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EGR 334 Thermodynamics Chapter 3: Section 6-8

EGR 334 Thermodynamics Chapter 3: Section 6-8 . Lecture 07: Enthalpy and Internal Energy. Quiz Today?. Today’s main concepts:. Define enthalpy, h Recognize that enthalpy and internal energy are both intensive state properties that can also be determined on the steam tables.

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EGR 334 Thermodynamics Chapter 3: Section 6-8

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  1. EGR 334 ThermodynamicsChapter 3: Section 6-8 Lecture 07: Enthalpy and Internal Energy Quiz Today?

  2. Today’s main concepts: • Define enthalpy, h • Recognize that enthalpy and internal energy are both intensive state properties that can also be determined on the steam tables. • Learn to pull values for enthalpy and internal energy from the steam tables. • Learn to use computer tools to find property values. • Use steam table values to solve 1st Law balance problems. Reading Assignment: • Read Chap 3: Sections 9-10 Homework Assignment: From Chap 3: 35, 41, 42, 45

  3. Last time we learned that if you know two of the three intensive properties: Temperature, T Specific Volume, v and Pressure, p ...then you could find the other properties at that state. Other intensive properties include: Enthalpy, h Internal energy, u andEntropy, s (which are also found on the steam tables. or or

  4. Sec 3.6.1 : Introducing Enthalpy Specific Internal Energy (u) : energy / mass Kinetic Energy translational rotational vibrational Potential Energy electric energy of atoms, molecules, or crystals Chemical Bonds Internal Energy (U) Enthalpy (H) Specific Enthalpy (h) : energy / mass Internal Energy + Work

  5. Saturated Steam Table Enthalpy, h For the time being, we’re going to still ignore this last quantity. Internal Energy, u Notice there are additional columns on the steam table that we ignored last time.

  6. What is enthalpy? Enthalpy is defined in terms of internal energy and the work of expansion as Intensive Form Extensive Form Your author indicates that this term is defined, simply because it’s convenient. This combination of terms will show up very frequently, especially when we begin working with control volumes. It’s a way to conveniently deal with the internal energy in combination with the energy associated with expansion or compression of the substance. Enthalpy is another formulation of energy and is commonly expressed in units of kJ or ft-lbf.

  7. Example 1: Determine property values Determine specific volume (v), internal energy (u), and enthalpy (h) of the following state properties of H20. a) p = 10 bar T = 179.9 deg C. x = 30% b) p = 10 bar T = 320 deg. C. c) p = 10 bar T = 450 deg C.

  8. Example 1: Determine property values Determine specific volume, internal energy, and enthalpy of the following state properties of H20. a) p = 10 bar T = 179.9 deg C. x = 30% from table A.3: at 10 bar the saturated temperature is 179.9 which means the substance is a mixture of liquid and vapor.

  9. From table A-3 the following values may be directly read: therefore:

  10. Example 1: Determine property values Determine specific volume, internal energy, and enthalpy of the following state properties of H20. b) p = 10 bar T = 320 deg. C. vapor therefore: v = 0.2678 m3/kg u = 2826.1 kJ/kg h = 3093.9 kJ/kg

  11. Example 1: Determine property values c) p = 10 bar T = 450 deg C. This also is a gas or superheated vapor and will need to be found by interpolating from table A-4.

  12. Linear Interpolation: * y2 yunknown * y1 x2 x1 xknown

  13. Your turn: Find the value for enthalpy at T = 400 deg. C and p = 50 bar at 400°C and p1=40 bar  h1 = 3213.6 kJ/kg at 400°C and p2=60 bar  h2 = 3177.2 kJ/kg = 3195.4 kJ/kg

  14. On Tests and Quizzes, you will be expected to pull out data from the printed steam tables. That’s not your only option when working homework. Computer Method 1: Web based Steam Table Calculator at: http://www.dofmaster.com/steam.html For: T = 400 deg C. p = 50 bar = 500 kPa Enthalpy per unit mass: h = 3195.4 kJ/kg

  15. Computer Method 2: Interactive Thermodynamics (IT) Software you can download to your computer from: http://www.wiley.com/college/moran IT is more than just a look up table. It is also an equation solver that can automatically look up properties values and attempt to find a solution using an iterative solver. h = 3195 kJ/kg

  16. Sec 3.5.2 : Saturation Tables Example: Use IT to solve this problem. A cylinder-piston assembly initially contains water at 3 MPa and 300oC. The water is cooled at constant volume to 200 oC, then compressed isothermally to a final pressure of 2.5 MPa. Find the specific volume and enthalpy at each of these three states.

