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Revision 2
June 2016
Thermodynamic
Processes
Student Guide
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Table of Contents
INTRODUCTION ..................................................................................................................... 2
TLO 1 FIRST LAW OF THERMODYNAMICS ............................................................................ 2
Overview .......................................................................................................................... 2
ELO 1.1 Thermodynamic Systems and Processes ........................................................... 3
ELO 1.2 Applying the First Law of Thermodynamics .................................................. 12
ELO 1.3 Identifying Process Paths on a T-s Diagram ................................................... 26
ELO 1.4 Thermodynamic Energy Balances on Major Components ............................. 37
ELO 1.5 Throttling Characteristics ................................................................................ 47
TLO 1 Summary ............................................................................................................ 54
TLO 2 THE TURBINE AND CONDENSER PROCESSES ........................................................... 56
Overview ........................................................................................................................ 56
ELO 2.1 Nozzle Characteristics ..................................................................................... 57
ELO 2.2 Condenser Design and Characteristics ............................................................ 64
ELO 2.3 Turbine Design and Characteristics ................................................................ 72
TLO 2 Summary ............................................................................................................ 86
THERMODYNAMIC PROCESSES SUMMARY .......................................................................... 87
KNOWLEDGE CHECK ANSWER KEY ...................................................................................... 1
ELO 1.1 Thermodynamic Systems and Processes ........................................................... 1
ELO 1.2 Applying the First Law of Thermodynamics for Open Systems ...................... 2
ELO 1.3 Identifying Process Paths on a T-s Diagram ..................................................... 3
ELO 1.4 Thermodynamic Energy Balances on Major Components ............................... 6
ELO 1.5 Throttling Characteristics .................................................................................. 8
ELO 2.2 Condenser Design and Characteristics .............................................................. 9
ELO 2.3 Turbine Design and Characteristics ................................................................ 13
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Thermodynamic Processes
Revision History
Revision
Date
Revision
Number
Purpose for Revision
Performed
By
11/6/2014
0
New Module
OGF Team
12/11/2014
1
Added signature of OGF
Working Group Chair
OGF Team
6/1/2016
2
Revised as part of the PPT
upgrade project
OGF Team
Rev 2
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.
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.
This module will focus on applying the first laws of thermodynamics to
analyze systems and processes and to describe the thermodynamic
processes associated with a steam turbine and condenser.
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. Describe the operation of the turbine and condenser processes.
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
equipment is not at its maximum efficiency and the best solution to return
maximum efficiency.
Rev 2
2
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Define the following as they apply to a thermodynamic process:
a.
b.
c.
d.
e.
f.
Open, closed, or isolated
Reversible (ideal) process
Irreversible (real) process
Adiabatic process
Isentropic process
Isenthalpic (throttling) process
2. Apply the First Law of Thermodynamics for open systems or
thermodynamic processes.
3. Identify the path(s) on a T-s diagram that represents the
thermodynamic processes occurring in a fluid system.
4. Given a defined system, perform energy balances on all major
components in the system.
5. Determine exit conditions for a throttling process.
ELO 1.1 Thermodynamic Systems and Processes
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.
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.
Rev 2
3
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.
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
Rev 2
4
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.
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.
Rev 2
5
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:
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:
or
Reactor Coolant
Figure: Steady Flow Systems
Rev 2
6
Knowledge Check (Answer Key)
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 (Answer Key)
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
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.
Rev 2
7
Reversible (Ideal) 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.
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 (Real) 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.
Rev 2
8
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.
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.
Rev 2
9
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
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.
Rev 2
10
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.
Isenthalpic (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:
(Where P = pressure and vel =
velocity)
These observations confirm the theory that:
Remember:
(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.
Rev 2
11
Knowledge Check (Answer Key)
Steam flowing through the main turbine control valve is
a(n) …?
A.
isenthalpic process
B.
isentropic process
C.
adiabatic process
D.
reversible process
ELO 1.2 Applying the First Law of Thermodynamics
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.
Rev 2
12
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.
Figure: Basic Energy Balance of the First Law of Thermodynamics
Therefore, in equation form, the balance appears as indicated on the figure
below:
Where:
•
•
= heat flow into the system (British Thermal Units per hour
[BTUs/hr])
= 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])
•
= kinetic energy into the system (ft-lbf/lbm)
•
= average velocity of fluid (feet per second [ft/sec])
•
Rev 2
gc = the gravitational constant (32.17 ft-lbm/lbf-sec2)
13
•
= potential energy of the fluid entering the system (ftlbf/lbm)
Where:
•
= 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)
•
= 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)
•
= kinetic energy out of the system (ft-lbf/lbm)
•
= potential energy of the fluid exiting the system (ftlbf/lbm)
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
transferred into and out of the control volume boundaries are accounted for
in the energy balance equation.
Rev 2
14
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. In the figure shown below, you see six
thermodynamic processes including:
1.
2.
3.
4.
5.
6.
Transfer heat into the system.
Transfer heat out of the system.
Perform work on the system.
Perform work by the system.
Transfer mass into the system.
Transfer mass out of the system.
We will focus on four of them related to the processes in our plants. They
are:
1. Transfer heat into the system (the steam generator process where
subcooled feedwater is pumped into the steam generator, heat added,
and saturated steam leaves).
2. Transfer heat out of the system (the condensing process where wet
vapor goes into the condenser, heat is removed, the subcooled
condensate goes out).
3. Perform work on the system (the pump process where subcooled
condensate goes in the system, work is don ON the fluid, and
subcooled feedwater goes out).
4. Perform work by the system (the turbine process where saturated
steam goes into the turbine, work is done BY the turbine, and wet
vapor exits).
Since our system is essentially closed, the processes related to transfer of
mass either in or out are (below) not applicable.
5. Transfer mass into the system.
6. Transfer mass out of the system.
Figure: Six Basic Processes of Steady Flow Systems
Rev 2
15
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
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.
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).
Rev 2
16
Enthalpy has been previously defined as
. 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:
Where:
•
= 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
(ft-lbf/lbm)
•
KEout = specific kinetic energy of working fluid leaving the system
(ft-lbf/lbm)
•
= rate of work done by the system (ft-lbf/hr)
•
= 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.
Rev 2
17
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.
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.
Rev 2
18
Analyzing Open Systems and Thermodynamic Processes
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.
