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# THERMODYNAMICS ENSC 2213

OK State

GPA 3.78

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This 0 page Class Notes was uploaded by Ms. Abe Bernier on Sunday November 1, 2015. The Class Notes belongs to ENSC 2213 at Oklahoma State University taught by Arland Johannes in Fall. Since its upload, it has received 27 views. For similar materials see /class/232955/ensc-2213-oklahoma-state-university in Engineering & Applied Science at Oklahoma State University.

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Date Created: 11/01/15

Road Map for Locating Fluid States in Property Tables I For Temperatures Below the Critical Temperature Given Temperature and Pressure T 1 Go to the saturated temperature table and determine if p gt pST gt Subcooled compressed liquid p pST gt Saturated state may be saturated vapor saturated liquid or twophase mixture p lt pST gt Superheated vapor OR Go to the saturated pressure table and determine if T lt Tsp gt Subcooled compressed liquid T Tsp gt Saturated state may be saturated vapor saturated liquid or twophase mixture T gt Tsp gt Superheated vapor Given Temperature and Speci c Volume T v Go to the saturated temperature table and determine if v ltVfT gt Subcooled compressed liquid v vfT gt Saturated liquid VfT lt v lt vgT gt Saturated twophase state v vgT gt Saturated vapor v gt vgT gt Superheated vapor Road Map for Locating Fluid States in Property Tables continued Given Pressure and Volume 19 v Go to the saturated pressure table and determine if V lt Vfp gt Subcooled compressed liquid V Vfp gt Saturated liquid Vfp lt Vlt Vgp gt Saturated twophase state V Vgp gt Saturated vapor V gt Vgp gt Superheated vapor II For Temperatures Above the Critical Temperature Use the superheated vapor table 111 Suggested Procedure A Make sure you are in the tables for the correct substance B Use the saturation tables temperature or pressure as described above to determine which table ie saturation superheated vapor compressed liquid to use to evaluate the desired properties C Check to make sure you retain the proper units on values read from the tables S1 or English Engineering D The above road map may be used for other properties e g internal energy u enthalpy h entropy s by simply replacing the specific volume V by the desired property eg replace V by u in all relationships above if the known information is uT or up ENSC 2213 T HERM 0D YNAMI CS ENSC 2213 THERMOD YNAMICS Chapter III Part 1 EVALUATING PROPERTIES A Johannes PhD PE Professor of Chemical Engineering School of Chemical Engineering Oklahoma State University Copies of Class Lecture Notes Portions of these notes were originally compiled by Professor Robert L Robinson Jr ENSC 2213 T HERM 0D YNAMI CS Chapter 111 GENERAL BEHAVIOR OF PURE SUBSTANCES We have seen in the previous chapter that to use the First Law requires us to know the properties e g internal energy etc of the substance under study A Obiectives 1 Understand the general qualitative behavior of pure substances 2 Provide the basis for evaluating quantitative numerical values of the properties of pure substances gtgtgt Sections 31 322 of your text ENSC 2213 T HERM 0D YNAMI CS THE PRESSUREVOLUME TEMPERATURE BEHAVIOR OF PURE SUBSTANCES 0 Most of the problems we shall deal with in this course will involve systems which contain pure uids in the liquid or vapor state 0 To solve these problems we must often know the values of certain properties pressure volume or density temperature etc of the pure substance 0 Knowledge of these properties must be preceded by a qualitative understanding of the relationship among pressure volume and temperature for pure uids 0 A satisfactory understanding of the material on this topic is essential to the understanding of most of the material which follows in this course ENSC 2213 T HERM 0D YNAMI CS Consider The Following Example 1 lbm z HZOquot T 300 F vosf In What form does the water exist at the conditions given Under the above conditions the water exists as 0925 lbm liquid and 0075 lbm vapor About 97 of the volume 0485 ft3 is occupied by the 0075 lbm of vapor and 3 by the liquid The pressure in the vessel is about 67 psia OI39 0075 1bm 0485 ft3 Vapor 0925 lbm0015 ft3 Liquid ENSC 2213 T HERM 0D YNAMI CS QUALITATIVE BEHAVIOR OF PURE SUBSTANCES We can gain an understanding of the nature of the relation among pressure volume and temperature for a pure uid by considering the behavior of l lbm of the uid con ned in a piston cylinder arrangement as shown below Piston l lbm of uid under the piston piston I PISTON 1 lbm of uid ENSC 2213 T HERM 0D YNAMI CS HEATING A FLUID AT CONSTANT PRESSURE Consider heating a pure substance like water at constant pressure from an initial liquid state room temperature 70 F and pressure 1 atm for water The process would proceed as shown a 1 1 grquot t 4 EUIECDGIAED EaTURMED LIQIJIIJVAPHSIH S T39u RETED SUPEHHEETE LIQUID LIQUID MIETIJFIE V F UR mama P 1 2 3 4 5 f X T1 T2 T2 T2 T5 V ENSC 2213 T HERM 0D YNAMI CS Constant Temperature Expansion Consider expanding a uid from an initial liquid state at high pressure while keeping the uid temperature constant The process would proceed as shown below l i a l ug hg 1 177 x 2 37 49 COMPRESSED SATURATED LIQUID LlOUID MIXTURE Lacunavapor Samaritan 5 VAPOR 5 1 B I SUPERHEAT E D WE POR ENSC 2213 T HERM 0D YNAMI CS THE PVT SURFACE The so called Pv diagram shown below is a two dimensional projection of what is really a three dimensional relationship We make the 2 d diagram Pv by plotting lines representing constant values of the third variable T the complete 3d surface is shown qualitatively below NOTE This gure and the following gures in this section are not drawn to scale The gures have been somewhat distorted in certain areas to facilitate the discussions of the various regions on the gures ENSC 2213 T HERM 0D YNAMI CS QUANTITATIVE VALUES OF PUREFLUID PROPERTIES To apply the energy balance in the analysis of engineering systems we need numerical values for thermodynamic properties A Obiectives 1 Establish the minimum number of properties required to x the equilibrium state of the system 2 Explore the relations that eXist among these system variables 3 Be able to locate andor calculate the required properties of uids of interest gtgtgt Sections 33 35 in your text ENSC 2213 T HERM 0D YNAMI CS B Tabulated Properties The pVT Surface I m cf 10 ENSC 2213 T HERM 0D YNAMI CS Prcmll rL39 39l rcmuru 5 I Critical Surl39nrzl Liquid MINI L V Wampum Tri39 u ninl V P F l cmpumtnurc I39I39III C39rirlIml I m hau n 11 Fri l I I I I E a 1 n I L39Iillt39ill ll paint I u I II II I 7777 2 1 Liquid I uamr I friplc lill luf I iquotf39ilrf ftm h Enli39iIimgmr j lt lIii Epwiliu mhm rw 1439 ENSC 2213 T HERM 0D YNAMI CS Temperature EOQCI 68 F 10 MP3 Liquid Vapor H014 bar 147 lhfxin Liquidvapor 1000C 2 1210F Speci c volume 12 ENSC 2213 T HERM 0D YNAMI CS Critical Liun pmnt a b c I a Meltmg Vapurizalinn t i I E aquot bquot Cquot Candle n SE11an a SOlld V 7 r alpm Suhlm latmn TUNE P011 3 239 V xx r H Tampel ature 13 ENSC 2213 T HERM 0D YNAMI CS Ina3 to 72154951 5 Units Fifie A 731MB A 2 main A 3 more 4 4 am A 5 Fillr 4 5 17153 A 7 T4544 A a mile A vet A 10 all A f7 74126 4 rz Taoa A 75 tabr A 24 ram 4 45 railc A 45 aira A 77 721512 A f8 Tal l39 A 7 robe A 20 MM A 27 firDr s A 22 MM 4 2 3 milEr l 24 5219 A 25 TilJr s A 226 Hilltr 4 2 7 Atomic or Molecular Weights and Critical Properties of Selected Elements and Compounds 816 Properties of Saturated Water LiquidVapor Temperature Table 817 Properties of Saturated Water Liquid Vapor Pressure Table 819 Properties of Superheated Water Vapor 821 Properties of Compressed Liquid Water 825 Properties of Saturated Water Solid Vapor Temperature Table 826 Properties of Saturated Refrigerant 22 Liquid Vapor Temperature Table 827 Properties of Saturated Refrigerant 22 Liquid Vapor Pressure Table 828 Properties of Superheated Re rigerant 22 Vapor 829 Properties of Saturated Refrigerant 134a Liquid Vapor Temperature Table 833 Properties of Saturated Refrigerant 134a LiquidVapor Pressure Table 834 Properties of Superheated Refrigerant 134a Vapor 835 Properties of Saturated Ammonia Liquidiapor Temperature Table 838 Properties of Saturated Ammonia liquid Vapor Pressure Table 839 Properties of Super heated Ammonia Vapor 840 Properties of Saturated Propane Liquid Vapor Temperature Table 844 Properties of Saturated Propane Liquid Vapor Pressure Table 845 Properties of Superheated Propane Vapor 846 Properties of Selected Solids and Liquids CD 0 and K 850 Ideal Gas Speci c Heats of Some Common Gases 851 Variation of Ep with Temperature