1. 1 Chapter 5 Mass and Energy Analysis of Control Volumes

2. 2 The First Law of Thermodynamics Open System (Control Volume)

3. 3 First low of thermodynamics for open Systems Reminder of an open System. Open system = Control volume It is a properly selected region in space. Mass and energy can cross its boundary.

4. 4 Control volume involves two main processes Steady flow processes. Fluid flows through the control volume steadily ( in = out). Its properties are experiencing no change with time at a fixed position. Unsteady flow processes. Fluid properties are changing with time.

5. 5 Mass balance for steady flow processes We already showed that for steady flow Many engineering devices involve a single stream (one inlet and one exit only).

6. 6 Energy Balance for Steady- Flow Systems 0(steady)

7. 7 Let us look at some common steady flow devices Only one in and one out More than one inlet and exit

8. 8 For single stream steady flow devices, the 1 st low becomes Often the change in kinetic energy and potential energy is small. Per unit mass

9. 9 Nozzles A nozzle is a device that increases the velocity of a fluid at the expense of pressure A1A1 A2A2

10. 10 A1A1 A2A2 Diffusers A diffuser is a device that slows down the velocity of a fluid causing an increase in its pressure

11. 11 Diffusers

12. 12 Nozzles and Diffusers (1 st low analysis) Is there work in this system? NO Is there heat transfer? In fact, it depends on the problem! Does the fluid change elevation? NO let us say: NO

13. 13 How can you find the mass flow rate in a nozzle? In a nozzle, enthalpy is converted into kinetic energy

14. 14 Example (5-4): Deceleration of Air in a Diffuser Air at 10 o C and 80 kpa enters the diffuser of a jet engine steadily with a velocity of 200 m/s. The inlet area of the diffuser is 0.4 m 2. The air leaves the diffuser with a velocity that is very small compare to the inlet velocity. Determine (1) The mass flow rate of the air and (2) The temperature of the air leaving the diffuser.

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17. 17 Example (5-5): Acceleration of Steam in a Nozzle Steam at 1.8 MPa and 400  C steadily enters a nozzle whose inlet area is 0.02 m 2 (0.2 ft 2.) The mass flow rate of the steam through the nozzle is 5 kg/s. Steam leaves the nozzle at 1.4 MPa with a velocity of 275 m/s. The heat losses from the nozzle per unit mass of the steam are estimated to be 2.8 KJ/kg. Determine: (a) the inlet velocity and (b) the exit temperature of the steam.

18. Solution of Example: Acceleration of Steam in a Nozzle Note that there is heat transfer (Q). So, you have to go back to the general form of the 1 st low for single stream devices and get the following:

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20. 20 Turbines A turbine is a device that produces work at the expense of temperature and pressure. As the fluid passes through the turbine, work is done against the blades, which are attached to a shaft. As a result, the shaft rotates, and the turbine produces work.

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24. 24 Compressors A compressor is a device that increases the pressure of a fluid by adding work to the system. Work is supplied from an external source through a rotating shaft.

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28. 28 Turbines and Compressors Is there work in this system? Yes! Is there heat transfer? Negligible because of insulation. Exception: Internal cooling in some compressors. Does the fluid change elevation? NO Does the kinetic energy change? Usually it can be ignored

29. 29 Example 5-6

30. 30 Example 5-6

31. 31 Example (5-7): Power Generation by a Steam Turbine The power output of an adiabatic steam turbine is 5 MW, and the inlet and the exit conditions of the steam are as indicated in the figure on the right. a) Compare the magnitude of  h,  ke, and  pe. b) Determine the work done per unit mass of the steam flowing through the turbine. c) Calculate the mass flow rate of the steam.

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34. Calculating Heat Transfer from a Steam Turbine Steam enters a turbine operating at steady state with a mass flow rate of 4600 kg/h. The turbine develops a power output of 1000 kW. At the inlet, the pressure is 60 bar, the temperature is 400 o C, and the velocity is 10 m/s. At the exit, the pressure is 0.1 bar, the quality is 0.9 (90%), and the velocity is 50 m/s. Calculate the rate of heat transfer between the turbine and surroundings, in kW. 34

35. Known: A steam turbine operates at steady state. The mass flow rate, power output, and states of the steam at the inlet and exit are known. Find: Calculate the rate of heat transfer. 35

36. To calculate the heat transfer rate, begin with the one-inlet, one-exit form of the energy rate balance for a control volume at steady state 36 State 2 is a two-phase liquid–vapor mixture, so with data from Table A-4 and the given quality

37. 37 The magnitude of the change in specific kinetic energy from inlet to exit is much smaller than the specific enthalpy change.

38. Solved Problem 1 A well-insulated turbine operating at steady state. Steam enters at 3 MPa, 400°C, with a volumetric flow rate of 85 m 3 /min. Some steam is extracted from the turbine at a pressure of 0.5 MPa and a temperature of 180°C. The rest expands to a pressure of 6 kPa and a quality of 90%. The total power developed by the turbine is 11,400 kW. Kinetic and potential energy effects can be neglected. Determine: (a) the mass flow rate of the steam at each of the two exits, in kg/h. (b) the diameter, in m, of the duct through which steam is extracted, if the velocity there is 20 m/s. 38

39. 39 Mass Balance

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42. 42 Throttling Valve A throttling valve reduces the fluid pressure. P 1 >P 2 It is small device and thus the flow through it may be assumed adiabatic (q=0) since there is neither sufficient time nor large enough area for any effective heat transfer to occur.