Thermodynamic Processes for Steam Generators, Turbines,
Condensers, and Pumps
Figure: Multiple Control Volumes in the Same System
The following discussions are based on the above figure:
Assumptions
•
Each process initially presented as IDEAL
– Minimal change to KE or PE
o Recall, 778 ft-lbf = 1 BTU
Rev 2
19
•
For heat transfer processes
– No work done ON or BY
– Insulated, so no heat losses
•
For work processes
– No heat lost or gained (no change in specific entropy)
•
Approximate values for specific enthalpy presented for each process
Steam Generator Process
•
Two flowpaths – (Primary flowpath, Secondary flowpath)
– Primary flowpath - heat transferred out of RCS
o Thot in, Tcold out
– Secondary flowpath – heat transferred into SG
o Subcooled feedwater in
– ≈ 420oF, ≈ 400 BTU/lbm
o Saturated steam out
– ≈ 1000 psia, ≈ 1200 BTU/lbm
•
Energy in (FW) plus heat added (SG) equals Energy out (steam)
– hfw + hSG = hstm
– 400 BTU/lbm + 800 BTU/lbm = 1200 BTU/lbm
Turbine Process
•
Inlet to the turbine process is the outlet of the SG process
– ≈ 1200 BTU/lbm (Energyin)
•
Work is done BY the system
– ≈ 400 BTU/lbm
•
Energy out based on condenser vacuum
– 1 psia or 28 in Hg equates to ≈ 800 BTU/lbm
•
Energy in (steam) minus work done (turbine) equals Energy out
(exhaust)
– hstm - hturbine = hexhaust
– 1200 BTU/lbm - 400 BTU/lbm = 800 BTU/lbm
Rev 2
20
Condenser Process
•
Inlet to the condenser process is the outlet of the turbine process
– ≈ 800 BTU/lbm (Energyin)
•
Heat is removed FROM the system
– ≈ 740 BTU/lbm
•
Energy out based on condenser vacuum
– Slightly subcooled condensate (based on 1 psia
backpressure)
o 92oF is ≈ 60 BTU/lbm
•
Energy in (exhaust) minus heat removed (condenser) equals Energy
out (condensate)
– hexhaust - hcondenser = hcondensate
– 800 BTU/lbm - 740 BTU/lbm = 60 BTU/lbm
Pump Process
•
Inlet to the pump process is the outlet of the condenser process
– ≈ 60 BTU/lbm (Energyin)
•
Work is done ON the system
– Pump(s) must not only make up for headloss but also raise
pressure to enter SG
– Enthalpy added by compression is minimal
– Most energy added by feedwater heating
•
Energy in (condensate) plus work done ON system (pump) AND
heat added to system (FW heating) equals Energy out (feedwater)
– hcondensate + hpump/FW heating = hfeedwater
– 60 BTU/lbm - 340 BTU/lbm = 400 BTU/lbm
Rev 2
21
Examples:
Example 1: Open System Control Volume
Tip
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
1. Divide by m since:
Where:
•
q = heat added to the system per pound (BTU/lbm)
•
w = work done by the system per pound (ft-lbf/lbm)
Rev 2
22
2. Use Joule’s constant,
known values.
for conversions and substitute
Note: The minus sign indicates heat out of the turbine.
3. Solve for work (w):
Rev 2
23
Example 3: Heat and Work
Tip
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.
Heat removed is a negative value and work done on a
system is a negative value.
Note
We must convert all of the energy values to the same
units.
Now,
substituting into and solving the equation gives:
Rev 2
24
Knowledge Check (Answer Key)
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.
A.
positive, negative, negative, positive
B.
negative, positive, positive, negative
C.
positive, negative, positive, negative
D.
negative, positive, negative, positive
Knowledge Check (Answer Key)
In an Open Steady Flow System, choose the energies that
are associated with the mass crossing the system
boundary.
Rev 2
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
25
ELO 1.3 Identifying Process Paths on a T-s Diagram
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
Iso is a constant or one.
Glossary
Figure: T-s Diagram With Rankine Cycle
Rev 2
26
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 where liquid is converted 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.
Figure: Typical Steam Plant Cycle
Rev 2
27
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.
An example using typical values follows (Values used in this example are
approximations and may vary at your plant. Also, it will be assumed that
the moisture separators are 100% efficient and the steam exiting is 100%
saturated steam).
Point 1:
 Subcooled feedwater at 1000 psia and 420oF enters SG
 Sensible and latent heat added from RCS (QA)
 Steam exits at ≈ 100% saturated steam
o Phase change
 Area under this curve is the total heat added
Rev 2
28
Point 2:
 100% saturated steam at 1000 psia enters turbine
o ≈ 1200 BTU/lbm
 Exits as a wet vapor
o ≈ 67% quality
o ≈ 800 BTU/lbm
 On IDEAL turbine, no Ds
 On REAL turbine, increase in s
o Less work out of steam, higher quality
Note: Regarding efficiencies, as an example, if the secondary cycle is about
33% efficient and 1200 BTU/lbm enter the turbine (heat added from RCS),
then about 400 BTU/lbm of work is done by the turbine, and 800 BTU/lbm
remain. 400/1200 = 0.33.
Also if the turbine is only getting about 1/3 of the energy out of the
saturated steam, then it stands to reason that the exhaust still has about 2/3
of the quality in it (started with 100% quality, ends with 67% quality).
Rev 2
29
Point 3:
 Turbine process can also be shown on a Mollier Diagram
 Find starting pressure (1000 psia)
 IDEAL work (no Ds)
o Draw line straight down to end pressure
 Condenser pressure of 1 psia
 REAL work (increase in s)
o Draw line down to end pressure, but at higher final enthalpy
 Still exhausts to 1 psia
Rev 2
30
Point 4:
 Turbine process can also be shown on a Mollier Diagram
 Find starting pressure (1000 psia)
 IDEAL work (no Ds)
o Draw line straight down to end pressure
 Condenser pressure of 1 psia
 REAL work (increase in s)
o Draw line down to end pressure, but at higher final enthalpy
 Still exhausts to 1 psia
Rev 2
31
Point 5:
 ≈ 67% quality wet vapor enters condenser
 Exits to hotwell as slightly subcooled condensate
o Called condensate depression
 Undergoes phase change
o Subcooling – lower efficiency
o Subcooling - better for pumps
 Heat rejected to Circ Water system
 Area under this curve is the total heat rejected
Note that the area between the “Heat Added” and the “Heat Rejected” is the
“work of the turbine”.
One of the ways to maximize efficiency of our thermodynamic cycle is to
maximize the DP of the turbine process. The higher the SG pressure and
the lower the backpressure (greater the vacuum), the more efficient the
cycle.
Another way to maximize efficiency is to minimize the amount of
subcooling in the condensate (condensate depression). This does, however,
make the pump operate closer to cavitation.
Rev 2
32
Point 6:
 Enters as subcooled condensate
o About 1 psia
 Exits to SG as subcooled feedwater
o About 1000 psia
 IDEAL work of pump
o All energy added as pressure
 Real work of pump
o Most added as pressure
o Some energy raises temperature
Knowledge Check (Answer Key)
Which one of the following will cause overall nuclear
power plant thermal efficiency to increase?
Rev 2
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.
33
Knowledge Check (Answer Key)
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 (Answer Key)
On a T-s diagram, if the temperature of the heat sink is
lowered, the efficiency of the cycle will _______ because
________.
A.
increase, more heat is rejected
B.
increase, less heat is rejected
C.
decrease, more heat must be added
D.
decrease, less heat must be added
Knowledge Check (Answer Key)
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 _____________ possible.
Rev 2
A.
as subcooled as practical; great
B.
as subcooled as practical; small
C.
close to saturation; great
D.
close to saturation; small
34
Knowledge Check (Answer Key)
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 (Answer Key)
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…
Rev 2
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.
35
Knowledge Check (Answer Key)
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…
A.
Increase because condensate depression will decrease.
B.
Decrease because condensate depression will increase.
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 (Answer Key)
Which one of the following actions result in a decrease in
overall nuclear power plant thermal efficiency?
Rev 2
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
36
ELO 1.4 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:
Rev 2
37
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 if needed.
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.
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.
Rev 2
38
Figure: Steam Generation System
Step 2:
Write the general energy equation simplified for this component.
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:
Step 4:
Substituting
Rev 2
, where cp = specific heat capacity (BTU/lbm-°F):
39
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:
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?
Solution:
Step 1:
Draw the system with boundaries.
