{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ch3 - EGN 33587 Thermo-Fluids-Heat Transfer Lecture 4...

Info iconThis preview shows pages 1–14. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 12
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 14
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: {________________ EGN 33587 Thermo-Fluids-Heat Transfer Lecture 4: Properties of a Pure Substance “Chapter 3 — Part 1” EGN 335a 411mm Flimls-Hpm’l'mtsfe: State Postulate - The number of independent properties needed to completely defifié”?fi§‘§t§t§bf a hermodynamic system is one more than the number of possibly relevant work interactions. - The state of a simple compressible system is completely specified by 2 independent, intensive properties. 2947335: - T'limm»Flmds-Hent'rmnsl:r State Postulate for Simple l- One plus one equals two. 1 . Therefore, for a simple system, two independent properties are needed to completely define the state of the EGN 7.1159 . Tillflom+1>|ids>HeflTnuzfr! Chapter 3 Properties of a Pure Substance EGNrslfiK A TmmibAFlinrernai Tunsfm Simple System *- A simple system is one where there is only one possibly relevant work interaction, eg. i expansion or compression work. EGN 435x — Themia EGN-315$ _ 711mm. rm Experiment - heat water at constant'pressure: 1 _Pl Experiment - heat water at constant pressure: I Pl V p2 156N439 _ humimnmxs mm ngm EGN .aasx . Fiemm»Fimxi(»Hur Tnms’i'r Experiment — heat water at constant pressure: P: 1 atm Superheated vapor p; 2 3 Saturated I Saturation Curve Compressed Uquid Specific Voiume, v aoNassx . Thumm-Fliiids-Hnn Trmisfuz Pressure Pressure Specific Volume Specific Voiume amass: —Tncn:m-F1ukinHui Tamra: EGN >325: . 'nu-mmnmk.Hmrrunqu ’l'lm I‘vlrL'I‘Sm-l’aue for a Real Substance Substance contracts Substance expands on freezmg on freezing EON-315K _ ThL-mioVFlvridereuTrans!“ mu m s mi 1m Spenlnomumc. imééha‘i'éiiuav. ’ ’ ’ ' Efllhulpyi Enlrnui “v.3” . A . .N'TKB' ‘ inn San 5.1. an, 5.x. . 5m. Yam» rem, liqulfl. 5n. uauid. mp“ aim, MINI, “c w- i P EGN .535» . 'nmnmrimi-wHi-m maul-x Saluraiiun pressure is the pressure an which me liquid and vapor phus K: are in equilibrium (ll :1 given temperature. Salumliml temperature is [he remperm’urc :1: which ilw liquid and \uprir plums “'0 in equilibrium Lil a given pressure The sulvxcriplfgr mud in ’l‘nlilcx Ami :le :‘\-5 refer; in the differ-ewe between liiL‘ Szuurulml vapor value 11ml the saturated liquid value region Thai is. cod J m m, mm :x mm m2 lulu as ilk“ \. .ipemhu‘c ur rruuuar mm. c». and lwcuum mu (l\ (he mum] pumi. EGNJI.W Equations of State ' Relation among three thermodynamic properties. i 9 Maybe: —algebraic —tabular —graphical | ES 56 7 'fiimnmfim<l5-Heqr mm" mum “an! mm m. swamp". 9m x...» mm. Haw a a BGN 4m . Tllmno-Fliuds-l‘lvm Tmuslu Quality and Saturated liquidJmpor Mimrre ”HUI? (”mum mm ll WUXX um}! , )‘ ( ¥A if??? e 111:1le +17" HI\' = in, l", +Hlul’ ”111’: ”12V? . + . . r: 7H Recall the delmnitm of qualit} .r ”7:; \=4— m m, + 111,, IIIy I11 — HI? III \‘=(l'.\)l",. in}, was a» mxwflwermut 4:; a F (’nmpresscd Liquid Water Table A cutictzmcc is \ilid to he :1 compressed liquid \vhcn the prcsmrc is greater than thr- saturation pru‘xmt: for HR: temperature. nu! Aer l u n e k-bkg lerkg~K "Ikg “my lurks lelkg Steps in Using Property Tables - 2 - Action depends on results of previous step: - if B or D (Saturated Liquid or Saturated Vapor) ECNAHR v'nlrnmv-Hmllnurul Tmlufrt i' Snpcrlwuml Writer mm- A substance is said to be superheated il’ihc glx’en temperature is greater ihun Ihc saturation tempcmmrc l’vr lltt‘ given pressure. EGN am A fiiemianlmdert-ll13:“!fo . . ' " ‘W Steps in Usmg Property Tables — 1 - Determine substance: . Determine fluid Condition A Compressed liquid Saturated liquid ’ Wet