Operator Generic Fundamentals Thermodynamic Units and Properties

Operator Generic Fundamentals Thermodynamic Units and Properties
K1.01 Convert between absolute and gauge pressure and vacuum scales. K1.02 Recognize the difference between absolute and relative (Kelvin) temperature scales. K1.03 Describe how pressure and level sensing instruments work. (covered in Sensors and Detectors) K1.04 Explain relationships between work, power, and energy. K1.05 Explain the law of conservation of energy. Operator Generic Fundamentals Thermodynamic Units and Properties

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% score on the following topics (TLOs): Describe thermodynamic properties and methods of measuring intensive and extensive properties. Explain the concepts of heat, work, and energy. This chapter includes some material previously taught in Basic Energy Concepts. Also information relating to DP level detectors is covered in – Sensors and Detectors. Depending on the order you teach GFES at your site, bank questions on DP level detectors might not be fair game yet. TLOs

Thermodynamic Properties
TLO 1 – Describe thermodynamic properties and methods of measuring intensive and extensive properties. 1.1 Define the following properties: specific volume, density, mass, weight, intensive, and extensive. 1.2 Define the thermodynamic properties of temperature and convert between the Fahrenheit, Celsius, Kelvin, and Rankine scales. 1.3 Define the thermodynamic properties of pressure and convert between pressure scales. The majority of the bank questions are related to ELO 1.3, based on different plants using variations of PSIA, inches of mercury absolute, inches mercury vacuum, etc. TLO 1

Properties and Definitions
ELO 1.1 – Define the following properties: specific volume, density, mass, weight, intensive, and extensive. Operators must be able to convert between units of measurement to ensure plant operating within established limits RCS leak rates, pump surveillances, etc. Some conversions provided at bottom of NRC Equation Sheet Operators must be able to convert between units of measurement to ensure plant operating within established limits Instrument readings may provide information in units different from those provided by a procedure For example, GPM versus mass flow rate (lbm/hr), or BTU/hr versus MW In this case, operator will be required to perform unit conversion ELO 1.1

American Engineering System Length Mass Time Inch Ounce Second* Foot* Pound* Minute Yard Ton Hour Mile Day NOTE: *Denotes standard unit of measure ELO 1.1

International System (SI) – MKS Units Length Mass Time Millimeter Milligram Second* Meter* Gram Minute Kilometer Kilogram* Hour Day NOTE: *Denotes standard unit of measure There are very little (if any) conversions between American/International standards required by the NRC. ELO 1.1

International System (SI) – CGS Units Length Mass Time Centimeter* Milligram Second* Meter Gram* Minute Kilometer Kilogram Hour Day NOTE: *Denotes standard unit of measure ELO 1.1

Prefix Symbol Power of 10 Example pico p 10-12 1 picosecond (ps) = seconds nano n 10-9 1 nanosecond (ns) = 10-9 seconds micro m 10-6 1 microsecond (ms) = 10-6 seconds milli 10-3 1 millimeter (mm) = 10-3 meters centi c 10-2 1 centimeter (cm) = 10-2 meters deci d 10-1 1 decigram (dg) = 10-1 grams hecto h 102 1 hectometer (hm) = 102 meters kilo k 103 1 kilogram (kg) = 103 grams mega M 106 1 megawatt (MW) = 106 watts giga G 109 1 gigawatt (GW) = 109 watts Some of these conversions might have already been done based on which GFE section you started with. For example, if Rx Theory you have already converted between PCM and Delta-k/K. PCM stands for Percent Milli Rho. Percent is 10-2, Milli is So, to convert between PCM and Delta-K/K you multiply by 10-5. Figure: Metric System Prefixes ELO 1.1

Typical Conversion Table
Unit English Units of Measurement Meter-Kilogram-Second (MKS) Units of Measurement Length 1 yard (yd) 12 inches (in.) 5,280 feet (ft) 1 (meter) m 1 in. = meter (m) = 1 ft = 1 mi = ft = m Time 60 seconds (sec) 3,600 sec = 1 minute (min) = 1 hour (hr) Mass 1 pound mass (lbm) 2.205 lbm 1 kilogram (kg) kg = 1 kg = 1,000 grams (g) Area 1 square foot (ft2) ft2 1 square yard (yd2) 1 square mile (mi2) = 144 in.2 = 1 square meter (m2) = 9 ft2 3.098 x 106 yd2 Volume 7.48 gallon (gal) 1 gal 1 liter (l) = 1 cubic foot (ft3) = l (liter) = 1,000 cubic centimeters (cm3) One standard conversion done in GFE is converting between Heat Transfer Rate (BTU/hr) and Megawatts (Mw). At the bottom of the NRC Equation Sheet is a conversion factor (1 Mw = 3.41 x 106 BTU/HR). Another one is converting between gallons per minute (GPM) and mass flow rate (lbm/hr). It requires multiplying the volumetric flow rate (gal/min) times the Density (lbm/ft3) and using the conversion factor (on Equation Sheet) of 1ft3 = 7.48 gal, as well as 60 min = 1 hr. ELO 1.1