Figure: Pump Returns Fluid from Heat Exchanger to Core
Rev 2
40
Step 2:
Write the general energy equation:
Step 3:
Simplify the equation.

Assume
, and neglect changes in PE and KE

where
is the rate of doing work by the pump
(vP)

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.
Step 4:
Arrange the correct terms:

Substituting the expression for work,

Using 0.01246 for specific volume:
we have:
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.
Rev 2
41
Thermodynamic Balance Across Heat Exchangers
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
, where cp = specific heat capacity:
For this example,
has been used to calculate the heat transfer
rate since no phase change has occurred. However,
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.
Rev 2
42
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
= pump work
= heat produced by the heat source
= heat transferred into steam generator
Rev 2
43
Steps 2 & 3: Write and simplify the equation:
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:
Step 4:
Arrange equation for the proper units.
Recall that 1 hp = 2,545 BTUs/hr. Therefore, converting to 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:
(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
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)
Rev 2
44
Determine the required cooling water flow rate.
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:
Steps 3 and 4:
Simplify the equation and arrange for required units.
Rev 2
45
All of the heat removed from the steam (stm) condensing and subcooling is
transferred to the circulating water (cw) system.
was required in this example since a phase change occurred.
when the steam was condensed to water. Since ΔT = 0 for a phase
change, Q = ṁcp𝛥T would not work. 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 (Answer Key)
Why can’t a “ΔT” formula be used in all heat transfer
systems in the plant?
Rev 2
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.
46
Knowledge Check (Answer Key)
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?
A.
101 Megawatts (Mw)
B.
126 Mw
C.
1,006 Mw
D.
1,258 Mw
Knowledge Check (Answer Key)
The rate of heat transfer between two liquids in a heat
exchanger will increase if the … (Assume specific heats
do not change.)
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.
ELO 1.5 Throttling Characteristics
Throttling is the process of restricting full flow through a restrictor, such as
an orifice or partially opened valve. The restriction causes a drop in fluid
pressure and a corresponding increase in velocity. This change takes place
without work interactions or changes in kinetic energy or potential energy.
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) during a throttling
process. This section will compare what we can observe with the above
theoretical assumptions to increase understanding of the theory of the ideal
throttling process.
Rev 2
47
In the figure below, we can observe that: Pin > Pout, velin < velout, (where P =
pressure and vel = velocity). Recall h = u + Pv (v = specific volume), so if
pressure decreases then specific volume must increase if enthalpy is to
remain constant (assuming u is constant). Because mass flow is constant,
the change in specific volume causes an increase in velocity.
The throttling process has a constant enthalpy with a large change in
entropy. The downstream fluid flow is somewhat turbulent from the
process.
Figure: Throttling Process by a Valve
Rev 2
48
Determining Downstream Properties Step-by-Step Table
Step
Action
1.
First, determine the condition upstream of the throttle or leak
(temperature, pressure (psia), quality or superheating). This is
usually given in the problem.
2.
Find the corresponding beginning point on the Mollier diagram
or in the steam tables.
3.
Determine the downstream pressure in psia. This is normally
given in the problem is some form.
4.
Draw a horizontal line from the initial condition point (constant
enthalpy) to the intersection of the constant pressure line for the
downstream pressure. The final condition is established by this
point (temperature, quality or superheating) (See diagram below
for explanation), or in the steam tables find the corresponding
enthalpy at the downstream pressure and determine other
properties required.
The diagram below shows step 4 above. Find initial point 1 and draw a
horizontal line until it intersects the downstream pressure which could be a
wet vapor under the dome (2) or superheated above the dome (3).
Figure: Throttling Process on a Mollier Diagram
Rev 2
49
Throttling Process Demonstration
In performing an analysis of the throttling process, we again assume steady
flow conditions (m1 = m2). We also select boundary locations sufficiently
away from the throttling location for flow to have returned to a stable,
uniform flow condition. With these conditions, we can analyze the process
as follows:
The elevation change from boundary 1 to boundary 2 is insignificant.
Inlet piping and outlet piping diameter are equal and there is no change in
fluid velocity.
There is no work done on or done by the fluid as it flows through the
throttle.
Assume insulation on the piping, so there is no heat transferred into or out
of the fluid.
This gives us the following results for a throttling process:
As shown in the figure on the next page, enthalpy remains constant while
entropy increases, the process does no work (J=joules), and no heat is
added. The result is a pressure drop and slight velocity increase.
Rev 2
50
Figure: Property Diagrams of a Throttling Process
Throttling can be beneficial, particularly in controlling flow rate to maintain
desired conditions in a system. However, the nature of the process (that is,
constant enthalpy) must be understood in order to recognize throttling
conditions. Failure to understand the downstream tailpipe temperature
indications of a Power Operated Relief Valve contributed to the events of
the Three Mile Island accident that changed the face of the nuclear industry
in the United States.
Ensure the students are comfortable working with throttling process
problems using the Mollier Diagram and the steam tables. NRC exams test
this topic heavily, as it is one of only two K/As in this chapter which have
related questions (value >2.5).
Rev 2
51
Example 1
A power-operated relief valve is stuck open at 2,200 psia in the pressurizer.
The valve is discharging to the pressurizer relief tank at 25 psig. What is
the temperature of the fluid downstream of the relief valve?
On the Mollier diagram, go to the 2,200-psia point on the saturation line.
Cross the constant enthalpy line (throttling is a constant enthalpy process) to
the 40 psia line (25 psig + 15 psi atmospheric = 40 psia). Follow that line
up to the saturation curve. The constant temperature line that ends at that
point on the curve establishes the temperature of the fluid. The temperature
is approximately 270F. The table below presents these steps in tabular
form.
Step
Action
1.
Determine the condition upstream of the throttle or leak
(temperature, pressure (psia), quality or superheating)
2.
Find the corresponding point on the Mollier diagram. (2,200
psia, saturated vapor)
3.
Determine the downstream pressure in psia. (25 psig = 40 psia)
4.
Go from the initial condition point along a horizontal line
(constant enthalpy) to the intersection with the constant pressure
line for the downstream pressure. This point establishes the final
condition. (temperature, quality or superheating)
Example 2
The RCS is operating at 2,185 psig. What would be the expected tailpipe
temperature of a leaking pressurizer safety valve assuming downstream
pressure is 35 psig? (Also, assume that the steam quality is 100 percent in
the pressurizer)
Solution:
From the Mollier diagram, the final condition is a mixture. Therefore, the
tailpipe temperature must be at the saturation temperature corresponding to
the pressure.
From steam tables,
Rev 2
52
Knowledge Check (Answer Key)
Which one of the following is essentially a constantenthalpy process?
A.
Throttling of main steam through main turbine steam
inlet valves
B.
Condensation of turbine exhaust in a main condenser
C.
Expansion of main steam through the stages of an ideal
turbine
D.
Steam flowing through an ideal convergent nozzle
Knowledge Check (Answer Key)
A nuclear power plant is maintained at 2,000 psia with
a pressurizer temperature of 636°F. A pressurizer
relief safety valve is leaking to a collection tank which
is being held at 10 psig. Which one of the following is
the approximate temperature of the fluid downstream
of the relief valve?
Rev 2
A.
280°F
B.
240°F
C.
190°F
D.