  17. State 1: p1 = 3 MPa= 30 bar and T1 = 300 deg C. i) Start with superheated vapor table A-4 ii) Interpolate between 280 and 320 deg C.

  18. State 2: v2 = v1 = 0.0811 m3/kg and T2 = 200 deg C. Recognize that this is in the liquid-vapor mixture range. Refer to Table A-2 i) Locate vg and vf at T = 200 deg C. ii) Determine quality, x2

  19. State 3: p3 = 2.5 MPa = 25 bar and T2 = T3= 200 deg C. Recognize that this is in the compressed liquid. Refer to Table A-5 i) Locate p3=25 bar and T3=200 deg C ii) Read the values directly.

  20. Repeat this work, but use IT instead. Results are the same as shown yesterday.

  21. Sec 3.6.2: Retrieving u and h Data Workout Problem a) Using Tables in Appendix: Given Ammonia at 12°C and v = 0.1217 m3/kg determine P, u and h. b) Using IT:

  22. Sec 3.6.2: Retrieving u and h Data Workout Problem a) Using Tables in Appendix: Given Ammonia at 12°C and v = 0.1217 m-3/kg determine P, u and h.

  23. Sec 3.6.2: Retrieving u and h Data Example: Notice that vf= 0.001608 and vg = 0.1923 Given Ammonia at 12°C and v=0.1217 m-3/kg determine P, u and h. Since: vf< v < vg the substance is a liquid-gas mixture Saturation Properties are: Tsat = 12 deg C. Psat= 658.9 kPa Quality if found using:

  24. Sec 3.6.2: Retrieving u and h Data Example: Given Ammonia at 12°C and v=0.1217 m-3/kg determine P, u and h. with • vf = 1.608x10-3, vg = 0.1923 m3/kg

  25. Sec 3.6.2: Retrieving u and h Data Example: Given Ammonia at 12°C and v=0.1217 m-3/kg determine P, u and h. Internal Energy: Enthalpy: Check:

  26. b) Using IT:

  27. Sec 3.8: Appling the Energy Balance using the Property Tables Workout Problem: (3.60) A rigid, insulated tank fitted with a paddle wheel is filled with water, initially a two phase liquid-vapor mixture at 20 psi, consisting of 0.07 lb of saturated liquid and 0.07 lb of saturated vapor. The tank contents are stirred by the paddle wheel until all of the water is saturated vapor at a pressure greater than 20 psi. Kinetic and potential energy effects are negligible. For the water, determine Pi = 20 psi mf = 0.07 lb ml = 0.07 lb • Volume occupied, in ft3, • Initial temperature, in °F, • Final pressure, in psi, • Work, in BTU

  28. Sec 3.8: Appling the Energy Balance using the Property Tables Example: (3.77) A system consisting of 1 kg or water undergoes a power cycle composed of the following processes. Process 1-2 : Constant-pressure heating at 10 bar from saturated vapor. • Process 2-3:Constant-volume cooling to P3 = 5 bar, T3 = 160°C Process 3-4: Isothermal compression with Q34 = -815.8 kJ • Process 4-1: Constant-volume heating Sketch the cycle on T-v and P-v diagrams. Neglecting kinetic and potential energy effects, determine the thermal efficiency.

