Presentation on theme: "Chapter 19: The Second Law of Thermodynamics"— Presentation transcript:
1Chapter 19: The Second Law of Thermodynamics Directions of thermodynamic processesIrreversible and reversible processesThermodynamic processes that occur in nature are all irreversibleprocesses which proceed spontaneously in one direction but notthe other.In a reversible process the system must be capable of being returnedto its original state with no other change in the surroundings.A reversible process proceeds slowly through equilibrium states.
2Heat engines Heat engine Any device that transforms heat partly into work or mechanicalenergy is called a heat engine.A quantity of matter inside the engine undergoes inflow and outflow ofheat, expansion and compression, and sometimes change of phase.This matter inside the engine is called the working substance of theengine.The simplest kind of engine to analyze is one in which the workingsubstance undergoes a cyclic process.All heat engines absorb heat from a source at a relatively hightemperature, perform some mechanical work, and discard or rejectsome heat at a lower temperature.For a cyclic process
3Heat engines Energy flow diagram for a heat engine Temperature of hot reservoirHeat flow from hot reservoirper cycleTemperature of cold reservoirHeat flow from cold reservoirper cyclewaste!Net heat Q absorbed per cycle:Net work done W per cycle:
4Heat engines (cont’d) Energy flow diagram for a heat engine (cont’d) Net heat Q absorbed per cycle:Net work done W per cycle:Thermal efficiency e:
6Internal-combustion engines Gasoline engine (a heat engine)First a mixture of air and gasoline vapor flows into a cylinder through an open intakevalve while the piston descends, increasing the volume of the cylinder from aminimum of V to a maximum of rV (r: compression ratio 8-10).At the end of intake stroke, the intake valve closes and the mixture is compressed,approximately adiabatically, back to volume V during the compression stroke.Then the mixture is ignited by the spark plug, and the heated gas expands,approximately adiabatically, back to volume rV, pushing on the piston-power stroke.
7Internal-combustion engines (cont’d) The Otto cyclecompression stroke(adiabatic compression)ignite fuel (heating atconstant volume)power stroke(adiabatic expansion)thermal efficiencyin Otto cycleReject heat to environment(cooling at constant volume)r=8,g=1.4->e=56%
8Internal-combustion engines (cont’d) The Otto cycle
9Internal-combustion engines (cont’d) The Diesel cyclecompression stroke(adiabatic compression)ignite fuel (heating atconstant pressure)power stroke(adiabatic expansion)Reject heat to environment(cooling at constant volume)
10Internal-combustion engines (cont’d) The Diesel cycle
11Refrigerators Refrigerator A refrigerator works as a reversed heat engine.
12Refrigerators Energy flow of a refrigerator Coefficient of performance A refrigerator takes heat from lowtemperature reservoir and gives itoff to high temperature reservoir.From the first law of thermodynamicsCoefficient of performanceCarnot refrigerator)
13Refrigerators Air conditioner A air conditioner works on exactly the same principle as a refrigerator.For air conditioners, the quantities ofgreatest practical importance are therate of heat removal (the heat currentH from the region being cooled) andthe power input P=W/t.H in Btu/h, P in wattsH/P (energy efficiency rating EER)in (Btu/h)/W. Note 1 W=3.413 Btu/h.EER~7-10.
14The second law of thermodynamics It is impossible for any system to undergo a process in which it absorbsheat from a reservoir at a single temperature and converts the heatcompletely into mechanical work, with the system ending in the samestate in which it began (Kelvin-Planck statement)The kinetic and potential (due to interactions between molecules)energies associated with random motion constitute the internal energy.When a body sliding on a surface comes to rest as a result of friction,the organized motion of the body is converted to random motion ofmolecules in the body and in the surface.Since we can’t control the motions of individual molecules, we can’tconvert this random motion COMPLTELY back to organized motion.It is impossible for any process to have as its sole result the transfer ofheat from a cooler to a hotter body (Clausius statement).