170°F
53
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:

Define the following terms as they relate to a thermodynamic
process:
o Open
o Closed
o Isolated
 A reversible process is a process that can be reversed resulting in
Define the following processes, which can be described by any of
the following terms:
o A reversible process is a process that can be reversed
resulting in no change in the system or surroundings.
o An irreversible process is a process that if reversed results in
a change to the system or surroundings.
o An adiabatic process is a process in which there is no heat
transfer across the system boundaries.
o An isentropic process is a process in which the entropy of the
system remains unchanged.
o A throttling process is a process in which enthalpy is
constant (h1 = h2), work = 0, and which is adiabatic, Q = 0.
2. Review ELO 1.2:

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.
Rev 2
54
3. Review ELO 1.3:
• 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.
4. Review ELO 1.4:


Review the steps for solving energy balance problems.
Review the use of the continuity equation:
• 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.
5. Review ELO 1.5:

Review the definition of the throttling process from a
thermodynamic perspective.
 Review the throttling process (and an example) as shown on a T-S
diagram.
Rev 2
55
Objectives
Now that you have completed this lesson, you should be able to do the
following:
1. Define the following as they apply to a thermodynamic process:
a.
b.
c.
d.
e.
f.
Open, closed, or isolated
Reversible (ideal) process
Irreversible (real) process
Adiabatic process
Isentropic process
Isenthalpic (throttling) process
2. Apply the First Law of Thermodynamics for open systems or
thermodynamic processes.
3. Identify the path(s) on a T-s diagram that represents the
thermodynamic processes occurring in a fluid system.
4. Given a defined system, perform energy balances on all major
components in the system.
5. Determine exit conditions for a throttling process.
TLO 2 The Turbine and Condenser Processes
Overview
There are many thermodynamic processes occurring continuously as a plant
operates; many of these processes are invisible to the operator. The plant
equipment incorporates these processes in the plant design to ensure the
plant operates properly. For instance, the operator can monitor the steam
properties at the inlet to the main turbine and the properties of steam at the
main turbine exhaust, but cannot see the thermodynamic process that occurs
within the turbine. The turbine extracts work from the steam (converting
the thermal energy of the steam into mechanical energy). Likewise, the
condenser takes exhaust steam from the turbine, extracts heat from it,
converting the vapor to liquid and causing the formation of a vacuum in the
condenser which increases the efficiency of the turbine and overall system.
In this case, the operator can monitor exhaust steam temperature, condenser
vacuum, and condensate temperature but cannot see the thermodynamic
process that is occurring.
Importance
These thermodynamic processes are important in understanding the
operation and design of the complete power plant and this module will
explain the processes.
Rev 2
56
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Describe the operation of nozzles to include the functions of nozzles
in flow restrictors and the functions of nozzles in air ejectors.
2. Describe the condensing process to include vacuum formation and
condensate depression
3. Explain the design of turbines, including the functions of nozzles,
fixed blading, moving blading, and the reason turbines are multistage.
ELO 2.1 Nozzle Characteristics
Introduction
A nozzle is a mechanical device used to change the energy of a working
fluid from one form to another for a specific purpose. The cross-sectional
area in a nozzle varies to control the flowrate, speed, direction, mass, shape,
and/or the pressure of the stream that emerges from them. Depending on
the type of nozzle, the kinetic energy of the fluid will increase or decrease
as it moves through the device. The figure below shows three of the
common types of nozzles.
Figure: Typical Nozzle Types
There are two types of nozzles: convergent and divergent. The convergent
nozzle cross-section narrows from a wide diameter to a smaller diameter in
the direction of the flow and accelerates the fluid. In the divergent nozzle
cross-section, the diameter expands and slows the fluid upon exit. A
convergent-divergent nozzle has a convergent section followed by a
divergent section and is often called a De Laval nozzle.
Rev 2
57
Theory of Nozzle Operation
We will use the General Energy Equation, with several simplifications, to
explain nozzle operation.
In a convergent nozzle, the piping at the exit is of a smaller diameter than
the entrance (A1 > A2).
Figure: Convergent Nozzle
The elevation change from entrance (1) to exit (2) is insignificant.
Inlet piping diameter is greater than outlet piping diameter. With steady
flow, outlet velocity is greater than inlet velocity.
There is no work done on or done by the fluid in the nozzle.
Assume that no heat transfers into or out of the fluid as it flows through the
nozzle.
Assume that there is no friction as the fluid flows through the nozzle.
Rev 2
58
Where:
= mass flow rate (lbm/sec)
A = cross-sectional flow area (ft2)
V = fluid velocity (ft/sec)
 = specific volume of fluid (ft3/lbm)
ρ = density of fluid (lbm/ft3)
Note: some texts use the symbol V with an overbar to denote velocity or
average velocity; in this module, we will use V or Vel to denote velocity.
The last equation above is termed the continuity equation for steady flow
processes. If we assume that the specific volume (and therefore density) is
constant (incompressible fluid) the velocity must increase if the crosssectional flow area decreases and vice versa. A converging nozzle increases
a fluid's velocity and kinetic energy at the expense of decreasing its
enthalpy (pressure). A diverging nozzle decreases the velocity and
increases pressure at its exit.
Application of the first law of thermodynamics shows that the change in KE
must balance with an opposite change in another stored energy form. With
the assumptions given, we can see that the Pv energy must decrease if KE is
increased. If we also assume that the fluid is incompressible (
) we
can see that the change in pressure (ΔP) is proportional to the change in KE.
The above equations show that a nozzle can exchange kinetic energy and
pressure volume energy as desired. For a convergent nozzle, the velocity
and kinetic energy increase and pressure decreases. The opposite effect
takes place for a divergent nozzle - the kinetic energy decreases as fluid
velocity decreases and Pv energy increases.
If the working fluid is steam, the specific volume is not constant since the
fluid is compressible. In this case we leave the internal energy term in the
equation and the result is:
Rev 2
59
The thermodynamic process of an ideal nozzle is adiabatic with no loss of
energy. (As defined by Princeton University: “In thermodynamics, an
adiabatic process or an isocaloric process is a thermodynamic process in
which no heat is transferred to or from the working fluid. The term
‘adiabatic’ literally means impassable.”) In a real nozzle, entropy will
increase slightly due to turbulence and friction losses.
Convergent nozzles accelerate subsonic fluids. Fluid moves through a
nozzle as a function of the differential pressure from the nozzle inlet to
outlet. The design shape of the nozzle will determine the critical pressure,
which is the outlet pressure that will cause supersonic flow in the throat of
the nozzle. In a convergent nozzle, the sonic flow creates turbulence and
backpressure that limits flow; this is termed choking flow. If the nozzle
critical pressure ratio (ratio of pressure that will cause sonic velocity to
pressure at inlet) is high enough, the flow will reach sonic velocity at the
narrowest point, the nozzle throat. In this situation, the nozzle is termed
choked. Increasing the nozzle pressure ratio above the critical pressure will
not increase the throat velocity. If the downstream flow is free to expand
smoothly as in a divergent nozzle, supersonic velocities can be reached.
Figure: Supersonic Flow through a Convergent-Divergent Nozzle
If the velocity reaches sonic conditions at the venturi, the increasing area of
the divergent nozzle causes the velocity to increase further and the pressure
to decrease further. If the velocity does not reach sonic conditions at the
venturi, then the velocity will begin decreasing and the pressure will begin
increasing.
