Thermodynamic Units and Properties PPT
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Operator Generic Fundamentals Thermodynamic Units and Properties © Copyright 2014 Operator Generic Fundamentals 2 Thermodynamic Units and Properties • Thermodynamics is a branch of natural science concerned with heat and its relation to energy and work • Defines macroscopic variables (such as temperature, internal energy, entropy, and pressure) that characterize materials and explains how they are related and the laws that govern how they change © Copyright 2014 Intro Operator Generic Fundamentals 3 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): 1. Describe thermodynamic properties and methods of measuring intensive and extensive properties. 2. Explain the concepts of heat, work, and energy. © Copyright 2014 Intro Operator Generic Fundamentals 4 Thermodynamic Properties TLO 1 – Describe thermodynamic properties and methods of measuring intensive and extensive properties. • Thermodynamic properties describe measurable characteristics of substance • Often used to identify substances or distinguish between two different or separate substances • Concerned with both thermal and mechanical properties of substances and their measurement • Operators must recognize the different types and their interrelationships in order to understand thermodynamics © Copyright 2014 TLO 1 Operator Generic Fundamentals 5 Enabling Learning Objectives for TLO 1 1. Define the following properties: specific volume, density, specific gravity, humidity, mass, weight, intensive, and extensive. 2. Define the thermodynamic properties of temperature and convert between the Fahrenheit, Celsius, Kelvin, and Rankine scales. 3. Define the thermodynamic properties of pressure and convert between pressure scales. © Copyright 2014 TLO 1 Operator Generic Fundamentals 6 Properties and Definitions ELO 1.1 – Define the following properties: specific volume, density, specific gravity, humidity, mass, weight, intensive, and extensive. • 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 – In this case, operator will be required to perform unit conversion © Copyright 2014 ELO 1.1 Operator Generic Fundamentals Measurement Systems • American Engineering System – From England – Currently only used in the United States • International System (SI) – Based on powers of ten – Used almost universally throughout the world © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 8 Properties and Definitions American Engineering System Length Inch Mass Ounce Time Second* Foot* Pound* Minute Yard Ton Hour Mile Day NOTE: *Denotes standard unit of measure © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 9 Properties and Definitions International System (SI) – MKS Units Length Millimeter Mass Milligram Time Second* Meter* Gram Minute Kilometer Kilogram* Hour Day NOTE: *Denotes standard unit of measure © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 10 Properties and Definitions International System (SI) – CGS Units Length Centimeter* Mass Milligram Time Second* Meter Gram* Minute Kilometer Kilogram Hour Day NOTE: *Denotes standard unit of measure © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 11 Properties and Definitions Prefix Symbol Power of 10 Example pico p 10-12 1 picosecond (ps) = 10-12 seconds nano n 10-9 1 nanosecond (ns) = 10-9 seconds micro m 10-6 1 microsecond (ms) = 10-6 seconds milli m 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 Figure: Metric System Prefixes © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 12 Typical Conversion Table Unit English Units of Measurement Length 1 yard (yd) Meter-Kilogram-Second (MKS) Units of Measurement = 0.9144 meter (m) 12 inches (in.) = 1 ft 5,280 feet (ft) = 1 mi 1 (meter) m = 3.281 ft Time 1 in. 60 seconds (sec) = 0.0254 m = 1 minute (min) Mass 3,600 sec 1 pound mass (lbm) = 1 hour (hr) 0.4535 kg 2.205 lbm = 1 kg 1 kilogram (kg) 1 square foot (ft2) = 1,000 grams (g) = 144 in.2 10.764 ft2 = 1 square meter (m2) 1 square yard (yd2) = 9 ft2 1 square mile (mi2) 7.48 gallon (gal) 3.098 x 106 yd2 = 1 cubic foot (ft3) 1 gal = 3.785 l (liter) 1 liter (l) = 1,000 cubic centimeters (cm3) Area Volume © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 13 Properties and Definitions • Conversions between units of measure performed using conversion factors • Conversion factor is ratio of two equivalent physical quantities expressed in different units © Copyright 2014 Steps for Converting Units: 1. Identify units given and units required 2. Write conversion factor relating units 3. Divide to obtain factor of 1 as a ratio (desired /current) 4. Multiply quantity by ratio 5. Multiple conversion factors may be required ELO 1.1 Operator Generic Fundamentals 14 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 πππ‘ππ(ππππ πππ‘ π’πππ‘π ) = 3.281 ππ‘(πππ ππππ π’πππ‘π ) • Step 3: Arrange equivalence ratio in appropriate manner: πππ ππππ π’πππ‘π ( ) ππππ πππ‘ π’πππ‘π 1= 3.281 ππ‘ 1π • Step 4: Multiply the quantity by the ratio: 3.281 ππ‘ 795 π 3.281 ππ‘ 795 π = = 795 × 3.281 ππ‘ 1π 1 1π = 2608.395 ππ‘ © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 15 Practice Unit Conversion Convert 2.91 square miles to square meters. • Step 1: Select equivalence relationship (no direct conversion, multiple conversions necessary) Square