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Operator Generic Fundamentals 193004 - Thermodynamic Processes © Copyright 2016 – Rev 2 Operator Generic Fundamentals 2 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. © Copyright 2016 – Rev 2 TLOs Operator Generic Fundamentals 3 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 © Copyright 2016 – Rev 2 ELOs Operator Generic Fundamentals 4 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. © Copyright 2016 – Rev 2 ELOs Operator Generic Fundamentals 5 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 © Copyright 2016 – Rev 2 ELO 1.1 Operator Generic Fundamentals 6 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 © Copyright 2016 – Rev 2 ELO 1.1 Operator Generic Fundamentals 7 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 © Copyright 2016 – Rev 2 ELO 1.1 Operator Generic Fundamentals 8 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 © Copyright 2016 – Rev 2 ELO 1.1 Operator Generic Fundamentals 9 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 © Copyright 2016 – Rev 2 ELO 1.1 Operator Generic Fundamentals 10 Steady Flow Equilibrium Process Figure: Steady Flow Systems © Copyright 2016 – Rev 2 ELO 1.1 Operator Generic Fundamentals 11 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 © Copyright 2016 – Rev 2 ELO 1.1 Operator Generic Fundamentals 12 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 © Copyright 2016 – Rev 2 ELO 1.1 Operator Generic Fundamentals 13 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 © Copyright 2016 – Rev 2 ELO 1.1 Operator Generic Fundamentals 14 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 © Copyright 2016 – Rev 2 ELO 1.1 Operator Generic Fundamentals 15 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 © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 16 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 © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 17 General Energy Equation Figure: General Energy Equation for the First Law of Thermodynamics © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 18 Thermodynamic Processes Figure: Six Basic Processes of Steady Flow Systems © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 19 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 © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 20 Boundary Example Figure: Open System Control Volume Concept for a Pump © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 21 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) © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 22 Heat Transferred Into or Out of System Figure: Heat and Work in a System © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 23 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 © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 24 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 © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 25 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 © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 26 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 © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 27 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 © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 28 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 © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 29 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 © Copyright 2016 – Rev 2 ELO 1.2 Operator Generic Fundamentals 30 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 © Copyright 2016 – Rev 2 ELO 1.3 Operator Generic Fundamentals 31 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 © Copyright 2016 – Rev 2 ELO 1.3 Operator Generic Fundamentals 32 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 © Copyright 2016 – Rev 2 1000 psia Figure: T-s Diagram ELO 1.3 Operator Generic Fundamentals 33 Moisture and quality on the Mollier diagram 0% Quality 100% All steam Sg 100% Moisture 0% All water Sf © Copyright 2016 – Rev 2 ELO 1.3 Operator Generic Fundamentals 34 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 © Copyright 2016 – Rev 2 ELO 1.3 Operator Generic Fundamentals 35 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 © Copyright 2016 – Rev 2 Real Ideal Figure: Mollier Diagram ELO 1.3 Operator Generic Fundamentals 36 • ≈ 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 © Copyright 2016 – Rev 2 ELO 1.3 Operator Generic Fundamentals 37 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 © Copyright 2016 – Rev 2 ELO 1.3 Operator Generic Fundamentals 38 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. © Copyright 2016 – Rev 2 ELO 1.3 Operator Generic Fundamentals 39 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 © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 40 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) © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 41 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) © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 42 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 © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 43 Steps for Solving Energy Balance Problems Step 1. 2. 3. 4. © Copyright 2016 – Rev 2 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. ELO 1.4 Operator Generic Fundamentals 44 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? © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 45 RCS Heat Transfer Process Step 1. Draw the system • Show what is given and what is asked for, or requested Figure: Steam Generation System © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 46 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 © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 47 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? © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 48 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 © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 49 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) © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 50 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 © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 51 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. © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 52 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 © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 53 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 © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 54 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. © Copyright 2016 – Rev 2 ELO 1.4 Operator Generic Fundamentals 55 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 © Copyright 2016 – Rev 2 ELO 1.5 Operator Generic Fundamentals 56 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 © Copyright 2016 – Rev 2 ELO 1.5 Operator Generic Fundamentals 57 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 © Copyright 2016 – Rev 2 ELO 1.5 Operator Generic Fundamentals 58 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 © Copyright 2016 – Rev 2 ELO 1.5 Operator Generic Fundamentals 59 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). © Copyright 2016 – Rev 2 ELO 1.5 Operator Generic Fundamentals 60 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 © Copyright 2016 – Rev 2 ELO 1.5 Operator Generic Fundamentals 61 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 270F. © Copyright 2016 – Rev 2 ELO 1.5 Operator Generic Fundamentals 62 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 © Copyright 2016 – Rev 2 ELO 1.5 Operator Generic Fundamentals 63 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. © Copyright 2016 – Rev 2 ELO 1.5 Operator Generic Fundamentals 64 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. © Copyright 2016 – Rev 2 ELO 1.5 Operator Generic Fundamentals 65 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. 267F, saturated B. 267F, superheated C. 312F, saturated D. 312F, superheated 0 is D. Correct answer © Copyright 2016 – Rev 2 ELO 1.5 Operator Generic Fundamentals 66 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. © Copyright 2016 – Rev 2 TLO 2 Operator Generic Fundamentals 67 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 © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 68 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 © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 69 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 ELO 2.1 Figure: Typical Nozzle Types Operator Generic Fundamentals 70 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. © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 71 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) © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 72 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 © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 73 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 © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 74 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 © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 75 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 © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 76 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) © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 77 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 © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 78 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 © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 79 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 © Copyright 2016 – Rev 2 ELO 2.1 Operator Generic Fundamentals 80 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” © Copyright 2016 – Rev 2 ELO 2.2 Operator Generic Fundamentals 81 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 © Copyright 2016 – Rev 2 ELO 2.2 Operator Generic Fundamentals 82 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 © Copyright 2016 – Rev 2 ELO 2.2 Operator Generic Fundamentals 83 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 © Copyright 2016 – Rev 2 ELO 2.2 Operator Generic Fundamentals 84 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 © Copyright 2016 – Rev 2 ELO 2.2 Operator Generic Fundamentals 85 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 © Copyright 2016 – Rev 2 4 °𝐹 ELO 2.2 Operator Generic Fundamentals 86 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 ELO 2.2 Operator Generic Fundamentals 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 © Copyright 2016 – Rev 2 ELO 2.2 Operator Generic Fundamentals 88 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. © Copyright 2016 – Rev 2 ELO 2.2 Operator Generic Fundamentals 89 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. © Copyright 2016 – Rev 2 ELO 2.2 Operator Generic Fundamentals 90 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 © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 91 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 © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 92 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 © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 93 Theory of Operation – Turbine • Two major types: – Impulse principle – Reaction principle Figure: Impulse Turbine Components © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 94 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. © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 95 Theory of Operation – Turbine Figures: Impulse Turbine Concepts © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 96 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 © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 97 Theory of Operation – Turbine Figure: Reaction Turbine Concepts © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 98 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 ELO 2.3 Operator Generic Fundamentals 99 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 © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 100 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 © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 101 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 © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 102 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 © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 103 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 © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 104 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 © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 105 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 © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 106 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 © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 107 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 © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 108 Major Components Figure: Turbine Diaphragm and Cross-Section © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 109 Major Components Figure: Actual Turbine Blading © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 110 Major Components © Copyright 2016 – Rev 2 ELO 2.3 Operator Generic Fundamentals 111 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