Transcript Thermodynamic Processes

Operator Generic Fundamentals
193004 - Thermodynamic Processes
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Terminal Learning 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 on the following TLOs:
1. Apply the first law of thermodynamics to analyze thermodynamic
systems and processes.
2. Describe the operation of the turbine and condensing processes.
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TLOs
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Thermodynamic Cycles TLO1 - ELOs
TLO 1 - Apply the first law of thermodynamics to analyze thermodynamic
systems and processes.
1.1 Define the following terms as they apply to a thermodynamic
process:
a. Open, closed, or isolated system
b. Reversible (ideal) process
c.
Irreversible (real) process
e. Adiabatic process
f.
Isentropic process
g. Isenthalpic (throttling) process
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Thermodynamic Processes - ELOs
1.2 Apply the First Law of Thermodynamics for open systems or
thermodynamic processes.
1.3 Identify the path(s) on a T-s diagram that represents the
thermodynamic processes occurring in a fluid system.
1.4 Given a defined system, perform energy balances on all major
components in the system.
1.5 Determine exit conditions for a throttling process.
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First Law of Thermodynamics
ELO 1.1 - Define the following terms as they apply to a thermodynamic
process: Open, closed, or isolated system, Reversible (ideal) process,
Irreversible (real) process, Adiabatic process, Isentropic process,
Isenthalpic (throttling) process.
• 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
Figure: Energy Balance Equals the First Law of Thermodynamics
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Thermodynamic 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
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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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
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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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Reversible (Ideal) Process
• A process where system and surroundings are returned to their
original condition before the process occurred
– No losses (change in entropy)
o Recall no change in entropy is called “isentropic”
• Reversible processes can be approximated but never matched by
real processes
– Steps can be done to minimize losses
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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
– An automobile engine does not give back the fuel it took to drive
up a hill as it coasts back down the hill
– Results in an increase in entropy (losses)
• Factors that make a process irreversible:
– Friction
– Heat transfer through a finite temperature difference
• Minimizing irreversibility
– Minimize ΔT
o Feedwater as close to saturation temperature
o Feedwater heating done in steps
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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 Process
• An isenthalpic process is one in which the enthalpy of the fluid
remains constant
– Throttling processes are isenthalpic
o Move from left to right on a Mollier diagram
• 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)
• Example(s)
– PORV leak to quench (PRT) tank
– Steam leak to atmosphere
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Analyzing Systems Using the First Law
of Thermodynamics
ELO 1.2 - Apply the First Law of Thermodynamics for open systems or
thermodynamic processes.
Thermodynamic Processes
• Transformation of a working fluid from one state to another
– Might be a phase change
• 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 Process
• Four Forms of Energy
– Potential energy, kinetic energy, internal energy, and flow energy
o Internal and flow energies combined into enthalpy
Figure: Basic Energy Balance of the First Law of Thermodynamics
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General Energy Equation
Figure: General Energy Equation for the First Law of Thermodynamics
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Thermodynamic Processes
Figure: Six Basic Processes of Steady Flow Systems
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Thermodynamic Processes
• Our four basic processes of our thermodynamic cycle are:
– Steam Generator Process
o Subcooled feedwater in, heat added, saturated steam out
– Turbine process
o Saturated steam in, work done BY steam, wet vapor out
– Condensing Process
o Wet vapor in, heat removed, subcooled condensate out
– Pump Process
o Subcooled condensate in, work done ON fluid, subcooled
feedwater out
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Boundary Example
Figure: Open System Control Volume Concept for a Pump
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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 on or by the system (ft-lbf/hr)
𝑄 = heat rate into or out of 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
• 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
• Almost all practical applications of the first law require an open
system analysis
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Open System Analysis
• Mass in motion has potential (PE), kinetic (KE), and internal energy
(U)
– There is another form of energy associated with fluid caused by its
pressure
– Flow energy (PV)
Figure: Multiple Control Volumes in the Same System
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Thermodynamic Process Assumptions
• Each process initially presented as IDEAL
– Minimal change to KE or PE
