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# 502 Class Note for ME 20000 at Purdue

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Date Created: 02/06/15

ME 200 Thermodynamics l Session 2 Fall 2008 Class information Prof S H Frankel Class time MWF 10301120AM Of ce ME 165 or Chaffee 125 MWF Email 7 rrarrlltelgurdue edu r rrarrlltel Stevengrnall corn Phone 7 76549471507 of ce 7 76540476067 cell Of ce Hours ME 165 MWF1130 1230 Course website httpengineeringpurdueedumeZOO Research website httpllristrettoecnpurdueedu Outline Course details Chapter 1 Getting started Introductory Concepts and Definitions Using thermodynamics De ning systems Describing systems and their behaviors Measuring mass length time and force Properties Speclflcvolume Pressure Temperature Problem solving methodology Course details En ineerin Course policy Thgrmodyngamics Read items 1 4 Textbookwebsite HW problems will be collected on Monday s Includes problems from previous M W F Only subset of problems will be graded Graded HWs returned to you usually within 1 week Read items 5 14 Schedule note exam datetimelocation Equation sheet bring to exam Basic Concepts What is thermodynamics Science of energy What is energy Ability to cause changes Thermodynamics from Greek word therme heat amp dynamis power heat to power Key principle is conservation of energy During interaction energy can change from one form to another but total amount of energy remains constant ie energy cannot be created or destroyed Potential energy of weight converted into thermal energy of water f ix Egiwhvnni l Joule apparatus 1 843 Laws m First law of thermodynamics A statement of conservation of energy principle Energy is a thermodynamic property tbd quantifies energy Second law of thermodynamics Energy has quality as well as quantity Actual processes occur in direction of decreasing quality of energy Example A cup of hot coffee left on a table eventually cools 55 but a cup of cool coffee left on a W 4 table never gets hot by itself degradation of high temperature energy why a degradation History Birth of thermodynamics as a science 1697 1712 atmospheric steam engine 1850s William Rankine Rudolf Clasius and Lord Kelvin Rankine first textbook in 1859 First ME 200 student falls asleep in class August 1860 Only two laws how hard a subject can this be Substances consist of large number of particles called molecules 5L N Two approaches 5 l Classical approach does not require 39 detailed knowledge of molecular motion but a little does not hurt 395 g Statistical approach considers 1 molecular motion in detail We shall pursue the classical approach but draw on molecular details for insight eg we will use to understand internal energy entropy and ideal gas aw Application areas Home electricgas range HVAC refrigerator humidifier pressure cooker water heater shower etc Automotive engines rockets jet engines conventional or nuclear power plants Even human body l l l l I httpwwwrollsrovcecomeducationschoolshow thinqs workiournev02flashhtml Systems Closed and open Thermodynamic system quantity of matter or a region of space chosen for study Mass or region outside system is called surroundings Real or imaginary surface that separates system from surroundings is boundary SURROUNDINGS BOUNDAR Y Boundary may be fixed eg rigid tank or moveable eg pistoncylinder device Closed system control mass Features Fixed amount of mass No mass can cross its boundary Energy in form of heat and work interactions can cross boundary Volume of closed system does not have to be fixed mass NO CLOSED SYSTEM m constant YES Open system control volume Properly selected region of space usually encloses device which involves mass flow ie compressor turbine or nozzle Both mass and energy can cross the boundary of a CV which is called a control surface CS Note Form of thermodynamic relations are different for open and closed systems so it is important to identify what type of system you are considering first to determine proper analysis Control volumes Imaginary boundary Real boundary gt VI r c F m i 1 l l I 1 1 1 Moving 1 CV gt 1 boundary I 1 a nozzle 1 1 1 1 f z l CV l I 1 Fixed 1 I a A control volume with real and b A control volume with fixed and imaginary boundaries moving boundaries Properties of a system Any characteristic of a system is a property ie pressure P temperature T volume V mass In etc Some properties are defined in terms of others ie density is mass per unit volume m 3 p kg m gt V Specific volume Note Pointwise property definitions based on concept of continuum ie spaces between molecules ignored no microscopic holes valid for most applications Intensive vs extensive properties Intensive independent of size of system ie T P or dens y Extensive depend on sizeextent of system ie m V E Does property change when system is divided in half Extensive properties per unit mass are called specific properties and are intensive ie specific volume 1 Extensive 12 value I 1 Intensive same value State Consider system