Properties of Pure Substances

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 Unit one – Properties of Pure Substances – 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 2 Thermodynamic Variables Thermodynamic System • 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 • 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 3 4 Thermodynamic State Ideal Gas Equations • Two independent intensive properties determine the state of a simple compressible system • Once the state of a system is determined all other properties are known • Thermodynamic processes are a continuous series of states known as a quasi-equilibrium process • 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 = nR T u • PV = (m/M)(MR)T = mRT PV = mRT • P(V/m) = Pv = RT Pv = RT 5 6 1 Z = 1 is ideal gas TR = T/Tcrit Why Use Ideal Gas? Figure 3-51 from Çengel, Thermodynamics An Engineer Approach, 2008 PR = P/Pcrit • 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) 7 Finding R 8 Ideal Gas Calculation • Table A-1, page 908: for water, R = 0.4615 kJ/kg·K = 0.4615 kPa·m3/kg·K 1kJ = 1 kN·m = 1 (kN/m2)·m3 = 1 KPa·m3 0.4615 kPa  m 3 (1273.15 K ) RT 58.756 m 3 kg  K v   P kg 10 kPa • Error is only 0.0017% at this point • Ideal gas often used for “permanent gases” such as air 9 Ideal Gas Calculation II P-v-T for Real Substances • Three phases: gas, liquid and solid • Look at gas and liquid • Thought experiment • 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) T PV  mR 300 psia 20 ft 3  2.6809 psia  ft 3 (20 lbm ) lbm  R 10 – Add heat to a liquid at constant pressure; measure T and v – After a while vapor starts to form  111.9 R  347.7 o F • 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) 11 12 2 Thought Experiment II Thought Experiment III • Repeat experiment with difference in variables • Hold temperature constant • Add heat and measure pressure, P, and specific volume v • Have same liquid-vapor transition • States 2 to 4 are liquid vapor transition – P is constant if T is constant – Liquid and vapor properties constant in transition only relative proportions change – Liquid and vapor have same properties (P, vf = 1/f, vg = 1/g, and T) during transition; only relative proportions change Figures 3-6 to 3-10 from Çengel, Thermodynamics An Engineering Approach, 2008 13 14 P-v Diagram for Water Thought Experiment Results 1.E+08 • A constant temperature line on P-v T = 50 C 1.E+07 Pressure (Pascals) – Looks like a hyperbola in the gas region – Has constant pressure during the transition from liquid to gas – Is nearly vertical in the liquid region • The points where the transition between liquid and gas take place for different temperatures: saturation curve • Results of “experiment” on next slide T = 100 C T = 200 C 1.E+06 T = 300 C Tcrit 1.E+05 Gas Liquid T = 500 C T = 1000 C Sat 1.E+04 Mixed 1.E+03 0.0001 0.001 15 0.01 0.1 1 10 100 1000 Specific Volume (cubic meters per kilogram) Critical P-v Diagram for Water Superheat P-v Diagram for Water 1.E+08 50,000 45,000 T = 200 C T = 300 C 1.E+07 Tcrit T = 500 C T = 1000 C Pressure (Pascals) Pressure (Pascals) Sat 40,000 Sat 35,000 T = 50 C 30,000 T = 100 C T = 200 C 25,000 T = 300 C 20,000 Tcrit 15,000 T = 500 C 10,000 T = 1000 C 5,000 1.E+06 0.001 0.01 0.1 Specific Volume (cubic meters per kilogram) 1 0 0 50 100 150 200 Specific Volume (cubic meters per kilogram) 3 What We Have to Do Finding Properties • Find all intensive thermodynamic properties when we are given only two intensive properties • When we have a mixture of liquid and gas T and P are not independent • For a single phase (liquid or gas) T and P are independent (can specify state) • In the mixed region we use vapor mass fraction, called quality, x mg v  vf v  v f • v and v are found f g  x  from either T or P mg  m f v g  v f v fg – Two independent, intensive properties define state • Here we will discuss only four intensive properties, P, v, T, and fraction of vapor in mixed region called the quality, x • Approach used here will be applied to finding other properties: internal energy, enthalpy, and entropy 19 20 Saturation Properties Quality Derived • Mixed region volume is sum of liquid and vapor volumes  mf mg m  mg V mg  vg  vf  vg  vf m m m m m P V  mg v g  m f v f • Use quality: x = mg/(mg + mf) = mg/m v Sat T = Tsat(P) P = Psat(T) Saturated liquid (f) and saturated vapor (g) properties can be found if P or T are known m  mg V mg  vg  v f  xvg  (1  x)v f  v f  xv g  v f  m m m Use quality, x, in mixed region • Will apply this later to other properties vf v x vg v  vf vg  v f v  v f  xv g  v f  21 22 Saturation Tables • See steam tables, pp 914–923 • Saturation tables – Table A-4 has T as lookup variable – Table A-5 has P as lookup variable – Same results in each table • Other variables besides specific volume have saturated liquid (f) and gas (g) • Example: what are P and v if T = 220oC and x = 0.25? 