Thermodynamic Principles ME 2233
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Date Created: 09/17/15
TurbinesCompressors Rotating blades and shafts either extract work from or perform work on the system resulting primarily in changes in system ressure Turbine extracts work Compressor inputs work Typical Assumptions When we say turbine or compressor we mean Adiabatic rate ofheat loss is small compared to War ss no changes with respect to time Ape 0 Ake 0 small relative to work term 1 inlet1 outlet TurbinesCompressors Continuity mm my m st 1 Law Me 7 AKB 7 0 SS r 2 Adiabatjc Combining W thl 504411 hgthE turbine hlthE compressor Examples Steam enters a turbine operating at steady state at 700 F and 600 psi andleaves at 0 5 psi with a quality of90 u The turbine develops 12000 hp and heat transfer from the turbine to the surroundings ccurs at a rate on 5x10 Btuhr Neglecting kinetic and potential energy changes determine the mass ow rate ofthe steam lbhr A wellinsulated compressortakes in air at60 EM 2 psi with a Volumetric ow rate of 1200 zrnin and compresses it to 500 F 120 psi Kinetic and potential energy changes are negligible Determine the compressorpower hp and the volumetric ow rate at the exit himin Pumps Compressor is usually used for gas systems while pump is usually used for liquid systems The analysis is the same W mat a he Pum assum tions When we say pump we mean All compressor assumptions w Incompressible uid Constant speci c heat ifyou need this mQATww Usually the pressure increase is large Often we will assume will justify this later so that WMWE E Examples A pump steadily delivers water through a hose terminated by a nozzle The exit ofthe nozzle has a diameter of6 cm andislocated 10 m above the pump inlet pipe which has a diameter of12 cm The pressure is equal to 1 bar at both the inlet and exit and the required by the pump is 1 5 kW Findthe mass ow rate kgs An oil pump operating at steady state delivers oil at a rate of12 lbs through a lin diameter pipe The oil is incompressible has a density of100 lb z and experiences a pressure rise from inlet to exit 39 There is no signi cant elevation difference between inlet and exit and the inlet kinetic energy is negligible Heat transfer in v hp increments determine the rating of the pump needed for this application Heat Exchangers The purpose ofa heat exchanger is to transfer heat from one working uid to another Contact mixer NonContact heat exchanger Typical assumptions When we say heat exchanger or mixer we mean Adiabatic W0 overall only work is ow work no changes with respect to time negligible small change in Velocity Heat Exchangels and Miers Continuity Z 39 e llets extlx lst Law Ape Ake 0 SS Adiabatic no work C ombining Z We Z Wight 0112er EMS Mixer 2 inlets 1 outlet Mass Conservation 2 2mg Inlet exits l5t Law lulu 2M 215 ml m1m2m3 ritlh1 ritth r513h3 Example A feedwater heater in a Vapor power plant operates at steady state with liquid entering at inlet 1 with T140 C p1 W or at T 200 C pl 7 bar enters at inlet 2 Saturated liquid water exits with a pressure p57 bar kinetic and potential energy effects determine the ratio of mass ow rates rnl rn Heat Exchanger 2 separated mass ows Mass Conservation 2ch 2mm Xmas Xmas mlet 2x12 inlet mt m1A m2A mA l5t Law 2 lm mash5 Z tzlhtzA marshals mlcrr cm m1AhlA m3Bh3B m2Ah2A m4Bh4B mAhlA h214 m3h43 h33 Example R134a eneters a heat exchanger in a refrigeration system operating at steady state as saturated vapor at o F and exits at 20 F with no 39 essure A s exiting at a 1ower temperature whi1e experiencing no pressure drop have e Vual mass ow rates Neriectinr kinetic and otenu39ai ener effects determine the exit temperature ofthe liquid stream F Control volume may also be applied to a single stream In this case the system is not adiabatic Q W101 h QA Q B rhAhl hz rh5h3 7114 Example uuu I am 39 at I 39 39 hnudng lled with electronic components At steady state water enters the jacket at 20 c and exits with a negligible change in pressure at a C quot L 39 receive 2 5 kW of electrical power There is no signi cant transfer by heat from the outer surface ofthe water jacket to the surroundin s andkinetic and otential ener effects can bei nored 39 39 39 rat in atria m in the limit on the temperature ofthe exiting water is met ENERGY ANALYSIS OF CYCLES FIRST LAW OF THERMODYNAMICS Conservation of energy for closed system Q W AE Thermodynamic Cycle Sequence of process paths returning to original state Y Property dr0 Cycle returns system to initial