1. Unit Three Quiz Solutions and Unit Four Goals Mechanical Engineering 370 Thermodynamics Larry Caretto February 25, 2003

2. 2 Outline Quiz Two and Three Solutions –Finding work as area under path –Find internal energy then Q = m  u + W Unit four – first law for ideal gases –Heat capacities, c v and c p are properties –For ideal gases du = c v dT and dh = c p dT, regardless of path –For ideal gases u = u(T) only and h = u + Pv = u + RT = h(T) only

3. In quiz two the final state lies along path and along 320 K iso- therm. Since this is not an ideal gas we have to use property tables Requires trial and error solution Unit four considers ideal gas behavior

4. 4 Quiz Three Solution Given: Neon in three-step process –T 1 = 280 K, V 1 = 1 m 3, P 1 = 200 kPa –1-2 is a linear path to P 2 = 700 kPa, V 2 = 0.08 m 3 –2-3 is constant volume with T 3 = 30 K –3-4 is constant pressure with V 4 = 0.04 m 3 Find the heat transfer, Q, using tables Find Q from first law: –Q =  U + W = m(u 4 – u 1 ) + W Work is (directional) area under path

5. 5 Path for This Process Work = area under path = trapezoid area plus rectangle area W = (P 1 + P 2 )(V 2 – V 1 )/2 + P 3-4 )(V 4 – V 3 )  V < 0 means work will be negative P V 1 2 3 4

6. 6 Finding the Answer Properties at the initial state –From T 1 = 280 K and P 1 = 200 kPa, find v 1 and m = V 1 /v 1 = 1.732 kg; also u 1 = h 1 – P 1 v 1 = 237 kJ/kg State 3 defined by T 3 = 80 K, v 3 = v 2 = V 2 /m –Find P 3 = P sat (30 K) = 223.8 kPa State 4 defined by P 4 = P 3, v 4 = V 4 /m –This is in mixed region with u 4 = 42.7 kJ/kg Q = (1.732 kg)(237 – 42.7) kJ/kg – 414 kJ = –751 kJ

7. 7 Unit Four Goals 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 3 –understand the heat capacities C x (e.g. C p and C v ) as dQ = C x dT in a “constant-x” process –use the property relations for ideal gases du = c v dT and dh = c p dT for any process

8. 8 Unit Four Goals Continued –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

9. 9 Unit Three Goals Continued –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 = c v dT, and a path equation (may be iterative) – use the equation c p - c v = R to find cp from cv (and vice versa), which also applies to equations; if cp = a + bT + cT 2, then c v = (a-R) + bT + cT 2

10. 10 Example Calculation Given: 10 kg of H 2 O at 100k Pa and 200 o C is expanded to 400 o C at constant pressure Find: Heat Transfer –using H2O tables –using ideal gas with constant heat capacity –using ideal gas with variable heat capacity First Law: Q =  U + W = m(u 2 – u 1 ) + W Path: W =  PdV = P 1-2 (V 2 – V 1 ) for constant pressure, P 1-2 = P 1 = P 2 u 2 – u 1 =  c v dT for ideal gas

11. 11 Using H 2 O Tables At T 1 = 200 o C and P 1 = 100 kPa, v 1 = 2.172 m 3 /kg and u 1 = 2658.1 kJ/kg At T 2 = 400 o C and P 2 = P 1 = 100 kPa, v 2 = 3.103 m 3 /kg and u 2 = 2967.9 kJ/kg W = P 1-2 (V 2 – V 1 ) = P 1-2 m(v 2 – v 1 ) = (10 kg)(100 kPa)(3.103 - 2.172) m 3 /kg = Q = m(u 2 - u 1 ) + W = (10 kg)(2967.9 - 2658.1) kJ/kg + 931 kJ = 4,029 kJ

12. 12 Ideal Gas Calculations Q =  U + W = m(u 2 – u 1 ) +  PdV Q = m(u 2 – u 1 ) + m  Pdv 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 c v (or c p = c v + R)

13. 13 Work – Ideal Gas Assumption At T 1 = 200 o C and P 1 = 100 kPa, v 1 = RT 1 /P 1 = (.4615 kJ/kg  K)(473.15 K)/(100 kPa) = 2.1836 m 3 /kg At T 2 = 400 o C and P 2 = P 1 = 100 kPa, v 2 = RT 2 /P 2 = (.4615 kJ/kg  K)(673.15 K)/(100 kPa) = 3.1066 m 3 /kg W = P 1-2 (V 2 – V 1 ) = P 1-2 m(v 2 – v 1 ) = (10 kg)(100 kPa)(3.106 - 2.184) m 3 /kg = 923 kJ

14. 14 Ideal Gas Internal Energy u 2 – u 1 =  c v (T)dT =  c p (T)dT - R  T Possible calculations for c v (or c p ) –Assume constant (easiest)  u = c  T –Integrate equation giving c v or c p as a function of temperature (Table A-2, p 827) –Use ideal gas tables giving u(T) and h(T) (Tables A-17 to A-26, pp 849-863) –Last two give molar properties

15. 15 Constant c v Ideal Gas Get c v = 1.4108 kJ/kg  K for water from Table A-2, p828  U = m  u = m  c v (T)dT = c v (T 2 – T 1 ) = mc v  T, if c v is constant Here,  U = mc v (T 2 - T 1 ) = (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

16. 16 Ideal Gas with c v (T) Use kelvins for temperature Molar enthalpy change = 7229.3 kJ/kmol Q = m  u+ W = (10 kg)(309 kJ/kg) + 931 kJ

17. 17 Ideal Gas Tables Find molar u(T) for H 2 O in Table A-23 on page 860 Have to interpolate to find u 1 = u(473.15 K) = 11,953 kJ/kmol and u 2 = 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

18. 18 Comparison of Results (kJ) Method UU WQ Tables 3,0989314,029 Const c v 2,8229233,745  c v (T)dT 3,0909234,013 Ideal gas tables 3,0749233,097

19. 19 Assuming c v Constant Assumption of constant heat capacity introduces about a 7% error Accounting for temperature variation of heat capacity reduces error to <= 0.8% 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 temper- ature range increases