上海交通大学：《热力学 Thermodynamics（I）》课程教学资源（课件讲义）Lecture 13_Equation of state and ideal gas model

上游充通大￥ SHANGHAI JIAO TONG UNIVERSITY Engineering Thermodynamics I Lecture 13 Spring,3/21/2018 Prof.,Dr.Yonghua HUANG 强 MAAMLMAMMAA http://cc.sjtu.edu.cn/G2S/site/thermo.html 1日G

Engineering Thermodynamics I Lecture 13 Spring, 3/21/2018 Prof., Dr. Yonghua HUANG http://cc.sjtu.edu.cn/G2S/site/thermo.html

The relationship among p,v,T of gases? Customary T uoden leeyJedns Solid Liquid pIbrT-P!oS Critical pqint Constant- P>Pat Vapor pressure line Per Critical point P<Pex Phase Vapor change Liquid Solid-vapor Specific volume Temperature Ver 上降文通大学 March 21,2018 2 SHANGHAI JIAO TONG UNIVERSITY

March 21, 2018 2 The relationship among p, v, T of gases? superheat vapor Customary liquid

Retrieving thermodynamic properties TABLE A-6 Superheated water T S Tables ℃ m3/kg kJ/kg kJ/kg kJ/kg-K P-0.01MPa(45.81℃) Sat. 14.670 2437.22583.9 8.1488 50 14.867 2443.32592.0 8.1741 100 17.196 2515.5 2687.5 8.4489 150 19.513 2587.92783.0 8.6893 200 21.826 2661.42879.6 8.9049 250 24.136 2736.1 2977.5 9.1015 300 26.446 2812.3 3076.7 9.2827 400 31.063 2969.33280.0 9.6094 Graphs 500 35.680 3132.93489.7 9.8998 600 40.296 3303.33706.3 10.1631 700 44.911 3480.83929.9 10,4056 How? 800 49.527 3665.44160.610.6312 Is it possible to have some simple relations among the properties that are sufficiently general and accurate? Equations RT a of State PV=RT 元-bD2 Computer software (EES,IT,Refprop...) they are based on EoS 上游充通大 March 21,2018 3 SHANGHAI JIAO TONG UNIVERSITY

March 21, 2018 3 Retrieving thermodynamic properties How? Tables Graphs Equations of State … Computer software (EES, IT, Refprop…) they are based on EoS pv RT  Is it possible to have some simple relations among the properties that are sufficiently general and accurate?

Ideal Gas Model ldeal gas as a collection of perfectly hard amssald Liquid spheres. Solid Critical all collisions between atoms or point p<<pc molecules are perfectly elastic and in which there are no inter- Triple point; VTT molecular attractive forces. Temperature infinitesimally small,taking up no volume themselves (point masses) Ideal gas equation of state iPU RT pV=mRT p而=RTpV=nRT 上降文通大学 March 21,2018 4 SHANGHAI JIAO TONG UNIVERSITY

March 21, 2018 4 Ideal Gas Model Ideal gas equation of state c p p T T  c • as a collection of perfectly hard spheres. • all collisions between atoms or molecules are perfectly elastic and in which there are no inter￾molecular attractive forces. • infinitesimally small, taking up no volume themselves (point masses) Ideal gas

Universal Gas Constant Measured data extrapolated to T 1 吧 zero pressure Volume per mole T 四 U=MU molecular T weight R卡 V2 T3 T p lim R R R= (kJ/kg-K) p→0 T M for all gases Universal gas constant =8.314kJ/kmol·K 上泽通大学 March 21,2018 5 SHANGHAI JIAO TONG UNIVERSITY

March 21, 2018 5 Universal Gas Constant p1 , v1 p2 , v2 p3 , v3 T T T for all gases  Universal gas constant Volume per mole R R M  (kJ/kg·K) molecular weight

Check the validity of Ideal Gas Model:Air R=287.06 J/(kg-K) TK D/atm v/m/kg Vexp/m'/kg err (% 300 0.84992 0.84925 0.02 300 10 0.084992 0.08477 0.26 300 100 0.0084992 0.00845 0.58 200 100 0.005666 0.0046 23.18 90 1 0.25498 0.24758 2.99 RT 287.06×300 Calculation V= 0.84992m3/kg p 101325 V-Vexp_0.84992-0.84925 Relative residual =0.02% 0.84925 上游充通大学 March 21,2018 6 SHANGHAI JIAO TONG UNIVERSITY