Steps for Converting Units: Identify units given and units required Select the equivalence relationship Arrange the ration in the appropriate manner such that (# desired /current) = 1 Multiply quantity by ratio Multiple conversion factors may be required Practice conversions using “railroad track” technique to ensure proper answers! ELO 1.1

Practice Unit Conversion
Convert 795 meters to feet: Step 1: Identify units given and units required (meters to feet) Step 2: Select equivalence relationship from conversion table: 1 𝑚𝑒𝑡𝑒𝑟(𝑔𝑖𝑣𝑒𝑛 𝑢𝑛𝑖𝑡𝑠)=3.281 𝑓𝑡(𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑢𝑛𝑖𝑡𝑠) Step 3: Arrange equivalence ratio in appropriate manner: ( 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑢𝑛𝑖𝑡𝑠 𝑔𝑖𝑣𝑒𝑛 𝑢𝑛𝑖𝑡𝑠 ) 1= 𝑓𝑡 1 𝑚 Step 4: Multiply the quantity by the ratio: 795 𝑚 𝑓𝑡 1 𝑚 = 795 𝑚 𝑓𝑡 1 𝑚 =795×3.281 𝑓𝑡 =2608 𝑓𝑡 ELO 1.1

Practice Unit Conversion
Reactor core thermal power is 1,800 MWth. Convert to BTU/hr. Units given are megawatts and units desired are BTU/hr 1 Mw = 3.41 x 106 BTU/hr 1,800 MW 3.41 x 106 1 MW 1,800 MW converts to x 109 BTU/hr ELO 1.1

Properties and Definitions - Mass & Weight
Mass (m) - measure of amount of material present in that body Weight (wt) - force exerted by that body when its mass is accelerated in a gravitational field Mass and weight related gc has same numerical value as acceleration of gravity 𝑤𝑡= 𝑚𝑔 𝑔 𝑐 Where: wt = weight (lbf) m = mass (lbm) g = acceleration due to gravity = ft/s gc = gravitational constant = lbm-ft/lbf-s2 This concept is used when converting back and forth between pressure and feet of water. Basically on Earth, 1 lbm = 1 lbf. ELO 1.1

Intensive vs. Extensive Properties
independent of mass and does not depend on how much of substance is present Temperature, pressure Extensive depends on mass (or how much of substance is present) Volume, weight For example Volume (V) is extensive Specific Volume (v) is intensive Volume/Mass – units are ft3/lbm If a quantity of matter in a given state is divided into two equal parts, each part will have same value of intensive property as original and half the value of extensive property ELO 1.1

Properties and Definitions – Specific Volume
Volume - amount of space a particular substance occupies Specific volume is total volume (V) of that substance divided by total mass (m) of that substance Specific volume values provided in Steam Table 𝑣= 𝑉 𝑚 Where: v = specific volume (ft3/lbm) V = volume (ft3) m = mass (lbm) ELO 1.1

Properties and Definitions - Density
Density (ρ) is total mass (m) of that substance divided by total volume (V) occupied by that substance Describes how much stuff is packed into specific volume Units of pound-mass per cubic feet (lbm/ft3) Density of a substance is reciprocal of its specific volume (ν) 𝜌= 𝑚 𝑉 = 1 𝑣 Where: ρ = density (lbm/ft3) m = mass (lbm) V = volume (ft3) v = specific volume (ft3/lbm) ELO 1.1

Properties and Definitions - Density
Density can be changed by changing pressure or temperature Increasing pressure increases density of a material Increasing temperature decreases density Change in density greater at higher temperatures Pressure effect greater on steam Liquids essentially incompressible Figure: Water Density Change versus Moderator Temperature ELO 1.1

Properties of Temperature
ELO 1.2 – Define the thermodynamic properties of temperature and convert between the Fahrenheit, Celsius, Kelvin, and Rankine scales. Related KA K Recognize the difference between absolute and relative (Kelvin) temperature scales.1.9, 2.0 Even though conversion factors between Fahrenheit and Celsius are provided on the NRC Equation Sheet, there are currently NO bank questions requiring the use of these conversions. There is one bank question in – Cycles relating to Fahrenheit and Rankine, but the conversion (460oF) is provided. Figure: Comparison of Temperature Scales ELO 1.2

