phase mixture at the outlet
(point 1) would pose technical problems.
Figure: Ideal Carnot Cycle
Rankine Cycle
Steam engines drove the industrial revolution, so early thermodynamic
developments centered on improving the performance of contemporary
steam engines. It was desirable to construct a steam cycle as close to
reversible as possible and would take better advantage of the characteristics
of steam than does the Carnot cycle. The Rankine cycle was developed as a
more practical version of the Carnot cycle.
The Rankine cycle, shown below in the figure, confines the isentropic
compression process to the liquid phase only (points 1 to 2). This
minimizes the amount of work required to attain operating pressures and
avoids the mechanical problems associated with pumping a two-phase
mixture. The compression process shown between points 1 and 2 is
exaggerated*. In reality, a temperature rise of only 1 °F occurs in
compressing water from 14.7 psig at a saturation temperature of 212 °F to
1,000 psig.
Rev 1
69
Figure: Rankine Cycle
*The constant pressure lines converge rapidly in the subcooled or
compressed liquid region and it is difficult to distinguish them from the
saturated liquid line without artificially expanding the constant pressure
lines away from it.
Like the Carnot cycle, in a Rankine cycle available and unavailable energy
on a T-s diagram is represented by the areas under the curves; therefore, the
larger the unavailable energy, the less efficient the cycle. The same loss of
cycle efficiency is seen when two Rankine cycles are compared as shown
below in the figure.
Figure: Rankine Cycle Efficiency Comparisons On a T-s Diagram
If the ideal turbine was replaced with a real turbine, the efficiency of the
cycle will be reduced. This is because the non-ideal turbine incurs an
increase in entropy increases the area under the T-s curve for the cycle.
However, the increase in the area of available energy (3-2-3') is less than
the increase in area for unavailable energy (a-3-3'-b).
70
Rev 1
Figure: Rankine Cycle With Real Versus Ideal Turbine
Rankine Cycle Efficiencies
Cycle efficiency compares by contrasting the amount of rejected energy to
available energy of both cycles. The comparison shows that Cycle b,
above, has less heat available for work and more rejected heat, making it
less efficient.
An h-s diagram also compares systems and helps determine their
efficiencies.
Typical Steam Cycle
A simplified version of the major components of a typical steam plant cycle
is shown below in the figure. This figure does not contain the exact detail
found at most power plants; however, it adequately demonstrates
understanding the basic operation of a power cycle.
Figure: Typical Steam Cycle
Rev 1
71
The typical steam cycle is comprised of the following processes:










1-2: Heat is added to the working fluid in the steam generator under a
constant pressure condition
2-3: Saturated steam from the steam generator is expanded in the
high-pressure (HP) turbine to provide shaft work output at constant
entropy
3-4: Moist steam from the HP turbine exit is dried and superheated in
the moisture separator reheater (MSR)
4-5: Superheated steam from the MSR is expanded in the lowpressure (LP) turbine to provide shaft work at constant entropy
5-6: Steam exhaust from the turbine is condensed in the condenser by
cooling water under a constant vacuum condition
6-7: Condensate is compressed as a liquid by the condensate pump
7-8: Condensate is preheated by the low-pressure feedwater heaters
8-9: Condensate is compressed as a liquid by the feedwater pump
9-1: Feedwater is preheated by the high-pressure heaters
1-2: Cycle starts again and heat is added to the working fluid in the
steam generator under a constant pressure condition
The typical steam cycle is shown below on the T-s diagram. The numbered
points on the cycle correspond to the numbered points on the above figure.
The Rankine cycle is an ideal cycle and does not exactly represent the real
processes in the plant but is a closer approximation than the Carnot cycle.
Real pumps and turbines would exhibit an entropy increase across them
when shown on a T-s diagram.
Figure: Rankine Steam Cycle (Ideal)
A T-s diagram of a cycle that closely approximates actual plant processes is
shown below in the figure. The pumps and turbines in this cycle are real
pumps and turbines and show an entropy increase across them. A small
72
Rev 1
amount of subcooling is evident in the condenser as demonstrated by the
small dip down to point 5.
Subcooling is the process of cooling condensed vapor beyond what is
required for the condensation process. This small amount of subcooling
decreases cycle efficiency because additional heat was removed from the
cycle to the cooling water as heat rejected. This additional heat rejected
must then be added back in the steam generator. Condenser subcooling
decreases the cycle’s efficiency and is sometimes required to have an
adequate suction head to prevent the condensate pumps from cavitating. By
controlling the temperature or flow rate of the cooling water to the
condenser, the operator directly affects the overall cycle efficiency.
Figure: Steam Cycle (Real)
The Mollier diagram plots and illustrates the energy transferred to or from
the steam during the cycle. The numbered points on the Mollier diagram
shown below correspond to the numbered points on the previous Rankine
cycle diagrams. The Mollier diagram is limited to plotting only the
saturated or superheated form of the working fluid; the liquid portion of the
steam cycle is not indicated on this type of diagram. The following
conditions are visible on the Mollier diagram:




Point 1: Saturated steam at 540 °F
Point 2: 82.5 percent quality at exit of HP turbine
Point 3: Temperature of superheated steam is 440 °F
Point 4: Condenser vacuum is 1 psia
The solid lines represent the conditions for ideal turbines as verified by the
fact that no entropy change shows across the turbines. The dotted lines
represent the path taken if real turbines were considered, in which case an
increase in entropy is evident.
Rev 1
73
Figure: Mollier Diagram
(Use student copy during lesson)
Causes of Inefficiency
In this section, we compare some of the types and causes for the
inefficiencies of real components and cycles to that of their ideal
counterparts.
Components
In real systems, a percentage of the overall cycle inefficiency is due to the
losses by the individual components. Turbines, pumps, and compressors all
behave non-ideally due to heat losses, friction and windage losses. All of
these losses contribute to the non-isentropic behavior of real equipment. As
previously explained, these losses are disclosed as an increase in the
system's entropy or amount of energy that is unavailable for use by the
cycle.
74
Rev 1
Cycles
Some compromises are made in real systems due to cost and other factors in
the design and operation of the cycle. In a large power generating station,
the condensers are designed to subcool the liquid by 8 °F to 10 °F. This
compromise allows the condensate pumps to pump without cavitation but
each degree of subcooling is energy that must be put back by reheating the
water. This energy used in reheating does no useful work, which decreases
cycle efficiency. Imperfect thermal insulation results in a heat loss to the
environment; again this is energy lost to the system and therefore
unavailable to do work. Both resistance to fluid flow and mechanical
friction in machines are other real world losses that result in decreased cycle
efficiency.
Secondary System Parameters
Operators should have a thorough understanding of how changing plant
parameters affect plant operation and overall efficiency. The loss of
efficiency causes a change in power output from the reactor. The
parameters that could affect thermodynamic efficiency are discussed below.
Increasing Steam Temperature at the Turbine Entrance

A higher turbine inlet steam temperature raises the available work that
can be extracted from the turbine.
 More work increases plant efficiency.
Increasing Feedwater Heating

Raising feedwater temperature to near Tsat for the existing steam
generator pressure increases the plant’s efficiency.
 Less energy from fission process is needed to raise feedwater to
operating temperatures.
 We must account for the energy added to the feedwater.
 Overall plant efficiency increases.
Increasing Condenser Vacuum

More work is extracted from the turbine due to the lower pressure in
the main condenser.
 Condensing steam increases condenser vacuum.
 Overall plant efficiency increases.
Increasing Circulating Water System Flow Rate

Increasing the circulating water system flow rate raises the differential
temperature between condensate and cooling medium.
 This reduces condenser temperature, lowering the pressure at the
exhaust of the turbine.
 Lower pressure at the exhaust of the turbine increases plant
efficiency.
Rev 1
75