o Recall, 778 ft-lbf = 1 BTU
• For heat transfer processes
– No work done ON or BY
– Insulated, so no heat losses
• For work processes
– No heat lost or gained (no change in specific entropy)
• Approximate values for specific enthalpy presented for each process
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Steam Generator Process
• Two flowpaths – (Primary flowpath, Secondary flowpath)
– Primary flowpath - heat transferred out of RCS
o Thot in, Tcold out
– Secondary flowpath – heat transferred into SG
o Subcooled feedwater in
– ≈ 420oF, ≈ 400 BTU/lbm
o Saturated steam out
– ≈ 1000 psia, ≈ 1200 BTU/lbm
• Energy in (FW) plus heat added (SG) equals Energy out (steam)
– hfw + hSG = hstm
– 400 BTU/lbm + 800 BTU/lbm = 1200 BTU/lbm
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Turbine Process
• Inlet to the turbine process is the outlet of the SG process
– ≈ 1200 BTU/lbm (Energyin)
• Work is done BY the system
– ≈ 400 BTU/lbm
• Energy out based on condenser vacuum
– 1 psia or 28 inHg equates to ≈ 800 BTU/lbm
• Energy in (steam) minus work done (turbine) equals Energy out
(exhaust)
– hstm - hturbine = hexhaust
– 1200 BTU/lbm - 400 BTU/lbm = 800 BTU/lbm
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Condenser Process
• Inlet to the condenser process is the outlet of the turbine process
– ≈ 800 BTU/lbm (Energyin)
• Heat is removed FROM the system
– ≈ 740 BTU/lbm
• Energy out based on condenser vacuum
– Slightly subcooled condensate (based on 1 psia backpressure)
o 92oF is ≈ 60 BTU/lbm
• Energy in (exhaust) minus heat removed (condenser) equals Energy
out (condensate)
– hexhaust - hcondenser = hcondensate
– 800 BTU/lbm - 740 BTU/lbm = 60 BTU/lbm
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Pump Process
• Inlet to the pump process is the outlet of the condenser process
– ≈ 60 BTU/lbm (Energyin)
• Work is done ON the system
– Pump(s) must not only make up for headloss but also raise
pressure to enter SG
– Enthalpy added by compression is minimal
– Most energy added by feedwater heating
• Energy in (condensate) plus work done ON system (pump) AND heat
added to system (FW heating) equals Energy out (feedwater)
– hcondensate + hpump/FW heating = hfeedwater
– 60 BTU/lbm - 340 BTU/lbm = 400 BTU/lbm
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Identifying Process Paths on a T-s Diagram
ELO1.3 - Identify the path(s) on a T-s diagram that represents the
thermodynamic processes occurring in a fluid system.
• Each of the previously discussed processes will be shown on a
typical T-s diagram
– Turbine process also shown on a h-s (Mollier) diagram
• Briefly look at REAL versus IDEAL
– Discussed in further detail in 19305 - Cycles
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Identifying Process Paths on a T-s Diagram
• Typical steam plant cycle
– Points 1 – 2: heat added to SG
– Points 2 – 3: work done by turbine
– Points 3 – 4: heat rejected by condenser
– Points 4 – 1: work done on system by pump(s)
SG
Figure: Typical Steam Plant System Cyclic Process
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SG Process - T-s Diagram
Temperature (T)
• Subcooled feedwater at 1000
psia and 420oF enters SG
• Sensible and latent heat added
from RCS (QA)
• Steam exits at ≈ 100%
saturated steam
– Phase change
1 psia
Specific Entropy (s)
• Area under this curve is the
total heat added
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1000 psia
Figure: T-s Diagram
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Moisture and quality on the Mollier diagram
0%
Quality
100%
All steam
Sg
100%
Moisture
0%
All water
Sf
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Turbine Process - T-s Diagram
Temperature (T)
• 100% saturated steam at 1000
psia enters turbine
– ≈ 1200 BTU/lbm
• Exits as a wet vapor
– ≈ 67% quality
1000 psia
1 psia
Ideal
Real
– ≈ 800 BTU/lbm
Specific Entropy (s)
• On IDEAL turbine, no Ds
Figure: T-s Diagram
• On REAL turbine, increase in s
– Less work out of steam,
higher quality
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Turbine Process - Mollier Diagram
• Turbine process can also be
shown on a Mollier Diagram
• Find starting pressure (1000
psia)
• IDEAL work (no Ds)
– Draw line straight down to
end pressure
o Condenser pressure of 1
psia
• REAL work (increase in s)
– Draw line down to end
pressure, but at higher final
enthalpy
o Still exhausts to 1 psia
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Real
Ideal
Figure: Mollier Diagram
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• ≈ 67% quality wet vapor enters
condenser
• Exits to hotwell as slightly
subcooled condensate
– Called condensate depression
Temperature (T)
Condenser Process - T-s Diagram
1000 psia
1 psia
• Undergoes phase change
– Subcooling – lower efficiency
Specific Entropy (s)
– Subcooling - better for pumps
Figure: T-s Diagram
• Heat rejected to Circ Water
system
• Area under this curve is the total
heat rejected
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Pump Process - T-s Diagram
• Enters as subcooled condensate
Temperature (T)
– About 1 psia
• Exits to SG as subcooled
feedwater
– About 1000 psia
1000 psia
Ideal
Real
1 psia
• IDEAL work of pump
Specific Entropy (s)
– All energy added as pressure
Figure: T-s Diagram
• Real work of pump
– Most added as pressure
– Some energy raises
temperature
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Identifying Process Paths on a T-s Diagram
Knowledge Check
A nuclear power plant is operating at 80 percent power with 5°F of
condensate depression in the main condenser. If the condensate
depression decreases to 2°F, the steam cycle thermal efficiency will
__________; and the condensate pumps will operate __________
cavitation.
