#### Transcript Thermodynamics

EGR 334 Thermodynamics Chapter 5: Sections 1-9 Lecture 21: Introduction to the 2nd Law of Thermodynamics Quiz Today? Today’s main concepts: • Understand the need for and the usefulness of the 2nd law • Be able to write and use the entropy balance • Be 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 10-11 Homework Assignment: Problems from Chap 5: 20, 43, 40, 56 Sec 5.1: Introducing the Second Law 3 The First Law of Thermodynamics in an energy balance and tells us the magnitudes energy flows. dE CV dt QCV WCV 2 V mh gz 2 It does not say anything about the spontaneous direction. Qout Qin Can we spontaneously cool the refrigerator by removing heat Can we spontaneously heat the house by removing heat from the environment? from the environment? Sec 5.1: Introducing the Second Law 4 We know intuitively, that the spontaneous direction for heat flow is from warm to cold. Qin Qout The 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 Sec 5.1: Introducing the Second Law 5 Sec 5.1: Introducing the Second Law The Second Law of Thermodynamics answers the following • Which 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 chaos •No perpetual motion machine 6 Sec 5.2: Statements of the Second Law 7 Clausius 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.” Q Thermal Reservoir: A body where energy exchange as heat does not effect the temperature. Kelvin-Planck Statement tells us that It is not possible to convert heat completely to work. Sec 5.2: Statements of the Second Law 8 Entropy Statement: “It is impossible for any system to operate in a way that entropy is destroyed.” [ ][ ] [ ] entropy within system = Net entropy transferred into the system + Entropy generated with system Note: Entropy can be generated (unlike mass) S sys (+ 0 -) Q T (+ 0 -) gen ( + or 0 ) Most processes do not operate at Ssys = 0, so generally, Ssys 9 For a Closed Systems: Energy Balance: E sys Q sys W sys dE sys Energy Rate Balance: d E sys dt W Q dt Q sys W sys Entropy Balance: S sys Q T Q g en T Entropy Rate Balance: d S sys dt Q T d S sys dt gen g en 10 Problem 5: 3 Classify the following processes of a closed system as possible, impossible, or indeterminate. -------------------------------------------------------------------------------------------------------Process A B C D E F G Entropy Change Entropy Transfer >0 <0 0 >0 0 >0 <0 S SYS Entropy change Entropy Production 0 if < 0 >0 >0 <0 if > 0 >0 if > 0 <0 <0 Q T Entropy transfer SYS Entropy production Possible, Impossible, or Indeterminate. Possible Possible Impossible Possible Possible Impossible Possible Sec 5.2: Statements of the Second Law An alternate way of looking at Entropy: Entropy, S, is a measure of the Disorder within a system 11 Sec 5.3: Identifying Irreversibility Sand 12 Reversible process: Both the system and surroundings can be returned to the initial state. Gas Gas Sand One grain of sand is removed. Sand The grain of sand is replaced. Gas Irreversible process: System and surroundings cannot be returned to the original state. Irreversibility can be internal to the system external to the system Rock Gas Remove rock Sec 5.3: Identifying Irreversibilities Typical sources of irreversibility: • Heat transfer due to a T • Unrestricted expansion • Spontaneous chemical reaction • Friction • Electric current producing heat due to resistance • Process with hysteresis • Inelastic deformation. 