1. CH4 Energy Analysis of Closed System 1

2. Objectives Examine the moving boundary work or P dV work commonly encountered in reciprocating devices such as automotive engines and compressors. Identify the first law of thermodynamics as simply a statement of the conservation of energy principle for closed (fixed mass) systems. Develop the general energy balance applied to closed systems. Define the specific heat at constant volume and the specific heat at constant pressure. Relate the specific heats to the calculation of the changes in internal energy and enthalpy of ideal gases. Describe incompressible substances and determine the changes in their internal energy and enthalpy. Solve energy balance problems for closed (fixed mass) systems that involve heat and work interactions for general pure substances, ideal gases, and incompressible substances. 2

3. Closed Systems 3

4. How does energy cross a closed system boundary? Energy can cross the boundaries of a closed system in the form of heat or work. If the energy transfer across the boundaries of a closed system is due to a temperature difference, it is heat; otherwise, it is work. 4

5. So, what do we mean by Heat? Heat is defined as the form of energy (thermal energy) that is transferred between two systems (or surroundings) by virtue of a temperature difference. Heat is energy in transition. It is recognized as it crosses the boundary of a system. 5 Hot Cold

6. No heat transfer =Adiabatic Process A process during which there is no heat transfer is called an adiabatic process. The adiabatic process should not be mixed with Isothermal process. A process could be an adiabatic process while the temperature of a system and surrounding are different. 6

7. Symbols of Heat Q or 1 Q 2 Definition: The amount of total heat transferred during a process from state 1 to state 2. Units: kJ or BTU q Heat transfer per unit mass. Units: kJ/kg or BTU/lb m Rate of heat transfer kJ/sec = kW 7

8. Sign convention Heat transferred to a system is positive. Heat transferred from a system is negative. 8

9. What do we mean by Work? Work is defined as the energy transfer associated with force acting through a distance. 9

10. Symbols of Work W or 1 W 2 The amount of work done by or on a system during a process from state 1 to state 2. Units: kJ or BTU w Work per unit mass. Units: kJ/kg or BTU/lb m Power: The work done by unit time. kJ/sec = kW 10

11. Sign convention Work done on a system by an external force acting in the direction of motion is negative. Work done by a system against an external force acting in the opposite direction of the motion is positive. 11

12. Characteristics of Heat and Work Both of them are boundary phenomena. That is they are recognized at the boundaries of a system as they cross them. Systems possess energy, not heat or work. Both are associated with a process, not a state. Both are path functions (i.e., their magnitudes depend on the path followed during a process as well as the end states). 12

13. What are Path Functions? Path functions have inexact differentials designated by the symbol , eg.  Q and  W. 13 Point functions have exact differentials designated by the symbol , eg.  V Properties are point functions, but heat and work are path functions.

14. Types of work Mechanical Work Moving boundary work (The most common form) Shaft work Spring work Electric Work An electric wire crossing the system boundaries, (W e =V I  t) J ( remember V I or I 2 R= Watt= J/s)  V =voltage  I=current   t =time 14

15. Two requirements for Mechanical work to exist 1. There must be a force on the boundary. Example: the expansion of a gas into an evacuated space (is not work since there is no force). 2. The boundary must move. Example: Fixed piston in the presence of external force (is not work since the boundary did not move). 15

16. 4-1 Moving Boundary Work The area under the process curve on a P-V diagram represents the boundary work. 16 If P is constant

17. Boundary Work is Path Dependent 17 The bigger the area under the P-V curve the bigger the work

18. Net Work done in a cycle The cycle shown in the figure (right) produces a net work output. This is because the work done by the system during the expansion process (area under path A) is greater than the work done on the system during the compression part of the cycle (area under path B), The difference between these two is the net work done during the cycle (the colored area). 18 W A > 0 since  V = V 1 –V 2 > 0 W B < 0 since  V = V 2 –V 1 < 0 W net = W A +W B = Area bounded by the curve

19. Car Engine Boundary Work, W b Overcome friction between the piston and the cylinder To Push the exhaust gas out of the way To rotate the Crankshaft 19

20. Example 4-1 20

21. Example 4-1 21

22. Moving Boundary Work for quasi- equilibrium process Strictly speaking the pressure in the integral is the pressure at the inner surface of the piston (P i ) It becomes equal to the pressure in the cylinder only if the process is quasi- equilibrium. Thus the entire gas is at the same pressure at any given time (slow process) A ds = dV 22

