Operator Generic Fundamentals Thermodynamic Processes

Operator Generic Fundamentals 193004 - Thermodynamic Processes
K1.01 Explain the relationship between real and ideal processes. K1.02 Explain the shape of the T-s diagram process line for a typical secondary system. K1.03 Describe the functions of nozzles in flow restrictors. K1.04 Describe the functions of nozzles in air ejectors. K1.05 Explain the function of nozzles fixed blading and moving blading in the turbine. K1.06 Explain the reason turbines are multistages. K1.07 Define turbine efficiency. K1.08 Explain the difference between real and ideal turbine efficiency. K1.09 Define pump efficiency. K1.10 Explain the difference between ideal and real pumping processes. K1.11 Describe the process of condensate depression and its effect on plant operation. (BANK QUESTIONS) K1.12 Explain vacuum formation in condenser processes. K1.13 Explain, the condensing process. K1.14 Explain the reduction of process pressure from throttling. K1.15 Determine the exit conditions for a throttling process based on the use of steam and/or water (BANK QUESTIONS) Operator Generic Fundamentals Thermodynamic Processes

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: Apply the first law of thermodynamics to analyze thermodynamic systems and processes. Describe the operation of the turbine and condensing processes. The first TLO relates to each of the four processes in our thermodynamic cycle. The second TLO takes a look at the “internals” of some of these processes. For example, looking at how a nozzle works within our turbine process. Nozzles are explained in further detail in – Sensors/Detectors. TLOs

Thermodynamic Processes 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: Open, closed, or isolated system Reversible (ideal) process Irreversible (real) process Adiabatic process Isentropic process Isenthalpic (throttling) process First Law of Thermodynamics states: Energy can be neither created nor destroyed, only altered in form Second Law of Thermodynamics states: No engine, actual or ideal, when operating in a cycle can convert all the heat supplied it into mechanical work–heat must be rejected ELOs

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. Several NRC bank questions relating to ELO Not understanding the downstream tailpipe temperature relating to a pressurizer PORV leaking was one of the issues at TMI. ELOs

First Law of Thermodynamics
ELO 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 No related KAs specific to this objective. Figure: Energy Balance Equals the First Law of Thermodynamics ELO 1.1

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 Keep in mind that when we discuss the processes (systems) of our thermodynamic cycle a “system” is not the ENTIRE system, just the boundary defined in our discussion. ELO 1.1

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 ELO 1.1

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 Most of our processes discussed in this chapter and the next chapter (unless mentioned otherwise) can be assumed to be a steady flow system. In that regards all we really need to look is the change in specific enthalpy of each process (specific work done). In real processes there might be losses. For example, we can assume that mass flow rate of steam equals mass flow rate of feedwater if SG level is constant. In the real world we will have losses in the SG process because of blowdown. Also, a generic assumption can be made that the heat transfer rate out of the RCS equals the heat transfer rate into the SG. In the real world there are ambient losses, heat addition from the RCPs, etc. The Heat Transfer chapter (193007) will discuss a secondary calorimetric that takes into account these “real” conditions. ELO 1.1

Types of Thermodynamic Systems
The RCS can be considered each type of system under certain operational conditions This chapter looks at each of the processes of our secondary thermodynamic cycle as CLOSED systems. An ISOLATED System does not really exist (except in discussions about IDEAL concepts). A better example would be how these new insulated mugs and bottles keep liquids hot/cold for finite periods of time (no mass or energy transferred). Figure: Reactor Coolant System a Type of Thermodynamic System ELO 1.1

Steady Flow Equilibrium Process
NOTE: Temperature is only constant assuming the only energy change is either the latent heat of vaporization or the latent heat of condensation. Since feedwater comes in subcooled and condensate is slightly subcooled, temperature does change slightly. Figure: Steady Flow Systems ELO 1.1

Reversible (Ideal) Process
A process where system and surroundings are returned to their original condition before the process occurred No losses (change in entropy) 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 ELO 1.1

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 Feedwater as close to saturation temperature Feedwater heating done in steps Recall that the RCS has a “finite” amount of heat to add at 100% rated thermal power. The higher the temperature of the feedwater entering the SG the higher the temperature (steam pressure) exiting the SG. The less the “sensible” heat required to add, the more efficient the process. These “efficiencies” will be discussed in greater detain in the next chapter ( – Cycles). ELO 1.1

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 work process ELO 1.1

Isenthalpic Process An isenthalpic process is one in which the enthalpy of the fluid remains constant Throttling processes are isenthalpic 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 Throttling process examples are explained in further detail later in this presentation. ELO 1.1

Analyzing Systems Using the First Law of Thermodynamics
ELO 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 Related KA K1.05 Explain the law of conservation of energy. 2.1* 2.1 Figure: Thermodynamic Process Shows Transformation of a Working Fluid from One State to Another ELO 1.2