However, the additional heat removed by more condensate depression
must be replenished.
Lowering Circulating Water System Inlet Temperature

By decreasing the circulating water system inlet temperature, the
differential temperature is increased between condensate and cooling
medium.
 Larger differential temperature results in larger heat transfer rate,
which lowers condenser temperature and pressure.
 Lower pressure at the exhaust of the turbine increases plant
efficiency.
 However, the additional heat removed by more condensate depression
must be replenished.
Reducing Condensate Depression

Any amount of condensate depression causes more required heat
addition to raise the condensate temperature to the operating
temperature.
 If less energy is rejected to the circulating water, the feedwater
requires less energy from the fission process to raise the temperature
to operating temperatures.
 Controlling condensate depression to the minimum required raises the
plant’s efficiency.
Remove Air and Non-Condensable Gases

Excessive air and non-condensable gases within the main condenser
minimize the heat transfer area.
 More energy is required to achieve the same cooling or achieving
insufficient condensate depression.
 Removal of non-condensable gases increases the plant’s efficiency.
Knowledge Check
Determine the condensate depression in a condenser
operating at 1 psia with a condensate temperature of 95
°F. Approximately:
76
A.
10 °F
B.
9 °F
C.
8 °F
D.
7 °F
Rev 1
Knowledge Check
Condensate depression is the process of …
A.
removing condensate from turbine exhaust steam.
B.
spraying condensate into turbine exhaust steam.
C.
heating turbine exhaust steam above its saturation
temperature.
D.
cooling turbine exhaust steam below its saturation
temperature.
Knowledge Check
Excessive heat removal from the low-pressure turbine
exhaust steam in the main condenser results in …
A.
thermal shock
B.
loss of condenser vacuum
C.
condensate depression
D.
fluid compression
Knowledge Check
The rate of heat transfer between two liquids in a heat
exchanger will increase if the … (Assume specific heats
do not change.)
Rev 1
A.
inlet temperature of the hotter liquid decreases by 20 °F.
B.
inlet temperature of the colder liquid increases by 20 °F.
C.
flow rates of both liquids decrease by 10 percent.
D.
flow rates of both liquids increase by 10 percent.
77
Knowledge Check
Which one of the following pairs of fluids undergoing
heat transfer in typical cross-flow design heat exchangers
yields the greatest heat exchanger overall heat transfer
coefficient? Assume comparable heat exchanger sizes
and fluid flow rates.
A.
Oil to water in a lube oil cooler
B.
Steam to water in a feedwater heater
C.
Water to air in a ventilation heating unit
D.
Water to water in a cooling water heat exchanger
Knowledge Check
A nuclear power plant is operating at 90 percent of rated
power. Main condenser pressure is 1.7 psia and hotwell
condensate temperature is 120 °F. Which one of the
following describes the effect of a 5 percent decrease in
cooling water flow rate through the main condenser?
A.
Overall steam cycle efficiency increases because the
work output of the turbine increases.
B.
Overall steam cycle efficiency increases because
condensate depression decreases.
C.
Overall steam cycle efficiency decreases because the
work output of the turbine decreases.
D.
Overall steam cycle efficiency decreases because
condensate depression increases.
Knowledge Check
Which one of the following actions decreases overall
nuclear power plant thermal efficiency?
78
A.
Reducing turbine inlet steam moisture content
B.
Reducing condensate depression
Rev 1
C.
Increasing turbine exhaust pressure
D.
Increasing temperature of feedwater entering the steam
generators
Knowledge Check
Which one of the following changes causes an increase
in overall nuclear power plant thermal efficiency?
A.
decreasing the temperature of the water entering the
steam generators
B.
decreasing the superheat of the steam entering the lowpressure turbines
C.
decreasing the circulating water flow rate through the
main condenser
D.
decreasing the concentration of non-condensable gases in
the main condenser
TLO 2 Summary
During this lesson, you learned about the Second Law of Thermodynamics,
which states that it is impossible to construct a device that operates within a
cycle that can convert all the heat supplied it into mechanical work.
Recognizing that even the most thermally and mechanically perfect cycles
must reject some heat defines thermodynamic power cycle efficiency. The
listing below provides a summary of sections in this TLO.
1. Review ELO 2.1 by asking students the following:
•
Planck's statement of the second law of thermodynamics, which is
that it is impossible to construct an engine that will work in a
complete cycle and produce no other effect except the raising of a
weight and the cooling of a heat reservoir.
•
Entropy is a measure of the unavailability of heat to perform work
in a cycle. This relates to the second law since the second law
predicts that not all heat provided to a cycle can be transformed
into an equal amount of work, some heat rejection must take place.
•
Second law of thermodynamics demonstrates that the maximum
possible efficiency of a system is the Carnot efficiency written as:
Rev 1
79
𝜂=
𝑇𝐻 − 𝑇𝐶
𝑇𝐻
2. Review ELO 2.2 by having student explain the following statements:
•
Maximum efficiency of a closed cycle can be determined by
calculating the efficiency of a Carnot cycle operating between the
same values of high and low temperatures.
•
Efficiency of a component can be calculated by comparing the work
produced by the component to the work that would have been
produced by an ideal component operating isentropically between
the same inlet and outlet conditions.
•
An isentropic expansion or compression process is represented as a
vertical line on a T-s or h-s diagram. A real expansion or
compression process looks similar, but is slanted slightly to the
right.
•
Maximizing the ΔT and ΔP between the source and the heat sink
ensures the highest possible cycle efficiency.
•
The second law of thermodynamics gives a maximum efficiency
limit (which is never reached in physical systems) that an ideal
thermodynamic system can perform. The efficiency is determined
by knowing the inlet and exit absolute temperatures of the overall
system and applying Carnot's efficiency equation.
•
Cycle efficiency = 1 − (𝑇 𝐶 ) (temperature in degrees R)
𝑇
𝐻
3. Review ELO 2.3 by having students plot their plant's steam cycle on a
T-s diagram.
•
A T-s diagram is frequently used to analyze energy transfer
system cycles. Work done by or on the system and heat added
to or removed from the system can be visualized on the T-s
diagram.
•
Use the following as a guideline:
𝜂=
𝑇𝐻 − 𝑇𝐶
𝑇𝐻
4. Review ELO 2.4 by having the students list the major components of
the steam cycle and discuss operating conditions that improve the
plant’s efficiency.
80
Rev 1
Figure: Typical Steam Cycle and T-s Diagram
•
Discuss conditions yielding improved cycle efficiency, shown below
in the table:
Improved Cycle Efficiency
Condition
Effect
Discussion
Superheating
More Efficient With
More Superheating
Increased heat added
results in more net work
from the system, even
though more heat is
rejected.
Moisture Separator
Reheater (MSR)
Use of MSR Has
Minor Effect On
Efficiency
More work is done by the
low-pressure (LP) turbine
since inlet enthalpy is
higher but more heat is
rejected.
The principle benefit of
MSR use is protection of
the final blading stages in
LP turbine from water
droplet impingement.
Feedwater Preheating More Efficient With
Less heat must be added
Feedwater Preheating from the heat source
(reactor) since the
feedwater enters the
steam generator closer to
saturation temperature.
Rev 1
81
Condition
Effect
Discussion
Condenser Vacuum
More Efficient With
Higher Vacuum
(Lower
Backpressure)
Net work output is higher
and heat rejection is
lower as condenser
pressure is lowered.
Condensate
Depression
More Efficient With
Minimal Condensate
Depression
Minimal condensate
depression reduces both
the amount of heat
rejected and the amount
of heat that must be
supplied to the cycle.
Steam
More Efficient At
At higher steam
Temperature/Pressure Higher Steam
temperature, the inlet and
Temperature/Pressure exit entropy from the
turbine are lower so less
heat is rejected.
Steam density increases
as pressure increases, so
more turbine work is
done.
Steam Quality
Enthalpy content
increases as moisture
content decreases and
more net work is done.
•
Hotwell is the area at the bottom of the condenser where the
condensed steam collects to pump back into the system feedwater.
•