A. increase; closer to
B. increase; farther from
C. decrease; closer to
D. decrease; farther from
Correct answer is A.
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Energy Balances on Major Components
ELO 1.4 - Given a defined system, Perform energy balances on all major
components in the system.
As previously discussed, boundaries can be set on any component.
Figure: Cyclic Process for Generating Electricity
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Heat Transfer Terms/Equations
• Each process can be looked at as the change in specific enthalpy
– Units – BTU/lbm
• If lbm known, then the work (W) or heat (Q) can be determined
– Units – BTU
• If the mass flow rate is known, then the heat transfer rate (𝑄) can be
determined
– Units – BTU/hr
• NRC Equation Sheet
– 𝑄 = 𝑚Dh (must be used for a phase change)
– 𝑄 = 𝑚cpDT (can be used with NO phase change)
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Heat Transfer Terms/Equations
• SG Process Equation
– 𝑄 𝑆𝐺 = 𝑚stm(hstm – hfw)
• Turbine Process Equation
– 𝑊 𝑇𝑢𝑟𝑏 = 𝑚stm(hstm – hexh)
• Condenser Process Equation
– 𝑄𝐶𝑜𝑛𝑑 = 𝑚exh(hexh – hcond)
• Pump Process Equation
– 𝑊 𝑃𝑢𝑚𝑝 = 𝑚fw(hfw – hcond)
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Analyzing Cyclic Process – Steam
Generator
• Heat transfer from RCS to steam generator (SG)
– Hot fluid (Thot) from the reactor heats feedwater across the SG
tubes to create steam
o Lower pressure in SG
– Colder Fluid (Tcold), with its energy removed, is pumped back to
the heat source for reheating
• For analysis purposes:
– 𝑄𝑅𝐶𝑆 = 𝑄𝑆𝐺
– DT method cannot be used on SG side
o No temperature change while adding latent heat
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Steps for Solving Energy Balance
Problems
Step
1.
2.
3.
4.
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Action
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.
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RCS Heat Transfer 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
• Average specific heat of the fluid is 1.5 BTU/lbm-°F
What is the heat transferred out of the RCS?
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RCS Heat Transfer Process
Step 1. Draw the system
• Show what is given and what is asked for, or requested
Figure: Steam Generation System
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RCS Heat Transfer 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
540 − 610 ℉
ℎ𝑟
𝑙𝑏𝑚−℉
10 𝐵𝑇𝑈
𝑄 = −1.45 × 10
ℎ𝑟
Note: Minus sign indicates heat out of RCS
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Heat Exchanger Process
• 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?
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Heat Exchanger Process
Step 1. Draw the system
• Show what is given and what is asked for, or requested
Figure: Heat Exchanger Analysis Shows Thermodynamic Balance
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Heat Exchanger 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
• No work is done
𝑚(ℎ𝑖𝑛 + 𝑃𝐸𝑖𝑛 + 𝐾𝐸𝑖𝑛) + 𝑄 = 𝑚(ℎ𝑜𝑢𝑡 + 𝑃𝐸𝑜𝑢𝑡 + 𝐾𝐸𝑜𝑢𝑡) + 𝑊
𝑄 = 𝑚(ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛)
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Heat Exchanger 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
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Condenser to Circ Water Process
• Steam flows through a condenser at 4.4 x 106 lbm/hr,
• Enters as saturated vapor at 104 °F (h = 1,106.8 BTU/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 BTU/lbm)
• Environmental requirements limit the Circulating Water exit
temperature to 77 °F (h = 45 BTU/lbm)
• Determine the required cooling water flow rate.
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Condenser to Circ Water Process
Step 1. Draw the system. Show what is given and what is asked for, or
requested.