Thermodynamic Process
Four Forms of Energy Potential energy, kinetic energy, internal energy, and flow energy Internal and flow energies combined into enthalpy IE – Internal Energy is noted by the term “u”. FE – Flow Energy is noted by the term “P-v” Energy IN must equal energy OUT U1 Pv1 U2 Pv2 Figure: Basic Energy Balance of the First Law of Thermodynamics ELO 1.2

General Energy Equation
The “General Energy Equation” is another term for “Bernoulli’s Equation”, which is provided on the NRC Equation Sheet. NOTE: The General Energy Equation is also sometimes called the “Continuity Equation”. Figure: General Energy Equation for the First Law of Thermodynamics ELO 1.2

Thermodynamic Processes
Figure: Six Basic Processes of Steady Flow Systems ELO 1.2

Thermodynamic Processes
Our four basic processes of our thermodynamic cycle are: Steam Generator Process Subcooled feedwater in, heat added, saturated steam out Turbine process Saturated steam in, work done BY steam, wet vapor out Condensing Process Wet vapor in, heat removed, subcooled condensate out Pump Process Subcooled condensate in, work done ON fluid, subcooled feedwater out The pump process is the most difficult one to analyze by calculations because most plants have various “numbers” of pumps (condensate, heater drain, feedwater). Plus, the energy added by the compression of the subcooled fluid is minimal compared to the energy added by feedwater heating. The first step to analyzing a process is to establish the boundary. The next slide shows an example. ELO 1.2

Boundary Example This same concept of establishing the boundary applies to each of our four processes in the secondary thermodynamic cycle (Note previous slide). Figure: Open System Control Volume Concept for a Pump ELO 1.2

Conservation of Energy
𝑚 ℎ 𝑖𝑛 + 𝑃𝐸 𝑖𝑛 + 𝐾𝐸 𝑖𝑛 + 𝑄 = 𝑚 ℎ 𝑜𝑢𝑡 + 𝑃𝐸 𝑜𝑢𝑡 + 𝐾𝐸 𝑜𝑢𝑡 + 𝑊 Where: 𝑚 = mass flow rate of working fluid (lbm/hr) ℎ 𝑖𝑛 = specific enthalpy of the working fluid entering the system (BTU/lbm) ℎ 𝑜𝑢𝑡 = specific enthalpy of the working fluid leaving the system (BTU/lbm) 𝑃𝐸 𝑖𝑛 = specific potential energy of working fluid entering the system (ft-lbf/lbm) 𝑃𝐸 𝑜𝑢𝑡 = specific potential energy of working fluid leaving the system (ft-lbf/lbm) 𝐾𝐸 𝑖𝑛 = specific kinetic energy of working fluid entering the system (ft-lbf/lbm) 𝐾𝐸 𝑜𝑢𝑡 = 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) ELO 1.2

Heat Transferred Into or Out of System
Figure: Heat and Work in a System ELO 1.2

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 Open system approach to First Law of Thermodynamics will be emphasized because it is more general ELO 1.2

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) The next few slides will look at the energies associated with each of our four main processes in our secondary thermodynamic cycle. Figure: Multiple Control Volumes in the Same System ELO 1.2

Thermodynamic Process Assumptions
Each process initially presented as IDEAL Minimal change to KE or PE 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 Keep in mind that these processes are being presented as IDEAL just to simplify each process. Actual losses, as well as ways to minimizes those losses, will be presented later in this chapter as well as in the next chapter. ELO 1.2

Steam Generator Process
Two flowpaths – (Primary flowpath, Secondary flowpath) Primary flowpath - heat transferred out of RCS Thot in, Tcold out Secondary flowpath – heat transferred into SG Subcooled feedwater in ≈ 420oF, ≈ 400 BTU/lbm Saturated steam out ≈ 1000 psia, ≈ 1200 BTU/lbm Energy in (FW) plus heat added (SG) equals Energy out (steam) hfw + qSG = hstm 400 BTU/lbm BTU/lbm = 1200 BTU/lbm The secondary flowpath is the one we will discuss. In IDEAL situations one can say the heat transferred out of the RCS equals the heat transferred into the SG. However, there are so many variables on both sides that this is impractical. For example, on the primary side what we really want to know is how much of the RCS heat is from the core. There is also RCP heat and pressurizer heat that must be subtracted. Also there are ambient losses that must be added in. On the secondary side there are losses from the SG process due to blowdown. All of these “real” situations will be discussed in Chapter – Heat Transfer. ELO 1.2

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 - wturbine = hexhaust 1200 BTU/lbm BTU/lbm = 800 BTU/lbm There are losses from the SG to the turbine as well as some steam flow directed to the MSRs. These changes will be discussed in greater detail in the next chapter. ELO 1.2

