1. Operator Generic Fundamentals 193005 – Thermodynamic Cycles
K1.01 Define thermodynamic cycle.
K1.02 Define thermodynamic cycle efficiency in terms of net work produced and energy applied.
K1.03 Describe how changes in secondary system parameter affect thermodynamic efficiency.
K1.04 Describe the moisture effects on turbine integrity and efficiency.
K1.05 State the advantages of moisture separators/repeaters and feedwater heaters for a typical steam cycle.
Operator Generic Fundamentals – Thermodynamic Cycles

2. Terminal Learning Objectives
At the completion of this training session, the trainee will demonstrate mastery of this topic by passing a written exam with a grade of ≥ 80 percent on the following TLO:
Apply the second law of thermodynamics to analyze real and ideal systems and components.
TLO 1 is not tested by the NRC, but the concepts help better under stand pressurizer concepts as well as cycle efficiency.
TLOs

3. Second Law of Thermodynamics
TLO 1 – Apply the second law of thermodynamics to analyze real and ideal systems and components.
1.1 Explain the Second Law of Thermodynamics using the term entropy.
1.2 Given a thermodynamic system, determine the:
Maximum efficiency of the system
Efficiency of the components within the system
1.3 Differentiate between the path for an ideal process and that for a real process on a T-s or h-s diagram.
1.4 Describe how individual factors affect system or component efficiency.
TLO

4. Second Law of Thermodynamics
ELO Explain the Second Law of Thermodynamics using the term entropy.
It is impossible to construct a device that operates in a cycle and produces no effect other than the removal of heat from a body at one temperature and the absorption of an equal quantity of heat by a body at a higher temperature.
Related KA K1.01 Explain the relationship between real and ideal processes. 1.8* 1.9*
Figure: Second Law of Thermodynamics for a Heat Engine
ELO 1.1

5. Second Law of Thermodynamics
Concept of second law is best stated using Kelvin & Planck’s description:
It is impossible to construct an engine that will work in a complete cycle and produce no other effect except the raising of a weight and the cooling of a heat reservoir
Summary of First and Second Laws:
“Not only can’t you get something from nothing, you can’t break even”.
The First Law represents the energy conversion that take place place in our processes.
The Second Law represents the “change in entropy”. To make the cycle more efficient, try and minimize the change in entropy in each of the four processes.
Figure: Kelvin-Planck’s Second Law of Thermodynamics
ELO 1.1

6. Second Law of Thermodynamics
The Second Law of Thermodynamics is needed because the First Law of Thermodynamics does not completely define the energy conversion process
The first law is used to relate and evaluate various energies involved in a process
No information about direction of process can be obtained by application of the first law
Figure: Heat Flow Direction
ELO 1.1

7. Second Law of Thermodynamics
Areas of application of second law is study of energy-conversion systems
The second law can be used to derive an expression for the maximum possible energy conversion efficiency taking those losses into account
The second law denies the possibility of completely converting into work all of the heat supplied to a system operating in a cycle
ELO 1.1

8. Second Law of Thermodynamics
The Second Law is used to determine the maximum efficiency of any process
A comparison can then be made between the maximum possible efficiency and the actual efficiency obtained
Figure: Rankine Cycle for a Typical Steam Plant
ELO 1.1

9. Entropy Entropy helps explain the Second Law of Thermodynamics
A measure of the unavailability of heat to perform work in a cycle
Change in entropy determines the direction in which a given process proceeds
The second law predicts that not all heat provided to a cycle can be transformed into an equal amount of work. Some heat rejection must take place
In a heat transfer process, minimizing the temperature at which the process occurs will reduce the change in entropy.
For example, the greater the heat transfer surface area, the smaller the Delta-T. The better the overall heat transfer coefficient (U), the lower the Delta-T. (Q-dot = UA Delta-T)
Figure: Entropy
ELO 1.1