Temperature Temperature is a intensive measure of amount of energy stored in an object of a standard mass Measure of average molecular kinetic energy of a substance The more molecular movement, the higher the temperature of the substance will be Relative measure of how "hot" or "cold" a substance Used to predict direction of heat transfer ELO 1.2

Temperature Scales Two temperature scales normally used for measurement purposes Fahrenheit (F) Celsius (C) Based on number of increments between freezing and boiling Celsius scale has 100 units Fahrenheit scale has 180 units Since both scales “relate” temperature of a substance to a recognized condition, they are referred to as relative temperature scales A way that may help remember degrees F to C: Draw a typical Y=MX+B graph. Defining “M” as slope (rise/run) and “B” as the Y-intercept Make the horizontal line degrees C and vertical line degrees F instead of X and Y Plot 2 known points (OC,32F) and (100C,212F) The “F-intercept” is shown to be 32. Then calculate “M” as 180/100 (9/5) So degrees C + 9/5 degrees F + 32 Figure: Boiling and Freezing Points of Water for Celsius and Fahrenheit Temperature Scales ELO 1.2

Temperature Scales Zero points on the scales are arbitrary
Celsius scale zero point is freezing point of water Fahrenheit scale zero point is coldest temperature achievable with a mixture of ice and salt water Temperature at which water boils 100 on Celsius scale 212 on Fahrenheit scale Mathematical relationships ℉= °𝐶 ℃=(℉−32) 5 9 ELO 1.2

Temperature Scales Relationship between scales represented by:
°𝑅=℉+460 °𝐾=℃+273 Figure: Comparison of Temperature Scales ELO 1.2

Scales of Pressure ELO 1.3 – Define the thermodynamic properties of pressure and convert between pressure scales. K1.01 Convert between absolute and gauge pressure and vacuum scales. (majority of NRC bank questions on this concept). Plants vary on how they recognize vacuum in the condenser. Some reference it as inches of mercury (inHg) and some absolute pressure (psia). You MUST be able to convert back and forth in order to answer these NRC bank questions. Figure: Pressure Scale Relationships ELO 1.3

Scales of Pressure Force exerted per unit area on boundaries of substance (or system) Pascal’s Principle – “pressure felt undiminished throughout” Collisions of molecules of substance with boundaries of system Hit walls of their container or system pushing outward Forces resulting from these collisions cause pressure exerted by a system on its surroundings Units lbf/in2 (psi) ELO 1.3

Scales of Pressure Scales use units of inches of H2O or Hg
Height of column of liquid provides a certain pressure that can be directly converted to force per unit area 0.491 psi = 1 inch of Hg NRC bank questions use 0.5 for simplicity 0.433 psi = 1 ft of water (based on lower temps – degrees) 14.7 psia = 408 inches of water 14.7 psia = 29.9 inches of Hg NRC bank questions use 15 psi and 30.0 in Hg Hg - mercury ELO 1.3

Scales of Pressure Absolute pressure (psia)
Relative to a perfect vacuum Gauge pressure (psig) Relative to atmospheric pressure (14.7 psi) Pressure gauges register zero when open to atmosphere Pressure below atmospheric designated as vacuum Perfect vacuum corresponds to absolute zero pressure All values of absolute pressure are positive Gauge pressures: Positive if above atmospheric pressure 𝑃 𝑎𝑏𝑠 = 𝑃 𝑎𝑡𝑚 + 𝑃 𝑔𝑎𝑢𝑔𝑒 Negative if below atmospheric pressure 𝑃 𝑎𝑏𝑠 = 𝑃 𝑎𝑡𝑚 − 𝑃 𝑣𝑎𝑐 ELO 1.3

Scales of Pressure Relationships between absolute, gauge, vacuum, and atmospheric pressures 15 30.0 30.0 15 NOTE: NRC Bank questions relate atmospheric (15 psi) and vacuum (30 inHg). These values really are 14.7 psi and inHg. Practical values (15 and 30) animated on with mouse click. Figure: Pressure Scale Relationships ELO 1.3

Inches Hg vs. PSIA- Relationships
Sum of inches Hg and inches Absolute equal 30 30 in Hg vacuum = 0 in absolute 28 in Hg vacuum = 2 in absolute Sum of PSIA and PSIV equals 15 15 psia = 0 psiv 1 psia = 14 psiv 2 inches for every pound 15 psia = 30 inches absolute 14 psiv = 28 in Hg vacuum Based on above: 28 in Hg vacuum = 1 psia (this is a key relationship used extensively!) NOTE that PSIV is not used in NRC bank questions. This is just provided to show the relationship. ELO 1.3