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) 92oF is ≈ 60 BTU/lbm Energy in (exhaust) minus heat removed (condenser) equals Energy out (condensate) hexhaust - qcondenser = hcondensate 800 BTU/lbm BTU/lbm = 60 BTU/lbm The concept of condensate depression (subcooling the condensate below Tsat for the condenser backpressure) will be discussed in a later slide. ELO 1.2

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 + wpump/FW heating = hfeedwater 60 BTU/lbm BTU/lbm = 400 BTU/lbm Calculations on the PUMP Process is difficult to analyze because the majority of energy added to the condensate is not from the pump itself, but from the feedwater heating. These values are only provided as approximations. ELO 1.2

Turbine Efficiency 𝜂 𝑡 = 𝑊 𝑡.𝑎𝑐𝑡𝑢𝑎𝑙 𝑊 𝑡.𝑖𝑑𝑒𝑎𝑙
𝜂 𝑡 = 𝑊 𝑡.𝑎𝑐𝑡𝑢𝑎𝑙 𝑊 𝑡.𝑖𝑑𝑒𝑎𝑙 𝜂 𝑡 = ℎ 𝑖𝑛 − ℎ 𝑜𝑢𝑡 𝑎𝑐𝑡𝑢𝑎𝑙 ℎ 𝑖𝑛 − ℎ 𝑜𝑢𝑡 𝑖𝑑𝑒𝑎𝑙 Where: 𝜂 𝑡 = turbine efficiency (no units) Wt.actual = actual work done by the turbine (ft-lbf) Wt.ideal = work done by an ideal turbine (ft-lbf) (hin – hout)actual = actual enthalpy change of the working fluid (BTU/lbm) (hin – hout)ideal = enthalpy change of the working fluid in an ideal turbine (BTU/lbm) Even though this equation is NOT on the NRC Equation Sheet, the only bank question that requires it use has a derivation of this formula provided in the STEM of the question. ELO 1.2

Turbine Efficiency Turbine efficiency is generally
60% to 80% for small turbines 90% for large turbines Turbine efficiency ηt is normally given by the manufacturer Permits actual work done to be calculated directly by multiplying turbine efficiency ηt by work done by an ideal turbine under the same conditions Figure: Entropy Diagram Measures System Efficiency ELO 1.2

Turbine Efficiency Ideal case is a constant entropy represented by a vertical line Actual turbine involves an increase in entropy The smaller the increase in entropy, the closer the turbine efficiency ηt is to 1.0 or 100% Figure: Entropy Diagram Measures System Efficiency ELO 1.2

Pump Work 𝐻 𝑜𝑢𝑡 − 𝐻 𝑖𝑛 = 𝑊 𝑝 𝑚 ℎ 𝑜𝑢𝑡 − ℎ 𝑖𝑛 = 𝑊 𝑝
A pump is designed to move the working fluid by doing work on it The increase in the enthalpy of the working fluid Hout - Hin equals the work done by the pump (Wp) on the working fluid Hout = enthalpy of the working fluid leaving the pump (BTU) Hin = enthalpy of the working fluid entering the pump (BTU) Wp = work done by the pump on the working fluid (ft-lbf) 𝑚 = mass flow rate of the working fluid (lbm/hr) hout = specific enthalpy of the working fluid leaving the pump (BTU/lbm) hin = specific enthalpy of the working fluid entering the pump (BTU/lbm) 𝑊 𝑝 = power of pump (BTU/hr) 𝐻 𝑜𝑢𝑡 − 𝐻 𝑖𝑛 = 𝑊 𝑝 𝑚 ℎ 𝑜𝑢𝑡 − ℎ 𝑖𝑛 = 𝑊 𝑝 ELO 1.2

Pump Efficiency An ideal pump provides a basis for analyzing the performance of actual pumps A pump requires more work because of unavoidable losses due to friction and fluid turbulence The work done by a pump equals the change in enthalpy across the actual pump 𝑊 𝑝.𝑎𝑐𝑡𝑢𝑎𝑙 = 𝐻 𝑜𝑢𝑡 − 𝐻 𝑖𝑛 𝑎𝑐𝑡𝑢𝑎𝑙 𝑊 𝑝.𝑎𝑐𝑡𝑢𝑎𝑙 = 𝑚 ℎ 𝑜𝑢𝑡 − ℎ 𝑖𝑛 𝑎𝑐𝑡𝑢𝑎𝑙 ELO 1.2

