First Law for Ideal Gases

Outline Unit Four – First Law for Ideal Gases • Quiz three solution • Unit four – first law for ideal gases – Heat capacities, cv and cp are properties – For ideal gases du = cvdT and dh = cpdT, regardless of path – For ideal gases u = u(T) only and h = u + Pv = u + RT = h(T) only – Solving first law problems with ideal gases 2 Unit Four Goals Unit Four Goals Continued • As a result of studying this unit you should be able to – describe the path for a process and determine the work with greater confidence than you had after completing unit three – understand the heat capacities Cx (e.g. Cp and Cv) as dQ = Cx dT in a “constant-x” process – use the property relations for ideal gases du = cvdT and dh = cp dT for any process – find changes in internal energy and enthalpy for an ideal gas where the heat capacity is constant or a function of temperature . – use ideal gas tables to find changes in internal energy and enthalpy where the heat capacities are functions of temperature – find internal energy changes for ideal gases as h = u + R T – convert results from a per-unit-mole basis to a per-unit-mass basis and vice versa 3 Unit Four Goals Continued 4 Why Use Ideal Gas? – be able to find other properties about a state when you know (or able to calculate) the internal energy or enthalpy – be able to work problems using the first law, Pv = RT, du = cv dT, and a path equation (may be iterative) – use the equation cp - cv = R to find cp from cv (and vice versa), which also applies to equations; if cp = a + bT + cT2, then cv = (aR) + bT + cT2 5 • Simple equations • Real gas behavior close to ideal gas for low pressure (or high temperature) • Want P low compared to Pc or T high compared to Tc – Have seen that ideal gas gives good results for P-v-T data when P < ~0.2Pc or T > ~2Tc – Will also see that this region gives good results for internal energy change, u 6 1 Z = 1 is ideal gas Heat Capacity TR = T/Tcrit Figure 3-51 from Çengel, Thermodynamics An Engineer Approach, 2008 TR > 2 or PR < 0.2  ideal gas PR = P/Pcrit – dq = cvdT at constant volume – dq = cpdT at constant pressure 7 Ideal Gases 8 Is cv for Constant Volume or Not? • For ideal gases, du = cvdT, dh = cpdT regardless of path • Note these differences • Depends on path and substance Find Equation – Path dependent heat transfer • dq = cvdT for constant volume only • dq = cpdT for constant pressure only – Path independent properties du and dh • u = U/m = cvdT, h = H/m = cpdT • h = u + Pv = u +RT for ideal gases – dh = du + RdT => cpdT = cvdT + RdT – cp = cv + R • Heat capacity, Cx = mcx, gives heat transfer dQ = CxdT in constant x process • Specific heat capacity, cx = Cx/m is a property • Use cv and cp for constant volume and pressure, respectively • For liquids and solids cv and cp are essentially the same dq cvdT dq cpdT du dh cvdT cpdT Path Constant volume Constant pressure Any Any Substance Any Any Ideal gas Ideal gas 9 Compare/Contrast Tables 10 Example Calculation • Given: 10 kg of H2O at 100 kPa, 200oC expanded to 400oC at constant pressure • Find: Heat Transfer • For any property relations we have: – First law: Q = U + W – Work: W = ∫pathPdV • So what is different? – using H2O tables (for comparison) – using ideal gas with constant heat capacity – using ideal gas with variable heat capacity – With property tables we use tables to find P-v-T relationships and internal energy – Ideal gases use