23. Moving Boundary Work for non- quasi- equilibrium process For non-quasi-equilibrium process, we still can compute the work if the pressure at the inner surface of the piston is used for P i. We can not speak of pressure of a system during non-quasi- equilibrium process. 23

24. Computing the work for some special processes Constant volume Constant pressure Isothermal Polytropic 24

25. Constant Volume Process 25 P V 1 2

26. Constant Pressure Process 26 P V 12 W

27. 27 Example 4-2: Boundary Work during a Constant- Pressure Process A floating frictionless piston-cylinder device contains 5 kg of water vapor at 400 kPa and 200 o C. Heat is now transferred to the steam until the temperature reaches 250 o C. Determine the work done by the steam during this process. (Answer: 121.7 kJ)

28. Example 4-2: Boundary Work during a Constant-Pressure Process… First determine the state 1 & 2 Find the specific volume for state 1 and 2 The W=P(V2-V1)=P.m(v2-v1) 28

29. Isothermal Process 29 P V 1 2 But for an ideal gas OR

30. 30 Example 4-3: Boundary Work during an Isothermal Process A piston-cylinder device initially contains 0.4 m 3 of air at 100 kPa and 80 o C. The air is now compressed to 0.1 m 3 in such a way that the temperature inside the cylinder remains constant. Determine the work done during this process. (Answer: -55.45 kJ)

31. Example 4-3: Boundary Work during an Isothermal Process 31

32. Polytropic Process 32 For General Gases, P and V are often related as

33. Lets go through that step by step 33 2 1

34. 34 But… And…. WbWb So… Polytropic Process V -n+1 = V/V n

35. 35 So far, we have not assumed an ideal gas in this derivation, if we do, then…. Ideal gas undergoing a Polytropic Process

36. Computing the work of a Gas expanding against a Spring 36 Spring just touching piston Spring exerting force on piston

37. Computing the Spring work alone 37 Pressure Volume W 1 2 Spring exerting force on piston 2 V1V2 No Spring exerting force on piston

38. Computing the Spring work alone 38 1

39. 39 Example 4-4: Expansion of a Gas against a Spring A piston-cylinder device contains 0.05 m 3 of gas initially at 200 kPa. At this state, a linear spring that has a spring constant of 150 kN/m is touching the piston but exerting no force on it. Now heat is transferred To the gas, causing the piston to rise and to compress the spring until the volume inside the cylinder doubles. If the cross-sectional area of the piston is 0.25 m 2, determine (a) the final pressure inside the cylinder, (b) the total work done by the gas, and c) the fraction of this work done against the spring to compress it.

40. Example 4-4: Expansion of a Gas against a Spring 40

41. Example 4-4: Expansion of a Gas against a Spring 41

42. Example 4-4: Expansion of a Gas against a Spring 42

43. Computing the Shaft work 43

44. Example: Power transmission by the shaft of a car 44 Determine the power transmission through the shaft of a car when the torque applied is 200 Nm and the shaft rotate at a rate of 4000rpm. (Answer: 83.8 kW)

45. 4-2 Energy Balance for Closed Systems Energy balance for any system undergoing any kind of process was expressed as (see Chap. 2) In a rate form Per Unit Mass 45

46. 4-2 Energy Balance for Closed Systems… For constant rates, the total quantities during a time interval Δt are related to the quantities per unit time Energy balance can also be expressed in the differential form as For a closed system undergoing a cycle, the initial and final states are identical, 46

47. A Cycle 47

48. For cycle in closed system the energy balance for a cycle in closed system can be expressed in terms of heat and work interactions as (cycle means you come back to the same point) Remember 48 ΔEΔE

49. The Energy Balance –First Law of Thermodynamic for closed system The net heat input is The net Work output is W = W net,out = W out -W in Obtaining a negative quantity for Q or W simply means that the assumed direction for that quantity is wrong and should be reversed. 49

50. First Law of Thermodynamic Relations for closed system 50

51. Example 4-5 51

52. Example 4-5 52

53. Example 4-5 53

54. Example 4-5 54

55. Example 4-6 55

56. 56

57. 57