Pump Efficiency Pump efficiency (ηp) is the ratio of the work required by the pump if it were an ideal pump Wp.ideal to the actual work required by the pump Wp.actual 𝜂 𝑝 = 𝑊 𝑝.𝑖𝑑𝑒𝑎𝑙 𝑊 𝑝.𝑎𝑐𝑡𝑢𝑎𝑙 Also shown as: 𝜂 𝑝 = 𝐹𝑙𝑜𝑤 𝐻𝑜𝑟𝑠𝑒𝑝𝑜𝑤𝑒𝑟 𝐵𝑟𝑎𝑘𝑒 𝐻𝑜𝑟𝑠𝑒𝑝𝑜𝑤𝑒𝑟 The efficiency of a pump can be represented by the words: “How much flow do I get out of the pump over how much energy is required to get this flow”. The purpose of a pump is to raise the pressure of the system to overcome the headloss in the system (or raise the pressure higher, in the case of feedwater pumps). The less efficient the pump, the less the pressure gets raised and the more the temperature is raised. Pumps are normally operated in a system where their normal operating point is as close to it’s most efficient operating point. ELO 1.2

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 Cycles Being able to draw or label a T-s diagram is not tested by the NRC, but having an understanding of the visual image of each process will assist in understanding how to make the process more efficient. ELO 1.3

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 system on the 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 ELO 1.3

SG Process - T-s Diagram
Subcooled feedwater at psia and 420oF enters SG Sensible and latent heat added from RCS (QA) Steam exits at ≈ 100% saturated steam Phase change Area under this curve is the total heat added Figure: T-s Diagram Specific Entropy (s) Temperature (T) 1000 psia 1 psia Values used in this example are approximations and may vary at your plant. Also, it will be assumed that the moisture separators are 100% efficient and the steam exiting is 100% saturated steam. Plant specifics can be mentioned as seen fit. ELO 1.3

Moisture and quality on the Mollier diagram
0% 100% All water Sf All steam Sg The Mollier also shows subcooled (compressed) and superheated areas, but most PWR problems are done under the saturation curve where the above line applies. Moisture 100% 0% ELO 1.3

Turbine Process - T-s Diagram
100% saturated steam at psia enters turbine ≈ 1200 BTU/lbm Exits as a wet vapor ≈ 67% quality ≈ 800 BTU/lbm On IDEAL turbine, no Ds On REAL turbine, increase in s Less work out of steam, higher quality Figure: T-s Diagram Specific Entropy (s) Temperature (T) 1000 psia 1 psia Ideal Real Remind students of the relationship between primary and secondary efficiency and some standard values that result. For example if the secondary cycle is about 33% efficient and 1200 BTU/lbm enter the turbine (heat added from RCS), then about 400 BTU/lbm of work is done by the turbine, and 800 BTU/lbm remain. 400/1200 = 0.33. Also if the turbine is only getting about 1/3 of the energy out of the saturated steam, then it stands to reason that the exhaust still has about 2/3 of the quality in it (started with 100% quality, ends with 67% quality). ELO 1.3

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 Condenser pressure of 1 psia REAL work (increase in s) Draw line down to end pressure, but at higher final enthalpy Still exhausts to 1 psia Figure: Mollier Diagram Ideal Real ELO 1.3

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 Figure: T-s Diagram Specific Entropy (s) Temperature (T) 1000 psia 1 psia Note that the area between the “Heat Added” and the “Heat Rejected” is the “work of the turbine”. One of the ways to maximize efficiency of our thermodynamic cycle (explained in greater detail in the next chapter) is to maximize the DP of the turbine process. The higher the SG pressure and the lower the backpressure (greater the vacuum), the more efficient the cycle. Another way to maximize efficiency is to minimize the amount of subcooling in the condensate (condensate depression). This does, however, make the pump operate closer to cavitation. ELO 1.3

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 Figure: T-s Diagram Specific Entropy (s) Temperature (T) 1000 psia 1 psia Ideal Real ELO 1.3

Energy Balances on Major Components
ELO Given a defined system, perform energy balances on all major components in the system. As previously discussed, boundaries can be set on any component. Instructor Note: This PowerPoint slide animates in full screen slide show view. However, hopefully rods are fully withdrawn at 100% power! Figure: Cyclic Process for Generating Electricity ELO 1.4

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 = 𝑚 Dh (must be used for a phase change) 𝑄 = 𝑚 cpDT (can be used with NO phase change) Before calculations can be performed related to a thermodynamic process, an understanding of heat/work terms as well as introduction of equations available need to be discussed. The two equations shown above are at the top left hand side of the NRC Equation Sheet. NOTE: The third equation, Q-dot = UA Delta-T, is designed to be used across a heat transfer surface (from RCS to SG, or from Condenser to CW) ELO 1.4

Heat Transfer Terms/Equations
SG Process Equation 𝑄 𝑆𝐺= 𝑚 stm(hstm – hfw) Turbine Process Equation 𝑊 𝑇𝑢𝑟𝑏= 𝑚 stm(hstm – hexh) Condenser Process Equation 𝑄 𝐶𝑜𝑛𝑑= 𝑚 exh(hexh – hcond) Pump Process Equation 𝑊 𝑃𝑢𝑚𝑝= 𝑚 fw(hfw – hcond) NOTE: Even though the Condensing and Pump processes are “negative” processes in our cycle, the mathematical equation is based on a positive value (large specific enthalpy minus small specific enthalpy) The pump process is not directly tested by the NRC in this section because it is difficult to analyze since the majority of the energy added in the “pump process” is not due to the pressure increase, but due to the temperature increase from feedwater heating. ELO 1.4

