1 EGR 334 Thermodynamics Chapter 5: Sections 1-9 Lecture 21:Introduction to the2nd Law of ThermodynamicsQuiz Today?
2 Today’s main concepts: Understand the need for and the usefulness of the 2nd lawBe able to write and use the entropy balanceBe able to predict the maximum possible efficiency and COP of power and refrigeration or heat pump cycles, respectively.Be able to provide several different expressions which explain the 2nd Law of Thermodynamics.Reading Assignment:Reread Chapter 5, Sections 10-11Homework Assignment:Problems from Chap 5: 20, 43, 40, 56
3 It does not say anything about the spontaneous direction. Sec 5.1: Introducing the Second LawThe First Law of Thermodynamics in an energy balance and tells us the magnitudes energy flows.It does not say anything about the spontaneous direction.QoutQinCan we spontaneously cool the refrigerator by removing heat from the environment?Can we spontaneously heat the house by removing heat from the environment?
4 Work needs to be done to move in the non-spontaneous direction Sec 5.1: Introducing the Second LawWe know intuitively, that the spontaneous direction for heat flow is from warm to cold.QinQoutThe fridge will warm.The house will cool.If TA > TB TA = TB.But, we have refrigeration and heating, so what is required to change the direction of heat flow?Work needs to be done to move in the non-spontaneous direction
6 The Second Law of Thermodynamics answers the following Sec 5.1: Introducing the Second LawThe Second Law of Thermodynamics answers the followingWhich direction the process will move spontaneously?What is equilibrium?What is the maximum efficiency?What parameters can be changed to move closer to the maximum efficiency?What is temperature?How can we measure u and h?Since the Second Law answers all these questions, there is not a single statement of the second law.Popular forms of the Second Law include:Everything degrades to chaosNo perpetual motion machine
7 Kelvin-Planck Statement Sec 5.2: Statements of the Second LawClausius Statement:“It is impossible for any system to operate in such a way that the sole result would be an energy transfer by heat from a colder to a hotter body.”Kelvin-Planck Statement“It is impossible for any system to operate in a thermodynamic cycle and deliver a net amount of energy by work to its surroundings while receiving heat transfer from a single thermal reservoir.”QThermal Reservoir:A body where energy exchange as heat does not effect the temperature.Kelvin-Planck Statement tells us thatIt is not possible to convert heat completely to work.
8 [ ] entropy within system = + Entropy Statement: Sec 5.2: Statements of the Second LawEntropy Statement:“It is impossible for any system to operate in a way that entropy is destroyed.”entropy within systemNet entropy transferred into the systemEntropy generated with system[ ]=+Note: Entropy can be generated (unlike mass)( )( )( + or 0 )Most processes do not operate at Ssys = 0, so generally, Ssys
9 For a Closed Systems:Energy Balance:Energy Rate Balance:Entropy Balance:Entropy Rate Balance:
10 Problem 5: 3 Classify the following processes of a closed system as possible, impossible, or indeterminate. ProcessEntropy ChangeEntropy TransferEntropy ProductionPossible,Impossible, orIndeterminate.A> 0if > 0 PossibleB< 0if < 0 CImpossibleDEif > 0 FGEntropychangeEntropytransferEntropyproduction
11 An alternate way of looking at Entropy: Sec 5.2: Statements of the Second LawAn alternate way of looking at Entropy:Entropy, S, is a measure of the Disorder within a system
12 Both the system and surroundings can be returned to the initial state. Sec 5.3: Identifying IrreversibilityGasSandGasSandReversible process:Both the system and surroundings can be returned to the initial state.GasSandOne grain of sand is removed.The grain of sand is replaced.Irreversible process:System and surroundings cannot be returned to the original state.GasRockRemove rockIrreversibility can beinternal to the systemexternal to the system
13 Typical sources of irreversibility: Heat transfer due to a T Sec 5.3: Identifying IrreversibilitiesTypical sources of irreversibility:Heat transfer due to a TUnrestricted expansionSpontaneous chemical reactionFrictionElectric current producing heat due to resistanceProcess with hysteresisInelastic deformation.Elastic hysteresis of an idealized rubber band. The area in the centre of the hysteresis loop is the energy dissipated as heatSince we cannot avoid all non- idealities, most of the time a real process is irreversible.In this class we are looking for the theoretical maximum, so we will model most processes as reversible.
14 Kelvin-Planck Statement Sec 5.4: Interpreting the Kelvin-Planck StatementKelvin-Planck Statement“It is impossible for any system to operate in a thermodynamic cycle and deliver a net amount of energy by work to its surroundings while receiving heat transfer from a single thermal reservoir.”For a single reservoir:< 0 : Internal irreversibility present= 0 : No internal irreversibilityThe inequality (<) is associated with irreversibility with the systemW
15 Sec 5.6: Second Law Aspects of Power Cycles Efficiency of a Power Cycle interacting with two reservoirs.If QC 0, then η 1.0 = 100%Second Law statements for power cycles (Carnot Corollaries):The thermal efficiency of an irreversible power cycle is always less than the thermal efficiency of a reversible power cycle when each operates between the same two thermal reservoirs.All reversible power cycles operating between the same two thermal reservoirs have the same maximum thermal efficiency.