miles to square yards to square feet to square meters 1 ππ 2 = 3.908 × 106 π¦π 2 1 π¦π 2 = 9 ππ‘ 2 10.764 ππ‘ 2 = 1 π2 • Step 2: Express relationship as ratio (desired units/present units): 3.098 × 106 π π π¦πππ 1= 1 π π ππππ • Step 3: Multiply original quantity by ratio: 2.91 π π πππππ © Copyright 2014 3.098 × 106 1 π π ππππ = 9.015 × 106 π π π¦πππ ELO 1.1 Operator Generic Fundamentals 16 Practice Unit Conversion (Continued) • Step 4: Repeat steps until value in desired units 9 π π ππ‘ 1= 1 π π π¦πππ 9 π π ππ‘ 9.015 × 106 π π π¦πππ = 8.114 × 107 π π ππ‘ 1 π π π¦πππ 1 π π πππ‘ππ 1= 10.764 π π ππ‘ 1 π π πππ‘ππ 7 8.114 × 10 π π ππ‘ 10.764 π π ππ‘ 8.114 × 107 π π ππ‘ 1 π π ππππ = 10.764 π π ππππ‘ 8.114 × 107 π π ππππ = = 7.538 × 106 π π πππ‘πππ 10.764 © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 17 Properties and Definitions Unit Gram (g) Kilogram (kg) Metric Ton (t) Pound-mass (lbm) 1 gram (g) 1 1 kilogram (kg) 1,000 0.001 10-5 2.2046 x 10-3 1 0.001 2.2046 1 metric ton (t) 106 1,000 1 2204.6 1 pound-mass (lbm) 453.59 0.45359 4.5359 x 10-4 1 1 slug 14.594 14.594 0.014594 32.174 Figure: Common Conversion Factors - Mass © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 18 Properties and Definitions Unit Centimeter Meter (cm) (m) Kilometer (km) Inch (in.) Foot (ft) 1 centimeter (cm) 1 0.01 10-5 0.3937 0.032808 6.2137 x 10-6 1 meter (m) 100 1 0.001 39.370 3.2808 1 kilometer (km) 105 1,000 1 39,370 3280.8 1 inch (in.) 2.5400 1 1 foot (ft) 30.480 0.02540 2.5400 x 105 0 0.30480 3.0480 x 10- 0.083333 1.5783 x 10-5 1 1.8939 x 10-4 5,280 1 12.000 4 1 mile (mi) 1.6093 x 105 1,609.3 1.6093 63,360 Mile (mi) 6.2137 x 10-4 0.62137 Figure: Common Conversion Factors - Length © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 19 Properties and Definitions Unit Second (s) Minute (min) Hour (hr) Day (d) Year (yr) 1 second (s) 1 0.017 2.7 x 10-4 1.16 x 10- 3.1 x 10-8 1 minute (min) 60 1 0.017 6.9 x 10-4 1.9 x 10-6 1 hour (hr) 3,600 60 1 4.16 x 10- 1.14 x 10-4 5 2 1 day (d) 86,400 1,440 24 1 year (yr) 3.15 x 107 5.26 x 105 8,760 1 2.74 x 10-3 365 1 Figure: Common Conversion Factors - Time © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 20 Properties and Definitions erg 1 erg (dyn cm) 1 joule (J) 1 kcal J 2.389 ο΄ 10-11 10-7 Btu 7.376 ο΄ 10-8 9.481 ο΄ 1011 107 2.389 ο΄ 10-4 1 1 kilocalorie 4.184 ο΄ 1010 1 4184 1 foot 1.356 ο΄ 107 3.239 ο΄ 10-4 1.356 pound-force (ft lbf) 1 Btu 1.055 ο΄ 1010 0.252 1055 1 electron volt ft lbf 0.7376 9.481 ο΄ 10-4 3087 3.968 1 1.285 ο΄ 10-3 778 1 1.602 ο΄ 10-12 3.83 ο΄ 10-23 1.602 ο΄ 10-19 1.182 ο΄ 10-19 1.52 ο΄ 10-22 Figure: Common Conversion Factors - Energy © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 21 Properties and Definitions BTU/hr hp W ft-lbf/sec 1 foot poundforce per second 1 horsepower (hp) 4.628 1.818 ο΄ 10-3 1.356 1 2545 1 746 550 1 Btu per hour 1 3.929 ο΄ 10-4 0.293 0.216 1 watt (W) 3.412 1.341 ο΄ 10-3 1 0.7376 Figure: Common Conversion Factors - Power © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 22 Properties and Definitions - Mass & Weight • Mass (m) of a body is measure of amount of material present in that body • Weight (wt) of a body is 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 – Is not acceleration of gravity – Is a dimensional constant that allows use of Newton’s Second Law of Motion with the English system of units © Copyright 2014 ELO 1.1 π€π‘ = ππ ππ Where: wt = weight (lbf) m = mass (lbm) g = acceleration due to gravity = 32.17 ft/s gc = gravitational constant = 32.17 lbm-ft/lbf-s2 Operator Generic Fundamentals 23 Practice Conversion Problem Reactor core thermal power is 1,800 MWth. Convert to BTU/hr. Units given are megawatts and units desired are BTU/hr 1 megawatt = 106 watts 1 watt = 3.412 BTU/hr 1,800 MW 106 W 3.412 Btu hr 1 MW 1W 1,800 MW converts to 6.142 x 109 BTU/hr © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 24 Properties and Definitions - Mass & Weight • Weight - force produced when mass of a body accelerated by a gravitational acceleration • Mass remains constant even if gravitational acceleration acting upon that body changes • Newton’s Second Law of Motion, force (F) = ma – a equals acceleration – On earth, an object has a certain mass and weight – If same object is placed in space away from earth’s gravitational field… • Mass remains same • Object now in a weightless condition © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 25 Properties and Definitions - Mass & Weight • English system uses pound-force (lbf) as unit of weight • Knowing that acceleration has units of ft/sec2 and using Newton’s second law, can determine units of mass are lbf-sec2/ft • For simplification, 1 lbf-sec2/ft called a slug – Basic unit of mass in English system • Slug almost meaningless unit for average individual • Unit of mass generally used is pound-mass (lbm) • To use lbm as unit of mass, must divide Newton’s second law by gravitational constant (gc) © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 26 Properties and Definitions - Mass & Weight Newton’s second law can be expressed: ππ πΉ= ππ πππ_ ππ‘ 32.17 = ππ _ 2 πππ π • Use of gc adapts Newton’s second law such that 1 lbf = 1 lbm at surface of earth • Only true at surface of earth where acceleration (a) due to gravity (g) is 32.17 ft/sec2 © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 