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 Lower pressure in SG Colder Fluid (Tcold), with its energy removed, is pumped back to the heat source for reheating For analysis purposes: 𝑄 𝑅𝐶𝑆= 𝑄 𝑆𝐺 DT method cannot be used on SG side No temperature change while adding latent heat Before calculations can be performed related to a thermodynamic process, an understanding of heat/work terms as well as introduction of equations available need to be discussed. ELO 1.4

Steps for Solving Energy Balance Problems
Action 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. ELO 1.4

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? ELO 1.4

Step 1. Draw the system Show what is given and what is asked for, or requested Figure: Steam Generation System ELO 1.4

Step 2 and 3. Write the equation and simplify as possible ṁ 𝑖𝑛 ℎ 𝑖𝑛 + 𝑃𝐸 𝑖𝑛 + 𝐾𝐸 𝑖𝑛 + 𝑄 = 𝑚 𝑜𝑢𝑡 ( ℎ 𝑜𝑢𝑡 + 𝑃𝐸 𝑜𝑢𝑡 + 𝐾𝐸 𝑜𝑢𝑡 )+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 ℎ 𝑖𝑛 + 𝑄 = 𝑚 ( ℎ 𝑜𝑢𝑡 ) 𝑄 = 𝑚 ( ℎ 𝑜𝑢𝑡 – ℎ 𝑖𝑛 ) Substituting, 𝑄 = 𝑚 𝑐 𝑝 𝛥𝛵 , where cp = specific heat capacity (𝐵𝑇𝑈 / 𝑙𝑏𝑚−℉): = 𝑚 𝑐 𝑝 𝑇 𝑜𝑢𝑡 – 𝑇 𝑖𝑛 = 1.38 × 𝑙𝑏𝑚 ℎ𝑟 𝐵𝑇𝑈 𝑙𝑏𝑚−℉ 540−610 ℉ 𝑄 =−1.45× 𝐵𝑇𝑈 ℎ𝑟 Note: Minus sign indicates heat out of RCS PE can be ignored even though the SG are tall it takes 778’ to = 1 BTU. KE is ignored because the flow rate is constant and the change in velocity is small. Mass flow rate is the same entering and leaving the SG. ΔT (change in temperature) equation used because no phase change on primary side of SG and Th and Tc are easily measured. For this “IDEAL” example, we can say that the “heat transferred out of the RCS” is equal to the “heat transferred into the SG”. ELO 1.4

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? ELO 1.4

Step 1. Draw the system Show what is given and what is asked for, or requested Figure: Heat Exchanger Analysis Shows Thermodynamic Balance ELO 1.4

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) PE can be ignored even though the SG are tall it takes 778’ to = 1 BTU. KE is ignored because the flow rate is constant and the change in velocity is small. Mass flow rate is the same entering and leaving the SG. ELO 1.4

Step 3 and 4. Make needed substitutions to ensure correct units are obtained. Substituting 𝑄 = 𝑚 𝑐 𝑝 𝛥𝑇, where cp = specific heat capacity: 𝑄 = 𝑚 𝑐𝑝 𝑇 𝑜𝑢𝑡 – 𝑇 𝑖𝑛 + 𝑊 𝑄 = 1.32 × 𝑙𝑏𝑚 ℎ𝑟 𝐵𝑇𝑈 𝑙𝑏𝑚–℉ – 542 °𝐹 𝑄 = 1.36 𝑥 𝐵𝑇𝑈 ℎ𝑟 PE can be ignored because it takes 778’ to = 1 BTU. KE is ignored because the flow rate is constant and the change in velocity is small from inlet to outlet. Mass flow rate is the same entering and leaving the pump, and water is incompressible. For this example, 𝑄 = 𝑚 𝑐 𝑝 𝛥𝑇 has been used to calculate the heat transfer rate since no phase change has occurred. However, 𝑄 = 𝑚 ℎ 𝑜𝑢𝑡 − ℎ 𝑖𝑛 could also have been used if the problem data included inlet and outlet enthalpies. NOTE: To determine the heat transfer rate in MW, simply divide the above number by 3.41 x 106 ELO 1.4

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. ELO 1.4

Step 1. Draw the system. Show what is given and what is asked for, or requested. Ask students how they would get enthalpy if not given. Use the saturated liquid enthalpy at the given temperature. Figure: Typical Single-Pass Condenser End View ELO 1.4

