© Copyright 2016Operator Generic Fundamentals Operator Generic Fundamentals 193004 - Thermodynamic Processes.

© Copyright 2016Operator Generic Fundamentals 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. 2 TLOs

© Copyright 2016Operator Generic Fundamentals Thermodynamic Cycles TLO1 - ELOs 1.1Define 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 3 TLO 1 - Apply the first law of thermodynamics to analyze thermodynamic systems and processes. ELOs

© Copyright 2016Operator Generic Fundamentals 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. 1.5Determine exit conditions for a throttling process. 4 ELOs

© Copyright 2016Operator Generic Fundamentals First Law of Thermodynamics The First Law of Thermodynamics is an energy balance in a defined system –Energy can be neither created nor destroyed, only altered in form –Referred to as the Conservation of Energy Principle 5 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. Figure: Energy Balance Equals the First Law of Thermodynamics ELO 1.1

© Copyright 2016Operator Generic Fundamentals 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 6 ELO 1.1

© Copyright 2016Operator Generic Fundamentals 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. 7 Figure: Types of Thermodynamic Systems ELO 1.1

© Copyright 2016Operator Generic Fundamentals 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 8 ELO 1.1

© Copyright 2016Operator Generic Fundamentals Types of Thermodynamic Systems The RCS can be considered each type of system under certain operational conditions 9 Figure: Reactor Coolant System a Type of Thermodynamic System ELO 1.1

© Copyright 2016Operator Generic Fundamentals 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 11 ELO 1.1

© Copyright 2016Operator Generic Fundamentals 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 12 ELO 1.1

© Copyright 2016Operator Generic Fundamentals 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 13 ELO 1.1

© Copyright 2016Operator Generic Fundamentals 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 (h 1 = h 2 ) –No work is done (W = 0) –The process is adiabatic (Q = 0) Example(s) –PORV leak to quench (PRT) tank –Steam leak to atmosphere 14 ELO 1.1

© Copyright 2016Operator Generic Fundamentals Analyzing Systems Using the First Law of Thermodynamics 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 15 ELO 1.2 - Apply the First Law of Thermodynamics for open systems or thermodynamic processes. ELO 1.2 Figure: Thermodynamic Process Shows Transformation of a Working Fluid from One State to Another

© Copyright 2016Operator Generic Fundamentals Thermodynamic Process Four Forms of Energy –Potential energy, kinetic energy, internal energy, and flow energy o Internal and flow energies combined into enthalpy 16 Figure: Basic Energy Balance of the First Law of Thermodynamics ELO 1.2

© Copyright 2016Operator Generic Fundamentals 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 19 ELO 1.2

© Copyright 2016Operator Generic Fundamentals 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 23 ELO 1.2

© Copyright 2016Operator Generic Fundamentals 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) 24 Figure: Multiple Control Volumes in the Same System ELO 1.2

© Copyright 2016Operator Generic Fundamentals 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 25 ELO 1.2

© Copyright 2016Operator Generic Fundamentals Steam Generator Process 26 ELO 1.2 Two flowpaths – (Primary flowpath, Secondary flowpath) –Primary flowpath - heat transferred out of RCS o T hot in, T cold out –Secondary flowpath – heat transferred into SG o Subcooled feedwater in –≈ 420 o F, ≈ 400 BTU/lbm o Saturated steam out –≈ 1000 psia, ≈ 1200 BTU/lbm Energy in (FW) plus heat added (SG) equals Energy out (steam) –h fw + h SG = h stm –400 BTU/lbm + 800 BTU/lbm = 1200 BTU/lbm

© Copyright 2016Operator Generic Fundamentals Turbine Process Inlet to the turbine process is the outlet of the SG process –≈ 1200 BTU/lbm (Energy in ) 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) –h stm - h turbine = h exhaust –1200 BTU/lbm - 400 BTU/lbm = 800 BTU/lbm 27 ELO 1.2

© Copyright 2016Operator Generic Fundamentals Condenser Process Inlet to the condenser process is the outlet of the turbine process –≈ 800 BTU/lbm (Energy in ) Heat is removed FROM the system –≈ 740 BTU/lbm Energy out based on condenser vacuum –Slightly subcooled condensate (based on 1 psia backpressure) o 92 o F is ≈ 60 BTU/lbm Energy in (exhaust) minus heat removed (condenser) equals Energy out (condensate) –h exhaust - h condenser = h condensate –800 BTU/lbm - 740 BTU/lbm = 60 BTU/lbm 28 ELO 1.2