for Selected Ideal Gases 852 Ideal Gas Properties of Air 853 Ideal Gas Properties of Selected Gases 855 Constants for the van der Waals Redlich Kwoog and Benedict Webb Rubin Equations of State 859 Thermochemical Properties of Selected Substances at 298 K and 1 atm 86o Standard Molar Chemical Exergy atquot k1 lkmol of Selected Substances at 298 K and p0 861 Logaritnms to the Base 10 of the Equilibrium Constant K 862 14 ENSC 2213 T HERM 0D YNAMI CS 1 Saturated Property Region Example Water Table A2 Temperature Table Table A3 Pressure Table 15 ENSC 2213 T HERM 0D YNAMI CS Tabies in 5 Units 817 7511714 42 1 Properties of Saturated Water Liquid Vapor Temperature Table 39Pm gmigang111quotr15 Speci c Volume Internal Energy Enlhulpy Entropy v 39 011111 111330 131110 kJkg kng K Sat Sat Sm Sat Sat S131 Sat Sat Press Liquid Vapor Liquid Vapor Liquid Evap Vapur Liquid Vapor Temp bar 04 X 03 19g 111 11g 11 hrg 11g Sf 3L DC 01 000611 10002 2116136 1100 23753 001125013 25014 00000 91562 01 3 439 000313 quot610001 157232quot 1677 23309 31673 24919quot 25037 006101 90514 4 s 539 5000372 391000391 147120 2097 23323 2098 24396 25106 00761quot 90257 639 000935 quot1390001 137734 312519 23336 2520quot 24372 25124 00912 9000311 001072 110113902 120917 iii3359 23864 33609 24325 25161 01212 3950132 3 10 001223 10004 106379 4200 23392 4201 24777 25193 01510 39003 10 11 001312 10004 99357 4620 23905 4620 24754 25216 01653 33765 11 12 001402 10005 93734 5041 23919 5041 24730 25234 01306 33524 12 13 001497 10007 33124 5460 23933 5460 24707 25253 01953 33235 13 14 001593 10003 32343 5379 23947 5330 24633 25271 02099 33043 14 15 001705 10009 77926 6299 23961 6299 24659 25239 02245 37314 15 39 16 V 0013113 10011 733332 6713 23974 6719 24636 25303 02390 37532 1625f 3917 3 j39001938 39391J00121f 3969044quot3923933 37133quot 24612 25326 02535 373513 j3917 13 39 002064 391001439 65038 7557 3924002 7553 24533 25344 02679 37123 13 3919 j03902193 39391390016 61293 2 quot 57976 24016 17977 quot 2456539 3925 362 02323 363971 quot 19 39 20 002339 10013 57791 3395 24029 3396 24541 25331 02966 36672 20 21 002437 10020 54514 3314 24043 3314 24513 25399 03109 36450 21 22 002645 10022 51447 9232 24057 9233 24494 25417 03251 36229 22 23 002310 10024 43574 9651 24070 9652 24470 25435 03393 36011 23 24 002935 10027 45333 10070 24034 10070 24447 25454 03534 35794 24 25 003169 10029 43360quot 10433 240935 10439 24423 25472 03674 35530 392539quot 26 003363 3100323 40994quotif 510906 241111 10907 24399 25490 03314 35367326 27 003567 1 10035 33774 011325 24125 11325 24376 25503 03954 35156 5127 23 quot03903732 100377 3936690 39117 42 324139 quot11743 39 243521 255265 04093quot349467 23 2939 51 004003 500402 3534733 391216O 241524121612 243235 25545 04231 84739 29 30 004246 10043 32394 12573 24166 12579 24305 25563 04369 34533 30 31 004496 10046 31165 12996 24130 12997 24231 25531 04507 34129 31 32 004759 10050 29540 13414 24193 13415 24257 25599 04644 34127 32 33 005034 10053 23011 13332 24207 131133 24234 25617 04731 33927 13 34 005324 10056 26571 14250 24220 14250 24210 25635 04917 33723 34 35 0056283 310060 25216 14667 24234 14663 2418639 25653 05053 33531 35 36 005947 10063 2394039 i 15 03539 242473 15036quot 24162 25671 05133 33336quot 06f 39 33 1006632 jquot100713912139602 15920 24274 15921 24115 25707 05453 3295039 38 40 20073344 100735 19523 16756 243017 16757 24067 25743 05725 32570 j4039 39 03903995399393ijquot10099 quotj15239539339 3939j1334439 24363 13345 23943 25332 06337 quot813964398j 4539jj 16 H206 ENSC 2213 T HERM 0D YNAMI CS 740 AZ Properties of Saturated Water Liquid Vapor Pressure Table Speci c Volume Internal Energy Enthalpy Entropy m3kg kJkg kJkg kJkg K Sat Sat Sat Sat Sat Sat Sat Sat Press Temp 39 Liquid Vapor Liquid Vapor Liquid Evap Vapor Liquid Vapor bar C v 14 km 11 s S39s quot 28967 34800 12145 24329 25544 04226 84746 36162 123739 39 V j1539939 quot256741 05210 83304 v 4151 18103 1 39 quot 325770 05926 82287 1 91010 4581 14674 19182 928 25847 06493 81502 020 6006 7649 25138 I 26097 08320 79085 030 6910 5229 28920 23361 26253 09439 77686 040 7587 3993 31753 23192 26368 10259 76700 050 8133 3240 34044 23054 26459 10910 75939 060 8594 2732 35979 22936 26535 11453 75320 070 8995 2365 37663 22833 26600 11919 74797 0180 9350 2087 39158 f 12329 74346 g090 9671 p 1869 1 40506 M A 12695 73949 1 71100 9963 1 1694 41736 f 39 1913026 73594 150 1114 1159 46694 f 14336 72233 quot1200 139 1202 08857 50449 019 27067 115301 71271 250 1274 07187 53510 25372 21815 27169 16072 70527 300 1336 06058 56115 25436 21638 27253 16718 69919 350 1389 05243 58395 25469 21481 27324 17275 69405 400 1436 04625 60431 25536 21338 27386 17766 68959 450 1479 04140 62225 25576 21207 27439 18207 68565 1 500 1519 03749 63968 27487 4 186073 68212 96003 1589 03157 66990 27568 1 19312 67600 7007 1650 02729 69644 4 27635 19922 67080 2800 1704 102404 72022 27691 20462 66628 39 900 1754 02150 74183 V 1 27739 20946 66226 100 1799 01944 76168 25836 20153 27781 21387 65863 150 1983 01318 84316 25945 19473 27922 23150 64448 200 2124 009963 90644 26003 18907 27995 24474 63409 250 2240 007998 18410 28031 25547 62575 006668 9 700 002737 800 002352 800 900 002048 900 100 001803 100 110 001599 27056 110 17 ENSC 2213 T HERM 0D YNAMI CS QUANTITATIVE VALUES OF PUREFLUID PROPERTIES 0 Adding heat to a subcooled liquid at constant pressure causes its temperature to increase this is known as sensible heating When the liquid reaches the saturation temperature adding more heat causes the formation of vapor but T remains constant 0 Throughout the vaporization T remains constant but an ever increasing portion of the liquid converts to vapor 0 Once all we have totally vaporized all the liquid the temperature can rise again with the addition of more energy by heat transfer 0 The heat added to cause the phase change at constant T is called latent heat 0 Note the molecules of the uid are either in the vapor gas phase g the liquid phase there is no inbetween phase we just have the two phases in equilibrium 18 ENSC 2213 T HERM 0D YNAMI CS Interpolations in Tables 1 Single Interpolation superheated region A4 Given Water p 200 MPa V 0185 m3kg Find T 2 Use linear interpolation At p 200 MPa from the property tables L V T T 01850 V given 600 T2 01996 V2 T T1 V V1 T2 T1 V2 V1 0139 TT1 VV1 T2 T1 V2 V1 19 ENSC 2213 T HERM 0D YNAMI CS Single Interpolation cont d T 2 E p constant A I I I I I I I I Known O1850 O1757 O1996 O1757 T 500 1600 500 T5389OC 20 ENSC 2213 T HERM 0D YNAMI CS Example page 90 Determine the speci c volume of water vapor at a state Where p 10 bar and T 215 C From Table A4 2409C 02275 39L a p 10 bur 710C 1 1113ng 200 02060 a 215 p 1 i2mm 020m 24quot 02275 zoo 2T5 zit WC v a v a 240 C 200 C V2150C VZOOOC 215200 24O200 3 dv m V2150C 02141E V2150C V2000C AT 07 Dr AJ S Way v v1 T T1 V 206O 215 200 v2 v1 T2 T1 2275 2060 240 200 21 ENSC 2213 T HERM 0D YNAMI CS 2 Saturation Property Interpolation If asked to nd a saturation property ega psat Tgiven or hg pgiven just interpolate linearly along the saturation line Example Problem Find uf for water at pS 82 bars From Table A3 pS bar uf kJkg E 80 13056 82 90 13505 p p1 u ul p2 p1 u2 u1 82 80 uf 13056 90 80 13505 13056 gt uf 82 bars uf 213146 kJkg 22 ENSC 2213 T HERM 0D YNAMI CS 3 Double Interpolation Substance Water Given T 380 C Neither 380 C nor 42 MPa and p 42 MPa appears in the tables Find V Step 1 Construct a table that brackets the given property values as shown below Table A4 Volumes 1 940 1242 1260 360 006788 004331 380 400 007341 004739 23 ENSC 2213 T HERM 0D YNAMI CS Double Interpolation cont d Substance Water Given T 380 C p 42 MPa Find V Step 2 Construct a T 380 C row or p 42 MPa column in the table Volumes I 1w w m 360 006788 004331 380 007065 004535 400 007341 004739 24 ENSC 2213 T HERM 0D YNAMI CS Double Interpolation cont d Substance Water Given T 380 C p 42 MPa Find V Step 3 lnterpolate at T 380 C or at 42 MPa as usual to get V Volumes I w w m 350 006f45 00il331 380 007f65 WOMTBS 400 007341 004739 Thus v 006812 m3kg 25 ENSC 2213 T HERM 0D YNAMI CS C The State Principle State Principle Any two independent properties are suf cient to x the state of a simple compressible system in equilibrium As you know by now the variables that describe the state of equilibrium at this stage of our study are T p V u This means u f Tp p f TN V f Tap u f TV T fpV and so on Are T and p always independent 26 ENSC 2213 T HERM 0D YNAMI CS D Behavior in the LiquidVapor Mixture Region Consider a vaporization process at constant pressure p p1 and thus constant temperature T T1 gt Vg Vf Vf Vg T and p constant Note that although the total volume V changes both speci c volumes