1. ME 200 L31: Review for Examination 3 ME 200 L31: Review for Examination 3 Thu 4/10/14 Examination 3 (L22 – L30) 6:30 – 7:30 PM WTHR 200, CL50 224, PHY 112 Kim See’s Office ME Gatewood Wing Room 2172 Please check HW and Examination Grades on Blackboard Please pick up all graded Home Work and Examinations from Class or Room 2172 after class today! https://engineering.purdue.edu/ME200/ ThermoMentor © Program Spring 2014 MWF 1030-1120 AM J. P. Gore gore@purdue.edu Gatewood Wing 3166, 765 494 0061 Office Hours: MWF 1130-1230 TAs: Robert Kapaku rkapaku@purdue.edu Dong Han han193@purdue.edurkapaku@purdue.eduhan193@purdue.edu

2. Examination 3 Extra Time Students 6-9 pm ME2063 Civil Engineering Banquet Students 6-7 pm ME1051 Can accommodate ~50 students. If you are not a CE banquet student go to your regular room Division 4 (Gore Students) WTHR 200 2

3. Isentropic Processes for Ideal Gases (Air) with Constant Specific Heat 3 T s 2 1 p2p2 v2v2 T2T2 s2s2 T1T1 p1p1 v1v1 s1s1 State 1 and State 2 are defined in the Figure by intersection of any two curves or lines passing through the points on the graph. The lines and curves passing through these points define additional states where one of the properties is constant. For example the vertical line is the isentropic line along which s is Constant. Tabulated as a function of T for air. in Table A22 for air. Tabulated as a function for air. of T in Table A22 for air.

4. Isentropic Processes for Ideal Gases with Constant Specific Heat 4 Constant specific heat assumption allows further simplification of the p-v relation for the isentropic process for ideal gases: Isentropic processes for ideal gases can be analyzed using pv=mRT and pv k = constant if specific heat is independent of temperature.

5. Isentropic Processes for Ideal Gases with Constant Specific Heat 5 T s n = k n = -1; p and v both increase and decrease together n = 1 v = Constant; n = ±∞ p = const; n = 0

6. Polytropic Processes for Ideal Gases with Constant Specific Heat 6 p v n = k n = 1 n = -1 n = -1; p and v both increase and decrease together v = Constant; n = ±∞ p = const; n = 0

7. Maximum performance measures (reversible processes) Power cycles Refrigeration cycles Heat pump cycles 7 Carnot efficiency

8. Example 8 A refrigeration cycle operating between two reservoirs receives energy Q C from the one at T C = 280 K and rejects energy Q H to one at T H = 320 K. For each of the following cases determine whether the cycle operates reversibly, irreversibly, or is impossible: –Q C = 1500 kJ, W cycle = 150 kJ. Impossible –Q C = 1400 kJ, Q H = 1600 kJ. Reversible –Q H = 1600 kJ, W cycle = 400 kJ. Irreversible –β = 5. Irreversible (β max = 280/40 =7)

9. Example: Use of Clausius Inequality Q H =1000 kJ, T H =500 K and Q C =600 kJ at T C : (a) 200 K, (b) 300 K, (c) 400 K. Find if each cycle is reversible, irreversible or ideal. Solution: Use the given Q H, Q C values to find work and ensure that the work produced does not result in a negative value for  cycle ∫

10. Example: Use of Clausius Inequality (b)  cycle = 0 kJ/K = 0 (a)  cycle = +1 kJ/K > 0 Irreversibilities present within systemNo irreversibilities present within system (c)  cycle = –0.5 kJ/K < 0 Impossible

11. Entropy is a Property ►Entropy is easier to understand if thought of as a property analogous to specific volume. It is defined as:

12. Entropy Change Calculations Q H =1000 kJ, T H =500 K and Q C =600 kJ at T C : (a) 200 K, (b) 300 K, (c) 400 K. Find if each heat transfer is reversible find entropy change for (the material that makes up) the reservoir. Solution: Use the given Q H, Q C and T H, T C values to find the entropy change for the reservoirs:

13. Examples: Entropy Change for the Reservoirs (Property of materials making up the reservoirs) (a) Entropy can decrease and increase! Net entropy change can be zero! (b) (c) Net entropy change can be negative! However, can’t continue that process in a cyclic manner! Since by definition reservoir properties are fixed, these changes must be compensated by reverse actions!

14. Entropy Change using Tables: Example 3 14 Given: 0.5 kg/s of steam at 280 o C, 20 bar is expanded in a turbine to 1 bar in a constant entropy process. Find: Find the work produced by the steam in kW and show the process on a T-s diagram. If the process was not a constant entropy process and resulted in saturated steam at 1 bar, find the decrease in work and increase in entropy in kW/K. Assumptions: Change in PE neglected, No heat transfer, work done on Turbine shaft, Steady state, Steady flow, Mass is conserved. StP,Thsx i20,2802976.46.6828- es1,99.632423.2-0.8883 e1,99.632675.57.3594- x=(s-s f )/(s g -s f )=(6.6828-1.3026)/(7.3594-1.3026)=5.3802/6.0568=0.8883 h es = 417.46+0.8883(2258)=2423.22 = 2976.4-2423.2=553.18 kJ/kg; =2976.4-2675.5=300.9 kJ/kg = 150.45 kW 276.59 kW

15. State 2a: Saturated State 1: 20 bar, 280 C On the T-s diagram drawn to scale State 1 and State 2 are close to each other as illustrated below. State 2s: Mixture

16. Entropy Change using Tables: Example 4 16 Given: Consider R134 throttled from p3 =120 lbf/in 2 to p4 =40 lbf/in 2. Find: Find the change in entropy of R134. Assumptions: Change in KE, PE neglected, No heat transfer, No work done other than flow work, Steady state, Steady flow, Mass is conserved. Adiabatic throttle with a pressure loss and phase change lead to increase in Entropy while keeping Enthalpy constant.