© Copyright 2016Operator Generic Fundamentals Pump Process Inlet to the pump process is the outlet of the condenser process –≈ 60 BTU/lbm (Energy in ) 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) –h condensate + h pump/FW heating = h feedwater –60 BTU/lbm - 340 BTU/lbm = 400 BTU/lbm 29 ELO 1.2

© Copyright 2016Operator Generic Fundamentals Identifying Process Paths on a T-s Diagram 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 30 ELO1.3 - Identify the path(s) on a T-s diagram that represents the thermodynamic processes occurring in a fluid system. ELO 1.3

© Copyright 2016Operator Generic Fundamentals 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) 31 ELO 1.3 Figure: Typical Steam Plant System Cyclic Process SG

© Copyright 2016Operator Generic Fundamentals SG Process - T-s Diagram Subcooled feedwater at 1000 psia and 420 o F enters SG Sensible and latent heat added from RCS (Q A ) Steam exits at ≈ 100% saturated steam –Phase change Area under this curve is the total heat added 32 ELO 1.3 Figure: T-s Diagram Specific Entropy (s) Temperature (T) 1000 psia 1 psia

© Copyright 2016Operator Generic Fundamentals Turbine Process - T-s Diagram 100% saturated steam at 1000 psia enters turbine –≈ 1200 BTU/lbm Exits as a wet vapor –≈ 67% quality –≈ 800 BTU/lbm On IDEAL turbine, no  s On REAL turbine, increase in s –Less work out of steam, higher quality 34 ELO 1.3 Figure: T-s Diagram Specific Entropy (s) Temperature (T) 1000 psia 1 psia Ideal Real

© Copyright 2016Operator Generic Fundamentals Turbine Process - Mollier Diagram Turbine process can also be shown on a Mollier Diagram Find starting pressure (1000 psia) IDEAL work (no  s) –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 35 ELO 1.3 Figure: Mollier Diagram Ideal Real

© Copyright 2016Operator Generic Fundamentals Condenser Process - T-s Diagram ≈ 67% quality wet vapor enters condenser Exits to hotwell as slightly subcooled condensate –Called condensate depression Undergoes phase change –Subcooling – lower efficiency –Subcooling - better for pumps Heat rejected to Circ Water system Area under this curve is the total heat rejected 36 ELO 1.3 Figure: T-s Diagram Specific Entropy (s) Temperature (T) 1000 psia 1 psia

© Copyright 2016Operator Generic Fundamentals Pump Process - T-s Diagram Enters as subcooled condensate –About 1 psia Exits to SG as subcooled feedwater –About 1000 psia IDEAL work of pump –All energy added as pressure Real work of pump –Most added as pressure –Some energy raises temperature 37 ELO 1.3 Figure: T-s Diagram Specific Entropy (s) Temperature (T) 1000 psia 1 psia Ideal Real

© Copyright 2016Operator Generic Fundamentals 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. 38 ELO 1.3

© Copyright 2016Operator Generic Fundamentals Energy Balances on Major Components As previously discussed, boundaries can be set on any component. 39 ELO 1.4 - Given a defined system, Perform energy balances on all major components in the system. Figure: Cyclic Process for Generating Electricity ELO 1.4

© Copyright 2016Operator Generic Fundamentals Steps for Solving Energy Balance Problems 43 ELO 1.4 StepAction 1.Draw the system with the boundaries. 2. Write the general energy equation and solve for the required information. 3. Determine which energies can be ignored to simplify the equation. 4. Make substitutions to ensure correct units are obtained if needed.

© Copyright 2016Operator Generic Fundamentals RCS Heat Transfer Process 44 ELO 1.4 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 10 8 lbm/hr Average specific heat of the fluid is 1.5 BTU/lbm-°F What is the heat transferred out of the RCS?

© Copyright 2016Operator Generic Fundamentals 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 10 8 lbm/hr. The c p of the fluid averages 1.47 BTU/lbm-°F. How much heat is being removed from the heat source? 47 ELO 1.4

© Copyright 2016Operator Generic Fundamentals Heat Exchanger Process 48 Figure: Heat Exchanger Analysis Shows Thermodynamic Balance ELO 1.4 Step 1. Draw the system Show what is given and what is asked for, or requested

© Copyright 2016Operator Generic Fundamentals Condenser to Circ Water Process Steam flows through a condenser at 4.4 x 10 6 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. 51 ELO 1.4

© Copyright 2016Operator Generic Fundamentals Condenser to Circ Water Process Step 1. Draw the system. Show what is given and what is asked for, or requested. 52 Figure: Typical Single-Pass Condenser End View ELO 1.4