Scales of Pressure Knowledge Check
A pressure gauge on a condenser reads 27 inches of mercury (Hg) vacuum. What is the absolute pressure corresponding to this vacuum? (Assume an atmospheric pressure of 15 psia.) 14.0 psia 13.5 psia 1.5 psia 1.0 psia Correct answer is C. Correct answer is C. NRC Bank Question – P273 Analysis: (visual on next slide) Recall the sum of inches vacuum and inches absolute equals 30. Therefore, 30 – 27 in Hg = 3 inches absolute. Since each pound equals 2 inches, then each inch equals 0.5 pounds. 3 in absolute x 0.5 psia/in absolute = 1.5 psia. ELO 1.3

Scales of Pressure Pabs = Patm – Pvacuum
Pabs = 15 psia – 27”Hg (1psia/2”Hg) Pabs = 15 psia – 13.5 psia = 1.5 psia Vacuum Absolute Absolute Figure: Gauge and Absolute Pressure Scale Relationship ELO 1.3

Scales of Pressure Pressure due to column of fluid is product of fluid’s density and height 𝑃= 𝜌𝑔𝑧 𝑔 𝑐 Where: P = pressure (lbf/ft2, lbf/in2, psi)  = density (lb/ft3) g = acceleration of gravity (32.2 ft/s2) z = height (ft, in) 𝑔 𝑐 = gravitational conversion constant (32.2 ft lbm/lbf s2) NOTE: the gc converts lbm to lbf ELO 1.3

PSI to Water Height Example
A water storage tank is vented to atmosphere. The tank is located at sea level and contains 100,000 gallons of water at 60°F. A pressure gauge at the bottom of the tank reads 9.0 psig. What is the approximate water level in the tank? z = 20.7 ft 𝑃= 𝜌𝑔𝑧 𝑔 𝑐 P4537 in NRC Exam Bank ELO 1.3

PSI to Water Height Example (cont.)
Recall the unit conversion from psi to ft of water 0.433 psi = 1 ft H2O Therefore, 1 psi = 2.3 ft H2O (1/0.433 = 2.3) Where does this come from? Unit conversion from previous equation Note: Since density vs temperature curve is relatively straight at low temps Density at standard temp/press of 62.4 can be used Only unit conversion required was 144 in2/ft2 9.0 x 2.3 = 20.7 ft; or, 9.0/0.433 = 20.7 ft P4537 in NRC Exam Bank Draw on board the Density vs Temperature curve explained in – Reactivity Coefficients to show how density is relatively constant for low temperatures. ELO 1.3

Scales of Pressure Knowledge Check – NRC Bank
Refer to the drawing of a tank with a differential pressure (D/P) level detector (see figure below). If the tank contains 30 feet of water at 60°F, what is the approximate D/P sensed by the detector? 7 psid 13 psid 20 psid 28 psid Correct answer is B. Correct answer is B. NRC Question P3673 Analysis: The volume of the tank is itself does not determine pressure due to the static column of water; the height of the tank affects pressure at the bottom of the tank. Note that atmospheric pressure acts on both sides of the D/P detector and thus can be neglected. Since the temperature is relatively low you can use the thumbrule of 1 ft water = psi. Therefore, 30 x = 13 psid ELO 1.3

Heat, Work, and Energy TLO 2 – Explain the concepts of heat, work, and energy. 2.1 State the First and Second Laws of Thermodynamics and how they relate to the conservation of energy. 2.2 Define the following thermodynamic properties: potential energy, kinetic energy, specific internal energy, specific P-V energy, specific enthalpy, and specific entropy. 2.3 Explain the relationship between work, energy, and power. 2.4 Define the following terms: heat, sensible heat, latent heat, specific heat, and super heat. TLO 2

Laws of Thermodynamics
ELO 2.1 – State the First and Second Laws of Thermodynamics and how they relate to the conservation of energy The processes of our secondary cycle are either a transfer of heat or a change in energy form K1.05 Explain the law of conservation of energy. ELO 2.1

Laws of Thermodynamics Introduction
First Law of Thermodynamics states: Energy can be neither created nor destroyed, only altered in form Several energy conversions discussed in Thermodynamics Velocity energy to pressure energy (water hammer) Flow energy to internal energy (headloss) Heat energy to mechanical energy (turbine) 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 Losses captured by term “entropy” Goal is to minimize the change in entropy Entropy and Specific Entropy are explained in the next section. ELO 2.1

Properties of Energy ELO 2.2 – Define the following thermodynamic properties: potential energy, kinetic energy, specific internal energy, specific P-V energy, specific enthalpy, and specific entropy. Understanding energy terms will help understand the General Energy equation Bernoulli’s Equation No related KAs to this ELO, however, understanding them will help answer several questions in the next few chapters. ELO 2.2

Properties of Energy Energy is capacity of a system to perform work
Forms of stored energy important in analysis of systems: Potential energy Due to height Kinetic energy Due to velocity Internal energy Due to temperature P-V (flow) energy Due to pressure ELO 2.2

Energy - Potential Energy of position
Potential energy will exist whenever an object which has mass has a position within a force field Objects in the earth's gravitational field Using English system units 𝑃𝐸= 𝑚𝑔𝑧 𝑔 𝑐 Where: PE = potential energy (ft-lbf) m = mass (lbm) z = height above some reference level (ft) g = acceleration due to gravity (ft/sec2) gc = gravitational conversion constant ft-lbm/lbf-sec2 ELO 2.2

Energy - Potential Determine the potential energy of 50 lbm of water in a storage tank 100 ft above the ground. 𝑃𝐸= 𝑚𝑔𝑧 𝑔 𝑐 𝑃𝐸 = 𝑙𝑏𝑚 𝑓𝑡 𝑠 𝑓𝑡 𝑓𝑡 _ 𝑙𝑏𝑚 𝑙𝑏𝑓 _ 𝑠 2 𝑃𝐸 = 5.00 × 𝑓𝑡 _ 𝑙𝑏𝑓 NOTE: Specific potential energy (pe) is the above value divided by 50 100 ft-lbf/lbm If desired, use the “railroad tracks” example of this calculation on the board ELO 2.2

Energy - Kinetic Work needed to accelerate a body from rest to its current velocity Body maintains this kinetic energy unless its speed changes Kinetic energy (KE) is energy that a body possesses as a result of its motion Using English system units Where: KE = kinetic energy (ft-lbf) m = mass (lbm) v = velocity (ft/sec) gc = gravitational conversion constant ft-lbm/lbf-sec2 ELO 2.2

Energy - Kinetic Determine the kinetic energy of 7 lbm of steam flowing through a pipe at a velocity of 100 ft/sec. 𝐾𝐸 = 𝑚 v 2 2𝑔 𝑐 𝐾𝐸 = 7 𝑙𝑏𝑚 𝑓𝑡 𝑠 𝑓𝑡 _ 𝑙𝑏𝑚 𝑙𝑏𝑓 _ 𝑠 2 𝐾𝐸=1.088 × 𝑓𝑡 _ 𝑙𝑏𝑓 If desired, use the “railroad tracks” example of this calculation on the board. ELO 2.2

Energy - Internal Potential and kinetic energy are macroscopic forms of energy Visualized in terms of position and velocity of objects Substances possess microscopic forms of energy including those due to: Rotation Vibration Translation Interactions among molecules of a substance None of these forms of energy can be measured or evaluated directly, but techniques have been developed to evaluate change in total sum of all these microscopic forms of energy ELO 2.2

Energy - Internal Microscopic forms of energy collectively called internal energy Represented by symbol U British Thermal Unit (BTU) also unit of heat Function of temperature Specific internal energy (u) of a substance is its internal energy per unit mass Total internal energy (U) divided by total mass (m) 𝑢= 𝑈 𝑀 Where: u = specific internal energy (Btu/lbm) U = internal energy (Btu) m = mass (lbm) ELO 2.2

Energy - Specific Internal
Example Determine the specific internal energy of 12 lbm of steam if the total internal energy is 23,000 Btu. 𝑢= 𝑈 𝑀 𝑢= 2.300× 𝐵𝑡𝑢 12 𝑙𝑏𝑚 𝑢=1.917× 𝐵𝑡𝑢 𝑙𝑏𝑚 ELO 2.2

Energy – Flow Energy (PV)
Energy arises from pressure (P) and volume (V) of a fluid Numerically equal to P x V A system where pressure and volume are permitted to expand performs work on its surroundings (flow work) Energy defined as capacity of a system to perform work Fluid under pressure has capacity to perform work P-V energy (flow energy) foot-pounds force (ft-lbf) Specific P-V energy of a substance is P-V energy per unit mass Equals total P-V divided by total mass m, OR Product of pressure P and specific volume v written as Pv ft-lbf/lbm ELO 2.2

