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Revision 2
June 2016
Thermodynamic
Cycles
Student Guide
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iii
Table of Contents
INTRODUCTION ....................................................................................................................... 2
TLO 1 COMPRESSION PROCESSES .......................................................................................... 2
Overview ........................................................................................................................... 2
ELO 1.1 Gas Laws ............................................................................................................ 4
ELO 1.2 Compression Process ........................................................................................ 11
TLO 1 Summary .............................................................................................................. 14
TLO 2 THE SECOND LAW OF THERMODYNAMICS ................................................................ 17
Overview ......................................................................................................................... 17
ELO 2.1 Thermodynamic Entropy .................................................................................. 18
ELO 2.2 Carnot’s Principle of Thermodynamics ............................................................ 22
ELO 2.3 Thermodynamics of Ideal and Real Processes ................................................. 29
ELO 2.4 Thermodynamic Power Plant Efficiency .......................................................... 31
TLO 2 Summary .............................................................................................................. 53
THERMODYNAMIC PROCESSES SUMMARY ............................................................................ 57
THERMODYNAMIC CYCLES KNOWLEDGE CHECK ANSWERS .................................................. 1
ELO 1.1 Gas Laws ............................................................................................................ 1
ELO 1.2 Compression Process .......................................................................................... 2
ELO 2.1 Thermodynamic Entropy .................................................................................... 2
ELO 2.2 Carnot’s Principle of Thermodynamics .............................................................. 3
ELO 2.3 Thermodynamics of Ideal and Real Processes ................................................... 4
ELO 2.4 Thermodynamic Power Plant Efficiency ............................................................ 4
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Thermodynamic Cycles
Revision History
Revision
Date
Version
Number
Purpose for Revision
Performed
By
11/7/2014
0
New Module
OGF Team
12/11/2014
1
Added signature of OGF
Working Group Chair
OGF Team
1/16/2016
2
Incorporated OGF Team
Changes
OGF Team
Rev 2
1
Introduction
We have not yet discussed processes performed by gases as we have
focused on the steam cycle, yet many applications of the use of gases are
occurring all the time during plant operation. The compression of a gas
results in different final states than the compression of a saturated vapor
such as steam. Gases follow laws that relate their volume, pressure, and
temperature unlike steam, which undergoes phase changes if temperature or
pressure varies sufficiently. These laws must be understood to ensure plant
equipment is maintained within design limits.
The second law of thermodynamics does not take into account the
feasibility of the process or the efficiency of the energy transformation
studied. The second law of thermodynamics allows us to determine the
maximum efficiency of the operating system so that design comparisons
maximize the system’s efficiency.
Objectives
At the completion of this training session, the trainee will demonstrate
mastery of this topic by passing a written exam with a grade of 80 percent
or higher on the following Terminal Learning Objectives (TLOs):
1. Explain compression processes and the laws associated with them.
2. Apply the Second Law of Thermodynamics to analyze real and
ideal systems and components.
TLO 1 Compression Processes
Overview
Gas is another working fluid used throughout the plant. Gas responds
differently to temperature, pressure, and volumetric changes than steam, so
it requires additional explanation.
A gas is a state of matter distinguished from the solid and liquid states by
the following:





Relatively low density and viscosity
Relatively great expansion
Contraction with changes in pressure and temperature
Ability to diffuse readily
Spontaneous tendency to distribute uniformly throughout any
container.
A vapor is often confused with a gas. Vapor has evaporated from a liquid
or solid such as water.
Rev 2
2
Most familiar gases are colorless and odorless and include the following:



The oxygen and nitrogen of the atmosphere
The bubbles of carbon dioxide that rise in a glass of soda
The helium gas that is used to fill balloons.
A few gases have characteristic color, such as:


Nitrogen dioxide is red-brown
Iodine vapor has a beautiful violet color
Anything that we can smell can exist in the gaseous state because our sense
of smell reacts only to gases.
Gases have many observable physical properties. They fill whatever space
is available, but applying pressure can compress them into a smaller
volume. Temperature affects them; they can expand and contract, or exert
different pressures, depending on the temperature. It is obvious from the
force of the wind on a stormy day that gases can flow readily from place to
place and that they have mass. However, gases are not very dense; a vessel
filled with air floats on the surface of a pond because the air is less dense
than the water.
Temperature, pressure, and volume must always be specified when gases
are discussed because of their interrelated effect. The quantitative
relationships among the temperature, pressure, and volume of a gas are
expressed in the gas laws, which were first explored in the eighteenth and
nineteenth centuries.
Compression and pressurization processes are very common in many types
of industrial plants. These processes vary from being the primary function
of a piece of equipment, such as an air compressor to an incidental result of
another process, such as filling a tank with water without first opening the
tank's vent valve. Operators maintain important plant parameters such as
the safety injection accumulators within required legal and design limits
using gas compression and expansion. Understanding the relationships
between temperature, pressure, and volume of gases are important to ensure
operating within required parameters.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Describe the ideal gas laws and how to solve for an unknown
pressure, temperature, or volume.
2. Describe the effects of pressure and temperature changes on confined
fluids.
Rev 2
3
ELO 1.1 Gas Laws
Introduction
Because of their interrelated effect, temperature, pressure, and volume must
always be specified when gases are discussed. The quantitative
relationships among the temperature, pressure, and volume of a gas are
expressed in the gas laws, which were first explored in the eighteenth and
nineteenth centuries.
The gas laws are useful because at low pressures all real gases behave like a
perfect gas. Monatomic gas behavior is very similar to perfect gas
behavior; the Ideal Gas Law is therefore accurate for predicting the gas
behavior. Accuracy will decrease with diatomic and polyatomic gases.
Still, the Ideal Gas Law is useful to develop the behavior of even these
gases with experimentally derived corrections made to produce the desired
accuracy.
Charles’s Laws
Charles’s law or the law of volumes is an experimental gas law that
describes how gases tend to expand when heated.
The figure below shows a piston and cylinder assembly filled with a gas at
absolute temperature (T1 = 300°K) and volume (V1 as shown). The piston is
free to move against a constant external pressure.
Figure: Charles’s Law for Constant Pressure
Adding heat causes the temperature of the gas to increase. The volume
increases and applies pressure against the piston causing the piston to move
outward as the gas temperature increases. The piston will continue to rise
until the cylinder pressure on the internal piston face equalizes to the
external pressure on the piston, restoring system equilibrium.
Rev 2
4
The initial and final pressures are the same but the absolute temperature (T2
= 600°K) is higher and volume (V2) is twice as large as (V1) since the
absolute temperature was doubled.
Repeating the process of adding heat to the gas, causing the piston to move
farther upward, and re-measuring the process variables of gas volume and
temperature, will lead to the following conclusion:
"The volume of a gas at constant pressure is directly proportional to the
temperature of the gas at low pressures.
Charles, as the result of experimentation also concluded that the "pressure
of a gas varies directly with temperature when the volume is held constant".
The mathematical expressions of Charles’s Law are:
𝑉1 𝑇1
𝑃1 𝑇1
=
𝑜𝑟
=
𝑉2 𝑇2
𝑃2 𝑇2
Boyle's Law
Now imagine that the piston and cylinder assembly without the heater. Gas
fills the cylinder to a gas at volume (V1), temperature (T1), and at an
absolute pressure (P1). The cylinder adds no heat to the gas, so the gas
temperature remains constant.
We move the piston physically to a new position by adding or removing
weights, as shown in the figure below. The volume (V2) and absolute
pressure (P2) are measured, and the procedure repeated. Examining the
measured variables, we develop the following conclusion about the gas:
Figure: Boyle's Law – V and P
Rev 2
5
Boyle's Law
"At low-pressures, the volume of a gas at constant temperature is inversely
proportional to the absolute pressure of the gas."
This statement is Boyle's Law, written mathematically as:
𝑃1 𝑉1 = 𝑃2 𝑉2 = 𝑃𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Combined Gas Law
Charles’s Law and Boyle's Law are valid for ideal gases and real gases in
the pressure range that the real gas behaves like an ideal gas. Therefore,
any real gas at low pressure will obey both these laws, stated as follows:
"For a given mass of any gas, the product of the absolute pressure and
volume occupied by the gas, divided by its absolute temperature, is a
constant."
The figure below shows how Boyle’s and Charles’s Laws relate to
compression and temperature increases of gases.
Figure: Combined Gas Law
This statement is the Combined Gas Law, written mathematically as:
𝑃1 𝑉1 𝑃2 𝑉2 𝑃𝑉
=
=
𝑇1
𝑇2
𝑇
NOTE: Pressure and temperature MUST be expressed in absolute values.
The P-T-V diagram on the next page shows all three relationships from the
above equation.
Rev 2
6
Figure: PTV Diagram for Combined Gas Law
Example
A compressor discharges into an air receiver, and cycles off when the
pressure in the receiver reaches 160 psia. During the compression, the
compressor added heat to the air and its temperature in the receiver is
140°F. Assuming no air loads are in service, at what temperature (°F)
should the compressor restart to maintain the receiver above 150 psia?
Solution:
Assuming an ideal gas:
𝑃1 𝑉1 𝑃2 𝑉2
=
𝑇1
𝑇2
The receiver volume is constant, therefore:
𝑃1 𝑃2
=
𝑇1 𝑇2
𝑇1 (°𝑅) = 460°𝐹 + 140℉ = 600°𝑅
𝑃1 = 160 𝑝𝑠𝑖
𝑃2 = 150 𝑝𝑠𝑖
160 𝑝𝑠𝑖 150 𝑝𝑠𝑖
=
600°𝑅
𝑇2
Rev 2
7
𝑇2 =
600 × 150
160
𝑇2 (°𝑅) = 562.5°𝑅
𝑇2 (°𝐹) = 562.5°𝑅 − 460° = 102.5℉
Ideal Gas Law
By combining the results of Charles’s and Boyle's experiments the
following is obtained using the specific volume (v = V/M):
𝑃𝑣
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑇
This constant is the ideal gas constant, designated by R0. Pressure, volume,
and temperature determines the state of an amount of gas, according to the
equation:
𝑃𝑣 = 𝑅0 𝑇
Where:
P = the absolute pressure (Pa)
v = the specific volume of the vessel containing n moles of gas
R0 = the ideal gas constant
T = the temperature in degrees Kelvin (°K)
Mole
It is common practice to discuss quantities of substances in terms of a
measurement called a mole. In order to define a mole, we must first define
another term, Avogadro's number. Avogadro's number is the number of
carbon atoms in 12 grams of carbon-12. The experimentally determined
value of Avogadro's number is 6.023 x 1023 atoms. One mole of any
substance is equal to the amount of that substance having Avogadro's
number of atoms.
Normally, atomic mass units (amu) quantify a substance's atomic mass
(atomic weight). One-twelfth (1/12) of the mass of a carbon-12 atom
defines one amu, which is equivalent to 1.6604 x 10-24 grams. One mole of
an element is equal to the atomic mass number of that element in grams.
1 𝑚𝑜𝑙𝑒 = 6.023 × 1023 𝑎𝑡𝑜𝑚𝑠
6.023 × 1023 𝑎𝑡𝑜𝑚𝑠 (12
Rev 2
𝑎𝑚𝑢
) = 72.27 × 1023 𝑎𝑚𝑢
𝑎𝑡𝑜𝑚𝐶12
8
𝑔
= 72.27 × 1023 𝑎𝑚𝑢 (1.6604 × 10−24 𝑎𝑚𝑢) when
𝑃𝑣
𝑇
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
= 12 𝑔𝑟𝑎𝑚𝑠
Using the element's atomic weight improves the accuracy of the calculation,
but the added accuracy is insignificant and is not usually required.
Simplifying the relationship yields:
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 =
𝑀𝑎𝑠𝑠(𝑔𝑟𝑎𝑚𝑠)
(𝐴𝑡𝑜𝑚𝑖𝑐 𝑀𝑎𝑠𝑠)(
𝑔𝑟𝑎𝑚𝑠
)
𝑚𝑜𝑙𝑒
, or, or mass = (# moles)(GMW)
An ideal gas was defined as one in which Pv/T = a constant under all
circumstances. Though no such gas exists, the fact that a real gas behaves
approximately like an ideal gas provides a specific target for theories for the
gaseous state.
Experimenters found the constant, in terms of the number of moles (n) of
gas in a sample, by making use of the fact that the molar volume of a gas at
standard temperature and pressure (STP) is 22.4 liters.
At STP: 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒(𝑇) = (0°𝐶 = 460°𝑅)
𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒(𝑃) = 1 𝑎𝑡𝑚
𝑉𝑜𝑙𝑢𝑚𝑒(𝑉) = (𝑛) (
Thus,
𝑃𝑣
𝑇
22.4 𝑙𝑖𝑡𝑒𝑟𝑠
)
𝑚𝑜𝑙𝑒
= 𝐾, 𝐾 = 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑃𝑉
22.4 𝑙𝑖𝑡𝑒𝑟𝑠
1
= (𝓃)(1 𝑎𝑡𝑚) (
)(
) = 𝑛𝑅
𝑇
𝑚𝑜𝑙𝑒
273°𝐾
The figure below shows the classic Ideal Gas Law expression:
Figure: Ideal Gas Law
Rev 2
9
The figure below expresses the mole-volume relationship of an ideal gas at
standard temperature and pressure:
Figure: Mole - Volume Relationship
We cannot refer to gases that do not obey this law as ideal. Water vapor
DOES NOT obey the ideal gas laws.
An ideal gas has properties that are constant throughout its mass and whose
molecular movements are not influenced by chemical reactions or external
forces.
There is no known ideal gas. The ideal gas equation is a good
approximation to real gases at sufficiently high temperatures and low
pressures; that is, PV = RT, where P is the pressure, V is the volume per
mole of gas, T is the temperature, and R is the gas constant.
At low pressures, all real gases behave like a perfect gas. The ideal gas law
is the most accurate for monatomic gases at high temperatures and low
pressures. This follows because the law neglects the size of the gas
molecules and the intermolecular attractions. Neglecting molecular size
becomes less important for larger volumes, i.e., for lower pressures. The
relative importance of intermolecular attractions diminishes with increasing
thermal kinetic energy.
Engineers use the ideal gas law because it is simple to use and approximates
real gas behavior. Most physical conditions of gases used by man fit this
description.
Knowledge Check (Answer Key)
According to Charles’s Law, at low pressure, the
_____ of a gas at constant _____ is directly
proportional to the temperature of the gas.
Rev 2
A.
density; pressure
B.
volume; pressure
C.
pressure; volume
D.
weight; volume
10
Knowledge Check (Answer Key)
Calculate the value of the missing property.
P1= 100 psia; P2 = ?
V1 = 50 ft3; V2 = 25 ft3
T1 = 60°F; T2 = 70°F
A.
233 psi
B.
210 psi
C.
204 psi
D.
200 psi
ELO 1.2 Compression Process
Introduction
The most common use of gas behavior is during the compression process
using ideal gas approximations. A compression process may occur at
constant temperature (ΔT = constant), constant volume (ΔV = constant), or
adiabatic (no heat transfer). The amount of work that results from it
depends upon the process, as brought out in the study of the first law of
thermodynamics. When using ideal gas assumptions, the compression
process results in work performed on the system, and is essentially the area
under a P-V curve. Maintaining constant temperature or maintaining
constant pressure results in different amounts of work as shown in the
figure below.
Figure: Pressure-Volume Diagram
Rev 2
11
Compressibility
A fluid is any substance that conforms to the shape of its container; a fluid
may be either a liquid or a gas.
A fluid is considered incompressible when the velocity of the fluid is
greater than one-third of the speed of sound for the fluid, or if the fluid is a
liquid. We assume that such a fluid has a constant density. The variation of
density of the fluid with changes in pressure is the primary factor
considered in deciding whether a fluid is incompressible.
Because compressible fluids experience density changes, their property
relationships vary more than incompressible fluids. In addition, it is easy to
determine the state of a liquid if you know its temperature and pressure.
The process becomes more difficult once the substance becomes a gas.
Figure: P-V Diagram for Gas
Constant Pressure Process
As shown on the above P-V diagram, the work done in a constant pressure
process is the product of the pressure and the change in volume.
𝑊1−2𝑎 = 𝑃(𝛥𝑉)
Constant Temperature Process
The work done in a constant temperature process is the product of the
temperature and the change in volume.
𝑊1−2𝑏 = 𝑇(𝛥𝑉)
Rev 2
12
Constant Volume Process
The work done in a constant volume process is the product of the volume
and the change in pressure.
𝑊1−2𝑐 = 𝑉(𝛥𝑃)
The above equation also applies to liquids. The power requirement for
pumps that move incompressible liquids (such as water) can be determined
by replacing the volume (V) with the product of the specific volume and the
mass.
Power Requirements
𝑊1−2𝑐 = 𝑚𝑣(𝛥𝑃)
Taking the time rate of change of both sides determines the power
requirements of the pump.
𝑊̇1−2𝑐 = 𝑚̇𝑣(∆𝑃)
Effects of Pressure Changes on Fluid Properties
The predominant effect of a pressure increase in a compressible fluid, such
as a gas, is an increase in the fluid density. A pressure in an incompressible
fluid will not result in a significant effect on the density. For example,
increasing the pressure of 100°F water from 15 psia to 15,000 psia will only
increase its density by approximately 6 percent. Therefore, in engineering
calculations, we assume that the density of incompressible fluids remains
constant.
Effects of Temperature Changes on Fluid Properties
An increase in temperature will tend to decrease the density of any fluid.
The effect of a temperature change will depend on whether the fluid is
compressible if the fluid is confined in a container of fixed volume.
A gas will respond to a temperature change in a manner predicted by the
ideal gas laws. A 5 percent increase in absolute temperature will result in a
5 percent increase in the absolute pressure.
If the fluid is an incompressible liquid in a closed container, a fluid
temperature increase will have a tremendously greater and potentially
catastrophic effect. The fluid tries to expand, but the walls of the container
prevent its expansion as the fluid temperature increases. Because the fluid
is incompressible, this results in a tremendous pressure increase for a
relatively minor temperature change. The change in specific volume for a
given change in temperature is not the same at various beginning
temperatures. Resultant pressure changes will vary.
Rev 2
13
A useful thumb rule for water is that pressure in a water-solid system will
increase about 100 psi for every 1°F increase in temperature.
Knowledge Check (Answer Key)
When can a fluid be considered incompressible?
A.
if it is liquid
B.
if it is steam but not flowing
C.
if it is a saturated vapor
D.
if it is a superheated steam
Knowledge Check (Answer Key)
A contained fluid is heated. The resulting change in
pressure will be…
A.
greater for an incompressible fluid.
B.
greater for a compressible fluid.
C.
the same for both fluids.
D.
the same for both fluids only if the volume is held
constant.
TLO 1 Summary
Review each ELO with the class by using good questioning techniques.
Example of an effective method for asking directed questions:





Rev 2
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
14
ELO 2.1
What is Charles’s Law, and what is the mathematical relationship?
The volume of a gas at constant pressure is directly proportional to the
temperature of the gas at low pressures.
𝑉1 𝑇1
𝑃1 𝑇1
=
𝑜𝑟
=
𝑉2 𝑇2
𝑃2 𝑇2
What is Boyle's Law, and what is the mathematical relationship?
At low-pressures, the volume of a gas at constant temperature is inversely
proportional to the absolute pressure of the gas
(𝑃1 )(𝑉1 ) = (𝑃2 )(𝑉2 ) = (𝑃3 )(𝑉3 ) = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
What is the mathematical expression of the combined law?
𝑃1 𝑉1 𝑃2 𝑉2 𝑃𝑉
=
=
𝑇1
𝑇2
𝑇
What pressures and temperatures do we use in solving the gas laws?
Temperature and Pressure MUST BE IN ABSOLUTE.
Explain the ideal gas law.
Ideal gases follow the above laws. A specific gas constant is used to
account for difference in gases atomic structure (R).
𝑃𝑣 = 𝑅𝑇
What is a mole of gas?
One mole of any substance is that amount having Avogadro's Number of
6.023 x 1023 atoms. We use moles to account for large volumes of gas in
the ideal gas equation.
Rev 2
15
ELO 2.2
When can we assume that a fluid is incompressible?
A fluid may be considered incompressible when the velocity of the fluid is
greater than one-third of the speed of sound for the fluid, or if the fluid is a
liquid.
What is the major parameter used to determine incompressibility?
Variation of fluid density with pressure changes is a primary factor
considered in deciding whether a fluid is incompressible. An increase in the
pressure of an incompressible fluid will not have a significant effect on the
fluid density. Increasing the pressure of 100°F water from 15 psia to 15,000
psia will only increase the water density by approximately 6 percent.
Explain how a compressible fluid responds to temperature changes.
For a fluid in a closed container of fixed volume, the effect of a temperature
change will depend on whether the fluid is compressible. If the fluid is a
compressible gas, it will respond to a temperature change in a manner
predicted by the ideal gas laws.
A 5 percent increase in absolute temperature will result in a 5 percent
increase in the absolute pressure.
If the fluid is an incompressible liquid in a closed container, an increase in
the temperature will have a tremendously greater and potentially
catastrophic effect.
Rule of thumb for water: pressure in a water-solid system will increase
about 100 psi for every 1°F increase in temperature.
Objectives
Now that you have completed this lesson, you should be able to:
1. Describe the ideal gas laws and explain how to solve for an
unknown pressure, temperature, or volume.
2. Describe the effects of pressure and temperature changes on
confined fluids.
Rev 2
16
TLO 2 The Second Law of Thermodynamics
Overview
The first law of thermodynamics requires a balance of the various forms of
energy as they pertain to the specified thermodynamic system and/or
control volume studied. However, this first law of thermodynamics does
not take into account the feasibility of the process or the efficiency of the
energy transformation studied. The second law of thermodynamics allows
us to determine the maximum efficiency of the operating system so that
design comparisons maximize the system’s efficiency.
The second law of thermodynamics states that it is impossible to construct a
device that operates within a cycle that can convert all the heat supplied it
into mechanical work. Recognizing that even the most thermally and
mechanically perfect cycles must reject some heat defines thermodynamic
power cycle efficiency. Maximizing the power cycle efficiency is a major
part of the operator's job. There are many important and interdependent
plant parameters affecting plant efficiency.
To produce the maximum electrical power output for the allowed core
thermal power input, the operator must continually monitor these
parameters and adjust plant conditions as necessary. Advancements in
sensing these key plant parameters give the operator real time data of actual
plant efficiency and core power levels, so the operator is able to make
immediate adjustments to maintain the plant within licensed limitations, set
by the Nuclear Regulatory Commission (NRC). The ability to relate current
plant conditions to the plant’s thermal efficiency is a fundamental operator
attribute. Precise control of plant parameters requires continual oversight
by the operator and adjustments as needed.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Explain the second law of thermodynamics using the term entropy.
2. Given a thermodynamic system, determine the:
a. Maximum efficiency of the system
b. Efficiency of the components within the system
3. Differentiate between the path for an ideal process and that for a real
process that for a real process on a T-s or h-s diagram.
4. Describe how individual factors affect system or component
efficiency.
Rev 2
17
ELO 2.1 Thermodynamic Entropy
Introduction
We must first understand that the first law of thermodynamics governs all
cycles when we examine thermodynamic cycles. The first law states energy
can be neither created nor destroyed, but only altered in form. This means
that all of the energy added to a cycle must be accounted for in its entirety.
The first law of thermodynamics places no restrictions on how conversions
from heat to work or vice versa take place or to what extent these
conversions may proceed, which is addressed by the second law of
thermodynamics.
The second law is based on experimental evidence and observations of
actual processes. It suggests that processes proceed in a certain direction
but not in the opposite direction. The second law, which implies that all
real processes are irreversible, governs all real processes. With the second
law of thermodynamics, the limitations imposed on any process can be
studied to determine the maximum possible efficiencies of such a process
then a comparison can be made between the maximum possible efficiency
and the actual efficiency achieved.
Energy-Conversion Systems
One of the areas of application of the second law is the study of energyconversion systems. For example, it is not possible to convert all the energy
obtained from a nuclear reactor into electrical energy. There must be losses
in the conversion process. As shown the figure below on the left as heat
flows from the heat source to the heat sink, it is capable of doing work. As
shown below in the figure on the right, not all of the heat transfers into
work. This is the second law: there must be rejected heat.
Figure: Second Law of Thermodynamics for a Heat Engine
Rev 2
18
Considering those losses, the second law of thermodynamics can be used to
derive an expression for the maximum possible energy conversion
efficiency. The second law denies the possibility of completely converting
into work all of the heat supplied to a system operating in a cycle, no matter
how perfectly designed the system may be. The restriction placed by the
second law requires that some of the heat supplied (QS) to the engine must
be rejected as heat (QR). The difference between the heat supplied and the
heat rejected is the net amount of work produced in the cycle (WNET). The
cycle efficiency is the percentage of energy input to a cycle that is
converted to net work output. The concept of the second law is best stated
using Max Planck's description:
Figure: Kelvin-Planck's Second Law of Thermodynamics
The first law of thermodynamics does not define the energy conversion
process completely. The first law relates to and evaluates the various
energies involved in a process. However, no information about the
direction of the process can be obtained by the application of the first law.
Early in the development of the science of thermodynamics, investigators
noted that while work could be converted completely into heat, the converse
was never true for a cyclic process. Certain natural processes were
observed always to proceed in a certain direction; for example, heat transfer
occurs from a hot to a cold body. The second law was developed as an
explanation of these natural phenomena.
Figure: Heat Flow Direction
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Entropy
The physical property of matter called entropy (S) explains the second law
of thermodynamics. The change in entropy determines the direction in
which a given process proceeds. Entropy also measures the unavailability
of heat to perform work in a cycle. The second law predicts that not all heat
provided to a cycle can be transformed into an equal amount of work; some
heat rejection must take place. The change in entropy is the ratio of heat
transferred during a reversible process to the absolute (abs) temperature of
the system.
∆𝑆 =
∆𝑄
(𝑓𝑜𝑟 𝑎 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠)
𝑇𝑎𝑏𝑠
Where:
•
∆𝑆 = the change in entropy of a system during some process
(BTU/°R [degrees Rankine])
•
∆𝑄 = the amount of heat added to the system during the process
(BTU) (∆Q is change in heat)
•
𝑇𝑎𝑏𝑠 = the absolute temperature at which the heat was transferred
(°R)
Figure: Entropy
Entropy (S) is a natural process that starts in one equilibrium state, ends in
another state, and will go in the direction that causes the entropy of the
system plus the environment to increase for an irreversible process and to
remain constant for a reversible process. Therefore, Sf = Si (reversible) and
Sf > Si (irreversible).
The second law of thermodynamics is also expressed as ΔS ≥ 0 for a closed
cycle. In other words, entropy must increase or stay the same for a cyclic
system; it can never decrease.
Rev 2
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Entropy is an extensive property of a system and like the total internal
energy or total enthalpy, may be calculated from specific entropies based on
a unit mass quantity of the system, so that S = ms. Values of the specific
entropy are tabulated along with specific enthalpy, specific volume, and
other thermodynamic properties of interest in the steam tables described in a
previous module.
The property of specific entropy is used advantageously as one of the
coordinates when representing a reversible process graphically. The area
under a reversible process curve on the T-s diagram represents the quantity
of heat transferred during the process.
Figure: T-s Diagram With Rankine Cycle
Reversible processes are often used in thermodynamic problems by
comparison to the real irreversible process to aid in a second law analysis.
Reversible processes can be depicted on diagrams such as h-s and T-s,
shown below in the figure. The dashes represent the intermediate state of
the fluid is not determined, only the beginning and end states are known. In
both real processes shown below, entropy increases, the dashed lines slant
to the right. The compression of the fluid by the pump and the expansion of
the fluid through the turbine both increase entropy compared to the ideal
case.
Figure: T-s and h-s Diagrams for Expansion and Compression Processes
Rev 2
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Knowledge Check (Answer Key)
The second law of thermodynamics can also be
expressed as ________ for a closed cycle.
A.
Sf = Si
B.
ΔS ≥ 0
C.
ΔT < 0
D.
ΔS < 0
ELO 2.2 Carnot’s Principle of Thermodynamics
Introduction
In 1824 Nicolas Léonard Sadi Carnot, a French military engineer and
physicist known as the father of thermodynamics, advanced the study of the
second law by using reversible processes that disclosed a principle
consisting of the following:
No engine can be more efficient (η) than a reversible (ideal) engine
operating between the same high temperature heat source and low
temperature heat sink.
 Efficiencies of all reversible engines operating between the same
constant temperature reservoirs are the same.
 Efficiency of a reversible engine depends only upon the temperatures
of the heat source and heat receiver.