43. 43 Throttling Valve Is there work in this system? NO Is there heat transfer? Usually it can be ignored Does the fluid change elevation? NO Does the fluid change velocity? Usually it can be ignored

44. 44 What happens to the fluid temperature a cross throttling Valves ?

45. 45 Throttling Valves (incompressible substance ) For incompressible substance (like water),  is constant and For incompressible substance only! “For Compressible substance (gas) the opposite T will drop ”

46. 46 Throttling Valves (compressible substance: Vapor) Example (5-8): Expansion of Refrigerant-134a in a Refrigerator Refrigerant-134a enters the capillary tube of a refrigerator as saturated liquid at 0.8 MPa and is throttled to a pressure of 0.12 MPa. Determine the quality of the refrigerant at the final state and the temperature drop during this process.

47. Solution of Example: Expansion of Refrigerant-134a in a Refrigerator Notice that T 2 <T 1

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49. Solved Problem 2: a steam turbine at steady state is operated at part load by throttling the steam to a lower pressure before it enters the turbine. Before throttling, the pressure and temperature are, respectively, 1.5 MPa and 320 C. After throttling, the pressure is 1 MPa. At the turbine exit, the steam is at.08 bar and a quality of 90%. Heat transfer with the surroundings and all kinetic and potential energy effects can be ignored. Determine (a) the temperature at the turbine inlet, in C. (b) the power developed by the turbine, in kJ/kg of steam flowing. 49

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52. 52 Throttling Valves (Compressible substance: an Ideal Gas) What happens if the gas is ideal? For ideal gases  h = Cp  TBut  h = 0 So…  T = 0 The inlet and outlet temperatures are the same!!!

53. 53 Throttling Valves: h1 = h2 (Compressible substance: an Ideal Gas) T2 =T1 Compressible substance: Real Gas T2 <<T1 Incompressible substance : Liquid T2>T1

54. 54 First Law: Q net (Q H -Q L ) = W net (W T -W P )

55. 55 Mixing Chamber Mixing two or more fluids is a common engineering process The mixing chamber does not have to be a distinct “chamber.” An ordinary T- elbow, or a Y-elbow in a shower, for example, serves as the mixing chamber for the cold- and hot-water streams as shown in the figure (Left).

56. 56 Mixing Chamber We no longer have only one inlet and one exit stream Is there any work done? No Is there any heat transferred? No Is there a velocity change? No Is there an elevation change? No

57. 57 Mixing Chamber Material Balance Energy balance

58. 58 Example(5-9): Mixing of Hot and Cold Waters in a Shower Consider an ordinary shower where hot water at 140 o F is mixed with cold water at 50 o F. If it is desired that a steady stream of warm water at 110 o F be supplied, determine the ratio of the mass flow rates of the hot to cold water. Assume the heat losses from the mixing chamber to be negligible and the mixing to take place at a pressure of 20 psia.

59. 59 Example(5-9): Mixing of Hot and Cold Waters in a Shower

60. 60 Example(5-9): Mixing of Hot and Cold Waters in a Shower

61. 61 Heat Exchanger A heat exchanger is a device where two moving fluids exchange heat without mixing.

62. 62 Heat Exchangers Your analysis approach will depend on how you define your system (Control Volume)

63. 63 Heat Exchangers System (a): entire HEX Mass Balance Divide into two separate streams with equal inlet and outlet flow rates Energy balance Two inlets Two outlets 1 2 34

64. 64 Heat Exchangers System (b): Single stream Mass Balance Considering one single stream with one inlet and one outlet flow rates Energy balance One inlet One outlet Plus heat transfer

65. 65 Example5-10: Cooling of Refrigerant-134a by Water Refrigerant-134a is to be cooled by water in a condenser. The refrigerant enters the condenser with a mass flow rate of 6 kg/min at 1 MPa and 70 o C and leaves at 35 o C. The cooling water enters at 300 kPa and 15 o C and leaves at 25 o C. Neglecting any pressure drop, determine (a) the mass flow rate of the cooling water required and (b) the heat transfer rate from the refrigerant to water.

66. 66 Example: Cooling of Refrigerant-134a by Water

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68. 68 Pipe and Duct Flow May involve Heat Loss May involve different kind of work K.E. & P.E. Changes are usually insignificant

69. 69 Example 5-11

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72. 72 Energy Analysis of Unsteady Flow Process Steady flow: no changes occur within the Control Volume Unsteady or transient: Processes involve changes within the control volume Usually involve filling or evacuating processes They may have boundary work

73. 73 Energy Analysis of Unsteady Flow Process Conservation of mass (mass balance))

74. 74 Energy Analysis of Unsteady Flow Process Conservation of Energy (Energy balance)) the uniform-flow process will be assumed, which involves the following idealization: The fluid flow at any inlet or exit is uniform and steady, and thus the fluid properties do not change with time or position over the cross section of an inlet or exit. If they do, they are averaged and treated as constants for the entire process. The changes inside the C.V. not on the boundaries

75. 75 Energy Balance IF K.E. and P.E. are negligible

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78. 78 W=P(V2-V1) W=P(m2v2-m1v1)