17. 17 State 4 State 3 T-s Diagram and Demonstration of Throttle Action; h-s diagram State 3 State 4

18. Entropy Change for Solids: Example 2 ft 3 of sand is heated from 32 o F to 70 o F find heat added and change in entropy. Assume: Sand is incompressible and has constant specific heat given in Table A-19E ρ=94.9 lbm/ft 3, c p = c= 0.191Btu/lbm- o F Solution:

19. Entropy Change for Liquids: Example 0.15 m 3 of water at 350 K is heated to 400 K find heat added and change in entropy considering constant specific heat at 375 K from Table A-19. Assume: Constant specific heat (and constant density) at 375 K from Table A-19. Solution:

20. Specific Entropy “s” (kJ/kg-K or BTU/lbm- o R) Change for Ideal Gases, Constant c p, c v ►Remember Chapter 3 (Property Relations). Divide by “m” kg

21. Specific Entropy “s” (kJ/kg-K or BTU/lbm- o R) Change for Ideal Gases, Variable c p, c v ►Remember Chapter 3 (Property Relations). Divide by “m” kg

22. Entropy Change for Air: Example 0.15 kg of Air at 700 K, 10 atm pressure is heated to 1200 K in a piston cylinder device at constant pressure, find heat added and change in entropy considering variable specific heat and properties from Table A-22. Solution: T, Ku, kJ/kgS o kJ/kg-K 700 512.332.57277 1200 933.333.17888

23. Entropy Change for Ideal Gases: Example 2 lbm of N 2 is heated at constant volume from 32 o F to 70 o F find heat added and change in entropy. Assume: Ideal Gas and variable specific heat with properties given in Table A23E Solution:

24. Entropy Generation using Tables: Example 1 24 Given: Steam at 120 o C, 0.7 bar is pressurized through a diffuser to 1 bar, 160 o C and negligible velocity. Find: Find the change in entropy of steam in kJ/kg-K and comment on whether the diffuser can be adiabatic and the resulting impact. Assumptions: Change in PE neglected, No heat transfer, No work done other than flow work, Steady state, Steady flow, Mass is conserved.

25. 25 State 1: 0.7bar, 100 C State 2: 1 bar, 160 C T-s Diagram and Diffuser Action State 2: 1 bar, h2>h1, s2>S1 State 1: 0., 7bar, 100 C

26. On the T-s diagram drawn to scale State 1 and State 2 2 1

27. Entropy Generation Calculation: Example 2 27 Given: 0.5 kg/s of steam at 280 o C, 20 bar is expanded in a turbine to 1 bar in a constant entropy process. If the process was not a constant entropy process and resulted in saturated steam at 1 bar, find the decrease in work and increase in entropy in kW/K. Find: Find the work produced by the steam in kW and show the processes on a T-s diagram. If the process was not a constant entropy process and resulted in saturated steam at 1 bar, find the decrease in work and increase in entropy in kW/K. Assumptions: Change in PE neglected, No heat transfer, work done on Turbine shaft, Steady state, Steady flow, Mass is conserved. StP,Thsx i20,2802976.46.6828- es1,99.632423.26.68280.8883 e1,99.632675.57.35941.0000 x=(s-s f )/(s g -s f )=(6.6828-1.3026)/(7.3594-1.3026)=5.3802/6.0568=0.8883; h es = 417.46 +0.8883 (2258) = 2423.22

28. State 2a: Saturated State 1: 20 bar, 280 C On the T-s diagram drawn to scale State 1 and State 2 are close to each other as illustrated below. State 2s: Mixture

29. Entropy Generation: Example 3 29 Given: Consider R134 throttled from p3 =120 lbf/in 2 to p4 =40 lbf/in 2. Find: Find the change in entropy of R134. Assumptions: Change in KE, PE neglected, No heat transfer, No work done other than flow work, Steady state, Steady flow, Mass is conserved. Adiabatic throttle with a pressure loss and phase change lead to increase in Entropy while keeping Enthalpy constant. Entropy is generated by fluid friction Or viscosity in this case, in spite of the process being (externally) adiabatic.

30. 30 State 4 State 3 T-s Diagram and Demonstration of Throttle Action; h-s diagram State 3 State 4

31. Entropy Generation: Example 4 31 Given: Consider R134 condensed from saturated vapor (state 2) to saturated liquid at p3 =120 lbf/in 2. Find: Find the change in entropy generation rate in the process of condensing R134. Assumptions: Change in KE, PE neglected, Heat transfer to a sink at 90.54 o F- δ and heat transfer to sink at 80.54 o F. No work done other than flow work, Steady state, Steady flow, Mass is conserved.

32. 32 State 2 State 3 T-s Diagram and Demonstration of Condenser Action; h-s diagram State 2 State 3 Sink Temperature

33. 33 Conservation Laws

34. Property Relations and Efficiency Definitions Cycle Efficiency Definitions