10. Entropy
Change in entropy is the ratio of heat transferred during a reversible process to the absolute temperature of the system
∆𝑆= ∆𝑄 𝑇 𝑎𝑏𝑠 (𝑓𝑜𝑟 𝑎 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠)
∆𝑆 = the change in entropy of a system during some process (𝐵𝑇𝑈/°𝑅)
∆𝑄 = the amount of heat added to the system during the process (BTU)
𝑇𝑎𝑏𝑠 = the absolute temperature at which the heat was transferred (°𝑅)
The second law can also be expressed as ΔS ≥ 0 for a closed cycle.
Entropy must increase or stay the same
It can never decrease
ELO 1.1

11. Entropy
Entropy is a extensive property of a system and may be calculated from specific entropies (S = ms)
Specific entropy (s) values are tabulated specific enthalpy and specific volume in the steam tables
Specific entropy is a property and is used as one of the coordinates when representing a reversible process graphically
Area under a reversible process curve on T-s diagram represents the quantity of heat transferred during the process
Figure: T-s Diagram With Rankine Cycle
ELO 1.1

12. Entropy
Note the “real” process on the T-s diagram results in more area under the curve
Thermodynamic problems, processes, and cycles are often investigated by substituting reversible processes for the actual irreversible process
Helpful because only reversible processes can be depicted on the diagrams
Actual or irreversible processes cannot be drawn since they are not a succession of equilibrium conditions
Only the initial and final conditions of irreversible processes are known
Figure: T-s and h-s Diagrams for Expansion and Compression Processes
ELO 1.1

13. Carnot’s Principle
ELO Given a thermodynamic system, determine the: Maximum efficiency of the system, Efficiency of the components within the system.
Related KAs K1.02 Define thermodynamic cycle efficiency in terms of net work produced and energy applied. 1.6* 1.8*; K1.08 Explain the difference between real and ideal turbine efficiency. 1.6* 1.7*
Figure: Carnot Cycle Representation
ELO 1.2

14. Carnot’s Principle
No engine can be more efficient than a reversible engine operating between the same high temperature and low temperature reservoirs
Efficiencies of all reversible engines operating between the same constant temperature reservoirs are the same
Efficiency of a reversible engine depends only on the temperatures of the heat source and heat receiver
Figure: Carnot’s Efficiency Principle
ELO 1.2

15. Carnot Cycle Reversible isothermal expansion (1-2: TH = constant)
Reversible adiabatic expansion (2-3: Q = 0, TH→TL)
Reversible isothermal compression (3-4: TL = constant)
Reversible adiabatic compression (4-1: Q = 0, TL→TH)
Note that the “net effect” in the change in specific entropy shown in the T-s diagram for this cycle equals ZERO.
Figure: Single Piston Carnot Engine Cycle
ELO 1.2

16. Carnot Cycle Heat added (QH) is the area under line 2-3
Heat rejected (QC) is the area under line 1-4
Difference between heat added and heat rejected is the net work (sum of all work processes), which is represented as the area of rectangle
The cycle consists of the following reversible processes.
1-2: adiabatic compression from Tcold to Thot due to work performed on fluid.
2-3: isothermal expansion as fluid expands when heat is added to the fluid at temperature TH.
3-4: adiabatic expansion as the fluid performs work during the expansion process and temperature drops from TH to TC.
4-1: isothermal compression as the fluid contracts when heat is removed from the fluid at temperature TC.
Figure: Carnot Cycle Representation
ELO 1.2

17. Carnot Cycle Shows max possible efficiency exists when either
TH is at its highest possible value
TC is at its lowest possible value
Carnot efficiency
Represent reversible processes
Upper limit of efficiency for any given system operating between the same two temperatures
However, all practical/real systems and processes are irreversible and the actual system will not reach this efficiency value
ELO 1.2