Figure: Typical Single-Pass Condenser End View
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Condenser to Circ Water 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
𝑚𝑠𝑡𝑚 ℎ𝑜𝑢𝑡 – ℎ𝑖𝑛 𝑠𝑡𝑚
𝑚𝑐𝑤 =
ℎ𝑜𝑢𝑡 – ℎ𝑖𝑛 𝑐𝑤
𝑐𝑤
𝐵𝑇𝑈
54–
1106.8
6 𝑙𝑏𝑚
𝑙𝑏𝑚
𝑚𝑐𝑤 = 4.4 × 10
𝑥
𝐵𝑇𝑈
ℎ𝑟
45 − 32
𝑙𝑏𝑚
8 𝑙𝑏𝑚
𝑚𝑐𝑤 = −3.56 × 10
ℎ𝑟
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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 BTU/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.
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Throttling Characteristics
ELO 1.5 – Describe the exit conditions for a throttling process.
• Process of restricting full flow through use of a restrictor such as an
orifice or partially opened valve
• Causes drop in fluid pressure and increase in velocity
• No work interactions or changes in kinetic energy or potential energy
• Throttling process has constant enthalpy with slight increase in
entropy
• Resulting flow in liquid is somewhat turbulent
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Throttling Process
• A process in which:
– No change in enthalpy from
state one to state two (h1=
h2 )
– No work is done (W = 0)
– Process is adiabatic (Q = 0)
• Process called “isenthalpic”
Figure: Throttling Process by a Valve
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Throttling Process
• Example - Ideal gas flowing through a valve in mid-position
– Pin > Pout, velin < velout
(where P = pressure and vel = velocity)
– Remember ℎ = 𝑢 + 𝑃𝑣 (v = specific volume), so if pressure
decreases then specific volume must increase if enthalpy is to
remain constant (assuming u is constant)
– Since mass flow is constant, the change in specific volume is
observed as an increase in velocity, verified by our observations
• Theory also states W = 0
– No "work" has been done by throttling process
• Finally, theory states that an ideal throttling process is adiabatic
– “Real" throttling process not ideal and will involve some heat
transfer
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Throttling Process
The elevation change from boundary 1 to
boundary 2 is insignificant.
𝑃𝐸1 = 𝑃𝐸2
Inlet piping and outlet piping diameter are equal
and there is no change in fluid velocity.
𝐾𝐸1 = 𝐾𝐸2
There is no work done on or done by the fluid
as it flows through the throttle.
𝑊𝐼𝑁 = 𝑊𝑂𝑈𝑇 = 0
The piping is insulated so there is no heat
transferred into or out of the fluid.
𝑄𝐼𝑁 = 𝑄𝑂𝑈𝑇 = 0
This gives us the following results for a throttling process:
𝑃1 𝜈1
𝑃2 𝜈2
+ 𝑢1 =
+ 𝑢2
𝐽
𝐽
ℎ𝑖𝑛 = ℎ𝑜𝑢𝑡
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Throttling Process
Determining Downstream Properties Step-by-Step Table
Step Action
1.
First, determine the condition upstream of the throttle or
leak (temperature, pressure (psia), quality, or superheating).
2.
Find the corresponding point on the Mollier diagram.
3.
Determine the downstream pressure in psia.
4.
Draw a horizontal line from the initial condition point
(constant enthalpy) until the constant pressure line for the
downstream pressure is reached. The final condition is
established by this point (temperature, quality, or
superheating).
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Throttling Process
The diagram below shows step 4 from the table. Find initial point 1 and
draw a horizontal line until it intersects the downstream pressure which
could be a wet vapor under the dome (2) or superheated above the
dome (3).
Figure: Throttling Process on a Mollier Diagram
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Throttling Process – Example 1
A power-operated relief valve is stuck open at 2,200 psia in the
pressurizer. The valve is discharging to the pressurizer relief tank at 25
psig. What is the temperature of the fluid downstream of the relief
valve?
Solution
On the Mollier diagram, go to the 2,200-psia point on the saturation
line. Cross the constant enthalpy line (throttling is a constant enthalpy
process) to the 40 psia line (25 psig + 15 psi atmospheric = 40 psia).
Follow that line up to the saturation curve. The constant temperature
line that ends at that point on the curve establishes the temperature of
the fluid. The temperature is approximately 270F.
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Throttling Process – Example 2
The RCS is operating at 2,185 psig. What would be the expected
tailpipe temperature of a leaking pressurizer safety valve assuming
downstream pressure is 35 psig? (Also, assume that the steam quality
is 100 percent in the pressurizer.)
Solution
𝑃1 = 2,185psig + 15psi = 2,200psia
𝑃2 = 35psig + 15psi = 50psia
From the Mollier diagram, the final condition is a mixture. Therefore the
tailpipe temperature must be at the saturation temperature
corresponding to the pressure.