16 As WCycle 0, then and ∞ Sec 5.7: Second Law Aspects of Refrigeration and Heat PumpsCOP of refrigeration and heat pumps cycles interacting with two reservoirsAs WCycle 0, then and ∞Such an arrangement would violate the Clausius Statement.Second Law statements for refrigeration:The coefficient of performancee of an irreversible refrigeration cycle is always less than the coefficient of performance of a reversible refrigeration cycle when each operates between the same two thermal reservoirs.All reversible refrigeration cycles operating between the same two thermal reservoirs have the same maximum coefficient of performance.
17 The Celsius Temperature Scale: Sec 5.8: Temperature ScalesThe Celsius Temperature Scale:0°C freezing point of water (at 1 atm)100°C boiling point of water (at 1 atm)The Fahrenheit Temperature Scale:0°F freezing point of water & NaCl solution (at 1 atm)100°F average normal human body temperature (at 1 atm)It is desirable to have a scale that is not dependent on a single substance.Phase changes of many substances allows extension of the scale based on properties
19 Use equation of state to relate T p Sec 5.8.2: Gas ThermometerGas Thermometer:Used instead of liquid since it will not change phase over a large range of temperatures.Use equation of state to relate T p
20 The driving force for heat transfer is a T Sec 5.8: Temperature ScalesThe driving force for heat transfer is a TThis causes a heat transfer Q TChooseThusThe ratio of the temperatures is equal to the ratio of the heat rejected and the heat absorbed.Used to define the Kelvin temperature scale.
21 Consider three heat engines , operating reversibly Sec 5.8: Temperature ScalesConsider three heat engines , operating reversibly12213332Therefore
22 Thus, ideally we want to maximize T = (TH - TC) Sec 5.9 : Maximum PerformanceWithFor a reversible cycleThe Carnot efficiencyincrease THdecrease TCTo increase η Thus, ideally we want to maximize T = (TH - TC)TH – limited by equipment costs(high p and high T)
23 What is the maximum efficiency of a power cycle operating between Sec 5.9 : Maximum PerformanceTypically, the cold reservoir is the atmosphere and large bodies of water and thus TC = 298 K, since it will cost too much to use a refrigerated reservoir.What is the maximum efficiency of a power cycle operating betweenTH = 745 Kand TC = 298 K
24 Sec 5.9 : Maximum Performance Example: (5.19) A power cycle operating at steady state receives energy by heat transfer at a rate of QH at TH=1000 K and rejects energy by heat transfer to a cold reservoir at a rate QC at TC = 300 K.For each of the following cases, determinewhether the cycle operates reversibly,irreversibly, or is impossible.Max system efficiency:a) QH = 500 kW, QC=100 kWb) QH = 500 kW, Wcycle = 250 kW,and QC = 200 kWImpossible: ηa > ηmaxImpossiblec) Wcycle = 350 kW, QC = 150 kWd) QH = 500 kW, QC = 200 kWReversible: ηc = ηmaxIrreversible: ηa < ηmax
25 The rate energy is rejected, in kW Sec 5.9 : Maximum PerformanceExample: (5.63) The refrigerator shown in the figure operates at steady state with a coefficient of performance of 4.5 and a power input of 0.8 kW. Energy is rejected from the refrigerator to the surroundings at 20°C by heat transfer from the metal coils whose average surface temperature is 28°C. DetermineThe rate energy is rejected, in kWThe lowest theoretical temperature within the refrigerator, in KThe maximum theoretical power, in kW, that could be developed from a power cycle operating between the coils and surroundings.W=0.8kWTH= 20°C= 293 KTC= ?=4.5
26 The rate energy is rejected, in kW Sec 5.9 : Maximum PerformanceExample: (5.63) DetermineW=0.8kWTH= 20°C= 293 KTC= ?The rate energy is rejected, in kWCOP of refrigeration cycle=4.5Work Energy Balance:Combining:
27 b) The lowest theoretical temperature within the refrigerator, in K Sec 5.9 : Maximum PerformanceExample: (5.63) DetermineW=0.8kWTH= 20°C= 293 KTC= ?b) The lowest theoretical temperature within the refrigerator, in KModeling as Reversible System with Maximum Possible COP = 4.5=4.5
28 Sec 5.9 : Maximum Performance Example: (5.63) DetermineW=?TH= 28°C= 301 KTH= 20°C= 293 Kc) The maximum theoretical power, in kW, that could be developed from a power cycle operating between the coils and surroundings.From part a) the rejected heat isMaximum Possible Power Cycle Efficiency:Note: Such a power cycle would operate between reservoir temperatures of 20 oC and 28 oC.Solve for possible Work done: