Transcript Thermodynamic Processes

Operator Generic Fundamentals
Thermodynamic Cycles
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Operator Generic Fundamentals
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Laws of Thermodynamics Introduction
The Laws simply state:
0.
1.
Two bodies in thermal equilibrium are at the same temperature
Energy can never be created or destroyed
∆𝐸 = 𝑞 + 𝑤
2. The total entropy of the universe must increase in every spontaneous
process
∆𝑆𝑡𝑜𝑡𝑎𝑙 = ∆𝑆𝑠𝑦𝑠𝑡𝑒𝑚 + ∆𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 > 0
3. The entropy (S) of a pure, perfectly crystalline compound at absolute zero
is zero
𝑆𝑇=0 = 0
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Intro
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Laws of Thermodynamics Introduction
• First Law of
Thermodynamics states:
– Energy can be neither
created nor destroyed,
only altered in form
• Second Law of
Thermodynamics states:
– No engine, actual or
ideal, when operating
in a cycle can convert
all the heat supplied it
into mechanical work–
heat must be rejected
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Intro
Operator Generic Fundamentals
Thermodynamic Cycles TLO1 - ELOs
TLO 1 - Apply the first law of thermodynamics to analyze thermodynamic
systems and processes.
1.1 Relate the following terms to open, closed, and isolated systems:
a. Thermodynamic surroundings
b. Thermodynamic equilibrium
c. Control volume
d. d. Steady state
1.2 Describe the following processes:
a. Thermodynamic process
b. Cyclic process
c. Reversible process
d. Irreversible process
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Thermodynamic Processes - ELOs
1.2 Describe the following processes (continued):
e. Adiabatic process
f.
Isentropic process
g. Throttling process (Isenthalpic)
1.3 Apply the First Law of Thermodynamics for open systems or cyclic
processes.
1.4 Identify the path(s) on a T-s diagram that represents the
thermodynamic processes occurring in a fluid system.
1.5 Given a defined system, perform energy balances on all major
components in the system.
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First Law of Thermodynamics
ELO 1.1 - Relate the following terms to Open, Closed, and Isolated
Systems: a. Thermodynamic surroundings, b. Thermodynamic
equilibrium, c. Control volume, d. Steady state
Figure: Energy Balance Equals the First Law of Thermodynamics
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First Law of Thermodynamics
• The First Law of Thermodynamics is an energy balance in a defined
system
– Energy can be neither created nor destroyed, only altered in form
– Referred to as the Conservation of Energy Principle
• For any system, energy transfer is associated with:
– Mass and energy crossing the control boundary
– External work and/or heat crossing the boundary
– Change of stored energy within the system
• Mass flow of fluid is associated with kinetic, potential, internal, and
flow energies, all affecting the overall energy balance of a system
– Exchange of external work and/or heat completes energy balance
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TLO 1
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Thermodynamic Systems and
Surroundings
• Thermodynamics involves the study of various systems
• A system is a collection of matter being studied, examples
– Water within one side of a heat exchanger
– Fluid inside a length of pipe
– Entire lubricating oil system for a diesel engine
• Determining the boundary to solve a thermodynamic problem for a
system depends on
– What information is known about the system
– What question is asked, or requested, about the system
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Types of Thermodynamic Systems
• A thermodynamic system is any three-dimensional region of space
bounded by one or more surfaces
• Bounding surfaces may be
– Real or imaginary
– At rest or in motion
• A boundary may change its size or shape
• The region of physical space outside the system’s selected
boundaries is called the surroundings or the environment
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Types of Thermodynamic Systems
Three system types:
• Isolated – completely
separated from its
surroundings. No mass
or energy cross its
boundaries.
• Closed – no mass
crosses its boundaries
but energy can cross the
boundaries.
• Open – has both mass
and energy crossing its
boundaries.
Figure: Types of Thermodynamic Systems
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Control Volume
• Control volume - fixed region in
space chosen for the
thermodynamic study of mass
and energy balances for flowing
systems
• A boundary may be a real or
imaginary envelope
• Control surface - is the
boundary of the control volume
Figure: Control Volume in an Open System
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Open System
• Typical analysis of an open
system assumes steady flow
conditions exist for the system.
Both mass and energy
cross the boundary of
the nozzle and the
atmosphere.
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Steady Flow System
• The following are constant
– Mass flow rates into and out of the system
– Physical properties of the working substance at any selected
location are constant with time
– Rate at which heat crosses the system boundary
– Rate at which work is performed is constant
• There is no accumulation of mass or energy within the control volume
• The properties at any point within the system are independent of time
• System equilibrium regards all possible changes in state
– The system is also in thermodynamic equilibrium
– For examples: if a gas that composes a system is in thermal
equilibrium, the temperature will be the same throughout the
entire system
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Types of Thermodynamic Systems
The RCS can be considered each type of system under certain
operational conditions
Figure: Reactor Coolant System a Type of Thermodynamic System
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Steady Flow Equilibrium Process
Figure: Steady Flow Systems
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First Law of Thermodynamics
Figure: Basic Energy Balance of the First Law of Thermodynamics
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Continuity Equation
Figure: Continuity Equation for the First Law of Thermodynamics
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Thermodynamic Surroundings,
Equilibrium, and Control Volume
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
Correct answer is C.
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Thermodynamic Processes
ELO 1.2 - Describe the following processes: a. Thermodynamic process,
b. Cyclic process, c. Reversible process, d. Irreversible process, e.
Adiabatic process, f. Isentropic process, g. Throttling process
(Isenthalpic)
Thermodynamic
Processes
• Transformation of
a working fluid
from one state to
another
• Evidence is a
change in one or
more fluid
properties
Figure: Thermodynamic Process Shows Transformation of a Working Fluid
from One State to Another
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Thermodynamic Processes
• Whenever one or more of the properties of a system changes, a
change in the state of the system occurs
• Thermodynamic process is the path of the succession of states
through which the system passes such as:
– Increasing fluid temperature while maintaining a constant
pressure
– Increasing confined gas pressure while maintaining a constant
temperature
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Thermodynamic Processes
Figure: Six Basic Processes of Steady Flow Systems
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Reversible Process
• A process that having taken place can be reversed, and doing so
leaves no change in either the system or its surroundings
– System and surroundings are returned to their original condition
before the process occurred
– A reversible process is a starting point for engineering study and
calculation
• Reversible processes can be approximated but never matched by
real processes
– One way to make a real process approximate a reversible
process is to carry out the process in a series of small steps
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Reversible Process
• Heat transfer may be considered reversible if it occurs due to a small
temperature difference between the system and its surroundings
– Transferring heat across a temperature difference of 0.00001 °F
appears to be more reversible than transferring heat across a
temperature difference of 100 °F
– By cooling or heating the system in a number of infinitesimal
steps, a reversible process can be approximated
• Although not practical for real processes, this method benefits
thermodynamic studies because the rate at which processes occur is
not important
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Irreversible Process
• An irreversible process is a process that cannot return both the
system and the surroundings to their original conditions
– System and surroundings will not return to their original conditions
if the process was reversed
– An automobile engine does not give back the fuel it took to drive
up a hill as it coasts back down the hill
• Factors that make a process irreversible:
– Friction
– Unrestrained expansion of a fluid
– Heat transfer through a finite temperature difference
– Mixing two different substances
– These factors are present in real, irreversible processes and
prevent those processes from being reversible
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Real Piston and Cylinder
If a small weight is removed from
the piston, it will probably not
move due to friction between the
piston and cylinder.
Additional weights would have to
be removed from the piston to
the shelf until the piston breaks
free and overcomes the
restraining friction forces.
Once it breaks free, the piston
will accelerate upward since the
pressure applied by the gas is
now more than sufficient to
balance the weight against it.
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Figure: Ideal Piston and Cylinder
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Ideal Piston and Cylinder
If a single metal shaving is
moved from the piston to the
shelf, since no friction exists,
the piston responds
immediately by moving upward
by a very small amount, ΔL
(change in length).
This process could be reversed
by placing weights from the
shelves onto the piston, which
would move the piston
downward to original position.
Figure: Ideal Piston and Cylinder
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Process Terminology
• Adiabatic
– No loss or gain of heat by the working fluid
• Isobaric
– Take place at constant pressure
• Isentropic
– No change in entropy
• Isenthalpic
– No change in enthalpy
• Isothermal
– Take place at constant temperature
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Adiabatic and Isentropic Processes
Adiabatic Process
• An adiabatic process is one where no heat transfers into or out of the
system
• System can be considered to be perfectly insulated
Isentropic Process
• An isentropic process is one in which the entropy of the fluid remains
constant
• This is true if the process the system goes through is reversible and
adiabatic
• Also called a constant entropy or an ideal process
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Isenthalpic Processes
• An isenthalpic process is one in which the enthalpy of the fluid
remains constant
• This will be true if the process the system goes through has
– No change in enthalpy from state one to state two (h1= h2)
– No work is done (W = 0)
– The process is adiabatic (Q = 0)
• Throttling processes are isenthalpic and therefore move from left to
right on a Mollier diagram
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Real and Ideal Processes
Knowledge Check
In what type of process does the fluid pass through various system
processes and then returns to the same state it began?
A. throttling process
B. isentropic process
C. adiabatic process
D. cyclic process
Correct answer is D.
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Analyzing Systems Using the First Law
of Thermodynamics
ELO 1.3 - Apply the First Law of Thermodynamics for open systems or
cyclic processes.
• Recall that energy in thermodynamic systems is composed of kinetic
energy (KE), potential energy (PE), internal energy (U), and flow
energy (PL); as well as heat and work processes.
• Energy balances can be expressed mathematically
Σ 𝑎𝑙𝑙 𝑒𝑛𝑒𝑟𝑔𝑖𝑒𝑠 𝑖𝑛 = Σ(𝑎𝑙𝑙 𝑒𝑛𝑒𝑟𝑔𝑖𝑒𝑠 𝑜𝑢𝑡) + Δ(𝑒𝑛𝑒𝑟𝑔𝑦 𝑠𝑡𝑜𝑟𝑒𝑑 𝑖𝑛 𝑠𝑦𝑠𝑡𝑒𝑚)
Σ𝐸𝑖𝑛 = Σ𝐸𝑜𝑢𝑡 + Δ𝐸𝑠𝑡𝑜𝑟𝑎𝑔𝑒
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Control Volume
Figure: Open System Control Volume Concept for a Pump
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Open System Control Volume for Multiple
Processes
Figure: Open System Control Volume for Multiple Processes
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Conservation of Energy
𝑚 ℎ𝑖𝑛 + 𝑃𝐸𝑖𝑛 + 𝐾𝐸𝑖𝑛 + 𝑄 = 𝑚 ℎ𝑜𝑢𝑡 + 𝑃𝐸𝑜𝑢𝑡 + 𝐾𝐸𝑜𝑢𝑡 + 𝑊
Where:
𝑚 = 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)
𝑊 = rate of work done by the system (ft-lbf/hr)
𝑄 = heat rate into the system (BTU/hr)
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Heat Transferred Into or Out of System
Figure: Heat and Work in a System
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Open System Analysis
• Heat and/or work can be directed into or out of the control volume
• Standard convention - net heat exchange assumed to be into system
and net work assumed to be out of system
• Isolated and closed systems are specialized cases of an open
system
– Closed system - no mass crosses boundary but work and/or heat
do
– Isolated system - Mass, work, and heat do not cross the boundary
(that is, the only energy exchanges taking place are within the
system)
• Open system approach to First Law of Thermodynamics will be
emphasized because it is more general
– Almost all practical applications of the first law require an open
system analysis
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Open System Analysis
• Two approaches exist in studying Thermodynamics:
– Control mass approach
– Control volume approach
• Control mass concept a clump of fluid is studied with its associated
energies
• Analyzer rides with the clump wherever it goes, keeping a balance of
all energies affecting the clump
• Control Volume is much more commonly used
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Open System Analysis
• The forms of energy that may
cross control volume boundary
include those associated with
mass (m) crossing the boundary
• Mass in motion has potential
(PE), kinetic (KE), and internal
energy (U)
– Flow is normally supplied
with some driving power
there is another form of
energy associated with fluid
caused by its pressure
– Form of energy is flow
energy (PV)
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Figure: Multiple Control Volumes in the Same System
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Open System Analysis
• In open system analysis U and PV terms occur so frequently that
another property enthalpy has been defined as h = U + PV
• This results in the expression being written as m (h + ke + PE)
• In addition to mass and energies, externally applied work (W), usually
designated as shaft work, is another form of energy that may cross
the system boundary
Example: Open System Control Volume
• Enthalpies of steam entering and leaving a steam turbine are 1,349
BTU/lbm and 1,100 BTU/lbm
• Estimated heat loss is 5 BTU/lbm of steam
• 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
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Open System Analysis
Step 1. Draw and define the
system boundaries
• Enthalpies entering and
leaving a steam turbine are
1,349 BTU/lbm and 1,100
BTU/lbm
• Heat loss is 5 BTU/lbm of
steam
• 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
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Figure: Open System Control Volume Concept
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Open System Analysis
Step 2. Write the General Energy Equations
𝑚𝑖𝑛 (ℎ𝑖𝑛 + 𝑃𝐸𝑖𝑛 + 𝐾𝐸𝑖𝑛 ) + 𝑞 = 𝑚𝑜𝑢𝑡 (ℎ𝑜𝑢𝑡 + 𝑃𝐸𝑜𝑢𝑡 + 𝐾𝐸𝑜𝑢𝑡 ) + 𝑤
Step 3. Simplify the General Energy Equation when possible
Divide by m since: 𝑚𝑖𝑛 = 𝑚𝑜𝑢𝑡 = 𝑚
(ℎ𝑖𝑛 + 𝑃𝐸𝑖𝑛 + 𝐾𝐸𝑖𝑛 ) + 𝑞 = (ℎ𝑜𝑢𝑡 + 𝑃𝐸𝑜𝑢𝑡 + 𝐾𝐸𝑜𝑢𝑡 ) + 𝑤
Where:
𝑞 = heat added to the system per pound (BTU/lbm)
𝑤 = work done by the system per pound (ft-lbf/lbm)
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Open System Analysis
Step 4. Convert for required terms
Use Joule’s constant 𝐽 = 778
known values.
𝐵𝑇𝑈
1,349
+
𝑙𝑏𝑚
𝑓𝑡–𝑙𝑏𝑓
𝐵𝑇𝑈
for conversions and substitute
6.5 𝐵𝑇𝑈
164 2
+
778 𝑙𝑏𝑚
2 32.17 778
𝐵𝑇𝑈
+ −5
=
𝑙𝑏𝑚
𝐵𝑇𝑈
262 2
𝐵𝑇𝑈
1,100
+ 0 𝑃𝐸𝑜𝑢𝑡 +
+𝑤
𝑙𝑏𝑚
2 32.17 778 𝑙𝑏𝑚
Note: The minus sign indicates heat out of the turbine.
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Open System Analysis
Solve for work (w):
𝐵𝑇𝑈
𝐵𝑇𝑈
𝐵𝑇𝑈
𝐵𝑇𝑈
−3
1,349
+ 8.3548 × 10
+ 0.5368
–5
=
𝑙𝑏𝑚
𝑙𝑏𝑚
𝑙𝑏𝑚
𝑙𝑏𝑚
𝐵𝑇𝑈
𝐵𝑇𝑈
1,100
= 1.37
+ 𝑤
𝑙𝑏𝑚
𝑙𝑏𝑚
𝐵𝑇𝑈
𝐵𝑇𝑈
1,344.54
= 1,101.37
+𝑤
𝑙𝑏𝑚
𝑙𝑏𝑚
𝐵𝑇𝑈
𝐵𝑇𝑈
𝑤 + 1,344.54
– 1,101.37
𝑙𝑏𝑚
𝑙𝑏𝑚
𝐵𝑇𝑈
𝑤 = 243.17
𝑙𝑏𝑚
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Open System Analysis
Example 2: Gas Filled Piston
• A system comprised of a certain mass of air is contained in a cylinder
fitted with a piston.
• The air expands from an initial state for which E1 = 70 BTUs to a final
state for which E2 = 20 BTUs.
• During the expansion, the air does 60 BTUs of work on its
surroundings.
• Find the amount of heat transferred to or from the system during the
process.
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Open System Analysis
Figure: Gas Filled Piston
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Open System Analysis
Step 2. Write the General Energy Equations
𝑚𝑖𝑛 (ℎ𝑖𝑛 + 𝑃𝐸𝑖𝑛 + 𝐾𝐸𝑖𝑛 ) + 𝑞 = 𝑚𝑜𝑢𝑡 (ℎ𝑜𝑢𝑡 + 𝑃𝐸𝑜𝑢𝑡 + 𝐾𝐸𝑜𝑢𝑡 ) + 𝑤
Step 3. Simplify the General Energy Equation when possible
𝑄 + 𝐸1 = 𝑊 + 𝐸2
𝑄 = 𝑊 + 𝐸2– 𝐸1
𝑄 = 60 + 20– 70 = 10 𝐵𝑇𝑈𝑠
The positive sign indicates that 10 BTUs of heat is added to the
system. Work was taken from the system. The total decrease in the
stored energy of the system equals the difference between the energy
added as heat and the energy removed as work.
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Open System Analysis
Example: 3 Heat and Work
Calculate the final E value for:
• A mass of water has an initial state 20 BTUs of energy
• 7, 780 ft-lbf of work is done on the water
• 3 BTUs of heat are removed from it.
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Open System Analysis
Step 1. Draw and define the system boundaries
Figure: Heat and Work in a Closed System
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Open System Analysis
Step 2. Write the General Energy Equations
𝑚𝑖𝑛 (ℎ𝑖𝑛 + 𝑃𝐸𝑖𝑛 + 𝐾𝐸) + 𝑞 = 𝑚𝑜𝑢𝑡 (ℎ𝑜𝑢𝑡 + 𝑃𝐸𝑜𝑢𝑡 + 𝐾𝐸𝑜𝑢𝑡 ) + 𝑤
Step 3. Simplify the General Energy Equation when possible
𝑄 + 𝐸1 = 𝑊 + 𝐸2
𝐸2 = 𝑄 + 𝐸1 – 𝑊
We must convert all of the energy values to the same units
1 𝐵𝑇𝑈
𝑊 = 7,780 𝑓𝑡– 𝑙𝑏𝑓
= 10 𝐵𝑇𝑈𝑠
778 𝑓𝑡– 𝑙𝑏𝑓
𝐸1 = 20 𝐵𝑇𝑈𝑠
𝑄 = −3 𝐵𝑇𝑈𝑠
𝐸2 = (−3) + 20 – (−10)
𝐸2 = 27 𝐵𝑇𝑈𝑠
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𝑊 = −10 𝐵𝑇𝑈𝑠
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Applying the First Law of Thermodynamics
Knowledge Check
In an Open Steady Flow System, choose the energies that are
associated with the mass crossing the system boundary.
A. kinetic energy, potential energy, internal energy, flow energy
B. work, kinetic energy, potential energy, heat
C. kinetic energy, heat, internal energy, flow energy
D. work, potential energy, internal energy, flow energy
Correct answer is A.
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Identifying Process Paths on a T-s Diagram
ELO1.4 - Identify the path(s) on a T-s diagram that represents the
thermodynamic processes occurring in a fluid system.
