Unit Four Quiz Solutions and Unit Five Goals

1. Unit Four Quiz Solutions and Unit Five Goals Mechanical Engineering 370 Thermodynamics Larry Caretto March 4, 2003

2. Outline • Quiz three and four solutions are similar • Finding work (area under path) and u • Q = m (ufinal – uinitial) + W • Unit five – open systems • View first law as a rate equation • Have mass crossing system boundaries • Flows across boundaries have several energy forms including internal (u) , kinetic and potential energy plus flow work (Pv)

3. Quiz Three Results • 21 students; max = 30; mean = 18.6 10 10 14 14 15 15 15 15 15 18 18 20 20 21 22 24 24 24 24 24 28 • Q = m(ufinal – uinitial) + W • To compute u = h – Pv • Use specific volume in m3/kg • convert P to kPa for h and u in kJ/kg

4. Quiz Three Path • Quiz three gave neon data near and in mixed region. • T1, P1, V1, P2, V2 • T3, V4 • The quiz three diagram is shown here. • This was not an ideal gas so we had to use property tables

5. Quiz Four Path • Quiz four has the same path with the same data items as quiz three. • T1, P1, V1, P2, V2 • T3, V4 • The quiz four path diagram is shown here. • This was an ideal gas so we use PV = mRT and du = cvdT

6. Quiz Four Solution • Given: Water in three-step process • T1 = 300oC, V1 = 1 m3, P1 = 100 kPa • 1-2 is a linear path to P2 = 300 kPa, V2 = 0.8 m3 • 2-3 is constant volume with T3 = 400oC • 3-4 is constant pressure with V4 = 0.4 m3 • Find the heat transfer, Q, using ideal gas • Find Q from first law (ideal gas with cv const): • Q = DU + W = m(u4 – u1) + W = mcv(T4 – T1) + W • Work is (directional) area under path

7. Path for This Process • Work = area under path = trapezoid area plus rectangle area • W = (P1 + P2)(V2 – V1)/2 + P3-4 (V4 – V3) • DV < 0 means work will be negative 2 P 3 4 1 V

8. Finding the Answer • Use properties at the initial state to find the mass • Find P3-4 = P3 = P4 from state 3 defined by T3 = 400oC and v3 = v2 = V2/m

9. Finding the Answer (cont’d) • Find T4 from m, P4 = P3 and V4 = 0.4 m3 • Now find heat and work

10. Finding the Answer (concluded)

11. Future Quizzes • Can use equation summary • Download from course web page (follow course notes link) • May have unannounced open book exams to allow use of tables • If you are late for a quiz you can • Come to class after quiz is over or • Start quiz and receive grade

12. Unit Five Goals • Topic is first law for open systems, i.e., systems in which mass flows across the boundary • Will look at general results and focus on steady-state systems. • As a result of studying this unit you should be able to • understand all the terms (and dimensions) in the first law for open systems:

13. Open System Concepts the useful work rate or mechanical power (ML2T-3) the mass flow rate (MT-1) the kinetic energy per unit mass (L2T-2) gz the potential energy per unit mass (L2T-2) total energy (ML2T-2) heat transfer rate (ML2T-3) rate of energy change (ML2T-3)

14. Unit Five Goals Continued • use the equation relating velocity, mass flow rate, flow area, A, and specific volume • use the mass balance equation

15. Flow Work • For open systems work is done on (or by) mass entering and leaving the system • Flow work is Pv times mass flow rate • Add this flow work to internal energy (times mass flow rate) • First law for mass flows has h = u + Pv (sum of internal energy plus flow work)

16. Unit Five Goals Continued • use the first law for open systems • use the steady-state assumptions and resulting equations

17. Steady-state equations • Steady-state first law for open systems • Steady-state mass balance for open systems

18. Unit Five Goals Continued • recognize that kinetic and potential energies are usually negligible • A 1oC temperature change in air (ideal gas with cp = 1.005 kJ/(kg∙K) has Dh = 1005J/kg • A similar kinetic energy change requires a velocity increase from zero to 45 m/s (~100 mph) • A similar potential energy change requires an elevation change of 102 m (336 ft)

19. Unit Five Goals Concluded • handle simplest case: steady-state, one inlet, one outlet (one mass flow rate), negligible changes in kinetic and potential energies • work with ratios q and w in simplest case

20. Example Calculation • Given: 10 kg/s of H2O at 10 MPa and 700oC renters a steam turbine; the outlet is at 500 kPa and 300oC. There is a heat loss of 400 kW. • Find: Useful work rate (power output) • Assumptions: Steady-state, negligible changes in kinetic and potential energies • Configuration: one inlet and one outlet First law

21. Getting the Answer • At Tin = 700oC and Pin = 10 MPa, hin = 3870.5 kJ/kg (p. 836) • At Tout = 300oC and Pout = 500 kPa, hout = 3061.6 kJ/kg • Heat loss is negative: Q = Qin - Qout

22. Review Ideal Gases • For ideal gases du = cvdT; dh = cpdT • Ideal gas DH = DU + RDT • May have molar DH data • Last week we looked at problem with H2O as an ideal gas, where T1 = 200oC and T2 = 400oC • How do we handle cv(T)?

23. Ideal Gas with cv(T) • Find a, b, c and d from Table A-2(c), p 827 • Use T1 = 473.15 K and T2 = 673.15 K • Molar enthalpy change = 7229.3 kJ/kmol

24. Getting Du from molar Dh • Use molar Dh just found, data on R and M, and DT = 200oC = 200 K

25. Ideal Gas Tables • Find molar u(T) for H2O in Table A-23 on page 860 • Have to interpolate to find u1 = u(473.15 K) = 11,953 kJ/kmol and u2 = u(673.15 K) = 17,490 kJ/kmol • Du = (17,490 kJ/kmol - 11,953 kJ/kmol) / (18.015 kg / kmol) = 307.4 kJ/kg