© Copyright 2016Operator Generic Fundamentals 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 10 7 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. 54 ELO 1.4

© Copyright 2016Operator Generic Fundamentals Throttling Characteristics 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 55 ELO 1.5 – Describe the exit conditions for a throttling process. ELO 1.5

© Copyright 2016Operator Generic Fundamentals Throttling Process A process in which: –No change in enthalpy from state one to state two (h 1 = h 2 ) –No work is done (W = 0) –Process is adiabatic (Q = 0) Process called “isenthalpic” 56 Figure: Throttling Process by a Valve ELO 1.5

© Copyright 2016Operator Generic Fundamentals The elevation change from boundary 1 to boundary 2 is insignificant. Inlet piping and outlet piping diameter are equal and there is no change in fluid velocity. There is no work done on or done by the fluid as it flows through the throttle. The piping is insulated so there is no heat transferred into or out of the fluid. Throttling Process 58 ELO 1.5

© Copyright 2016Operator Generic Fundamentals StepAction 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). Throttling Process Determining Downstream Properties Step-by-Step Table 59 ELO 1.5

© Copyright 2016Operator Generic Fundamentals Figure: Throttling Process on a Mollier Diagram 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). 60 ELO 1.5

© Copyright 2016Operator Generic Fundamentals 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. 61 ELO 1.5

© Copyright 2016Operator Generic Fundamentals 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. 63 ELO 1.5

© Copyright 2016Operator Generic Fundamentals Throttling Characteristics 64 ELO 1.5 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 2016Operator Generic Fundamentals 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 Correct answer is D. 65 ELO 1.5 0

© Copyright 2016Operator Generic Fundamentals Thermodynamic Systems & Processes 2.1Describe the operation of nozzles to include functions of nozzles in flow restrictors and functions of nozzles in air ejectors. 2.2Describe the condensing process to include vacuum formation and condensate depression. 2.3Explain the design of turbines to include the functions of nozzles, fixed blading, moving blading and the reason turbines are multistage. 66 TLO 2 – Describe the operation of the turbine and condensing processes. TLO 2

© Copyright 2016Operator Generic Fundamentals ELO 2.1 Nozzle Characteristics 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 67 ELO 2.1 – Describe the operation of nozzles to include functions of nozzles in flow restrictors and functions of nozzles in air ejectors.

© Copyright 2016Operator Generic Fundamentals 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 –Shape –Pressure 68 Figure: Typical Nozzle Types ELO 2.1

© Copyright 2016Operator Generic Fundamentals 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 69 ELO 2.1 Figure: Typical Nozzle Types

© Copyright 2016Operator Generic Fundamentals Nozzles – Theory of Operation Uses simplified general energy equation to explain nozzle operation In a convergent nozzle, (A 1 > A 2 ) 70 The elevation change from entrance (1) to exit (2) is insignificant. Inlet piping diameter is greater than outlet piping diameter. With steady flow, outlet velocity is greater than inlet velocity. There is no work done on or done by the fluid in the nozzle. The nozzle is perfectly insulated so no heat is transferred into or out of the fluid. Assume that there is no friction as the fluid flows through the nozzle. ELO 2.1

© Copyright 2016Operator Generic Fundamentals 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 73 ELO 2.1

© Copyright 2016Operator Generic Fundamentals 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 74 Figure: Convergent-Divergent Venturi Tube for Flow Measurement ELO 2.1

© Copyright 2016Operator Generic Fundamentals Air Ejectors Pump-like device with no moving parts or pistons that utilizes high- pressure 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 75 ELO 2.1

© Copyright 2016Operator Generic Fundamentals 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 76 Figure: Simple Air Ejector (Jet Pump) ELO 2.1

© Copyright 2016Operator Generic Fundamentals 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 77 Figure: Two-Stage Steam Jet Ejector ELO 2.1

© Copyright 2016Operator Generic Fundamentals 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 78 ELO 2.1 Figure: Two-Stage Steam Jet Ejector

© Copyright 2016Operator Generic Fundamentals 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 79 ELO 2.1

© Copyright 2016Operator Generic Fundamentals Condenser Design/Operation 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” 80 ELO 2.2 – Describe the condensing process to include vacuum formation and condensate depression. ELO 2.2

© Copyright 2016Operator Generic Fundamentals 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. 81 Figure: Typical Single-Pass Condenser ELO 2.2

© Copyright 2016Operator Generic Fundamentals Main Condenser Steam is condensed –Latent heat of condensation Specific volume decreases drastically Creates a low pressure, maintaining vacuum Increases plant efficiency 82 Figure: Typical Single-Pass Condenser End View ELO 2.2 To SG Through Feedwater Heaters