state No net change in system energy Qcycle chcle POWER CYCLES Conversion of thermal energy to useful work Qin cannot be converted perfectly to W cycle Qout Will M be present Ichcle 1 77 power Qin Qin Thermal ef ciency Hot Body Cold Body REFRIGERATION CYCLE Work is used to extract heat from cold body QOut Will be greater than Qin COP 3 g Qm Q re 1 W Q Q cycle out m Cold Body HEAT PIHVIP CYCLE Work is used to supply heat to hot body Qout will be greater than Qin COPI 2 Qoul Qout hp chycle Qout Qin Hot Body Wc cleQout39Qin Cold Body Example The thermal ef ciency of a power cycle operating as shown in the gure is 30 and Qom650 MJ Determine the net work developed and the heat transfer Q each in MI in Hot Body cycleQinQout Cold Body Example A power cycle has a thermal efficiency of 35 and generates electricity at a rate of 100 MW The electricity is valued at 008kWh Based on the cost of fuel the cost to supply Qin is 450GJ For 8000 hours of operation annually determine in 35 a the value of electricity generated per year b the annual fuel cost Example A gas undergoes a thermodynamic cycle consisting of three processes Process 12 compression with pVconst from p1l bar V1l6 m3 to V2O2 m3 UZU1O Process 23 constant pressure to V3V1 Process 31 constant volume UlU33549 kJ There are no signi cant changes in kinetic or potential energy Determine the heat transfer and work for Process 23 in kJ Is this a power cycle or a refrigeration cycle Example problem Known My coffee is cold room temperature m025 kg 1 microwave it which transfers 70 k of radiation heat transfer to the liquid 1 also stir it vigorously adding 10 Id of energy paddle wheel work What is the change in internal energy What is the temperature change this is what I really care about Analysis State Principle The eguilibrium state of a simple pure substance is xed by specifying m independent intensive 1 r0 erties Equilibrium state balance of all forces no gradients in system Pure uniform and constant chemical composition homogenous Simple only one relevant quasiequilibrium work mode Most important case is simple compressible system for which only quasiequilibrium work mode is WjpdV IMPORTANT two intensive properties fix the state but a single extensive property usually mass must also be known to specify the size of the system Important consequences of state principle We know calculated were given measured p and T gt can look up v and u v and T gt can look up p and u u and T gt can look up p and v p and u gt can look up v and T this would be the case for the cup of coffee etc Three ways of looking up these property relationships 1 Graph pvT surface 2 Table property tables 3 Equation equation of state Today we will look at the graphical technique THE pvT SURFACE Threedimensional representation of equation of state for simple compressible substance pfT V Best representation of phase transformations solid liquidgas shows one two and threephase regions along with saturation states between them Phase Changes Sublimation solidgas Evaporation or condensation liquidgas Melting or freezing solidliquid i ummnl prcx39xure line Pn shure iC rilicul 5 a Q TgtZ T Triple puinl I39lt 739 Tcnqwl39ulurc Speci c mlum 1h 1 F39 12 liL39iTsurl ucc and projections for 21 substancc that L UI III39HCb un frcu ing a 39I hrccdimcnxinnnl View Iv Phase diagram 0171 diagram Phase Change Process Consider phase change process at constant pressure ie pure water in sealed cylinder A Mass P1atmT20 C Compressed liquid B Mass P1atmT100 C Saturated liquid 1 11111111111 V Increases Slightly Q Phase Change Process P1atmT100 C Liquidvapor mixture V increases dramatically P1atmT100 C Saturated Vapor V increases Mass E P1atmT300 C Superheated vapor V increases as about VT h Phase Change Process Plot T vs Volume for constant pressure phase change process 350 300 250 200 150 temperature C 100 v a v b v c V d v e volume mno Note T and p are NOT independent properties for a liquidvapor mixture 2phase region Critical pninl Saturated vapor a Vapor 5 5 e v gt uE E I Speci c vulume Different types of 2D diagrams Phase Diagram pT diagram demonstrates that at all phase change locations saturation curves pressure and temperature not independent pv Diagram Most commonly used diagram for 1 rocess Visualization Clear re resentation of boundary work W I pdV T v Diagram Usually replaced by T s diagrams later which allow heat transfer Visualization similar to work Visualization in pv diagram Could also construct uT diagrams vu diagrams etc Important De nitions