March 21, 2018 6 Check the validity of Ideal Gas Model: Air R= 287.06 J/(kg·K) 287.06 300 3 0.84992 m / kg 101325 RT v p  Calculation    Relative residual = exp exp 0.84992 0.84925 0.02% 0.84925 v v v    T/K p/atm v/ 3 m /kg vexp/ 3 m /kg err（%） 300 1 0.84992 0.84925 0.02 300 10 0.084992 0.08477 0.26 300 100 0.0084992 0.00845 0.58 200 100 0.005666 0.0046 23.18 90 1 0.25498 0.24758 2.99

Ideal gas model-properties Generally for gases:u and h~f(T,p)or f(T,v) Ideal gas:temperature dependent only pU=RT→ Accuracy OK uu(T) h=h(T)=u(T)+RT 上游充通大学 March 21,2018 7 SHANGHAI JIAO TONG UNIVERSITY

March 21, 2018 7 Ideal gas model - properties Generally for gases: u and h ~ f(T, p) or f(T, v) Ideal gas: temperature dependent only Accuracy OK

Example 13.1 Air as an ideal gas air undergoes a thermodynamic cycle consisting of three processes. Process 1-2: Assume:ideal gas Process 2-3: expansion P2=2 bars Process 3-1: (p) compression v=C T=C Given:T=300K,p=1bar; P2=2bar Find:T2,V3 P1=1 bar p=C 600K 300K 上游通大学 March 21,2018 8 SHANGHAI JLAO TONG UNIVERSITY

March 21, 2018 8 Example 13.1 Air as an ideal gas air undergoes a thermodynamic cycle consisting of three processes. Process 1–2: Process 2–3: expansion Process 3–1: compression v T p Given: T1=300K, p1=1bar; p2=2bar Find: T2 , v3 Assume: ideal gas

Solution P2=2 bars Process1→2 State 1 = RT U=C po-RT P1=I bar p=C P RT 600K 300K 2 bars (300K)=600K p At state 3: v3=RT3/p37 RT2 T Mp 2 3T3=T2 kJ 8.31 kmol·K kg 1P3=P1 28.97 ol 3 =1.72m3/kg 上游充通大 March 21,2018 9 SHANGHAI JIAO TONG UNIVERSITY

March 21, 2018 9 Solution State 1 State 2 1 1 1 p v RT  2 2 2 p v RT  Process 12 At state 3: 2 3 T p 3 1

Example 13.2 A gas meter shows the gas consumption is 68.37 m3,the (time)average gage pressure 44 mmH2O,average temperature is 17 C,the atmosphere pressure is 751.4 mmHg.Determine, 1)gas consumption in standard m3(标方：0℃，1atm); 2)If the average pressure drops to 30 mmH2O,gas consumption in standard m3 for the same gas meter reading; 3)If the average temperature is 30 C,what happens? Solution:1)the pressure is not high,gas could be considered as ideal gas. m=p'=当 RT RTo →=BIy-7514×13.32+44×981Pa 273.15K ×68.37m3 Po T 101325Pa (273.15+17)K =63.91m3 上游充通大 March 21,2018 10 SHANGHAI JIAO TONG UNIVERSITY

March 21, 2018 10 A gas meter shows the gas consumption is 68.37 m3 , the (time) average gage pressure 44 mmH2O, average temperature is 17 ℃, the atmosphere pressure is 751.4 mmHg. Determine, 1）gas consumption in standard m3 (标方: 0℃, 1 atm); 2）If the average pressure drops to 30 mmH2O, gas consumption in standard m3 for the same gas meter reading; 3）If the average temperature is 30 ℃, what happens? 1）the pressure is not high, gas could be considered as ideal gas. 0 0 0 pV p V m RT RT   Solution：     1 0 3 0 0 1 3 751.4 133.32 44 9.81 Pa 273.15 K 68.37 m 101 325 Pa 273.15 17 K 63.91 m p T V V p T          Example 13.2