vf and vg remains constant for xed values of p T 27 ENSC 2213 T HERM 0D YNAMI CS E Volumes of LV Mixtures Consider the system containing mass m of a L V mixture vk mg p K M mmgmf gt F Vf mf The total volume V is VVgVf mgvgmfvfmgvgmmgvf m In m m De ne V E Vm vg g jvf In vxvg1Xvf X 2 Quality mass fraction vapor mg m so X v vf Vm vf 9 Where V Vm speci c volume of system 28 ENSC 2213 T HERM 0D YNAMI CS Some Example Propertv Problems 1 Example 1 Fill in the table below for the substance Refrigerant 134a T F Ppsia JI bm State a 80 80 b 200 025 c 120 015 d 1507 280 Use tables AlOE A12E 29 ENSC 2213 T HERM 0D YNAMI CS Example 2 Given Two kg of H20 are con ned in a rigid volume of 01 m3 at 205 0C V 01 m3 m2kg 0 Solution System Water 1 Determine the State of the System at 205 C the properties of water are as shown in the gure P 1723 NIPa 2050C V 0001164 vf vg 011521 m3kg 30 ENSC 2213 T HERM 0D YNAMI CS Now V Vm 012 005 m3kg So vfltvltvg gt2phase 2 Calculate the System Pressure p p 21723 MPa sat d 3 Calculate the System Quality v vf 005 0001164 X vg vf 011521 0001164 x 0428 or 428 Mass fraction vapor 4 Calculate Volume Vapor Vmgvg mfvf fl total vapor liquid volume volume volume m v m v VolumeVap0rxlOO XIOO V mv VolumeVap0r xvg vxlOO 31 ENSC 2213 T HERM 0D YNAMI CS Thus V01 Vapor xvg x100 V 1xquotf V01 Liquid x100 v Thus O428O1152 V01 Vapor x100 005 987 32 ENSC 2213 T HERM 0D YNAMI CS F ENTHALPY For closed systems the internal energy u or U is used in most engineering calculations For open systems and certain other types of problems the internal energy U and the product of the volume times the pressure occurs frequently So we de ne a quantity called enthalpy which combines these two quantities H E U pV Or per unit mass hupV Or per unit mole h a p The units of enthalpy are the same as u ie kJkg for SI or BtulbIn for English engineering units note 1 Btu 77817 ft lbf 10551 kJ Values for H or h can be found in the same tables as u The speci c internal energy u or the speci c enthalpy h for a saturated mixture is calculated the same way as speci c volume 33 ENSC 2213 T HERM 0D YNAMI CS u 1 Xuf Xug uf Xug uf The increase in internal energy on vaporization 11g Ur is often denoted by Ufg likewise for enthalpy h l Xhf th hf Xhg hf The increase in enthalpy during vaporization hg hf is often denoted by hfg Example Calculate h from u and pV data and compare With table value Consider water at a state xed by a pressure of 147 lbfin2 psia and a temperature of 250 F From table A4E V 2842 ft3lbm u 10915 Btulbm and h 11688 Btulbm h u pV B u f 3 3 1441112 lBtu 10915 147 XZS lz E4 l 1ft2 778ft1bf h 10915 773 11688 Btulbm 34 ENSC 2213 T HERM 0D YNAMI CS G Reference States In all calculations involving energy u h and s are not obtained by direct measurement but are calculated based on variables we can measure like p T and v This requires that we de ne a reference state usually where the quantity is equal to zero The value of the reference state does not matter since we calculate changes when we do calculations ie u2u1 For water the reference state is saturated liquid at 00l C where Uf is set to 00 H Specific Heats or heat capacity cV and cp cV and cp are intensive properties and are de ned as au ah cV c 6T V p 8T p The units of cV or cp are kJkgK or kJkmolK or in English units BtulbmoR or BtulbmoloR We define a term k as the speci c heat ratio or C k P C V 35 ENSC 2213 T HERM 0D YNAMI CS ENSC 2213 THERMOD YNAMICS AJ Johannes PhD PE Professor of Chemical Engineering School of Chemical Engineering Oklahoma State University Class Lecture Notes Portions of these notes were originally compiled by Professor Robert L Robinson Jr and Others ENSC 2213 T HERM 0D YNAMI CS Chapter I INTRODUCTORY CONCEPTS amp DEFINITIONS Topics For Discussion gtgtgt Sections 1119 of your text 0 Objectives 0 Units and Dimensions 0 Pressure Hydrostatic Pressure ManometersBarometers Temperature 0 Engineering Design and Analysis Obiectives The objective of Chapter I in the course is to introduce you to several fundamental concepts and de nitions that are used in the study of engineering thermodynamics ENSC 2213 T HERM 0D YNAMI CS Ass1gnment Read Chapter 1 Concepts and De nitions In This Chapter System Pure Substance Surroundings Units Closed System Density Open System Control Volume Speci c Volume Property Pressure Process Temperature Equilibrium Pressure Scales Extensive Property Temperature Scales Intensive Property Manometers Phase Units and Dimensions 1 Basic Dimensions Time Length Mass Temperature ENSC 2213 T HERM 0D YNAMI CS 2 Fundamental Units Note i means has the units of SI Units Systeme International d Unites Time i Second s Length i Meter m Mass i Kilogram kg BG Units British Gravitational System Time i Second s Length i Feet ft Mass i Slug slug English Engineering System Time i Second s Length i Feet ft Mass i Pound Mass lbm 3 Derived Units Acceleration Force Newton s 2nd Law of Motion states F on ma ENSC 2213 T HERM 0D YNAMI CS or Fmagc a The SI system m 2 kg a i ms2 F i Newton N so N E force required to accelerate 1 kg by 1 ms2 so 1 N 1 kg 1 ms21 kg mszN or gclkgmN s2 SI b BG Units British Gravitational System m i slug a i fts2 F i lbf Note lbf in text 1 lbf E force required to accelerate l slug by l fts2 SO 1 Halo 1 slug 1 fts2l slug ftszlbf OI gc 1 slug 11113f s2 BG ENSC 2213 T HERM 0D YNAMI CS c The English Engineering System Time i 5 Length i it Mass i 1bm or 1b in our textbook 1 lbf E force required to accelerate 1 1bm by 32174 fts2 SO 1 1bf 1 lbm32174 ftSZgc 1b ft 01quot gc 232174 m 2 lbfs SO Fz Where go gc32174lbmft or 1 kgm 1 Slug or lbfs2 N 32 lbfs2 ENSC 2213 T HERM 0D YNAMI CS d Summary of Units Q E Eg ih Time s s s Length m ft ft Mass kg Slug lbm Force N lbf lbf gc 1kgmN s2 lslug ftlbfs2 32174 113m ftlbfs2 Temp 0C K OF 0R OF OR Note In Thermo Book lbm is written as lb In Fluids Book lbf is written as lb ENSC 2213 T HERM 0D YNAMI CS e Example A tool box weighs 10 lbf on Earth where the acceleration due to gravity is 32174 fts2 determine the mass m of the tool box in lbm Determine the mass of the tool box on the Moon where the gravity is about 16 of Earth 1 Determine the mass m of the tool box System the tool box Tool Box 1F Important Note Weight Force exerted by gravity Since we know the weight and the value of gravity in 10 1bf and g 32174 ftS2 we can calculate the mass using Newton s Law Wt F gc m Wtgc 10113 321741bmft s2 g lbfs2 32174ft m 2100 lbm ENSC 2213 T HERM 0D YNAMI CS Since mass is constant and not a function of gravity the mass on Earth is the same as the mass on the Moon so HIE HIM lbm 2 the mass of the box in slugs Important Note Weight Force exerted by gravity Since we know the weight and the value of gravity on Earth 10 1bf and g 32174 ftsz we can calculate the mass in slugs using Newton s Law WtFmggC WtgC 10 lbfllslugft s2 g lbf s2 32174 ft m 01 m 0311 slugs 1 slug 321741bm ENSC 2213 T HERM 0D YNAMI CS Pressure 1 Definition Force Pressure E unlt area 01 F P E A 2 Units N P 2 m Derived unit pascal Pa lPazll2 m 10 ENSC 2213 T HERM 0D YNAMI CS 3 Magnitude of the Pascal Consider atmospheric barometric pressure patm patm 1 atm 34 ft H20 760 mm Hg 1471bfin2 105Pa102 kPa 01 MPa 4 Exact pressure equalities conversions 1 atm 760 mm Hg 14696 lbfin2 101325 Pa 101325 kPa 0101325 MPa 5 The Standard Atmosphere One standard atmosphere 1 atm is de ned as the pressure generated by 760 millimeters of mercury at 0 C under standard gravity of 980665 ms2 11 ENSC 2213 T HERM 0D YNAMI CS 6 Types of Pressure Absolute pressure p 01 pabs Gage pressure pgage Atmospheric pressure patm also termed ambient pressure or barometric pressure 0 Vacuum pressure pvac Pgage Z P 39 patm pvac Patm 39 P In English Engineering nomenclature p i psia pgage i psig Where psia denotes lbfin2 absolute psig denotes lbfin2 gage 12 ENSC 2213 T HERM 0D YNAMI CS 6 Typical Bourdon Tube Pressure Gage EHiptical metal Pointer Bourdon tube Pinion gear quot71 O 9 W Suppmt 13 l Q3 Linkage M quot fl Gas at pressure p 13 ENSC 2213 T HERM 0D YNAMI CS P gage Atmospheric pressure p vacuum Zero pressure Zero pressure 14 ENSC 2213 T HERM 0D YNAMI CS 8 Pressures Exerted By Fluids Hydrostatic Pressures 1 3 quot quot Ejgtgt Note Density is mass or moles per unit volume m P Vt Force Balance F2 2 F1 mggc but m th 2 Ah p so F2 F1 Ahpggc or F2 F1 ApghgC NOW Divide by A to get p2 p1 pghgc 15 ENSC 2213 T HERM 0D YNAMI CS a How to Avoid Problems with Units Example Problem on Use of Units For a manometer Apng g0 Given 2 lgmcc g 981ms2 L 2 150m Find Ap in kPa 16 ENSC 2213 THERMOD YNAMICS Solution By de nition 1 Pa E 1 g0 AP lgm 981m 15cm 1st 104cm2 cm3 52 kgm m2 X kg lPam2 lkPa 103 gm N 103 Pa 147 kPa 17 ENSC 2213 