27 Properties and Definitions - Mass & Weight Example: • Using πΉ = Solution: ππ ππ ππ‘ π ππ 2 1 πππ = πππ_ ππ‘ 32.17 πππ_ π ππ 2 1 πππ = 1 πππ (an equality) • Prove 1 lbf = 1 lbm on earth © Copyright 2014 1 πππ ELO 1.1 32.17 Operator Generic Fundamentals 28 Properties and Definitions - Volume • Volume is amount of space a particular substance occupies expressed in units of cubic feet or cubic meters • Often of more concern is specific volume which measures amount of volume occupied by a unit mass of a substance • Specific volume is total volume (V) of that substance divided by total mass (m) of that substance – Has units of cubic feet per pound-mass (ft3/lbm) © Copyright 2014 ELO 1.1 π π£= π Where: v = specific volume (ft3/lbm) V = volume (ft3) m = mass (lbm) Operator Generic Fundamentals 29 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) © Copyright 2014 π 1 π= = π π£ Where: ρ = density (lbm/ft3) m = mass (lbm) V = volume (ft3) v = specific volume (ft3/lbm) ELO 1.1 Operator Generic Fundamentals 30 Properties and Definitions - Density • • • • Density can be changed by changing pressure or temperature Increasing pressure will always increase density of a material Increasing temperature generally decreases density Effect of pressure and temperature on densities of liquids and solids relatively small in contrast, density of gases strongly affected by pressure © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 31 Properties and Definitions - Gravity • Measure of relative density of a substance as compared to density of water at a standard temperature – Physicists use 39.2oF (4oC) as standard, engineers use 60oF • SI units, density of water is 1.00 g/cm3 at standard temperature • Specific gravity for a liquid has same numerical value as its density in units of g/cm3 • Varies with temperature, specific gravities must be specified at particular temperatures © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 32 Properties and Definitions - Gravity ππππ’ππ ππππππππ πΊπππ£ππ‘π¦ π. πΊ. = ππ€ππ‘ππ Where: rfluid = density of fluid being measured (lbm/ft3) Rwater = density of water at standard temperature (40oF) and pressure (14.7 psia) (62.4 lbm/ft3) © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 33 Properties and Definitions - Humidity • Amount of moisture (water vapor) in air • Absolute humidity or relative humidity – Absolute humidity is mass of water vapor divided by a unit volume of air (grams of water/cm3 of air) – Relative humidity is amount of water vapor present in air divided by maximum amount that air could contain at that temperature o Expressed as a percentage o 100% if air is saturated with water vapor o 0% if no water vapor is present in air at all © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 34 Intensive and Extensive Properties • Intensive properties are independent of amount of mass – Temperature, pressure, specific volume, and density • Extensive property varies directly with mass – Mass and total volume • 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 © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 35 Properties and Definitions Knowledge Check Which one of the following is an example of an extensive thermodynamic property? A. Temperature B. Pressure C. Volume D. Density Correct answer is C. © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 36 Properties and Definitions Knowledge Check A reactor core's thermal power is 2,000 megawatts thermal (MWth). Convert this to BTU/hr. A. 6.824E9 BTU/hr B. 6.824E12 BTU/hr C. 2.93E9 BTU/hr D. 2.93E12 BTU/hr Correct answer is A. © Copyright 2014 ELO 1.1 Operator Generic Fundamentals 37 Properties of Temperature ELO 1.2 – Define the thermodynamic properties of temperature and convert between the Fahrenheit, Celsius, Kelvin, and Rankine scales. Figure: Comparison of Temperature Scales © Copyright 2014 ELO 1.2 Operator Generic Fundamentals 38 Temperature • Temperature is a measure of amount of energy stored in an object – 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 © Copyright 2014 ELO 1.2 Operator Generic Fundamentals 39 Temperature Scales • Two temperature scales normally used for measurement purposes – Fahrenheit (F) – Celsius (C) • Based on number of increments between freezing and boiling point of water at standard atmospheric pressure – 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 © Copyright 2014 ELO 1.2 Figure: Boiling and Freezing Points of Water for Celsius and Fahrenheit Temperature Scales Operator Generic Fundamentals 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 9 β= °πΆ + 32 5 5 β= 9 © Copyright 2014 β − 32 ELO 1.2 Operator Generic Fundamentals 41 Temperature Scales • Relationship between scales represented by: °π = β + 460 °πΎ = β + 273 Figure: Comparison of Temperature Scales © Copyright 2014 ELO 1.2 Operator Generic Fundamentals 42 Temperature Scales • When working thermodynamic problems, more convenient to work with scales where temperatures are all positive numbers • Absolute temperature scales have only positive values – Kelvin (K) scale corresponds to Celsius scale – Rankine (R) scale corresponds to Fahrenheit scale • Zero points on both absolute scales represent same physical state – No molecular motion of individual atoms © Copyright 2014 ELO 1.2 Operator Generic Fundamentals 43 