not undergoing any change measure properties of system this defines state A system at 2 states Equilibrium Thermodynamics deals with equilibrium states Equilibrium implies a balance ie no unbalanced driving forces If isolated from surroundings system remains unchanged Thermal equilibrium implies temperature same throughout eg see below for before and after ll No temperature differentials system temperature can be described by a single number in equilibrium Mechanical no change in pressure Chemical no change In composntlon Thermodynamic all aspects of system in equilibrium Process path Any change system undergoes from one equilibrium state to another is called a process Series of states through which a system passes during a process is called path of process Property A Advantage Only one value for each State 2 property describes entire system eg one value of P T etc State of system represented by processpath single point on plot w properties as coordinates State 1 Path can only be drawn if processes Pr t B Oper y proceed In eqUIIIbrIum manner Quasiequilibrium process When process proceeds so that the system remains infinitesimally close to an equilibrium state at all times we call is quasiequilibrium QE process Process occurs slow enough to allow system to adjust properties win system so that one part of system doesn t change any faster than other parts QE process is approximation to real process useful because a easy to evaluate and b deliverrequire mOStleaSt work Higher p etc near piston Pressure density etc change uniformly lower p near cylinder head Slow compression QE process Fast compression non QE process PV property diagram P Illustrates states Finalstate and process path To connect states with line must be QE process Why Process path Initial state 1 Hlt lt Ii I I I I i I I I 2 1 More processes and cycles Other important processes involve particular constant properties such as isothermal T isobaric P isochoric V A system is said to have undergone a cycle if it returns to its initial state at end of process ie initial and final states of cycle are same E E State postulate How many properties must be speci ed to x the state of a system We need only specify a certain number of properties to x the state of system The exact number depends on type of system eg how complex is it For a system that only involves QE compressionexpansion eg a simple compressible system state is completely speci ed bytwo independent intensive properties Two properties independent if one can be varied wo changing other eg T P away from phase change Dimensions and units Physical quantities characterized by dimensions mass m orvelocity V etc Arbitrary magnitudes assigned to dimensions are units kg ms etc Basic dimension mass m length L time t temperature T are chosen as primary or fundamental dimensions Others velocity V energy E volume V expressed in terms of primary dimensions are called secondary or derived dimensions Units Two sets of units still used a Englisn uscsy no math oasis arbltl ary confusll lg dlfflcuit to learn e g A grade Woes a Sl a decimal oase simple loglcal SI system ofunits m L t T are primary with base units of kg m s K respectively Force is a derived dimension in SI via Newton s 2 law d 1 L L F saw Fla lMlHa MlH Force unit 1 N newton 1 kg 1 ms2 1 kgms2 1 Newton is force required to accelerate 1 kg mass at rate of 1ms2 More on units English system of units F m L t T are primary Force is not derived through Newton s law force and mass independently defined But since force can be derived from Newton s law the system is overdetermined hence FiiiMiL So we must write Newton s law as F ziimV with g 32174mmJt gc d lbf s Dimensional constant gc Still more on units If instead we define a force unit as derived from Newton s law with proportionality constant of 1 then force is secondary dimension Then we could simply define 1be 32174lbm 2 S For mconst Fma In order to have dimensional homogeneity same units for all terms in equation gc must be introduced such that if one applies 1 lbf to 1 lbm in standard gravity ie g32174 st2 so Fma is satisfied Hence 1be M g0 lbm fts2 321744 2 g 32174 s lbf Unit conversions 1 lbm 045359 kg 1 ft 03048 m 1 N 1 kg msZ 1 lbf 32174 lbm ftsz azlmls2 mzlkg F1N m 321741bm a1ft32 F11bf Weight Weight is gravitational force applied to body W m g N g is local gravitational acceleration constant 981 ms2 32174 fts2 at SL Mass of body is constant with location but weight changes with local gravitational acceleration eg gmoon1l69eanh etc Example An object occupies a volume of 25113 and weighs 20 lbf at a location where the acceleration of gravity is 310 s2 Determine its weight in lbf and its average density in lbm 3 on the moon where g557 s2 V25 3W201bfg310 s2 lbmi 20W 3217472 W W 7 S 2081b mg m g 311m2 llbf m 1 32174lb39quot S W mg 2081bm557 x2 361bf pmV2081bm25 3 08321bm 3 Work and more Work is a form of energy defined loosely as force times distance SI unit is newtonmeter Nm called Joule 1J1Nm 1kJ10quot3J English unit