T HERM 0D YNAMI CS b Example of Hydrostatic Pressure a What pressure will you feel if you swim at a depth of 10 feet in a pool where the water density is 63 lbmft3 b Would the pressure he the same if you were swimming in the salty water of the Paci c Ocean Justify your answer Atmosphere W 10 ft of water L 18 ENSC 2213 T HERM 0D YNAMI CS Solution System swimmer h p h p2p1amp pzzpi39l39i C C lbf 3 p 10 ft patm gc but patm 1 p10ft 14696 i112 1ft2 322 11ms210 2 S 144 in2 322 lbm ft 63 lbrn ft3 Thus pressure increases about 045 psift of depth p 10 ft psia 19 ENSC 2213 T HERM 0D YNAMI CS b The pressure felt Will be greater because the salty water is denser N 651bm p salt water N 3 ft Thus 65 10 p10 14696 1921 psia compared to 1907 psia for fresh water 20 ENSC 2213 T HERM 0D YNAMI CS c Manometers p0 p1pgh1 p0 192 pgh2 SO P2 101 Pgh1 h2 or P2 131 PgL 21 ENSC 2213 T HERM 0D YNAMI CS d General Manometer Equation Reference plane T i h L 3 h2 i JLWEI Po Consider the quotwater mercuryquot manometer shown p0 p1refplane h3 h2pwg thHgg p0 2 p2 refp1ane h3 h1pwg hlpHgg p2 p1refp1ane bl 112pr 112 h1pHgg L p2 p1refp1ane pHg pw g C 22 ENSC 2213 T HERM 0D YNAMI CS e Alternative Approach to Manometer Equation Reference plane T i h L 3 h2 i JLWEI Po Add pressure as you go down in a uid column and subtract as you go up as follows p1refplane h3 h2pwg thHgg hlpHgg h3 h1pwg 2 p2 refplane p2 p1refp1ane hl h2pwg h2 h1pHgg p2 p1 refplane pHg pw Lg 23 ENSC 2213 T HERM 0D YNAMI CS f Barometers Mercury vapor patm 100 33 T 39 Liquid mercury pm 2 1359 gcm3 The manometer formula states pgh patm ptop h patm O i go 01 pgh atm c This is the barometer equation 24 ENSC 2213 T HERM 0D YNAMI CS g Temperature Scales III 1 C Oquot K m L C R r b v V p m 0 c o r f 3 rx Hquot v KO 1 Steam pmm x 3 g c 393 Tillple poml E Q Q mrm b H mm b g a N H C rr 5 39 m c r 39 3 quotI C 0 ri lce pomt m c s m ox quotI 39139 395 quotE 392 U a f E a E 5 at U 22 L In K 9 A q rquot 3 z r 4 U H M 53 13 3 I o l Abmhlle zero Conversion of Temperature Temperature T may be expressed in units of K 0C 0F and OR Where TK T C 27315 T R T F 4597 T R 18 TK T C 59T F 32 T F 95 T C 32 25 ENSC 2213 T HERM 0D YNAMI CS Properties Of Fluids Weight W E downward force on a body due to gravity W Fgravity mggC in BG amp EE units W i lbf note for earth s standard gravity g 32174 fts2 m 322 fts2 In SI units weight W i N note earth s standard gravity in si units is g 9807 ms2 z 981 ms2 Density p E mass per unit volume p mVt Units slugsft3 kgm3 lbmft3 Speci c Volume v E volume per unit mass v Vtm Note v 1 p Speci c Weight y E weight per unit volume 7 g units lbfft3 or Nm3 26 ENSC 2213 T HERM 0D YNAMI CS Speci c Gravity Density of a Fluid Density of Water at Standard Conditions 0 7 Or so 0 yw W 4 C 39201 I gt 194 slugsft3 1000 kgm3 624 lbmft3 SGE Pressure P E normal force per unit area P Forcearea Units lbfft2 or Nm2 Using The Hydrostatic Equation PyhPO P P Where h is de ned as the pressure head h E If we use gage pressure Remember Gage Pressure Pabsolute Pgage Patmospheric P020 27 ENSC 2213 T HERM 0D YNAMI CS And P yh Gage P Let s do an example O p e n v T 1 7 ft G a sol i n e i 1 39 3939 l 3 2 4 Example 21 Because of a leak in a buried gasoline storage tank water has seeped in to a depth shown in the gure above If the speci c gravity of the gasoline is 068 determine the pressure at the gasolinewater interface and at the bottom of the tank Express in units of psf gage psia psig and feet of water So Leaky tank We want to nd P1and 2 Knowns Gasoline SG 068 28 ENSC 2213 T HERM 0D YNAMI CS Assume Incompressible Fluid at rest For gasoline v 50va 068624j g 1b Ygasoline P1 ygasoline h P0 1bf P1 424 3x170ftP0 ft P1 721 PO If P0 O gage 7211b 1b P1 ft2501 144 1 ft2 29 ENSC 2213 T HERM 0D Y NAMI CS P 721 1 11 f 1b 6 t VHzo 624 f ft3 Note Atm Pressure Standard P0 14696 16171112 abs s 147 psia P0 21 161bfft2 abs 21 16 psf abs P0 101326 Nm2 Abs P0 O psig B Find P2 P2 VHZO 39hH20 P1 P2 624lb f3ft7211 908psf ft3 ft gage 30 ENSC 2213 T HERM 0D YNAMI CS MER C UR Y BAR OME T ER Atmospheric Pressure ls Usually Measured With A Mercury Barometer Consider A Glass Tube Initially Filled With Mercury We Place The Tube In A Dish Of Mercury And Let The System Come To Equilibrium pvapor A A A A l 1 tm HUN will Mercury At Equilibrium patm pvapor 39 z 0 Very Small For Hg For Standard Atmospheric Pressure 1013 kPa The Height Of The Column Of Mercury ls pm 1013kPa 31 ENSC 2213 T HERM 01 Y NAMI CS lbf or since YHg 847 F lbf and Pam le6 h 2498 ft 299 inches of Hg Note If uid is water h z 34 ft Evangelista Torricelli 1644 Manometrv A Standard Technique For Measuring Pressure Involves The Use Of Liquid Columns In Vertical Or Inclined Tubes Pressure Measuring Devices Based On This Technique Are Called Manometers Three Common Types Of Manometers Include l Piezometer Tube 2 U Tube Manometers 3 InclinedTube Manometers 32 ENSC 2213 T HERM 0D YNAMI CS l Piezometer Tube The Simplest Type Of Manometer Open LIMITATIONS i pAgtpatm 2 pA SHOULD BE SMALL m 3 FLUID IN THE CONTAINER MUST BE A LIQUID The Governing Equation For Manometers Is The Hydrostatic Equation PZYhP0 Apply This To The Piezometer Tube WHERE PA YIhI y1 y Of Liquid In The Container h1 Height Of Liquid In The Vertical Tube P0 patm abS 0 Gage Pressure pA p1 Points 1 And A Are At The Same Elevation 33 ENSC 2213 T HERM 0D YNAMI CS 2 UT UBE MANOME T ER This Type Of Manometer Overcomes The Dif culties Noted With The Piezometer Tube ope MANOMETER FLUID IS CALLED GAGE FLUID 7 T i IF p A IS LARGE USE HEAVY 1 GAGE FLUID Hg 1 2 o 3 IF pA IS SMALL USE LIGHT GAGE FLUID H20 fluid firy The Governing Eq Is Equation Is The Hydrostatic Equation P Yh P0 Apply this Equation To The UTube Manometer By Starting At One End Of The System And Work Your Way Around To The Other End Starting At Point A PA Yihi Yzhz O Gage Pressure 34 ENSC 2213 T HERM 0D YNAMI CS Note Pa P1 Same Elevation In A Continuous Fluid P2 P3 Same As Above P Increases As We Move Vertically Down P Decreases As We Move Vertically Up 39 PA Y2h2 Y1h1 UTUBE MANOMETERS Example Using A UTube Manometers To Measure Pressure Differences In The Figure Below Find P A PB 1 l 3 Solution Start At Point A Sequentially Add gong down And Subtract going up Head Until Point B Is Reached PA Y1h1 39 Y2h2 39 Y3h3 2 PB 35 ENSC 2213 T HERM 0D Y NAMI CS or PA PB 2 Y2h2 Y3h3 Y1h1 Remember p 0 V E g0 Note 0 Flow 39 as 39 3 b F Flow nozzle 3 InclinedTube Manometers Normally Used To Measure Small Pressure Changes 36 ENSC 2213 T HERM 0D YNAMI CS Starting At Point A Sequentially Add gong down And Subtract going up Head Until Point B Is Reached pA l ylhl YZEZ 9 Y3h3 133 or PA PB 2 3252 SIN 9 Y3h3 Y1h1 InclinedTube Manometers Are Often Used To Measure AP In Gases If A amp B Contain A Gas yl amp V3 z 0 Very Small Compared To Manometer Fluid pA pB Pa P ESQ O E b Y22 1D 1 2 YZSING Note As E2 T As 9 l Pressure Measuring Devices Manometers Are Not Well Suited For Measuring Very High Pressures Or Rapidly Changing Pressure The Alternative Devices Generally Operate On The Idea That When A Pressure Acts On An Elastic Structure The Structure Will Deform This Deformation Can Then Be Related To The Magnitude Of Pressure 37 ENSC 2213 T HERM 0D YNAMI CS Electronic Devices Pressure Transducers Converts The Pressure Into An Electrical Output Diaphragm U 38 ENSC 2213 T HERM 0D YNAMI CS Diaphragm I stop Armature Electrical connections Diaphragm gt Link pin Beam strain gages deposited on beam 39 ENSC 2213 T HERM 0D YNAMI CS ENSC 2213 THERMOD YNAMICS Chapter 3 Part 2 VOLUMETRIC PROPERTIES OF IDEAL GASES AJ Johannes PhD PE Professor of Chemical Engineering School of Chemical Engineering Oklahoma State University Copies of Class Lecture Notes Portions of these notes were originally compiled by Professor Robert L Robinson Jr ENSC 2213 T HERM 0D YNAMI CS VOLUMETRIC PROPERTIES OF IDEAL GASES A Obiectives 1 Develop the techniques for calculating the volumetric and thermodynamic properties of ideal gases 2 Determine When the assumption of ideal gas behavior is justi ed 3 Solve example problems using ideal gas behavior gtgtgt Sections 311 313 in your text ENSC 2213 T HERM 0D YNAMI CS B Volumetric pVT Properties The Ideal Gas Equation of State For an ideal gas the relationship among pressure volume and temperature are given by the following relations Molar Basis Mass Basis W ZnET pV z mRiT V V n pV ZRiT pVZlT VV m E 831434 kN m kg molK 3 2 10732 m lb mol R ENSC 2213 T HERM 0D YNAMI CS Since pV ZnET ZmRiT WC 866 R121 01 M 2 molecular weight m 11 Values of the gas constant R of Selected Elements and Compounds Subsluncu Chemical Formula R klkg K R Blur lb quot39R Air 02970 006855 Ammuniu NH 04882 01 1662 Argon