vapor (Sat. Mixture) Saturated vapor Superheated vapor EGN >335: _ “ituflDrF‘lldSVHEM Trmslux lf Saturated (B or D) - Assume the first known property is pressure it The value of the second known property (except temperature) e<actly matches the saturated liquid or saturated vapor value of that property. ‘ The condition of thefluid is saturated liquid or saturated vapor respectively, EGN 7335‘ , fivemm-nmm») lv-m Traits!" lf Saturated (B or D) ° Value of second known property matches the saturated liquid or saturated vapor value corresponding to known pressure or temperature. Go to appropriate saturation table and read all properties directly from table, either saturated liquid or saturated vapor values. saw-335x _ ThmnanlIudsJ-Iear Tum!" WetVapor(C) ' Value of second known property is between saturated liquid and saturated vapor values corresponding to known pressure or temperature. - Calculate quality from second known property and corresponding saturation values. EGNJJSK A Tlxmnu-Flnids-HeatTwusfu: Steps in Usrng Property Tables -4 ‘ l - Action depends on results of previous step: ' -w__._l ' If B or D - if C ~ if E(Superheated Vapor) swabs» A ThumvahMls-HenrTwist" Steps in Using Property Tables g . Action depends on results of previous step: ° ltBorD - if C (wet’vapor region) Eowaasx _ Thzuno»F]nle$~HMvTmm‘fi'X Wet Vapor (cont’d) Quality is the mass fraction of the mixture that is vapOI v—vu—u ~hs—s gfgfgfgf EGN 7335!! A 'l‘lmum-Flluds-Hent Ttwsirz Superheated Vapor (E) - lfthe value of the second known property is greater than the saturated vapor value corresponding to the known pressure or temperature. ° Go to appropriate superheat table and read other properties, using 5 l l linear interpolation as necessary. Ecru”: , T'Iwnm-Rmdsfinl Twist}! ; uper eats T 0C a C m3/ kg 1(J/ kg kJ/ kg kJ/ l, = 1.4 bars = rm MP3 (1,. = 41.91%) 0 ms ‘ 161.52 0.1179 162.50 (1.1235 157.59 177.87 179.01 134.97 0.I289 I72.94 1 190.99 1 0.134] 1 711.211 0.3197 183.67 0. 1449 189.17 (1.1511? 194.72 0.1553 2110.218 (1.1005 206.011 0.1707 217.74 0.1809 229.67 197.015 203.23 2119.06 215.75 222. 12 228.55 241.64 255.00 sax ma . 'nwmm-Fhmlsvh‘lnrTwistsr 1 1 1 0.7102 0.7147 0.7378 0.7602 0.7821 0.8035 0,824.1 0.3447 0.3648 (1.8844 11.9225 0.9593 0.0922 1164.20 180.81) 1 0.0925 i164.” 131.03 0.0991 0.11134 0.11176 0.1118 0.1160 0.1201 (1.1241 0 1322 (LMUZ l . 194.35 172.37 190.21 177.77 183.23 196.38 202.61) 215.23 200.02 221.64 205.75 217.47 229.45 228.12 241.27 254.69 1 188.77 § 306.89 1 1 l 1 1 (1.7054 (1.7181 _ (1.7408 0.76.10 0.7846 0.8057 0.8263 0.8464 0.8662 0.9045 0.91114 Compressed Liquid - 1fthe value of the second property is less than the saturated liquid value corresponding to the known temperature or pressure, the, fluid is compressed liquid, - 1f compressedliquid tables are available, they may be used to determine prope’ties in a manner similar to superheat tables. eon 1m 7 Timmnzmweur Tamera 1 ‘ Steps in Using Property Tables I; V e Action depends on results of previous step: .- if B or D 1- if C . if E e if A- compressed liquid EGNV'USS A Wanna-Fluids +12." Tmmru ..,. . ..... W. . \L If compressed tables are not available, use theapproximation that, in the compressed liquid region, v, u, h, and s are functions of temperature only. Then the saturation table (pressure or Temperature) may be used to approximate properties. scream . T‘hrmmelu-risJ‘lm Music: "”3 1‘: 7 Mill. \\\'1 mumnud umuumtud liquid 40% 57% Lima; mic. 13 5 1-1 i; = 135711—— 2 kg 3:5”? . I diagra It. i 3 Mumum -_- it; . I'm 7 mm ”I > 9 y = 4}: 0.05 7'” M mx‘islnrc hc y. awn-m .|.~ then, m: quaiuy i5 .