© Copyright 2016Operator Generic Fundamentals Main Condenser Condensate Depression As the condensate falls toward the hotwell, it subcools (goes below T SAT ) as it comes in contact with tubes lower in the condenser The amount of subcooling is the condensate depression T SAT – T HOTWELL = the amount of condensate depression 83 Figure: T-s Diagram for Typical Condenser ELO 2.2

© Copyright 2016Operator Generic Fundamentals 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 T in = 75 °F and T out = 97 °F Assume c p for the condensate and the circulating water is 1 BTU/lbm- °F Steam mass flow rate in the condenser is 8 x 10 6 lbm/hr and the circulating water is 3.1 x 10 8 lbm/hr 84 ELO 2.2

© Copyright 2016Operator Generic Fundamentals 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. 88 ELO 2.2

© Copyright 2016Operator Generic Fundamentals 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. 89 ELO 2.2

© Copyright 2016Operator Generic Fundamentals Turbine Design and Characteristics 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 90 ELO 2.3 – Explain the design of turbines to include the functions of nozzles, fixed blading, moving blading, and benefits of multistage turbines. ELO 2.3

© Copyright 2016Operator Generic Fundamentals Figure: Rankine Cycle on an h-s Diagram 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 3 o 91 ELO 2.3

© Copyright 2016Operator Generic Fundamentals 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 92 ELO 2.3

© Copyright 2016Operator Generic Fundamentals 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. 94 ELO 2.3

© Copyright 2016Operator Generic Fundamentals 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 96 ELO 2.3

© Copyright 2016Operator Generic Fundamentals 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 98 Figure: Basic Impulse Turbine ELO 2.3

© Copyright 2016Operator Generic Fundamentals 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 99 Figure: Impulse/Reaction Turbine Comparison ELO 2.3

© Copyright 2016Operator Generic Fundamentals 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 100 Figure: Basic Reaction Turbine ELO 2.3

© Copyright 2016Operator Generic Fundamentals 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 101 ELO 1.2 ELO 2.3 Figure: Impulse/Reaction Turbine Comparison

© Copyright 2016Operator Generic Fundamentals 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 102 ELO 2.3 Figure: Impulse/Reaction Turbine Comparison

© Copyright 2016Operator Generic Fundamentals 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 103 Figure: Steam Property Variation in Simple Impulse Turbine ELO 2.3

© Copyright 2016Operator Generic Fundamentals 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 104 Figure: Steam Property Variation In Actual Impulse Turbine ELO 2.3

© Copyright 2016Operator Generic Fundamentals 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 105 Figure: Steam Property Variation in a Reaction Turbine ELO 2.3

© Copyright 2016Operator Generic Fundamentals 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 106 ELO 2.3 Figure: Steam Property Variation in a Reaction Turbine

© Copyright 2016Operator Generic Fundamentals 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 107 ELO 2.3

© Copyright 2016Operator Generic Fundamentals 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. 111 Summary

© Copyright 2016Operator Generic Fundamentals NRC KA to ELO Tie 112 KA #KA StatementROSROELO K1.01Explain the relationship between real and ideal processes.1.81.91.1* K1.02Explain the shape of the T-s diagram process line for a typical secondary system.1.71.91.3 K1.03Describe the functions of nozzles in flow restrictors.1.9 2.1 K1.04Describe the functions of nozzles in air ejectors.2.0 2.1 K1.05Explain the function of nozzles fixed blading and moving blading in the turbine.1.61.72.3 K1.06Explain the reason turbines are multistages.1.51.72.3 K1.07Define turbine efficiency.1.6 1.2 K1.08Explain the difference between real and ideal turbine efficiency.1.61.71.2 K1.09Define pump efficiency.1.3 1.2* K1.10Explain the difference between ideal and real pumping processes.1.3 1.2* K1.11Describe the process of condensate depression and its effect on plant operation.2.42.52.2 K1.12Explain vacuum formation in condenser processes.2.22.32.2 K1.13Explain the condensing process.2.22.32.2 K1.14Explain the reduction of process pressure from throttling.2.12.31.5 K1.15Determine the exit conditions for a throttling process based on the use of steam and/or water2.8 1.5 * K1.01, K1.09, and K1.10 are discussed in further detail in 193005 - Cycles

© Copyright 2016Operator Generic Fundamentals Operator Generic Fundamentals 193004 - Thermodynamic Processes.

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© Copyright 2016Operator Generic Fundamentals Operator Generic Fundamentals 193004 - Thermodynamic Processes.

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