Convergent-divergent nozzles can therefore accelerate fluids that have
choked in the convergent section to supersonic speeds. This convergentdivergent process is more efficient than allowing a convergent nozzle to
expand supersonically externally. The shape of the divergent section also
ensures that the direction of the escaping gases is directly backwards, as any
sideways component would not contribute to thrust.
Rev 2
60
Flow Restrictors
If steam generator piping includes a convergent-divergent nozzle as shown
in the figure below, it could provide a method of measuring flow by using
the ∆P between the inlet and outlet of the convergent section. In case of a
main steam line rupture, nozzles will limit (choke) flow of steam to limit
pipe whip and impingement damage. Controlling flow of steam is
important in this instance due to power excursion created by a suddenly
large steam demand.
Figure: Convergent-Divergent Venturi Tube for Flow Measurement
Steam Jet Air Ejectors
An air ejector is a pump-like device, with no moving parts that utilizes
high-pressure steam to compress vapors or gases as illustrated in the figure
below. High-pressure steam enters a convergent nozzle; the steam exits the
nozzle at increased velocity and decreased pressure. The nozzle creates a
low-pressure area that will draw in the fluids around it and mix those with
the high velocity fluid in the throat of the device. The mixture passes
through a divergent section to decrease the velocity of the mixture and
increase its discharge pressure. A jet pump works similarly, except water
functions as both the driving fluid and the entrained fluid. This allows the
jet pump to increase the mass of liquid being pumped without additional
pumps or electrical power.
Figure: Simple Air Ejector (Jet Pump)
Rev 2
61
Steam enters a convergent-divergent nozzle at a relatively high pressure and
low velocity in a steam jet air ejector shown below. The convergent nozzle
increases the velocity of the steam to sonic velocity (1) and then to
supersonic velocity in the divergent section (2) as it enters into the suction
chamber. The supersonic velocity results in the lowest pressure of the
steam in the nozzle (2). The condenser feeds the suction chamber. The
low-pressure area will draw more fluid from around the nozzle into the
throat of the diffuser, entraining the drawn fluid with the driving fluid. The
decreasing area slows the supersonic velocity back to sonic and increases
the pressure (3) as the fluid moves through the convergent-divergent
diffuser section. The divergent section of the diffuser then drops the
velocity to subsonic and increases the pressure enough to discharge to
atmosphere or greater (4). Use of steam at a pressure between 200 psi and
300 psi as the high-pressure fluid enables a single-stage air ejector to draw a
vacuum of about 26 inches Hg.
The low-pressure area in the suction chamber draws air and noncondensable gases into the nozzle. Momentum transfers from the steam to
the air and non-condensable gases and they become "entrained" in the steam
flowing through the air ejector.
The mixture of steam and gas reaches sonic velocity in the suction chamber.
Sonic velocity is the speed of sound in that substance. The nozzle design
allows the steam to accelerate to super-sonic velocity resulting in a lower
pressure at the suction chamber. As the air and non-condensable gases exit
the suction chamber, the nozzle maintains the low-pressure suction area,
which allows more air and non-condensable gases to enter from the suction
line. As the mixture enters the diffuser section, the diverging nozzle slows
down the flow while increasing its pressure. The figure below shows a
cross-section of a steam jet air ejector with pressure and velocity levels
along the section.
Figure: Steam Jet Air Ejector Operation
Rev 2
62
Vacuums of 29 inches Hg require two stages of air ejection.


First Stage - suction located on top of the condenser
Second Stage - suction comes from the first stage diffuser
The exhaust steam from the second stage passes through an air ejector
condenser cooled by condensate to condense the steam, as shown in the
figure below. The air ejector condenser also preheats the condensate
returning to the boiler.
Figure: Two-Stage Steam Jet Ejector
Assume steady flow conditions from the entrance into the first stage air
ejector to the exit at the second stage air ejector, which means no mass is
lost or gained through the nozzle and mass flow rate is constant.
Knowledge Check (Answer Key)
In what type of nozzle will flow be limited to sonic?
Rev 2
A.
Convergent
B.
Divergent
C.
Convergent-Divergent
D.
Divergent-Convergent
63
ELO 2.2 Condenser Design and Characteristics
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
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).
Rev 2
64
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,
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.
As can be seen on the T-s diagram on the next page, condensate depression
decreases the plant’s operating efficiency because the subcooled condensate
is reheated in the steam generator, requiring more heat or energy from the
heat source. Condensate depression decreases overall cycle efficiency.
Excessive condensate depression also allows increased absorption of air by
the condensate and accelerated oxygen corrosion of plant materials.
Rev 2
65
Figure: T-s Diagram for a 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,
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.
Rev 2
66
Example:
Determine the quality of the steam entering a condenser operating at:
• 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
• Steam mass flow rate in the condenser is 8 x 106 lbm/hr and the
circulating water is 3.1 x 108 lbm/hr
Solution:
Find
of the circulating water:
6.82 x 109 BTU/hr represents the necessary to condense the steam and
subcool it to 4°F below saturation temperature. Therefore, the necessary
to subcool the condensate is:
This number is insignificant compared to the total , and therefore will not be
considered. From the steam tables, saturated liquid at 1 psia:
Using
Solving for
:
Therefore:
Rev 2
67
Solving for
:
Using
Solving for :
Knowledge Check (Answer Key)
What is the approximate value of condensate
depression in a steam condenser operating at 2.0 psia
with a condensate temperature 115ᴼF?
Rev 2
A.
9ᴼF
B.
11ᴼF
C.
13ᴼF
D.
15ᴼF
68
Knowledge Check (Answer Key)
Main turbine exhaust enters a main condenser and
condenses at 126°F. The condensate is cooled to 100°F
before entering the main condenser hotwell. Assuming
main condenser vacuum does not change, which one of
the following would improve the thermal efficiency of
the steam cycle?
A.
Increase condenser cooling water flow rate by 5
percent.
B.
Decrease condenser cooling water flow rate by 5
percent.
C.
Increase main condenser hotwell level by 5 percent.
D.
Decrease main condenser hotwell level by 5 percent
Knowledge Check (Answer Key)
Determine the condensate depression in a condenser
operating at 1 psia with a condensate temperature of 95
°F. Approximately:
Rev 2
A.
10 °F
B.
9 °F
C.
8 °F
D.
7 °F
69
Knowledge Check (Answer Key)
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 (Answer Key)
Excessive heat removal from the low-pressure turbine
exhaust steam in the main condenser results in …
Rev 2
A.
thermal shock
B.
loss of condenser vacuum
C.
condensate depression
D.
fluid compression
70
Knowledge Check (Answer Key)
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 (Answer Key)
Which one of the following actions decreases overall
nuclear power plant thermal efficiency?
Rev 2
A.
Reducing turbine inlet steam moisture content
B.
Reducing condensate depression
C.
Increasing turbine exhaust pressure
D.
Increasing temperature of feedwater entering the steam
generators
71
Knowledge Check (Answer Key)
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
ELO 2.3 Turbine Design and Characteristics
Introduction
A turbine is a device used to convert the energy in high-pressure steam into
rotating kinetic energy. The two steps to accomplish this are as follow:
1. The P is the energy of the steam and is converted to kinetic energy
2. A portion of the steam's kinetic energy is imparted to the turbine rotor
giving it kinetic energy
Large turbines used in nuclear power plants generally have multiple turbine
components: one high-pressure turbine mounted on the same shaft as two or
three low-pressure turbines. The turbine is termed a "tandem" unit when all
turbine units share a common shaft. The generator is mounted at the end of
that shaft.