Temperature Scale Conversion What is Rankine equivalent of 80oC? • Rankine (R) scale corresponds to Fahrenheit scale, therefore convert C to F first! Solution: 9 β= °πΆ + 32 5 = 9 5 80 + 32 = 176°πΉ °π = β + 460 = 176 + 460 = 636°π © Copyright 2014 ELO 1.2 Operator Generic Fundamentals 44 Temperature Scale Conversion What is Kelvin equivalent of 80oF? Solution: β= 5 = 9 5 9 β − 32 80 − 32 = 26.7β °πΎ = β + 273 = 26.7 + 273 = 299.7°πΎ © Copyright 2014 ELO 1.2 Operator Generic Fundamentals 45 Temperature Scale Conversion Practice Question The water in the reactor coolant system returning to the reactor is 550.4β. What is this temperature in degrees Celsius, Kelvin and Rankine? Solution 5 β= 9 5 β − 32 = 9 550.4 − 32 = 288β °πΎ = β + 273 = 288 + 273 = 561°πΎ °π = β + 460 = 550.4 + 460 = 1010.4°π © Copyright 2014 ELO 1.2 Operator Generic Fundamentals 46 Properties of Temperature Knowledge Check The low-temperature condition at which all molecular or atomic motion ceases is referred to as . A. reference point B. freeze point C. absolute zero D. reference zero Correct answer is C. © Copyright 2014 ELO 1.2 Operator Generic Fundamentals 47 Properties of Pressure ELO 1.3 – Define the thermodynamic properties of pressure and convert between pressure scales. Figure: Pressure Scale Relationships © Copyright 2014 ELO 1.3 Operator Generic Fundamentals 48 Properties of Pressure • Force exerted per unit area on boundaries of substance (or system) • 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 • Pressure frequently expressed in units of lbf/in2 (psi) in English System of Measurement © Copyright 2014 ELO 1.3 Operator Generic Fundamentals 49 Properties 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 – 0.433 psi = 1 ft of water – 14.7 psia = 408 inches of water – 14.7 psia = 29.9 inches of mercury – 1 inch of mercury = 25.4 millimeters of mercury • In SI units, pressure given in pascals (N/m2) © Copyright 2014 ELO 1.3 Operator Generic Fundamentals 50 Properties 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 • Measure difference between pressure of fluid to which they are connected and that of surrounding air SI units = 1 Atmosphere = column of mercury 760 mm in height. How many inches of mercury is that equal to?? 760 mm mercury = 29.92 inches of mercury = 14.7 psi © Copyright 2014 ELO 1.3 Operator Generic Fundamentals 51 Properties of Pressure • Pressure below atmospheric designated as vacuum • Perfect vacuum corresponds to absolute zero pressure – All values of absolute pressure are positive – Negative value would indicate tension which is considered impossible in any fluid • Gauge pressures: – Positive if above atmospheric pressure – Negative if below atmospheric pressure © Copyright 2014 ELO 1.3 Operator Generic Fundamentals 52 Properties of Pressure • Relationships between absolute, gauge, vacuum, and atmospheric pressures Figure: Pressure Scale Relationships © Copyright 2014 ELO 1.3 Operator Generic Fundamentals 53 Properties of Pressure Example 1 Practice 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.) A. 14.0 psia B. 13.5 psia C. 1.5 psia D. 1.0 psia Correct answer is C. © Copyright 2014 ELO 1.3 Operator Generic Fundamentals 54 Properties of Pressure Pabs = Patm – Pvacuum Pabs = 15 psia – 27”Hg (1psia/2.03”Hg) Pabs = 15 psia – 13.5 psia = 1.5 psia Figure: Gauge and Absolute Pressure Scale Relationship © Copyright 2014 ELO 1.3 Operator Generic Fundamentals 55 Properties of Pressure • Relationships between absolute, gauge, vacuum, and atmospheric pressures depicted as follows: 2 Atmospheres © Copyright 2014 ELO 1.3 Operator Generic Fundamentals 56 Properties of Pressure ππππ = πππ‘π + ππππ’ππ ππππ = πππ‘π − ππ£ππ Where: Patm = atmospheric or barometric pressure Pgauge = gauge pressure Pvac = vacuum © Copyright 2014 ELO 1.3 Operator Generic Fundamentals 57 Properties of Pressure Pressure due to column of fluid is product of fluid’s density and height πππ§ π= ππ Where: P = pressure (lbf/ft2, lbf/in2, psi) r = density (lb/ft3) g = acceleration of gravity (32.2 ft/s2) z = height (ft, in) ππ = gravitational constant (32.2 ft lbm/lbf s2) © Copyright 2014 ELO 1.3 Operator Generic Fundamentals 58 Properties of Pressure Practice Example 3 Solution How deep can a diver descend in ocean water (density = 64 lbm/ft3) without damaging his watch, which will withstand an absolute pressure of 80 psia? (P = density x height) 80 ππ ππ = 14.7 ππ ππ + ππππ’ππ ππππ’ππ = 80 ππ ππ − 14.7 ππ ππ ππππ’ππ = 65.3 ππ ππ ππππ’ππ = ππππ ππ‘π¦ × βπππβπ‘ ππππ’ππ βπππβπ‘ = ππππ ππ‘π¦ πππ ππ2 65.3 2 × 144 2 ππ ππ‘ βπππβπ‘ = πππ 64 3 ππ‘ βπππβπ‘ = 1.47 × 102 ππ‘ Assume: – Patm = 14.7 psia – density = 64 lbm/ft3 ππππ = πππ‘π + ππππ’ππ © Copyright 2014 ELO 1.3 Operator Generic Fundamentals 59 Properties of Pressure Practice Example 4 Solution What is the absolute pressure at the bottom of a swimming pool 6 feet deep that is filled with fresh water? (P = density x height) ππππ = πππ‘π + ππππ’ππ = 14.7 + ππ» ππππ ππ‘π¦ × βπππβπ‘ = 14.7 + Assume: – Patm = 14.7 psia – density = 62.4 lbm/ft3 πππ 6 ππ‘. ππ‘ 3 ππ2 144 2 ππ‘ 62.4 = 14.7 + 2.6 ππππ = 17.3 ππ ππ ππππ = πππ‘π + ππππ’ππ © Copyright 2014 ELO 1.3 Operator Generic Fundamentals 60 Properties of Pressure Knowledge Check Convert the pressures and fill in the blanks below: 1. 40 inches Hg (absolute) = _____ psia or ______ psig 2. 