18. Carnot Cycle
Efficiency (η) of the cycle is the ratio of the net work of the cycle to the heat input to the cycle.
Expressed by equation:
𝜂= 𝑄 𝐻 − 𝑄 𝐶 𝑄 𝐻 = 𝑇 𝐻 − 𝑇 𝐶 𝑇 𝐻
=1− 𝑇 𝐶 𝑇 𝐻
Where:
𝜂 = cycle efficiency
𝑇 𝐶 = designates the low-temperature reservoir (˚R)
𝑇 𝐻 = designates the high-temperature reservoir (˚R)
°R must be used in calculation
ELO 1.2

19. Carnot Cycle
The most important aspect of the second law is the determination of maximum possible efficiencies obtained from a power system
Actual efficiencies will always be less than this maximum
System losses (friction, windage, turbulence, etc.)
Systems are not reversible
ELO 1.2

20. Carnot Cycle Example: Actual Versus Ideal Efficiency
The actual efficiency of a steam cycle is 18.0%. The facility operates from a steam source at 340 °F and rejects heat to atmosphere at 60 °F. Compare the Carnot efficiency to the actual efficiency.
35% as compared to 18.0% actual efficiency
ELO 1.2

21. Carnot Cycle Example: Actual Versus Ideal Efficiency
Figure: Real Process Cycle Compared to Carnot Cycle
ELO 1.2

22. Carnot Cycle
The Second Law of Thermodynamics gives an upper limit for how efficiently a thermodynamic system can perform
Determining that efficiency:
Know the inlet and exit temperatures of the overall system
Apply Carnot’s efficiency equation using those temperatures in absolute degrees
ELO 1.2

23. Carnot Cycle Knowledge Check
The theoretical maximum efficiency of a steam cycle is given by the equation:
Effmax = (1 - Tout/Tin) x 100%
where Tout is the absolute temperature for heat rejection and Tin is the absolute temperature for heat addition. (Fahrenheit temperature is converted to absolute temperature by adding 460°F.)
A nuclear power plant is operating with a stable steam generator pressure of 900 psia. What is the approximate theoretical maximum steam cycle efficiency this plant can achieve by establishing its main condenser vacuum at 1.0 psia?
35 percent
43 percent
65 percent
81 percent
Correct answer is B.
Correct answer is B
NRC Bank Question – P2778
Analysis:
To calculate maximum theoretical thermal efficiency, one must determine the two temperatures, convert to degrees Rankine, and solve the given equation:
At 900 psia, saturation temperature is 532ºF. Adding 460º, this equals 992 R. This is Tin; the steam generators are the input to the system.
At 1 psia, saturation temperature is 102ºF. Adding 460º, this equals 562 R. This is Tout.
Effthmax = (1 - Tout/Tin) X 100% = (1 – 562 / 992) x 100% = 43%
ELO 1.2

24. Thermodynamics of Ideal and Real Processes
ELO Differentiate between the paths for an ideal process and for a real process on a T-s or h-s diagram
Related KA K1.01 Explain the relationship between real and ideal processes. 1.8* 1.9*; K1.02 Explain the shape of the T-s diagram process line for a typical secondary system. 1.7* 1.9;
Figure: T-s Diagram Shows Paths of Processes
ELO 1.3

25. Diagrams of Ideal and Real Processes
Any ideal thermodynamic process can be drawn on a T-s or an h-s diagram
isentropic
A real process that approximates the ideal process can be shown
Usually with dashed lines
Figure: T-s Diagram Shows Paths of Processes
ELO 1.3

26. Diagrams of Ideal and Real Processes
In an ideal process, the entropy will be constant
Turbine process is an “expansion” process
Real expansion or compression process slants slightly toward the right
Entropy increases
Smaller change in enthalpy
Real processes are irreversible
These isentropic processes will be represented by vertical lines on either T-s or h-s diagrams, since entropy is on the horizontal axis and its value does not change
Real expansion or compression process slants slightly toward the right. Entropy increases from the start to the end of the process
Real expansion process (turbine) results in a smaller change in enthalpy – less work is done by the real turbine
Real compression process (pump) - more enthalpy must be added during the pump process - more work done on the system
Reversible process indicate a maximum work output for a given input, which can be compared to real processes for efficiency purposes
Figure: h-s Diagram Shows Expansion and Compression Paths
ELO 1.3

27. Thermodynamic Cycle Efficiency
ELO Describe how individual factors affect system or component efficiencies
Related KAs K1.03 Describe how changes in secondary system parameter affect thermodynamic efficiency ; K1.04 Describe the moisture effects on turbine integrity and efficiency ; K1.05 State the advantages of moisture separators/repeaters and feedwater heaters for a typical steam cycle
NOTE: There are NO bank questions relating to pumps in this chapter and pump operation has no direct effect on cycle efficiency.
All of the NRC exam bank questions on this chapter are tied to this KA ( K1.03 Describe how changes in secondary system parameter affect thermodynamic efficiency )
Figure: Typical Steam Cycle
ELO 1.4

28. Real vs. Ideal Cycle Efficiency
Efficiency of a Carnot cycle solely depends on the temperature of the heat source and the heat sink
To improve efficiency increase the temperature of the heat source and decrease the temperature of the heat sink
For a real cycle the heat sink is limited by the fact that the earth is our final heat sink, and therefore, is fixed at about 60 °F (520 °R)
Most plants notice an increase in plant efficiency when the heat sink temperature drops in the winter
ELO 1.4

29. Real vs. Ideal Cycle Efficiency
Heat source is limited to the combustion temperatures of the fuel to be burned
Fossil fuel cycles upper limit is ~ 3,040 °F (3,500 °𝑅)
Not attainable due to the metallurgical restraints of the SGs,
Limited to about 1,500 °F (1,960 °𝑅) for a maximum heat source temperature
Using these limits to calculate the maximum efficiency attainable by an ideal Carnot cycle gives the following:
𝜂= 𝑇 𝑆𝑂𝑈𝑅𝐶𝐸 − 𝑇 𝑆𝐼𝑁𝐾 𝑇 𝑆𝑂𝑈𝑅𝐶𝐸 = 1,960 °𝑅 −520 °𝑅 1,960 °𝑅 =73.5%
Our actual steam temperature (heat source) is based on saturation pressure that the SG is at. For a pressure of 1000 psia, Tsat is about 545oF, which equates to 1005oR.
ELO 1.4

30. Real vs. Ideal Cycle Efficiency
Heat Rejection
73% efficiency is not possible
Energy added to a working fluid during the Carnot isothermal expansion is given by qs
Not all of this energy is available for use by the heat engine since a portion of it (qr) must be rejected to the environment
ELO 1.4

31. Real vs. Ideal Cycle Efficiency
Operating less than 1,962 °R is less efficient
The change in entropy corresponds to a measure of the heat rejected by the engine
Recall: Entropy is a measure of the energy unavailable to do work
Developing materials capable of withstanding the stresses above 1,962 °R increases the energy available for use by the plant cycle
Figure: Entropy Measures Temperature as Pressure and Volume Increase
ELO 1.4

32. Real vs. Ideal Cycle Efficiency
Actual available energy (area )
Less than half of the ideal Carnot cycle (area 1-2'-4) operating between the same two temperatures
Fossil plants are about 40% efficient
Nuclear plants are around 33%
Fossil fuel power cycle – 2,000 psia maximum, heat addition at temperature of 1,962 °𝑅 is not possible
Water requires heat addition in a constant pressure process
Average temperature for heat addition is below the maximum 1,962 °𝑅
Figure: Carnot Cycle Versus Typical Power Cycle Available Energy
ELO 1.4

33. Real vs. Ideal Cycle Efficiency
Carnot steam cycle on a T-s diagram
A great deal of pump work is required to compress a two phase mixture of water and steam from point 1 to the saturated liquid state at point 2
Pump cavitation would exist
Condenser designed to produce a two-phase mixture at the outlet
Would pose technical problems
Figure: Ideal Carnot Cycle
ELO 1.4