Ar 12032 004973 Carme dioxide CO U 889 004513 Carbml mmluxidc CO 02968 107090 Helium He ill R69 149613 llydmgcn H1 4 240 098512 Methane CH4 05 I83 112382 Nilrl gt3n N3 13963 HWUQU 39nych 0 12598 01116206 Water H30 146M H lUZl Sumter R values arc calculalml in terms of lhc universle gas constant R 3314 kakmol I 1936 Bunquot lhmul quotR and llhc molecular weight M provided in Table Arl Lia ng R R39M Eq 325 ENSC 2213 T HERM 0D YNAMI CS C Information on Ideal Gas Behavior 0 For an ideal gas Z 1 more latter 0 An ideal gas represents a uid in which molecules do not interact 0 Applicability to real substances depends on the pressure and temperature of interest w on the substance 0 It works satisfactorily for gases like N2 02 air He CO etc near room temperature and below a few hundred kPa 0 We shall develop a method to estimate the error involved in using the ideal gas assumption ENSC 2213 T HERM 0D YNAMI CS D Example Applications of the Ideal Gas Law Example 1 A rigid tank contains 5 lbmol of N2 If T 36OOR and p 30 psia What is the volume of the tank Ir E 5 lbmol Solution 5 N2 5 System contents of tank E i 360 R Assuming ideal gas Z 1 V ZnET p 51bmol 1545ftlbf 360 R in2 ft2 1b moloR 301bf 1441112 6438 ft3 ENSC 2213 T HERM 0D YNAMI CS 83l4kJkmolK Remember E 1545ftlbf1b moloR 1986 Btulb moloR Example 2 A vessel contains H20 at 400 C and 10 MPa If the volume of the vessel in 01 m3 What is the mass of H20 H2O 400 C E and 10 MPa ENSC 2213 T HERM 0D Y NAMI CS Solution E E 5 H2O 4000C 5 System H20 5 and 10 MPa 5 a Assuming ideal gas behavior Z l m amp Z5R5T M lOMPa 106N 01m3 kmolK18kg Md 1 5 1 MPam2 8314kJ kmol103 Nm 6732K5 3276 kg b Using H20 property tables Steam Tables m VV V6732 K lOMPa 00264 m3kg m 0100264 378 kg 2 Ideal gas law With Zl is wrong by 14 ENSC 2213 T HERM 0D YNAMI CS 0 600 173 180 24 05 00 0 C 200 100 100 kPa Percentage of error viable videalvvtble X 100 invowed 10 kpa in assuming steam to be an ideal gas and the region where steam can be 0 8 kpa 39eated as an ideal gas with less than 1 0 39 percent error 0001 001 01 1 10 100 v m31 ENSC 2213 T HERM 0D YNAMI CS E How Can I Decide When to Use the Ideal Gas Law Rationale The assumption of ideal gas behavior makes the calculation of pvT and thermodynamic properties u h s very easy Caveat The range of applicability of the ideal gas behavior assumption is very limited How to Decide The Principle of Corresponding States can help de ne the range of Applicability of ideal gas behavior 10 ENSC 2213 T HERM 0D YNAMI CS 1 The Principle of Corresponding States Consider the pV diagrams below V What do you notice about these These data can be reported as Z E pQET pVnET pVmRiT to yield ice 11 ENSC 2213 T HERM 0D YNAMI CS Where Z is the Compressibility Factor of the gas For an ideal gas Z B V 10 RT so p r will apply Consider the behavior of methane ethane and propane at equal conditions of p and T commmt FEM It 39 b quot msme kw V Compressibility Factors of Metha39negi Ethane and Propane at 140 asa Function of Pressure 12 ENSC 2213 T HERM 0D YNAMI CS The behaviors of the gases are not identical at equal values of T and p Consider Methane Ethane Propane TC K 191 305 370 pC bars 464 488 427 Let s compare the gases at conditions of equal distance from their critical points expressed as quotwquot ea NJ 39 n strum I 0 atrium 39 39 V calnaming 39 391 i tquot Var I 4 darlv 1rquot 39 am my 7 l Compressibility 39Factqrsl of Methane Ethane sland Propane u a Function of Reduced Pressure Vand Reduced Temperature 39 quot 39 39 f v 39 39 39 39 I 15 ENSC 2213 T HERM 0D YNAMI CS The Principle of Corresponding States says All substances behave identically in terms of their reduced properties expressed as the ratio of the property to the critical property Consider the following behavior 1 5655th a quot i 1939 39 quot C1 e a 39 V 18 7 1 2 a z m I I E L l H 013 N Legend 13 I Melhzme El lsopcnrane 393 Ethylene 9 n Hcpmnc A Ethane c5 Nitrogen 4 3 Propane 395 Carbon dioxide D n Bumnc 0 Water L3 Average curve based rm dam on hydrocurlmm Uquot U I 7 7 7 7 7 l 35 I ll I5 3 25 3H 35 40 43 SJ 55 6 15 7 Reduced pressure 11R 14 ENSC 2213 T HERM 0D YNAMI CS 2 The Compressibility Factor Chart The Z Factor Chart can be used to estimate when the ideal gas law is applicable For example if 09SZSll then the error in using pV nlT will be less than 10 A working version of the Z chart is given in the Appendix of your text It relates Z Compressibility Factor PR 2 1013C Reduced Pressure TR TTC Reduced Temperature V2 FETC pc 2 Pseudoreduced Volume 15 ENSC 2213 T HERM 0D YNAMI CS Reduced pressure pg c o o o o o quot393 0 0 O O o c iHAd 2 10192 mqussmdwoo 16 17 lyAd 2 Home Amgqlssadeoo 120 110 100 120 l 3 110 l quot 100 f7 0 M Xx K k I s t K 4 080 070 060 30 40 50 60 70 39 30 90 100 Reduced pressure pn T HERM 0D YNAMI CS ENSC 2213 ENSC 2213 T HERM 0D YNAMI CS 3 Example Problem As a new engineer with Ole ns International Inc you have been assigned to their Production Operations Group Your rst assignment deals With the use of propylene at their research complex A research project requires that 2 lbm of propylene be con ned in a pistoncylinder apparatus held at a temperature of 230 F The problem is to determine the volume to Which the propylene can be compressed at 230 F if the maximum safe operating pressure in the pistoncylinder is 840 psia Please solve the problem expressing the volume in units of ft3 Solution System 2 lbm propylene Given m 2 lbm T 23 O F P 840 psia Find V ft3 18 ENSC 2213 T HERM 0D YNAMI CS Solution a Assume ideal gas behavior Z 1 V nlT p mETMp M 4208 lbm lbmol for propylene V 2 lbm 1073 psia 3 lbmol R 690 R 4208 lbm lbmol 840 psia V 0149 ft3 Was ideal gas assumption okay Cheek using Z Chart For propylene Table AlE TC657 R pc456 atm Thus TR21223O4602L05 TC 657 pR p 840 ps1a 2125 pC 456 atml47psiaatm l9 ENSC 2213 T HERMODYNAMICS From the Z Chart Z 045 v 039 Thus VZZmRT O4521073690 019 E Mp 4208840 or V11quch 2O391073657 0195 3 Mp 4208456147 C 4 How to Use the Z Chart The Z ohart relates TRpRVU Z TR T absolute T pR L absolute p C pC V 5pc Vpc Vpc Vpc R ETC RiTC nETC mRiTC 20 ENSC 2213 T HERM 0D YNAMI CS Given Find E Tp V0rV TRpR Zor vi TV0rV p TRV Z or pR p pV0r V T pRV Z or TR T Thus the Z Chart relates p V and T for gases just like the superheated vapor tables do since Z W pV p pV nET mRiT ET RiT 21 ENSC 2213 T HERM 0D YNAMI CS THERMODYNAMIC PROPERTIES OF IDEAL GASES Topics for Discussion 0 Internal Energy and Enthalpy of Ideal Gases Ideal Gas Speci c Heat 0 Sources of Ideal Values Ideal Gas Property Tables Examples gtgtgt Sections 312 314 in your text 22 ENSC 2213 T HERM 0D YNAMI CS A Enthalpy and Internal Energy of Ideal Gases Experimental Fact The internal energy of an ideal gas does not depend on the volume density of the gas 1 Specific Heats In general CV E 811 aTV and cV E fTp or fTV orfpV Where cV constantvolume speci c heat or heat capacity For an ideal gas u 72 f V Thus C in du V 6T V dT or o du 0V 2 dT Note 0 signi es ideal gas your book does not use this nomenclature 23 ENSC 2213 T HERM 0D YNAMI CS so separate variables to solve this linear 1st order DE du e dT and for an ideal gas 0 f T0nly so 2 T2 0 f du r chT T2 0 u2 u1 J eVdT Where u2 uevaluated T2 and 111 uevaluated Tl If e constant 112 u1 CT2 T1 24 ENSC 2213 T HERM 0D YNAMI CS 2 Ideal Gas Enthalpy h E upV uRiT Thus h f T0nly NOW constant pressure ah Cp 8 T becomes p spe01 c heat 0 dh a and dh c dT Thus T2 h2 h1 L EC is constant h2 h1 CT2 T1 25 ENSC 2213 T HERM 0D YNAMI CS Relation between C and c h u pV u RT db duRdT c dT e dTRdT so cp CVR Ar Ne He 7 I 7 I 0 3000 4000 5000 Temperature R 39 I 1000 2000 Temperature K 26 ENSC 2213 T HERM 0D YNAMI CS B Sources of c and c 1 Table A20 or 20E tabulates information for some common gases 6th ed of MampS C f k CV kgK k kJ kgK at 300K Example for air c 1005 2 Table A 21E gives ESR Tin R E gt fpz ocBTyT25T3 ET4 Example for carbon dioxide KP 2401 4853 x 10 3T 2O39x10 6T2 0343xio 9T3 27 ENSC 2213 T HERM 0D YNAMI CS Sources of CV and Cp Continued 3 Ideal Gas Property Tables Someone integrated the data for us and prepared a table For Example Table A23 for N2 T K EkJkmo1 EkJkmo1 300 6 229 8 723 400 500 28 ENSC 2213 T HERM 0D YNAMI CS Ideal Gas Specific Heats of Some Common Gases kJkg K cp ct k 6p c k c c k Temp Air Nitrogen N2 Oxygen C 2 250 1003 0716 1401 1039 0742 1400 0913 0653 1398 250 300 1005 0718 1400 1 1039 0743 1400 0918 0658 1395 300 350 1008 0721 1398 1041 0744 1399 0928 0668 1389 350 400 1013 0726 1395 1044 0747 1397 0941 0681 1382 400 450 1020 0733 1391 1049 0752 1395 0956 0696 1373 450 500 1029 0742 1387 1056 0759 1391 1 0972 0712 1365 500 550 1040 0753 1381 1065 0768 1387 0988 0728 1358 