Recall that a T-s or h-s diagram of a steam cycle can illustrate the work
performed by the turbine. On the h-s diagram below, real turbine work is
shown by the line from Point 2 to 3'.
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72
Turbine Design Section
Note
The following information relating to turbine design is not
tested by the NRC. This material is merely provided to
see various turbine “designs” that can be used to maximize
the efficiency of the turbine. Since the turbine process
takes the high energy steam and converts it to high
velocity steam, reviewing turbine internals can help with
that understanding.
There are two basic types of turbines:


impulse turbines
reaction turbines
The name impulse comes from the fact that high velocity steam strikes the
turbine blading (an impulse) and causes the turbine rotor to turn. Reaction
turbines use nozzle energy conversion in the moving blades to create
rotation force.
The figure on the next page illustrates how steam enters at the steam chest
in the center of the turbine and flows axially in both directions through the
turbine multiple stages. Each turbine stage consists of a fixed nozzle block
where the steam gains kinetic energy before impinging on the moving
blades. A stage is comprised of one set of nozzles and moving blades.
Because the high kinetic energy of the steam strikes the moving blades and
causes the blades to rotate, this is termed an impulse turbine.
Rev 2
73
Figure: Impulse Turbine Components
Theory of Operation
A steam turbine can be defined as a form of heat engine in which the input
energy of the steam is converted into useful work output in two distinct
steps:
1. The available thermal energy of steam is converted into kinetic
energy by expansion through a nozzle.
2. The resultant steam jet impinges against blades attached to the steam
turbine shaft causing the shaft to turn, thereby converting the kinetic
energy into useful work.
Impulse Principle
The figure below illustrates the four distinct steps in the conversion of the
thermal energy of steam into mechanical energy:
1. Steam in an impulse turbine passes through stationary nozzles
2. Stationary nozzles convert some of the thermal energy contained in
the steam (indicated by its pressure and temperature) into kinetic
energy (velocity).
3. Stationary nozzles also direct the steam flow onto the blades of the
turbine wheel.
4. As they rotate, the blades and moving wheel convert kinetic energy of
steam into mechanical rotational energy. In turn, the shaft or rotor
turns a generator or an attached power coupling.
Power plant turbines normally consist of multiple stages with a stage being
one set of rotating blades and nozzles. The multistage approach allows
more energy extraction from the steam and conversion to kinetic energy.
Rev 2
74
Figure: Impulse Turbine Concepts
Reaction Principle
Steam passes through a row of "fixed blades" which act as nozzles to
expand the steam or decrease pressure in a reaction turbine. This process
increases the steam's velocity. The fixed blades also direct the highvelocity steam into the "moving blades", which are almost identical in shape
to the fixed blades as shown in the figure below:
Figure: Reaction Turbine Concepts
The difference between an impulse turbine and a reaction turbine is that
when the steam flows into the reaction turbine’s moving blades, the moving
blades act as nozzles. The moving nozzles convert more thermal energy of
the steam into kinetic energy than the static nozzles of the impulse turbine.
Rev 2
75
The kinetic energy imparts a rotational force on the movable nozzles and
turbine shaft as the steam accelerates through the nozzles. A reaction
turbine exhibits a drop in both pressure and velocity across the moving
blades as opposed to a drop in velocity alone in the impulse turbine. The
pressure drop across the moving blades provides available energy or
reaction principle. Changing the direction of the steam flow in the impulse
turbine blades allows additional energy extraction from the steam or
reaction principle.
Impulse Turbine
The impulse turbine consists of two basic elements:

A fixed nozzle to convert the steam energy or thermal energy into the
kinetic energy
 A rotor consisting of blades mounted on a disk to absorb the kinetic
energy of the steam jet to convert it into rotary motion.
The figure below shows a basic impulse turbine.
Figure: Basic Impulse Turbine
When the high velocity steam strikes the turbine blade, some of the steam's
velocity energy converts into an impulse force acting on the turbine blade.
The blade is mounted on a wheel, and the force will rotate in the direction
of the impulse force.
Rev 2
76
Newton's first law states that a force is required to change either the speed
or the direction of a body in motion. A greater force acts on the blade if the
blade curves in such a manner as to cause the jet of steam to reverse its
direction. The torque developed in a wheel with reversing blades is nearly
twice as great as the force developed in a wheel with flat blades.
Turbine blading should be designed to convert as much of the kinetic
energy of the steam leaving the nozzle into work as is practical. Turbine
blades from large turbines are often called "buckets" because their size and
concavity resembles a bucket.
There are several nozzles directing the steam against the blades or buckets
to turn the rotor. See the left half of the figure below as reference. As
steam exits the first row of blades or rotor, it still has a high velocity and is
still capable of doing more work. A second row of stationary blades may
direct the steam toward a second row of moving blades, if designed with a
second stage, thus converting more of the steam’s kinetic energy into work.
Figure: Impulse/Reaction Turbine Comparison
The figure on the next page shows a two-stage impulse turbine, where the
shape of the vanes in the stationary diaphragm redirect steam exiting from
the first stage to the moving blades of the second stage.
Rev 2
77
Figure: Reaction Turbine Moving and Stationary Blades
Reaction Turbine
A reaction turbine differs from the impulse turbine in that nozzles mounted
on the disk replace the blades of the impulse turbine. In the reaction turbine
steam expands in the moving nozzles shown below
Figure: Basic Reaction Turbine
Rev 2
78
Steam enters the unit and flows to the nozzles. Heat energy converts to
kinetic energy and produces a reactive force when it expands through the
nozzles from a high pressure to a low pressure. The reactive force will
cause rotation opposite to the direction of the steam jet.
This principle of creating a reaction force is identical to the principal of a
rocket engine. Fixed blades direct steam through moving blades attached to
the turbine shaft in a reaction turbine, as seen in the right half of the
Impulse/Reaction Turbine Comparison figure above. As steam passes
through each set of moving blades, a reaction force occurs that is opposite
in direction to the flow of steam. This reaction force causes the turbine
shaft to turn. One stage is a combination of one set of moving blades and
one set of stationary blades. The moving blades of the reaction turbine act
like nozzles. Between each two rows of moving blades are fixed blades that
function as nozzles. Velocity increases while pressure decreases as steam
passes through them.
Figure: Impulse/Reaction Turbine Comparison
The reaction turbine has all the advantages of the impulse turbine, plus
greater efficiency.
Rev 2
79
Turbine Characteristics
As the kinetic energy in the steam converts into work moving the blades of
a reaction turbine, there is a drop in the steam’s velocity and pressure. This
is one major difference between the impulse turbine and the reaction
turbine. In the impulse turbine, there is no pressure drop across the blades,
only across the nozzle. The reaction turbine has a pressure drop across each
set of blades.
Impulse Turbine Characteristics
The upper portions of the two figures below show a cross-section with the
nozzle and blade arrangement for a simple impulse and actual impulse
turbine. The lower section of each figure shows the variation in steam
properties along the cross-section during the conversion from potential
energy to kinetic energy, and then to work.