is Btu amount of energy required to raise temperature of 1 lbm of water at 68F by 1F 1Btu 1055 kJ Dimensional homogeneity don t add apples and oranges every term in an equation must have same units eg E 25 kJ 7 kJkg 1O Temperature Thermal equilibrium implies equality of temperature I 0m law of thermodynamics if two bodies are in thermal equilibrium w a 3rd body they are in thermal equilibrium with each other Replace 3rd body with thermometer and you have basis for measuring temperature Hence 2 bodies are in thermal equilibrium if they have the same T wo bringing them into contact Temperature scales two approaches First approach Material properties change with temperature in a measurable and predictable way ie volume of Hg in glass tube Actual temperature define by fixing two temps and assigning numbers Celcius scale 00 is temp when water freezes at 1 atm and 1000 as temp of boiling water at 1 atm and divide by 100 centigrade Fahrenheit scale 32F212F resp as ice point and steam point TF18TC32 Temperature scales cont Second approach Independent of properties of any substance based on 2 d law tbd Thermodynamic or absolute temp scale or Kelvin scale ie unit is Kelvin K Assign reference T as 27316K as triple point TP of water point where SLV phases coexist Freezing point of water is 001K below TP so 27315K TC TK 27315 Rankine scale TR18TK so 1K18R TP of water is 49169R TFTR 45967 11 Comparison of temperature scales C K 0F R quot quot Boiling 10000 37315 21200 67157 point of water at 1 mm 001 27316 3202 49169 Triple poml of water 27315 0 45967 0 Absolute yero Pressure WK EL Consider fluid liqgas at rest W De ne pressure P lim FA Molecular collisions AeA with test surface Rotating surface to new orientation and measuring P will get same answer ie isotropic at a point equal in all directions for fluid at rest Pressure in fluid does increase with depth due to weight of fluid negligible for gas RgtE within same uid P increases More on pressure If fluid is in motion net forcearea can be divided into 3 mutually perpendicular stresses FA one normal and 2 tangential shear wrt surface Magnitude of stress is a function of surface orientation when fluid is in motion fluid dynamics We assume for our problems normal stress at a p0int isapprOXImately pressure w or wo fluid motion 12 Units absolute and gage pressure SI units of pressure are forcearea are Nm2 called Pascal Pa 1 Pa 1 Nm2 1 kPa 10quot3 Pa 1 MPa10quot6 Pa 1 bar10quot5 Pa 01 MPa 100 kPa 1 atm 101325 Pa 101325 kPa 101325 bar English lbfinquot2 or psi and 1 atm 14696 psi Actual pressure at a given position is absolute pressure measured relative to absolute zero pressure vacuum no molecules no collisions Most pressure measuring devices are calibratd so that zero pressure is Patm which gives the gage pressure Absolute gage and vacuum pressures Pgage Pabs Patm for Pabs gt Patm Pvac Patm Pabs for Pabs lt Patm In thermo relations P is usually Pabs unless otherwise P noted by g subscript for gage me PMJ1 P Absolute Absolute vacuu m Vacuum If gage shows Pg100psig absolute pressure is PabsPgPatm1001471147psia Measuring pressure Manometer device used to measure small to moderate pressure changes Glassplastic Utube containing fluid such as H20 alcohol Hg Heavy fluids used for larger pressure differentials to keep device small We wish to measure pressure P of gas in tank rt Neglecting weight of gas P same everywhere in tank for gas and at position 1 also PP1P2 Consider fluid column of height h and draw FBD 13 Manometer calculation FBD of fluid column at equilibrium balance of forces says ZF0 AR APazmmg APmpVg APmpAhg 2131 Pam AP Pg1 pgh kPa Barometer meas atm pressure r PP B arm PC 0 Pm pgh kPa Problem solving technique Step 1 Problem statement In your own words state problem list key information given and quantities to be found Step 2 Schematic Draw realistic sketch of physical system choose your system list key information on figure indicate massenergy interactions with surroundings check for constant properties Step 3 Assumptions State your assumptions needed to simply and solve problem assume reasonable values for missing data 14 Problem solving technique cont Step 4 Physical laws 7 Apply releyant pnysical laws to cnosen system e g cons energy reduce to simplest form using assumptions for your system Step 5 Properties 7 Determine unknown properties at known states from property relations or tables list properties separately and indicate source Step 6 Calculations 7 Substitute known quantities into Slmpllfled relations and perform computations to find unknowns watcn Lll lltS round results as appropriate Step 7 Reasoning veri cation and discussion a cneck to make sure results make sense conclusions recommendations rtnink orsolution as engineering analysisi Summary System closed oropen properties eg T p change during a process from one equilibrium state to another resulting in less potential to do useful work all the while conserving energy 15

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