Critical point saturated liquid and vapor lines meet no differentiation between liquid and gas uid maximum pressure for separate liquid state Critical temperature TC temperature at the critical point Critical pressure PC pressure at the critical point maximum P and T for which liquid and vapor coexist Saturation temperature Tsat the temperature at which a phase change takes place at a given pressure Saturation pressure PM the pressure at which a phase change takes place at a given temperature Compressed liq we subcooled liquid liquid in which the pressure is greater than the saturation pressure or equivalently the temperature is lower than the saturation temperature Superheated vapor vapor in which the temperature is greater than the saturation temperature Isotherm curve of constant temperature Isobar curve of constant pressure g 3 mvapor x mliquid mvapor Vapor Dome vaporliquid twophase region Triple line SolidLiquidVapor equilibrium a single point triple point on PT diagram Property Tables Water Saturated water liquidvapor Tables A2 A3 Superheated water Table A4 Compressed water supercooled Table A5 Saturated water solidvapor Table A6 Refrigerant 22 Tables A7 A8 A9 Refrigerant 134a Tables AlO Al l A12 Ammonia 39l39ables Al3 Al4 AlS Propane Tables Al6 Al7 A18 Ideal Gases Tables A20 through A23 British units Table A2E through A23E The equilibrium state of a simple pure substance is xed by specifying 1m independent intensive properties Property Tables 1 Superheated vapor fp T 2 Compressed liquid f19T 3 Liquidvapor region 0f Procedure for using tables 1 Go to the saturation table 2phase for the correct substance 2 Determine the phase of the substance 3 Go to the correct phase table for the correct substance 4 Look up the property Examples 1 What is the internal energy of water at p1 bar and T200 C 2 What is the pressure of water at T200 0C and v04249 m3kg 3 Liquid water and steam coexist at p30 bar What is the temperature and internal energy 4 Why can t you make a good cup of coffee on top of Mt Everest Interpolation Most otten the state of the system does not tall exactly on a table grid point What is the internal energy of water at p3 bar and T180 C Could use value at T200 0C closest grid point which yields l2 error UK interp01ate between gr1d p01nts Dy assum1ng that the missing data is linear uC 1u0 We know u1 T1 and u2 T2 the two closest grid pOIHIS We want u3 T3 u1CT1u0 u2CT2u0 Subtracting uluz CTlTz C ulu2T1T2 u0 ulT1u1u2T1T2 uzT2u1u2T1T2 u3 T3u1u2T1T2u1T1u1u2T1T2 u3 ulu2T3T1T1T2 u1 On all homework always interpolate if necessary On quizzes and exams use the closest grid point if you do not have time to interpolate TWOPHASE REGION For a given temperature there is only one pressure For a given pressure there is only one temperature v and u and h and s used later are de ned at the saturated liquid and saturated vapor point vf function of T g p only speci c volume if all of the substance is liquid vg function of T g p only speci c volume if all of the substance is vapor Recall that the quality X tells us how much mass of vapor versus liquid is in our system This is often our second property required to X the state Thus use tables to evaluate vg and vf for twophase mixture the average speci c volume is g Vf g mtot mtot v vf v l xvf xvg v vf xvg vf v vf xvfg Similarly for any mixture property Ym Yf xYg Yf Yf fog Calculation of quality from properties x W Yfg Examples 1 What is the internal energy of water at p25 bar and x05 2 What is the internal energy of water at T180 0C and x03 3 What is the quality of water with v 01 m3kg at T150 C 4 What is the temperature of water at p30 bar and u2000 kJkg u3000 kJkg INTERNAL ENERGY and ENTHALPY Recall that total energy E KE PE U Internal energy Measure of energy stored at the molecular level UUTVm uuTvUm Enthalpy HUpV hupvhTv What are the units Enthalpy is a completely derived quantity It is not necessary but the term upv shows up so often especially in open system analysis that h is a time saVing quantity to tabulate Values of u and it listed relative to a reference state For simple compressible system the differential form of the first law of thermodynamics is Q Vz where W pd V gt 5Q dU pd V Thus for a constant volume process QzAU For a constant pressure process QzAH Example 1 What is the enthalpy Btulbm of water at 30 lbfin2 and 420 0F First What phase is it in Double interpolation Neither data point falls on a grid point thus we must interpolate in two directions EXAMPLE Given System consisting of 2 