34. Ideal Rankine Cycle QA – red + green area QR – green area
WT – red area
The greater the QA and the less the QR
The greater the WT
Operational analogy:
The greater the steam pressure, and,
The lower the backpressure
The more work of the turbine
Figure: Rankine Cycle
An example shown on the next slide.
ELO 1.4

35. Figure: Rankine Cycle Efficiency Comparisons on a T-s Diagram
Ideal Rankine Cycle
Rankine Cycle Efficiencies
Cycle A has less heat rejected than Cycle B
Better vacuum, for example
Both have same amount of heat added
Cycle A
Same heat added
Less heat rejected
More work of turbine
More efficient
Figure: Rankine Cycle Efficiency Comparisons on a T-s Diagram
ELO 1.4

36. Real vs Ideal Rankine Cycle
The ideal turbine replaced with a real turbine
The efficiency is reduced
The non-ideal turbine incurs an increase in entropy
The increase in the area of 3-2-3‘ < the increase in area of a-3-3'-b
Also, condensate slightly subcooled
More sensible heat added to FW
Green line
For simplification this diagram only showed the difference between IDEAL and REAL for the turbine process. The IDEAL vs REAL Pump process has been shown in an earlier slide. The added green line shows the REAL pump process (slight change in specific entropy). Even though it raises the specific enthalpy of the fluid (higher temperature of fluid), it still requires a larger pump to get the required pressure increase to make up for the headloss.
Figure: Rankine Cycle With Real Versus Ideal Turbine
ELO 1.4

37. Rankine Cycle – Steam Plant
Processes that comprise the steam cycle:
1-2: Heat added in the SG under a constant pressure
2-3: Saturated steam is expanded in high pressure (HP) turbine to provide shaft work
3-4: HP exhaust is dried and superheated in the MSR
Figure: Typical Steam Cycle
ELO 1.4

38. Rankine Cycle – Steam Plant
4-5: Superheated steam in LP turbine provides shaft work
5-6: Exhaust from the turbine is condensed under a constant vacuum
6-7: Condensate is compressed as a liquid by the condensate pump
Figure: Typical Steam Cycle
ELO 1.4

39. Rankine Cycle – Steam Plant
7-8: Condensate is preheated by the low pressure feedwater heaters
8-9: Condensate is compressed as a liquid by the feedwater pump
9-1: Feedwater is preheated by the high- pressure heaters
1-2: Cycle starts again
Figure: Typical Steam Cycle
ELO 1.4

40. Ideal Rankine Cycle – Steam Plant
Differences between IDEAL and REAL:
Ideal pumps and turbines do not exhibit an increase in entropy; real pumps and turbines will
No condensate subcooling as point 6 is on the saturation line; condensate actually slightly subcooled (condensate depression)
Figure: Typical Steam Cycle
Figure: Rankine Steam Cycle (Ideal)
ELO 1.4

41. Rankine Cycle Steam Plant – Real T-s
Additional heat rejected in condenser must then be made up for in the SG
Real pumps and turbines would exhibit an entropy increase across them
Subcooling decreases cycle efficiency, but aids in preventing condensate pump cavitation
Keep in mind that from a cycle efficiency standpoint, subcooling is BAD. For a given amount of heat available from the RCS to be added to the secondary, the lower the temperature of the feedwater entering the SG, the lower the temperature (pressure) exiting the SG. A lower steam pressure means less change in specific enthalpy of the turbine. This can be shown on a Mollier Diagram.
Figure: Steam Cycle (Real)
ELO 1.4

42. Component Effects on Cycle Efficiency
All of the NRC exam bank questions on this chapter are tied to this portion of ELO-1.4.
Each component discussed on subsequent slides
Some efficiency impacts are by design, some controllable
ELO 1.4