20 ft of water (gauge) = _______ psig or _______ psia 3. 13 psiv = _______ psia or ______ inches of Hg (vacuum) 4. 28 inches of Hg (vacuum) = _______ psia or _______ psiv 5. 5 inches of water (gauge) = _______ ft of water (gauge) or _______ psig A. (1) 19.6, 4.9; (2) 8.7, 23.4;(3) 1.7, 26.4; (4) 1.0, 13.7;(5 ) 0.42, 0.18 B. (1) 19.4, 4.9; (2) 8.7, 22.4;(3) 1.6, 24.4; (4) 1.0, 13.7;(5) 0.42, 0.18 C. (1) 20.6, 4.9; (2) 9.7, 23.4;(3) 1.6, 26.4; (4) 1.0, 13.7;(5) 0.42, 0.18 D. (1) 19.6, 4.9; (2) 8.7, 22.3;(3) 1.7, 26.4; (4) 1.0, 13.7;(5) 0.42, 0.18 Correct answer is A. © Copyright 2014 ELO 1.3 Operator Generic Fundamentals 61 Properties of Pressure Knowledge Check – NRC Bank Refer to the drawing of two water storage tanks with four differential pressure (D/P) level detectors. The tanks are identical and are being maintained at 2 psig overpressure, 60F, and the same constant water level. The tanks are located within a sealed containment structure that is being maintained at standard atmospheric pressure. All level detectors have been calibrated and are producing the same level indication. If a ventilation malfunction causes the containment structure pressure to decrease to 13 psia, which detectors will produce the highest level indications? A. 1 and 2 B. 3 and 4 C. 1 and 4 D. 2 and 3 © Copyright 2014 Correct answer is D. ELO 1.3 Operator Generic Fundamentals 62 TLO 1 Summary 1. Define the following properties: – Mass (m) measure of amount of material present in that body – Weight (wt) of a body is force exerted by that body when its mass is accelerated in a gravitational field – Specific volume is total volume of substance divided by total mass – Density is total mass divided by total volume – Specific gravity is measure of relative density compared to water density at standard temperature – Humidity is amount of moisture in air – Intensive properties are those independent of amount of mass – Extensive properties vary directly with mass © Copyright 2014 TLO 1 Operator Generic Fundamentals 63 TLO 1 Summary 2. Define the thermodynamic properties of temperature and convert between the Fahrenheit, Celsius, Kelvin, and Rankine scales. – Temperature is measure of molecular activity of a substance – Pressure is measure of force per unit area exerted on boundaries of a substance (or system) – Relationship between Fahrenheit, Celsius, Kelvin, and Rankine temperature scales o Absolute zero = -460°F or -273°C o Freezing point of water = 32°F or 0°C o Boiling point of water = 212°F or 100°C – Conversions between scales can be made using the following: 9 5 5 9 °πΉ = 32 + ( )°πΆ °πΆ = (°πΉ − 32)( ) °π = °πΉ + 460 °πΎ = °πΆ + 273 © Copyright 2014 TLO 1 Operator Generic Fundamentals 64 TLO 1 Summary 3. Define the thermodynamic properties of pressure and convert between all pressure scales. – Relationships between absolute pressure, gauge pressure, and vacuum ππππ = πππ‘π + ππππ’ππ ππππ = πππ‘π − ππ£ππ – Converting between pressure units can be done using: 14.7 psia = 408 inches of water 14.7 psia = 29.9 inches of mercury 1 inch of mercury = 25.4 millimeters of mercury 1 millimeter of mercury = 103 microns of mercury © Copyright 2014 TLO 1 Operator Generic Fundamentals 65 Heat, Work, and Energy TLO 2 – Explain the concepts of heat, work, and energy. • This lesson will identify and explain different forms of energy, examine conversion of one form of energy to another, and explain concepts and terminology related to energy, work, and power. 1. Define the following thermodynamic properties: potential energy, kinetic energy, specific internal energy, specific P-V energy, specific enthalpy, and entropy. 2. Explain the relationship between work, energy, and power. 3. Define the following terms: heat, latent heat, sensible heat, unit used to measure heat, specific heat, and super heat. © Copyright 2014 TLO 2 Operator Generic Fundamentals 66 Properties of Energy ELO 2.1 – Define the following thermodynamic properties: potential energy, kinetic energy, specific internal energy, specific P-V energy, specific enthalpy, and entropy. • Heat and work are two ways in which energy can be transferred across system boundaries © Copyright 2014 ELO 2.1 Operator Generic Fundamentals 67 Properties of Energy • Energy is capacity of a system to perform work or produce heat • Forms of stored energy important in analysis of systems: – Potential energy – Kinetic energy – Internal energy – P-V (flow) energy © Copyright 2014 ELO 2.1 Operator Generic Fundamentals 68 Potential Energy • 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 πππ§ ππΈ = ππ 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 constant 32.17 ft-lbm/lbf-sec2 © Copyright 2014 ELO 2.1 Operator Generic Fundamentals 69 Properties of Energy • For most practical engineering calculations, acceleration due to gravity (g) numerically equal to gravitational constant (gc) – Therefore, potential energy (PE) in foot-pounds-force numerically equal to mass (m) in pounds-mass times height (z) in feet above some reference level © Copyright 2014 ELO 2.1 Operator Generic Fundamentals 70 Energy - Potential Determine the potential energy of 50 lbm of water in a storage tank 100 ft above the ground. πππ§ ππΈ = ππ ππ‘ 2 100 ππ‘ π ππ‘ _ πππ 2 32.17 πππ _ π 50.0 πππ ππΈ = 32.17 3 ππΈ = 5.00 × 10 ππ‘ _ πππ © Copyright 2014 ELO 2.1 Operator Generic Fundamentals 71 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 © Copyright 2014 mv 2 KE ο½ 2g c Where: KE = kinetic energy (ft-lbf) m = mass (lbm) v = velocity (ft/sec) gc = gravitational constant 32.17 ft-lbm/lbf-sec2 ELO 2.1 Operator Generic Fundamentals 72 Energy - Kinetic • Determine the kinetic energy of 7 lbm of steam flowing through a pipe at a velocity of 100 ft/sec. 2 ππ£ πΎπΈ = 2ππ ππ‘ π ππ‘ _ πππ 2 32.17 πππ _ π 7 πππ πΎπΈ = 2 100.0 2 3 πΎπΈ = 1.088 × 10 ππ‘ _ πππ © Copyright 2014 ELO 2.1 Operator Generic Fundamentals 73 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 © Copyright 2014 ELO 2.1 Operator Generic Fundamentals 74 Energy - Internal • Microscopic forms of energy collectively called internal energy – Represented by symbol U – British thermal unit (Btu) also unit of heat • Specific internal energy (u) of a substance is its internal energy per unit mass – Total internal energy (U) divided by total mass (m) π π’= π Where: u = specific internal energy (Btu/lbm) U = internal energy (Btu) m = mass (lbm) © Copyright 2014 ELO 2.1 Operator Generic Fundamentals 75 Energy - Specific Internal Example Determine the specific internal energy of 12 lbm of steam if the total internal energy is 23,000 Btu. π’= π π 2.300 × 104 π΅π‘π’ π’= 12 πππ 103 π΅π‘π’ π’ = 1.917 × πππ © Copyright 2014 ELO 2.1 Operator Generic Fundamentals 76 Energy - Pressure-Volume • 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 © Copyright 2014 ELO 2.1 Operator Generic Fundamentals 77 Energy - Pressure-Volume Example: Determine the specific P-V energy of 15 lbm of steam at 1000 psi in an 18 ft3 tank. ππ ππ£ = π 3 πππ 2 18 ππ‘ ππ 15.0 πππ 1,000 ππ£ = 2 ππ 144 2 ππ‘ ππ‘ _ πππ ππ£ = 1.73 × 10 πππ 5 © Copyright 2014 ELO 2.1 Operator Generic Fundamentals 78 Energy - Enthalpy • Enthalpy (H) is a property of a substance, like pressure, temperature, and volume, but cannot be measured directly • Enthalpy (H) is measure of energy content of the fluid due to its temperature, pressure, and volume • Specific enthalpy (h) defined as: β = π’ + ππ Where: u = specific internal energy (Btu/lbm) of system being studied P = pressure of system (lbf/ft2) ν = specific volume (ft3/lbm) of system © Copyright 2014 ELO 2.1 Operator Generic Fundamentals 79 Energy - Enthalpy • Enthalpy given with respect to some reference value • Specific enthalpy of water is zero at .01oC and normal atmospheric pressure – Change in specific enthalpy (Δh) and not the absolute value that is important in practical problems – Steam tables include values of enthalpy as part of the information tabulated © Copyright 2014 ELO 2.1 Operator Generic Fundamentals 80 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 © Copyright 2014 ELO 2.1 Operator Generic Fundamentals 81 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 © Copyright 2014 Figure: Entropy of Ice and Water ELO 2.1 Operator Generic Fundamentals 82 Energy - Entropy 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) © Copyright 2014 ELO 2.1 βπ βπ = ππππ βπ βπ = ππππ Operator Generic Fundamentals 83 Properties of Energy Knowledge Check ___________ is the measure of energy content of the fluid due to its temperature, pressure, and volume. A. Entropy B. Kinetic energy C. Enthalpy D. Specific internal energy Correct answer is C. © Copyright 2014 ELO 2.1 Operator Generic Fundamentals 84 Work, Energy, and Power ELO 2.2 – Explain the relationship between work, energy, and power. • Purpose of a nuclear-powered generating station is to transfer thermal energy produced in nuclear fuel to the turbine-generator, where thermal energy is converted into mechanical work and then electrical energy • Work is 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 problems © Copyright 2014 ELO 2.2 Operator Generic Fundamentals 85 Work, Energy, and Power • Work is: – A form of energy, but energy in transit – Not a property of a system – Process done by or on a system, but a system contains no work • For mechanical systems, defined as action of a force on an object through a distance π = πΉπ Where: W = work (ft-lbf) F = force (lbf) d = displacement (ft) © Copyright 2014 ELO 2.2 Operator Generic Fundamentals 86 Work, Energy, and Power • Two types are mechanical and flow • Mechanical work is mechanical energy in transition and is equal to a force acting through a distance Figure: Pressure Scale Relationships © Copyright 2014 ELO 2.2 Operator Generic Fundamentals 87 Work, Energy, and Power • Flow is work required to maintain continuous study flow of fluid • Also force through a distance • When a volume of fluid forced past a boundary in a pipe, as shown in figure, fluid performs work • Flow work equivalent to force acting through a distance (such as length) • Since πΉππππ = ππ΄ (ππππ π π’ππ × ππππ), πππππ€ = ππ΄πΏ • Since ππππ’ππ = π΄πΏ (ππππ × πππππ‘β), πππππ€ = ππ Figure: Pipe Boundary Volume for Flow Energy and Related Formulas © Copyright 2014 ELO 2.2 Operator Generic Fundamentals 88 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. ππππ = πΉπ π = (150 πππ)(30.0 ππ‘) π = 4.50 × 103 ππ‘ _ πππ © Copyright 2014 ELO 2.2 Operator Generic Fundamentals 89 Work, Energy, and Power Example A force of 25 newtons pushes an object 10 meters. How much work is done? ππππ = πΉπ = (25 πππ€π‘πππ )(10 π) = 250 πππ€π‘ππ_ πππ‘πππ (πππ’πππ ) © Copyright 2014 ELO 2.2 Operator Generic Fundamentals 90 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) © Copyright 2014 ELO 2.2 Operator Generic Fundamentals 91 Work, Energy, and Power • Power is rate at which work is done or work per unit time • In American Engineering System, power expressed in terms of horsepower ππ‘ _ πππ 1 βπ = 550.0 π • In SI System, power expressed in terms of watts π½ππ’ππ 1 πππ‘π‘ π = 1 π © Copyright 2014 ELO 2.2 Operator Generic Fundamentals 92 Work, Energy, and Power • Power is rate at which work is done or work per unit time π€πππ ππππ πππ€ππ = π‘πππ ππππ’ππππ • One horsepower is equivalent to 550 ft-lbf/s and 745.7 watts πΉπ£ π= 500 πΉπ π= π‘ Where: Where: P = power (W or ft-lbf/s) P = power (hp) F = force (N or lbf) F = force (lbf) d = distance (m or ft) v = velocity (ft/s) t = time (sec) © Copyright 2014 ELO 2.2 Operator Generic Fundamentals 93 Power Example A pump provides a flow rate of 10,000 liters per minute (lpm). The pump does 1.5 x 108 Joules of work every 100 minutes. What is the power of the pump? Solution π€πππ ππππ πππ€ππ = π‘πππ ππππ’ππππ πππ€ππ = 1.5 × 108 π 100 πππ 1 πππ 60 π ππ πππ€ππ = 25,000 πππ‘π‘π When using above equations, you must either assume force and velocity constant, or that average values of force and velocity used © Copyright 2014 ELO 2.2 Operator Generic Fundamentals 94 Power Example A boy rolls a ball with a steady force of 1 lbf, giving the ball a constant velocity of 5 ft/s. What is the power used by the boy in rolling the ball? Solution πΉπ£ π= 500 1 πππ π= 5 ππ‘ π ππ 550 π = 9.0 × 10−3 βπ © Copyright 2014 ELO 2.2 Operator Generic Fundamentals 95 Energy and Power Equivalences • 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: – 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 © Copyright 2014 ELO 2.2 Operator Generic Fundamentals 96 Energy and Power Equivalences • Joule showed quantitatively a direct correspondence between mechanical and thermal energy • Experiments showed: – One kilocalorie = 4,186 joules – One British thermal unit (Btu) = 778.3 ft-lbf • Engineering equivalences: – 1 ft-lbf = 1.286 x 10-3 Btu = 3.766 x 10-7 kW-hr – 1 Btu = 778.3 ft-lbf = 2.928 x 10-4 kW-hr – 1 kW-hr = 3.413 x 103 Btu = 2.655 x 106 ft-lbf – 1 hp-hr = 1.980 x 106 ft-lbf © Copyright 2014 ELO 2.2 Operator Generic Fundamentals 97 Energy and Power Equivalences • Mechanical equivalent of heat often denoted by J, Joule’s constant ππ‘ _ πππ π½ = 778 π΅π‘π’ • Other useful conversions, see formula sheet 1 π΅π‘π’ = 778 ππ‘ _ πππ π΅π‘π’ 1 ππ = 3.41 × βπ π΅π‘π’ 3 1 βπ = 2.54 × 10 βπ 106 © Copyright 2014 ELO 2.2 Operator Generic Fundamentals 98 Energy, Work, and Power Equivalences Energy – ability to do work Work – measures completed task Power – measures amount of work over time Power defined as time rate of doing work and is equivalent to rate of energy transfer © Copyright 2014 ELO 2.2 Operator Generic Fundamentals 99 Energy, Work, and Power Equivalences Practice Problem Solution An electric motor lifts a 1500 lb elevator 25 feet in 30 seconds. Assuming no friction losses, what is the horsepower required of the motor? πΉππππ × π·ππ π‘ππππ πππ€ππ = ππππ 1,500 πππ × 25 ππππ‘ = 30 π ππ ππ‘ _ πππ = 1,250 π ππ Since 1 βπ = 550 ππ‘ _ πππ 1,250 π ππ x ππ‘ _ πππ π ππ 1 βπ = 2.27 βπ ππ‘ _ πππ 550 π ππ © Copyright 2014 ELO 2.2 Operator Generic Fundamentals 100 Energy, Work, and Power Equivalences Knowledge Check A 600 lbm casting is lifted 4 feet to the bed of a milling machine. How much work is done? A. 240 ft-lbs B. 150 ft-lbs C. 2,400 ft-lbs D. 1,500 ft-lbs Correct answer is C. © Copyright 2014 ELO 2.2 Operator Generic Fundamentals 101 Properties of Heat ELO 2.3 – Define the following terms: heat, latent heat, sensible heat, unit used to measure heat, specific heat, and super heat. • Many thermodynamic analyses involve transfer of heat between systems and/or substances via thermodynamic processes and cycles • Heat is energy in transition caused by a difference in temperature © Copyright 2014 ELO 2.3 Operator Generic Fundamentals 102 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 © Copyright 2014 ELO 2.3 Operator Generic Fundamentals 103 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) – Negative value for heat indicates heat is removed from system by its surroundings (-Q) – Contrast with work - positive when energy is transferred from the system and negative when transferred to the system © Copyright 2014 ELO 2.3 Operator Generic Fundamentals 104 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) • Specific heat not used for q since specific heat used for another parameter – Quantity represented by q referred to as heat transferred per unit mass π π= π Where: q = heat transferred per unit mass (Btu/lbm) Q = heat transferred (Btu) m = mass (lbm) © Copyright 2014 ELO 2.3 Operator Generic Fundamentals 105 Properties of Heat Example Solution • Determine heat transferred per unit mass if 1,500 Btu’s are transferred to 40 lbm of water π π= π 1,500 π΅π‘π’ π = 40.0 πππ π΅π‘π’ π = 37.5 πππ © Copyright 2014 ELO 2.3 Operator Generic Fundamentals 106 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 • Everyone is familiar with physical phenomena that when a substance is heated, its temperature increases, and when it is cooled, its temperature decreases © Copyright 2014 ELO 2.3 Operator Generic Fundamentals 107 Latent Heat • Amount of heat added to or removed from a substance to produce a change in phase • When latent heat added, no temperature change occurs • Two types of latent heat – Latent heat of fusion is amount of heat added or removed to change phase between solid and liquid – Latent heat of vaporization is amount of heat added or removed to change phase between liquid and vapor o Sometimes called latent heat of condensation © Copyright 2014 ELO 2.3 Operator Generic Fundamentals 108 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 © Copyright 2014 ELO 2.3 Operator Generic Fundamentals 109 Properties of Heat - Specific Heat • Different substances are affected to different magnitudes by addition of heat • Specific heat is a measure of heat energy required to increase temperature of a unit quantity of a substance by a certain temperature interval © Copyright 2014 ELO 2.3 Operator Generic Fundamentals 110 Properties of Heat - Specific Heat π πΆπ = βπ π πΆπ = πβπ πΆπ = π βπ Where: Cp = heat capacity at constant pressure (Btu/°F) Cp = specific heat at constant pressure (Btu/lbm-°F) Q = heat transferred (Btu) q = heat transferred per unit mass (Btu/lbm) m = mass (lbm) βπ = 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 © Copyright 2014 ELO 2.3 Operator Generic Fundamentals 111 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 © Copyright 2014 ELO 2.3 Operator Generic Fundamentals 112 Properties of Heat Specific Heat Example Solution • How much heat is required to raise the temperature of 5 lbm of water from 50oF to 150oF? π πΆπ = πβπ π = ππΆπ βπ π = (5 πππ) Assume specific heat (cp) for water is constant at 1.0 Btu/lbm-oF 1.0 π΅π‘π’ (150β − 50β) _ πππ β 1.0 π΅π‘π’ π = (5 πππ) (100β) πππ_ β π = 5.0 × 102 π΅π‘π’ © Copyright 2014 ELO 2.3 Operator Generic Fundamentals 113 Properties of Heat Knowledge Check Which of the following must be added to or removed from a substance to produce a temperature change? A. Latent heat B. Specific heat C. Sensible heat D. Thermal heat Correct answer is C. © Copyright 2014 ELO 2.3 Operator Generic Fundamentals 114 TLO 2 Summary 1. Define the following thermodynamic properties. – Potential energy (PE) – energy of position – Kinetic energy (KE) – energy a body possesses as a result of its motion – Specific internal energy (u) of a substance – its internal energy per unit mass – Specific enthalpy β = π’ + ππ – Entropy – measure of inability to do work for a given heat transferred © Copyright 2014 TLO 2 Operator Generic Fundamentals 115 TLO 2 Summary 2. Explain the relationship between work, energy, and power. – Power is time rate of doing work o Equivalent to rate of energy transfer o Has units of energy per unit time – Power equivalences o 1 ππ‘ _ πππ/sec = 4.6263 π΅π‘π’/βπ = 1.356 × 10−3 ππ o 1 π΅π‘π’/βπ = 0.2162 ππ‘ _ πππ/sec = 2.931 × 10−4 ππ o 1 ππ = 3.413 × 103 π΅π‘π’/βπ = 737.6 ππ‘ _ πππ/sec – Horsepower related to foot-pounds-force per second (ft-lbf/sec) by following relationship: o 1 βπ = 550.0 ππ‘ _ πππ/sec © Copyright 2014 TLO 2 Operator Generic Fundamentals 116 TLO 2 Summary 3. Define the following terms: heat, latent heat, sensible heat, unit used to measure heat, specific heat, and super heat. – Heat described as energy in transit o Occurs on molecular level, result of temperature differences o Unit of heat is British thermal unit (Btu) – Latent heat is amount of heat added or removed to produce only a phase change – Sensible heat is heat added or removed that causes a temperature change – Specific heat is ratio of heat (Q) added to or removed from a substance to resulting change in temperature (βT), referred to as substance's heat capacity (Cp) – Super heat is number of degrees a vapor is above saturation temperature at a specific pressure © Copyright 2014 TLO 2 Operator Generic Fundamentals 117 Module Summary • This module introduced new terms and thermodynamic properties • Thermodynamic property is parameter describing a physical condition of a substance • When state is defined by two independent intensive properties, other properties are implied • Properties are divided into two classifications, extensive and intensive – Extensive properties are mass dependent and include examples such as mass and volume – Intensive properties are independent of total mass and include examples such as temperature, pressure, and specific volume © Copyright 2014 Summary Operator Generic Fundamentals 118 Module Summary • Temperature is a measure of average kinetic energy of atoms of a substance, measured in degrees • Four temperature scales: Fahrenheit, Celsius, Rankine, and Kelvin • Pressure is a force per unit area acting on a fluid, usually measured in psig, psiv, psid, in. Hg abs., or in. Hg vac. – Absolute pressure given in units of psia • Specific volume is amount of space a unit of mass occupies and measured in cubic feet per pound mass – Inverse of specific volume is density © Copyright 2014 Summary Operator Generic Fundamentals 119 Module Summary Now that you have completed this module, you should be able to do the following: 1. Describe thermodynamic properties and methods of measuring intensive and extensive properties. 2. Explain the concepts of heat, work, and energy. © Copyright 2014 Summary Operator Generic Fundamentals