13 Elastic hysteresis of an idealized rubber band. The area in the centre of the hysteresis loop is the energy dissipated as heat Since we cannot avoid all nonidealities, 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. Sec 5.4: Interpreting the Kelvin-Planck Statement 14 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.” For a single reservoir: < 0 : Internal irreversibility present W cycle 0 = 0 : No internal irreversibility The inequality (<) is associated with irreversibility with the system W Sec 5.6: Second Law Aspects of Power Cycles 15 Efficiency of a Power Cycle interacting with two reservoirs. W cycle Q in Q in Q out Q in 1 Q out 1 Q in 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. Q Cold Q Hot 1 QC QH Sec 5.7: Second Law Aspects of Refrigeration and Heat Pumps 16 COP of refrigeration and heat pumps cycles interacting with two reservoirs Q out W cycle QC Q H QC & Q in W cycle As 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. QH Q H QC Sec 5.8: Temperature Scales The 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 17 Sec 5.8: Temperature Scales 18 Sec 5.8.2: Gas Thermometer Gas 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 19 Sec 5.8: Temperature Scales 20 The driving force for heat transfer is a T This causes a heat transfer Q T QC Q H Thus TC , T H reversible QC Q H Choose TC TH TC TH reversible The ratio of the temperatures is equal to the ratio of the heat rejected and the heat absorbed. Used to define the Kelvin temperature scale. Q T 273 . 16 Q TP reversible Sec 5.8: Temperature Scales 21 Consider three heat engines , operating reversibly 1 QC 1 QH T H , T C QC 2 f T C , T I QI QH 3 QI 3 2 QH QI QC Q I Therefore QH QC 2 1 f T H , T I 3 f T H , T I f T C , T I f T H f T C TH TC Sec 5.9 : Maximum Performance With Q C 22 TC Q TH H reversible The Carnot efficiency max 1 To increase η increase TH decrease TC Thus, ideally we want to maximize T = (TH - TC) TH – limited by equipment costs (high p and high T) TC TH For a reversible cycle Sec 5.9 : Maximum Performance Typically, 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 between T 298 TH = 745 K 1 C 1 60% and TC = 298 K TH 745 23 Sec 5.9 : Maximum Performance 24 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. Max system efficiency: For each of the following cases, determine T 300 m ax 1 C 1 0 .7 0 whether the cycle operates reversibly, TH 1000 irreversibly, or is impossible. a) QH = 500 kW, QC=100 kW a 1 QC 1 QH Impossible: b) QH = 500 kW, Wcycle = 250 kW, and 100 0.80 500 c QH W cycle W cycle Q C 350 350 150 Reversible: ηc = ηmax W cycle Q H Q C 250 kW ηa > ηmax b ? c) Wcycle = 350 kW, QC = 150 kW W cycle QC = 200 kW 0.70 500 200 300 kW Impossible d) QH = 500 kW, QC = 200 kW d 1 QC QH 1 200 0.60 500 Irreversible: ηa < ηmax Sec 5.9 : Maximum Performance Example: (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. Determine (a) The rate energy is rejected, in kW (b) The lowest theoretical temperature within the refrigerator, in K (c) The maximum theoretical power, in kW, that could be developed from a power cycle operating between the coils and surroundings. TH= 20°C= 293 K W=0.8kW =4.5 TC= ? 25 Sec 5.9 : Maximum Performance 26 Example: (5.63) Determine (a) The rate energy is rejected, in kW TH= 20°C= 293 K W=0.8kW COP of refrigeration cycle max Q in W cycle =4.5 TC= ? Work Energy Balance: W cycle Q H Q C Q C Q H W cycle Combining: m ax Q H W cycle W cycle Q H W cycle 1 m ax 0.8 kW 1 4.5 4.4 kW Sec 5.9 : Maximum Performance 27 Example: (5.63) Determine b) The lowest theoretical temperature within the refrigerator, in K Modeling as Reversible System with Maximum Possible COP = 4.5 m ax TC TC TH= 20°C= 293 K W=0.8kW =4.5 TC= ? TC T H TC m ax 1 m ax 4 .5 1 4 . 5 TH 293 0 . 82 293 240 K 33 C Sec 5.9 : Maximum Performance 28 Example: (5.63) Determine c) The maximum theoretical power, in kW, that could be developed from a power cycle operating between the coils and surroundings. TH= 28°C= 301 K W=? From part a) the rejected heat is Q H 4 . 4 kW TH= 20°C= 293 K Maximum Possible Power Cycle Efficiency: m ax 1 TC W cycle TH Note: Such a power cycle would operate between reservoir temperatures of 20 oC and 28 oC. QH Solve for possible Work done: T W cycle Q H 1 C TH 293 (4.4 kW ) 1 (4.4 kW ) 0.0266 0.117 kW 301 29 End of Lecture 21 Slides