lb of water undergoes the following cycle l2 adiabatic compression from p1200 psi X1 90 to p2600 psi T2500 C F 23 isothermal process with W23 2362 Btu to p3 200 psi 31 isobaric process Find Net work and heat transfers Assumptions 1 closed system 2 AKEAPEO Solution SPECIFIC HEATS u uTv du dT dv 3T v av T duzcvdT j dv 3V T a CV W V cV is called speci c heat hhTpupv dh dT dp 5T p 3P T cp is also called speci c heat SPECIFIC HEATS Notes 1 cV and cp are properties of a singlephase substance just like u v h etc They can be a function of temperature and pressure 2 speci c heat can be characterized as a measure of energy required to change the temperature of a substance 3 the ratio of speci c heats is often used in calculations IMPORTA speci c heats are only specify energy changes of a substance due to temperature A system s energy can also change because of pressure or phase changes and these must be considered separately For a simple compressible closed system 1 Constant Volume du cvdT dv 3V T AU 2 m If CVdT 2 Constant Pressure dh cpdTj dp 51 T T2 AH m L cpdT NOTE you must substitute the function cVT or cpT and carefully integrate Table A2021 What if the speci c heat is constant LIQUID AND SOLID PROPERTIES Compressed liquid tables are not available for most substances However pressure has only a small effect on the internal energy or speci c volume of solids and liquids Therefore u and v for a compressed liquid are essentially equal to the values for a saturated liquid at the same temperature Thus vTpz MT uTpzufT Convince yourself using water tables hTpz ufTpvfT hTpz hfTvfTp psatTl INCOMPRESSIBLE SUBSTANCES We can further assume that v constant see the water tables Implication equal speci c heats Table A19 8h au gt 5flgflcwdT Often incompressible substances especially solids will be assumed to have a constant c AuchT AhcATvAP Example Problems A quantity of water is at 15 MPa and 100 C Evaluate the speci c volume m3kg and the speci c enthalpy kJkg using a Data from Table A5 b Saturated liquid data from Table AA A wellinsulated copper tank of mass 13 kg contains 4 kg of liquid water Initially the temperature of the copper is 27 C and the temperature of the water is 50 C An electrical resistor of negligible mass transfers 100 kJ of energy to the contents of the tank The tank and its contents come to equilibrium What is the nal temperature C THE GENERAL GAS Compressibility factor Z de ned as Z 1i RT RT If we know Z we have a relationship between pressure speci c volume and temperature for the gas Z is plotted versus reduced temperature and reduced pressure PR NINFN TR Figure 312 and Fig Al from text plots Z as a function of PR and TR Pseudoreduced speci c volume is de ned as This is also available on Figs Al 2 3 Thus if two reduced properties are known the others can be found consequence of the state principle W39henisZ l THE IDEAL GAS When pressure is small or temperature is large Z is about one ideal gas assumption implies Z1 pRlt 003 and TRgt 10 2 1 within1 pRltl0andTRgt20 Z1within2 THE IDEAL GAS LAW Molar Basis Equations of state relating p v and T pV NRT p pressure Nmz or psia V volume m3 or N N number of moles kmol or lbmol T temperature K or R i universal gas constant E 8314 kJkmolK R 1986 Btulbmol R THE IDEAL GAS LAW Mass Basis pszT Note that is gas speci c Gas Constant R 1 M Mmolec ar weight amu kgkmol lbmlbmol Other forms in pRT pVmRT We Will show in detail next lecture Internal energy and enthalpy are only a function of temperature for an ideal gas Ideal gas tables Table A2023 are thus tabulated versus only temperature Question Why is enthalpy hupv not a function ofpressure for an ideal gas We have now covered all of the tables and charts that you Will use in ME 2233 Example Given Tank containing 01 m3 of steam at 400 C and 10 MPa Find Mass of steam using a ideal gas law b compressibility chart c superheated vapor es Which is correct Example Problems A rigid tank contains 0 5 kg of 02 initiany at 33 bar and 150 K The gas is heated and the pressure increases to 40 bar Determine the volume ofthe tank m3 andthe nal temperature K Two lbmol of c2114 initia11y at 213 psi and 512 Ris compressed at constant pressure in a pistoncylinder assernb1y For the gas w 800 Btu Determine the nal temperature as IDEAL GAS INTERNAL ENERGY For an ideal gas internal energy is a function oftemperature alone Recall u T du dr jm W V 5v T d11chT dv 5v T a Foranideal gas 0 based on experiments aquot 7 Independent ofprocess d11chT Au chT i IDEAL GAS ENTHAPLY Recall h hT