• When a system changes properties (temperature, pressure, or
volume) from one value to another as a consequence of work or heat
or internal energy exchange, its fluid goes through a process
• In some processes, the relationships between pressure, temperature,
and volume are specified as the fluid changes thermodynamic state
• Most common processes are those in which the temperature,
pressure, or volume holds constant during the process
• Classified as isothermal, isobaric, or isovolumetric processes
– Iso means constant or one
• If the fluid passes through various processes and then eventually
returns to the same state with which it began, the system is said to
have undergone a cyclic process
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Identifying Process Paths on a T-s Diagram
• One such cyclic process used is the
Rankine cycle
• ab: Liquid is compressed with no
change in entropy (by an ideal
pump)
• bc: Constant pressure transfer of
heat in the boiler
– Heat added to compressed
liquid, 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)
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Figure: T-s Diagram With Rankine Cycle
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Identifying Process Paths on a T-s Diagram
• These are individual
processes that fluid must go
through before completing an
entire cycle
• Typical steam plant system
– Heat source (Reactor) to
produce the thermal
energy (QA)
– Steam generator to
change the thermal energy
into steam energy
– Steam turbine to extract
work from the steam (WT)
© Copyright 2014
Figure: Typical Steam Plant System Cyclic Process
ELO 1.4
Operator Generic Fundamentals
54
Identifying Process Paths on a T-s Diagram
• Typical steam plant system
– Pumps to transfer fluid back
to the heat source (WP)
– Heat sink rejects unused
heat and condenses steam
(QR)
• The steam plant in its entirety
is a large closed system
• Each component may be
thermodynamically analyzed as
an open system as the fluid
passes through it
© Copyright 2014
Figure: Typical Steam Plant System Cyclic Process
ELO 1.4
Operator Generic Fundamentals
55
Identifying Process Paths on a T-s Diagram
Typical Steam Plant Cycle
• Heat is supplied to a steam
generator (boiler) where
liquid converts to steam or
vapor (5 – 1)
• Vapor expands adiabatically
in the turbine to produce a
work output (1 – 2)
• Vapor leaving the turbine
enters the condenser where
heat is removed and vapor is
condensed into a liquid state
(2 – 3)
© Copyright 2014
Figure: Rankine Cycle for a Typical Steam Plant
ELO 1.4
Operator Generic Fundamentals
56
Identifying Process Paths on a T-s Diagram
Typical Steam Plant Cycle
• The condensation process is a
heat-rejection mechanism for
the cycle (area under 2 – 3)
• Saturated liquid is delivered to
condensate pump and then the
feed pump where its pressure
is raised to the saturation
pressure, corresponding to the
steam generator temperature
(3 – 5)
Figure: Rankine Cycle for a Typical Steam Plant
• High-pressure liquid is
delivered to the steam
generator, where the cycle
repeats itself
© Copyright 2014
ELO 1.4
Operator Generic Fundamentals
57
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% to 99.9%.
C. Bypassing a feedwater heater during normal plant operations.
D. Increasing condenser pressure to 2 psia from 1 psia.
Correct answer is B.
© Copyright 2014
ELO 1.4
Operator Generic Fundamentals
58
Energy Balances on Major Components
ELO 1.5 - Given a defined system, Perform energy balances on all major
components in the system.
Boundaries can be set on any component
Figure: Cyclic Process for Generating Electricity
© Copyright 2014
ELO 1.5
58
Operator Generic Fundamentals
59
Analyzing Cyclic Process – Steam
Generator
• The process of dissipating the energy created by the heat source is
of primary importance
• This process takes place in the steam generator, which is a twophase heat exchanger
– Hot fluid (TH) from the reactor passes through primary side of the
steam generator where its energy is transferred to the secondary
side of the heat exchanger to create steam
– Colder Fluid (TC), with its energy removed, leaves the primary
side and is pumped back to the heat source for reheating
• The analysis can be performed on the secondary side as the heat
transferred from the primary system must equal the heat transferred
into the secondary system
– Change in temperature cannot be used because a phase change
occurs
© Copyright 2014
ELO 1.5
Operator Generic Fundamentals
60
Steps for Solving Energy Balance
Problems
Step
Action
1.
Draw the system with the boundaries
Write the general energy equation and solve for
the required information.
Determine which energies can be ignored to
simplify the equation.
Make substitutions to ensure correct units are
obtained if needed.
2.
3.
4.
© Copyright 2014
ELO 1.5
60
Operator Generic Fundamentals
61
Analyzing Cyclic Process
Steam Generator Analysis – Primary Side
• Fluid from heat source enters the steam generator at 610 °F and
leaves at 540 °F
• Flow rate is approximately 1.38 x 108 lbm/hr
• If specific heat of the fluid is taken as 1.5 BTU/lbm-°F
What is the heat transferred out of the steam generator?
© Copyright 2014
ELO 1.5
Operator Generic Fundamentals
62
Analyzing Cyclic Process
Step 1. Draw the system
• Show what is given and what is asked for, or requested
Figure: Steam Generation System
© Copyright 2014
ELO 1.5
Operator Generic Fundamentals
63
Analyzing Cyclic Process
Step 2 and 3. Write the equation and simplify as possible
ṁ𝑖𝑛 ℎ𝑖𝑛 + 𝑃𝐸𝑖𝑛 + 𝐾𝐸𝑖𝑛 + 𝑄 = 𝑚𝑜𝑢𝑡 (ℎ𝑜𝑢𝑡 + 𝑃𝐸𝑜𝑢𝑡 + 𝐾𝐸𝑜𝑢𝑡 ) + Ẇ
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:
𝑚 ℎ𝑖𝑛 + 𝑄 = 𝑚(ℎ𝑜𝑢𝑡 )
𝑄 = 𝑚(ℎ𝑜𝑢𝑡 – ℎ𝑖𝑛 )
Substituting, 𝑄 = 𝑚𝑐𝑝 𝛥𝛵 , where cp = specific heat capacity (𝐵𝑇𝑈 / 𝑙𝑏𝑚−℉):
= 𝑚 𝑐𝑝 𝑇𝑜𝑢𝑡 – 𝑇𝑖𝑛
𝐵𝑇𝑈
8 𝑙𝑏𝑚
= 1.38 × 10
1.5
ℎ𝑟
𝑙𝑏𝑚−℉
10 𝐵𝑇𝑈
𝑄 = −1.45 × 10
ℎ𝑟
© Copyright 2014
540 − 610 ℉
Note: Minus sign
indicates heat out
of heat exchanger
ELO 1.5
Operator Generic Fundamentals
64
Analyzing Cyclic Process
Pumps
•
•
•
•
A pump returns the fluid from the SG to the reactor core.
Flow rate through the pump is 3.0 x 107 lbm/hr.
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?
© Copyright 2014
ELO 1.5
Operator Generic Fundamentals
65
Analyzing Cyclic Process
Step 1. Draw the system
• Show what is given and what is asked for, or requested
?
Figure: Pump Returns Fluid From Heat Exchanger to Core
© Copyright 2014
ELO 1.5
Operator Generic Fundamentals
66
Analyzing Cyclic Process
Step 2. Write the general energy equation.
𝑚(ℎ𝑖𝑛 + 𝑃𝐸𝑖𝑛 + 𝐾𝐸𝑖𝑛 ) + 𝑄 = 𝑚(ℎ𝑜𝑢𝑡 + 𝑃𝐸𝑜𝑢𝑡 + 𝐾𝐸𝑜𝑢𝑡 ) + 𝑊
Step 3. Simplify the equation.
Assume 𝑄 = 0 and neglect changes in PE and KE
=
𝑚 ℎ
𝑚 ℎ𝑜𝑢𝑡 + 𝑊
𝑖𝑛
+
)
𝑊 = 𝑚(ℎ𝑖𝑛 ℎ𝑜𝑢𝑡 where 𝑊 is the rate of doing work by the pump
ℎ𝑖𝑛 = 𝑢𝑖𝑛 + ѵ𝑃𝑖𝑛
ℎ𝑜𝑢𝑡 = 𝑢𝑜𝑢𝑡 + ѵ𝑃𝑜𝑢𝑡
(ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 ) = 𝑈𝑖𝑛 − 𝑈𝑜𝑢𝑡 + ѵ𝑃𝑖𝑛 − ѵ𝑃𝑜𝑢𝑡 = 𝛥𝑈 + ѵ𝑃𝑖𝑛 − ѵ𝑃𝑜𝑢𝑡
© Copyright 2014
ELO 1.5
Operator Generic Fundamentals
67
Analyzing Cyclic Process
Step 3 and 4. Simplify and arrange the terms
Since no heat is transferred, ΔU = 0 and the specific volume out of the pump is
the same as the specific volume entering since water is incompressible.
(ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 ) = ѵ 𝑃𝑖𝑛 − 𝑃𝑜𝑢𝑡
Step 4. Arrange the correct terms:
Substituting the expression for work, 𝑊 = 𝑚 ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 we have:
𝑊 = 𝑚ѵ 𝑃𝑖𝑛 − 𝑃𝑜𝑢𝑡
Using 0.01246 for specific volume:
3
𝑓𝑡
𝑙𝑏𝑚
10
0.01246
−90 𝑝𝑠𝑖𝑎
ℎ𝑟
𝑙𝑏𝑚
𝑊 = 3.0 ×
𝑓𝑡– 𝑙𝑏𝑓
778 𝐵𝑇𝑈
7
2
𝑖𝑛
144 2
𝑓𝑡
𝐵𝑇𝑈
6
𝑊 = −6.23 × 10
© Copyright 2014
ℎ𝑟
or -2,446 hp
ELO 1.5
Operator Generic Fundamentals
68
Analyzing Cyclic Process
Heat Exchangers
•
•
•
•
The temperature leaving the heat source is 612 °F
The temperature 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 BTU/lbm-°F.
How much heat is being removed from the heat source?
© Copyright 2014
ELO 1.5
Operator Generic Fundamentals
69
Analyzing Cyclic Process
Step 1. Draw the system
• Show what is given and what is asked for, or requested
Figure: Heat Exchanger Analysis Shows Thermodynamic Balance
© Copyright 2014
ELO 1.5
Operator Generic Fundamentals
70
Analyzing Cyclic Process
Step 2 and 3. Write the general energy equations and reduce the terms
as appropriate. The PE and KE energies are small compared to other
terms and may be neglected.
𝑚(ℎ𝑖𝑛 + 𝑃𝐸𝑖𝑛 + 𝐾𝐸𝑖𝑛 ) + 𝑄 = 𝑚(ℎ𝑜𝑢𝑡 + 𝑃𝐸𝑜𝑢𝑡 + 𝐾𝐸𝑜𝑢𝑡 ) + 𝑊
𝑄 = 𝑚 ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛
© Copyright 2014
ELO 1.5
Operator Generic Fundamentals
71
Analyzing Cyclic Process
Step 3 and 4. Make needed substitutions to ensure correct units are
obtained.
Substituting 𝑄 = 𝑚𝑐𝑝 𝛥𝑇, where cp = specific heat capacity:
𝑄 = 𝑚 𝑐𝑝 𝑇𝑜𝑢𝑡 – 𝑇𝑖𝑛 + 𝑊
𝐵𝑇𝑈
8 𝑙𝑏𝑚
𝑄 = 1.32 × 10
. 1.47
612 – 542 °𝐹 + 0
ℎ𝑟
𝑙𝑏𝑚– ℉
10 𝐵𝑇𝑈
𝑄 = 1.36 𝑥 10
ℎ𝑟
For this example, 𝑄 = 𝑚𝑐𝑝 𝛥𝑇 has been used to calculate the heat
transfer rate since no phase change has occurred. However, 𝑄 =
𝑚 ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 could also have been used if the problem data included
inlet and outlet enthalpies.