Figure: Carnot's Efficiency Principle
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Carnot Cycle Guidelines
The Carnot cycle can best be described using an ideal frictionless thermally
isolated piston operating between a constant heat source and heat sink. A
P-v and T-s diagram shown below illustrates the cycle as the heat source is
applied to the piston, causing a reversible isothermal expansion between
point 1 and 2. The piston then moves doing an amount of work (w1-2) due
to the isothermal (constant temperature) expansion of the gas, shown on the
figure as the line between points 1-2.
The gas is allowed to finish expanding adiabatically between point 2 and 3
and an amount of work (w2-3) is done, shown as the line between points 2-3.
Next, the heat sink is applied to the piston, and a reversible isothermal
compression of the gas occurs between points 3 and 4. The piston is used to
compress the gas and an amount of heat (q3-4) is transferred to the heat sink
through the cylinder head. This isothermal compression requires some
amount of work (w3-4) to be done on the piston, shown as the line between
points 3-4.
In process 4, the cylinder is removed from the heat sink. The piston returns
to its initial state by undergoing adiabatic compression requiring some
amount of work (w4-1). The cycle is completed when the cylinder is again
placed in contact with the heat source, shown in the figure below as the line
between points 4-1:
Figure: Single Piston Carnot Engine Cycle
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During the Carnot cycle just described, a certain amount of heat and work
were added or removed from the system. Work is done on the system when
the piston travels into the cylinder and compresses the gas. Work is done
by the system as the gas expands to force the piston out of the cylinder.
Heat is added to the system to cause the piston to move outward (𝑄𝐴 =
𝑞1−2 ). Heat is added, entropy increases, and the process line goes from left
to right. Heat is removed from the system as the piston is compressed in the
final portion of the cycle (𝑄𝑅 = 𝑞3−4 ). Entropy decreases, and the process
line goes from right to left as heat is removed from the system. Two ideal
assumptions are made that result in the Carnot cycle having the highest
possible efficiency:
1. Both work processes occur with no friction and thus there is no
change in entropy.
2. The heat addition and heat rejection occur with no change in the
temperature of the working fluid. Therefore, the temperature
difference (T) between the working fluid and the heat source and the
heat sink remains constant.
We define thermodynamic cycle efficiency by analyzing the energy output
or work (W) produced compared to the energy input (QA). The greater the
percentage of energy input converted to work, the greater the cycle
efficiency.
The figure below shows a Carnot cycle representation. The heat input (QH)
is the area under line 2-3. The heat rejected (QC) is the area under line 1-4.
The difference between the heat added and the heat rejected is the net work
(sum of all work processes), represented as the area of rectangle 1-2-3-4.
Figure: Carnot Cycle Representation
In a perfectly efficient cycle, all of the energy put into the cycle converts to
a useful work output. However, as stated previously, heat must be rejected
for the cycle to be continuous.
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The efficiency (η) of the cycle is the ratio of the net work of the cycle to the
heat input to the cycle. This ratio can be expressed by the following
equation:
𝜂=
𝑊𝑛𝑒𝑡
ℎ𝑒𝑎𝑡 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑
𝜂=
(𝑄𝐴 − 𝑄𝑅 )
𝑄𝐴
𝜂=
(𝑄𝐻 − 𝑄𝐶 ) (𝑇𝐻 − 𝑇𝐶 )
=
𝑄𝐻
𝑇𝐻
𝑇
= 1 − (𝑇 𝐶 )Where:
𝐻
•
η = cycle efficiency
•
TC = designates the low-temperature reservoir (°R)
•
TH = designates the high-temperature reservoir (°R)
This equation shows that the maximum possible efficiency exists when TH
is at its highest possible value or when TC is at its lowest value. The above
represents an upper limit of efficiency for any given system operating
between the same two temperatures since all practical systems and
processes are irreversible. The system's maximum possible efficiency
would be that of a Carnot cycle, but because Carnot cycles represent
reversible processes, the real system cannot reach the Carnot efficiency
value. Thus, the Carnot efficiency serves as an unattainable upper limit for
any real system's efficiency. The following example demonstrates the
above principles.
Example 1: Carnot Efficiency
An inventor claims to have an engine that receives 100 BTUs of heat and
produces 25 BTUs of useful work when operating between a source at 140
°F and a receiver at 0 °F. Is the claim a valid claim?
Solution 1:
𝑇ℎ = 140 ℉ + 460 = 600 °𝑅
𝑇𝑐 = 0 ℉ + 460 = 460 °𝑅
𝜂=
600 − 460
× 100 = 23.3%
600
Claimed efficiency = 25/100 = 25 percent; this exceeds the Carnot
efficiency value. Therefore, the claim is invalid.
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The second law determines the maximum possible efficiencies obtained
from a power system. Actual efficiencies will always be less than this
maximum. Real systems have losses, such as friction, that are not
reversible and that preclude real systems from obtaining the maximum
possible efficiency. An illustration of the difference that may exist between
the ideal and actual efficiency is presented in the figure below and in the
following example:
Example 2: Actual Versus Ideal Efficiency
The actual efficiency of a steam cycle is 18.0 percent. The facility operates
from a steam source at 340 °F and rejects heat to atmosphere at 60 °F.
Compare the Carnot efficiency to the actual efficiency.
Figure: Real Process Cycle Compared to Carnot Cycle
Solution:
Solve for the Carnot maximum efficiency:
𝑇𝑐
𝜂 = 1−( )
𝑇ℎ
𝜂 = 1−
60 + 460
340 + 460
𝜂 = 1−
520
800
𝜂 = 1 − 0.65
𝜂 = 35%
Therefore, ideal Carnot efficiency of 35 percent compared to 18.0 percent
actual efficiency.
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The second law equations are treated in much the same manner as the first
law equations. An isolated, closed, or open system used in the analysis
depends on the types of energy that cross the boundary. The open system
analysis is still the more general case, with the closed and isolated systems
being special cases of the open system. The approach used to solve second
law problems is similar to that used in the first law analysis.
A control volume using the second law is shown below in the figure. In this
diagram, the fluid moves through the control volume from the inlet section
to the outlet section while work is delivered externally to the control
volume. We assume that the boundary of the control volume is at some
environmental temperature and that all of the heat transfer (Q) occurs at this
boundary.
Entropy is a property that may be transported with the flow of the fluid into
and out of the control volume, just like enthalpy or internal energy. The
entropy flow into the control volume resulting from mass transport, 𝑚̇𝑖𝑛 𝑠𝑖𝑛 ,
and the entropy flow out of the control volume is 𝑚̇𝑜𝑢𝑡 𝑠𝑜𝑢𝑡 , assuming that
the properties are uniform at sections in and out. Entropy may also be
added to the control volume because of heat transfer at the boundary of the
control volume.
Figure: Control Volume for Second Law Analysis
A simple demonstration of the use of this form of system in second law
analysis gives the student a better understanding of its use.
Rev 2
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Example 3: Open System Second Law
Steam enters the nozzle of a steam turbine with a velocity of 10 ft/sec at a
pressure of 100 psia and temperature of 500 °F. At the nozzle discharge,
the pressure and temperature are one (1) atmospheric pressure (atm) at 300
°F. What is the increase in entropy for the system if the mass flow rate is
10,000 lbm/hr?
Solution:
𝑚̇𝑠𝑖𝑛 + ∆𝑠 = 𝑚̇𝑠𝑜𝑢𝑡
Where:
•
∆𝑠 = entropy added to the system
•
∆𝑠 = 𝑚̇(𝑠𝑜𝑢𝑡 − 𝑠𝑖𝑛 )
•
sin = 1.7088 BTUs/lbm-°R (from steam tables)
•
sout = 1.8158 BTUs/lbm-°R (from steam tables)
•
∆𝑠
•
∆𝑠
•
∆𝑠 = 10,000(0.107)
•
∆𝑠 = 1,070
𝑚̇
𝑚̇
𝐵𝑇𝑈
= 𝑠𝑜𝑢𝑡 − 𝑠𝑖𝑛 = 1.8158 − 1.7088 𝑙𝑏𝑚-°𝑅
𝐵𝑇𝑈
= 0.107 𝑙𝑏𝑚-°𝑅
𝐵𝑇𝑈
𝑙𝑏𝑚-°𝑅
= 𝑒𝑛𝑡𝑟𝑜𝑝𝑦 𝑎𝑑𝑑𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚
The second law of thermodynamics gives a maximum efficiency limit
(which is never reached in physical systems) that an ideal thermodynamic
system can perform. The efficiency is determined by knowing the inlet and
exit absolute temperatures of the overall system (one that works in a cycle)
and applying Carnot's efficiency equation.
Rev 2
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Knowledge Check (Answer Key)
The steam generators produce dry saturated steam at
1,000 psig. The main condenser is operating with zero
subcooling at 1 psia. What is the maximum efficiency
obtainable?
A.
44 percent
B.
34 percent
C.
28 percent
D.
23 percent
Knowledge Check (Knowledge Check)
Determine the Carnot Efficiency of a steam engine that is
supplied with saturated steam at 300 psia and exhausts to
atmosphere…
A.
44 percent
B.
56 percent
C.
42 percent
D.
35 percent
ELO 2.3 Thermodynamics of Ideal and Real Processes
Introduction
It is convenient to arrange the various thermodynamic processes on a
property diagram in evaluating the various cycles present in a nuclear power
plant. The most common set of coordinates used is a plot of temperature
versus specific entropy, a T-s diagram. Using this type of diagram, we can
analyze the various processes that take place and how these processes effect
the entire cycle as well as the amount of heat and work, both of which occur
during the processes. Any ideal thermodynamic process can be drawn as a
path on a property diagram. A real process that approximates the ideal
process can also be represented on the same diagrams, usually by dashed
lines.
Rev 2
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Carnot Cycle Guidelines
Entropy is constant in an ideal expansion or compression process.
Isentropic processes are represented by vertical lines on T-s and h-s
diagrams, shown below in the figures. A real expansion and real
compression process operating between the same pressures as the ideal
process are shown by dashed lines and will slant slightly toward the right,
since the entropy increases from the start to the end of the real process.
All real processes are irreversible. It is helpful to compare real processes to
ideal processes in system design. The reversible process indicates a
maximum work output for a given input, which compares to real work
output for efficiency purposes. The h-s diagram clearly shows that the real
expansion process (turbine) results in a smaller change in enthalpy, meaning
less energy is extracted in the real turbine than the ideal turbine. The figure
below also shows that more enthalpy must be added during the compression
(pump) process, meaning more energy must be added to the real system
than the ideal system. These differences in enthalpy across various
components tell us how efficient the real process compares to the ideal
process of the Carnot cycle.
Figure: T-s and h-s Diagrams for Expansion and Compression Processes
Rev 2
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Knowledge Check (Answer Key)
Why are real processes shown with dotted lines on
property diagrams?
A.
They occur faster than real processes.
B.
The value of entropy during the process is not
determined.
C.
The entropy values during the process are the same as the
real process until the outlet from the process.
D.
You would not be able to distinguish between real and
ideal processes if the real process was a solid line.
ELO 2.4 Thermodynamic Power Plant Efficiency
Introduction
The actual construction of the steam cycle in nuclear power plants is
considerably more complex than the basic cycles that we have covered so
far, including numerous required auxiliary systems and instrumentation and
control equipment. However, adding this detail to the steam cycle does not
contribute significantly to an understanding of the energy transfer
characteristics of the overall plant. A simplified steam cycle for a typical
steam-electric plant is shown below in the figure. Notice that the cycle
shown includes two components we have not yet discussed, the Moisture
Separator Reheater (MSR) and the Feedwater Preheater. We discuss both
of these components and their effect on cycle efficiency later in this
module.
To analyze a complete power plant steam cycle, it is first necessary to
analyze the elements that make up the cycle. Although specific designs
differ, there are three basic types of elements in power cycles: (1) turbines,
(2) pumps, and (3) heat exchangers. Each of these three elements imparts a
characteristic change in the properties of the working fluid.
Typical Steam Cycle
Previously we calculated system efficiency by knowing the temperature of
the heat source and the heat sink. It is also possible to calculate the
efficiencies of each individual component by comparing the actual work
produced by the component to the work that produced by an ideal
component operating isentropically between the same inlet and outlet
conditions.
Rev 2
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Figure: Typical Steam Cycle
Steam Turbine Efficiency Guidelines
A steam turbine extracts energy from the working fluid (steam) to do work
in the form of rotating the turbine shaft. The steam works as it expands
through the turbine. The shaft work converts to electrical energy by the
generator. In the application of the first law general energy equation to a
simple turbine under steady flow conditions, demonstrates that the decrease
in the enthalpy of the working fluid Hin - Hout equals the work done by the
working fluid in the turbine (Wt).
Figure: Turbine Work
Rev 2
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𝐻𝑖𝑛 − 𝐻𝑜𝑢𝑡 = 𝑊𝑡
𝑚̇(ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 ) = 𝑤̇𝑡
Where:
•
Hin = enthalpy of the working fluid entering the turbine (BTU)
•
Hout = enthalpy of the working fluid leaving the turbine (BTU)
•
Wt = work done by the turbine (ft-lbf)
•
𝑚̇ = mass flow rate of the working fluid (lbm/hr)
•
hin = specific enthalpy of the working fluid entering the turbine
(BTU/lbm)
•
hout = specific enthalpy of the working fluid leaving the turbine
(BTU/lbm)
•
𝑤̇𝑡 = power of the turbine (BTU/hr)
Ideal Versus Real Turbine
The calculation of turbine work using only the enthalpy change is valid
because the change of kinetic and potential energy and the amount of heat
lost by the working fluid while in the turbine are negligible. These
assumptions are valid for most practical applications. However, to apply
these relationships, one additional definition is necessary. In any ideal case,
the working fluid does work reversibly by expanding at constant entropy.
In an ideal turbine, the entropy of the working fluid entering the turbine Sin
equals the entropy of the working fluid leaving the turbine.
𝑆𝑖𝑛 = 𝑆𝑜𝑢𝑡 or 𝑠𝑖𝑛 = 𝑠𝑜𝑢𝑡
Where:
•
Sin = entropy of the working fluid entering the turbine (BTU/°R)
•
Sout = entropy of the working fluid leaving the turbine (BTU/°R)
•
sin = specific entropy of the working fluid entering the turbine
(BTU/lbm-°R)
•
sout = specific entropy of the working fluid leaving the turbine
(BTU/lbm-°R)
An ideal turbine performs the maximum amount of work theoretically
possible, and therefore provides a basis for analyzing the performance of
real turbines.
Rev 2
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Because of friction losses in the blades, steam leakage past the blades and to
a lesser extent mechanical friction, a real turbine does less work than an
ideal turbine. Turbine efficiency (ηt), is defined as the ratio of the actual
work done by the turbine (Wt.actual) to the work that would be done by the
turbine if it were an ideal turbine (Wt.ideal). It is shown that Wi is larger than
Wa, meaning more enthalpy is extracted from the steam in an ideal turbine
operating between the same temperatures (pressures) than in a real turbine.
Figure: h-s Diagram for Ideal and Real Turbines
An actual turbine does less work than an ideal turbine because of factors
such as:

Friction losses in the turbine blades

Steam leakage past the turbine blades

Mechanical friction
Turbine efficiency (ηt) is the ratio of the actual work done by the turbine
Wt.actual to the work that would be done by the turbine if it were an ideal
turbine Wt.ideal
Generally, turbine efficiency is 60% - 80% for small turbines and 90% for
large turbines such that:
𝜂𝑡 =
𝑊𝑡.𝑎𝑐𝑡𝑢𝑎𝑙
𝑊𝑡.𝑖𝑑𝑒𝑎𝑙
𝜂𝑡 =
(ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 )𝑎𝑐𝑡𝑢𝑎𝑙
(ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 )𝑖𝑑𝑒𝑎𝑙
Where:
•
𝜂𝑡 = turbine efficiency (no units)
Rev 2
34
•
Wt.actual = actual work done by the turbine (ft-lbf)
•
Wt.ideal = work done by an ideal turbine (ft-lbf)
•
(hin – hout)actual = actual enthalpy change of the working fluid
(BTU/lbm)
•
(hin – hout)ideal = actual enthalpy change of the working fluid in an ideal
turbine (BTU/lbm)
Figure: Entropy Diagram Measures System Efficiency
A vertical line on the T-s diagram is a constant entropy ideal process.
Entropy increases in the actual turbine process. The smaller the increase in
entropy, the closer the turbine efficiency (ηt) is to 1.0 or 100 percent.
If an ideal turbine became a real turbine, its output would decrease due to
losses such as friction, windage, moisture, and tip leakage. To raise the
turbine back to its original output, the turbine steam supply valves would
open, increasing the mass flow rate of the steam going into the turbine. The
work of the turbine increases to overcome the losses. Opening the steam
supply valves decreases steam generator pressure and adding more heat
raises the steam generator pressure back to its original value. However,
since the heat added is greater than the increase in work from the turbine,
the cycle efficiency decreases.
Rev 2
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Pump Efficiency Guidelines
A pump performs work on a system's working fluid to overcome the head
loss and keep the fluid moving. Like the turbine, the application of the first
law general energy equation to a simple pump under steady flow conditions
results in the work of the pump (Wp), equals the change in working fluid
enthalpy across the pump ( Hout - Hin).
Figure: Work Done by the Pump Equals Change in Enthalpy Across the
Ideal Pump
𝐻𝑜𝑢𝑡 − 𝐻𝑖𝑛 = 𝑊𝑝
𝑚̇(ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 ) = 𝑊̇𝑝
Where:
•
Hout = enthalpy of the working fluid leaving the pump (BTU)
•
Hin = enthalpy of the working fluid entering the pump (BTU)
•
Wp = work done by the pump on the working fluid (ft-lbf)
•
𝑚̇ = mass flow rate of the working fluid (lbm/hr)
•
hout = specific enthalpy of the working fluid leaving the pump
(BTU/lbm)
•
hin = specific enthalpy of the working fluid entering the pump
(BTU/lbm)
•
𝑊̇𝑝 = power of pump (BTU/hr)
Real Versus Ideal Pump
As in the turbine, the kinetic and potential energy changes and the heat lost
by the working fluid while in the pump are negligible. These are valid
assumptions along with the assumption that the working fluid is
incompressible. For the ideal case, it can be shown that the work done by
the pump (Wp) equals the change in enthalpy across the ideal pump.
Rev 2
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𝑊𝑝.𝑖𝑑𝑒𝑎𝑙 = (𝐻𝑜𝑢𝑡 − 𝐻𝑖𝑛 )𝑖𝑑𝑒𝑎𝑙
𝑊̇𝑝.𝑖𝑑𝑒𝑎𝑙 = 𝑚̇(ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 )𝑖𝑑𝑒𝑎𝑙
Where:
•
Wp = work done by the pump on the working fluid (ft-lbf)
•
Hout = enthalpy of the working fluid leaving the pump (BTU)
•
Hin = enthalpy of the working fluid entering the pump (BTU)
•
𝑊̇𝑝 = power of pump (BTU/hr)
•
𝑚̇ = mass flow rate of the working fluid (lbm/hr)
•
hout = specific enthalpy of the working fluid leaving the pump
(BTU/lbm)
•
hin = specific enthalpy of the working fluid entering the pump
(BTU/lbm)
The ideal pump provides a basis for analyzing the performance of actual
pumps, which requires more work because of unavoidable losses due to
friction and fluid turbulence. The work done by a pump Wp equals the
change in enthalpy across the actual pump.
𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙 = (𝐻𝑜𝑢𝑡 − 𝐻𝑖𝑛 )𝑎𝑐𝑡𝑢𝑎𝑙
𝑊̇𝑝.𝑎𝑐𝑡𝑢𝑎𝑙 = 𝑚̇(ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 )𝑎𝑐𝑡𝑢𝑎𝑙
Pump Efficiency
Now, consider a real pumping process opposed to the ideal pumping
process. In the real pumping process with friction taken into account, the
entropy increases across the pump. Recall that entropy increases in all real
processes. Because the work of the pump (WPUMP) increases to make up for
the frictional losses, the net work (WNET) decreases. Since the net work
decreases, the overall cycle efficiency also decreases.
Pump efficiency (ηp) is the ratio of the work required by the pump if it were
an ideal pump (Wp.ideal) to the actual work required by the pump (Wp.actual).
𝜂𝑝 =
𝑊𝑝.𝑖𝑑𝑒𝑎𝑙
𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙
Also shown as:
𝜂𝑝 = 𝐹𝑙𝑜𝑤
Rev 2
𝐻𝑜𝑟𝑠𝑒𝑝𝑜𝑤𝑒𝑟
𝐵𝑟𝑎𝑘𝑒 𝐻𝑜𝑟𝑠𝑒𝑝𝑜𝑤𝑒𝑟
37
Example:
A pump operating at 75 percent efficiency has an inlet specific enthalpy of
200 BTU/lbm. The exit specific enthalpy of the ideal pump is 600
BTU/lbm. What is the exit specific enthalpy of the actual pump?
Solution:
Using the equation above:
𝜂𝑝 =
𝑊𝑝.𝑖𝑑𝑒𝑎𝑙
𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙
𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙 =
𝑊𝑝.𝑖𝑑𝑒𝑎𝑙
𝜂𝑝
(ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 )𝑎𝑐𝑡𝑢𝑎𝑙 =
ℎ𝑜𝑢𝑡.𝑎𝑐𝑡𝑢𝑎𝑙 =
ℎ𝑜𝑢𝑡.𝑎𝑐𝑡𝑢𝑎𝑙 =
(ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 )𝑖𝑑𝑒𝑎𝑙
𝜂𝑝
(ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 )𝑖𝑑𝑒𝑎𝑙
+ ℎ𝑖𝑛.𝑎𝑐𝑡𝑢𝑎𝑙
𝜂𝑝
(600
𝐵𝑇𝑈
𝐵𝑇𝑈
− 200
)
𝑙𝑏𝑚
𝑙𝑏𝑚 + 200 𝐵𝑇𝑈
0.75
𝑙𝑏𝑚
ℎ𝑜𝑢𝑡.𝑎𝑐𝑡𝑢𝑎𝑙 = 533.3
𝐵𝑇𝑈
𝐵𝑇𝑈
+ 200
𝑙𝑏𝑚
𝑙𝑏𝑚
ℎ𝑜𝑢𝑡.𝑎𝑐𝑡𝑢𝑎𝑙 = 733.3 𝐵𝑇𝑈/𝑙𝑏𝑚
Pump efficiency (ηp) relates the minimum amount of work theoretically
possible to the actual work required by the real pump. However, the work
required by a pump is normally only an intermediate form of energy.
Usually, a motor or turbine runs the pump. Pump efficiency does not
account for losses in this motor or turbine. An additional efficiency factor,
motor efficiency (ηm) is the ratio of the actual work required by the pump to
the electrical energy input to the pump motor, when expressing both in the
same units.
Rev 2
38
𝜂𝑚 =
𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙
𝑊𝑚.𝑖𝑛 𝐶
Where:
•
𝜂𝑚 = motor efficiency (no units)
•
Wp.actual = actual work required by the pump (ft-lbf)
•
Wm.in = electrical energy input to the pump motor per kilowatt hour
(kWh)
•
C = conversion factor = 2.655 x 106 ft-lbf/kWh
Motor Efficiency Guidelines
Like pump efficiency, motor efficiency is always less than 1.0 or 100
percent for an actual pump motor. The combination of pump efficiency and
motor efficiency relates the ideal pump to the electrical energy input to the
pump motor.
𝜂𝑚 𝜂𝑝 =
𝑊𝑝.𝑖𝑑𝑒𝑎𝑙
𝑊𝑚.𝑖𝑛 𝐶
Where:
•
𝜂𝑚 = motor efficiency (no units)
•
𝜂𝑝 = pump efficiency (no units)
•
𝑊𝑝.𝑖𝑑𝑒𝑎𝑙 = ideal work required by the pump (ft-lbf)
•
𝑊𝑚.𝑖𝑛 = electrical energy input to the pump motor (kWh)
•
C = conversion factor = 2.655 x 106 ft-lbf/kWh