Step 2, 3, and 4. Write the equation and simplify as possible convert to required units 𝑄 𝑐𝑜𝑛𝑑 =− 𝑄 𝑐𝑤 Steps 3 and 4. Simplify the equation and arrange for required units. 𝑚 𝑒𝑥ℎ ℎ 𝑜𝑢𝑡 – ℎ 𝑖𝑛 𝑐𝑜𝑛𝑑 = 𝑚 ℎ 𝑜𝑢𝑡 – ℎ 𝑖𝑛 𝑐𝑤 𝑚 𝑐𝑤 = 𝑚 𝑒𝑥ℎ ℎ 𝑜𝑢𝑡 – ℎ 𝑖𝑛 𝑐𝑜𝑛𝑑 ℎ 𝑜𝑢𝑡 – ℎ 𝑖𝑛 𝑐𝑤 𝑚 𝑐𝑤 =4.4× 𝑙𝑏𝑚 ℎ𝑟 𝑥 54– 𝐵𝑇𝑈 𝑙𝑏𝑚 45−32 𝐵𝑇𝑈 𝑙𝑏𝑚 𝑚 𝑐𝑤 =−3.56× 𝑙𝑏𝑚 ℎ𝑟 ELO 1.4

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 Related KAs K1.14 Explain the reduction of process pressure from throttling ; K1.15 Determine the exit conditions for a throttling process based on the use of steam and/or water ELO 1.5

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 ELO 1.5

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 Ensure the students are comfortable working with throttling process problems using the Mollier Diagram and the steam tables. NRC exams test this topic heavily, as it is one of only two K/As in this chapter which have related questions (value >2.5). ELO 1.5

Throttling Process 𝑃𝐸1 = 𝑃𝐸2 𝐾𝐸1 = 𝐾𝐸2
This gives us the following results for a throttling process: 𝑃 1 𝜈 1 𝐽 + 𝑢 1 = 𝑃 2 𝜈 2 𝐽 + 𝑢 2 ℎ 𝑖𝑛 = ℎ 𝑜𝑢𝑡 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. 𝑊 𝐼𝑁 = 𝑊 𝑂𝑈𝑇 = 0 The piping is insulated so there is no heat transferred into or out of the fluid. 𝑄 𝐼𝑁 = 𝑄 𝑂𝑈𝑇 = 0 ELO 1.5

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). ELO 1.5

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 ELO 1.5

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. NOTE: ALL pressure values in the Steam Tables are in PSIA! Be sure to convert. Also, in this example since you finish under the dome of the Mollier Diagram you know you are at saturation temperature for the pressure (40 psia). That value can easily be looked up in Table 2. This is a TMI example problem! ELO 1.5

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 𝑃 1 = 2,185psig + 15psi = 2,200psia 𝑃 2 = 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, 𝑇 𝑠𝑎𝑡 =281°𝐹 ELO 1.5

Knowledge Check Which one of the following is essentially a constant enthalpy process? Steam flowing through an ideal convergent nozzle Condensation of turbine exhaust in a main condenser Expansion of main steam through the stages of an ideal turbine Throttling of main steam through a main turbine steam inlet valve Correct answer is D. Correct answer is D NRC Bank Question: P1777 Analysis: A. WRONG. Steam (assuming subsonic) flowing through an ideal convergent nozzle has a lower pressure (thus lower enthalpy) as it travels through to the throat of the nozzle, and therefore a higher velocity. B. WRONG. As turbine exhaust is condensed, the enthalpy lowers. C. WRONG. The enthalpy of the steam is transferred to work through the stages of an ideal turbine. This lowers the enthalpy. D. CORRECT. Main steam passing through turbine inlet valves is a throttling process, and therefore, a constant enthalpy process. 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... decrease because the entropy of the safety valve discharge will decrease during the pressurizer pressure decrease in this range. decrease because the enthalpy of the safety valve discharge will decrease during the pressurizer pressure decrease in this range. increase because the safety valve discharge will become more superheated during the pressurizer pressure decrease in this range. 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. Correct answer is C NRC Question P7610 Analysis: This process is called “isenthalpic”, by moving straight across a given enthalpy on the Mollier Diagram. Both conditions (1,000 psia and 700 psia) will intersect with the 40 psia constant pressure line above the saturation curve. As pressure decreases to 700 psia, the substance will become MORE superheated. Therefore, Choice “C” is correct Another sanity check is if the hg of your starting pressure is greater than the hg of your ending pressure, then the solution must be SUPERHEATED. If the hg of your starting pressure is less than the hg of your ending pressure, then the solution must be a WET VAPOR. psia is BTU/lbm (superheated) 700 psia is BTU/lbm (more superheated) 40 psia is BTU/lbm ELO 1.5