\' 1— y = 1—005 = 0.95 ,3... zmgiutmg TuMc A-g U =11, tun“ "lid = 1257-6 + 095<25305— 1257.6) Nuiicc mm wu muid him: Ind! ":11, +.ru,,_, Lucuk- mm 1 on His 'llr. P4: and l’-'/ diagram» Laid: sum: 4 an the 1—12 ‘. I": 7' MKL fl — IOWC ‘ me A4. A: f~ H3017, Pm: — unpyzzmzd nqazxd. Apprmxmmu xoiuxinn: = 418.94g kg llEll,;.H.,1m(- Emu-135x A 11\mwaF|Im1r»HuA(TmnSrH ‘ T fir; 4.: 0 460 500 Using llk‘ above mhic. form Ihc i'nilmvmg mrinx 460— 450 _ u — 2978.0 500— 450 _ 3073.4 — 2978!) Id kg u = 2997.1 Locale \knc (: (m the T-v, P-r. and I"-7’diugrum\. lame—335K — ‘fiwmm-Fimtis-Hmr Tum-x P recess Lines - p-v Diagram eisobaric - isometric - Isothermal - ,rAdiabatic EGN-JBSK A T‘.)e:mmfiwls~}hn( Transfer (1) L. :3 a) m (D L. D. Specific Volume ~‘anr Tums!" EGN am . Tlriumo-Fhm\s-HI1l Tmnsfi'x Isobaric Processes P2>P1 Pressure Constant p1 Specific Voiume am .33va V Thnmw-fimxis-Hm mink : Isometric Processes Isothermal Processes [ . Constant v1 Pressure Constant T Specific Volume EGN .3353 _ 11mm Jainism... Trunk! Specific Volume Eon-335x , Thauua-H\id§»Hn:ll Tmni'et Example Adiabatic Processes Saturated liquid water contained in a closed, £9551 ' tank is cooled to a final state where the temperature is 50°C, and the masses of saturated vapor and liquid are 003 and 1999.97 kg respectively. Determine: The initial temperature [00], and the volume of the tank [m3]. ‘ Sketch the process line on a p-v diagram relative E to the saturation curve. Pressure Specific Volume EON-3A“ _ memmhmdsmm Transfer Example - Solution — p-V diagram 1 Example - Solution - I Pressure State 1: 1’ State 2: Saturated T = 50°C liquid mI : 1999.97 kg T1 = ??°C mV = 0.03 kg V = ?? m3 , Specific Volume Example - Solution -= Equations: Example — Solution — Numerical ’ values x2: mv 2mm; =15x10“5‘ ml+mv 199997+ 0.03 v2 = vf +x(vg —vf) =i.021x10‘3 +(15x10—5)(12.032~i.021><10_3) 3 21.193x 10431 kg ideal Gas Equations of State : Example — Solution — Numerical values A fluid far enough from its saturation curve that it is not likely to condense into the liquid phase in any process is generally referred to as a gas. _ ”1v _. Viv? - m, my i99997+ 0.03 VZ :v/ +x(vg~vf) =1021x10'3 +(15x10‘5)(12.032—l.021x10'3) 3 =1.193x10"3 5’: k g & from Table A-firead Vf= 1.193x10'3 at T = 220°C ‘ Therefore: x2 1.5x 10’5 1F the temperature is high enough and the pressure is low enough, we consider it to be an ideal gas. I! \leh» 7 g V = (2000 kg)(l.l93xlO'3 m3/kg) 2" Leia: Sam. Vigié‘ wJM, $33. w? aim-2f main» damage: . ldealGas - Quantitative for an Ideal Gas ' i‘ . . . pv = constant 1F a fluid is a gas and T T/Tc > 2 or NH < DJ The constant is called the “gas constant”, and is given the symbol “R”. we consider it to be far _ enough from the saturation Each different gas has curve to behave as an ideal a different value of R. gas. imam» » TlmnumFluiilx-Hui megm :3er 7 11mm: 7mm: Hun Tmnsf“ Universal Gas Constant Equal numbers of molecules behave similarly. Yield same value of gas constant, called universal gas constant, Ru. Values in front of text: Ru = 8.314 kJ/kgmol K, 1545 ft 1bf/1bmolR, 1.986 Btu/1bmol R. EGN 7.1m 7 'l'lwmmrFluirlyHear Tnmclrr I Combining the previous ideas, we may generate several versions of the Ideal Gas Equation. 10 r—‘—._—*—w Relation between R and RU A mol is an Achardo’s number of molecules, a mass equal to the molar mass of the. gas in some mass unit, eg. kg, 1b, etc. Therefore: 12 EGN 733$» , Thvmvahmler-w Tnmfvr :Compressibility factor ~ a measure of deviation from ideal gas behavior pv = ZRT z = team we? Z = Vactual/Videal Videal = RT/p EGN ms» , Themlthuv—lshmi’rmmfvr émmflfi 1 gfiifimw Mwammi flaw? a g; kg gig”? E a M23559? 5;» PS 5‘ WW” *3"? ”f; 9.255%?«5? Q ‘1’???» PN-gflfiwmfiaw‘? * .Magflg Mgw 5. 5;.ng ”45%me mfigafii 5555?“? “g”??? g W m 5 m {:5 m5 at g» jay {“1 £3"? “65%“? a “ W i ”3%“; mm M;? £3 NN‘ 5:: W} W 5*; > I 51%} ‘ f? Na 5% W é N? {fig}??? 55:) 5% “Nammmflm.Wm {'5 s} ...
View Full Document

{[ snackBarMessage ]}