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  1. Energy in thermal processes PHY231

  2. Where are we going? • We learnt about ideal gases • And we learnt that temperature is proportional to the average kinetic energy of particles in a gas • (monoatomic gas) • We learnt about transfer of thermal energy (heat) • Changing the thermal energy of a system can • change its temperature • be linked to a phase change • Be linked to mechanical work: Engine, heat pump… !!  Thermodynamics PHY231

  3. Work in thermodynamic processes Work >0 • During a process where the Pressure P and Volume V of a system are changed, the work done on the system is the area under the PV curve • An arrow indicates the direction of the process • Arrow points right (V increases): the work done on the system is negative • Arrow points left (V decreases): The work done on the system is positive Work <0 i f PHY231 Vf Vi

  4. More (on) work • W>0, the system is gaining energy, • W<0, the system is losing energy and provides work to the outside,. (engine mode) • Limiting cases in the PV space • Isobaric process (P constant) • Isovolumetric process (V constant) Pf Pi PHY231 Vi,f

  5. A piston is pushed down and pulled up on a volume of gas. The following four diagrams show the pressure P and volume V during the processes. • Which one of the following has the largest increase in energy ? A) B) C) D) PHY231

  6. Work is the area under the PV curve • Arrow to the right (V increases) W<0 • Arrow to the left (V decreases) W>0 A) B) C) D) PHY231 6

  7. Pressure conversion unit table PHY231

  8. First Law of Thermodynamics • If Q is the energy transferred to a system by heat and W the work done on it, the change in the system’s internal energy DU is given by the sum • The change in internal energy is related to the position of the atoms/molecules and to their kinetic energy • W and Q >0 means the system gains energy • W and Q <0 means the system loses energy PHY231

  9. Compressed gas • A gas is compressed at a constant pressure of 0.800 atm from a volume of 9.00 L down to 2.00 L. In the process 400 J of energy leaves the gas by heat. • What is the work done on the gas and the change of its internal energy? PHY231

  10. Thermal processes • The change of internal energy for an ideal gas is given by • Cv is the molar specific heat at constant volume • For monoatomic gases Cv= 3R/2, diatomic Cv= 5R/2

  11. Partial summary • First law of thermodynamics • W given by area under PV diagram • P=constant then • Internal energy change is • W and Q > 0 mean the internal energy increases • W and Q <0 mean the internal energy decreases • Also remember ideal gas law

  12. Moving forward • Consider the two important equations • DU=Q+W • PV=nRT • These equations tell us that: • The change in a gas’ internal energy can be due to heat and work simultaneously • The three main quantities P, V and T are not independent and can vary simultaneously • So, in general, the changes for an ideal gas can be quite complicated… • To simplify things, processes where one of the variable is kept constant are usually considered • There are four important basic processes

  13. Four basic processes • Isobaric process (P=constant) • Adiabatic process (Q=0) • Isovolumetric process (V=constant) • Isothermal process (T=constant) • These are simplification of a general process for which none of the quantities is kept constant PHY231

  14. Isobaric process • In this process, the pressure P is constant • The heat transferred to an ideal gas is given by • Cp is the molar specific heat at constant pressure • For most gases R=8.31 J/(K.mol) PHY231

  15. Adiabatic process - • No heat enters or leaves the system • The first law simplifies to • As a result • With the definition of the adiabatic index g PHY231

  16. Specific heat table • For most gases Cp-Cv=R PHY231

  17. Isovolumetric process • In isovolumetric process the volume is constant • It follows that the work on the gas in this case is always zero • The first law simplifies to • Thus in this case • Isovolumetric process is sometimes also called isochoric process PHY231

  18. Isothermal process • In this process the temperature is kept constant. Since DU is linked to the change in temperature it follows that • And then • T=constant and • The work on the gas can be shown to be given by

  19. Summary Advice: copy that table on your equation sheet PHY231

  20. Visualizing the four processes in PV space Isobaric Isothermal Isovolumetric Adiabatic

  21. PV diagram with T=constant lines plotted Isobaric Isovolumetric Isothermal Adiabatic

  22. Isothermal process:(additional materialnot needed forhomework nor exam)

  23. HW problem • Adiabatic process, find V2 knowing P1, V1 and P2 PHY231

  24. PV diagram • A quantity of monatomic ideal gas goes through the following process. Calculate the work, the change in internal energy, and the heat associated with this process PHY231