Real Versus Ideal Cycle Efficiency Guidelines
In the preceding sections, we discussed the Carnot cycle, cycle efficiencies,
and component efficiencies. In this section, we apply this information to
compare and evaluate various ideal and real cycles. This determines how
modifying a cycle affects the cycle's available energy that can be extracted
for work.
A Carnot cycle's efficiency depends solely on the temperature of the heat
source and the heat sink. To improve a cycle's efficiency, all we have to do
is increase the temperature of the heat source and decrease the temperature
of the heat sink. In the real world, the ability to do this is limited by the
following constraints.
Rev 2
39
For a real cycle, the heat sink is limited by the fact that the earth is our final
heat sink, and therefore is fixed at about 60 °F (520 °R).
The heat source is limited to the combustion temperatures of the fuel burned
or the maximum limits placed on nuclear fuels by their structural
components (pellets, cladding, etc.). In the case of fossil fuel cycles, the
upper limit is ~ 3,040 °F (3,500 °R). However, even this temperature is not
attainable due to the metallurgical restraints of the boilers; and therefore,
they are limited to about ~ 1,500 °F (~ 1,960 °R) for a maximum heat
source temperature.
Using these limits to calculate the maximum efficiency attainable by an
ideal Carnot cycle gives the following:
𝜂=
𝑇𝑆𝑂𝑈𝑅𝐶𝐸 − 𝑇𝑆𝐼𝑁𝐾 1,960 °𝑅 − 520 °𝑅
=
= 73.5%
𝑇𝑆𝑂𝑈𝑅𝐶𝐸
1,960 °𝑅
This calculation indicates that the Carnot cycle, operating with ideal
components under real world constraints, should convert almost threequarters of the input heat into work. This ideal efficiency is beyond the
present capabilities of any real systems.
Heat Rejection
By analyzing a T-s diagram, we will understand why an efficiency of 73
percent is not possible for real components.
The energy added to a working fluid during the Carnot isothermal
expansion is given by qs. Not all of this energy is available for use by the
heat engine since a portion of it (qr) must be rejected to the environment.
This is given by:
𝑞𝑟 = 𝑇𝑜 𝛥𝑠 in units of BTU/lbm
Where:
•
To = the average heat sink temperature of 520 °R
The available energy (A.E.) for the Carnot cycle may be given as:
•
𝐴. 𝐸. = 𝑞𝑠 − 𝑞𝑟
Substituting the above equation for qr gives:
𝐴. 𝐸. = 𝑞𝑠 − 𝑇𝑜 𝛥𝑠 in units of BTU/lbm, (Δs is change in absolute entropy)
and equals the area of the shaded region labeled available energy in the
figure on the next page between the temperatures 1,962 °R and 520 °R.
Rev 2
40
Figure: Entropy Measures Temperature as Pressure and Volume Increase
Typical Power Cycle
A typical power cycle employed by a fossil fuel plant as shown on the
above below. The working fluid is water, which places certain restrictions
on the cycle. If we wish to limit ourselves to operation at or below 2,000
psia, it is readily apparent that constant heat addition at our maximum
temperature of 1,962 °R is not possible (2 to 4). The nature of water and
certain elements of the process controls require us to add heat in a constant
pressure process instead (1-2-3-4). Because of this, the average temperature
where we add heat is far below the maximum allowable material
temperature
Rev 2
41
Figure: Carnot Cycle Versus Typical Power Cycle Available Energy
The actual available energy (area under the 1-2-3-4 on the above curve) is
less than half of what is available from the ideal Carnot cycle (area under 12'-4) operating between the same two temperatures. Typical thermal
efficiencies for fossil plants are 40 percent while nuclear plants have
efficiencies of 31 percent. These numbers are less than half the maximum
thermal efficiency of the ideal Carnot cycle calculated earlier.
The figure on the following page shows a proposed Carnot steam cycle
superimposed on a T-s diagram. Several problems make it undesirable as a
practical power cycle. Significant pump work is required to compress a
two-phase mixture of water and steam from point 1 to the saturated liquid
state at point 2. Cavitation in the pump would occur if the inlet fluid was at
saturation. Third, a condenser designed to produce a two-phase mixture at
the outlet (point 1) would pose technical problems.
Rev 2
42
Figure: Ideal Carnot Cycle
Rankine Cycle
Steam engines drove the industrial revolution, so early thermodynamic
developments centered on improving the performance of contemporary
steam engines. It was desirable to construct a steam cycle as close to
reversible as possible and would take better advantage of the characteristics
of steam than does the Carnot cycle. The Rankine cycle was developed as a
more practical version of the Carnot cycle.
The Rankine cycle, shown below in the figure, confines the isentropic
compression process to the liquid phase only (points 1 to 2). This
minimizes the amount of work required to attain operating pressures and
avoids the mechanical problems associated with pumping a two-phase
mixture. The compression process shown between points 1 and 2 is
exaggerated*. In reality, a temperature rise of only 1 °F occurs in
compressing water from 14.7 psig at a saturation temperature of 212 °F to
1,000 psig.
Figure: Rankine Cycle
Rev 2
43
*The constant pressure lines converge rapidly in the subcooled or
compressed liquid region and it is difficult to distinguish them from the
saturated liquid line without artificially expanding the constant pressure
lines away from it.
Like the Carnot cycle, in a Rankine cycle available and unavailable energy
on a T-s diagram is represented by the areas under the curves; therefore, the
larger the unavailable energy, the less efficient the cycle. The same loss of
cycle efficiency is seen when two Rankine cycles are compared as shown
below in the figure.
Figure: Rankine Cycle Efficiency Comparisons On a T-s Diagram
If the ideal turbine was replaced with a real turbine, the efficiency of the
cycle will be reduced. This is because the non-ideal turbine incurs an
increase in entropy increases the area under the T-s curve for the cycle.
However, the increase in the area of available energy (3-2-3') is less than
the increase in area for unavailable energy (a-3-3'-b).
Figure: Rankine Cycle With Real Versus Ideal Turbine
Rev 2
44
Rankine Cycle Efficiencies
Cycle efficiency compares by contrasting the amount of rejected energy to
available energy of both cycles. The comparison shows that Cycle b,
above, has less heat available for work and more rejected heat, making it
less efficient.
An h-s diagram also compares systems and helps determine their
efficiencies.
Typical Steam Cycle
A simplified version of the major components of a typical steam plant cycle
is shown below in the figure. This figure does not contain the exact detail
found at most power plants; however, it adequately demonstrates
understanding the basic operation of a power cycle.
Figure: Typical Steam Cycle
The typical steam cycle is comprised of the following processes:







Rev 2
1-2: Heat is added to the working fluid in the steam generator under a
constant pressure condition
2-3: Saturated steam from the steam generator is expanded in the
high-pressure (HP) turbine to provide shaft work output at constant
entropy
3-4: Moist steam from the HP turbine exit is dried and superheated in
the moisture separator reheater (MSR)
4-5: Superheated steam from the MSR is expanded in the lowpressure (LP) turbine to provide shaft work at constant entropy
5-6: Steam exhaust from the turbine is condensed in the condenser by
cooling water under a constant vacuum condition
6-7: Condensate is compressed as a liquid by the condensate pump
7-8: Condensate is preheated by the low-pressure feedwater heaters
45



8-9: Condensate is compressed as a liquid by the feedwater pump
9-1: Feedwater is preheated by the high-pressure heaters
1-2: Cycle starts again and heat is added to the working fluid in the
steam generator under a constant pressure condition
The typical steam cycle is shown below on the T-s diagram. The numbered
points on the cycle correspond to the numbered points on the above figure.
The Rankine cycle is an ideal cycle and does not exactly represent the real
processes in the plant but is a closer approximation than the Carnot cycle.
Real pumps and turbines would exhibit an entropy increase across them
when shown on a T-s diagram.
Figure: Rankine Steam Cycle (Ideal)
A T-s diagram of a cycle that closely approximates actual plant processes is
shown below in the figure. The pumps and turbines in this cycle are real
pumps and turbines and show an entropy increase across them. A small
amount of subcooling is evident in the condenser as demonstrated by the
small dip down to point 5.
Subcooling is the process of cooling condensed vapor beyond what is
required for the condensation process. This small amount of subcooling
decreases cycle efficiency because additional heat was removed from the
cycle to the cooling water as heat rejected. This additional heat rejected
must then be added back in the steam generator. Condenser subcooling
decreases the cycle’s efficiency and is sometimes required to have an
adequate suction head to prevent the condensate pumps from cavitating. By
controlling the temperature or flow rate of the cooling water to the
condenser, the operator directly affects the overall cycle efficiency.
Rev 2
46
Figure: Steam Cycle (Real)
The Mollier diagram plots and illustrates the energy transferred to or from
the steam during the cycle. The numbered points on the Mollier diagram
shown below correspond to the numbered points on the previous Rankine
cycle diagrams. The Mollier diagram is limited to plotting only the
saturated or superheated form of the working fluid; the liquid portion of the
steam cycle is not indicated on this type of diagram. The following
conditions are visible on the Mollier diagram:




Point 1: Saturated steam at 540 °F
Point 2: 82.5 percent quality at exit of HP turbine
Point 3: Temperature of superheated steam is 440 °F
Point 4: Condenser vacuum is 1 psia
The solid lines represent the conditions for ideal turbines as verified by the
fact that no entropy change shows across the turbines. The dotted lines
represent the path taken if real turbines were considered, in which case an
increase in entropy is evident.
Rev 2
47
Figure: Mollier Diagram
(Use student copy during lesson)
Causes of Inefficiency
In this section, we compare some of the types and causes for the
inefficiencies of real components and cycles to that of their ideal
counterparts.
Components
In real systems, a percentage of the overall cycle inefficiency is due to the
losses by the individual components. Turbines, pumps, and compressors all
behave non-ideally due to heat losses, friction and windage losses. All of
these losses contribute to the non-isentropic behavior of real equipment. As
previously explained, these losses are disclosed as an increase in the
system's entropy or amount of energy that is unavailable for use by the
cycle.
Rev 2
48
Cycles
Some compromises are made in real systems due to cost and other factors in
the design and operation of the cycle. In a large power generating station,
the condensers are designed to subcool the liquid by 8 °F to 10 °F. This
compromise allows the condensate pumps to pump without cavitation but
each degree of subcooling is energy that must be put back by reheating the
water. This energy used in reheating does no useful work, which decreases
cycle efficiency. Imperfect thermal insulation results in a heat loss to the
environment; again this is energy lost to the system and therefore
unavailable to do work. Both resistance to fluid flow and mechanical
friction in machines are other real world losses that result in decreased cycle
efficiency.
Secondary System Parameters
Operators should have a thorough understanding of how changing plant
parameters affect plant operation and overall efficiency. The loss of
efficiency causes a change in power output from the reactor. The
parameters that could affect thermodynamic efficiency are discussed below.
Increasing Steam Temperature at the Turbine Entrance

A higher turbine inlet steam temperature raises the available work that
can be extracted from the turbine.
 More work increases plant efficiency.
Increasing Feedwater Heating

Raising feedwater temperature to near Tsat for the existing steam
generator pressure increases the plant’s efficiency.
 Less energy from fission process is needed to raise feedwater to
operating temperatures.
 We must account for the energy added to the feedwater.
 Overall plant efficiency increases.
Increasing Condenser Vacuum

More work is extracted from the turbine due to the lower pressure in
the main condenser.
 Condensing steam increases condenser vacuum.
 Overall plant efficiency increases.
Increasing Circulating Water System Flow Rate

Increasing the circulating water system flow rate raises the differential
temperature between condensate and cooling medium.
 This reduces condenser temperature, lowering the pressure at the
exhaust of the turbine.
 Lower pressure at the exhaust of the turbine increases plant
efficiency.
Rev 2
49

However, the additional heat removed by more condensate depression
must be replenished.
Lowering Circulating Water System Inlet Temperature

By decreasing the circulating water system inlet temperature, the
differential temperature is increased between condensate and cooling
medium.
 Larger differential temperature results in larger heat transfer rate,
which lowers condenser temperature and pressure.
 Lower pressure at the exhaust of the turbine increases plant
efficiency.
 However, the additional heat removed by more condensate depression
must be replenished.
Reducing Condensate Depression

Any amount of condensate depression causes more required heat
addition to raise the condensate temperature to the operating
temperature.
 If less energy is rejected to the circulating water, the feedwater
requires less energy from the fission process to raise the temperature
to operating temperatures.
 Controlling condensate depression to the minimum required raises the
plant’s efficiency.
Remove Air and Non-Condensable Gases

