Thermodynamic Properties - Thermodynamics - Lecture Slides, Slides of Thermodynamics

Download Thermodynamic Properties - Thermodynamics - Lecture Slides and more Slides Thermodynamics in PDF only on Docsity! Thermodynamic Properties August 26, 2010 ME 370 – Thermodynamics 1 Unit one – Properties of Pure Substances Larry Caretto Mechanical Engineering 370 Thermodynamics August 26, 2010 2 Outline • Extensive, E, m, intensive, T, P, , and specific, e = E/m, variables •  = m/V = 1/v => density = 1/(sp. vol.) • Look at P-v-T data for real substances • Ideal gases PV = nRuT = mRT – Engineering gas constant: R = Ru / M – With v = V/m, Pv = RT • Use of V or v = V/m gives difference between PV = mRT and Pv = RT 3 Thermodynamic Variables • Extensive variables (volume, mass, energy) depend on size of system • Intensive variables (pressure, temperature, density) do not • Specific variables are ratio of an extensive variable to mass (e.g. specific volume, v = V/m) • General notation, if E is an extensive variable, e = E/m is specific variable 4 Thermodynamic System • Basic unit of analysis defined by boundaries where all interactions occur • State of system defined by properties • Interaction of system with surroundings – Can interchange energy and mass – Open/closed systems: Have/Do not have mass addition or removal • Open system called control volume – Isolated system: constant energy and mass – All interactions across boundaries 5 Thermodynamic State • Two independent intensive properties determine the state of a simple compressible system • Once the state of a system is deter- mined all other properties are known • Thermodynamic processes are a continuous series of states known as a quasi-equilibrium process 6 Ideal Gas Equations • From chemistry: PV = nRuT • Ru = Universal gas constant = 8.314 kJ/kmol·K = 8.314 kPa · m3/kmol·K = 10.73 psia ·ft3/lbmol ·K • Engineering gas constant, R = Ru/M • Definition of mol, n = m / M • PV = (m/M)(MR)T = mRT • P(V/m) = Pv = RT PV = nRuT PV = mRT Pv = RT Thermodynamic Properties August 26, 2010 ME 370 – Thermodynamics 2 7 PR = P/Pcrit TR = T/Tcrit Figure 3-51 from Çengel, Thermodynamics An Engineer Approach, 2008 Z = 1 is ideal gas 8 Why Use Ideal Gas? • Simple equations • Real gas behavior close to ideal gas for low pressure (or high temperature) • Want P low compared to Pc • Example calculation: find specific volume of water at 1,000oC and 10 kPa • Compare to property table value of v = 58.758 m3/kg for water (page 918) 9 Finding R 1kJ = 1 kN·m = 1 (kN/m2)·m3 = 1 KPa·m3 10 Ideal Gas Calculation • Table A-1, page 908: for water, R = 0.4615 kJ/kg·K = 0.4615 kPa·m3/kg·K kg m kPa K Kkg mkPa P RT v 3 3 756.58 10 )15.1273( 4615.0    • Error is only 0.0017% at this point • Ideal gas often used for “permanent gases” such as air 11 Ideal Gas Calculation II • A volume of 20 ft3 at a pressure of 300 psia contains 20 lbm of Helium; what is the temperature? • T = PV / (mR) • R = 2.6809 psia•ft3/lbm•R (page 958)    FR Rlb ftpsia lb ftpsia mR PV T o m m 7.3479.111 6809.2 )20( 20300 3 3     12 P-v-T for Real Substances • Three phases: gas, liquid and solid • Look at gas and liquid • Thought experiment – Add heat to a liquid at constant pressure; measure T and v – After a while vapor starts to form • Heat addition increases volume by converting liquid to gas while T is constant (at constant P) – Finally have only gas that expands readily • Ideal gas has v/T = const (at constant P) Thermodynamic Properties August 26, 2010 ME 370 – Thermodynamics 5 25 What is v for x = 0.25? • Found from Table A-4 with lookup T – P = Psat(T = 220oC) = 2319.6 kPa – vf(T = 220oC) = 0.001190 m3/kg – vg(T = 220oC) = 0.086094 m3/kg gffgf fg f xvvxvvxvv vv vv x     )1()( kg m kg m xvvxv gf 33 086094.0 )25.0( 001190.0 )25.01()1(  kg m v 3022416.0  26 What if P and T are Given? • If these are independent properties then they define a state • The “superheat” (gas) tables for water (Table A-6, pp 918 – 921) have a set of tables for P with rows of data for v • Data are also given for other properties • Basic question: Given T and P, how do we know if we have a liquid or a gas? 27 28 Finding State with Given T and P v P Sat T = T1 < T2 T = T2 >T1 Psat(T1) Psat(T2) vf vg Gas when T > Tsat(P) or P < Psat(T) Liquid