Condensate depression is the amount the condensate in a condenser
that is cooled below saturation (degrees subcooled).
•
Condensers operate at a vacuum to ensure the temperature (and thus
the pressure) of the steam is as low as possible.
•
Causes of decreased efficiency include the following:
—
—
—
82
More Efficient At
Higher Steam
Quality
Presence of friction
Heat losses
Cycle inefficiencies
– Subcooling
– Tsat of the steam generator
Rev 1
—
Turbine service lifetime is affected by moisture impingement
on the blades and other internal parts
— Removing as much moisture from the steam limits moisture
content at every stage of the turbine
— Feedwater heater is a power plant component used to preheat
water delivered to a steam generating boiler
— Moisture separator reheaters improve the plant’s efficiency
and are used to avoid the erosion corrosion and droplet
impingement erosion in the LP turbine, to remove moisture,
and to superheat the steam
Objectives
Now that you have completed this lesson, you should be able to do the
following:
1. Explain the second law of thermodynamics using the term entropy.
2. Given a thermodynamic system, determine the:
a. Maximum efficiency of the system
b. Efficiency of the components within the system
3. Differentiate between the path for an ideal process and that for a real
process on a T-s or h-s diagram.
4. Describe how individual factors affect system or component
efficiency.
Thermodynamic Processes Summary
In this module, you learned about applying the First and Second Laws of
Thermodynamics to processes, systems, diagram principles, and energy
balances on major components within a nuclear power generation plant or
facility.
During this lesson, you learned about the First Law of Thermodynamics,
which states that energy can be neither created nor destroyed, but only
altered in form. The energy forms may not always be the same but the total
energy in the system remains constant. You learned about open, closed,
isolated, and steady flow systems. You studied processes including
thermodynamic, cyclic, reversible, irreversible, adiabatic, isentropic, and
isenthalpic.
All of the energies entering and leaving the control volume boundary, any
work done on or by the control volume, and any heat transferred into and
out of the control volume boundaries are in the energy balance equation.
The Second Law of Thermodynamics states that it is impossible to construct
a device that operates within a cycle that can convert all the heat supplied it
into mechanical work. Recognizing that even the most thermally and
mechanically perfect cycles must reject some heat defines thermodynamic
power cycle efficiency.
Rev 1
83
You studied entropy—a measure of the unavailability of heat to perform
work in a cycle—to explain the second law, and that the change in entropy
determines the direction a given process will proceed.
The Carnot cycle represents an upper limit of efficiency for any given
system operating between the same two temperatures since all practical
systems and processes are irreversible. The system's maximum possible
efficiency would be that of a Carnot cycle, but because Carnot cycles
represent reversible processes, the real system cannot reach the Carnot
efficiency value. Thus, the Carnot efficiency serves as an unattainable
upper limit for any real system's efficiency. You studied the Rankine cycle
is an ideal cycle where no increase in entropy occurs as work is done on and
by the system.
You learned about the thermodynamics of ideal and real systems by
arranging the various thermodynamic processes on a property diagram to
evaluate the various cycles present in a nuclear power plant. The most
common set of coordinates used is a plot of temperature versus specific
entropy is a T-s diagram. You also studied an h-s diagram, which compares
systems and determines their efficiencies, and a Mollier diagram, which
plots and illustrates the energy transferred to or from the steam during the
cycle.
Now that you have completed this module, you should be able to
demonstrate mastery of this topic by passing a written exam with a grade of
80 percent or higher on the following TLOs:
1. Apply the First Law of Thermodynamics to analyze
thermodynamic systems and processes.
2. Apply the Second Law of Thermodynamics to analyze real and
ideal systems and components.
84
Rev 1