From steam tables, 𝑇𝑠𝑎𝑡 = 281°𝐹
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Throttling Characteristics
Knowledge Check
Which one of the following is essentially a constant-enthalpy process?
A. Throttling of main steam through main turbine steam inlet valves
B. Condensation of turbine exhaust in a main condenser
C. Expansion of main steam through the stages of an ideal turbine
D. Steam flowing through an ideal convergent nozzle
Correct answer is A.
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Throttling Characteristics
Knowledge Check – NRC Bank
A pressurizer safety valve is leaking by, allowing the 100 percent quality saturated steam
from the pressurizer to enter the discharge pipe, which remains at a constant pressure of
40 psia. Initial safety valve discharge pipe temperature is elevated but stable. Assume
no heat loss occurs from the safety valve discharge pipe.
Upon discovery of the leak, the reactor is shut down, and a plant cooldown and
depressurization are commenced. Throughout the cooldown and depressurization, 100
percent quality saturated steam continues to leak through the pressurizer safety valve.
As pressurizer pressure decreases from 1,000 psia to 700 psia, the safety valve
discharge pipe temperature will...
A. decrease because the entropy of the safety valve discharge will decrease during
the pressurizer pressure decrease in this range.
B. decrease because the enthalpy of the safety valve discharge will decrease
during the pressurizer pressure decrease in this range.
C. increase because the safety valve discharge will become more superheated
during the pressurizer pressure decrease in this range.
D. remain the same because the safety valve discharge will remain a saturated
steam-water mixture at 40 psia during pressurizer pressure decrease in this
range.
Correct answer is C.
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Throttling Characteristics
Knowledge Check – NRC Bank
A heatup and pressurization of a reactor coolant system (RCS) is in
progress following a maintenance shutdown. The RCS pressure is
1,000 psia with a steam bubble (100 percent quality saturated steam) in
the pressurizer. Pressurizer power-operated relief valve (PORV)
tailpipe temperature has been steadily rising. The PORV downstream
pressure is 40 psia.
Which one of the following will be the approximate PORV tailpipe
temperature and phase of the escaping fluid if a PORV is leaking by?
A. 267F, saturated
B. 267F, superheated
C. 312F, saturated
D. 312F, superheated
0 is D.
Correct answer
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Thermodynamic Systems & Processes
TLO 2 – Describe the operation of the turbine and condensing processes.
2.1 Describe the operation of nozzles to include functions of nozzles in
flow restrictors and functions of nozzles in air ejectors.
2.2 Describe the condensing process to include vacuum formation and
condensate depression.
2.3 Explain the design of turbines to include the functions of nozzles,
fixed blading, moving blading and the reason turbines are
multistage.
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Nozzle Characteristics
ELO 2.1 – Describe the operation of nozzles to include functions of
nozzles in flow restrictors and functions of nozzles in air ejectors.
• Nozzle - Mechanical device designed to control characteristics of fluid
flow as it exits or enters enclosed chamber or pipe via orifice
• Depending on type of nozzle, kinetic energy of fluid will increase or
decrease as it moves through device
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Nozzle Characteristics
• Often a pipe or tube of varying
cross-sectional area, can be
used to direct or modify flow of
a fluid (liquid or gas)
• Used to control emergent
stream:
– Rate of flow
– Speed
– Direction
– Mass
Figure: Typical Nozzle Types
– Shape
– Pressure
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High Velocity Nozzles
• Change energy of a fluid from
one form to another
• Increase kinetic energy at
expense of pressure and
internal energy
• Convergent nozzle narrowing
down from a wide diameter to a
smaller diameter in direction of
flow
• Divergent expanding from a
smaller diameter to a larger one
• De Laval nozzle has convergent
section followed by divergent
section and often called a
convergent-divergent nozzle
© Copyright 2016 – Rev 2
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Figure: Typical Nozzle Types
Operator Generic Fundamentals
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Nozzles – Theory of Operation
• Uses simplified general energy equation
to explain nozzle operation
• In a convergent nozzle, (A1 > A2)
The elevation change from entrance (1) to exit (2) is
insignificant.
𝑃𝐸1 = 𝑃𝐸2
Inlet piping diameter is greater than outlet piping
diameter. With steady flow, outlet velocity is greater
than inlet velocity.
There is no work done on or done by the fluid in the
nozzle.
The nozzle is perfectly insulated so no heat is
transferred into or out of the fluid.
𝐾𝐸2 > 𝐾𝐸1
𝑊𝐼𝑁 = 𝑊𝑂𝑈𝑇 = 0
𝑄𝐼𝑁 = 𝑄𝑂𝑈𝑇 = 0
Assume that there is no friction as the fluid flows through 𝑈1 = 𝑈2
the nozzle.
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Nozzles – Theory of Operation
𝑚1 = 𝑚2
𝑚=
𝐴v
= 𝜌𝐴v
𝜈
𝜌1 𝐴1 v1 = 𝜌2 𝐴2 v2 = 𝜌𝑥 𝐴𝑥 v𝑥
• Where:
𝑚1 = mass flow rate (lbm/sec)
A = cross-sectional flow area (ft2)
v = fluid velocity (ft/sec)
 = specific volume of fluid
(ft3/lbm)
ρ = density of fluid (lbm/ft3)
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Nozzles – Theory of Operation
• First law shows that the change in KE must be balanced by opposite
change in another stored energy form
• Pv energy must decrease if KE increased
• If fluid incompressible:
(𝑣1 = 𝑣2)
𝐾𝐸1 + 𝑃1𝑣 = 𝐾𝐸2 + 𝑃2𝑣
𝐾𝐸2 – 𝐾𝐸1 = 𝑃1𝑣 – 𝑃2𝑣
𝐾𝐸2 – 𝐾𝐸1 = (𝑃1 – 𝑃2)𝑣
Figure: Supersonic Flow Through Convergent-Divergent Nozzle
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Flow Restrictors
• Nozzles can be used as flow restrictor to limit flow
• Reduces flow area while allowing normal flow
• Used in power plant in main steam piping near or at SG exit
• In case of a main steam line rupture, nozzles will limit (choke) flow of
steam to limit pipe whip and impingement damage
• Controlling flow of steam is important in this instance due to power
excursion created by a suddenly large steam demand
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Flow Restrictors
• A properly designed flow restrictor offers little pressure drop from inlet
to outlet at normal steam flows
• Large pressure drop internal to device at narrowest point
• Pressure drop can be measured by instrumentation to determine flow
rate
Figure: Convergent-Divergent Venturi Tube for Flow Measurement
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Air Ejectors
• Pump-like device with no moving parts or pistons that utilizes highpressure steam to compress vapors or gases
– Creates a vacuum in any vessel or chamber connected to the
suction inlet
– Essentially jet pump or eductor
• High-pressure fluid flows through nozzle
• Fluid being pumped flows around nozzle, into throat of diffuser
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Air Ejectors
• High-velocity fluid enters diffuser where its molecules strike other
molecules
• Molecules in turn carried along with high-velocity fluid out of diffuser
– Creates low-pressure area around mouth of nozzle
• Low-pressure area will draw more fluid from around nozzle into throat
of diffuser
Figure: Simple Air Ejector (Jet Pump)
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Air Ejectors
• Steam pressure between 200 psi and 300 psi enables single-stage
air ejector to draw a vacuum of about 26 inches Hg
• For better vacuums, multiple-stage ejector used
Figure: Two-Stage Steam Jet Ejector
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Air Ejectors
• Normally consist of two suction stages
• First stage suction located on top of condenser
• Second stage suction comes from diffuser of first stage
Figure: Two-Stage Steam Jet Ejector
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Air Ejectors
• Exhaust steam from second stage must be condensed
• Accomplished by an air ejector condenser that is cooled by
condensate
• Air ejector condenser preheats condensate returning to boiler
• Two-stage air ejectors capable of drawing vacuums to 29 inches Hg
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Condenser Design/Operation
ELO 2.2 – Describe the condensing process to include vacuum formation
and condensate depression.
• Condensing process is a phase change
• Heat transferred to Circ Water system
• Large change in specific volume
• Water pumped must be subcooled to prevent cavitation
– However, excessive Subcooling and condensate depression
lowers cycle efficiency
• “Normally” a change in circ water flow will change vacuum
– Some bank questions note “no change in vacuum”
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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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Main Condenser
• Steam is condensed
– Latent heat of
condensation
• Specific volume decreases
drastically
• Creates a low pressure,
maintaining vacuum
• Increases plant efficiency
To SG
Through
Feedwater
Heaters
Figure: Typical Single-Pass Condenser End View
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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-s Diagram for Typical Condenser
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Condenser Example Problem
Determine the quality of the steam entering a condenser operating at:
• 1 psia
• 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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Condenser Example Problem (cont’d)
• Find 𝑄 of the circulating water: 𝑄 = 𝑚𝑐𝑝 𝛥𝑇
1 𝐵𝑇𝑈
=𝑚
𝑙𝑏𝑚−°𝐹
97 °𝐹 − 75° 𝐹
108 𝑙𝑏𝑚
𝐵𝑇𝑈
= 3.1 ×
22
ℎ𝑟
𝑙𝑏𝑚
𝐵𝑇𝑈
9
= 6.82 × 10
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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Condenser Example Problem (cont’d)
• 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 2016 – Rev 2
𝑄
=
𝑚
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87
Condenser Example Problem (cont’d)
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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Condensate Depression
Knowledge Check
What is the approximate value of condensate depression in a steam
condenser operating at 2.0 psia with a condensate temperature of
115°F?