© Copyright 2014
ELO 1.5
Operator Generic Fundamentals
72
Analyzing Cyclic Process
Overall Plant
• A steam generating facility is studied as a complete system
• The heat produced by the heat source is 1.36 x 1010 BTU/hr
• The heat removed by the heat exchanger (steam generator) is 1.361
x 1010 BTU/hr
What is the required pump power to maintain a stable temperature?
© Copyright 2014
ELO 1.5
Operator Generic Fundamentals
73
Analyzing Cyclic Process
Step 1. Draw the system.
• Show what is given and what is asked for, or requested.
𝑊𝑝 = pump work
𝑄𝑐 = heat produced by
the heat source
𝑄𝑆𝐺 = heat transferred
into steam generator
© Copyright 2014
Figure: Pump Power of a Steam Generating Facility
ELO 1.5
Operator Generic Fundamentals
74
Analyzing Cyclic Process
Steps 2 and 3. Write and simplify the equation.
𝑚(ℎ + 𝑃𝐸 + 𝐾𝐸) + 𝑊𝑝 + 𝑄𝑐 = 𝑄𝑆𝐺 + 𝑚 + 𝑃𝐸 + 𝐾𝐸
For a closed system, the mass entering and leaving the system is zero,
therefore, 𝑚 is constant. The energy entering and leaving the system is
zero, and you can assume that the KE and PE are constant so that:
𝑄𝑐 + 𝑊𝑝 = 𝑄𝑆𝐺
𝑊𝑝 = 𝑄𝑆𝐺 − 𝑄𝑐
𝐵𝑇𝑈
10 𝐵𝑇𝑈
= 1.361 × 10
– 1.36 × 10
ℎ𝑟
ℎ𝑟
7 𝐵𝑇𝑈
= 1.0 × 10
ℎ𝑟
Step 4. Arrange equation for the proper units.
10
Recall that 1 hp = 2,545 BTU/hr. Therefore, converting to hp:
𝑊𝑝 = 3,929 ℎ𝑝
© Copyright 2014
ELO 1.5
Operator Generic Fundamentals
75
Analyzing Cyclic Process
Phase Changes
• Steam flows through a condenser at 4.4 x 106 lbm/hr,
• Enters as saturated vapor at 104 °F (h = 1,106.8 BTUs/lbm), and
• Exits at the same pressure as subcooled liquid at 86 °F (h = 54
BTUs/lbm).
• Cooling water temperature is 64.4 °F (h = 32 BTUs/lbm).
• Environmental requirements limit the Circulating Water exit
temperature to 77 °F (h = 45 BTUs/lbm)
Determine the required cooling water flow rate.
© Copyright 2014
ELO 1.5
Operator Generic Fundamentals
76
Analyzing Cyclic Process
Step 1. Draw the system. Show what is given and what is asked for, or
requested.
Figure: Typical Single-Pass Condenser End View
© Copyright 2014
ELO 1.5
Operator Generic Fundamentals
77
Analyzing Cyclic Process
Step 2, 3, and 4. Write the equation and simplify as possible convert to
required units
𝑄𝑠𝑡𝑚 = −𝑄
𝑐𝑤
Steps 3 and 4. Simplify the equation and arrange for required units.
𝑚𝑠𝑡𝑚 ℎ𝑜𝑢𝑡 – ℎ𝑖𝑛 𝑠𝑡𝑚 = 𝑚 ℎ𝑜𝑢𝑡 – ℎ𝑖𝑛 𝑐𝑤
𝑚𝑠𝑡𝑚 ℎ𝑜𝑢𝑡 – ℎ𝑖𝑛 𝑠𝑡𝑚
𝑚𝑐𝑤 =
ℎ𝑜𝑢𝑡 – ℎ𝑖𝑛 𝑐𝑤
𝐵𝑇𝑈𝑠
54–
1106.8
6 𝑙𝑏𝑚
𝑙𝑏𝑠
𝑚𝑐𝑤 = 4.4 × 10
ℎ𝑟 45 − 32 𝐵𝑡𝑢𝑠
𝑙𝑏𝑚
8 𝑙𝑏𝑚
𝑚𝑐𝑤 = 3.67 × 10
ℎ𝑟
© Copyright 2014
ELO 1.5
Operator Generic Fundamentals
78
Energy Balances on Major Components
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. 1006 Mw
D. 1258 Mw
Correct answer is D.
© Copyright 2014
ELO 1.5
Operator Generic Fundamentals
79
Summary
ELO 1.1
• First Law of Thermodynamics - energy can be neither created nor
destroyed, only altered in form.
• Energy into the system equals the energy leaving the system.
• The amount of energy transferred across a heat exchanger depends
on the temperature of the fluid entering the heat exchanger from both
sides and the flow rates of these fluids.
• A T-s diagram can be used to represent thermodynamic processes.
• Thermodynamic system - collection of matter and space with its
boundaries defined in such a way that the energy transfer across the
boundaries can be best understood.
• Surroundings are everything not in the system being studied.
© Copyright 2014
TLO 1 Summary
Operator Generic Fundamentals
80
Summary
ELO 1.1 (continued)
• System groups:
– Isolated system - neither mass nor energy can cross the
boundaries
– Closed system - only energy can cross the boundaries
– Open system - both mass and energy can cross the boundaries
• Control volume - fixed region of space studied as a thermodynamic
system.
• Steady state - condition where the properties at any given point within
the system are constant over time. Neither mass nor energy are
accumulating within the system.
© Copyright 2014
TLO 1 Summary
Operator Generic Fundamentals
81
Summary
ELO 1.2
• Thermodynamic process - is the succession of states that a system
passes through
• Cyclic process - fluid passes through various processes and then returns
to the same state in which it began
• Reversible process - a process that can be reversed, resulting in no
change in the system or surroundings
• Irreversible process - a process that, if reversed, would result in a change
to the system or surroundings
• Adiabatic process - a process in which there is no heat transfer across the
system boundaries
• Isentropic process - a process in which the entropy of the system remains
unchanged
• Throttling process - a process in which enthalpy is constant h1 = h2, work =
0, and which is adiabatic, Q = 0.
© Copyright 2014
TLO 1 Summary
Operator Generic Fundamentals
82
Summary
ELO 1.3
• Define: Thermodynamic cycle - 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:
– Supply of energy from a source (steam generator).
– Conversion of some of the energy to work in a turbine.
– Rejection of most of the remaining steam energy to a heat sink
(condenser).
– Condensed steam (liquid water) is pumped back to the source to
start the cycle again.
© Copyright 2014
TLO 1 Summary
Operator Generic Fundamentals
83
Summary
ELO 1.4
• Describe the two primary classes of thermodynamic cycles
– Power cycles
– Heat pump cycles
• Thermodynamic cycle efficiency is the ratio of net work or energy
output of the system divided by the heat energy added to the system
• 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.
• Heat engine performs the conversion of heat energy to mechanical
work by exploiting the temperature gradient between a hot source
and a cold sink
© Copyright 2014
TLO 1 Summary
Operator Generic Fundamentals
84
Summary
ELO 1.5
• 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
– Simplifies to changes in enthalpy normally
© Copyright 2014
TLO 1 Summary
Operator Generic Fundamentals
85
Summary
Review Question – NRC Exam Bank
Feedwater 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 feedwater temperature lowers the temperature at
which heat is rejected in the condenser.
D. less power is required by the feedwater pumps to pump the
warmer feedwater.
Correct answer is A.
© Copyright 2014
TLO 1 Summary
Operator Generic Fundamentals
86
Summary
Review Question – NRC Exam Bank
Turbine X and turbine Y are ideal steam turbines that exhaust to a
condenser at 1.0 psia. Turbine X is driven by saturated steam (100%
quality) at 900 psia. Turbine Y is driven by superheated steam at 500
psia and 620 °F.
The greatest amount of work is being performed by turbine ______,
and the greatest moisture content exists in the exhaust of turbine
______.
A. X; Y
B. X; X
C. Y; Y
D. Y; X
1
Correct answer is D.
© Copyright 2014
TLO 1 Summary
Operator Generic Fundamentals
87
Second Law of Thermodynamics
TLO 2 – Apply the second law of thermodynamics to analyze real and
ideal systems and components.
Figure: Entropy of Ice and Water
© Copyright 2014
TLO 2
Operator Generic Fundamentals
88
TLO 2
2.1 Explain the Second Law of Thermodynamics using the term
entropy
2.2 Given a thermodynamic system, determine the:
a. Maximum efficiency of the system
b. Efficiency of the components within the system
2.3 Differentiate between the path for an ideal process and that for a
real process on a T-s or h-s diagram
2.4 Describe how individual factors affect system or component
efficiency
© Copyright 2014
TLO 2 - ELOs
Operator Generic Fundamentals
89
Second Law of Thermodynamics
ELO 2.1 - Explain the Second Law of Thermodynamics using the term
entropy.
It is impossible to
construct a device that
operates in a cycle and
produces no effect other
than the removal of heat
from a body at one
temperature and the
absorption of an equal
quantity of heat by a
body at a higher
temperature.