Pv = RT and du = cvdT • Use data sources for cv as average value, cv as a function of temperature and integrate • Find u(T) in ideal gas tables 11 • First Law: Q = U + W = m(u2 – u1) + W • Path: W = pathPdV = P1-2 (V2 – V1) for constant pressure, P1-2 = P1 = P2 • u2 – u1 = cvdT for ideal gas 12 2 Using H2O Tables Ideal Gas Calculations 200oC • At T1 = and P1 = 100 kPa, v1 = 2.1724 m3/kg and u1 = 2658.2 kJ/kg • At T2 = 400oC and P2 = P1 = 100 kPa, v2 = 3.1027 m3/kg and u2 = 2968.3 kJ/kg • W = P1-2 (V2 – V1) = P1-2 m(v2 – v1) = (10 kg)(100 kPa)(3.1027 - 2.1724) m3/kg = 931 kPa·m3 = 931 kJ • Q = m(u2 - u1) + W = (10 kg)(2968.3 – 2658.2) kJ/kg + 931 kJ = 4,029 kJ Q = U + W = m(u2 – u1) + pathPdV Q = m(u2 – u1) + mpathPdv PV = mRT Pv = RT We use PV = mRT to determine mass and specific volume from P and T • The work calculation does not depend on assumptions about cv (or cp = cv + R) • Find R = 0.4615 kJ/kg·K = 0.4615 kPa·m3/kg·K in Table A-1, page 908 • • • • 13 Work – Ideal Gas Assumption • At T1 = 200oC and P1 = 100 kPa, v1 = RT1/P1 = (.4615 kJ/kgK)(473.15 K)/(100 kPa) = 2.1836 m3/kg • At T2 = 400oC and P2 = P1 = 100 kPa, v2 = RT2/P2 = (.4615 kJ/kgK)(673.15 K)/(100 kPa) = 3.1066 m3/kg • W = P1-2 (V2 – V1) = P1-2 m(v2 – v1) = (10 kg)(100 kPa)(3.1066 - 2.1836) m3/kg = 923 kPa·m3 = 923 kJ 14 Ideal Gas Internal Energy • u2 – u1 = cv(T)dT = cp(T)dT - RT • Possible calculations for cv (or cp) – Assume constant (easiest) u = cvT (Table A-2(a), page 909 or Table A-2(b), page 910) – Integrate equation for cv or cp as a function of temperature (Table A-2(c), p 911) – Use ideal gas tables giving u(T) and h(T) (Tables A-17 to A-26, pp 934-947) – Last two give molar properties (except air) 15 • Divide by molar mass, M, to get per-unit-mass property values 16 17 18 3 Constant cv Ideal Gas • Get cv = 1.4108 kJ/kg·K at 300 K (26.85oC) for water from Table A-2(a), p 909 (no data for water in Table A-2(b), page 910) • U = mu = m cv(T)dT = mcv(T2 – T1) = mcvT, if cv is constant • Here, U = mcv(T2 - T1) = (10 kg) (1.4108 kJ/kg·K )(673.15 K - 473.15 K) = 2,822 kJ • Q = U + W = 2,822 kJ + 923 kJ = 3,745 kJ, a 7% error compared to actual properties 19 Ideal Gas with cv(T) T2 T2 T1 T1 20 Ideal Gas Tables for Water • T is in kelvins • Need to interpolate for values at T = 200oC = 473.15 K • u(200oC) = 11,953 kJ/kmol h   c p (T )dT   a  bT  cT 2  dT 3 dT b 2 T2  T12   c3 T23  T13   d4 T24  T14  2 a,b,c,d data from Table u h  R T h u     RT A-2(c), page 911 M M M Details not shown here  a T2  T1   • Use kelvins for temperature • Molar enthalpy change = 7229.3 kJ/kmol • u = (7229.3 kJ/kmol) / (18.015 kg/kmol) (.4615 kJ/kgK)(200 K) = 309.0 kJ/kg 21 Ideal Gas Tables 22 Comparison of Results (kJ) • Find molar u(T) for H2O in Table A-23 on page 946 • Have to interpolate to find the values of u1 = u(473.15 K) = 11,953 kJ/kmol and u2 = u(673.15 K) = 17,490 kJ/kmol • U = (10 kg)(17,490 kJ/kmol - 11,953 kJ/kmol) / (18.015 kg / kmol) = 3,074 kJ • Q = U + W = 3,074 kJ + 923 kJ • Q = 3,997 kJ 23 Method U W Q Tables 3,098 931 4,029 Const cv 2,822 923 3,745 cv(T)dT 3,090 923 4,013 Ideal gas tables 3,074 923 3,997 24 4 Ideal gases: du = cvdT Assuming cv Constant • This does not depend on path • We just computed the Q = U + W for a constant pressure path of an ideal gas • We used U = mu = m∫cvdT to compute the internal energy change • We would use the same equation regardless of the path between state 1 and state 2 • Assumption of constant heat capacity introduces about a 7% error for this problem • Accounting for temperature variation of heat capacity reduces error to <= 0.8% • Next page shows figure 4-24, page 177 – Constant heat capacity assumption is best for noble gases (e.g., argon, neon) and reasonable for diatomic molecules at ambient temperatures – Assumption worsens as the temperature range increases 25 26 Variation In-class exercise • Monatomic gases have constant cp • Diatomic gases have moderate variation with T • Triatomic gases have larger variation with T • What is cv? • 2 kg of air initially at 100 kPa and 300 K heated to 1200 K. Find the heat transfer for each of the following processes: Figure 4-24 from Çengel, Thermodynamics An Engineering Approach, 2008 – Constant volume – Constant pressure – A straight line path to a pressure of 250 kPa • Use air tables to find the change in internal energy • How will path affect U? 27 28 In-class Exercise Solution Exercise Solution II • Path will not affect U mRT1 (2 kg )(300 K ) P   P1 100 kPa 0.2870 kPa  m 3  1.722 m 3 kg  K V1  – For an ideal gas, u is a function of temperature only. Since T is the same, U = mu will be the same. – From air tables, Table A-17, pp 934-935 T = 1200 K • Look at constant T = 300 K V volume path first • For constant volume path, W = 0 so Q = U = 1438.48 kJ • u(300 K) = 214.07 kJ/kg • u(1200 K) = 933.33 kJ/kg • mu = (2 kg)(719.24 kJ/kg) = 1438.48 kJ – Q = W + 1438.48 kJ 29 30 5 Exercise Solution III • For constant pressure, V2 = V1(T2/T1) = 6.888 m3 Exercise Solution IV P • W = P1-2(V2 – V1) = (100 kPa)(6.888 m3 – 1.722 m3) = 516.6 kJ • For straight line path, V2 = V1(P1T2/P2T1) = 2.7552 m3 T = 1200 K T = 300 K P • W = (P1 + P2)(V2 – V1) / 2 = (100 kPa + 250 kPa)(2.7552 m3 – 1.722 m3) = 180.81 kJ V • For constant pressure path, Q = U + W = 1438.48 kJ + 516.66 kJ = 1955.14 kJ T = 1200 K T = 300 K V • For this straight-line path, Q = U + W = 1438.48 kJ + 180.81 kJ = 1619.29 kJ 31 32 In-Class Exercise In Class Exercise II • Air, initially at 100 psia, 1000 R and 1 ft3 undergoes an expansion following the polytropic path PV1.3 = constant to a final temperature of 500 R. Find the heat transfer. • For a polytropic path the work is given by the following equation W P2V2  P1V1 1 n • Given data: – Initial state: T1 = 1000 R, P1 = 100 psia, V1 = 1 ft3 – Path: PV1.3 = constant • P1V11.3 = P2V21.3 – Final state: T2 = 500 R – Substance: Air – Property relations: ideal gas • PV = mRT and du = cvdT • Air tables give integral of cvdT 33 34 In Class Exercise III In Class Exercise IV • Equations and data: – First Law: Q = U + W = m(u2 – u1) + W – W = path PdV – W = ( P2V2 – P1V1) / (1 - n) for polytropic path, PVn = constant – PV = mRT – du = cvdT (or use air tables for u, A-17E, pp 982–983 for engineering units) – For air, R = 0.3704 psia·ft3/lbm·R = 0.06855 Btu/lbm·R (Table A-1E, page 958) 35 • From air tables, u1 = u(1000 R) = 172.43 Btu/lbm and u2 = u(500 R) = 85.20 Btu/lbm P1V1 (100 psia )(1 ft 3 )   0.2700 lbm RT1 0.3704 psia  ft 3 (1000 R) lbm  R P2V2  P1V1 mRT2  mRT1 mRT2  T1  W   1 n 1 n 1 n 0.06855 Btu 0.2700 lb m  500 R  1000 R  lbm  R   30.845 Btu 1  1.3 m 36 6 In Class Exercise V • Q = U + W = m(u2 – u1) + W  85.20 Btu 172.43 Btu    30.845 Btu  Q  0.2700 lbm  lbm  lbm  Q = 7.29 Btu • Assuming constant cv = 0.174 Btu/lbm·R at mean temperature of 750 R = 290.3oF from Table A-2E(b), page 960 gives Q = 7.36 Btu 37 7