A. 9°F
B. 11°F
C. 13°F
D. 15°F
Correct answer is B.
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Condenser Cycle Efficiency
Knowledge Check
Main turbine exhaust enters a main condenser and condenses at
126°F. The condensate is cooled to 100°F before entering the main
condenser hotwell. Assuming main condenser vacuum does not
change, which one of the following would improve the thermal
efficiency of the steam cycle?
A. Increase condenser cooling water flow rate by 5 percent.
B. Decrease condenser cooling water flow rate by 5 percent.
C. Increase main condenser hotwell level by 5 percent.
D. Decrease main condenser hotwell level by 5 percent.
Correct answer is B.
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Turbine Design and Characteristics
ELO 2.3 – Explain the design of turbines to include the functions of
nozzles, fixed blading, moving blading, and benefits of multistage turbines.
• A steam turbine converts thermal energy of steam into mechanical
energy to perform useful work
– At a power generation facility, steam turbines typically drive
electrical generators to produce electricity
– Steam turbines may also be used as a prime mover for a
compressor or pump
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Turbine Review in h-s Diagram
• Recall that work performed by the turbine can be illustrated on T-s or
h-s of a steam cycle
• On the h-s diagram below, idea turbine shown by line at point 2 to 3
– Real turbine work shown by line from point 2 to 3o
Figure: Rankine Cycle on an h-s Diagram
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Theory of Operation – Turbine
• Potential energy of steam converted into useful work in two steps
– Expansion through a nozzle converts available energy of steam
into kinetic energy
o Individual nozzle process similar to turbine process
– Uses REAL and IDEAL analogy
– Steam jet directed against blades attached to shaft, causing shaft
to turn, therefore converting kinetic energy into useful work
o Enthalpy (pressure and temperature) drops
o Kinetic energy increases
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Theory of Operation – Turbine
• Two major types:
– Impulse principle
– Reaction principle
Figure: Impulse Turbine Components
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Theory of Operation – Turbine
• Impulse principle
– Thermal energy of steam converted into mechanical energy in
essentially four steps:
1. Steam passes through stationary nozzles.
2. Stationary nozzles convert some of the steam’s thermal
energy (indicated by its pressure and temperature) into
kinetic energy (velocity).
3. The nozzles direct the steam flow into blades mounted on the
turbine wheel.
4. The blades and moving wheel convert the kinetic energy of
the steam into mechanical energy in the form of movement of
the turbine wheel and shaft or rotor.
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Theory of Operation – Turbine
Figures: Impulse Turbine Concepts
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Theory of Operation – Turbine
• Reaction principle
– Newton’s third law of motion: for every action, there is an equal
but opposite reaction
– Both “fixed blades” and “moving blades” act as nozzles to convert
steam’s thermal energy into kinetic energy
o Indicated by a decrease in pressure and temperature, and an
increase in velocity
– Decreases area of blade tip accelerates steam
– Additional pressure drop across moving blades provides
additional energy (reaction principle) for work
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Theory of Operation – Turbine
Figure: Reaction Turbine Concepts
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Impulse Turbine
• Two basic elements:
– Fixed nozzle
o Converts steam’s thermal
energy into kinetic energy
– Rotor
o Blades attached to rotor
disk absorb kinetic energy
of steam jet to convert it
into rotary motion
• Steam causes an impulse force
when it hits the turbine blades
© Copyright 2016 – Rev 2
Figure: Basic Impulse Turbine
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Impulse Turbine
• Blading designed to convert
maximum steam’s kinetic
energy into work
• Curved blades cause steam jet
to reverse direction, resulting in
a greater impulse force
• Blades often referred to as
“buckets,” alluding to their size
and concavity
• In an actual turbine, several
nozzles direct steam against
blades (or buckets) to turn rotor
Figure: Impulse/Reaction Turbine Comparison
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Reaction Turbine
• Nozzles attached to rotor disk
• Steam expands in moving
nozzles converting steam’s
thermal energy into kinetic
energy and producing a reactive
force
• Reactive force causes rotation
opposite to direction of steam
jet
Figure: Basic Reaction Turbine
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Reaction Turbine
• In actual reaction turbine, steam
directed by fixed blades through
moving blades attached to
turbine shaft
• One set moving blades and one
set stationary blades called a
stage
• Reaction turbine has all the
