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5 Exergy

5 Exergy. 5.1 Introduction. Quantity. Evaluation of heat. Quality. Exergy and Anergy. Full convertible energy: mechanical Partial convertible energy: heat Unconvertible energy: environmental. Exergy: useful work potential; available energy;

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5 Exergy

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  1. 5Exergy 5.1 Introduction Quantity Evaluation of heat Quality Exergy and Anergy Full convertible energy: mechanical Partial convertible energy: heat Unconvertible energy: environmental

  2. Exergy: useful work potential; available energy; The maximum useful work a system can delivered from a specified state to the state of its environment in theory. Anergy: unavailable energy, unconvertible energy Conditions for definition of exergy （1）based on environment; exergy of environmental energy is zero （2）reversible process （3）there is no other heat resource in the process.

  3. unbalance Type Chemical Chemical potential Physical  Temperature and pressure Kinetic  Velocity Position  Position Concentration Distribution 5.2 Calculation of exergy

  4. Unbalance Source Electricity voltage Water level Hydraulic Wind pressure Pneumatic Difference between interior and surface Wave 

  5. 5.2.1 Work resource Electricity, mechanical energy, pneumatic energy, hydraulic energy, can be converted to work entirely. Exergy of work resource = its total energy

  6. T T0 5.2.2 Heat exergy Potential work of heat W0

  7. For constant temperature heat resource

  8. T0 For finite heat resource

  9. Influential factors: • Heat quantity • Heat resource temperature • Environmental temperature T0 If the system absorbs heat, it absorbs exergy; If the system discharges heat, it discharges exergy;

  10. Example 1kg air with temperature of 200℃ was cooled to 40℃. Please calculate the heat exergy. The specific heat of air is cp=1.004kJ/(kg.K). the environmental temperature is T0=25℃。

  11. Solution ：

  12. T0 W0 T 5.2.3 Low temperature heat exergy Potential work of heat at a temperature below the environmental temperature

  13. For constant temperature

  14. For finite heat resource

  15. Influential factors: • Heat quantity • Heat resource temperature • Environmental temperature T0 If the system absorbs heat, it discharges exergy; If the system discharges heat, it absorbs exergy;

  16. 5.2.4 Inner energy exergy For closed system From initial state to final state

  17. For A→B→O The dead work to resist environment The maximum work

  18. For m kg substance The maximum available work

  19. Example Please calculate the inner energy exergy of air in the state of 1MPa and 50℃. The environmental pressure is p0=0.1MPa，the temperature is T0=25℃ and the specific heat is cv=0.716kJ/(kg·K)。

  20. 5.2.5 Enthalpy exergy For steady flow system From initial state to final state

  21. For A→B→O For m kg substance The maximum available work

  22. 开系工质物理

  23. Example Please compare the enthalpy exergy value of saturated steam of 0.5MPa with 5MPa. The environmental state is p0=0.1MPa and T0=20℃。

  24. Solution : Look up properties in the table of wateror steam h0=84 kJ/kg s0=0.2963kJ/(kg·K) h1=2747.5 kJ/kg s1=6.8192 kJ/(kg·K) h2=2794.2 kJ/kg s2=5.9735 kJ/(kg·K)

  25. ex1 =(h1－h0)－T0(s1－s0) =(2747.5－84)－293×(6.8192－0.2963) =752.3kJ/kg ex2=(h2－h0)－T0(s2－s0) =(2791.2－84)－293×(5.9735－0.2963) =1046.8kJ/kg

  26. TA Q H.E T0 5.3 Exergy loss Exergy loss caused by temperature difference

  27. TA Q TB Q H.E T0

  28. Exergy loss Entropy production

  29. T H.E T0

  30. 温差传热引起的损失

  31. Exergy loss caused by temperature difference

  32. If discharging temperature is The whole exergy loss

  33. 5.4 exergy equation

  34. For many streams Technical work Heat exergy Enthalpy exergy Exergy loss

  35. 5.5 Exergy efficiency 1 2

  36. Example The high temperature is TH=1800K and the low temperature is the environmental temperature that is T0=290K. A heat engine absorbs heat at T1=900K and discharges heat at T2=320K. The engine efficiency is 70% of that of corresponding carnot cycle。If each kilogram substance absorbs heat 100kJ，please calculate: (1)the practical work of heat engine; (2)the heat exergy at given temperature; (3)the entropy production and exergy loss of each process; (4)the entropy increase of isolated system and the whole exergy loss.

  37. TH = 1800 K T1 = 900 K R W T2 = 320 K T0 = 290 K

  38. TH T1 2 1 T2 3 4 T0 s

  39. Solution: (1) The carnot cycle works between T1 and T2 Practical work of engine

  40. Discharged heat (2) Heat exergy Heat exergy at 1800K Heat exergy at 900K

  41. Heat exergy at320K (3) Entropy production and exergy loss Entropy production caused by temperature difference in absorbing heat process Exergy loss or

  42. Friction work loss Entropy production Exergy loss 0r

  43. Entropy production caused by temperature difference in discharging heat process Exergy loss 0r

  44. (4)Entropy increase of isolated system 0r

  45. The whole exergy loss 0r

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