Excessive air and non-condensable gases within the main condenser
minimize the heat transfer area.
 More energy is required to achieve the same cooling or achieving
insufficient condensate depression.
 Removal of non-condensable gases increases the plant’s efficiency.
Knowledge Check (Answer Key)
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.
50
Knowledge Check (Answer Key)
Which one of the following pairs of fluids undergoing
heat transfer in typical cross-flow design heat exchangers
yields the greatest heat exchanger overall heat transfer
coefficient? Assume comparable heat exchanger sizes
and fluid flow rates.
A.
Oil to water in a lube oil cooler
B.
Steam to water in a feedwater heater
C.
Water to air in a ventilation heating unit
D.
Water to water in a cooling water heat exchanger
Knowledge Check (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?
Rev 2
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.
51
Knowledge Check (Answer Key)
Which one of the following actions decreases overall
nuclear power plant thermal efficiency?
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
Knowledge Check (Answer Key)
Which one of the following changes causes an increase
in overall nuclear power plant thermal efficiency?
Rev 2
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
52
TLO 2 Summary
During this lesson, you learned about the Second Law of Thermodynamics,
which states that it is impossible to construct a device that operates within a
cycle that can convert all the heat supplied it into mechanical work.
Recognizing that even the most thermally and mechanically perfect cycles
must reject some heat defines thermodynamic power cycle efficiency. The
listing below provides a summary of sections in this TLO.
1. Review ELO 2.1 by asking students the following:
•
Planck's statement of the second law of thermodynamics, which is
that it is impossible to construct an engine that will work in a
complete cycle and produce no other effect except the raising of a
weight and the cooling of a heat reservoir.
•
Entropy is a measure of the unavailability of heat to perform work
in a cycle. This relates to the second law since the second law
predicts that not all heat provided to a cycle can be transformed
into an equal amount of work, some heat rejection must take place.
•
Second law of thermodynamics demonstrates that the maximum
possible efficiency of a system is the Carnot efficiency written as:
𝜂=
𝑇𝐻 − 𝑇𝐶
𝑇𝐻
2. Review ELO 2.2 by having student explain the following statements:
•
Maximum efficiency of a closed cycle can be determined by
calculating the efficiency of a Carnot cycle operating between the
same values of high and low temperatures.
•
Efficiency of a component can be calculated by comparing the work
produced by the component to the work that would have been
produced by an ideal component operating isentropically between
the same inlet and outlet conditions.
•
An isentropic expansion or compression process is represented as a
vertical line on a T-s or h-s diagram. A real expansion or
compression process looks similar, but is slanted slightly to the
right.
•
Maximizing the ΔT and ΔP between the source and the heat sink
ensures the highest possible cycle efficiency.
Rev 2
53
•
The second law of thermodynamics gives a maximum efficiency
limit (which is never reached in physical systems) that an ideal
thermodynamic system can perform. The efficiency is determined
by knowing the inlet and exit absolute temperatures of the overall
system and applying Carnot's efficiency equation.
•
Cycle efficiency = 1 − (𝑇 𝐶 ) (temperature in degrees R)
𝑇
𝐻
3. Review ELO 2.3 by having students plot their plant's steam cycle on a
T-s diagram.
•
A T-s diagram is frequently used to analyze energy transfer
system cycles. Work done by or on the system and heat added
to or removed from the system can be visualized on the T-s
diagram.
•
Use the following as a guideline:
𝜂=
𝑇𝐻 − 𝑇𝐶
𝑇𝐻
4. Review ELO 2.4 by having the students list the major components of
the steam cycle and discuss operating conditions that improve the
plant’s efficiency.
Figure: Typical Steam Cycle and T-s Diagram
•
Rev 2
Discuss conditions yielding improved cycle efficiency, shown below
in the table:
54
Improved Cycle Efficiency
Condition
Effect
Discussion
Superheating
More Efficient With
More Superheating
Increased heat added
results in more net work
from the system, even
though more heat is
rejected.
Moisture Separator
Reheater (MSR)
Use of MSR Has
Minor Effect On
Efficiency
More work is done by the
low-pressure (LP) turbine
since inlet enthalpy is
higher but more heat is
rejected.
The principle benefit of
MSR use is protection of
the final blading stages in
LP turbine from water
droplet impingement.
Feedwater Preheating More Efficient With
Less heat must be added
Feedwater Preheating from the heat source
(reactor) since the
feedwater enters the
steam generator closer to
saturation temperature.
Condenser Vacuum
More Efficient With
Higher Vacuum
(Lower
Backpressure)
Net work output is higher
and heat rejection is
lower as condenser
pressure is lowered.
Condensate
Depression
More Efficient With
Minimal Condensate
Depression
Minimal condensate
depression reduces both
the amount of heat
rejected and the amount
of heat that must be
supplied to the cycle.
Rev 2
55
Condition
Effect
Discussion
Steam
More Efficient At
At higher steam
Temperature/Pressure Higher Steam
temperature, the inlet and
Temperature/Pressure exit entropy from the
turbine are lower so less
heat is rejected.
Steam density increases
as pressure increases, so
more turbine work is
done.
Steam Quality
More Efficient At
Higher Steam
Quality
Enthalpy content
increases as moisture
content decreases and
more net work is done.
•
Hotwell is the area at the bottom of the condenser where the
condensed steam collects to pump back into the system feedwater.
•
Condensate depression is the amount the condensate in a condenser
that is cooled below saturation (degrees subcooled).
•
Condensers operate at a vacuum to ensure the temperature (and thus
the pressure) of the steam is as low as possible.
•
Causes of decreased efficiency include the following:
—
—
—
—
—
—
—
Rev 2
Presence of friction
Heat losses
Cycle inefficiencies
– Subcooling
– Tsat of the steam generator
Turbine service lifetime is affected by moisture impingement
on the blades and other internal parts
Removing as much moisture from the steam limits moisture
content at every stage of the turbine
Feedwater heater is a power plant component used to preheat
water delivered to a steam generating boiler
Moisture separator reheaters improve the plant’s efficiency
and are used to avoid the erosion corrosion and droplet
impingement erosion in the LP turbine, to remove moisture,
and to superheat the steam
56
Objectives
Now that you have completed this lesson, you should be able to do the
following:
1. Explain the second law of thermodynamics using the term entropy.
2. Given a thermodynamic system, determine the:
a. Maximum efficiency of the system
b. Efficiency of the components within the system
3. Differentiate between the path for an ideal process and that for a real
process on a T-s or h-s diagram.
4. Describe how individual factors affect system or component
efficiency.
Thermodynamic Processes Summary
In this module, you learned about applying the First and Second Laws of
Thermodynamics to processes, systems, diagram principles, and energy
balances on major components within a nuclear power generation plant or
facility.
During this lesson, you learned about the First Law of Thermodynamics,
which states that energy can be neither created nor destroyed, but only
altered in form. The energy forms may not always be the same but the total
energy in the system remains constant. You learned about open, closed,
isolated, and steady flow systems. You studied processes including
thermodynamic, cyclic, reversible, irreversible, adiabatic, isentropic, and
isenthalpic.
All of the energies entering and leaving the control volume boundary, any
work done on or by the control volume, and any heat transferred into and
out of the control volume boundaries are in the energy balance equation.
The Second Law of Thermodynamics states that it is impossible to construct
a device that operates within a cycle that can convert all the heat supplied it
into mechanical work. Recognizing that even the most thermally and
mechanically perfect cycles must reject some heat defines thermodynamic
power cycle efficiency.
You studied entropy—a measure of the unavailability of heat to perform
work in a cycle—to explain the second law, and that the change in entropy
determines the direction a given process will proceed.
The Carnot cycle represents an upper limit of efficiency for any given
system operating between the same two temperatures since all practical
systems and processes are irreversible. The system's maximum possible
efficiency would be that of a Carnot cycle, but because Carnot cycles
represent reversible processes, the real system cannot reach the Carnot
efficiency value. Thus, the Carnot efficiency serves as an unattainable
upper limit for any real system's efficiency.
Rev 2
57
You studied the ideal Rankine cycle, which is an ideal cycle where no
increase in entropy occurs as work is done on and by the system.
You learned about the thermodynamics of ideal and real systems by
arranging the various thermodynamic processes on a property diagram to
evaluate the various cycles present in a nuclear power plant. The most
common set of coordinates used is a plot of temperature versus specific
entropy is a T-s diagram. You also studied an h-s diagram, which compares
systems and determines their efficiencies, and a Mollier diagram, which
plots and illustrates the energy transferred to or from the steam during the
cycle.
Now that you have completed this module, you should be able to
demonstrate mastery of this topic by passing a written exam with a grade of
80 percent or higher on the following TLOs:
1. Apply the First Law of Thermodynamics to analyze
thermodynamic systems and processes.
2. Apply the Second Law of Thermodynamics to analyze real and
ideal systems and components.
Rev 2
58
Thermodynamic Cycles Knowledge Check Answer Key
Thermodynamic Cycles Knowledge Check Answers
ELO 1.1 Gas Laws
Knowledge Check
According to Charles’s Law, at low pressure, the
_____ of a gas at constant _____ is directly
proportional to the temperature of the gas.
A.
density; pressure
B.
volume; pressure
C.
pressure; volume
D.
weight; volume
Knowledge Check
Calculate the value of the missing property.
P1= 100 psia; P2 = ?
V1 = 50 ft3; V2 = 25 ft3
T1 = 60°F; T2 = 70°F
Rev 2
A.
233 psi
B.
210 psi
C.
204 psi
D.
200 psi
1
Thermodynamic Cycles Knowledge Check Answer Key
ELO 1.2 Compression Process
Knowledge Check
When can a fluid be considered incompressible?
A.
if it is liquid
B.
if it is steam but not flowing
C.
if it is a saturated vapor
D.
if it is a superheated steam
Knowledge Check
A contained fluid is heated. The resulting change in
pressure will be…
A.
greater for an incompressible fluid.
B.
greater for a compressible fluid.
C.
the same for both fluids.
D.
the same for both fluids only if the volume is held
constant.
ELO 2.1 Thermodynamic Entropy
Knowledge Check
The second law of thermodynamics can also be
expressed as ________ for a closed cycle.
Rev 2
A.
Sf = Si
B.
ΔS ≥ 0
C.
ΔT < 0
D.
ΔS < 0
2
Thermodynamic Cycles Knowledge Check Answer Key
ELO 2.2 Carnot’s Principle of Thermodynamics
Knowledge Check
The steam generators produce dry saturated steam at
1,000 psig. The main condenser is operating with zero
subcooling at 1 psia. What is the maximum efficiency
obtainable?
A.
44 percent
B.
34 percent
C.
28 percent
D.
23 percent
Knowledge Check
Determine the Carnot Efficiency of a steam engine that is
supplied with saturated steam at 300 psia and exhausts to
atmosphere…
Rev 2
A.
44 percent
B.
56 percent
C.
42 percent
D.
35 percent
3
Thermodynamic Cycles Knowledge Check Answer Key
ELO 2.3 Thermodynamics of Ideal and Real Processes
Knowledge Check
Why are real processes shown with dotted lines on
property diagrams?
A.
They occur faster than real processes.
B.
The value of entropy during the process is not
determined.
C.
The entropy values during the process are the same as the
real process until the outlet from the process.
D.
You would not be able to distinguish between real and
ideal processes if the real process was a solid line.
ELO 2.4 Thermodynamic Power Plant Efficiency
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.
4
Thermodynamic Cycles Knowledge Check Answer Key
Knowledge Check
Which one of the following pairs of fluids undergoing
heat transfer in typical cross-flow design heat exchangers
yields the greatest heat exchanger overall heat transfer
coefficient? Assume comparable heat exchanger sizes
and fluid flow rates.
A.
Oil to water in a lube oil cooler
B.
Steam to water in a feedwater heater
C.
Water to air in a ventilation heating unit
D.
Water to water in a cooling water heat exchanger
Knowledge Check
A nuclear power plant is operating at 90 percent of rated
power. Main condenser pressure is 1.7 psia and hotwell
condensate temperature is 120 °F. Which one of the
following describes the effect of a 5 percent decrease in
cooling water flow rate through the main condenser?
Rev 2
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.
5
Thermodynamic Cycles Knowledge Check Answer Key
Knowledge Check
Which one of the following actions decreases overall
nuclear power plant thermal efficiency?
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
Knowledge Check
Which one of the following changes causes an increase
in overall nuclear power plant thermal efficiency?
Rev 2
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
6