© Copyright 2014
Figure: Second Law of Thermodynamics for a Heat Engine
ELO 2.1
89
Operator Generic Fundamentals
90
Second Law of Thermodynamics
Concept of second law is best stated using Kelvin & Planck’s
description:
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
Figure: Kelvin-Planck’s Second Law of Thermodynamics
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
91
Second Law of Thermodynamics
• The Second Law of Thermodynamics is needed because the First
Law of Thermodynamics does not completely define the energy
conversion process
• The first law is used to relate and evaluate various energies involved
in a process
– No information about direction of process can be obtained by
application of the first law
Figure: Heat Flow Direction
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
92
Second Law of Thermodynamics
• Limitations imposed on any process can be studied to determine
– Maximum possible efficiencies of such a process and then
– Comparison can be made between the maximum possible
efficiency and actual efficiency achieved
• Areas of application of second law is study of energy-conversion
systems
• The second law can be used to derive an expression for the
maximum possible energy conversion efficiency taking those losses
into account
• The second law denies the possibility of completely converting into
work all of the heat supplied to a system operating in a cycle
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
93
Second Law of Thermodynamics
• The Second Law is used to determine the maximum efficiency of any
process
• A comparison can then be made between the maximum possible
efficiency and the actual efficiency obtained
Figure: Rankine Cycle for a Typical Steam Plant
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
94
Entropy
• Entropy was introduced to help explain the Second Law
– Change in this property determines the direction in which a given
process proceeds
– A measure of 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
• Change in entropy is the ratio of heat transferred during a reversible
process to the absolute temperature of the system
Figure: Entropy
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
95
Entropy
• Change in entropy is the ratio of heat transferred during a reversible
process to the absolute temperature of the system
∆𝑄
∆𝑆 =
(𝑓𝑜𝑟 𝑎 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠)
𝑇𝑎𝑏𝑠
∆𝑆 = the change in entropy of a system during some process (𝐵𝑇𝑈/°𝑅)
∆𝑄 = the amount of heat added to the system during the process (BTU)
𝑇𝑎𝑏𝑠 = the absolute temperature at which the heat was transferred (°𝑅)
• The second law can also be expressed as ΔS ≥ 0 for a closed cycle.
• Entropy must increase or stay the same for a cyclic system; it can
never decrease.
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
96
Entropy
• Entropy is a extensive property of a system and may be calculated
from specific entropies (S = ms)
– Specific entropy (s) values are tabulated specific enthalpy and
specific volume in the steam tables
• Specific entropy is a property and is used as one of the coordinates
when representing a reversible process graphically
• Area under a reversible process curve on T-s diagram represents the
quantity of heat transferred during the process
Figure: T-s Diagram With Rankine Cycle
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
97
Entropy
• Thermodynamic problems, processes, and cycles are often
investigated by substituting reversible processes for the actual
irreversible process
– Helpful because only reversible processes can be depicted on the
diagrams
• Actual or irreversible processes cannot be drawn since they are not a
succession of equilibrium conditions
• Only the initial and final conditions of irreversible processes are
known
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
98
Entropy
• Actual or irreversible processes cannot be drawn since they are not a
succession of equilibrium conditions
• Only the initial and final conditions of irreversible processes are
known
Figure: T-s and h-s Diagrams for Expansion and Compression Processes
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
99
Entropy
Knowledge Check
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
Correct answer is B.
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
100
Carnot’s Principle
ELO 2.2 - Given a thermodynamic system, Determine the:
a. Maximum efficiency of the system, b. Efficiency of the components
within the system
Figure: Carnot Cycle Representation
© Copyright 2014
ELO 2.2
100
Operator Generic Fundamentals
101
Carnot’s Principle
• No engine can be more efficient
than a reversible engine
operating between the same
high temperature and low
temperature reservoirs
• Efficiencies of all reversible
engines operating between the
same constant temperature
reservoirs are the same
• Efficiency of a reversible engine
depends only on the
temperatures of the heat source
and heat receiver
© Copyright 2014
ELO 2.2
ELO 2.1
Figure: Carnot’s Efficiency Principle
Operator Generic Fundamentals
102
Carnot Cycle
Reversible isothermal expansion (1-2: TH = constant)
Reversible adiabatic expansion (2-3: Q = 0, TH→TL)
Reversible isothermal compression (3-4: TL = constant)
Reversible adiabatic compression (4-1: Q = 0, TL→TH)
Figure: Single Piston Carnot Engine Cycle
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
103
Carnot Cycle
• Heat input (QH) is the area
under line 2-3
• Heat rejected (QC) is the
area under line 1-4
• Difference between heat
added and heat rejected is
the net work (sum of all work
processes), which is
represented as the area of
rectangle 1-2-3-4
Figure: Carnot Cycle Representation
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
104
Carnot Cycle
• Shows max possible efficiency exists when either
– TH is at its highest possible value
– TC is at its lowest possible value
• Carnot efficiency
– Represent reversible processes
– Upper limit of efficiency for any given system operating between
the same two temperatures
– However, all practical/real systems and processes are irreversible
and the actual system will not reach this efficiency value
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
105
Carnot Cycle
• Efficiency (η) of the cycle is the ratio of the net work of the cycle to
the heat input to the cycle.
• Expressed by equation:
𝑄𝐻 − 𝑄𝐶
𝑇𝐻 − 𝑇𝐶
𝜂=
=
𝑄𝐻
𝑇𝐻
𝑇𝐶
=1−
𝑇𝐻
Where: 𝜂 = cycle efficiency
𝑇𝐶 = designates the low-temperature reservoir (˚R)
𝑇𝐻 = designates the high-temperature reservoir (˚R)
°R must be used in calculation
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
106
Carnot Cycle
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?
𝑇ℎ = 140 ℉ + 460 = 600 °𝑅
𝑇𝑐 = 0 ℉ + 460 = 460 °𝑅
600 − 460
𝜂=
× 100 = 23.3%
600
• Claimed efficiency 25 BTUs/100 BTUs = 25%
– Therefore, the claim is invalid
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
107
Carnot Cycle
• The most important aspect of the second law is the determination of
maximum possible efficiencies obtained from a power system
– Actual efficiencies will always be less than this maximum
– System losses (friction, windage, turbulence, etc.) and the fact
that systems are not reversible preclude us from obtaining
maximum possible efficiency
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
108
Carnot Cycle
Example 2: Actual Versus Ideal Efficiency
• The actual efficiency of a steam cycle is 18.0%. 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.
 Tc 

 Th 
  1  
  1
60  460
340  460
520
800
  1  .65
  35%
  1
• 35% as compared to 18.0% actual efficiency
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
109
Carnot Cycle
Example 2: Actual Versus Ideal Efficiency
Figure: Real Process Cycle Compared to Carnot Cycle
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
110
Carnot Cycle
• An open system analysis was performed using the First Law of
Thermodynamics
• Second law problems are treated the same way
– An isolated, closed, or open system used in the analysis depends
on the types of energy that cross the boundary
• Open system analysis using the second law equations is the more
general case
– Closed and isolated systems are special cases of the open
system
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
111
Carnot Cycle
• Fluid moves through control
volume from inlet to outlet while
work is delivered external to the
control volume
• Assume the control volume
boundary is at some
environmental temperature and
that all of the heat transfer (Q)
occurs at this boundary
• Entropy is a property
– May be transported with the
flow of the fluid into and out
of the control volume
© Copyright 2014
ELO 2.2
Figure: Control Volume for Second Law Analysis
Operator Generic Fundamentals
112
Carnot Cycle
• Entropy flow into the control
volume resulting from mass
transport is 𝑚𝑖𝑛 s 𝑖𝑛
• Entropy flow out of control
volume is 𝑚𝑜𝑢𝑡 s 𝑜𝑢𝑡
– Assuming that properties
are uniform at sections in
and out
• Entropy may also be added to
control volume because of heat
transfer at control volume
boundary
© Copyright 2014
Figure: Control Volume for Second Law Analysis
ELO 2.2
Operator Generic Fundamentals
113
Carnot Cycle
Example: 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 1 atm at
300 °F.
What is the increase in entropy for the system if the mass flow rate is
10,000 lbm/hr?
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
114
Carnot Cycle
Solution:
𝑚𝑠𝑖𝑛 + 𝑝 = 𝑚𝑠𝑜𝑢𝑡 , where 𝑝 = entropy added to the system
𝑝 = 𝑚 𝑠𝑜𝑢𝑡 − 𝑠𝑖𝑛
𝐵𝑇𝑈
sin = 1.7088
(𝑓𝑟𝑜𝑚 𝑠𝑡𝑒𝑎𝑚 𝑡𝑎𝑏𝑙𝑒𝑠)
𝑙𝑏𝑚– °𝑅
𝐵𝑇𝑈
𝑠𝑜𝑢𝑡 = 1.8158
(𝑓𝑟𝑜𝑚 𝑠𝑡𝑒𝑎𝑚 𝑡𝑎𝑏𝑙𝑒𝑠)
𝑙𝑏𝑚– °𝑅
𝑝
𝐵𝑇𝑈
= 𝑠𝑜𝑢𝑡 − 𝑠𝑖𝑛 = 1.8158 − 1.7088
𝑚
𝑙𝑏𝑚– °𝑅
𝑝
𝐵𝑇𝑈
= 0.107
𝑚
𝑙𝑏𝑚– °𝑅
𝑝 = 10,000 0.107
𝐵𝑇𝑈
𝑝 = 1,070
= 𝑒𝑛𝑡𝑟𝑜𝑝𝑦 𝑎𝑑𝑑𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚
𝑙𝑏𝑚– °𝑅
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
115
Carnot Cycle
• The Second Law of Thermodynamics gives an upper limit (which is
never reached in physical systems) for how efficiently a
thermodynamic system can perform
• Determining that efficiency is as simple as knowing the inlet and exit
temperatures of the overall system (one that works in a cycle) and
applying Carnot’s efficiency equation using those temperatures in
absolute degrees
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
116
Carnot Cycle
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
Correct answer is A.
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
117
Thermodynamics of Ideal and Real Processes
ELO 2.3 - Differentiate between the paths for an ideal process and for a
real process on a T-s or h-s diagram
Figure: T-s Diagram Shows Paths of Processes
© Copyright 2014
ELO 2.3
Operator Generic Fundamentals
118
Diagrams of Ideal and Real Processes
• Any ideal thermodynamic
process can be drawn as a
path on a property
diagram, such as a T-s or
an h-s diagram
• A real process that
approximates the ideal
process can also be
represented on the same
diagrams
– Usually with the use of
dashed lines
© Copyright 2014
Figure: T-s Diagram Shows Paths of Processes
ELO 2.3
Operator Generic Fundamentals
119
Diagrams of Ideal and Real Processes
• In an ideal process involving
either a reversible expansion
or a reversible compression,
the entropy will be constant
• These isentropic processes
will be represented by vertical
lines on either T-s or h-s
diagrams, since entropy is on
the horizontal axis and its
value does not change
Figure: h-s Diagram Shows Expansion and Compression Paths
© Copyright 2014
ELO 2.3
Operator Generic Fundamentals
120
Diagrams of Ideal and Real Processes
• Real expansion or
compression process slants
slightly toward the right.