550 600 1051 0764 1376 1075 0778 1382 1003 0743 1350 600 650 1063 0776 1370 1086 0789 1376 1017 0758 1343 650 700 1075 0788 1364 1098 0801 1371 1031 0771 1337 700 750 1087 0800 1359 1110 0813 1365 1043 0783 1332 750 800 1099 1 0812 1354 1121 0825 1360 1054 0794 1327 800 900 1121 0834 1344 1145 0849 39 1349 1074 0814 1319 900 1000 1 1142 0855 1336 71167 0870 1341 1090 0830 1313 1000 Temp Carbon Carbon 1 Temp K 1 dioxide co2 monoxide co Hydrogen H K 250 0791 0602 1314 1039 0743 1400 14051 9927 1416 250 300 0846 0657 11288 1 104039 0744 1399 14307 10183 1405 300 3 50 0895 0706 1268 1043 39 0746 1398 14427 10302 1400 350 400 0939 0750 1252 1047 0751 1395 14476 10352 1398 400 1 450 0978 0790 1239 1054 0757 1392 14501 10377 1398 450 500 39 1014 0825 1229 1063 10767 1387 14513 10389 1397 500 550 v 1046 0857 1220 1075 0778 1382 14530 10405 1396 550 600 1075 0886 1213 1087 0790 1376 14546 10422 1396 600 650 1102 0913 39 1207 1100 0803 1370 1457139 10447 39 1395 650 700 1126 0937 1202 V 1113 0816 1364 14604 10480 1394 700 750 1148 0959 1197 1126 0829 1358 14645 10521 1392 750 800 1169 0980 1193 39 0842 1 1353 14695 10570 1390 800 900 39 1204 1015 1186 1163 0866 1343 14822 10698 1385 900 1000 1234 1045 1181 1185 0888 1335 14983 10859 1380 1000 Source Adapted from K Wark Thermodynamics 4th ed McGrawHill New York 1983 as based on Tables of Thermal Properties of Gasesquot NBS Circular 564 1955 29 ENSC 2213 T HERM 0D YNAMI CS Table A2l Variation of C with Temperature for Selected Ideal Gases if a 3T yT2 5T3 874 T is in K equations valid from 300 to 1000 K Gas a 1x103 39 yx106 6x109 7 egtlt1012 CO v 3710 1619 39 3692 2032 r 0240 CC 2401 8735 6607 r 2002 0 Hz 3057 2677 5810 5521 1812 H20 4070 7 1108 4152 296439 0807 O 3626 1878 7055 6764 39 2156 N I 3675 1208 2324 quot0632 39 0226 Air quot 3653 7 39 1337 3294 1913 02763 so2 3267 5324 0684 5281 2559 CH4 3826 3979 24558 7 22733 6963 Csz 39 39 1410 19057 24501 16391 4135 C2H4 1426 11383 1 7989 16254 6749 Monatomic gasesa 25 0 0 0 0 a For monatomic gases such as He Ne and Ar 5 is constant over a wide temperature range andis very nearly equal to 52 R 7 t Source Adapted fromK Wark Thermodynamics 4th ed McGrawHill New York 1983 as based on NASA SP273 US Government Printing Of ce Washington DC 1971 r 30 ENSC 2213 T HERM 0D YNAMI CS C Example 1 kg of H20 undergoes a process State 1 gt State 2 T14OOK gt T29OOK P1O1MPa gt P2O1MPa Calculate Ah a Using ideal gas speci c heat b Using ideal gas properties table c Using the Steam Tables Solution System 1 kg of H20 a dh c dT 2 T2 j dh cdT h2 h1 f2ocBTyT25T3ET4dT 31 ENSC 2213 T HERM 0D YNAMI CS From Table A21 h2 h1 w407900 400 13910 9002 4002 1802 210 4152 2964 31069003 4003 109 9004 5004 03983 9005 4005 500 Ah 10250 kJkg 32 ENSC 2213 T HERM 0D YNAMI CS b Using the ideal gas properties table For H20 use Table A23 E1400K 13356kJkmol E2900K 31828kJkmol E2 E1 31828 13356 2 18472kJkmol 18472 k k mol k mol l h h 2 1 1802kg Ah 10251 kJkg Remember h fp for ideal gas 33 ENSC 2213 T HERM 0D YNAMI CS c Using the Steam Tables State 1 gt State 2 T1 400 K 126850C gt T2 900 K 626850C P1O1MPa gt P2O1MPa h1 27302 kJkg 120 27167 12685120 hl 27166 1268 hl 160120 27962 27166 160 27962 h1 27302 kJkg 440 3361 T 62685440 1 30326 6268 112 500 440 31316 30326 500 3488i 112 33409 kJkg 1 12 1 11 34 ENSC 2213 T HERM 0D YNAMI CS D Summarv of Ideal Gas h and u Values T2 1 h2 h139r cng Exact relation if C is known as function of T 2 h2 h1 c T2 T1 Approximate relation OK if A c3 zconst from T1 to T2 B T1 z T2 and c is evaluated near T1 or T2 C Only approximate answer is desired 3 Tables of h and u Exact relationships for ideal gases 35 ENSC 2213 T HERM 0D YNAMI CS THE FIRST LAW OF THERMODYNAMICS FOR CLOSED SYSTEMS REVISITED A Topics For Discussion 1 Obtaining Properties for use in the First Law 2 Example Closed System First Law Problems gtgtgt Read 25 and 313315 of your text 36 ENSC 2213 T HERM 0D YNAMI CS B Example Problems 1 Example 1 A rigid vessel of 20 liter volume contains water at 90 C with a quality of 50 The vessel is cooled to 10 C What is the heat transferred in this process Process Sketch V 20 l V 20 1 H20 gt H20 90 C le Z 2 10 C X 05 X pV Diagram P 37 ENSC 2213 T HERM 0D YNAMI CS Solution System H20 contained in the vessel Mass balance m2 m1 Energy balance 1st law analysis 0 0 0 imQ VEH1IHU2 1111Qyl v Properties Calculations State 1 T 90 C X 05 111 l X1uf1 X1ug1 uf X1ug uf From Table A2 u1 O537685O524945 u1 13857 kJkg 38 ENSC 2213 T HERM 0D YNAMI CS Speci c Volume Enthalpy Entropy m3kg kJkg kJkg kJkg K 7 Temp Press Sat Sat Sat Sat Sat Sat Sat Sat Temp 39 C bars Liquid Vapor Liquid Vapor Liquid Evap Vapor Liquid Vapor C 11 X Us u as hf hfg 13 Sf 88 I 50 1235 10121 1 12032 20932 24435 20933 23827 25921 7038 80763 V 50 55 1576 10146 9568 23021 24501 23023 23707 26009 7679 79913 55 60 1994 10172 7671 2511 24566 25113 23585 26096 8312 79096 60 65 2503 10199 6197 27202 24631 27206 23462 26183 8935 78310 65 70 3119 10228 5042 29295 24696 29298 2333 8 26268 9549 77553 70 75 3858 10259 4131 31390 24759 31393 2321 4 26 53 1 0155 7 824 75 80 14739 54 10291 3407 33486 24822 3349 2308 8 26437 10753 76122 80 85 5783 10325 2828 355 84 24884 35590 22960 26519 39 11343 75445 85 90 7014 10360 2361 37685 24945 37692 22832 26601 11925 74791 90 95 8455 10397 1982 39788 25006 39796 22702 26681 12500 74159 95 V1 1 X1Vf1 leg1 3 0510360X10 052361 1181 m3 kg m 2 V1 V1 sm 6 V1 mV1 3 3 3 20 hter 10 cm m kg m 00169 kg 71118115 Aral Continued Internal Energy liter 106cm3 1181 m3 39 ENSC 2213 T HERM 0D YNAMI CS State 2 T 10 C and V2Vf 1181 0001 2 00111 X2 V sz 10638 0001 g2 u2 1 X2uf2 X2ug2 O98894200 001 1 123 892 6805 kJkg So IQ mu2 u1O01696805 13857 1Q 222kJ 2 40 ENSC 2213 T HERM 0D YNAMI CS 2 Example 2 Air is con ned in a pistoncylinder apparatus as shown below The top of the piston is attached to a linear spring Which is in the unstressed position at the initial condition of the system Heat is added until the pressure of the air doubles at Which time the volume has increased by 50 A What is the nal temperature B How much heat is added p2 Straight line for p linear spring P1 CD 41 ENSC 2213 T HERM 0D YNAMI CS Solution System Air in a pistoncylinder Mass balance m2 m1 Energy balance 1st law analysis 0 mu2 u1g 2 Z1V2 V121Q2 1W2 or the 1st Law reduces to mu2 u11Q2 1W2 Where V2 1W2 L1 pdv ltp1p2gtltV2 Vlgt Why 42 ENSC 2213 T HERM 0D YNAMI CS 1W2 2p12p115V1 V1p1V1 105N Md 2 3 75kJ m bar 10 Nm 1W2 1 32bar 05m3 n 11 p2VZ so T2 2T1 p2V2 RT1 RT2 prl p11 1220 F T210046O R2p11 5V156OX31680 R 2897 jg bar 05 m3 kg pr1 Mpr1 kmol RT1 RT1 8314 k 560 K kmolK 18 43 ENSC 2213 T HERM 0D YNAMI CS m289720518kgbarm3 105N k 8314560 kJ mZbar103Nm m112kg 1Q2 mu2u1 1W2 u1222OkJkg IOOOF 311K u27022kJkg 1220 F933K 1Q2 1127022 2220 75 1Q2613kJ lt Signs on 53 OK 44 ENSC 2213 T HERM 0D YNAMI CS 39TK3th and ukJkg s kJkg K 39 391 when As 39 01 when As 0 u T h u s p v quot 1707 1 450 45180 32262 211161 5775 2236 1j 1512 quot 460 46202 232997 213407 6245 21114 V I 470 47224 33732 215604 6742 2001 g 39 1 48039 148249 34470 217760 7268 1895 490 549274 35208 2198765 7824 151917 39 979 500 50302 35949 221952 8411 1706 155848 39 8878 510 51332 36692 223993 9031 1621 2159634 8080 520 52363 37436 225997 9684 1541 163279 7380 530 53398 38184 227967 1037 1467 165055 7061 540 54435 38934 229906 1110 1397 6761 550 55474 39686 231809 1186 1331 6479 560 56517 40442 233685 1266 1270 6212 570 57559 41197 235531 1350 1212 7 5960 580 58604 41955 237348 1438 1157 g5723 590 59652 42715 239140 1531 1106 5498 600 60702 43478 240902 1628 1058 5286 39 610 61753 44242 242644 1730 1012 5084 620 62807 45009 244356 1836 9692 4894 630 63863 45778 246048 1984 9284 39 4541 640 64922 46550 247716 2064 8899 4222 650 65984 47325 249364 2186 85 34 3934 I 660 67047 48101 250985 2313 8189 3 3672 670 68114 48881 252589 2446 7861 f 39 3434 680 69182 49662 254175 2585 7550 3 3215 690 70252 50445 255731 2729 7256 7 39 400 40038 28616 199194 3806 3016 700 71327 51233 257277 2880 6976 410 41112 29343 201699 4153 2833 710 72404 52023 258810 3038 6707 420 42126 30069 204142 4522 2666 720 73482 52814 260319 3202 6453 430 43143 30799 206533 4915 2511 730 74562 53607 261803 3372 6213 440 44161 31530 208870 5332 2368 740 75644 54402 263280 3550 5982 1 p and Dr data39for use with Eqs 643 and 644 respectively 45 ENSC 2213 T HERM 0D YNAMI CS 47 ENSC 2213 T HERM 0D YNAMI CS ENSC 2213 THERMOD YNAMICS Chapter 2 Energy and the First Law of Thermodynamics AJ Johannes PhD PE Professor of Chemical Engineering School of Chemical Engineering Oklahoma State University Copies of Class Lecture Notes Portions of these notes were originally compiled by Professor Robert L Robinson Jr ENSC 2213 