  29. End of Slides for Lecture 07

  30. Sec 3.8: Appling the Energy Balance using the Property Tables Workout Problem: (3.60) A rigid, insulated tank fitted with a paddle wheel is filled with water, initially a two phase liquid-vapor mixture at 20 psi, consisting of 0.07 lb of saturated liquid and 0.07 lb of saturated vapor. The tank contents are stirred by the paddle wheel until all of the water is saturated vapor at a pressure greater than 20 psi. Kinetic and potential energy effects are negligible. For the water, determine Pi = 20 psi mf = 0.07 lb ml = 0.07 lb • Volume occupied, in ft3, • Initial temperature, in °F, • Final pressure, in psi, • Work, in BTU

  31. Sec 3.8: Appling the Energy Balance using the Property Tables • Volume occupied, in ft3, • Initial temperature, in °F, • Final pressure, in psi, • Work, in BTU Pi = 20 psi mf = 0.07 lb ml = 0.07 lb

  32. Sec 3.8: Appling the Energy Balance using the Property Tables • Volume occupied, in ft3, • Initial temperature, in °F, • Final pressure, in psi, • Work, in BTU Pi = 20 psi mf = 0.07 lb ml = 0.07 lb xi = 0.5 a) b) T = Tsat = 227.96 °F

  33. Sec 3.8: Appling the Energy Balance using the Property Tables • Volume occupied, in ft3, • Initial temperature, in °F, • Final pressure, in psi, • Work, in BTU Pi = 20 psi mf = 0.07 lb ml = 0.07 lb xi = 0.5 c) Interpolate values to find P2 = 42 psi & u2 = 1093 BTU/lb

  34. Sec 3.8: Appling the Energy Balance using the Property Tables Example: (3.77) A system consisting of 1 kg or water undergoes a power cycle composed of the following processes. Process 1-2 : Constant-pressure heating at 10 bar from saturated vapor. • Process 2-3:Constant-volume cooling to P3 = 5 bar, T3 = 160°C Process 3-4: Isothermal compression with Q34 = -815.8 kJ • Process 4-1: Constant-volume heating Sketch the cycle on T-v and P-v diagrams. Neglecting kinetic and potential energy effects, determine the thermal efficiency.

  35. Sec 3.8: Appling the Energy Balance using the Property Tables Example: (3.77) A system consisting of 1 kg or water undergoes a power cycle composed of the following processes. Process 1-2 : Constant-pressure heating at 10 bar from saturated vapor. • Process 2-3:Constant-volume cooling to P3 = 5 bar, T3 = 160°C Process 3-4: Isothermal compression with Q34 = -815.8 kJ • Process 4-1: Constant-volume heating Sketch the cycle on T-v and P-v diagrams. Neglecting kinetic and potential energy effects, determine the thermal efficiency.

  36. Sec 3.8: Appling the Energy Balance using the Property Tables Example: (3.77) A system consisting of 1 kg or water undergoes a power cycle composed of the following processes. Process 1-2 : Constant-pressure heating at 10 bar from saturated vapor. • Process 2-3:Constant-volume cooling to P3 = 5 bar, T3 = 160°C Process 3-4: Isothermal compression with Q34 = -815.8 kJ • Process 4-1: Constant-volume heating Sketch the cycle on T-v and P-v diagrams. Neglecting kinetic and potential energy effects, determine the thermal efficiency.

  37. Sec 3.8: Appling the Energy Balance using the Property Tables Example: (3.77) A system consisting of 1 kg or water undergoes a power cycle composed of the following processes. Process 1-2 : Constant-pressure heating at 10 bar from saturated vapor. • Process 2-3:Constant-volume cooling to P3 = 5 bar, T3 = 160°C

  38. Sec 3.8: Appling the Energy Balance using the Property Tables Example: (3.77) A system consisting of 1 kg or water undergoes a power cycle composed of the following processes. Process 3-4: Isothermal compression with Q34 = -815.8 kJ • Process 4-1: Constant-volume heating

  39. Sec 3.8: Appling the Energy Balance using the Property Tables Example: (3.77) A system consisting of 1 kg or water undergoes a power cycle composed of the following processes. Process 1-2 : Constant-pressure heating at 10 bar from saturated vapor. • Process 2-3:Constant-volume cooling to P3 = 5 bar, T3 = 160°C Process 3-4: Isothermal compression with Q34 = -815.8 kJ • Process 4-1: Constant-volume heating Sketch the cycle on T-v and P-v diagrams. Neglecting kinetic and potential energy effects, determine the thermal efficiency.

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