when T < Tsat(P) or P > Psat(T) 29 General Rules • When we are given T and P we have a single phase. Is it gas or liquid? – It is a gas if T > Tsat(P) or P < Psat(T) – It is a liquid if T < Tsat(P) or P > Psat(T) – “Superheat” tables have Tsat by P – Can guess superheat and check • Example: R134a at 0oC and P = 100 kPa, 400 kPa • Guess superheat and find at 100 kPa and 0oC, v = 0.21630 m3/kg; at 400 kPa there is no 0oC entry • Data from page 929 30 Thermodynamic Properties August 26, 2010 ME 370 – Thermodynamics 6 31 Results • v(0oC, 100 kPa) = 0.21630 m3/kg • What is v(0oC, 100 kPa)? Page 929 • Why are there no data at T = 0oC and P = 400 MPa (= 0.4 MPa)? 32 General Rules Continued • The example did not find a 0oC entry at 400 kPa because this is a liquid – The header 0.4 MPa (Tsat = 8.91oC confirms that T = 0oC < Tsat = 8.91oC • Approximation for compressed liquid: v(T,P)  vf(T) Note it’s vf(T) – To get answer for 400 kPa and 0oC look at saturation table for T = 0oC, A-11, page 926 – v(0oC,400 kPa)  vf(0oC) = 0.0007723 m3/kg 33 Two More Problems • What is the specific volume for refrigerant 134a (R-134a) when the pressure is 90 psia and the temperature is 240oF – R-134a tables in english units: pp. 976-979 – Guess that this is gas and look in “superheated” tables on page 979 – v = 0.77796 ft3/lbm at given T and P • What is P when T = 240oF and v = 0.77796 ft3/lbm? 90 psia 34 Review Last Problem • What is the specific volume for refrigerant 134a (R-134a) when the pressure is 90 psia and the temperature is 240oF – R-134a tables in english units: pp. 978-81 – What if we could not guess that this is gas or made a wrong guess? • Use general rule that gas occurs when P < Psat(T) or T > Tsat(P) • For this problem T = 240oF > Tsat(90 psia) = • Also P = 90 psia < Psat(240oF) = 72.83oF 138.79 psia 35 Find State Given v and (T or P) v P Sat T = Tsat(P) P = Psat(T) vf vg Find vf and vg from T or P Liquid if v < vf Gas if v > vg Otherwise mixed Mixed Liquid Gas 36 Given T-v(or P-v); Find P(or T) • Determine state: find vf and vg from saturation tables for given T (or P) – If v > vg it is a gas; use superheat table – If v < vf it is a liquid; see below – If vf < v < vg it is in mixed region. Can find x • For liquid to use v  vf(T) we must be given v-P: find T for which vf(T)= vgiven • Compressed liquid tables for water Thermodynamic Properties August 26, 2010 ME 370 – Thermodynamics 7 37 Interpolation • Can use interpolation function on your calculator if you have one • Basic idea is that you have a table with data pairs (x1, y1) and (x2, y2) • You want to find a value of y for some value of x between x1 and x2 • Simplest idea is to assume a straight line between x1 and x2 38 Interpolation II x y y1 x2x1 y2 x Want y for this x 1 1 12 12 xx yy xx yy      • For a straight line any two points on the line can give slope y  1 12 12 1 xx xx yy yy     Interpo- lated y 39 Interpolation III • Equation for a straight line between data pairs (x1, y1) and (x2, y2) is   12 12 1 1 1 12 12 1 xx yy xx yy orxx xx yy yy          • Second form is easier to remember – It says that the slope between any point (x,y) and the point (x1, y1) is the same as the slope between (x1, y1) and (x2, y2) • First form used for computations 40 Example Problem • For H2O at P = 2 MPa, what is T when v(m3/kg) = (a) 0.00115, (b) 0.01, (c)0.2? • Table A-5, p 917 At 2000 kPa, vf = 0.001177 m3/kg, vg = 0.099587 m 3/kg – Thus (a) is liquid, (b) is mixed, (c) is gas • For compressed liquid state (a), correct T is one for which vf(T)= 0.00115 m 3/kg – Extract from Table A-4, next slide, shows that this vf is between T = 195oC and T = 200oC 41 Data and Interpolation • We want to find T for v = 0.00115 m3/kg – Between 195oC and 200oC • For this interpolation x1 = 0.001149 m3/kg, x2 = 0.001157 m3/kg, y1 = 195oC and y2 = 200 oC Ckg m kg m kg m kg m CC CT o oo o 6.195001149.00115. 001149.001157. 195200 195 33 33         42 Example: (b) Mixed, (c) Gas • (b) for mixed region, T = Tsat(2 Mpa) = 212.38oC; can find quality Ckg m kg m kg m kg m CC CT o oo o 6.60119962.2. 19962.22326. 600700 600 33 33          %96.80896.0 001177.099587. 001177.01. 33 33        kg m kg m kg m kg m vv vv x fg f • (c) v = .2 m3/kg at P = 2 MPa found on p. 921 between T = 600oC and 700oC