T HERM 0D YNAMI CS ENERGY AND THE FIRST LAW OF THERMODYNAMICS CHAPTER 2 WORK AND HEAT TRANSFER A Obiectives The purpose of this Section is to 0 Introduce the concept of energy 0 Develop some of the relationships for m and heat transfer needed to apply the principle of conservation of energy Our discussions in this chapter are limited to closed systems gtgtgt Sections 21 224 amp 24 243 of your text ENSC 2213 T HERM 0D YNAMI CS B The System A system is anything we Choose to isolate for our study 0 The system is de ned by de ning its boundaries 0 A closed system is one Who39s mass is constant and no mass crosses the boundary of the system System Boundary The System ENSC 2213 T HERM 0D YNAMI CS C Possible Interactions with a SVstem closed system W E We 0 Change in velocity AV 0 Change in elevation AZ 0 Work done W3 W3 WE 0 Heat transferred Q 1 1 ENSC 2213 T HERM 0D YNAMI CS D Concept of Work Work is de ned as follows 82 1VVZ J FbdSb where F force on the system S displacement of the system distance Note the above equation contains an implied sign convention ie work is positive when F and AS are in the same direction More on this topic later ENSC 2213 T HERM 0D YNAMI CS 1 Mechanical Work on a System SYStem Body Forces e g f gravity mg T dS R Ehh39 Boundary Forces 1W2 fid f d8 E d8 23 ds ENSC 2213 T HERM 0D YNAMI CS 1W2E2Ed W 140 STAT 35 2 mg f R ds 1 g0 f2Rds SZ SI 1 g0 but 212majdsz 1 gc 1 gc 2 If dvz szVdV 1 gc dt 1 gc In 2 2 2gCV2 V1 1 Thus EszS 2 s1 my v12 ENSC 2213 THERMOD YNAMICS m m iRdS g 82 81 go 2gC Chan e in Work of Change 111 K gt GraV1tatlonal 1116 1C Boundary P 39 1 Fnerov Forces otentia Energy In thermodynamics we deal with work done by forces at the system boundaries So in thermodynamic terms 1w2 APE AKE Here 1W2 is the work done ON the system ENSC 2213 T HERM 0D YNAMI CS 2 Thermodynamic Work 5W 2 III Fb de W work done m the system Fb force exerted by surroundings on the system at the system boundary dxb distance of boundary movement at location Where Fb is exerted 8W Fb dxb if Fb and dxb are in opposite directions 8W Fb dxb if Fb and de are in the same direction ENSC 2213 T HERM 0D YNAMI CS Thoughts On Energy 0 Energy Can Be Stored Within Systems In Various Forms 0 Energy Can Be Converted From One Form To Another 0 Energy Can Be Transferred Between Systems For Closed Systems 0 Energy Is Transferred By Work And Heat Transfer 0 The Total Amount Of Energy Is Conserved In All Conversions And Transfers Sign Conventions W gtOWork Done By The System W ltOWork Done On The System ENSC 2213 T HERM 0D YNAMI CS T dx 39I I quotI System but szpbA sz 1W2 i Ib pbAde 1 but AdxbdV V2 so 1W2 V 1 note that Sign applies Why ENSC 2213 T HERM 0D YNAMI CS 1 Calculation of pV Work For expansioncontraction of a uid system we nd 1W2 K2 pdV consider the integral X2 2 L de l we must know y fX to do the integration eg y 3X2 4X 7 Thus to evaluate 1W2 K2 pdV we must know the relationship between p and V ie p fV ENSC 2213 T HERM 0D YNAMI CS 2 Example 1 A gas contained in a pistoncylinder assembly undergoes an expansion Compare the work performed by the gas for the following two processes a A process where pb is constant b A process where an constant polytropic process ENSC 2213 T HERM 0D YNAMI CS Solution System the gas a IW2 2 pdez pdV or lvvz p2219 Sinceplzpzpz b in this part an constant 191 szzn IN constant constant n p n constant 0 V V2 V2 constant 1W2 2 L1 L1 1W2 pzvz p1V1 1 n gtforn 1 ENSC 2213 T HERM 0D YNAMI CS 3 Example 2 constant I E 520 Id 5 temperature quotquotquotquotquotquotquotquot a i i i E 200C E E kPa E Calculate the work done by the oxygen during the compression process shown above ENSC 2213 T HERM 0D YNAMI CS Solution System 01 kg of 02 V2 1W2 L1 w ideal gas III T or PV HRT constant 2 p1V1 V2 p1V1 E 1W2 L1 plvlln V1 bUt p E n 2 V M R p1V1 WhCI39C RI V2 V1 p1 p2 at constant T Closed system and p 1W2 mRiTln 1 p2 01kg kmol 8314kJ 2932K 1n g 92 Id 500 32kg kmolK ENSC 2213 T HERM 0D YNAMI CS 4 A Closed System 0 N0 mass crosses the boundary 0 m1m2m 0r Amm2 m10 ENSC 2213 T HERM 0D YNAMI CS 5 Some Closed System pV Work Expressions Process Path Work 0 p const pV2 V1 pzabV pavg V2V1 or P linealquot in V 12l31 P2 V2 V1 0 an const n 721 p2V2 p1V1 Polytropic 1 11 0 Ideal gas mRiT lnV2V1 T const Conclusion 1W2 expression depends on the process path We must know the process path to integrate pdV ENSC 2213 T HERM 0D YNAMI CS F Comments on Work Work De nition 1W2 i f2 FdX For pV work V 1W2 L1 13de 1W 2 is if the system does work on surroundings 1W2 is if surroundings does work on the system Thermodynamic work occurs only at the system boundary Positive work decreases the energy of the system 10 ENSC 2213 T HERM 0D YNAMI CS G Is Any Work Done To decide if work is done in a process observe the system boundaries and ask 1 Does the system boundary move 2 Do electrical leads cross the boundary 3 Do shafts pulley belts drive chains etc cross the boundary If yes work may be done If no 39 no work is done 11 ENSC 2213 T HERM 0D YNAMI CS H Energy Transfer as Heat Heat is de ned as energy in transit across the system boundary due to a temperature gradient at the system boundary Candle H m Sign convention 1Q2 is if energy is added to the system 1Q2 is if energy is removed from the system Positive heat transfer increases the energy of the system 12 ENSC 2213 T HERM 0D YNAMI CS 31 State 1 T1 p1 1quot3 1 other preps 4 v2 22 btate 2 I T2 132 V2 other r0 quot z 0 p pb 13 ENSC 2213 T HERM 0D YNAMI CS A Objectives The purpose of Part III is to Develop the First Law 0 Understand the subdivision of types of energy 0 Introduce the concept of internal energy Present example applications of the First LaW gtgtgt Sections 23 25 of your text ENSC 2213 T HERM 0D YNAMI CS B Inventing the First Law of Thermodvnamics Let s conduct an experiment on the behavior of energy in closed systems The Apparatus J 1 Force Gauge Cii Cooling Water i at a ll Elect1031 Heater Results of Experiment A D A ZQ4OH P ZW20kJ ENSC 2213 T HERM 0D YNAMI CS Example for Experiment A patm ppjston Step 1 p 200 kPa Q p 200 kPa T 300 K heatin T 450 K V005 m3 g V 015 m3 V2 1W2 L1 pdVpV2 V1 200 kPa 015 005m3 k m3 kPa 20 kJ ENSC 2213 T HERM 0D Y NAMI CS Step 2 p 200 kPa Q p 150 kPa VO15 m3 V O15 m3 T 450 K 000mg T 3375 K Q 40 kJ measured for steps 1 plus 2 W 0 Why ENSC 2213 T HERM 0D YNAMI CS EXptNO 2Q m Egg 2W A 40 20 20 B 55 35 20 C 11 9 20 Z 15 35 20 After many experiments we conclude ZQ ZW constant for 1 gt 2 or 1Q 2 1W 2 constant ENSC 2213 T HERM 0D YNAMI CS C De nition of System EnergL 1Q2 1W2 constant independent of path from to This suggests that 1Q2 1W2 has characteristics of a system property So we de ne E2 E11Q2 1W2 OI39 dE ESQ8W K First Law of Thermodynamics Closed System A G p B V 1Q2 depends on the path 1Q 72 1Q2B 1W2 depends on the path 1W2A 7t 1W213 b t u 1Qz1vv2A 1Q21W2B E2E1 ENSC 2213 T HERM 0D YNAMI CS D Subdivision of System Energy Let us break down total energy into components E E PE KE U This leads to 1 E2 E1 mgZ2 Z1 mV22 V12U2 U1 and the rst law becomes closed system 1 U2 U1 mV22 V12 mgZ2 Z11Q21W2 llere U total internal energy of the system 7 velocity of the system 2 elevation of the system m mass of the system g acceleration due to gravity ENSC 2213 T HERM 0D YNAMI CS E Some De nitions System What we specify to study Surroundings Everything external to the system Closed System A system across Whose boundaries no mass crosses Boundary The surface Which separates the system from the surroundings Properties Characteristics of a system e g pressure temperature mass volume etc Which have numerical magnitudes at a given time independent of the history of the system ENSC 2213 T HERM 0D YNAMI CS Some De nitions continued Extensive Property a property Which depends on the size of the system eg mass volume energy Intensive Property a property Which is independent of the mass of the system e g density speci c volume Phase a region of mass in a system Which is of uniform composition and structure e g all liquid gas or solid Note Different phases in a system are set apart by identi able boundaries e g the interface between a gas or vapor and a liquid phase ENSC 2213 T HERM 0D YNAMI CS F What is Internal EnergL Let s add energy to a rigid vessel lled with water Water 1Q 2 T1 P1 T2 P2 Candle 39 Intuitively the water has more energy after heating T19 T2 but AKEV Vfo APEmgz2 zl 0 Thus the increased energy must be stored internally in the water It s at the molecular level in the form of increased translational rotational Vibrational etc energy of the water molecules ENSC 2213 T HERM 0D YNAMI CS G Internal Energy U Internal energy is a property ie it depends only on the state of the system