43. Steam Generator Efficiency
Moisture separators remove moisture content to minimize turbine blade friction
Higher the steam pressure, the higher the steam cycle efficiency
For a given amount of RCS heat, this is a function of the feedwater temperature entering the SG
Higher steam pressure will mean lower steam enthalpy, however, still results in greater change in enthalpy (work of turbine)
ELO 1.4

44. Turbine Effect on Cycle Efficiency
Turbine efficiency (ηt) is the ratio of the actual work done by the turbine Wt.actual to the work that would be done by the turbine if it were an ideal turbine Wt.ideal
Most turbines designed/rated at approximately 90% efficiency
Efficiency improved by minimizing friction
Steam friction – by removing moisture
Mechanical friction – use of oil lift pumps
ELO 1.4

45. MSR Effect on Cycle Efficiency
Removes moisture prior to LP turbine and raises enthalpy by superheating steam
Points 2 to 3
Results in greater change in enthalpy in turbine
Does take some mass flow rate from the SG prior to entering HP turbine
However, net effect of using MRS is an INCREASE in cycle efficiency
NOTE: Efficiency can increase or decrease with MSRs, but the main reason for using MSRs is to minimize impingement in the final stages of the LP turbine blading. Generically, the NRC looks at the use of the MSR as an “increase” in efficiency.
Figure: Mollier Diagram
ELO 1.4

46. Condenser Effect on Cycle Efficiency
Removal of non-condensable gasses increases cycle efficiency
Greater vacuum (lower backpressure)
Less Subcooling (for a given vacuum) increases cycle efficiency
Less sensible heat required to be added in SG
ELO 1.4

47. Circ Water Effect on Cycle Efficiency
Lower the circ water temperature, INCREASES the cycle efficiency
Assuming the vacuum improves
Raising the circ water flow rate, INCREASES the cycle efficiency
If the above effects only increase condensate depression/subcooling (same vacuum)
Cycle efficiency would DECREASE
ELO 1.4

48. Pump Effect on Cycle Efficiency
Pump efficiency is the ratio of Fluid Horsepower/Brake Horsepower
Mechanical to electrical work
Pumps have no direct effect on cycle efficiency
Pumps do require subcooled liquid to pump
More subcooled the hotwell condensate, the farther away from pump cavitation
However, the less efficient the cycle
No NRC bank questions on pump effects on cycle efficiencies, just their requirement for adequate net positive suction head (NPSH).
ELO 1.4

49. FW Heating Effect on Cycle Efficiency
LP and HP Feedwater Heaters raise the temperature of the feedwater entering the SG
Higher the temperature, the less sensible heat added, the higher the cycle efficiency
Heating is added in “stages” to maximize the efficiency of the heat added
Minimal DT across each FW heater stage
Negative effect is it lowers the mass flow rate going through the turbines
However, net effect of FW heating is it INCREASES cycle efficiency
No NRC bank questions on pump effects on cycle efficiencies, just their requirement for adequate net positive suction head (NPSH).
ELO 1.4

50. Improved Cycle Efficiency
Condition
Effect
Discussion
Superheating
More Efficient With More Superheating
Increased heat added results in more net work from the system, even though more heat is rejected.
Moisture Separator Reheater (MSR)
Use of MSR Has Minor Effect On Efficiency
More work is done by the low-pressure (LP) turbine since inlet enthalpy is higher but more heat is rejected.
The principle benefit of MSR use is protection of the final blading stages in LP turbine from erosion by water droplet impingement.
Feedwater Preheating
More Efficient With Feedwater Preheating
Less heat must be added from the heat source (reactor) since the feedwater enters the steam generator closer to saturation temperature.
Condenser Vacuum
More Efficient With Higher Vacuum (Lower Backpressure)
Net work output is higher and heat rejection is lower as condenser pressure is lowered.
Keep in mind that some of these “conditions” come at a cost, but generically they ALL result in an increase in cycle efficiency. For example:
The MSR’s raise the enthalpy of steam entering the LP turbine which means more work of the turbine. However, steam going to the MSR from the SG does not go to the HP turbine (less mass flow rate)
Feedwater heating raises the temperature of the “pump” process meaning less sensible heat must be added in the SG. However, since this uses extraction steam, there is less steam mass flow rate going through the LP and HP turbines.
ELO 1.4