Energy - Pressure-Volume
Example Determine the specific P-V energy of 15 lbm of steam at 1000 psi in an 18 ft3 tank. 𝑃𝑣= 𝑃𝑉 𝑚 𝑃𝑣 = 1,000 𝑙𝑏𝑓 𝑖 𝑛 𝑓 𝑡 𝑙𝑏𝑚 𝑖 𝑛 2 𝑓 𝑡 2 𝑃𝑣 = 1.73 × 𝑓𝑡 _ 𝑙𝑏𝑓 𝑙𝑏𝑚 In a later Thermo chapter we will combine the kinetic energy, potential energy, and flow energy (system pressure) to determine the “total pressure” of a system. If desired, use the “railroad tracks” example of this calculation on the board. ELO 2.2

Energy - Enthalpy Enthalpy (H) is measure of energy content of the fluid due to its temperature, pressure, and volume Combination of internal and flow energies (H = U + PV) Specific enthalpy (h) defined as: ℎ = 𝑢 + 𝑃𝜈 Where: u = specific internal energy (Btu/lbm) of system being studied P = pressure of system (lbf/ft2) ν = specific volume (ft3/lbm) of system The specific enthalpy of the saturated steam entering the turbine and the wet vapor exiting the turbine is a function of the “specific work” done by the turbine. This concept is explained in further detail in a future Thermo chapter. ELO 2.2

Energy - Entropy Measure of inability to do work for a given heat transferred Quantifies energy of a substance that is no longer available to perform useful work Represented by S Property of a substance like pressure, temperature, volume, and enthalpy Steam tables include values of specific entropy (s = S/m) as part of information tabulated Specific entropy (s) property is of no real value, but the Ds is ELO 2.2

Energy - Entropy Like enthalpy, entropy cannot be measured directly
Entropy of a substance is given with respect to some reference value – specific entropy of water is zero at 32oF (492oR) Change in specific entropy (Δs), not absolute value, important in practical problems Figure: Entropy of Ice and Water ELO 2.2

Energy - Entropy ∆𝑆= ∆𝑄 𝑇 𝑎𝑏𝑠 ∆𝑠= ∆𝑞 𝑇 𝑎𝑏𝑠 Where:
∆𝑆= ∆𝑄 𝑇 𝑎𝑏𝑠 ∆𝑠= ∆𝑞 𝑇 𝑎𝑏𝑠 Where: ΔS = change in entropy of a system during some process (Btu/oR) ΔQ = amount of heat transferred to or from system during process (Btu) Tabs = absolute temperature at which heat was transferred (oR) Δs = change in specific entropy of a system during some process (Btu/lbm -oR) Δq = amount of heat transferred to/from system during process (Btu/lbm) ELO 2.2

Work, Energy, and Power ELO 2.3 – Explain the relationship between work, energy, and power. Nuclear power plants transfer thermal energy produced in nuclear fuel into mechanical work of the turbine-generator, and then electrical energy Work is the applied force to move a mass, multiplied by distance that mass was moved; power is rate of doing work (work done per unit time) Each term is related and must be understood to solve thermodynamic process equations ELO 2.3

Work, Energy, and Power Work – measures completed task
Energy – ability to do work Power – measures amount of work over time Power defined as time rate of doing work and is equivalent to rate of energy transfer ELO 2.3

Work, Energy, and Power Work is: A form of energy in transit
Not a property of a system Two types or work are mechanical and flow For mechanical systems, defined as action of a force on an object through a distance 𝑊=𝐹𝑑 Where: W = work (ft-lbf) F = force (lbf) d = displacement (ft) ELO 2.3

Work, Energy, and Power Example
Determine the amount of work done if a force of 150 lbf is applied to an object until it has moved a distance of 30 feet. 𝑊𝑜𝑟𝑘=𝐹𝑑 𝑊 = (150 𝑙𝑏𝑓)(30.0 𝑓𝑡) 𝑊 = 4.50 × 103 𝑓𝑡 _ 𝑙𝑏𝑓 In some cases various energy processes (heat and mechanical) might need to be converted into the same term. There is a relationship between work (ft-lbf) and heat (BTU). 1 BTU = 778 ft-lbf. This conversion is provided on the NRC Equation Sheet. ELO 2.3

Work, Energy, and Power Flow is work required to maintain continuous steady movement of fluid in a system Also force through a distance Flow work equivalent to force acting through a distance (such as length) to maintain the flow 𝑊 𝑓𝑙𝑜𝑤 =𝐹𝐷 Since 𝐹𝑜𝑟𝑐𝑒=𝑃𝐴 (𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒×𝑎𝑟𝑒𝑎), 𝑊 𝑓𝑙𝑜𝑤 =𝑃𝐴𝐿 Since 𝑉𝑜𝑙𝑢𝑚𝑒=𝐴𝐿 (𝑎𝑟𝑒𝑎×𝑙𝑒𝑛𝑔𝑡ℎ), 𝑊 𝑓𝑙𝑜𝑤 =𝑃𝑉 Figure: Pipe Boundary Volume for Flow Energy and Related Formulas ELO 2.3