Thus U m X u Where u speci c internal energy and ufTporfTVorfPV For the twophase vapor g liquid f system Umgugmfuf but mmgmf J Jk so uxug1Xuf Where X fraction of total mass in the vapor phase SO mg u uf x m ug uf ENSC 2213 T HERM 0D YNAMI CS U1 T1 p1 V1 State 1 U2 T2 p2 V2 W E11Q2 1W2 3 E2 or E2 E11Q2 1vvz Or 1 U2 U1 mV22 V12mgZ2 Z11Q21W2 process path state properties dependent ENSC 2213 T HERM 0D YNAMI CS I Example Applications of the First Law Example 1 A rigid vessel of 002 m3 volume contains water at 90 C The vessel is cooled to 10 C What is the heat transferred in this process Process Water is cooled in a rigid vessel as shown below V1O02m3 Q2 VZZ H2O H20 90 100C 111 13857 kJkg u2 6805 kJkg v1 181 m3kg ENSC 2213 T HERM 0D YNAMI CS Solution System H20 contained in the vessel Mass Balance m1 2 m2 Energy Balance First Law Analysis AKE APE AU1Q2 1W2 Why U2U11Q2 mu2mu11Q2 or mu2u11Q2 NOWVmVs0 3 V VI l1181m ENSC 2213 T HERM 0D YNAMI CS Solution continued Then 00169 kg 680513857 kJ 1Q2 2 kg 222 k 0 Heat transfer to the system is negative so the system loses energy by heat transfer 0 Note we must know properties of steam in State 1 and State 2 to evaluate Q This is covered in the next section of the course 10 ENSC 2213 T HERM 0D YNAMI CS Example 2 As illustrated below a well insulated tank is divided into two parts by a weightless frictionless partition Initially part A 100 ft3 volume was full of steam at 650 F and 450 psia and part B was empty evacuated The pin holding the partition in place was then removed which resulted in a nal pressure of 60 psia Determine a the amount of work performed by the steam and b the change in the internal energy of the steam Solution V2 a System steam so IW2 2 I pde Now What is doing the work and Doing work against what 1W2 E O Because pb 0 there is no resistance to do work against ENSC 2213 T HERM 0D YNAMI CS Solution continued b System steam Mass Balance m1 2 m2 Energy Balance First Law Analysis AKEAPEAU1Q2 1W2 For our process AU 0 Alternate Solution b System entire contents of apparatus both sides of the piston Mass Balance m1 m2 Energy Balance First Law Analysis AKEAPEAU1Q2 1W2 Again for our process AU 0 Conclusion A gas expanding into a vacuum does no work 3 ENSC 2213 T HERM 0D YNAMI CS Example 3 Gas is contained in a pistoncylinder assembly As shown in the gure on the next page the piston is restrained by a linear spring The initial pressure and volume of the gas are 150 kPa and 0001 m3 respectively and the spring touches the piston but exerts no force at the initial position The gas is heated until the volume is tripled and the pressure is 1000 kPa If the pressure exerted by the piston and the spring is given by the following relation p 50 425000v 0001 p in kPa and Vin m3 a How is the above equation consistent with p1 150 kPa v1 00011113 b Determine the work done by the gas during the expansion c Would the gas do more or less work if the spring was removed ENSC 2213 T HERM 0D YNAMI CS Solution System the gas V2 Part A and B 1W2 I pdV 131 P patm ppiston pspring p 150 425000V 0001 1W2 2150425000 V 000l dV 2 V2 1W2 150V 425000V7 425V V1 0003 1W2 2125002 275V 0001 115 kNm 115 k ENSC 2213 T HERM 0D YNAMI CS More easily done when p is linear in V 1W2 Pavg AV 2150 1000kPa 0003 0001m3 115 kJ b The gas will do lei work because the resistance offered L5 16 ie V2 V2 1W2 L pdV L 150dV150V2 V1 0139 1W2 1500003 0001 03 kJ P Wu o 0 Chapter 2 o W Eguatjons O V p Sp 60 weight v riggc 524 lbm z 1000 kgm3 or 1702 Vi P401150 17 NEW Vi linear lsz ini1n Wgt o Wlt0 ON system oooow UIrngu gmfuf uufxugu or h or V Fuuf uguf o FVVfVgVf Vapoi mgvg hq pv enthalpy p Jp TRTT c V V RT p pseudo reduced Volume m am QWmzhzg22Z 2Vzz rhihig2iV2Vi Wmhh2 adiabatic 0 PE0 Q mpVA or VAV Turbine TABLE 34 Valual cl um Mall on eonum rhirhzp ltpi 2 W7 r h 412 gt0 Value of 1 Units 3311 x 1m eragmol K calgmol 10 831 m moi K mmpCompressor 0082057 ma amo 8 rh rh gt 32057 cm at amu gig m 6236 39 mm sltrmo1 x 2 00848 kgcm1 menrum 0 9959 mm Hg IOlbmo 19 H m K 19869 mil moi 11 Heat Exchangers 7305 x 104 hp hrlbmnl Q0 memz pzgtp W o 5319 x 1w kw hrlbmol m o 1 39atm imammoi QIZ T IG Z39hI equnxhz 555 mm HamOM am hi 10731 psi 3lbmol 11 1545 Ibi lbmol m 1351 x m lbrinlb nol quot11 Oxygen gas is stored in a tank with a Volume of 5 ft3 at a temperature of 400 R and pressure of 4000 psia Determine the mass in the tank in lbm Justify your calculations and any assumptions you make Ammonia is stored in a tank having a Volume of21 m3 Determine the pressure in kPa andthe mass ofammonia in the tank as sumingsaturated liquid at ODC Water is enclosed in a container at 75 C Ifthe speci c Volume ofthe water in the container is 24 mZkg determine the quality Three lbm of water is contained in a rigid container having a Volume of 45 Initially the pressure is measured to be p 350 psia The container is then cooled until the pressure drops to p2100 psia a What is the initial intemal energy u of the water in Btu lb b Find the quality C n L L Ir of of 60 lbfin2 and temperature of 631 Determine the internal energy of steam at a pressure of 237 bar and a temperature of AJ has a piston cylinder system that contains air at 640 F 294 psia and an initial volume of 7000 ft3 The air is heated at constant pressure until the volume is doubled Assume ideal gas calculate the work heat transfer and change in internal energy on multiple exams Water is enclosed in a container at 120 bar If the internal energy of the water in the container is 22015 kJkg determine the quality Determine the specific volume of steam in ft3lbmat a pressure of 450 lbfin2 and a temperature of 491 F Determine the specific volume of propane at 60 F and 10 psia Nitrogen gas is stored in a tank with a volume of 5ft3 at a temperature of 261 R and a pressure of 2000 psia Determine the mass in the tank in lbm Steam enters an adiabatic Q0 turbine at 10 MPa and 500 C and leaves at 10KPa with a quality of x9 KEPE0 Determine the mass flow rate for a power output of 5MW You have been asked to finish the design of a steam turbine The design calls for steam to flow into the turbine through an inlet pipe having a diameter of 02 m At the inlet the velocity is 100ms at conditions of 14 MPa and 600 C The steam exits the turbine through a pipe of diameter 08 m at conditions of 500 kPa and 180 C PEKEQ0 Calculate power output of the turbine in kW Butane gas is stored in a tank with a volume of 12 m3 at a temperature of 165 C and a pressure of 19 bar Determine the mass in the tank in kg Steam enters the turbine at 1600 lbfin and 1000 F The adiabatic Q0 turbine operates at steady state 25 of the steam entering is extracted midway through the expansion 2 process in the turbine at 160 lbfin and 450 F The rest of the steam exits as saturated vapor at 1 lbfinz The turbine develops a power output of 910 BTUhr KEPE0 Determine mass flow rate entering the turbine Given Temperature and Pressure T P Go to the saturated temperature table and determine if p gtpsT Subcooled compressed liquid p psT Saturated state may be saturatedvapor saturated liquid or twophasernixture p ltpsT Superheated vapor T ltTsp Subcooled compressed liquid T Tsp Saturated state may be saturatedvapor saturated liquid or twophasernixture T gtTsp Superheated vapor Given Temperature and Speci c Volume T v Go to the saturated temperature table and determine if V ltVfT Subcooled compressed liquid V VfT Saturated liquid VfT ltV ltVgT Saturated twophase state V VgT Saturated vapor V gtVgT Superheated Vapor Given Pressure and Volume p v Go to the saturated pressure table and determine if V ltVfp Subcooled compressed liquid V VfP Saturated liquid VfP ltVltVgp Saturated twophase state V Vgp Saturated vapor V gtVgp Superheated vapor Nitrogen gas is stored in a tank with a volume of5 ft3 at a temperature of 400 R and a pressure of 4000 psia Determine the mass in the tank in lbm Three pound mass of water is contained in a rigid container having a volume of 45 ft3 Initially the pressure is measure to be pl 450 psia The container is the cooled until the pressure 100 psia a What is the specific internal energy 111 in BTU1bquot What is the final quality How much total heat was transferred 35 You have been asked to nish the design ofa steam turbine The design callsfor steam to ow into the turbine through an inlet pipe haVing a diameter of 02 m At theinlet the velocity is 100 ms at conditions of 14 MPa and 600 C The steam exits the turbine through a pipe of diameter 08m at conditions of 500 kPa and 180 C Determine a Mass ow rate through turbine in kgh b Velocity of steam at exit in ms c Q0 calculate power output in kW

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