Entropy increases from the
start to the end of the process
• Real expansion process
(turbine) results in a smaller
change in enthalpy – less work
is done by the real turbine
• Real compression process
(pump) - more enthalpy must
be added during the pump
process - more work done on
the system
© Copyright 2014
Figure: h-s Diagram Shows Expansion and Compression Paths
ELO 2.3
Operator Generic Fundamentals
121
Diagrams of Ideal and Real Processes
• Real processes are irreversible
• Compare real processes to
ideal processes during system
design
• Reversible process indicate a
maximum work output for a
given input, which can be
compared to real processes for
efficiency purposes
Figure: h-s Diagram Shows Expansion and Compression Paths
© Copyright 2014
ELO 2.3
Operator Generic Fundamentals
122
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.
Correct answer is B.
© Copyright 2014
ELO 2.3
Operator Generic Fundamentals
123
Thermodynamic Power Plant Efficiency
ELO 2.4 - Describe how individual factors affect system or component
efficiencies
Figure: Typical Steam Cycle
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
124
Power Plant Components
• Three basic types of components in
power cycles
– Turbines
– Pumps
– Heat exchangers
• Each of these components
produces a characteristic change in
the properties of the working fluid
• It is possible to calculate efficiencies
of each individual component
– Compare actual work produced
by the component to the work
produced by an ideal component
© Copyright 2014
ELO 2.4
Figure: Typical Steam Cycle
Operator Generic Fundamentals
125
Power Plant Components
Steam Turbine
• Designed to extract energy from
steam and use it to do work in
the form of rotating the turbine
shaft
• Steam works as it expands
through the turbine
• The shaft work is then
converted to electrical energy
by the generator
Figure: Turbine Work
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
126
Power Plant Components
First law, general energy equation - the decrease in the enthalpy of the
working fluid Hin - Hout equals the work done by the turbine (𝑊𝑡 )
𝐻𝑖𝑛 − 𝐻𝑜𝑢𝑡 = 𝑊𝑡
𝑚 ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 = 𝑤𝑡
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)
𝑚 = 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)
𝑤𝑡 = power of turbine (BTU/hr)
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
127
Power Plant Components
• An ideal turbine provides a basis for analyzing the performance of
turbines
• An ideal turbine performs the maximum amount of work theoretically
possible for the plant’s operating conditions
𝑆𝑖𝑛 = 𝑆𝑜𝑢𝑡
𝑠𝑖𝑛 = 𝑠𝑜𝑢𝑡
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)
© Copyright 2014
ELO 2.4
Figure: Mollier Diagram
Operator Generic Fundamentals
128
Power Plant Components
Ideal versus Real
Turbine
• KE and PE changes and
heat lost by the working
fluid in the turbine are
negligible
• In an ideal case, the
working fluid does work
reversibly by expanding at a
constant entropy
• Entropy of the working fluid
entering the turbine: Sin
equals the entropy of the
working fluid leaving the
turbine Sout
© Copyright 2014
Figure: h-s Diagram for Ideal and Real Turbine
ELO 2.4
Operator Generic Fundamentals
129
Power Plant Components
• An actual turbine does less work because of
– Friction losses in the blades
– Leakage past the 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
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
130
Power Plant Components
𝑊𝑡.𝑎𝑐𝑡𝑢𝑎𝑙
𝜂𝑡 =
𝑊𝑡.𝑖𝑑𝑒𝑎𝑙
ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 𝑎𝑐𝑡𝑢𝑎𝑙
𝜂𝑡 =
ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 𝑖𝑑𝑒𝑎𝑙
Where:
𝜂𝑡 = turbine efficiency (no units)
Wt.actual = actual work done by the turbine (ft-lbf)
Wt.ideal = work done by an ideal turbine (ft-lbf)
(hin – hout)actual = actual enthalpy change of the working fluid (BTU/lbm)
(hin – hout)ideal = actual enthalpy change of the working fluid in an ideal
turbine (BTU/lbm)
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
131
Power Plant Components
• Turbine efficiency ηt is
normally given by the
manufacturer
– Permits actual work done
to be calculated directly
by multiplying turbine
efficiency ηt by work done
by an ideal turbine under
the same conditions
• Turbine efficiency is
generally
– 60% to 80% for small
turbines
– 90% for large turbines
© Copyright 2014
Figure: Entropy Diagram Measures System Efficiency
ELO 2.4
Operator Generic Fundamentals
132
Power Plant Components
• The actual and idealized
performances of a turbine
may be conveniently
compared using a T-s
diagram
– Ideal case is a constant
entropy represented by a
vertical line
– Actual turbine involves an
increase in entropy
– The smaller the increase
in entropy, the closer the
turbine efficiency ηt is to
1.0 or 100%
© Copyright 2014
Figure: Entropy Diagram Measures System Efficiency
ELO 2.4
Operator Generic Fundamentals
133
Power Plant Components
• A pump is designed to move the working fluid by doing work on it
• In the application of the first law general energy equation to a simple
pump under steady flow conditions, the increase in the enthalpy of
the working fluid Hout - Hin equals the work done by the pump (Wp) on
the working fluid
𝐻𝑜𝑢𝑡 − 𝐻𝑖𝑛 = 𝑊𝑝
𝑚 ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 = 𝑊𝑝
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)
𝑚 = 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)
𝑊𝑝 = power of pump (BTU/hr)
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
134
Pumps
Real Versus Ideal Pump
• Assume the change in the KE, PE and heat losses of the working
fluid while in the pump are negligible
• It is also assumed that the working fluid is incompressible
• For the ideal case, it can be shown that the work done by the pump
equals the change in enthalpy across the ideal pump
Figure: Work Done by the Pump Equals Change In Enthalpy Across the Ideal Pump
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
135
Pumps
𝑊𝑝.𝑖𝑑𝑒𝑎𝑙 = 𝐻𝑜𝑢𝑡 − 𝐻𝑖𝑛
𝑊𝑝.𝑖𝑑𝑒𝑎𝑙 = 𝑚 ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛
𝑖𝑑𝑒𝑎𝑙
𝑖𝑑𝑒𝑎𝑙
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)
𝑊𝑝 = power of pump (BTU/hr)
𝑚 = 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)
Figure: Work Done by the Pump Equals Change In Enthalpy Across the Ideal Pump
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
136
Pumps
• An ideal pump provides a basis for analyzing the performance of
actual pumps
• A pump requires more work because of unavoidable losses due to
friction and fluid turbulence
• The work done by a pump equals the change in enthalpy across the
actual pump
𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙 = 𝐻𝑜𝑢𝑡 − 𝐻𝑖𝑛
𝑎𝑐𝑡𝑢𝑎𝑙
𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙 = 𝑚 ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛
© Copyright 2014
ELO 2.4
𝑎𝑐𝑡𝑢𝑎𝑙
Operator Generic Fundamentals
137
Pumps
Pump Efficiency
• 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
𝑊𝑝.𝑖𝑑𝑒𝑎𝑙
𝜂𝑝 =
𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
138
Pumps
Example:
• A pump operating at
75% 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?
𝑊𝑝.𝑖𝑑𝑒𝑎𝑙
𝜂𝑝 =
𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙
ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛
ℎ𝑜𝑢𝑡.𝑎𝑐𝑡𝑢𝑎𝑙
𝑎𝑐𝑡𝑢𝑎𝑙
=
ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛
𝜂𝑝
ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛
=
𝜂𝑝
𝑖𝑑𝑒𝑎𝑙
𝑖𝑑𝑒𝑎𝑙
+ ℎ𝑖𝑛.𝑎𝑐𝑡𝑢𝑎𝑙
ℎ𝑜𝑢𝑡.𝑎𝑐𝑡𝑢𝑎𝑙
𝐵𝑇𝑈
𝐵𝑇𝑈
600
− 200
𝐵𝑇𝑈
𝑙𝑏𝑚
𝑙𝑏𝑚
=
+ 200
75
𝑙𝑏𝑚
𝐵𝑇𝑈
𝐵𝑇𝑈
ℎ𝑜𝑢𝑡.𝑎𝑐𝑡𝑢𝑎𝑙 = 533.3
+ 200
𝑙𝑏𝑚
𝑙𝑏𝑚
ℎ𝑜𝑢𝑡.𝑎𝑐𝑡𝑢𝑎𝑙 = 733.3 𝐵𝑇𝑈𝑠/𝑙𝑏𝑚
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
139
Pumps
• Pump efficiency (ηp) relates work required by an ideal pump to actual
work required by the pump
– Minimum amount of work theoretically possible to actual work
required by pump
• Pump efficiency does not account for losses in the prime mover such
as a motor or turbine
• Motor efficiency (ηm) is the ratio of actual work required by the pump
to the electrical energy input to the pump motor
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
140
Motors
Motor Efficiency
𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙
𝜂𝑚 =
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 (kWh)
C = conversion factor = 2.655 x 106 ft-lbf/kWh
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
141
Motors
Motor Efficiency
• Like pump efficiency (ηp), motor efficiency (ηm) is always less than 1.0
or 100% 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
𝑊𝑝.𝑖𝑑𝑒𝑎𝑙
𝜂 𝑚𝜂 𝑝 =
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
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
142
Heat Exchangers
• Designed to transfer heat between two working fluids
• Several heat exchangers are used in power plant steam cycles
– Steam generator or boiler - heat source used to heat and vaporize
feedwater
– Condenser – turbine exhaust steam is condensed before being
returned to steam generator
– Numerous smaller heat exchangers are used throughout the
steam cycle
• Factors determining rate of heat transfer
– Mass flow rates of the fluids flowing through the heat exchanger
– Temperature difference between the two fluids
• Subcooling is the process of cooling condensed vapor beyond what is
required for the condensation process.
© Copyright 2014
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Operator Generic Fundamentals
143
Heat Exchangers
• Factors determining rate of heat transfer
– Mass flow rates of the fluids flowing through the heat exchanger
– Temperature difference between the two fluids
𝑚1 ℎ𝑜𝑢𝑡,1 − ℎ𝑖𝑛,1 = −𝑚2 (ℎ𝑜𝑢𝑡,2 − ℎ𝑖𝑛,2 )
Figure: Typical Parallel and Counter-Flow Heat Exchangers Different Flow Regimes and Associated
Temperature Profiles in a Double-Pipe Heat Exchanger
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ELO 2.4
Operator Generic Fundamentals
144
Heat Exchangers
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.