P dh dr j dp W p 7 7 5h dh dT 0P laid De nition h u pv However pv RT Thus h u RT Enthalpy is function of temperature only dhduRdT For an ideal gas enthalpy is a function oftemperature alone IDEAL GAS INTEGRAL EVALUATION Three methods to evaluate changes in u h Au chT Ah CPdT Constant Speci c Heat Assumption TzuT1 CVT2 T1 hTzThTr CFT2 T1 Convenient but inaccurate for large temperature changes Only use for small AT To minimize error evaluate the speci c heats at an average temperature T T T2 2 avg IDEAL GAS ENTHAPLY dh chT RdT dh CV mar dh cpdT T2 Ah IT cpdT As shown c cv R IDEAL GAS INTEGRAL EVALUATION 2 Perform integration with knon function of cv and cp Table A2 cp tabulated but cv is not 7 2 4 s cp7Ra TyT 5T 5T Then 2 Ahja TyTz5T4gTsdT Tl Note to get Au Au Ahi Apv Ahi RAT AuhrhinRTrTi OrcvcpR IDEAL GAS INTEGRAL EVALUATION T uT um chT 7rd T hT hm IcpdT TM hm Oat Tm 0 K The reference point has been selected as Au ma 7 um Ah mm 7 mm Thus 3 Use tabular data Tables A17 to A25 given as Examples 1 What is the internal energy ofnitrogen at 300 K note difference in units for gases other than air I 2 What is the change in enthalpy H air 7 m 300 K to 400 K a use the tables b use a constant C P What about from 300 K to 1000 K 2000 K Which is better EXAMPLE Given Air in pistoncylinder device is heated at constant pressure Find Final temperature and heat transfer Schematic and Given Data Assumptions 1 ideal as 2 quasistatic process 3 AKE Analysis Example Two uninsulated rigid tanks contain air Initially tank A holds 1 lb of air at 1440 R andtank B holds 2 lb of air at 900 R The initial A p opened Eventually the contents come to equilibrium at the temperature ofthe sunoundings 520 R Assuming ideal gas rressure Vsi We are nished with closed system analysis Elements ofthis analysis Definitions of systems states and properties Mass is conserved constant Energy is conserved AuAketAchqw State principle 2 intensive properties x state vwa Tables and graphs for evaluating properties Now you are faced with a multistep complicated problem Today s Ob39ec ive How do all ofthese pieces t together How do I actually solve a problem using all ofthe things that have been discussed in class Examples One lb of air undergoes a power cyc1e consisting ofthe following processes 12 constant volume from p20 Ibfinl 1500 R to T2820 R 23 adiabatic expansion to v31 4v2 31 constantpressure compression Sketch pv diagram determine a p2 b T3 c ef ciency ofthe cyc1e Procedure for solving multistep problems BE GOAL ORIENTED 7 How do I get to the answer 1 What amI supposedto find What do I already know How much ana1ysis is necessary Can I simp1y draw a pv diagram 39 stem What Is m 2 How many states are involved on Make a table ofproperties for each state may be unnecessary as more experience is gained Ea for me to give partia1 credit Draw a pv diagram ifyou can p1ot it the states are xed m or V 4 Does each state need to be xed to solve this problem How will you x each state 2 intensive properties required Look for information about process paths const Pressure volume etc T is may require the rst law or ancillary equations eg WipdV Some exceptions to the rule a IfI am only interested in internal energy for an ideal gas I only need temperature 1 property and may not be given pressure b Likewise for an incompressible substance we may only be given temperature what assumption will you make 5 Once the states are suf ciently xed you can evaluate any properties that are required Evaluate these properties using the correct technique a Incompressible assumption vv AucAT b 2 phase substances 7 Table c Superheated vapor 7 Table d Ideal gas 7 Equation and table also cp cv simpli cations e Compressible gas 7 Z chart Unless speci cally asked to do so do not waste time evaluating properties that are not needed 6 Ifyou must evaluate energy terms START WITH FULL FIRST LAW EQUATIONH AU APe AKe Q 7 w closed system Which ofthese terms can you evaluate What do you do with the remaining terms Are the assumptions you have made reasonable This equation can be applied to each process or to the entire cycle QyWy 7 Plug intoyour equations using the properties from the table 8 Have you solved for everything required ef ciencies etc Put a box aroundyour answer Ideal Gas Processes A special case occurs when PVRT for isothermal processes Pvconst Polytropic nl constmRT Thus isothermal ideal gas is another way of telling you polytropic nl Example Find the work required to compress air isothermally 300 K from 1m3 to 05 m3