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? 267F, saturated 267F, superheated 312F, saturated 312F, superheated Correct answer is D. Correct answer is D NRC Question P4040 Analysis: This process is called “isenthalpic”, noted by moving straight across a given enthalpy on the Mollier Diagram. With saturated steam at 1,000 psia, the initial (and final) enthalpy (approximately 1203 Btu/lbm) can be found on the Mollier diagram using the 1,000 psia constant pressure line. The constant enthalpy line intersects 40 psia at between the 300ºF and 350ºF. Note that this is above the saturation temperature for 40 psia, and outside the “dome”. ELO 1.5

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. TLO 2

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 Related KAs Nozzles: K1.03 Describe the functions of nozzles in flow restrictors. 1.9* 1.9*; K1.04 Describe the functions of nozzles in air ejectors Nozzles are only presented here as an understanding of how the turbine can be more efficient. Bank questions on nozzles are provided in – Sensors and detectors chapter. ELO 2.1

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 Figure: Typical Nozzle Types ELO 2.1

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 Figure: Typical Nozzle Types ELO 2.1

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. 𝐾𝐸2 > 𝐾𝐸1 There is no work done on or done by the fluid in the nozzle. 𝑊𝐼𝑁 = 𝑊𝑂𝑈𝑇 = 0 The nozzle is perfectly insulated so no heat is transferred into or out of the fluid. 𝑄𝐼𝑁 = 𝑄𝑂𝑈𝑇 = 0 Assume that there is no friction as the fluid flows through the nozzle. 𝑈1 = 𝑈2 ELO 2.1

𝑚 1 = 𝑚 2 𝑚 = 𝐴v 𝜈 =𝜌𝐴v 𝜌 1 𝐴 1 v 1 = 𝜌 2 𝐴 2 v 2 = 𝜌 𝑥 𝐴 𝑥 v 𝑥 Where: 𝑚 1 = 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) ELO 2.1

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 ELO 2.1

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 ELO 2.1

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 ELO 2.1

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 ELO 2.1

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) ELO 2.1

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 ELO 2.1

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 ELO 2.1

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 ELO 2.1

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” K1.11 Describe the process of condensate depression and its effect on plant operation. K1.12 Explain vacuum formation in condenser processes. K1.13 Explain the condensing process. There are several bank questions on K1.11. ELO 2.2

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. ANSWER: D. Figure: Typical Single-Pass Condenser ELO 2.2

To SG Through Feedwater Heaters
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 ELO 2.2

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 ANSWER: D. Figure: T-s Diagram for Typical Condenser ELO 2.2

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 This example and the next three slides showing the solution(s) may be hidden or not used based on your needs. However, it is good at ensuring the student knows where to start when trying to answer process questions. ELO 2.2

Condenser Example Problem (cont’d)
Find 𝑄 of the circulating water: 𝑄 = 𝑚 𝑐 𝑝 𝛥𝑇 = 𝑚 1 𝐵𝑇𝑈 𝑙𝑏𝑚−°𝐹 °𝐹−75° 𝐹 = 3.1× 𝑙𝑏𝑚 ℎ𝑟 22 𝐵𝑇𝑈 𝑙𝑏𝑚 =6.82× 𝐵𝑇𝑈 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 = 𝑚 𝑐 𝑝 𝛥𝑇 = 8× 𝑙𝑏𝑚 ℎ𝑟 1 𝐵𝑇𝑈 𝑙𝑏𝑚−°𝐹 4 °𝐹 =3.2× 𝐵𝑇𝑈 ℎ𝑟 ELO 2.2

This number is insignificant compared to the total 𝑄 , and therefore will not be considered. From the steam tables, saturated liquid at 1 psia: ℎ 𝑓 =69.73 𝐵𝑇𝑈 𝑙 𝑏 𝑚 ℎ 𝑓𝑔 =1,036.1 𝐵𝑇𝑈 𝑙 𝑏 𝑚 ℎ 𝑔 =1,105.8 𝐵𝑇𝑈 𝑙 𝑏 𝑚 Using 𝑄 = 𝑚 𝛥ℎ: Solving for 𝛥ℎ: 𝛥ℎ= 𝑄 𝑚 Therefore: ℎ 𝑠𝑡𝑚 − ℎ 𝑐𝑜𝑛𝑑 = 𝑄 𝑚 ELO 2.2

Solving for ℎ 𝑠𝑡𝑚 : ℎ 𝑠𝑡𝑚 = 𝑄 𝑚 + ℎ 𝑐𝑜𝑛𝑑 ℎ 𝑠𝑡𝑚 = 6.82× 10 9 𝐵𝑇𝑈 ℎ𝑟 8× 10 6 𝑙 𝑏 𝑚 ℎ𝑟 𝐵𝑇𝑈 𝑙 𝑏 𝑚 = 𝐵𝑇𝑈 𝑙 𝑏 𝑚 Using ℎ 𝑠𝑡𝑚 = ℎ 𝑓 +𝑋 ℎ 𝑓𝑔 Solving for 𝑋: 𝑋= ℎ 𝑠𝑡𝑚 − ℎ 𝑓 ℎ 𝑓𝑔 𝑋= 𝐵𝑇𝑈 𝑙 𝑏 𝑚 −69.73 𝐵𝑇𝑈 𝑙 𝑏 𝑚 1,036.1 𝐵𝑇𝑈 𝑙 𝑏 𝑚 𝑋=0.823 or 82.3% steam quality ELO 2.2