Figure: Steam Property Variation in Simple Impulse Turbine
In a simple impulse turbine, the steam enters the nozzle with a maximum
pressure and a minimum velocity. The steam velocity and volume increase
while the pressure decreases. Steam leaves the nozzle at peak velocity and
enters the moving blade section of the turbine, where its kinetic energy
converts to useful work. There is no expansion of steam in the moving
blades of the turbine; therefore, the steam pressure and volume are constant
across the blade section through the exhaust. However, the velocity of the
steam greatly decreases in passing through the blades during the conversion
of the steam jet kinetic energy (a direct function of velocity) to work.
Rev 2
80
Turbines of this type have a very high relative speed because the maximum
efficiency occurs when the velocity of the blades is one-half the velocity of
the steam jet leaving the nozzle.
The figure below shows a cross-section in an actual impulse turbine, with
the steam properties graphed along the cross-section.
Figure: Steam Property Variation in Actual Impulse Turbine
In an impulse turbine, a set of stationary blades directs the steam toward a
second row of moving blades that can extract more work. The figure above
shows a cross-section of such a turbine and the change in steam properties.
Steam leaving the first set of moving blades enters a second row of
stationary blades that redirect the flow of steam. Steam leaves the second
row of fixed blades with no changes in pressure or volume, and enters a
second row of moving blades where the steam velocity decreases as the
steam performs more work.
Reaction Turbine Characteristics
The figure below illustrates a cross-section and the variation in steam
properties as steam passes through two stages of a reaction turbine. As in
the impulse turbine, there are fixed nozzles between the rows of moving
blades. The stationary nozzles, shaped like a blade, are slightly convergent,
permitting an expansion of steam.
Rev 2
81
Figure: Steam Property Variation in a Reaction Turbine
The steam jet enters a set of stationary nozzles, where steam velocity and
volume increase as the steam expands before impinging upon the first row
of moving blades. The moving blades are also nozzle-shaped to permit a
further reduction of steam pressure and increase in velocity. From the
velocity curve, it appears that a reduction in velocity takes place in the
moving blades. This is because the moving blades also have their own
velocity at this point and the curve above shows the absolute velocity. The
absolute velocity is the relative difference between the moving blade and
steam velocities.
After the steam leaves the first row of moving blades, the steam enters
another set of nozzle-shaped stationary blades where the steam pressure
decreases further, and the jet velocity increases. This set of stationary
blades directs the steam flow to the second row of moving blades. Turbines
use a large number of stages between inlet and exhaust conditions in order
to limit the pressure drop across any stage. The small pressure drop in the
nozzles results in a low steam velocity per stage. Thus, in general the
velocity of reaction turbine blades is less than the velocity of impulse
blades.
Each successive set of blading is larger. The thermal energy contained in
the steam is being converted to mechanical energy (rotation of the turbine)
as steam flows through the turbine. As a result, the temperature and
pressure of the steam will decrease. The specific volume of the steam
increases as the steam pressure decreases. The later stages of the turbine are
physically larger than earlier stages to ensure that the amount force acting
on the turbine blades remains relatively constant from stage to stage.
Rev 2
82
Figure: Actual Turbine Blading
The turbine can extract energy by successive pressure drops, in which case
the turbine is termed pressure-compounded. Alternatively, the turbine can
extract energy by successive velocity decreases, in which case it is termed
velocity-compounded. A pressure-compounded turbine must consist of two
or more stages since pressure decreases occur only through nozzles.
However, a velocity-compounded turbine may consist of only one stage: a
nozzle followed by a set of moving blades, a set of fixed blades and another
set of moving blades. The figure below shows this arrangement, which is
sometimes referred to as a Curtis Wheel.
Figure: Pressure – Velocity in a Curtis Stage
Rev 2
83
The most important motive force in a reaction turbine is the jet-like thrust
that results when the steam expands through the tail of the teardrop shaped
blades. The reaction turbine differs from an impulse turbine in that the
casing has no nozzles between successive sets of blades, but has fixed
(immovable) reaction blades very similar to those on the rotor. They serve
to redirect steam as well as increase its velocity.
Nozzle Diaphragms
Nozzle diaphragms, illustrated in figure A below, are installed to admit
steam to the rotating blades of each stage of a pressure-compounded
impulse turbine.
The diaphragms contain nozzles that admit steam in an arc of a circle
around the blades. Diaphragms that only admit steam to certain quadrants
of the circle are "partial arc admission diaphragms". Diaphragms that have
nozzles extending around the entire circle of blades are "full arc admission
diaphragms".
Because of the pressure drop that exists across each diaphragm, steam
pressure leaks across the diaphragm and along the rotor. A labyrinth
packing ring (similar to the shaft gland packing) located in a groove in the
inner periphery of the diaphragm, shown in figure B below, minimizes
pressure leakage. Any leakage through the inner periphery of the
diaphragm reduces the amount of the steam thermal energy converted to
mechanical energy and, therefore, reduces the work developed by the stage.
Decreasing leakage is another reason to use multiple stages that reduce the
pressure drop across any individual stage. The right side figure below
shows the placement of these rings, which are installed in sections and are
spring-backed to hold them together and in place.
Figure: Turbine Diaphragm and Cross-Section
Rev 2
84
Knowledge Check (Answer Key)
What is the function of fixed blades in a reaction
turbine?
A.
Maintain steam flow direction
B.
Act as nozzles and expand the steam (decrease
pressure)
C.
Increase steam pressure before impacting next movable
blade.
D.
Change direction of steam flow to impact buckets in
next stage.
Knowledge Check (Answer Key)
Which of the following is NOT a reason turbines
consist of multiple stages?
Rev 2
A.
There is a smaller pressure drop across each stage
B.
Allows more energy to be extracted from the steam
C.
Equalizes axial thrust on the shaft
D.
Accommodates for the expansion of the steam through
the turbine
85
TLO 2 Summary
Review each ELO with the class by using good questioning techniques.






Example of an effective method for asking directed questions:
I have a question...I will select someone to respond"
Ask the question
Pause
Select an individual to answer the question
Ensure that everyone heard and understood the response, amplify and
re-state as necessary
ELO 2.1
What is a nozzle? A nozzle is a mechanical device designed to control the
characteristics of a fluid flow.
How do nozzles work? Nozzles change the energy of a fluid from one form
to another frequently the goal is to increase the kinetic energy of the
flowing medium at the expense of its pressure and internal energy.
Describe a convergent nozzle. Convergent nozzle is a narrowing down
from a wide diameter to a smaller diameter in the direction of the flow.
Describe a divergent nozzle. Divergent is an expanding from a smaller
diameter to a larger one.
What are other uses of nozzles? A nozzle can serve as a flow restrictor to
limit flow, create a differential pressure for flow measurement, or create
high velocities for use in a turbine.
Explain how an air ejector works. The nozzle in an air ejector lowers
pressure and increases velocity. Supersonic flow creates very low pressures
to maintain the heat sink. Diffuser recovers pressure and slow velocity.
ELO 2.2
What is the purpose of a condenser?
What thermodynamic processes occur within a condenser?
What is condensate depression and how is it controlled?
How does condensate depression effect overall system efficiency and why?
ELO 2.3
What at the two major designs of turbines? Two turbine types: impulse and
reaction
Why do turbines normally have multiple stages? Turbines normally consist
of multiple stages to allow more energy extraction from the steam.
Rev 2
86
How does a reaction turbine extract more energy from the steam? Reaction
turbine steam passes through a row of "fixed blades" which act as nozzles
and expand the steam (decrease pressure)
Increases steam's velocity and directs it into the "moving blades."