  25. Work done during the process PHY231

  26. DU during the process PHY231

  27. Q during the process PHY231

  28. Let’s build a heat engine… • What’s a heat engine? • Basically it’s an energy converter • Takes energy from a heat source • Convert part of it into mechanical energy (= work) • Cyclic process (closed path in PV diagram) • Second law of thermodynamic says it’s impossible to convert all the heat into mechanical energy… • This means some (wasted) energy will be released in the exhaust through heat • Second Law of Thermodynamics • No heat engine operating in a cycle can absorb energy from a reservoir and use it entirely for performance of an equal amount of work

  29. Heat engine concept • Schematically a heat engine looks like this • Conservation of energy (what comes in the engine leaves the engine): Qhot=|Qcold|+|Weng| 1) Heat (Qhot) is introduced into the engine from a hot reservoir Qh 2) Work is done by the engine (the engine loses that energy ) 3) Some heat (Qcold) is expelled by the engine to a cold reservoir

  30. Heat engine cycle • An example of heat engine cycle in PV diagram • Enclosed area is the work provided to the engine • W<0 since the engine loses energy through work Path B to C Isothermal T=constant Path A to B Isovolumetric V=constant Path C to A Isobaric P=constant

  31. Basics properties of heat engines • Cyclic process • Starting point and end points are the same • DU for a complete cycle is 0 PHY231

  32. Heat engine - thermal efficiency • For a heat engine, the thermal efficiency e gives an idea of how much work is done with respect to the heat supplied • For an ideal engine (= Carnot engine explained later) Tc = Temperature of the cold reservoir Th = Temperature of the hot reservoir T in K !!!!

  33. Examples • What is the efficiency of a heat engine working with 20 kJ heat supply and 10 kJ heat exhaust? • What is the maximum efficiency of a heat engine operating between 20 ºC and 100 ºC

  34. Heat engine – full example • Let’s go back to our heat engine example using a monatomic ideal gas Path A to B Isovolumetric V=constant Path B to C Isothermal T=constant Reminder: 1 L = 10-3 m3 1 atm = 101,325 Pa R=8.31 J/(mol.K) Monatomic ideal gas Cv=3R/2 Cp=5R/2 Path C to A Isobaric P=constant PHY231

  35. Before starting • This is a fully worked example illustrating all the concepts of the chapter • You are not supposed or expected to know the following formula • Especially, the end results are specific to this example and NOT general formulas applicable to all heat engines • Nonetheless, you are encouraged to look at the following to gain some insight • Much simpler problems are asked from you in homework-set 11 • You are expected to know the Carnot-engine (also called ideal engine) results PHY231

  36. PHY231

  37. PHY231

  38. PHY231

  39. Summary results

  40. Numerically (non SI units) PHY231

  41. Remark about the work W • You must remember that the work is the area under the PV curve. Let’s look back at our results and picture them graphically PHY231

  42. Remark about the work W done on the gas • When we sum the last two we get the initial area as we should (note that the WBC <0 because V increases whereas WCA>0 because V decreases) + = (If we were reversing the directions of all arrows then W would become 6 L.atm) PHY231

  43. Carnot engine • The Carnot engine is an ideal engine and is the most efficient engine possible • Carnot Theorem • No real engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs • The Carnot cycle uses 2 isothermals at Tc and Th and 2 adiabatic processes • Thermal efficiency of a Carnot engine is fixed by the temperature of the two reservoirs (Tc<Th) T in K !!!!

  44. Refrigerators and Heat pumps • These are engine working in reverse • (same as engine but with arrows reverse directions in PV diagram) Engine (W<0) Refrig./heat pump (W>0)

  45. Coefficient of performance (COP) • Refrigerator • Refrigerator uses work energy to take some heat away from the cold reservoir • Heat pump • Heat pump uses work energy to add-up heat to the hot reservoir • Both options need work because heat never flows naturally from the cold to the hot reservoir…

  46. Engine or fridge/heat pump ? • A device work along the cycle ABC. Is it an engine or a refrigerator/heat pump ? • A) Engine • B) Refrigerator/heat pump • C) Both • D) Neither PHY231

  47. A) Engine • B) Refrigerator/heat pump • C) Both • D) Neither WBA>0 WAC<0 WCB=0 W=WBA+WAC+WCB>0

  48. PV diagram • On a PV diagram, 2 curves are plotted. • Both start at (P1, V1) and both end at the same increased volume (V2). One curve is for an isothermal process; the other is for an adiabatic process. Except for the common starting point, which curve is the upper one? • A) Isothermal • B) Adiabatic • C) Adiabatic and isothermal are the same curve • D) The curves cross in the middle PHY231

  49. A) Isothermal • B) Adiabatic • C) Adiabatic and isothermal are the same curve • D) The curves cross in the middle PHY231