© Copyright 2014
Figure: Typical Parallel and Counter-Flow Heat Exchangers
Different Flow Regimes and Associated Temperature
Profiles in a Double-Pipe Heat Exchanger
ELO 2.4
Operator Generic Fundamentals
145
Main Condenser
• Condenses turbine wet vapor exhaust.
• Rejected heat transfers to the environment by the circulating water
flowing through the condenser tubes.
• Condensate, the liquid formed, is subcooled slightly during the
process.
Figure: Typical Single-Pass Condenser
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Operator Generic Fundamentals
146
Main Condenser
• Steam is condensed
– Latent heat of
condensation
• Specific volume
decreases drastically
• Creates a low pressure,
maintaining vacuum
• Increases plant efficiency
Figure: Typical Single-Pass Condenser End View
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Operator Generic Fundamentals
147
Main Condenser
Condensate Depression
• As the condensate falls toward the hotwell, it subcools (goes below
TSAT) as it comes in contact with tubes lower in the condenser
• The amount of subcooling is the condensate depression
• TSAT – THOTWELL = the amount of condensate depression
Figure: T-v Diagram for Typical Condenser
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Operator Generic Fundamentals
148
Condensate Depression
Practice Question
Condensate temperature in the hotwell is 108 °F and the condenser
pressure is 1.5 psia. What is the approximate amount of condensate
depression?
A. 10 °F
B. 8 °F
C. 6 °F
D. It is at saturation and there is no depression
Correct answer is B.
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Operator Generic Fundamentals
149
Condenser Example Problem
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
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Operator Generic Fundamentals
150
Condenser Example Problem (cont’d)
Find 𝑄 of the circulating water:
1 𝐵𝑇𝑈
=𝑚
𝑙𝑏𝑚 −°𝐹
𝑄 = 𝑚𝑐𝑝 𝛥𝑇
97 °𝐹 − 75° 𝐹
108 𝑙𝑏𝑚
𝐵𝑇𝑈
= 3.1 ×
22
ℎ𝑟
𝑙𝑏𝑚
𝐵𝑇𝑈
= 6.82 × 109
ℎ𝑟
6.82 x 109 BTU/hr represents the 𝑄 necessary to condense the steam and
subcool it to 4 °F below saturation temperature. Therefore, the 𝑄 necessary to
subcool the condensate is:
𝑄 = 𝑚𝑐𝑝 𝛥𝑇
106 𝑙𝑏𝑚
1 𝐵𝑇𝑈
= 8×
ℎ𝑟
𝑙𝑏𝑚− °𝐹
𝐵𝑇𝑈
= 3.2 × 107
ℎ𝑟
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4 °𝐹
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Operator Generic Fundamentals
151
Condenser Example Problem (continued)
This number is insignificant compared to the total 𝑄, and therefore will
not be considered. From the steam tables, saturated liquid at 1 psia:
ℎ𝑓 = 69.73
𝐵𝑇𝑈
𝑙𝑏𝑚
ℎ𝑓𝑔 = 1,036.1
𝐵𝑇𝑈
𝑙𝑏𝑚
ℎ𝑔 = 1,105.8
𝐵𝑇𝑈
𝑙𝑏𝑚
Using 𝑄 = 𝑚𝛥ℎ:
Solving for 𝛥ℎ:
𝛥ℎ =
𝑄
𝑚
Therefore:
ℎ𝑠𝑡𝑚 − ℎ𝑐𝑜𝑛𝑑
© Copyright 2014
𝑄
=
𝑚
ELO 2.4
Operator Generic Fundamentals
152
Condenser Example Problem (continued)
Solving for ℎ𝑠𝑡𝑚 :
ℎ𝑠𝑡𝑚 =
𝑄
+ ℎ𝑐𝑜𝑛𝑑
𝑚
ℎ𝑠𝑡𝑚
𝐵𝑇𝑈
𝐵𝑇𝑈
ℎ𝑟
=
+ 69.73
𝑙𝑏
𝑙𝑏𝑚
8 × 106 𝑚
ℎ𝑟
𝐵𝑇𝑈
= 922.23
𝑙𝑏𝑚
6.82 × 109
Solving for 𝑋:
ℎ𝑠𝑡𝑚 − ℎ𝑓
𝑋=
ℎ𝑓𝑔
𝐵𝑇𝑈
𝐵𝑇𝑈
922.23
− 69.73
𝑙𝑏𝑚
𝑙𝑏𝑚
𝑋=
𝐵𝑇𝑈
1,036.1
𝑙𝑏𝑚
𝑋 = 0.823 or 82.3% steam quality
Using ℎ𝑠𝑡𝑚 = ℎ𝑓 + 𝑋ℎ𝑓𝑔
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ELO 2.4
Operator Generic Fundamentals
153
Real vs. Ideal Cycle Efficiency
• Efficiency of a Carnot cycle solely depends on the temperature of the
heat source and the heat sink
– To improve efficiency increase the temperature of the heat source
and decrease the temperature of the heat sink
• 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)
• Most plants notice an increase in plant efficiency when the heat sink
temperature drops in the winter
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ELO 2.4
Operator Generic Fundamentals
154
Real vs. Ideal Cycle Efficiency
• Heat source is limited to the combustion temperatures of the fuel
to be burned
– Fossil fuel cycles upper limit is ~ 3,040 °F (3,500 °𝑅)
– Not attainable due to the metallurgical restraints of the boilers,
– 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
𝑇𝑆𝑂𝑈𝑅𝐶𝐸 − 𝑇𝑆𝐼𝑁𝐾 1,960 °𝑅 − 520 °𝑅
𝜂=
=
= 73.5%
𝑇𝑆𝑂𝑈𝑅𝐶𝐸
1,960 °𝑅
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Operator Generic Fundamentals
155
Real vs. Ideal Cycle Efficiency
Heat Rejection
• 73% efficiency is not possible
• 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
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Operator Generic Fundamentals
156
Real vs. Ideal Cycle Efficiency
𝑞𝑟 = 𝑇𝑜 𝛥𝑠
Measured in units of BTU/lbm
Where:
To = average heat sink temperature of 520 °R
Available Energy (A.E.) for Carnot cycle
• Equation:
𝐴. 𝐸. = 𝑞𝑠 − 𝑞𝑟
• Substituting equation for qr gives:
𝐴. 𝐸. = 𝑞𝑠 − 𝑇0 Δ𝑠
• Measured in units of BTU/lbm
• Between the temperatures 1,962 °F and 520 °R
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Operator Generic Fundamentals
157
Real vs. Ideal Cycle Efficiency
• Operating at a temperature of
less than 1,962 °R is less
efficient
• Developing materials capable of
withstanding the stresses above
1,962 °R increases the energy
available for use by the plant
cycle
• Entropy is a measure of the
energy unavailable to do work
• If the temperature of the heat
sink is known, then the change
in entropy corresponds to a
measure of the heat rejected by
the engine
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ELO 2.4
Figure: Entropy Measures Temperature as Pressure
and Volume Increase
Operator Generic Fundamentals
158
Real vs. Ideal Cycle Efficiency
• Fossil fuel power cycle – 2,000 psia
maximum, heat addition at
temperature of 1,962 °𝑅 is not
possible
• Water requires heat addition in a
constant pressure process
– Average temperature for heat
addition is below the maximum
1,962 °𝑅
• Actual available energy (area 1-2-34) is less than half of the ideal
Carnot cycle (area 1-2'-4) operating
between the same two temperatures
• Fossil plants are on the order of
40% efficient while nuclear plants
are around 31%
© Copyright 2014
ELO 2.4
Figure: Carnot Cycle Versus Typical Power Cycle
Available Energy
Operator Generic Fundamentals
159
Real vs. Ideal Cycle Efficiency
• Carnot steam cycle on a T-s
diagram
• A great deal of 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
• Isentropic compression
probably results in some pump
cavitation in the feed system
• Condenser designed to
produce a two-phase mixture
at the outlet (point 1) would
pose technical problems
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Figure: Ideal Carnot Cycle
ELO 2.4
Operator Generic Fundamentals
160
Rankine Cycle
• T-s diagram shows available
and unavailable energy
represented by the areas under
the curves
• The energy available for work is
the difference between the total
Q added and Q rejected
• The larger the unavailable (Q
rejected) energy, the less
efficient the cycle
Figure: Rankine Cycle
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Operator Generic Fundamentals
161
Rankine Cycle
Rankine Cycle
Efficiencies
• Comparing two Rankine
cycles on T-s diagram
• The amount of rejected
energy to available energy of
one cycle can be compared to
another cycle
• The most efficient has the
least amount of unavailable
energy
• It can be seen that Cycle A
has less rejected heat and
more energy for work
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Figure: Rankine Cycle Efficiency Comparisons on a T-s Diagram
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Operator Generic Fundamentals
162
Rankine Cycle
• Isentropic compression process
to the liquid phase only
– Points 1 to 2
• Minimizes the amount of pump
work
• Avoids the mechanical
problems associated with
pumping a two-phase mixture
• 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: Isentropic Compression Process
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Operator Generic Fundamentals
163
Rankine Cycle
• Compression process shown in
between points 1 and 2 is
greatly exaggerated
– Constant pressure lines
converge in subcooled or
compressed liquid region
– Difficulty distinguishing the
constant pressures lines
from the saturated liquid line
• The processes that comprise
the cycle are:
– 1-2: Liquid is compressed
with no change in entropy
(by ideal pump)
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Figure: Isentropic Compression Process
ELO 2.4
Operator Generic Fundamentals
164
Rankine Cycle
• The processes that comprise
the cycle are:
– 2-3: Heat added to the
system raises the liquid to
saturation temperature and
further to saturated steam at
constant pressure.
– 3-4: Constant entropy
expansion with shaft work
output (in an ideal turbine)
– 4-1: Constant pressure
transfer of heat in the sink.
Unavailable heat is rejected
to the atmospheric sink.
© Copyright 2014
Figure: Isentropic Compression Process
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Operator Generic Fundamentals
165
Rankine Cycle
• The ideal turbine is
replaced with a real turbine.
• The efficiency is reduced.
• The non-ideal turbine
incurs an increase in
entropy, which increases
the area under the T-s
curve for the cycle.
• But 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).