advantages of impulse turbine
plus greater efficiency
ELO 1.2
Figure: Impulse/Reaction Turbine Comparison
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Turbine Characteristics
• Theoretically, reaction turbine
differs from an impulse turbine
because moving blades of
reaction turbine act as nozzles
• Major difference between
impulse turbine and reaction
turbine is how steam is
expanded
– In impulse turbine, pressure
drop only across nozzle
– Reaction turbine has
pressure drop across each
set of blades
Figure: Impulse/Reaction Turbine Comparison
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Impulse Turbine Characteristics
• Steam enters nozzle with a
maximum pressure and a
minimum velocity
• In nozzle, steam velocity and
volume increase, pressure
decreases
• Exits nozzle at peak velocity
and impacts moving blades
where its kinetic energy is
converted to useful work
Figure: Steam Property Variation in Simple Impulse Turbine
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Actual Impulse Turbine Characteristics
• Steam leaving first set of
moving blades permitted to
enter a second row of
stationary blades that redirect
flow of steam
• Steam leaves fixed blades
with no change in properties
and enters a second row of
moving blades where its
velocity is reduced, and so
more work is done
Figure: Steam Property Variation In Actual Impulse Turbine
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Reaction Turbine Characteristics
• Fixed nozzles provided
between rows of moving blades
• Stationary nozzles shaped like
convergent nozzles permitting
steam expansion
– Steam’s pressure reduced,
velocity and volume
increased as it expands
Figure: Steam Property Variation in a Reaction Turbine
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Turbine Characteristics
• Moving blades are nozzle
shaped to permit further
expansion of steam
• When velocity is plotted on a
curve, absolute velocity is used
– Absolute velocity is relative
difference between moving
blade and steam velocities
Figure: Steam Property Variation in a Reaction Turbine
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Nozzle Diaphragms
• Contain nozzles, which admit steam to each stage of rotating blades
• Diaphragms types:
– Partial admission diaphragms
o Admit steam in an arc of a circle
– Full admission diaphragms
o Have nozzles extending around entire circle of blades
• Labyrinth packing rings minimize leakage of steam across diaphragm
and along rotor
– Placed in a groove in inner periphery of diaphragm
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ELO 2.3
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Major Components
Figure: Turbine Diaphragm and Cross-Section
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Major Components
Figure: Actual Turbine Blading
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Operator Generic Fundamentals
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Major Components
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Operator Generic Fundamentals
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Module Review
Knowledge Check
A nuclear power plant is operating at 90 percent of rated power. Main
condenser pressure is 1.69 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 on steam cycle
thermal efficiency?
A. Efficiency will increase because condensate depression will
decrease.
B. Efficiency will increase because the work output of the main
turbine will increase.
C. Efficiency will decrease because condensate depression will
increase.
D. Efficiency will decrease because the work output of the main
turbine will decrease.
Correct answer is D.
© Copyright 2016 – Rev 2
Summary
Operator Generic Fundamentals
112
NRC KA to ELO Tie
KA #
KA Statement
RO SRO ELO
K1.01 Explain the relationship between real and ideal processes.
1.8
1.9
1.1*
K1.02 Explain the shape of the T-s diagram process line for a typical secondary system.
1.7
1.9
1.3
K1.03 Describe the functions of nozzles in flow restrictors.
1.9
1.9
2.1
K1.04 Describe the functions of nozzles in air ejectors.
2.0
2.0
2.1
K1.05 Explain the function of nozzles fixed blading and moving blading in the turbine.
1.6
1.7
2.3
K1.06 Explain the reason turbines are multistages.
1.5
1.7
2.3
K1.07 Define turbine efficiency.
1.6
1.6
1.2
K1.08 Explain the difference between real and ideal turbine efficiency.
1.6
1.7
1.2
K1.09 Define pump efficiency.
1.3
1.3
1.2*
K1.10 Explain the difference between ideal and real pumping processes.
1.3
1.3
1.2*
K1.11 Describe the process of condensate depression and its effect on plant operation.
2.4
2.5
2.2
K1.12 Explain vacuum formation in condenser processes.
2.2
2.3
2.2
K1.13 Explain the condensing process.
2.2
2.3
2.2
K1.14 Explain the reduction of process pressure from throttling.
2.1
2.3
1.5
K1.15 Determine the exit conditions for a throttling process based on the use of steam and/or water
2.8
2.8
1.5
* K1.01, K1.09, and K1.10 are discussed in further detail in 193005 - Cycles
© Copyright 2016 – Rev 2
Operator Generic Fundamentals