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? 9°F 11°F 13°F 15°F Correct answer is B. Correct answer is B. NRC Bank Question – P1376 Analysis: A main condenser pressure of 26 inches Hg vacuum corresponds to 2.0 psia. Saturation temperature at 2.0 psia is 126ºF. Therefore, if condensate temperature is 115ºF, then the condensate is approximately 11ºF subcooled (below saturation temperature). ELO 2.2

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? Increase condenser cooling water flow rate by 5 percent. Decrease condenser cooling water flow rate by 5 percent. Increase main condenser hotwell level by 5 percent. Decrease main condenser hotwell level by 5 percent. Correct answer is B. Correct answer is B. NRC Bank Question – P3876 Analysis: Main turbine exhaust enters the condenser at a saturation temperature of 126ºF (saturation pressure of 2 psia). If the condensate is cooled to 100ºF, this corresponds to 26ºF of subcooling (also known as condensate depression). Condensate depression decreases the overall plant efficiency because more sensible heat must be added to reach saturated conditions in the steam generators. However, the advantage of having a small degree of condensate depression is to ensure adequate net positive suction head (NPSH) for the main condensate pumps and prevent cavitation. With 26ºF of condensate depression, lowering condenser cooling water will result in less heat transfer, and therefore a higher condensate temperature. This results in higher plant efficiency because not as much sensible heat will need to be added to raise the feedwater in the steam generators to saturation temperature. The reason Choice “B” is also correct is because the STEM states “Assuming main condenser vacuum does not change…”. If vacuum remains the same, lowering condenser cooling water (Circ Water) flow will result in less condensate depression, which improves thermodynamic cycle efficiency. ELO 2.2

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. Functions of nozzles, blading, and multi-stage pressure drops are all designed to maximize the efficiency of the turbine This section provides a brief overview of turbine design Related KAs K1.05 Explain the function of nozzles fixed blading and moving blading in the turbine. K1.06 Explain the reason turbines are multistages. There are no bank questions on these KA’s. It is presented to better understand how turbine design can improve turbine efficiency. ELO 2.3

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 ELO 2.3

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 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 Enthalpy (pressure and temperature) drops Kinetic energy increases ELO 2.3

Impulse principle Thermal energy of steam converted into mechanical energy in essentially four steps: Steam passes through stationary nozzles. Stationary nozzles convert some of the steam’s thermal energy (indicated by its pressure and temperature) into kinetic energy (velocity). The nozzles direct the steam flow into blades mounted on the turbine wheel. 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. ELO 2.3

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 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 ELO 2.3

Turbine Efficiency Main concepts of turbine design are to maximize efficiency: Not one large pressure drop (use multistages) Impulse and/or Reaction stages Reaction blading – increase velocity imparted on blading Sealing steam – minimize steam losses (more steam to impart on blades) Blade design – smooth blades, less friction Minimize moisture – damage to blading ELO 2.3

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? Efficiency will increase because condensate depression will decrease. Efficiency will increase because the work output of the main turbine will increase. Efficiency will decrease because condensate depression will increase. Efficiency will decrease because the work output of the main turbine will decrease. Correct answer is D. Correct answer is D NRC Bank Question - P2476 Analysis: Note that the original main condenser conditions are at saturation: Saturation pressure for 120ºF is 1.69 psia. No initial condensate depression exists. Therefore, any reduction in main condenser cooling water flow rate will result in less heat transfer in the main condenser, which reduces vacuum. Lower condenser vacuum will result in less work output of the main turbine, and lower plant efficiency. Summary

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 1.3 K1.03 Describe the functions of nozzles in flow restrictors. 2.1 K1.04 Describe the functions of nozzles in air ejectors. 2.0 K1.05 Explain the function of nozzles fixed blading and moving blading in the turbine. 1.6 2.3 K1.06 Explain the reason turbines are multistages. 1.5 K1.07 Define turbine efficiency. 1.2 K1.08 Explain the difference between real and ideal turbine efficiency. K1.09 Define pump efficiency. K1.10 Explain the difference between ideal and real pumping processes. 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. K1.13 Explain the condensing process. K1.14 Explain the reduction of process pressure from throttling. K1.15 Determine the exit conditions for a throttling process based on the use of steam and/or water 2.8 K1.01 and K1.10 are discussed in further detail in Cycles

Operator Generic Fundamentals Thermodynamic Processes

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