51. Improved Cycle Efficiency
Condition
Effect
Discussion
Condensate Depression
More Efficient With Minimal Condensate Depression
Minimal condensate depression reduces both the amount of heat rejected and the amount of heat that must be supplied to the cycle.
Steam Temperature/ Pressure
More Efficient At Higher Steam Temperature/ Pressure
At higher steam temperature, the inlet and exit entropy from the turbine are lower so less heat is rejected.
Steam density increases as pressure increases, so more turbine work is done.
Steam Quality
More Efficient At Higher Steam Quality
Enthalpy content increases as moisture content decreases and more net work is done.
ELO 1.4

52. Thermodynamic Power Plant Efficiency
Knowledge Check
To achieve maximum overall nuclear power plant thermal efficiency, feedwater should enter the steam generator (SG)__________ and the pressure difference between the SG and the condenser should be as __________ as possible.
close to saturation; great
close to saturation; small
as subcooled as practical; great
as subcooled as practical; small
Correct answer is A.
Correct answer is A.
NRC Bank Question – P478
Analysis:
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. Excessive condensate depression causes more air absorbed in condensate and accelerates oxygen corrosion of secondary plant materials.
Therefore, to maximize plant efficiency, feedwater should enter the steam generator as close to saturation as possible. If the water were highly subcooled, more sensible heat must be added to reach saturated conditions in the steam generators.
Having a large pressure difference between the steam generator and condenser enhances thermal efficiency. For a given Mollier diagram, note the increased turbine work (due to larger change in enthalpy) for the condenser at a 28” Hg vacuum (1 psia absolute) compared to the condenser at atmospheric pressure (14.7 psia).
Also note, that as you continue to increase steam pressure (after the hump around 500 psia), the specific enthalpy decreases slightly. However, based on the slope of the backpressure line the change in specific enthalpy increases as SG pressure increases.
ELO 1.4

53. Thermodynamic Power Plant Efficiency
Knowledge Check
A nuclear power plant was initially operating at steady-state 90 percent reactor power when extraction steam to the feedwater heaters was isolated. With extraction steam still isolated, reactor power was returned to 90 percent and the plant was stabilized.
Compared to the initial main generator MW output, the current main generator MW output is...
lower, because the steam cycle is less efficient.
higher, because the steam cycle is less efficient.
lower, because more steam heat energy is available to the main turbine.
higher, because more steam heat energy is available to the main turbine.
Correct answer is A.
Correct answer is A.
NRC Bank Question – P3378
Analysis:
Isolating steam to a high pressure feedwater heater will lower plant efficiency; the feedwater will now
enter the steam generator at a lower temperature (further from saturated conditions). More sensible heat must
be added to raise the feedwater to saturated conditions.
Recall that efficiency of the plant is equal to the work of the turbine divided by the heat added from the RCS (Nplant = Mwe/Mwth).
Since losing feedwater heating is a loss of cycle efficiency, if reactor power is returned back to original value, Turbine power must be lower.
ELO 1.4

54. NRC KA to ELO Tie KA # KA Statement RO SRO ELO K1.01
Define thermodynamic cycle.
1.6
1.7
2.1
K1.02
Define thermodynamic cycle efficiency in terms of net work produced and energy applied.
1.8
2.2, 2.3
K1.03
Describe how changes in secondary system parameter affect thermodynamic efficiency.
2.5
2.6
2.4
K1.04
Describe the moisture effects on turbine integrity and efficiency.
2.3
K1.05
State the advantages of moisture separators/repeaters and feedwater heaters for a typical steam cycle.
1.9