Objectives
Now that you have completed this lesson, you should be able to:
1. Describe the operation of nozzles to include the functions of nozzles
in flow restrictors and the functions of nozzles in air ejectors.
2. Describe the condensing process to include vacuum formation and
condensate depression
3. Explain the design of turbines, including the functions of nozzles,
fixed blading, moving blading, and the reason turbines are multistage.
Thermodynamic Processes Summary
This chapter examined various processes that occur in a plant; many of
these processes are invisible to the operator. For instance, we examined
how nozzles located within the plant piping systems perform their design
function of converting fluid energy into the desired fluid properties. We
looked at how nozzles can create increased mass flow rates as in the jet
pumps or create large differential pressures to choke flow if needed. The
design of the nozzle throat determines whether the fluid reaches sonic or
supersonic velocity and how the properties of fluids changed when they
were supersonic. We saw that instrumentation determines flow rates based
on the inherent pressure drop in nozzles.
We then looked at the use of nozzles within the turbine to create the desired
pressure or velocity compounding to maximize the work of the turbine. The
use of the nozzle to create high velocity steam flow to impinge on the
moving part of an impulse turbine was shown to be less efficient than using
nozzle shaped movable blades to create a reaction turbine.
Certain processes are termed throttling processes. We can easily determine
downstream fluid properties in throttling processes, because throttling
processes do no work and are isenthalpic. This means that the enthalpy
entering the process is the same as the enthalpy exiting the process.
Entropy will increase slightly through the process.
We then looked at specific processes that took place in the turbine and
condenser along with some design features of each. We learned of the
thermodynamic processes that took place in each, some of the parameters an
operator can use to evaluate those processes (since they are “invisible” to
the operator) as well as some ways that overall system efficiency is effected
by some of those parameters.
Rev 2
87
Objectives
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. Describe the operation of the turbine and condenser processes.
Rev 2
88
Thermodynamic Processes Knowledge Check Answer Key
Knowledge Check Answer Key
ELO 1.1 Thermodynamic Systems and Processes
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
Knowledge Check
Steam flowing through the main turbine control valve is
a(n) …?
Rev 2
A.
isenthalpic process
B.
isentropic process
C.
adiabatic process
D.
reversible process
1
Thermodynamic Processes Knowledge Check Answer Key
ELO 1.2 Applying the First Law of Thermodynamics for Open
Systems
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.
A.
positive, negative, negative, positive
B.
negative, positive, positive, negative
C.
positive, negative, positive, negative
D.
negative, positive, negative, positive
Knowledge Check
In an Open Steady Flow System, choose the energies that
are associated with the mass crossing the system
boundary.
Rev 2
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
2
Thermodynamic Processes Knowledge Check Answer Key
ELO 1.3 Identifying Process Paths on a T-s Diagram
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
On a T-s diagram, if the temperature of the heat sink is
lowered, the efficiency of the cycle will _______ because
________.
Rev 2
A.
increase, more heat is rejected
B.
increase, less heat is rejected
C.
decrease, more heat must be added
D.
decrease, less heat must be added
3
Thermodynamic Processes Knowledge Check Answer Key
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.
A.
as subcooled as practical; great
B.
as subcooled as practical; small
C.
close to saturation; great
D.
close to saturation; small
Knowledge Check
Feed water heating increases overall nuclear power plant
thermal efficiency because…
Rev 2
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.
4
Thermodynamic Processes Knowledge Check Answer Key
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 2
A.
Increase because condensate depression will decrease.
B.
Decrease because condensate depression will increase.
C.
Increase because the work output of the main turbine will
increase.
D.
Decrease because the work output of the main turbine
will decrease.
5
Thermodynamic Processes Knowledge Check Answer Key
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
ELO 1.4 Thermodynamic Energy Balances on Major
Components
Knowledge Check
Why can’t a “ΔT” formula be used in all heat transfer
systems in the plant?
Rev 2
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.
6
Thermodynamic Processes Knowledge Check Answer Key
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?
A.
101 Megawatts (Mw)
B.
126 Mw
C.
1,006 Mw
D.
1,258 Mw
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 2
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.
7
Thermodynamic Processes Knowledge Check Answer Key
ELO 1.5 Throttling Characteristics
Knowledge Check
Which one of the following is essentially a constantenthalpy process?
A.
Throttling of main steam through main turbine steam
inlet valves
B.
Condensation of turbine exhaust in a main condenser
C.
Expansion of main steam through the stages of an ideal
turbine
D.
Steam flowing through an ideal convergent nozzle
Knowledge Check
A nuclear power plant is maintained at 2,000 psia with
a pressurizer temperature of 636°F. A pressurizer
relief safety valve is leaking to a collection tank which
is being held at 10 psig. Which one of the following is
the approximate temperature of the fluid downstream
of the relief valve?
Rev 2
A.
280°F
B.
240°F
C.
190°F
D.
170°F
8
Thermodynamic Processes Knowledge Check Answer Key
Knowledge Check
In what type of nozzle will flow be limited to sonic?
A.
Convergent
B.
Divergent
C.
Convergent-Divergent
D.
Divergent-Convergent
ELO 2.2 Condenser Design and Characteristics
Knowledge Check
What is the approximate value of condensate
depression in a steam condenser operating at 2.0 psia
with a condensate temperature 115ᴼF?
Rev 2
A.
9ᴼF
B.
11ᴼF
C.
13ᴼF
D.
15ᴼF
9
Thermodynamic Processes Knowledge Check Answer Key
Knowledge Check
Main turbine exhaust enters a main condenser and
condenses at 126°F. The condensate is cooled to 100°F
before entering the main condenser hotwell. Assuming
main condenser vacuum does not change, which one of
the following would improve the thermal efficiency of
the steam cycle?
A.
Increase condenser cooling water flow rate by 5
percent.
B.
Decrease condenser cooling water flow rate by 5
percent.
C.
Increase main condenser hotwell level by 5 percent.
D.
Decrease main condenser hotwell level by 5 percent
Knowledge Check
Determine the condensate depression in a condenser
operating at 1 psia with a condensate temperature of 95
°F. Approximately:
Rev 2
A.
10 °F
B.
9 °F
C.
8 °F
D.
7 °F
10
Thermodynamic Processes Knowledge Check Answer Key
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 …
Rev 2
A.
thermal shock
B.
loss of condenser vacuum
C.
condensate depression
D.
fluid compression
11
Thermodynamic Processes Knowledge Check Answer Key
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?
Rev 2
A.
Reducing turbine inlet steam moisture content
B.
Reducing condensate depression
C.
Increasing turbine exhaust pressure
D.
Increasing temperature of feedwater entering the steam
generators
12
Thermodynamic Processes Knowledge Check Answer Key
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
ELO 2.3 Turbine Design and Characteristics
Knowledge Check
What is the function of fixed blades in a reaction
turbine?
Rev 2
A.
Maintain steam flow direction
B.
Act as nozzles and expand the steam (decrease
pressure)
C.
Increase steam pressure before impacting next movable
blade.
D.
Change direction of steam flow to impact buckets in
next stage.
13
Thermodynamic Processes Knowledge Check Answer Key
Knowledge Check
Which of the following is NOT a reason turbines
consist of multiple stages?
Rev 2
A.
There is a smaller pressure drop across each stage
B.
Allows more energy to be extracted from the steam
C.
Equalizes axial thrust on the shaft
D.
Accommodates for the expansion of the steam through
the turbine
14