15The second law of thermodynamics (cont’d) Proof : Kelvin-Planck statement = Clausius statementimpossibleimpossibleA workless refrigerator violates2nd law (K-P statement). If it existed,it could be used to make a 100%efficient engine, which violate K-Pstatement.(b) An impossible 100% efficient engineviolates C statement. If it existed, itcould be used to make a worklessrefrigerator, which violates C state-ment.K-P statement C-statement
16The second law of thermodynamics (cont’d) The conversion of work to heat, as in friction or viscous fluid flow,and heat flow from hot to cold across a finite temperature gradient,are irreversible processes.Two equivalent statements of the 2nd law state that these processescan be only partially reversedGases, for example, always seep spontaneously through an openingfrom a region of high pressure to a region of low pressure.The 2nd law is an expression of the inherent one-way aspect of thisand many other irreversible processes.To have the maximum efficiency of a heat engine, we must avoidall irreversible processes Answer: Carnot cycleIn any process in which the temperature of the working substanceof the engine is intermediate between TH and TC , there must beNO heat transfer between the engine and either reservoir becausesuch heat transfer could not be reversible.Any process in which the temperature of the working substancechanges must be adiabatic.As heat flow through a finite temperature drop is an irreversible,during heat transfer, there must be NO finite temperature difference.
18The Carnot cycle The Carnot cycle : ideal gas approximation isothermal expansionisothermalexpansionAdiabatic processes:thermal efficiencyof a Carnot engine
19The Carnot cycle (cont’d) The Carnot refrigeratorA reversed Carnot engine = a Carnot refrigeratorcoefficient of performance of a Carnot refrigeratorWhen the temperature difference is small, K is much larger thanunity, in which case a lot of heat can be pumped from the lowerto the higher temperature with only a little expenditure of work.
20The Carnot cycle (cont’d) The Carnot cycle and the second lawMore efficientthan a CarnotengineimpossibleA Carnot refrigeratorSince each step in the Carnotcycle is reversible, the entirecycle may be reversed. If yourun the entire backward, theengine becomes a refrigerator.
21The Carnot cycle (cont’d) The Carnot cycle and the second law (cont’d)The refrigerator does work W=-|W|,takes in heat QC from the cold reservoir,and expels hear |QH| to the hot reservoir.The superefficient heat engine expels heat|QC|, but to do so, it takes in a greateramount of heat QH+D. Its work output isthen W+D.The net effect of the two machines togetheris to take a quantity of heat D and convertit completely into work, which violates the2nd law.No engine can be more efficient than a Carnot engine.All Carnot engines operating between the same two temperatureshave the same efficiency, irrespective of the nature of the workingsubstance.
22Entropy Entropy and disorder Entropy provides a quantitative measure of disorder.Define the infinitesimal entropy change dS during an infinitesimalreversible process at absolute temperature T as:If a total amount of heat Q is added during a reversible isothermalprocess at absolute temperature T, the total entropy change is givenby:Higher temperature means greater randomness of motionAdding heat Q causes a substantial fractional increase in molecularmotion and randomness
23Entropy (cont’d) Entropy and disorder (cont’d) Generalization of definition of entropy change is to include ANYreversible process leading from one state to another, whetherit is isothermal or not:Let us represent the process as a series of infinitesimal reversiblesteps. During a step, an infinitesimal quantity of heat dQ is addedto the system at absolute temperature T. Then the change ofentropy for the entire process is:entropy change in areversible processThe change in entropy does not depend on the path leading fromthe initial to the final state but is the same for all possible processes.Since entropy is a function only of the current state of the system,we can compute entropy change in irreversible processes using theabove formula. ( we need to invent a path connecting the given initialand final state that consist entirely of reversible, equilibrium processesand compute the total entropy change as for the actual path).
24Entropy (cont’d) Entropy in cyclic processes For any reversible process(e.g. Carnot cycle):The entropy of a system ina given state is independentof the path taken to get there,and is thus a state variable.