Work, Energy, and Power In dealing with work in relation to energy transfer systems, it is important to distinguish between work done by the system on its surroundings and work done on the system by its surroundings Work is done BY the system when used to turn a turbine and thereby generate electricity in a turbine-generator (+ Work) Work is done ON the system when a pump is used to move working fluid from one location to another (– Work) Wturb and Wpump are two of our four processes in our thermodynamic cycle and will be discussed in greater detail in a future chapter. The other two processes are Qin (SG) and Qout (condenser) ELO 2.3

Work, Energy, and Power Units of various forms of energy are different but equivalent Potential, kinetic, internal, P-V, work, and heat may be measured in numerous basic units Three types of units used to measure energy: Mechanical units such as foot-pound-force (ft-lbf) Thermal units such as British thermal unit (Btu) Electrical units such as watt-second (W-sec) In the mks and cgs systems (not testable): Mechanical units of energy are joule (j) and erg Thermal units are kilocalorie (kcal) and calorie (cal) Electrical units are watt-second (W-sec) and erg ELO 2.3

Work, Energy, and Power Power is the rate at which work is done, or work per unit time 𝑃𝑜𝑤𝑒𝑟= 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 Power has units of energy per unit time mechanical units of power are ft-lbf/s or ft-lbf/hr and hp thermal units are BTU/hr electrical units are watts (W) or kilowatts (kW = 103 W) ELO 2.3

Work, Energy, and Power One horsepower is equivalent to 550 ft-lbf/s
𝑃= 𝐹𝑑 𝑡 Where: P = power (W or ft-lbf/s) F = force (lbf) d = distance (ft) t = time (sec) 𝑃= 𝐹v 550 P = power (hp) v = velocity (ft/s) ELO 2.3

Work, Energy, and Power Mechanical equivalent of heat often denoted by J, Joule’s constant 𝐽=778 𝑓𝑡 _ 𝑙𝑏𝑓 𝐵𝑡𝑢 Other useful conversions, see formula sheet 1 𝐵𝑡𝑢=778 𝑓𝑡 _ 𝑙𝑏𝑓 In other words, 1 BTU of heat can do 778 ft-lbf of work 1 𝑀𝑊=3.41× 𝐵𝑡𝑢 ℎ𝑟 1 ℎ𝑝=2.54× 𝐵𝑡𝑢 ℎ𝑟 Review the various conversions provided at the bottom of the NRC Equation Sheet ELO 2.3

Properties of Heat ELO 2.4 – Define the following terms: heat, sensible heat, latent heat, specific heat, and super heat. Heat is energy in transition caused by a difference in temperature Sample heat transfer process Steam Generator Related KAs Define the following terms: K Latent heat of vaporization K Specific heat 2.3* 2.3 ELO 2.4

Properties of Heat Heat is energy in transit
Transfer of energy as heat occurs at molecular level as a result of a temperature difference Symbol Q used to denote heat Unit of heat is the British thermal unit (Btu) Specifically called 60 degree Btu since it is measured by a one degree temperature change from 59.5oF to 60.5oF ELO 2.4

Properties of Heat Amount of heat transferred depends upon path
Important to distinguish between heat added to a system from its surroundings and heat removed from a system to its surroundings Positive value for heat indicates heat is added to system by its surroundings (+Q) Steam Generator Negative value for heat indicates heat is removed from system by its surroundings (-Q) Condenser Contrast with work - positive when energy is transferred from the system and negative when transferred to the system ELO 2.4

Properties of Heat q indicates heat added to or removed from a system per unit mass Equals total heat (Q) added or removed divided by mass (m) Quantity represented by q referred to as heat transferred per unit mass 𝑞= 𝑄 𝑚 Where: q = heat transferred per unit mass (Btu/lbm) Q = heat transferred (Btu) m = mass (lbm) The term “Specific heat” not just used for q since specific heat “capacity” is also mentioned. However, some NRC bank questions refer to the change in BTU/lbm in the condensing process as “specific heat removed”. This term is also (and mostly) used for the specific heat capacity (cp) of a substance. ELO 2.4

Properties of Heat Example
Determine heat transferred per unit mass if 1,500 Btu’s are transferred to 40 lbm of water Solution 𝑞= 𝑄 𝑚 𝑞 = 1,500 𝐵𝑡𝑢 40.0 𝑙𝑏𝑚 𝑞 = 37.5 𝐵𝑡𝑢 𝑙𝑏𝑚 ELO 2.4