23 • Use Table A-4, Page 915, T = 220oC • Find P = 2319.6 kPa, vf = 0.001190 m3/kg and vg = 0.086094 m3/kg 24 4 What is v for x = 0.25? What if P and T are Given? • Found from Table A-4 with lookup T • 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? – P = Psat(T = 220oC) = 2319.6 kPa – vf(T = 220oC) = 0.001190 m3/kg – vg(T = 220oC) = 0.086094 m3/kg x v  vf vg  v f  v  v f  x(vg  v f )  (1  x)v f  xvg v  (1  x)v f  xvg  (1  0.25) 0.001190 m3 0.086094 m3  (0.25) kg kg v 0.022416 m 3 kg 25 26 Finding State with Given T and P Sat T = T1 < T2 T = T2 >T1 Psat(T1) Psat(T2) P Liquid when T < Tsat(P) or P > Psat(T) Gas when T > Tsat(P) or P < Psat(T) vf 27 v vg 28 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 29 30 5 Results General Rules Continued • v(0oC, 100 kPa) = 0.21630 m3/kg • What is v(0oC, 100 kPa)? Page 929 • 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) • Why are there no data at T = = 400 MPa (= 0.4 MPa)? 0oC and P – 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 31 Two More Problems 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. 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 = 0.77796 ft3/lbm? 240oF and v = 90 psia Sat T = Tsat(P) P = Psat(T) vf • Use general rule that gas occurs when P < Psat(T) or T > Tsat(P) • For this problem T = 240oF > Tsat(90 psia) = 72.83oF • Also P = 90 psia < Psat(240oF) = 138.79 psia 34 Given T-v(or P-v); Find P(or T) • Determine state: find vf and vg from saturation tables for given T (or P) • 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 Gas v – R-134a tables in english units: pp. 978-81 – What if we could not guess that this is gas or made a wrong guess? – 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 P Find vf and vg from T or P Liquid if v < vf Gas if v > vg Otherwise mixed Mixed • What is the specific volume for refrigerant 134a (R-134a) when the pressure is 90 psia and the temperature is 240oF 33 Find State Given v and (T or P) Liquid 32 vg 35 36 6 Interpolation Interpolation II • 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 y2 Interpolated y y y1 x Interpolation III x Want y for this x y2  y1 y  y1  x2  x1 x  x1 y  y1  y2  y1 x  x1  x2  x1 38 Example Problem • Equation for a straight line between data pairs (x1, y1) and (x2, y2) is y2  y1 x  x1  or x2  x1 x2 x1 37 y  y1  • For a straight line any two points on the line can give slope y • 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 m3/kg y  y1 y2  y1  x  x1 x2  x1 – Thus (a) is liquid, (b) is mixed, (c) is gas • Second form is easier to remember • For compressed liquid state (a), correct T is one for which vf(T)= 0.00115 m3/kg – 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) – Extract from Table A-4, next slide, shows that this vf is between T = 195oC and T = 200oC • First form used for computations 39 Data and Interpolation 40 Example: (b) Mixed, (c) Gas • (b) for mixed region, T = Tsat(2 Mpa) = 212.38oC; can find quality • We want to find T for v = 0.00115 m3/kg – Between 195oC and 200oC v  vf • For this interpolation x1 = 0.001149 m3/kg, x2 = 0.001157 m3/kg, y1 = 195oC and y2 = 200oC 200 o C  195o C  .00115 m 3  .001149 m 3   195.6o C T  195 C  3 3 kg kg    .001149 m .001157 m kg kg o 41 .01 m 3 kg x  v g  v f .099587 m3  .001177 m kg 3 kg  .001177 m  0.0896  8.96% 3 kg • (c) v = .2 m3/kg at P = 2 MPa found on p. 921 between T = 600oC and 700oC T  600 o C  700o C  600 o C  .2 m3  .19962 m3   601.6o C 3 3 kg kg   .22326 m  .19962 m kg kg 42 7 Compressed Liquid Table More on Compressed Liquid • Not a common table, use here to check approximation that v(P,T)  vf(T) • Table A-7, page 922, gives v(50 MPa, 100oC) = 0.0010201 m3/kg • Table A-4, page 914, gives vf(100oC) = 0.001043 m3/kg • Error is about 2% even for a pressure difference of 49.9 MPa between the actual pressure and saturation pressure • Check approximation that v(P,T)  vf(T) for previous problem: v = 0.00115 m3/kg and P = 2 MPa • Do not have data for compressed liquid at 2 MPa in table on page 922 43 44 – First entry is at 5 MPa • Have to use double interpolation and interpolation with saturation points – Interpolate between Psat and 5 MPa Interpolations Table 2 MPa • We are looking for v = 0.00115 m3/kg which is between 180oC and 200oC in compressed liquid table at 5 MPa • We have to find data at these temperatures and 2 MPa Compressed liquid data shown here Looking for v = 0.00115 m3/kg at P = 2 MPa No data for P < 5 M Pa Use saturated liquid data for lower pressure At 180oC, vf = 0.001127 m3/kg at P = 1.0021 MPa At 200oC, vf = 0.001157 m3/kg at P = 1.5538 MPa Interpolate to get v at 2 MPa at 180oC, 200oC 46 – Because 5 MPa is the first table entry we use saturation as the lower data point vt(T) = v(T, Psat) v(T , 2 MPa)  v f (T )  v(T ,5 MPa)  v f (T ) 5 MPa  Psat (T ) 2 MPa  Psat (T ) 45 Interpolations II Interpolations III • Apply formula for T = 180oC and 200oC 3 v(180 C , 2 MPa)  .001127 m o 3 .001125 m kg  3  .001127 m kg kg 2 MPa  1.0021 MPa   .001126501 m 5 MPa  1.0021 MPa kg 3 v( 200 C , 2 MPa)  .001157 m o 3 kg • Now interpolate between results for two different T’s to find T at desired v 3 200o C  180o C .00115648 m 3 .001126501 m3  kg kg  .00115 m3 .001126501 m 3     195.7 o C  kg kg    3  .001157 m 3 kg kg 2 MPa  1.5538 MPa   .00115648 m 5 MPa  1.5538 MPa kg .001153 m T  180o C  47 • Previous result using v(T,P) ≈ vf(T) was 195.6oC 48 8