25Entropy (cont’d) Entropy in irreversible processes Consider a combined system with a Carnot engine anda cyclic system.heat reservoir TRCarnotengineCarnotcyclecombined cyclic systemTCyclicsystem
26Entropy (cont’d) Entropy in irreversible processes Consider a combined system with a Carnot engine anda cyclic system.heat reservoir TRCarnotengineIf WC > 0, a cyclic device exchangingheat with a single heat reservoir andproducing an equivalent amount of work.- Violation of K-P statement of 2nd law-combined cyclic systemTCyclicsystem-the 2nd law-The equalityis true for areversible process.Clausius inequality
27Entropy (cont’d) Entropy in irreversible processes (cont’d) Processes: aAb and aBb reversibleProcess : aCb irreversibleTA reversible heat transfer causes changesin entropy of both the system and thereservoir (at least one needed except foradiabatic process)In an irreversible heat transfer process,there is a finite source of energy insteadof an energy reservoir, and the temperaturedifference during the heat transfer process is finite.ACBSFor an isolated systemdQ=0, so dS>=0.(> for irreversible and = for reversible process)
28Entropy (cont’d) Entropy and the second law When all systems taking part in a process are included, the entropyeither remains constant or increases.ORNo process is possible in which the total entropy decreases, when allsystems taking part in the process are included.
29ExercisesProblem 1A Carnot engine takes 2000 J of heat from a reservoir at 500 K, doessome work, and discards some heat to a reservoir at 350 K.How much work does it do, how much heat is discarded, and whatis the efficiency?SolutionThe heat discarded by the engine is:From the first law, the work done by the engine is:The thermal efficiency is:
30ExercisesProblem 2One kilogram of water at 0oC is heated to 100oC. Compute itschange in entropy.SolutionIn practice, the process described would be done irreversibly, perhapsby setting a pan of water on an electric range whose cooking surface ismaintained at 100oC. But the entropy change of the water depends onlyon the initial and final states of the system, and is the same whetherthe process is reversible or not. Hence we can imagine that the temp.of the water is increased reversibly in a series of infinitesimal steps,in each of which the temp is raised by an infinitesimal amount dT.
31Example: Refrigerator A refrigerator pumps heat from the inside of the freezer (-5°C) to the room (25°C). What is the maximum coefficient of performance?i.e. 8.9 Joules of heat would be pumped from the freezer for every Joule of work done by the compressor. (typical K = 3-5)
32Exercises Problem 3 What is the entropy change in a free expansion process,when the volume isdoubled.SolutionThe work done by n moles of ideal gas in an isothermal expansionfrom V1 to V2 is:Therefore the entropy change for n=1 is:
33ExercisesProblem 4A physics student immerses one end of a copper rod in boiling water atat 100oC and the other end in an ice-water mixture at 0oC. The sides of therod are insulated. After steady-state conditions have been achieved in therod, kg of ice melts in a certain time interval. For this time interval findthe entropy change of the boiling water; (b) the entropy change of theice-water mixture; (c) the entropy change of the copper rod; (d) the totalentropy change of the entire system.Solution(a) DS=Q/T=-mLf/T=-(0.160 kg)(334x103 J/kg)/( K)=-143 J/K.(b) DS=Q/T=mLf/T=(0.160 kg)(334x103 J/kg)/( K)= 196 J/K.(c) From the time equilibrium has been reached, there is no net heatexchange between the rod and its surroundings, so the entropychange of the copper rod is zero.(d) 196 J/K-143 J/K=53 J/K.
34ExercisesProblem 5An experimental power plant at the Natural Energy Lab generateselectricity from the temperature gradient of the ocean. The surface anddeep-water temperatures are 27oC and 6oC, respectively. (a) What isthe maximum theoretical efficiency of this power plant? (b) If the powerplant is to produce 210 kW of power, at what rate must heat be extractedfrom the warm water? At what rate must heat be absorbed by the coldwater? (c) The cold water that enters the plant leaves it at a temperatureof 10oC. What must be the flow rate of cold water through the system?Solution(a)(b)(c)