Properties of Heat - Sensible Heat
Heat added to or removed from a substance to produce a change in its temperature Units of heat often defined in terms of changes in temperature they produce Basically the heat added to a subcooled liquid Feedwater entering the SG, for example ELO 2.4

Properties of Heat - Latent Heat
Amount of heat added to or removed from a substance to produce a change in phase When latent heat added/removed, no temperature change occurs Three types of latent heat Latent heat of vaporization - heat added or removed to change phase between liquid and vapor Removal part normally called latent heat of condensation Latent heat of fusion - heat added or removed to change phase between solid and liquid Not applicable to our thermodynamic processes Latent heat of sublimation - heat added or removed to change phase between solid and vapor ELO 2.4

Properties of Heat - Specific Heat
Ratio of heat (Q) added to or removed from a substance to change in temperature (ΔT) produced called heat capacity (Cp) of substance Heat capacity of a substance per unit mass called specific heat (cp) of substance Cp and cp apply when heat is added or removed at constant pressure ELO 2.4

Properties of Heat - Specific Heat
𝐶 𝑝 = 𝑄 ∆𝑇 𝑐 𝑝 = 𝑄 𝑚∆𝑇 𝑐 𝑝 = 𝑞 ∆𝑇 Where: Cp = heat capacity at constant pressure (Btu/°F) cp = specific heat capacity at constant pressure (Btu/lbm-°F) Q = heat transferred (Btu) q = heat transferred per unit mass (Btu/lbm) m = mass (lbm) ∆𝑇 = temperature change (°F) One lbm of water raised 1°F and one Btu of heat added Implies specific heat (cp) of water is 1 Btu/lbm-°F cp of water equal to 1 Btu/lbm-°F at 39.1°F NOTE: Specific Heat Capacity of fluids is relatively close to 1.0 at low temperatures (note the density vs temperature curve). However, at higher temperatures, Specific Heat Capacity much greater than 1 (about 1.4 at 580oF). Examples on how to calculate Specific Heat capacity is provided in – Steam. ELO 2.4

Properties of Heat Specific Heat Example
How much heat is required to raise the temperature of 5 lbm of water from 50oF to 150oF? Assume specific heat (cp) for water is constant at 1.0 Btu/lbm- oF Solution 𝐶 𝑝 = 𝑄 𝑚∆𝑇 𝑄=𝑚 𝐶 𝑝 ∆𝑇 𝑄 =(5 𝑙𝑏𝑚) 1.0 𝐵𝑡𝑢 𝑙𝑏𝑚 _ ℉ (150℉ −50℉) 𝑄=(5 𝑙𝑏𝑚) 1.0 𝐵𝑡𝑢 𝑙𝑏𝑚 _ ℉ (100℉) 𝑄=500 𝐵𝑡𝑢 ELO 2.4

Properties of Heat - Super Heat
Number of degrees above saturation temperature at a specific pressure From previous discussions on heat and work, similarities evident: Heat and work both transient phenomena Systems never possess heat or work, but either or both may occur when a system undergoes a change of energy state Both heat and work are boundary phenomena in that both are observed at boundary of system Both represent energy crossing system boundary Superheat is explained in further detail in the next chapter. Table 3 of the Steam Tables is the Superheat table. ELO 2.4

Units and Properties Summary
Relationships between inHg vac, inHg abs, and psia Majority of bank questions on this concept Relationship between psi and ft H20 Recall psi = 1 ft H2O (at low temps) DP Level Detector concepts discussed in – Sensors and Detectors The first and second laws of thermodynamics Thermodynamic properties related to energy (PE, KE, internal energy, enthalpy and entropy) The relationship of work, energy, and power Types of heat (sensible, latent, specific, and super) Superheat is explained in further detail in the next chapter. Table 3 of the Steam Tables is the Superheat table. Summary

NRC KA to ELO Tie KA # KA Statement RO SRO ELO K1.01
Convert between absolute and gauge pressure and vacuum scales. 2.5 2.7 1.3 K1.02 Recognize the difference between absolute and relative (Kelvin) temperature scales. 1.9 2.0 1.2 K1.03 Describe how pressure and level sensing instruments work. 2.6 NOTE K1.04 Explain relationships between work, power, and energy. 2.2 2.3 K1.05 Explain the law of conservation of energy. 2.1 K1.03 is covered (and tested) in Sensors and Detectors – Units and properties NRC exam bank questions on DP level detectors will be discussed in – Sensors and Detectors.

Operator Generic Fundamentals Thermodynamic Units and Properties

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Operator Generic Fundamentals Thermodynamic Units and Properties

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