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Figure: Rankine Cycle With Real Versus Ideal Turbine
ELO 2.4
Operator Generic Fundamentals
166
Rankine Cycle
• h-s diagram shows that
substituting non-ideal
components in place of ideal
components in a cycle, results
in a reduction in the cycle’s
efficiency
• a change in enthalpy (h) always
occurs when work is done or
heat is added or removed in an
actual cycle (non-ideal)
• This deviation from an ideal
constant entropy process
(vertical line on the diagram)
allows the inefficiencies of the
cycle to be seen on a h-s
diagram
© Copyright 2014
Figure: h-s Diagram Shows Using Non-Ideal Components
Reduces Cycle’s Efficiency
ELO 2.4
Operator Generic Fundamentals
167
Rankine Cycle – Steam Plant
• Processes that comprise
the steam cycle:
– 1-2: Heat is added to
the working fluid in the
steam generator under
a constant pressure
condition
– 2-3: Saturated steam is
expanded in high
pressure (HP) turbine to
provide shaft work
output at a constant
entropy
– 3-4: Moist steam from
the exit of HP turbine is
dried and superheated
in the moisture
separator reheater
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Figure: Typical Steam Cycle
ELO 2.4
Operator Generic Fundamentals
168
Rankine Cycle – Steam Plant
– 4-5: Superheated
steam from MSR is
expanded in the low
pressure (LP) turbine
to provide shaft work at
a 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
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Figure: Typical Steam Cycle
ELO 2.4
Operator Generic Fundamentals
169
Rankine Cycle – Steam Plant
– 7-8: Condensate is
preheated by the Low
Pressure feedwater
heaters
– 8-9: Condensate is
compressed as a liquid
by the feedwater pump
– 9-1: Feedwater is
preheated by the HighPressure heaters
– 1-2: Cycle starts again
- heat is added to the
working fluid in the
steam generator under
a constant pressure
condition
© Copyright 2014
Figure: Typical Steam Cycle
ELO 2.4
Operator Generic Fundamentals
170
Rankine Cycle – Steam Plant
Figure: Typical Steam Cycle
© Copyright 2014
Figure: Rankine Steam Cycle (Ideal)
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170
Operator Generic Fundamentals
171
Rankine Cycle Steam Plant – Ideal T-s
• Rankine Ideal Cycle
• Ideal pumps and turbines do not exhibit an increase in entropy
• No condensate subcooling as point 6 is on the saturation line
Figure: Rankine Steam Cycle (Ideal)
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Operator Generic Fundamentals
172
Rankine Cycle Steam Plant – Real T-s
• This additional heat rejected
must then be made up for in
the steam generator
• Real pumps and turbines
would exhibit an entropy
increase across them
• Subcooling decreases cycle
efficiency but aids in
preventing condensate pump
cavitation
Figure: Steam Cycle (Real)
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Operator Generic Fundamentals
173
Rankine Cycle – Steam Plant
• A Mollier Diagram is a plot of
the conditions existing for water
in vapor form
• Point 1: Saturated steam at
540 °F
• Point 2: 82.5% quality at exit
of HP turbine
• Point 3: Temperature of
superheated steam is 440 °F
• Point 4: Condenser vacuum is
psia
Figure: Mollier Diagram
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Operator Generic Fundamentals
174
Causes of Inefficiency
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
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
Causes of Inefficiency
Cycles
Real system compromises are made due to cost and other factors
in the design and operation of the cycle
• Condensers are designed to subcool the liquid by 8-10 °F. This
subcooling allows the condensate pumps to pump without cavitation.
– But, each degree of subcooling is energy that must be put back
by reheating the water, and this heat (energy) does no useful
work and therefore increases the cycle’s inefficiency.
• Heat loss to the environment, such as thin or poor insulation.
– Again, this is energy lost to the system and therefore unavailable
to do work.
• Friction from resistance to fluid flow and mechanical friction in
machines
– Converts fluid energy into heat that is not available for work
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Operator Generic Fundamentals
176
Improved Cycle Efficiency
Condition
Superheating
Moisture
Separator
Reheater
(MSR)
Feedwater
Preheating
Condenser
Vacuum
© Copyright 2014
Effect
More Efficient With
More Superheating
Discussion
Increased heat added results in more
net work from the system, even though
more heat is rejected.
Use of MSR Has Minor More work is done by the low-pressure
Effect On Efficiency
(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.
More Efficient With
Less heat must be added from the heat
Feedwater Preheating source (reactor) since the feedwater
enters the steam generator closer to
saturation temperature.
More Efficient With
Net work output is higher and heat
Higher Vacuum (Lower rejection is lower as condenser
Backpressure)
pressure is lowered.
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Operator Generic Fundamentals
177
Improved Cycle Efficiency
Condition
Condensate
Depression
Effect
More Efficient With
Minimal Condensate
Depression
Steam
Temperature/
Pressure
More Efficient At
Higher Steam
Temperature/
Pressure
Discussion
Minimal condensate depression
reduces both the amount of heat
rejected and the amount of heat that
must be supplied to the cycle.
At higher steam temperature, the inlet
and exit entropy from the turbine are
lower so less heat is rejected.
Steam Quality More Efficient At
Higher Steam Quality
© Copyright 2014
Steam density increases as pressure
increases, so more turbine work is
done.
Enthalpy content increases as moisture
content decreases and more net work
is done.
ELO 2.4
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Operator Generic Fundamentals
178
Other Operational Considerations
• Minimize the number of auxiliaries running to those necessary to
support the power output level
• Minimize the amount of steam generator blowdown.
• Fix steam leaks
• Fix air leaks into the condenser
• Operate air ejector condensers, gland seal condensers, and
blowdown heat exchangers to recover as much heat as possible
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Operator Generic Fundamentals
179
Thermodynamic Power Plant Efficiency
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.
Correct answer is B.
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Operator Generic Fundamentals
180
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.)
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.
Correct answer is D.
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Operator Generic Fundamentals
181
TLO 2 Summary
ELO 2.1 – Ask students to explain the following:
• Planck’s statement of the Second Law of Thermodynamics is:
– It is impossible to construct an engine that works 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 =
© Copyright 2014
𝑇𝐻 −𝑇𝐶
𝑇𝐻
TLO 2 Summary
Operator Generic Fundamentals
182
TLO 2 Summary
ELO 2.2 – Ask students to 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 high- and low-temperatures.
• Efficiency of a component can be calculated by comparing the work
produced by 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 will be represented
as a vertical line on a T-s or h-s diagram. A real expansion or
compression process will look similar, but will be slanted slightly to
the right.
• The 2nd law of thermodynamics gives maximum efficiency limit
(never reached in physical systems) that an ideal thermodynamic
system can perform. Efficiency is determined by knowing inlet and
exit absolute temperatures of overall system and applying Carnot's
efficiency equation.
© Copyright 2014
TLO 1 Summary
Operator Generic Fundamentals
183
TLO 2 Summary
ELO 2.2 – Ask students to explain the following statements:
• Cycle efficiency = 1 −
𝑇
𝐶
𝑇
(temperature in degrees R)
𝐻
ELO 2.3 Review T-s diagrams for real and Ideal systems, identify major
cycle processes on the diagrams.
Figure: Real Process Cycle Compared to Carnot Cycle
© Copyright 2014
TLO 2 Summary
Operator Generic Fundamentals
184
TLO 2 Summary - ELO 2.4
Condition
Superheating
Effect
Discussion
More Efficient With
More Superheating
Moisture
Use of MSR Has Minor
Separator
Effect On Efficiency
Reheater (MSR)
Feedwater
Preheating
© Copyright 2014
More Efficient With
Feedwater Preheating
Increased heat added results in more
net work from the system, even though
more heat is rejected.
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.
Less heat must be added from the heat
source (reactor) since the feedwater
enters the steam generator closer to
saturation temperature.
TLO 2 Summary
184
Operator Generic Fundamentals
185
TLO 2 Summary - ELO 2.4
Condition
Condenser
Vacuum
Condensate
Depression
Steam
Temperature/
Pressure
Steam Quality
© Copyright 2014
Effect
Discussion
More Efficient With
Higher Vacuum (Lower
Backpressure)
More Efficient With
Minimal Condensate
Depression
Net work output is higher and heat
rejection is lower as condenser
pressure is lowered.
Minimal condensate depression
reduces both the amount of heat
rejected and the amount of heat that
must be supplied to the cycle.
More Efficient At Higher At higher steam temperature, the inlet
Steam Temperature/
and exit entropy from the turbine are
Pressure
lower so less heat is rejected.
Steam density increases as pressure
increases, so more turbine work is
done.
More Efficient At Higher Enthalpy content increases as moisture
Steam Quality
content decreases and more net work
is done.
TLO 2 Summary
185
Operator Generic Fundamentals
186
TLO 2 Summary
ELO 2.4 Review
• 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:
—
Presence of friction
—
Heat losses
—
Cycle inefficiencies
—
Subcooling
—
Tsat of the steam generator
© Copyright 2014
TLO 2 Summary
Operator Generic Fundamentals
187
TLO 2 Summary
ELO 2.4 Review
• 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
© Copyright 2014
TLO 2 Summary
Operator Generic Fundamentals
188
TLO 2 Summary
Now that you have completed this TLO, 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.
© Copyright 2014
TLO 2 Summary
Operator Generic Fundamentals
189
Module 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.
• The First Law of Thermodynamics states that energy can be neither
created nor destroyed, but only altered in form.
• 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 (heat must be rejected).
• Entropy is a measure of the unavailability of heat to perform work in a
cycle, the change in entropy determines direction a given process will
proceed.
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Summary
Operator Generic Fundamentals
190
Module Summary
• The Carnot cycle represents an upper limit of efficiency for any given
system operating between the same two temperatures. Carnot cycles
represent reversible processes, therefore real systems cannot reach
the Carnot efficiency value.
• Carnot efficiency serves as an unattainable upper limit for any real
system's efficiency.
• Rankine cycle is an ideal cycle where no increase in entropy occurs
as work is done on and by the system.
• Thermodynamic processes can be arranged on a property diagram to
evaluate the cycles present in a nuclear power plant.
• Most common set of coordinates used is a plot of temperature versus
specific entropy is a T-s diagram.
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Summary
Operator Generic Fundamentals
191
Module Summary
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:
• Apply the First Law of Thermodynamics to analyze thermodynamic
systems and processes.
• Apply the Second Law of Thermodynamics to analyze real and ideal
systems and components.
© Copyright 2014
Operator Generic Fundamentals